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4 4 3 3 0431 0431 Reinforced Concrete Design Reinforced Concrete Design Welcome to Instructor: Mongkol JIRAVACHARADET Instructor: Mongkol JIRAVACHARADET School of Civil Engineering School of Civil Engineering Suranaree University of Technology Suranaree University of Technology Lecture 1 - Introduction กกกก, 4 (2550) , กก , 2 (2540), กก , 5 (2539),

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Page 1: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

443304310431 Reinforced Concrete DesignReinforced Concrete Design

Welcome to

Instructor: Mongkol JIRAVACHARADETInstructor: Mongkol JIRAVACHARADET

School of Civil EngineeringSchool of Civil Engineering

Suranaree University of TechnologySuranaree University of Technology

Lecture 1 - Introduction

��ก�����������ก���������

ก����ก������ก���� �����ก, � ���������� 4 (2550)

���� � ��������, ���� ��������������� ����

���!���"�������������ก���� �����ก ���� #ก"����,� ���������� 2 (2540), � $�ก����%����&�'�(��$)��

���!���"�������������ก���� �����ก ���� #��&�����*�+���,� ���������� 5 (2539), � $�ก����%����&�'�(��$)��

Page 2: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

TEXTBOOKSTEXTBOOKS

Reinforced Concrete: Mechanics and Design, 5th EditionJames G. MacGregor, James K. Wight, Prentice Hall, 2009.

Design of Concrete Structures, 13th EditionArthur H. Nilson, David Darwin, Charles W. Dolan, McGraw-Hill, 2003.

Reinforced Concrete: A Fundamental Approach, 6th EditionEdward G. Nawy, Prentice Hall, 2009.

Building Code Requirements for Structural Concrete,ACI318-08,American Concrete Institute, 2005.

TA683.2 W53 2009TA444 N38 2009

Page 3: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

TA683.2 N55 2004 TA683.2 H365 2005

TA683.2 M39 2009 TA683.2 R48 2008

Page 4: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Course ObjectivesCourse Objectives

More than just trial and error, design is based on built up experience

as well as a solid background in analysis and an understanding of the

parameters affecting a good design solution.

��������ก �ก� ���������ก���ก� ���������� �� ������ก �

�� ������ก

�� !�������������� "��ก� �����#���� ���$�����ก����� �� ���

���ก ��� ������ก

�� !���%�������&�!'�&�(��ก� ��ก�����ก�� ������ก�������

���ก ��� ������ก

Conduct of Course

Design Projects 20 %

Midterm Exam 40 %

Final Exam 40 %

Page 5: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Grading Policy

Final Score Grade

100 - 90 A89 - 85 B+84 - 80 B79 - 75 C+74 - 70 C69 - 65 D+64 - 60 D59 - 0 F

*�',��ก-.�������&��(�"� /0������'�����'��)�+����������(��

WARNINGS !!!

1)1) Participation expectedParticipation expected,, check check 80%80%

2)2) Study in groups but submit work on your ownStudy in groups but submit work on your own

3)3) No Copying of ProjectNo Copying of Project

4)4) Submit Project at the right place and timeSubmit Project at the right place and time

5)5) Late Project with penalty Late Project with penalty 30%30%

6)6) No make up quizzes or examsNo make up quizzes or exams

Page 6: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Reinforced Concrete Design (RC Design)Reinforced Concrete Design (RC Design)

• Specifications, Loads, and Design Methods

• Strength of Rectangular Section in Bending

• Shear and Diagonal Tension

• Design of Stairs, Double RC Beam, and T-Beam

• Analysis and Design for Torsion

• Design of Slabs: One-way, and Two-way

• Bond and Achorage

• Design of Column, and Footing

• Serviceability

Content:Content:

� Structural Design Concept

� Mechanical Properties of Concrete

� Steel Reinforcement

� Reinforced Concrete Structures

Reinforced Concrete Design

Lecture 1 : IntroductionLecture 1 : Introduction

Topics Covered

Page 7: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Structural Design ConceptStructural Design Concept

Structural Engineering

�1�$������($ �'(*�ก��

�"����� ��������)�&�2+�*���&�����3�-���*�+�',��3'�����)�&�����%� ����(��)�+��&����&��"�

��1��+�������������)�&�����%����(��)�+��&��%3ก+��

*�� ก� %�����1��*�+��*����������)')�&�����*�2�����ก���"����2�����

���4 ����4����)'����� (2�������1�������+��)

Page 8: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Structural Design Concept

�������� Stability

�������� Safety

�������� Serviceability

��������Economy

��������Environment

LIFE-CYCLE OF STRUCTURE

Lesscompetitors

Traditionalactivities

New Design

Construction

Maintenance / Repairs /Renovation

Removal / Failure

Page 9: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

��������������/&� ���������

�+������� ��+��/�������� ���20 �.�. 2553

Page 10: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Concrete & Steel PropertiesConcrete & Steel Properties

What is Concrete?

Concrete is a mix of :

Water Cement Ratio (W/C) :

Low W/C

High StrengthLow Workability

High W/C

Low StrengthHigh Workability

0.3 0.7

Optimal ratios obtainedby trial and experience

Page 11: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

30cm

∅ 15 cm

ASTM BS15 cm

15 cm

15 cm

( ) ( )0.85c cASTM BSf f′ ′≅

Compressive Strength of Concrete

cf ′ compression test of standard cylinder at 28 days

Normal used: 210, 240, 280, 320 kg/cm2

High strength: 350 - 700 kg/cm2

�.�.�. ��������� �.�. 2522 : < 150 kg/cm2

200

250

300

350

400

450

500

Com

pres

sive

str

engt

h,kg

f/cm

2

0.4 0.5 0.6 0.7

Water-cement ratio, by weight

Air-entrained concrete

Non-air-entrained concrete

For type Iportland cement

Effect of water-cement ratio on 28 days compressive strength

Page 12: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Tensile Strength of Concrete

- Greatly affects cracking in structures.

- Tensile strength is about 10-15% of compressive strength.

Splitting Tensile Test (ASTM C496):

L

P

P

D2

ct

Pf

LDπ=

2

2

1.59 1.86 kgf/cm for normal-weight concrete

1.33 1.59 kgf/cm for light-weight concrete

ct c

ct c

f f

f f

′≈ −

′≈ −

แรงกด

���ก����������)�+�3�

แรงดึง

���ก��������0�)�+�+��

Page 13: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Standard Beam Test (ASTM C78):

r

Mcf

I= = Modulus of rupture

7.5 psi 2.0 kscr c cf f f′ ′= =

Practical choice for design purposes

Tensile Strength in Flexure

P

Stress-Strain Relationship of Concrete

0.003≈

Ultimate strainASTM

ε

σ

εεεεcu

Initial modulus

0.5 cf ′

Secant modulusat = Ec0.5 cf ′

cf ′

Page 14: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Concrete & Steel Strength-Deformations

REINF.ROD

CONCRETE

L

∆L

εc = ∆L/L = εs

Strain

σCompression Steel

Tension

Concrete

ε1

fs

fc1

ε2

fy

εy

fc2

ε3

fy

εcm

f’c

εcu

FailureStrain

fy

0.85f’c

Concrete: 1.533 psic c cE w f ′=

lb/ft3 psi

1.54, 270 kscc c cE w f ′=

t/m3 ksc

315,100 ksc for 2.32 t/mc c cE f w′= =

62.04 10 kscsE = ×Steel:

Modulus of elasticity

Page 15: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Concrete Weight

Plain concrete = 2.323 t/m3

Steel = 7.850 t/m3

Reinforced concrete = 2.400 t/m3

Lightweight concrete = 1.6 - 2.0 t/m3

Steel Reinforcment

Round Bar (����กก����������)

SR24: Fy = 2,400 ksc, Fu = 3,900 ksc

Deformed Bar (����ก$%&&%&�)

SD30: Fy = 3,000 ksc, Fu = 4,900 ksc

SD40: Fy = 4,000 ksc, Fu = 5,700 ksc

SD50: Fy = 5,000 ksc, Fu = 6,300 ksc

Page 16: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Stardard Reinforcing Bar Dimension and Weight

RB6

RB9

DB12

DB16

DB20

DB25

DB28

DB32

0.28

0.64

1.13

2.01

2.84

4.91

6.16

8.04

0.222

0.499

0.888

1.58

2.23

3.85

4.83

6.31

1.89

2.83

3.77

5.03

5.97

7.86

8.80

10.06

BAR SIZE(mm)

AREA(cm2)

WEIGHT(kg/m)

PERIMETER(cm)

Reinforced ConcreteReinforced Concrete

Page 17: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Reinforced Concrete (RC) Structures

Steel bars

Concrete

Section A-A

PA

A Steel bars

compression zonetension zone

Neutral axis

Concrete: high compressive strength but low tensile strength

Steel bars: embedded in concrete (reinforcing)provide tensile strength

Steel and Concrete in Combination

(1) Bond between steel and concrete prevents slip of the steel bars.

(2) Concrete covering prevent water intrusion and bar corrosion.

(3) Similar rate of thermal expansion,

Concrete: 0.000010 - 0.000013

Steel: 0.000012

Page 18: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

WHY Reinforced Concrete?

� Concrete is cheaper than steel

� Good combination of Concrete & Steel

� Durability from concrete covering

� Continuity from monolithic joint

Disadvantages of RC

� Construction time

� Concrete Quality Control

� Cracking of Concrete

Page 19: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Column

Typical Structure

1st Floor

2nd Floor

Beam Joist

Spandrelbeam

Wall footingSpread footing

Typical Structure

Foundation(Footing)

Spandrelbeam

Pier

Column

Floor slab

Main beam(Girder)

Page 20: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

รอยแตกราวเนื่องจากการดัด

รอยแตกราวเนือ่งจากการเฉือน

คาน

แรงดึง

แรงอัด

รอยแตกราว

บริเวณเกิดแรงดึงสูงสุด

บริเวณเกิดแรงอัดสูงสุด

Page 21: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

1/6 W

2/6 W

3/6 W

4/6 W

5/6 W

W

ขนาดแรงดึง kPa

การขยายตัวของแรงดึง ตามการแอนตัวของคานจากน้ําหนักที่เพิ่มขึ้น

แรงดึงสูงสุด

แรงดึงต่ําแรงอัด

ลูกศรแสดงทิศทางของแรงดึง ณ จุดตางๆ

Page 22: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

ระนาบของรอยราวที่เปนไปได ซึ่งตองตั้งฉากกบัทิศทางของแรงดึง

ลูกศรแสดงทิศทางของแรงดึงในเนื้อคอนกรีต

รอยแตกราวเนื่องจากการดัดตัวของคาน

เหลก็เสนรับแรงดึง

Page 23: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

รอยแตกราวเนือ่งจากการเฉือน

เหล็กปลอกท่ีใชรับแรงเฉือน

รอยแตกหลังคาน

รอยแตกใตทองคาน

Page 24: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

ฐานรากทรุดรอยแตกใตทองคาน

รอยแตกหลังคาน

การแตกราวที่ผนัง เปนอาการของการทรุดตัวโครงสราง

Page 25: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Reinforced Concrete DesignReinforced Concrete Design

Lecture 2 - Specification, Loads andDesign Methods

�� Structural Design ProcessStructural Design Process

�� Building CodesBuilding Codes

�� Working Stress DesignWorking Stress Design

�� Strength Design MethodStrength Design Method

�� Dead Load & Live LoadDead Load & Live Load

�� Load Transfer in Structure Load Transfer in Structure

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

ArchitecturalFunctional Plans

DesignDesign ProcessProcess

Select StructuralSystem

Trial Sections,Assume Selfweight

Analysis for internalforces in member

Member Design

Acceptable?Redesign

NG

Final Design& Detailing

OK

Design LoopDesign Loop

Page 26: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

SpecificationsSpecifications

Developed by organizations such as AISC, ACIASCE, and EIT

Recommendations of good practice based onthe accepted body of knowledge

NOT legally enforceable

OrganizationsOrganizations

EIT = Engineering Institute of Thailand

ASCE = American Society of Civil Engineers

AASHTO = American Association of State Highwayand Transportation Officials

UBC = Uniform Building Code

BOCA = Building Officials & Code Administrators

ACI = American Concrete Institute

Page 27: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Building CodesBuilding Codes

���ก������ �������������� ก�� ��

Minimum requirements to protect the public

- �.�.�. ����� ����� 2522

- �����""��#ก���$� �����

- $%��""��#

Design Loads

Dead Loads - stationary loads of constant magnitude

Live Loads - moving loads or loads that vary in magnitude

Page 28: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Caused by the weight of structure

Include both the load bearing and non-load

bearing elements in a structure

Generally can be estimated with reasonable

certainty

�&�����ก����ก���� (Dead Load)

�&�����ก��/�ก0�/�������12

���������� kg/m3

���ก�������� �ก 2,400���ก��� ��� 2,320��� 500-1,200�� �ก 7,850

�������������� kg/m2

ก������� ���!" 14ก������� #�$%�&����' 50�� �ก��( ��, * ก�� 5

���������� 10-30���� 5����ก!""#$�"% 180-360����ก!""#$(�)"� 100-200

Page 29: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Tributary area = 0.5SL sq.m

Load on beam = 0.5wSL kg/m

Floor load = w kg/sq.m

LoadLoad from from PrecastPrecast Concrete SlabConcrete Slab

S

L

Example: Example: CPACCPAC Hollow Core Slab Hollow Core Slab HC100HC100

600 mm

100 mm

SLAB WEIGHT 296 KG/M2

PC WIRE 6 ∅ 4 MM.

SPAN 4 M.

LIVE LOAD 300 KG/M2

4 m

L : Beam span

w = ? kg/m

Page 30: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

Floor Loads

Snow and Ice: 50 - 200 kg/sq.m.

Traffic Load & Pedestrian Load for Bridges

Impact Loads

Lateral Loads: Wind & Earthquake

�&������&�����กก����ก3�����ก3� ((Live LoadLive Load))

�&�����ก����ก3����6&������ก�ก������ ��� 6 (�.�. 2527) �.�.. ���������� �.�. 2522

������� �!"�#$"��%&������� '#"�(#)*�'#�ก+�(kg/m2)

(1) ����

(2) ก� ����������� ก���

(3) �����ก���� ������� � ��� ���� ! ������"#

(4) ����%&" �'ก%&"���()���ก���� ��)�� ����ก ���%�# %�*���� +,��-��.,���������

(5) �! �ก� 0 �

(6) (ก) ���2-)�3 �4" ,������%&" �'ก%&"���()������ก��2-)�3#�"-����� "-����� ������� %�*�������

(,) �����&� �� +� )4������- ,����)�� ����ก ���%�#�! �ก� %�*0 �

30

100

150

200

250

300

300

Page 31: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

�&�����ก����ก3����6&������ (�0�)ก�ก������ ��� 6 (�.�. 2527) �.�.. ���������� �.�. 2522

������� �!"�#$"��%&������� '#"�(#)*�'#�ก+�(kg/m2)

(7) (ก) ��� �������- � ��7�*)�# ���#���� 8��������7�*)�# �����4 � �����( �����#���������#�����9�������ก:��&� �3 �������9�ก�� � �3

(,) �����&� �� +� )4������- ,�����2-)�3 #�"-����� "-����� %�*�������

(8) (ก) ����- � ���ก�< �-�-08�2=3 ��>9� ��3 ���� �����ก��#����-#�3 �����ก:���ก��%�*�����

(,) �����&� �� +� )4������- ,����� �������- � ����7�*)�# ��7�*)�# ���#���� 8���� �����#�� %�*���#��

(9) �����ก:�� �����,�������#���������#��

400

500

500

600

(10) ���9�������ก:��&�����ก�7�4

500

800

Wind LoadsWind Loads

��� ��� ก���*�ก�0���!�1��'�23103(ก���*�#�4���3�# ��� �ก53ก�65������

25.0 Vq ρ=

��7 � q = stagnation pressure ���3�# � (กก./�.2)

V = basic wind speed ������8� ��7)#9�#��3�� ����!:� 10 ��$� (ก�./=�.)

ASCE 7-98200483.0 VKq =

K = �>ก�$��?!*�'������!:��7 #�� $"��+�ก 10 ��$�

Page 32: Welcome to 430431 Reinforced Concrete Design · TEXTBOOKS Reinforced Concrete: Mechanics and Design , 5 th Edition James G. MacGregor, James K. Wight, Prentice Hall, 2009. Design

���ก"4 10 5010 < h < 20 8020 < h < 40 120#กก"4 40 160

����!:������ '#"�(��� �(��$�) (กก./$�.�.)

WIND DIRECTION

Windwardside

Leew

ard

side

0 m

10 m

20 m

30 m

Step wind loading

��� �$�� �.�.. ���������� �.�. 2522

��� 19 ��ก�������������ก ����������� ��� �������� ����������������ก ���� �������ก!��������"#$��"�� ���������������ก��� %ก&� ���� �"�$ ��ก����'(�� �'�����'������('�"�$ ��!�������'��"��()��

50(8) �� ���&#'��'&�ก����������'�'/0���� ��"����$�

40(7) �� ���ก��'&�ก����������'�'/0�

30(6) �� �������'&�ก����������'�'/0�

20(5) �� �������'&�ก����������'�'/0�

10(4) �� ����$��'&�ก����������'�'/0�

0(3) �� �������'&�ก����������'�'/0�

0(2) �� ����7����'&�ก����������'�'/0�

0(1) ����������'�'/0�

���ก� ������ก���ก�������������������������

ก��������ก�� ����

����������$���8 ����)�� %$ ��)�� %$ �����$%' ���$%' 8989:;�<= ��>&�� �= �����9���� ������ �%"���ก��$ �����&�'�����ก#�����"= ����&�ก������"= ����9'�����������ก��� %ก&��"#$��"�� %ก ��

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Early 1900s: WSD was mainly used.

aci 318

ACI 318-56: USD was first introduced.

ACI 318-63: Treated WSD and USD on equal basis.

ACI 318-71: Based entirely on strength approach (USD)WSD was small part called Alternate Design Method (ADM).

ACI 318-77: ADM moved to Appendix AUSD was called Strength Design Method.

Building Code Requirements forStructural Concrete (ACI318-XX)and Commentary (ACI318R-XX)

ACI 318-95: Unified Design was introduced in Appendix B

ACI 318-05

ACI 318-83: ADM moved to Appendix B

ACI 318-89: ADM back to Appendix A

ACI 318-99: Limit State at Failure Approach was introduced

aci 318Building Code Requirements for

Structural Concrete (ACI318-XX)and Commentary (ACI318R-XX)

ACI 318-02: Change load factor to 1.2DL + 1.6LL

ACI 318-08

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Reinforced Concrete

Design MethodsWorking Stress Design

(WSD)

Ultimate Strength Design(USD)

Limit State Design(LSD)

Performance-based Design(PBD)

ACI: Alternate Design Method

Stress fromservice load

Allowable stressFa

Concrete: Fa = 0.45f’c (ACI and "��.),= 0.375 f’c (�.�.�. "��#�� 2522)

Steel: Fa = 0.50Fy

- Design under service load condition

�#7���0��8�������� (Working Stress Design : WSD)

- Apply F.S. to strength of materials forallowable stress level Fa

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Disadvantages of WSDDisadvantages of WSD::

-- Inability to deal with groups of loads where one loadInability to deal with groups of loads where one load

increases at a rate different from that of the others.increases at a rate different from that of the others.

-- Not account for the variability of the resistances Not account for the variability of the resistances and loadsand loads

-- Lack of any knowledge of the level of saftyLack of any knowledge of the level of safty

F.S. is not known explicitlyF.S. is not known explicitly

�#*+ก,�����-��. = Ultimate Stress Design (USD)

Design Strength Required Strength (U)≥

�#7�ก����� (Strength Design Method : SDM)

- Factored load condition = Structure is about to fail

(Ultimate load = �/,����ก(����ก�-��. )

- Apply F.S. in design via:

- Load factors (> 1.0)

- Strength reduction factors (< 1.0)

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Dead Load Factor = 1.4

Live Load Factor = 1.7

Factored Load = 1.4 DL + 1.7 LL

Service Load = DL + LL

Required Strength (U) = Load Factors × Service load

= Factored Load

= �������ก��ก� ���

Load Factors

General:

U = 1.4 DL + 1.7 LL

Wind Load:

U = 0.75(1.4 DL + 1.7 LL+1.7W)

U = 1.05DL + 1.275W

Lateral Earth Pressure:

U = 1.4 DL + 1.7 LL+1.7H

U = 0.9DL + 1.7H

Factored Load Combinations

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Strength reduction factor (φφφφ) :

Bending φ = 0.90

Shear and Torsion φ = 0.85

Compression φ = 0.70 or 0.75

Strength Reduction Factor = factor that account for

(1) Variations in material strengths and dimensions

(2) Inaccuracies in the design equations

(3) Degree of ductility and required reliability of member

(4) Importance of member in the structure

Nominal Strength (N) = Strength of a member calculated usingStrength Design Method.

Strength Reduction Factors

Load Transfer in StructureLoad Transfer in Structure

Floor loads

Slab + Dead loadRoof + Dead load

Snow, Rain, Windand Construction load

Column + Dead load

Wall load

Beam + Dead load

Foundation

Wind load

SoilEarthquake

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Bending in Beam 1

� Floor Framing System

� Load Transferred to Beam from Slab

� ACI Moment and Shear Coefficients

� Location of Reinforcement

� Beam Design Requirements

Reinforced Concrete DesignReinforced Concrete Design

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Columns

Stair

Stringer

Floor beam or Girder

Joist

Spandrel

Floor Framing SystemFloor Framing System

Layout of Beams and Columns

- Occupancy requirements

- Commonly used beam size

- Ceiling and services requirements

To transfer vertical loads on the floor to the beams and columns in amost efficient and economical way

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Loading on BeamsLoading on Beams

Tributary area = Area for which the beam is supporting

One-way Floor System (m =S/L < 0.5)wS kg/m

B1 Loading

Load from B1

B3 Loading

B1 = Secondary Beam

B3 = Primary Beam

If span of B3 is too large, more secondary beam may be used.

C1

B1

B2

B3

SL

Floor load w kg/m2

Tributary area

Tributary area = 0.5SL sq.m

Load on beam = 0.5wSL kg/m

Floor load = w kg/sq.m

Precast Concrete Slab

C1B2

B3 S

L

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Long span (AB):

Floor load = w kg/sq.m

Tributary area = SL/2 - S2/4 = sq.m

Load on beam kg/m

m

mS 2

4

2

2

3

3

2mwS

Short span (BC):

Floor load = w kg/sq.m

Tributary area = S2/4 sq.m

Load on beam = wS/4 wS/3 kg/m

Two-way Slab

45o 45o

45o 45o

A

D C

B

S

L

Span ratio m = S/L

B C B C

������������������ ���� ���� �������������� ���������� 5050 ����������������������

��������� 11 �� � �����������ก���� ���� �������ก��� !��� �� ������ � �� ����� � 1 ������ ��� !����� � 2-4 �� ���� �%�&�'�ก(����%����)�(��*� �����+,�ก�-��& ,��./��!�%ก��!� �����,��,���ก�� �)�&�)�,ก��!�����,�* 10 �,��

���%ก��*-� ���(���(��-����)���1�2����,��� ��� �ก(�34 ��,�'����� 10.30 �. 3�� ���&�' 2 ,(.�. 2547

�����9� �1��./���,�ก����,ก��9� &�')�(��*�� � 8 "�#$����%

��&����ก'�'�� 2��&������ก������;�,(;�� ���4��� ��� �(�

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CONTINUOUS BEAMS AND SLABSCONTINUOUS BEAMS AND SLABS

Methods of Analysis:Methods of Analysis:

w

L

w

L

w

L

w

L

SHEAR:SHEAR:

MOMENT:MOMENT:

-- Exact analysis: Exact analysis: slopeslope--deflection, moment distributiondeflection, moment distribution

-- Approximate analysis: Approximate analysis: ACI shears and moments coefficientsACI shears and moments coefficients

-- Computer: Computer: MicroFEAPMicroFEAP, Grasp, , Grasp, SUTStructorSUTStructor, , STAAD.ProSTAAD.Pro, SAP2000, SAP2000

������������������� �����������������ก��������������� �!�������"#� ���������� �����������������ก��������������� �!�������"#� ���������

����������������::

1)1) ����������� ������������ � 22 ������������������

2)2) �������������ก�����������������������ก���������� ������������ก�������������ก�����ก�������������ก�������������ก�����ก� 2020%%

3)3) ��!�"�#�ก $�%��"��%�&�'��(ก������!�"�#�ก $�%��"��%�&�'��(ก����

4)4) �"�#�ก)�����ก��"�#�ก)�����ก� 33 �����&��"�#�ก!���(ก����������&��"�#�ก!���(ก�����

5)5) &��+&������,�ก-����.���/��#0��������&��+&������,�ก-����.���/��#0��������

ACI Approximated Coefficients for Moments and ShearsACI Approximated Coefficients for Moments and Shears

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�������� 13.1 �������+ ,� ���12&���3�0%�����%��4�5�&� ACI

(ก) ���������ก

1) �������,��

- �,������&�2�&�����������ก�!)(��&���! wuln2/11

- �,������&�2�&�#,�&��.�2�&�����ก�ก�!)(��&���! wuln2/14

2) ������3 wuln2/16

(�) ���������

1) ����+,!����&!&ก�&�)(��&���!��3�� �ก

- ��2�&�� 2 ���� wuln2/9

- ��2�&����กก��� 2 ���� wuln2/10

2) ����+,!����&!�&�)(��&���!��3&2�< wuln2/11

(�) ��������� (���)

3) ����+,!����&!�&�)(��&���!�(ก #��%"�#��!

- =2�����������������ก� 3.00 �. wuln2/12

- �������&���%��%�?�%�&��%��&�� > 8 wuln2/12

4) ����+,!����&!3�&�)(��&���!��������#,�&��.�2�&�����ก�!)(��&���!

- ��2�&)(��&���!��.���&! wuln2/24

- ��2�&)(��&���!��.�%� wuln2/16

(�) ��������

1) ���12&����&!&ก�&�)(��&���!��3 �ก 1.15 wuln/2

2) ���12&����&!�&�)(��&���!��&2�< wuln/2

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(a) ��������������กก����������

)(��&���!��.���&!: 1/24 1/14

�,������&�2�&�����������: 0 1/11

)(��&���!��.�%�: 1/16 1/14 1/10 1/11 1/16 1/11 1/11

���&!(Spandrel)

)(��&���!��.���&!

)(��&���!��.�%�

��� ����!"�ก#���$

)(��&���!)(��&���! )(��&���!

�%���.)(��&���!

���%�#ก

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(b) �������������������

)(��&���!��.���&!: 1/24 1/14

�,������&�2�&�����������: 0 1/11

)(��&���!��.�%�: 1/16 1/14 1/9 1/9 1/14 1/16

(c) &�'����� ��(���ก)� 3 ����

1/12 1/14 1/12 1/12 1/16 1/12 1/12

(d) ���*+��,������)-��������กก��� 8 �������,������)-������

1/12 1/14 1/12 1/12 1/16 1/12 1/12

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Ex3.1: A two span beam is supported by spandrel beams at the outer edges and by a column in the center. Dead load (including beam weight) is 1.5 t/m and live load is 3 t/m on both beams. Calculate all critical service-load shear forces and bending moments for the beams. The torsional resistance of the spandrel beam is not sufficient to cause restraint of beam ABC at the masonry walls.

MasonryWall

MasonryWall

A B C

CLD CL

E

6 m 6.5 m

Check conditions (a) Loads are uniformly distributed,

(b) LL/DL = 3/1.5 = 2 < 3,

(c) (L2 – L1)/L1 = (6.5 – 6)/6 = 0.083 < 0.2

B’ B’’

Bending Moments MAB = -4.5(6)2/24 = -6.75 t-m, MBA = -4.5(6.25)2/9 = -19.5 t-m,

MCB = -4.5(6.5)2/24 = -7.92 t-m, MBC = -4.5(6.25)2/9 = -19.5 t-m,

MD = 4.5(6)2/11 = 14.7 t-m, ME = 4.5(6.5)2/11 = 17.3 t-m

MasonryWall

MasonryWall

A B C

CLD CL

E

6 m 6.5 m

B’ B’’

Shear Forces

VA = 4.5(6)/2 = 13.5 tons, VB’ = 1.15(4.5)(6)/2 = 15.5 tons,

VC = 4.5(6.5)/2 = 14.6 tons, VB’’ = 1.15(4.5)(6.5)/2 = 16.8 t-m

Reactions

RA = VA = 13.5 tons,

RB = VB’ + VB’’ = 15.5 + 16.8 = 32.3 tons,

RC = VC = 14.6 tons

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Location of ReinforcementLocation of Reinforcement

•• Simply supported beamSimply supported beam

Concrete cracks due to tension, and as a result, reinforcement iConcrete cracks due to tension, and as a result, reinforcement is requireds required

where flexure, axial loads, or shrinkage effects cause tensile swhere flexure, axial loads, or shrinkage effects cause tensile stresses.tresses.

tensile stresses and cracks aredeveloped along bottom of the beam

BMD

longitudinal reinforcement is placedclosed to the bottom side of the beam

PositiveMoment

Location of ReinforcementLocation of Reinforcement

• Cantilever beam

- Top bars

- Ties and anchorageto support

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•• Continuous beamContinuous beam

Location of ReinforcementLocation of Reinforcement

Location of ReinforcementLocation of Reinforcement

• Continuous beam with 2 spans

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Behavior of Beam under Load

Working Stress Condition

cf f ′<cε

T = As fs

C

w

L

cε 2.0r cf f f ′< =

cf f ′<Elastic Bending (Plain Concrete)

s yε ε<

T = As fs

C

Brittle failure mode

s yf f<

s yε ε≥

T = As fs

C

Ductile failure mode

s yf f=

Crushing

εcu= 0.003

εc < 0.003

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Beam Design Requirements

1) Minimum Depth (for deflection control)

onewayslab

L/24 L/28 L/10L/20

BEAM L/18.5 L/21 L/8L/16

2) Temperature Steel (for slab)

As

t

bSR24: As = 0.0025 bt

SD30: As = 0.0020 bt

SD40: As = 0.0018 bt

fy > 4,000 ksc: As = 0.0018� 4,000 bt

fy

3) Minimum Steel (for beam)

AsAs min = 14 / fy

To ensure that steel not fail before first crack

5) Bar Spacing> 4/3 max. aggregate size

4) Concrete Covering

���+ก���

stirrup���+ก���ก

Durability and Fire protection

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Bending in Beam 2

Reinforced Concrete DesignReinforced Concrete Design

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

� Working Stress Design (WSD)

� Practical Design of RC Beam

� Analysis of RC Beam

� Double Reinforcement

WSD of Beam for Moment

Assumptions:

1) Section remains plane

2) Stress proportioned to Strain

3) Concrete not take tension

4) No concrete-steel slip

Modular ratio (n):

62.04 10 134

15,100s

c c c

En

E f f

×= = ≈

′ ′

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Effective Depth (d) : Distance from compression face to centroid of steel

d

d

compression face

b

kd

sε s s

s s s

T A f

f E ε

=

=

N.A.

c c cf E ε=C

jd

Cracked transformed section

strain condition force equilibrium

Equilibrium ΣFx= 0 :

Compression = Tension

1

2 c s sf b kd A f=

Compression in concrete: 1

2 cC f b kd=

Tension in steel:s sT A f=

s s

s s s

T A f

f E ε

=

=

N.A.

c c cf E ε=C

jd

kd

Reinforcement ratio: /sA bdρ =

2c

s

f

f k

ρ= 1

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Strain compatibility:

1

/

/ 1

1

c

s

c c

s s

c

s

kd k

d kd k

f E k

f E k

f kn

f k

εε= =

− −

=−

=−

2

d

kd

Analysis: know ρ find k 1 2 ( )22k n n nρ ρ ρ= + −

Design: know fc , fs find k 21

1

c

sc s

c

n fk

fn f fn f

= =+ +

Example : = 150 ksc , fs = 1,500 ksccf ′

13410.94 11 (nearest integer)

150

0.375(150) 56 ksc

10.291

1,5001

11(56)

= = ⇒

= =

= =+

c

n

f

k

Allowable Stresses

20.33 60 kg/cmc cf f ′= ≤

Plain concrete:

SR24: fs = 0.5(2,400) = 1,200 ksc

SD30: fs = 0.5(3,000) = 1,500 ksc

SD40, SD50: fs = 1,700 ksc

Steel:

20.375 65 kg/cmc cf f ′= ≤

Reinforced concrete:

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Resisting Moment

M

T = As fs

jd

kd/3

1

2 cC f k b d=

Moment arm distance : j d

3

kdjd d= −

13

kj = −

Steel:s sM T jd A f jd= × =

Concrete: 2 21

2 cM C jd f k j b d Rb d= × = =

1

2 cR f k j=

Design Step: known M, fc, fs, n

1) Compute parameters

1

1 s c

kf n f

=+

1 / 3j k= −1

2 cR f k j=

45

50

55

60

65

fc(kg/cm2)

R (kg/cm2)

fs=1,200(kg/cm2)

fs=1,500(kg/cm2)

fs=1,700(kg/cm2)

6.260

7.407

8.188

9.386

10.082

n

12

12

11

11

10

5.430

6.463

7.147

8.233

8.835

4.988

5.955

6.587

7.608

8.161

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Design Parameter k and j

45

50

55

60

65

fc(kg/cm2)

fs=1,200(kg/cm2)

fs=1,500(kg/cm2)

fs=1,700(kg/cm2)

0.310

0.333

0.335

0.355

0.351

n

12

12

11

11

10

k j

0.897

0.889

0.888

0.882

0.883

0.241

0.261

0.262

0.280

0.277

k j

0.920

0.913

0.913

0.907

0.908

k j

0.265

0.286

0.287

0.306

0.302

0.912

0.905

0.904

0.898

0.899

1) For greater fs , k becomes smaller → smaller compression area

2) j ≈ 0.9 → moment arm j d ≈ 0.9d can be used in approximation

design.

2) Determine size of section bd2

Such that resisting moment of concrete Mc = R b d 2≥ Required M

Usually b ≈ d / 2 : b = 10 cm, 20 cm, 30 cm, 40 cm, . . .

d = 20 cm, 30 cm, 40 cm, 50 cm, . . .

3) Determine steel area

s s ss

MM A f jd A

f j d= → =From

4) Select steel bars and Detailing

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�������� ก.1 � ������� ���������ก����������� � ��� , ��.2

Bar Dia.Number of Bars

1 2 3 4 5 6

RB6

RB9

DB10

DB12

DB16

DB20

DB25

0.283

0.636

0.785

1.13

2.01

3.14

4.91

0.565

1.27

1.57

2.26

4.02

6.28

9.82

0.848

1.91

2.36

3.53

6.03

9.42

14.73

1.13

2.54

3.14

4.52

8.04

12.57

19.63

1.41

3.18

3.93

5.65

10.05

15.71

24.54

1.70

3.82

4.71

6.79

12.06

18.85

29.45

�������� ก.3 !����"ก ��#����$�%��!� �������&� ACI

Member

One-way slab

Beam

Simplesupported

One-endcontinuous

Both-endscontinuous

Cantilever

L/20

L/16

L/24

L/18.5

L/28

L/21

L/10

L/8

L = span length

For steel with fy not equal 4,000 kg/cm2 multiply with 0.4 + fy/7,000

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Example 3.2: Working Stress Design of Beam

w = 4 t/m

5.0 m

Concrete: fc = 65 kg/cm2

Steel: fs = 1,700 kg/cm2

From table: n = 10, R = 8.161 kg/cm2

Required moment strength M = (4) (5)2 / 8 = 12.5 t-m

Recommended depth for simple supported beam:

d = L/16 = 500/16 = 31.25 cm

USE section 30 x 50 cm with steel bar DB20

d = 50 - 4(covering) - 2.0/2(bar) = 45 cm

Moment strength of concrete:

Mc = R b d2 = 8.161 (30) (45)2

= 495,781 kg-cm

= 4.96 t-m < 12.5 t-m NG

TRY section 40 x 80 cm d = 75 cm

Mc = R b d2 = 8.161 (40) (75)2

= 1,836,225 kg-cm

= 18.36 t-m > 12.5 t-m OK

Steel area: 25

cm 8.1075908.0700,1

105.12=

×××

==jdf

MA

ss

Select steel bar 4DB20 (As = 12.57 cm2)

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Alternative Solution:

From Mc = R b d2 = required moment M

bR

Md

R

Mdb =⇒=2

For example M = 12.5 t-m, R = 8.161 ksc, b = 40 cm

cm 88.6140161.8

105.12 5

=××

=d

USE section 40 x 80 cm d = 75 cm

Revised Design due to Self Weight

From selected section 40 x 80 cm

Beam weight wbm = 0.4 × 0.8 × 2.4(t/m3) = 0.768 t/m

Required moment M = (4 + 0.768) (5)2 / 8 = 14.90 < 18.36 t-m OK

Revised Design due to Support width

5.0 m span

30 cm30 cmColumn width 30 cm

4.7 m clear span

Required moment:

M = (4.768) (4.7)2 / 8

= 13.17 t-m

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Practical Design of RC Beam

B1 30x60 Mc = 8.02 t-m, Vc = 6.29 t.fc = 65 ksc, fs = 1,500 ksc, n = 10

k = 0.302, j = 0.899, R = 8.835 ksc

b = 30 cm, d = 60 - 5 = 55 cm

Mc = 8.835(30)(55)2/105 = 8.02 t-m

Vc = 0.29(173)1/2(30)(55)/103

= 6.29 t

w = 2.30 t/m

5.00

Loaddl 0.43wall 0.63slab 1.24w 2.30

M± = (1/9)(2.3)(5.0)2 = 6.39 t-m

As± = 6.39×105/(1,500×0.899×55)

= 8.62 cm2

As± = 8.62 cm2 (2DB25)

V = 5.75 t ([email protected] St.)

B2 40x80 Mc = 19.88 t-m, Vc = 11.44 t.

w = 2.64 t/m

8.00 5.00

w = 2.64 t/m

SFD8.54 9.83

12.58 3.37

BMD+13.81

-16.17

+2.15

As13.65

4DB25

2.13

3DB25

15.99

2DB25

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GRASP Version 1.02

B11-B12

-47.7369.700.0081.47-92.256

-52.6131.27-92.256.59-28.265

-38.9244.96-28.2625.88-46.354

-43.3440.54-46.3520.75-37.973

-39.3644.52-37.9717.36-53.422

-50.8433.04-53.4239.0301

Fy.j [Ton]Fy.i [Ton]Mz.j [T-m]Mz.pos [T-m]Mz.i [T-m]Membe

r

Analysis of RC BeamAnalysis of RC Beam

Given: Section As , b, d Materials fc , fs

Find: Mallow = Moment capacity of section

STEP 1 : Locate Neutral Axis (kd)

( )22k n n nρ ρ ρ= + −

1 / 3j k= −

where Reinforcement ratiosAbd

ρ = =

62.04 10 134

15,100s

c c c

En

E f f

×= = ≈

′ ′

kd

d

εc

εs

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STEP 2 : Compute Resisting Moment

Concrete: 2

2

1dbjkfM cc =

Steel: djfAM sss =

Under reinforcement is preferable because steel is weaker

than concrete. The RC beam would fail in ductile mode.

If Mc > Ms , Under reinforcement Mallow = Ms

If Mc < Ms , Over reinforcement Mallow = Mc

From Example 3.2: Analysis for Bending Strength of Beam Section

w = 4 t/m

5.0 m

Concrete: fc = 65 kg/cm2

Steel: fs = 1,700 kg/cm2

40 cm

80 c

m

Design Section

4DB20(As = 12.57 cm2)

Effective depth, d = 80 – 2.5(cover) – 2.0/2 = 76.5 cm

Reinforcement ratio, ρ = As/bd = 12.57/(40×76.5) = 0.00411 ρn = 0.0411

= + −ρ ρ ρ22k n ( n) n = × + −22 0.0411 (0.0411) 0.0411

k = 0.249 j = 1 – 0.249/3 = 0.917

Required moment, M = (4.768) (4.7)2 / 8 = 13.17 t-m

= = × × × × × =2 2 51 165 0 249 0 917 40 76 5 10 17 37 t-m

2 2c cM f k j bd . . . / .

512 57 1 700 0 917 76 5 10 14 99 t-ms s sM A f j d . , . . / .= = × × × = CONTROL

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Double Reinforcement (Double RC)

When Mreq’d > Mallow

- Increase steel area- Enlarge section

- Double RConly when no choice

εs

εc

M

T = As fs

C = fc k b d12

As

A’sε’sd’

T’ = A’s f’s

As1 fs

As2 fs

��������������� ������������ ��ก������������� ��ก���������

T = As fs

C = fckbd12

T’ = A’s f’s

T1 = As1 fs

C = fckbd12

T2 = As2 fs

T’ = A’s f’s

jd d-d’

21

1

1

2c c

s s

M M f kjbd

A f jd

= =

=

2

2 ( )

( )

c

s s

s s

M M M

A f d d

A f d d

= −

′= −

′ ′ ′= −

Steel area As = 1c

ss

MA

f jd= + 2 ( )

cs

s

M MA

f d d

−=

′−

M = M1 + M2

Moment strength

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Compatibility Conditionεc

d

d’

kd ε’s

εs

s

s

d kd

kd d

εε

−=

′ ′−

From Hook’s law: εs= Es fs, ε’ s= Es f’ s

s s s

s s s

E f f d kd

E f f kd d

−= =

′ ′ ′−

1s s

k d df f

k

′−′ =

�.�.�. ก���� �� 21s s

k d df f

k

′−′ =

������� ��ก�������������� ( A’ s )

T2 = As2 fs

T’ = A’s f’s

d-d’

Force equilibrium [ ΣFx=0 ]

T’ = T2

A’s f’s = As2 fs

Substitute 21s s

k d df f

k

′−′ =

2

1 1

2s s

kA A

k d d

−′ =

′−

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���� � ��ก������ ( k )

εc

d

d’

kd ε’ s

εs

Compression = Tension

c sC C T′+ =

1

2 c s s s sf bkd A f A f′ ′+ =

Substitute 2 ,1

ss s

Ak d df f

k b dρ

′′−′ ′= =

1, s

s c

Akf n f

k bdρ

−= =

( ) ( )222 2 2 2d

k n n nd

ρ ρ ρ ρ ρ ρ′ ′ ′ ′= + + + − +

Example 3.4 Design 40x80 cm beam using double RC

fc = 65 ksc, fs = 1,700 ksc,

n = 10, d = 75 cm

k = 0.277, j = 0.908, R = 8.161 ksc

w = 6 t/m

5.0 m

Required M = (6.768) (5)2 / 8 = 21.15 t-m

Beam weight wbm = 0.4 × 0.8 × 2.4(t/m3) = 0.768 t/m

Mc = Rbd2 = 8.161(40)(75)2/105 = 18.36 t-m < req’d M Double RC

52

1

18.36 1015.86 cm

1,700 0.908 75c

ss

MA

f jd

×= = =

× ×

52

2

(21.15 18.36) 102.34 cm

( ) 1,700 (75 5)c

ss

M MA

f d d

− − ×= = =

′− × −

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Tension steel As = As1 + As2 = 15.86 + 2.34 = 18.20 cm2

USE 6DB20 (As = 18.85 cm2)

Compression steel

22

1 1 1 1 0.2772.34 4.02 cm

2 2 0.277 5 / 75s s

kA A

k d d

− −′ = = × × =

′− −

USE 2DB20 (As = 6.28 cm2)

6DB20

2DB20

0.80

m

0.40 m

Double RC : Analysis for Bending Strength of Beam Section

w = 6 t/m

5.0 m

Effective depth, d = 80 – 2.5(cover) – 2.0/2 = 76.5 cm

Reinforcement ratio, ρ = As/bd = 12.57/(40×76.5) = 0.00411

k = 0.222 j = 1 – 0.222/3 = 0.926

Required moment, M = (6.768) (4.7)2 / 8 = 18.69 t-m

40 cm

80 c

m

Design Section

6DB20(As = 18.85 cm2)

2DB20(A’s = 6.28 cm2)

Concrete: fc = 65 ksc

Steel: fs = 1,700 ksc

n = 10

( )6 28 40 76 5 0 00205sA / bd . / . .ρ′ ′= = × =

2 5 2 0 2 3 5 cmd . . / .′ = + =

( ) ( )222 2 2 2d

k n n nd

ρ ρ ρ ρ ρ ρ′ ′ ′ ′= + + + − +

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Strain compatibility:

/

1 / 1 1

εε= = → = → =

− − − −c c c c

s s s s

f E fkd k k kn

d kd k f E k f k

d

kd

ε ′s

′d

Assume fc = 65 ksc:

(1 ) /s cf nf k k= − = 10 × 65 (1 – 0.222 ) / 0.222

= 2,278 ksc > 1,700 ksc NG

Use fs = 1,700 ksc:

/ (1 )c sf f k n k= − = 1,700 × 0.222 / (10 × (1 – 0.222))

= 48.5 ksc OK< 65 ksc

2 ( / ) / (1 )s sf f k d d k′ ′= − −

= 2 × 1,700 (0.222 – 3.5/76.5) / (1 – 0.222)

= 770 ksc OK< 1,700 ksc

21

12c cM M f k j bd= =

Moment Strength: M = M1 + M2

2 5148 5 0 222 0 926 40 76 5 10

2. . . . /= × × × × ×

= 11.67 t-m

2 ( )s sM A f d d′ ′ ′= − 56.28 770(76.5 3.5) /10= × − = 3.53 t-m

M = 11.67 + 3.53 = 15.2 t-m < [ Mreq’d = 18.69 t-m ] NG

REVISE DESIGN ???

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Bending in Beam 3

� Strength Design Method (SDM)

� Beam behavior under increasing load

� Nominal Moment Strength (Mn)

� Design Procedure

Reinforced Concrete DesignReinforced Concrete Design

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Ultimate Stress Design (USD)

Advantage of SDM over WSD:

1) Consider mode of failure

2) Nonlinear behavior of concrete

3) More realistic F.S.

4) Ultimate load prediction ≅ 5%

5) Saving (lower F.S.)

Strength Design Method (SDM)Strength Design Method (SDM)

Working Stress Design (WSD): early 1900s → early 1960s

Strength Design Method (SDM) is more realistic for safety and reliability at the strength limit state.

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Strength Design Method (SDM)Strength Design Method (SDM)

Design Strength ≥ Required Strength (U)

Design Strength = Strength Reduction Factor (φ) × Nominal Strength (N)

Strength Reduction Factor ( φφφφ) accounts for

(1) Variations in material strengths and dimensions

(2) Inaccuracies in the design equations

(3) Ductility & reliability of the members

(4) Importance of the member in the structure

Nominal Strength = Strength of a member calculated by using SDM

Required Strength = Load factors × Service Load

REQUIRED STRENGTH (U)REQUIRED STRENGTH (U)

Load Combinations:

U = 1.4(D + F)

U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W)

U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)

U = 1.2D + 1.0E + 1.0L + 0.2S

U = 0.9D + 1.6W + 1.6H

U = 0.9D + 1.0E + 1.6H

where

D = dead load, E = earthquake, F = fluid pressure, H = lateral soil pressure,

L = live load, Lr = roof live load, R = rain load, S = snow load, T = temp. load,

U = required strength to resist factored load, and W = wind load

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Required Strength for Simplified Load CombinationsRequired Strength for Simplified Load Combinations

Loads

Dead (D) and Live (L)

Required Strength

1.4D

1.2D + 1.6L + 0.5Lr

Dead, Live and Wind (W) 1.2D + 1.6Lr + 1.0L

1.2D + 1.6Lr + 0.8W

1.2D + 1.6W + 1.0L + 0.5Lr

0.9D + 1.6W

Dead, Live and Earthquake (E) 1.2D + 1.0L + 1.0E

0.9D + 1.0E

In Thailand, we still use: U = 1.4D + 1.7L

Strength Reduction Factors Strength Reduction Factors φφφφφφφφ in the Strength Design Methodin the Strength Design Method

0.90Tension-controlled sections

0.700.65

Compression-controlled sectionsMembers with spiral reinforcementOther reinforced members

0.75Shear and torsion

0.65Bearing on concrete

0.85Post-tensioned anchorage zones

0.75Struts, ties, nodal zones and shearing areas

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Behavior of Concrete Beam under increasing load

As

h

b

As

d

ctεfs

fc

fct

Before crack

As

After crack

sε fs

fc

Working Stress State

cuε

sε fs

fc

Strength Limit State

Overload

Ultimate strain

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Nominal Moment Strength ( Mn )

b

d

As

εcu = crushing strain

εs

N.A.

fc’

x

T = As fy ( for εs > εy )

C = k fc’ bx

T = As fy

C = 0.85 fc’ a b

0.85 fc’

a = β1xa/2

d – a/2

Equivalent Stress Distribution(Whitney stress block)

[ΣFx = 0] C = T

0.85 fc’ a b = As fy

ρ= =

′ ′0.85 0.85s y y

c c

A f f da

f b f

Equivalent Stress Distribution(Whitney stress block)

ρρ

= − ′

2 11.7

yn y

c

fM f bd

f

T = As fy

C = 0.85 fc’ a b

0.85 fc’

a = β1xa/2

d – a/2

ρ = − = − ′

( / 2)2(0.85)

yn s y

c

f dM T d a A f d

f

Flexural resistance factor Rn : = 2n nM R bd

ρρ

= − ′

11.7

yn y

c

fR f

f

=′0.85

y

c

fm

fModular ratio: ρ ρ

= −

11

2n yR f m

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for fc’ ≤ 280 ksc, β =1 0.85

for fc’ > 280 ksc, β′ −

= − ≥

1

2800.85 0.05 0.65

70cf

Reinforcement Ratio ρρρρ :

ρ ρ = −

11

2n yR f mFrom ρ ρ− + =2 2 2 0y y nm f f R

ρ

= − −

211 1 n

y

mRm f

cf ′280

0.85

0

0.65

560

210 0.85

cf ′ 1β

240 0.85

280 0.85

320 0.82

350 0.80

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Balance Steel Ratio ( ρρρρb )

��������ก�� ��������������ก�������ก���ก����� ��� !����ก��

T = As fy

d

baf85.0C c′=

εcu = 0.003

εs = εy = fy / Es

cd

Concrete crushing

Steel yielding

From strain condition,

From force equilibrium, [ΣFx = 0]

cu

cu y

cd

ε=

ε + εcu

cu y

c d ε

= ε + ε

C = T

c s y0.85 f ab A f′ = c 1 y0.85 f cb f bd′ β = ρ

����"� εcu = 0.003

c cub 1

y cu y

0.85 ff

′ ερ = β ε + ε

Balance Steel Ratio:

εy = fy / 2.04×106��

cb 1

y y

0.85 f 6,120f 6,120 f

′ρ = β +

��������ก�� ��#���$%��&��������ก����� ρρρρ = As/bd

ρρρρ = ρρρρb

���������balance condition

εcu

εy

������� ก��กOver RC

ρρρρ > ρρρρbεcu

εs < εy

������� ก����Under RC

εcu

ρρρρ < ρρρρb

εs > εy

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��$�'"�� ก����$%��&��������ก����� (Under RC)

30 cm

50 c

m

2DB25

cf 240 ksc′ =

β1 = 0.85

fy = 4,000 ksc

b0.85 240 6,120

0.854,000 6,120 4,000

× ρ = × × +

= 0.0262

As = 2×4.91 = 9.82 cm2

sA 9.820.0073

bd 30 45ρ = = =

×ρ < ρb Under RC

Steel Yield : fs = fy

������� ก����Under RC

εcu

ρρρρ < ρρρρb

εεεεs > εεεεy

C = T

0.85×240×0.85×c×30 = 9.82×4,000

c = 7.55 cm

c 1 s y0.85 f cb A f′ β =

Strain Condition:

d

c

εcu = 0.003

εεεεs = ?

s

cu

d cc

ε −=

ε

εs = (45 – 7.55) / 7.55 × 0.003 = 0.0149

εy = fy / Es = 4,000 / 2.04e6 = 0.00196

εεεεs > εεεεy Steel Yield : fs = fy Under RC

ACI 318-08 Section 10.5 : Minimum reinforcement of flexural members

10.5.1 – At every section of a flexural member where tensile reinforcement is required, As provided shall not be less than that given by

cs, min

y

0.8 fA b d

f

′=

and not less than 14 b d / f y To prevent concrete first crack

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��$�'"�� ก����$%��&��������ก����� (Over RC)

cf 240 ksc′ =

β1 = 0.85

fy = 4,000 ksc

b0.85 240 6,120

0.854,000 6,120 4,000

× ρ = × × +

= 0.0262

As = 8×4.91 = 39.28 cm2

ρ > ρb Over RC

Steel NOT Yield : fs < fy

C = T

0.85×240×0.85×c×30 = 39.28 fs

2 unknowns: c and fs ?

30 cm

50 c

m

8DB25

39.280.0312

30 42ρ = =

×

������� ก��กOver RC

ρρρρ > ρρρρbεcu

εεεεs < εεεεy

c 1 s s0.85 f cb A f′ β =

1

Strain Condition:

dc

εcu = 0.003

εεεεs = fs/Es

s

cu

d cc

ε −=

ε

ss cu

s

f d cE c

− ε = = ε

s42 c

f 6,120c− =

1

5,202 c2 + 240,393.6 c – 10,096,531.2 = 0

>> roots([5202 240393.6 -10096531.2]) ans = -72.8530 26.6412

MATLAB:

c = 26.6 cm

fs = 3,543 ksc

425,202 39.28 6,120

− = ×

cc

c

fs < fy

Steel NOT Yield Over RC

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ρρρρ = 0.5 ρρρρmax = 0.375 ρρρρb

ACI 318-08: Section 10.3 – General principles and re quirements

10.3.5 – For flexural members, a net tensile strain εεεεt in extreme tension steel shall not be less than 0.004.

For conservative design, we may use

ACI Code before 2002, ρρρρmax = 0.75 ρρρρb

From

2n nM R bd=

yn y

c

fR f 1

1.7 f

ρ = ρ − ′

If we use ρρρρmax Rn,max Mn,max

where Mn,max is the maximum moment

capacity of the section

�������� ก.5 ��������ก��������"����������*�+������� �

f’c(ksc)

ρmin ρb ρmax m Rn,max(ksc)

180

210

240

280

320

350

0.0035

0.0035

0.0035

0.0035

0.0035

0.0035

0.0197

0.0229

0.0262

0.0306

0.0338

0.0360

0.0147

0.0172

0.0197

0.0229

0.0253

0.0270

26.1

22.4

19.6

16.8

14.7

13.4

47.62

55.55

63.49

74.07

82.46

88.36

������� fy = 4,000 ก.ก./��.2

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0 0.005 0.01 0.015 0.02 0.0250

10

20

30

40

50

60

70

80

Strength Curve (Rn vs. ρρρρ) for SD40 Reinforcement

Coe

ffici

ent o

f res

ista

nce

Rn

(kg/

cm2 )

Reinforcement ratio ρ = As /bd

f’c = 180 ksc

f’c = 210 ksc

f’c = 240 ksc

f’c = 280 ksc

Upper limit at 0.75ρb

��������ก������������ก��

�ก���ก��ก�������

��������ก ρρρρb �� !�����ก"!� 1/m ����(�$��ก�$%�) �����'�

Required moment from load = Mu

Design Moment Strength = MnuM

2nR bd= u

n 2

MR

bd=

φ

From n y1

R f 1 m2

= ρ − ρ

where y

c

fm

0.85 f=

�()�� 2y y nmf 2f 2R 0ρ − ρ + =

2y y n y

y

2f 4 f 8mR f

2mf

± −ρ = n

y

2mR11 1

m f

= ± −

n

y

2mR11 1

m f

ρ = − −

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STEP 1 ���ก�������� ก�����*������(��+��,�-!"� min maxρ ρ ρ≤ ≤

min max

1

1

14 / 0.75

0.85 6,120

6,120

0.85 ; 280 ksc

2800.85 0.05 ;280 560 ksc

70

0.65 ; 560 ksc

y b

cb

y y

c

cc

c

f

f

f f

f

ff

f

ρ ρ ρ

ρ β

β

= =

′= +

′ ≤ ′− ′= − < ≤

′ >

Conservative design select ρρρρ = 0.5= 0.5= 0.5= 0.5ρρρρmax = 0.375 ρρρρb

,�-����ก����ก�&&������ ������ก����������ก

STEP 2

onewayslab L/24 L/28 L/10L/20

BEAM L/18.5 L/21 L/8L/16

���ก.��� ��*��%���ก�� b d 2

��ก 2n nM R bd= 2 n

n

Mbd

R= u

n

MR

n y1

R f 1 m2

= ρ − ρ

����� y

c

fm

0.85 f=

STEP 3 ���ก d ��ก "��/ก h *����(����0�����ก������12��ก����!�%�"

���ก b ≈≈≈≈ d/2

ก�����ก.�������%�� ���(,-��.�% ������. �-!� 30x50 ��.

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STEP 4 ������� !� ρρρρ %��.�������%��*�����ก (b, d)

��ก 2n nM R bd= n

n 2

MR

bd= u

2

M

bd=

φ

n

y

2mR11 1

m f

ρ = − −

STEP 5 %�"������������ ก"!���$!,�-!"� ρρρρmin < ρρρρ < ρρρρmax ����)�! ?

4�� ρρρρ < ρρρρmin ,��,-� ρρρρ = ρρρρmin

4�� ρρρρ > ρρρρmax ,���0���.�������%�� ��" ���"�,��!

STEP 6 ���"�0�'�*���� ก As = ρρρρbd ��"���ก.����(����"��� ก�����

STEP 7 %�"����ก��������%�� y2n y

c

fM f bd 1

1.7 f

ρ = ρ − ′

uM≥

φ

����bf85.0

fAa

c

ys

′=

−=

2a

dfAM ysnuM

≥φ

c

y

f85.0

fm

′=�����

�������� ก.4 � ��ก ����������������(��.)

����ก����� �������ก����������� ����� ����

���������2 3 4 5 6 7 8

DB12

DB16

DB20

DB25

DB28

16.9

17.3

17.7

18.2

18.8

A B C D

20.6

21.4

22.2

23.2

24.4

24.3

25.5

26.7

28.2

30.0

28.0

29.6

31.2

33.2

35.6

31.7

33.7

35.7

38.2

41.2

35.4

37.8

40.2

43.2

46.8

39.1

41.9

44.7

48.2

52.4

3.7

4.1

4.5

5.0

5.6

A = 4 ��. �(�����ก9�" ��ก��%4/��� ก��ก

B = 9 ��. �� ก��กC = 1.9 ��.D = -!��"!���(�"!���� ก = db ���� 2.5 ��.

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Example 2.5 Design B1 in the floor plan shown below.

Slab thickness = 12 cm

LL = 300 kg/m2

= 280 kg/cm2

Steel: SD40

cf ′B1

B2

8.00

4.00

2.00

5.00 3.00

Reaction at B2’s ends = wL/2 = (2,331+504)(4)/2 = 5,670 kg

Slab DL = 0.12(2,400) = 288 kg/m2

Ultimate load = 1.4(288) + 1.7(300) = 913.2 kg/m2

2913.2(4) 913.2(3) 3 0.752,331 kg/m

3 3 2

−+ =

Load on B2 =

B2 weight (assume section 30× 50 cm) = 1.4(0.3)(0.5)(2,400) = 504 kg/m

Load on B1:

B1

B2:5,670 kg

2,350 kg/m 1,826 kg/m

5,670 kg

5.00 m 3.00 m

5,670 kg

913.2 kg/m913.21,437 kg/m

B1 weight: simply support min. depth = 800/16 = 50 cm

Try section 30 × 60 cm, wu = 1.4(0.3)(0.6)(2400) = 605 kg/m

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Mmax = 2,431(8.0)2/8

= 19,448 kg-m

1

max

524(5)(5 / 2)819 kg

8819(3) 2,456 kg-m

R

M

= =

= =

1,826 + 605 = 2,431 kg/m

8.00

5.00

2,350-1,826=524 kg/m

3.00R1

5.00

5,670 kg

3.00

max

5,670(5.0)(3.0)

810,631 kg-m

M =

=

Mu = 19,448 + 2,456 + 10,631 = 32,535 kg-m

Max. moment on B1:

USE DB20: d = 60 - 4 - 2.0/2 - 0.9 = 54 cm

0.85(280) 6120(0.85) 0.0306

4,000 6120 4000bρ = = +

ρmax= 0.75ρb = 0.75(0.0306) = 0.0230

2 2

4,00016.81

0.85 0.85(280)

32,535(100)41.32

0.9(30)(54)

y

c

un

fm

f

MR

bdφ

= = =′

= = =

ρmin = 14/fy = 14/4,000 = 0.0035

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21Required 1 1

1 2(16.81)(41.32)1 1

16.81 (4,000)

0.0114

n

y

m R

m fρ

= − −

= − −

=

ρρρρmin = 0.0035 < ρρρρ = 0.0114 < ρρρρmax = 0.0230 OK

0.60

0.30

6DB20BUT 6DB20 need bmin = 35.7 cm NG

Home work: redesign section

As = ρbd = 0.0114(30)(54) = 18.51 cm2

USE 6DB20 (As = 18.85 cm2)

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� Tension Steel Location

� Analysis of RC Beam

� Strength of Doubly Reinforced Beam

� Compression Steel Yield Condition

� Design of Double RC Beams

� Investigation of Double RC Beams

Asst.Prof.Dr.Mongkol JIRAVACHARADET

S U R A N A R E E

UNIVERSITY OF TECHNOLOGY

INSTITUTE OF ENGINEERING

SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Bending in Beam 4

Tension Steel Position in Beam

w

L

+ Mmax = wL2/8

Bending Moment Diagram

d

Effective depth

Compression face

Centroid ofsteel area

Elastic curve

Need reinforcement

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����������� ������������

L L

wL2/8

wL2/14 wL2/14d

d d

L L<< L

������ �����������������ก��ก��

d

d d

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������ ������������ !"���#���� �$�

Small -M

Critical sectionat face of supports

��%����ก��ก��$����#��#��ก���$�&��!��ก

- ����ก���� � � �������������

- ����ก���ก�����������������

#�� ก#������� :

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ก����������!��ก�$�&���ก#���#(��")*��

3 m 2 m 6 m

w w w = 1 t/m

1.04 t-m

-3.48 t-m

2.93 t-m

2DB20 7DB20 6DB20

2DB20

2DB20

A A

7DB20

2DB20

B B

2DB20

6DB20

C C

A

A

B

B

C

C

3 m 2 m 6 m

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Analysis of Single RC Beam (Tension steel yield)Analysis of Single RC Beam (Tension steel yield)

εcu = 0.003

c

d – c

εs

C = 0.85 f’ca b

T = As fs

a

dd – a/2

cus cu

s

c d cd c c

ε − = → ε = ε − ε

s y y s s yIf f /E f f ε > ε = → =

Check by ρ < ρb

[C = T] 0.85 f’c a b = As fy

s y1

c

A fa c a /

0.85 f b= → = β

s s s y

d cf E 6,120 f

c−

= ε = ≤

Mn = As fy (d – a/2)

Example 6.1 – Moment Strength of Single RC Beam

3DB25As = 14.73 cm2

30 cm

60 cm

f’c = 240 ksc, fy = 4,000 ksc

d = 60 – 4 – 0.9 – 2.5/2 = 52.6 cm

ρ = 14.73/(30x52.6) = 0.00933

จากตารางที ่ก.5 ρmin < ρ < ρmax

Assuming εεεεs > εεεεy T = As fy = 14.73 x 4.0 = 58.9 ton

s y

c

A fa

0.85f b=

′58.9

9.62 cm0.85 0.24 30

= =× ×

c = a / β1 = 9.62 / 0.85 = 11.32 cm

s cu

d cc− ε = ε

52.6 11.320.003 0.0109

11.32− = =

yy 6

s

f 4,0000.00196

E 2.04 10ε = = =

×εεεεs > εεεεy OK

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Nominal Moment Strength

Mn = As fy (d – a/2) = 58.9 (52.6 – 9.62/2)

= 2,815 ton-cm = 28.2 ton-m Ans

a = 9.62 cmC = 58.9 ton

T = 58.9 ton

n.a.c = 11.32 cm

εcu = 0.003

d = 52.6 cm

εy

εs = 0.0109

Tension, Compression and Balance FailuresTension, Compression and Balance Failures

εcu = 0.003

εs > εy

Tension Failure

εcu = 0.003

εs > εy → fs = fy

εcu = 0.003

εy = fy/Es ≈ 0.002

Balanced Failure

εcu = 0.003

εs = εy → fs = fy

εcu = 0.003

Compression Failure

εcu = 0.003

εs < εy → fs < fy

εs < εy

ρρρρ = ρρρρbρρρρ < ρρρρb ρρρρ > ρρρρb

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Compression FailuresCompression Failures

εcu = 0.003

εs

c

d

εy

s

cu

d cc

ε −=

ε

s cu s

d cf E

c−

= ε

s y

d cf 6,120 f

c− = <

a

0.85 f’c

C = 0.85 f’c b a

T = As fs

C = T

0.85 f’c b β1 c = As fs

c 1 s0.85 f b c 6,120 A (d c) / c′ β = − Solve for c

n s s

aM A f d

2

= −

Example 6.2 – Moment Strength of Single RC Beam #2

f’c = 240 ksc, fy = 4,000 ksc

d = 60 – 4 – 0.9 – 5.25 = 49.9 cm

ρ = 49.09/(30x49.9) = 0.0328

จากตารางที่ ก.5 ρ > [ ρb = 0.0262 ]

x = (4x1.25+4x6.25+2x11.25)/10 = 5.25 cm

4 cm

เหล็กปลอก 9 mm

d = ?

2.5 cm

x

10DB25As = 49.09 cm2

30 cm

60 cm

∴ Tension steel not yield : fs < fy

c 1 s0.85 f b c 6,120 A (d c) / c′ β = −

0.85×240×30×0.85×c = 6,120×49.09(49.9-c)/c

c2 + 57.8c – 2,882 = 0 → c = 32.1 cm

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a = β1 c = 0.85×32.1 = 27.3 cm

n s s

a 27.3M A f d 49.09 3.394 49.9 /100

2 2

= − = × −

= 60.4 ton-m Ans.

s

d cf 6,120

c−

=

49.9 32.16,120 3,394 ksc

32.1−

= =

Strength of Doubly Reinforced BeamStrength of Doubly Reinforced Beam

b

dh

d’

A’s

As

a

εcu = 0.003

εsT=As fy

x sε ′

cf ′85.0

0.85c cC f ba′=s s sC A f′ ′ ′=

= +1 2nM M M

Moment: Force:′= + = +1 2 c sT T T C C

′ ′= − + −

( )2c s

aC d C d d

′ ′ ′ ′= − + −

0.85 ( )2c s s

af ba d A f d d

′ ′ ′= +0.85s y c s sA f f ba A f

= +1 2s s sA A A

′= = =1 1 0.85s y c cT A f C f ba

′ ′ ′= = =2 2s y s s sT A f C A f

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ก����"�$��#"���#������!��ก��������� b

dh

d’

A’s

As

εcu = 0.003

εs

x sε ′

0.003 0.003 1s

x d d

x xε

′ ′− ′ = = −

Compression steel yield condition:2,040,000

y ys y

s

f f

Eε ε

′ ≥ = =

1From 0.85c s s y c s sT C C A f f b x A fβ′ ′ ′ ′= + → = +

( ) ( )1 10.85 0.85

s s y y

c c

A A f f dx

f b f

ρ ρ

β β

′ ′− −= =

′ ′

( ) ( )0.85 0.85

ρ ρ′ ′− −= =

′ ′s s y y

c c

A A f f da

f b f

Compression Steel Yield Compression Steel Yield

βρ ρ

′ ′′− ≥ −

10.85 6,1206,120

c

y y

f df d f

Stress in compression steel( )

βε

ρ ρ

′ ′′ ′= = − ≤ ′−

10.856,120 1 c

s s s yy

f df E f

f d

Double RC balance steel ratio sb b

y

f

fρ ρ ρ

′′= +

ε ε′ ′ = − ≤

0.003 1s y

dx

( )β

ρ ρ

′ ′− ≥ ′−

10.850.003 1

2,040,000yc

y

ff df d

Single RC balance steel ratio

ρ ρ ρ′

′= +max 0.75 sb

y

ff

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If comp. steel not yield:

1From 0.85c s s y c s sT C C A f f b x A fβ′ ′ ′ ′= + → = +

s y s yf fε ε′ ′< → <

10.85s y s s

c

A f A fx

f b β

′ ′−=

′ 0.85s y s s

c

A f A fa

f b

′ ′−=

Comp. Steel Yield: ( )( ) / 2 ( )n s s y s yM A A f d a A f d d′ ′ ′= − − + −

Comp. Steel NOT Yield: ( )( ) / 2 ( )n s y s s s sM A f A f d a A f d d′ ′ ′ ′ ′= − − + −

Example 1: Determine resisting moment of double RC beam with d = 50 cm,b = 40 cm, d’ = 6 cm, comp. steel 2DB20 (A’s = 6.28 cm2) and ten. steel 8DB25

(As = 39.27 cm2) use f’c = 240 ksc, fy = 4,000 ksc

50 c

m

40 cm

2DB20

8DB25

239.27 6.28 32.99 cms sA A′− = − =

32.990.0165

40 50ρ ρ′− = =

×

���������� �ก�����������ก������

10.85 6,1206,120

c

y y

f df d f

βρ ρ

′ ′′− ≥ −

0.85 0.85 240 6 6,1200.0150

4,000 50 6,120 4,000× × ×

= × −

Since 0.0165 0.0150,ρ ρ′− = > comp. steel yield 4,000 kscs yf f′ = =

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�� �ก���������������� :

1

0.85 6,120 0.85 240 6,1200.85 0.0262

6,120 4,000 6,120 4,000c

by y

ff f

ρ β ′ ×

= = = + +

���!�� �ก�������ก����"��ก������ :

max

4,0000.75 0.75(0.0262) 0.0074 0.0271

4,000s

by

ff

ρ ρ ρ′

′= + = + = [ 0.0196]ρ> = OK

( ) 32.99 4,00016.17 cm

0.85 0.85 240 40s s y

c

A A fa

f b

′− ×= = =

′ × ×

ก# �����$���%�& : ( ) ( )/ 2 ( )n s s y s yM A A f d a A f d d′ ′ ′= − − + −

39.27 4,000 (50 16.17 / 2) 6.28 4,000 (50 6)= × × − + × × −

= 6,633,030 kg-cm = 66.33 t-m Ans

Example 2: Repeat Ex.1 by changing reinforcing steel to comp. steel 2DB25 (A’s = 9.82 cm2) and ten. steel 6DB25 (As = 29.45 cm2) usef’c = 240 ksc, fy = 4,000 ksc

50 c

m

40 cm

2DB25

6DB25

229.45 9.82 19.63 cms sA A′− = − =

19.630.0098

40 50ρ ρ′− = =

×

Comp. steel not yield s yf f′∴ <

< 0.0150 (from Ex.1)

10.856,120 1

( )c

sy

f df

f dβ

ρ ρ

′ ′′ = − ′−

0.85 0.85 240 66,120 1 2,871 ksc

0.0098 4,000 50× × ×

= − = × ×

First trial: Comp. steel yieldassumption

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1

29.45 4,000 9.82 2,87112.92 cm

0.85 0.85 240 40 0.85s y s s

c

A f A fx

f b β

′ ′− × − ×= = =

′ × × ×

12.92 60.003 0.003 0.0016

12.92s

x dx

ε′− − ′ = = =

2,040,000 0.0016 3,264 kscs s sf E ε′ ′= = × = 2,871 ksc≠

Trial loopof f’s

f’s x2,871 12.92

3,264 12.37

3,152 12.52

3,187 12.47

3,175 OK

max

3,1750.75 0.75(0.0262) 0.0049

4,000s

by

ff

ρ ρ ρ′

′= + = +

[ ]0.0235 0.0196 ρ= > = OK

29.45 4,000 9.82 3,17510.62 cm

0.85 0.85 240 40s y s s

c

A f A fa

f b

′ ′− × − ×= = =

′ × ×

( ) ( )2n s y s s s s

aM A f A f d A f d d ′ ′ ′ ′ ′= − − + −

(29.45 4,000 9.82 3,175) (50 10.62 / 2) 9.82 3,175 (50 6)= × − × × − + × × −

5,242,969 kg-cm 52.43 t-m= = Ans

Moment strength: ( )0.852n c s s

aM f ab d A f d d ′ ′ ′ ′= − + −

( )10.610.85 240 10.61 40 50 9.82 3178 50 6

2

5,242,736 kg-cm 52.43 t-m

nM

= × × × − + × −

= = Ans

210.85 ( ) 0c s y s s cuf b x A f x A E x dβ ε′ ′ ′− + − =

26936 57702 360588 0x x− − = x = 12.48 cm

12.48 66120 3,178 ksc

12.48sf− ′ = =

a = 10.61 cm

Alternative method: Comp. steel not yield

1From 0.85c s s y c s sT C C A f f b x A fβ′ ′ ′ ′= + → = +

s sE ε ′

0.003 1s s

dE

′ ′ = −

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Design Procedure of Double RC BeamDesign Procedure of Double RC Beam

STEP 1: Moment strength from single RC beam

1 1 1

121 1

Choose 0.75 /

11.7

s sb s

yn y

c

A A A bd

fM f bd

f

ρ

ρρ

≤ ⇒ =

= − ′

STEP 2: Addition moment strength required

2 1/n u nM M Mφ= −

STEP 3: Addition tension steel As2

2 2 2( ) ( )n s yM T d d A f d d′ ′= − = −

STEP 4: Total tension steel As = As1 + As2

STEP 5: Stress in compression steel

11, /

0.85s y

c

A fa x a

f bβ= =

0.003 6,120s s y

x d x df E f

x x

′ ′− − ′ = = ≤

STEP 6: Compression steel

2s y s sA f A f′ ′=

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Example 3: Determine As and A’s required. MLL = 32 t-m, MDL = 18 t-m

f’c = 240 ksc, fy = 4,000 ksc

40 cm

50 c

m

60 c

m

d’ = 6 cm

A’s

As

a

εcu = 0.003

εs

Cc

T

xC’s

sε ′

cf ′85.0

Mu = 1.4 (18) + 1.7 (32) = 80 t-m

Mn = Mu/φ = 80/0.9 = 89 t-m

��� ��ก����� ��� �!��ก������"��� #�����������ก�����$�

As1 = 0.75ρbbd = 0.75(0.0262)(40)(50) = 39.3 cm2( )

121 1

2

11.7

0.0197 4.00.0197 4.0 40 50 1 /100 63.6 t-m

1.7 0.24

yn y

c

fM f bd

f

ρρ

= − ′

× = × × × − = ×

� %���&�ก Mn ����#��ก��%� 89 t-m ��กก�'� Mn1 &$��#���������ก������� �(%���(��ก�����

�� ��� �!"$� ����ก

Mn2 = Mn - Mn1 = 89 - 63.9 = 25.4 t-m

����ก��������$�����#��ก��(��:

222

25.4 10014.4 cm

( ) 4.0(50 6)n

sy

MA

f d d

×= = =

′− −

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��������ก�����$��������: As = As1 + As2 = 39.3 + 14.4 = 53.7 cm2

����ก�+��!��ก�$�&��������,� 7DB32(As = 56.3 cm2)

��&���� '����* ���������

1

1

39.3 4.019.3 cm,

0.85 0.85 0.24 40

/ 19.3/ 0.85 22.7 cm

s y

c

A fa

f b

x a β

×= = =

′ × ×

= = =

222.7 66,120 6,120 4,500 kg/cm

22.7s y

x df f

x

′− − ′ = = = >

����ก������������ก

2 22 14.4 cm USE 3DB25 ( 14.73 cm )s y s s sf f A A A′ ′ ′= → = = =

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Design of TDesign of T--BeamBeam

� Effective Flange Width

� Strength of T-Sections

� Maximum Steel in T-Beams

� T-Beams Design

Reinforced Concrete DesignReinforced Concrete Design

Asst.Prof.Dr.Mongkol JIRAVACHARADET

S U R A N A R E E

UNIVERSITY OF TECHNOLOGY

INSTITUTE OF ENGINEERING

SCHOOL OF CIVIL ENGINEERING

Moment Strength of Concrete SectionsMoment Strength of Concrete Sections

d

b

As

����ก����� ����ก� ����������������������������

��� ����ก� ���������� (� ��������� �)

Rectangular Sections :

T Beams :

d

b

bw

As

- �������� ����ก� ����������

- #���� ����ก� ����������

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Increase Increase BBendingending SStrengthtrength by by UUsingsing FFlangelange AArearea

N.A.

bf

As steel area canbe increased easily

T beams in a oneT beams in a one--way beamway beam--andand--slab floorslab floor

bw

t

SLABFLANGE

WEB

bw

bEbE

h

s = span

s0 = clear span

Built-in T Section

- ����������ก ����������������ก��

- ��������� ������������ก �!�����

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Effective Flange Effective Flange WidtWidth ( h ( bbEE ))

0.85 cf ′

b

Theoretical stress distribution

0.85 cf ′

bE

Simplified rectangular stress distribution

Built-in T-section

Determine Determine EffectiveEffective Flange Flange WidtWidth ( h ( bbEE ))

$�%������� ����������� ��%��ก%����&ก�����'$��(�)#����� �������ก��

E w

0

L / 4

b b 16t

s

= ≤ + ≤

bE

bws0

t bE

w

E w

w 0

b L /12

b b 6t

b s / 2

≤ +

= ≤ + ≤ +

Built-in L-section

Isolated T-section

E w

w

b 4 b

t b / 2

bE

t

bw

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Continuous TContinuous T--BeamBeam

- -

+ + +Bending MomentDiagram

Compression Areain Sections

A

A

Section A-A

B

B

Section B-B

C

C

Section C-C

?

Case 2: Compression area inflanges only

Behave as a rectangularsection: width = bE

bE

Midspan section: A-A

Case 1: Compression area inflanges and web

Behave as a compositeT-section

As

AsSupport section: B-B

Compression area in web(flanges cracked)

Behave as a rectangularsection: width = bw bw

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T = As fs

fc

Strength of TStrength of T--section (WSD)section (WSD)

Case 1: Compression area in flanges and web ( kd > t )

N.A.

bE

td

As

bw

εc

εs

kdA1A2/2 A2/2

Separate compression area into A1 and A2

1

1

2 c wC f b kd=Compression on A1

( )2

2

2 c E w

kd tC f b b t

kd

−= −Compression on A2

C2

C1

fc(kd-t)/kd

From ΣFx = 0, T = C1 + C2

( )1 2

2 2s s c w c E w

kd tA f f b kd f b b t

kd

−= + −

Define: ,s s

E c

A fm

b d fρ = =

( )1 2

2 2E c c w c E w

kd tb d m f f b kd f b b t

kdρ

−= + −

( ) ( ) ( )2 22 2 0w E w E E wb kd t b b kd m b d kd b b tρ+ − − − − =

Solve quadratic equation for kd

12c c

tM f bt jd

kd

= −

Concrete:

s s sM A f jd=Steel:

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Case 2: Compression area in flanges only ( kd < t )

bE

td

As

bw

T = As fs

fcC

εc

εs

kd

Behave as a rectangularsection: width = bE

[ ] 1

2 c E s sC T f b kd A f= =

[ ] 2Check s s

c E

A fkd t

f b= ≤

Strength of TStrength of T--section (SDM)section (SDM)

N.A.

bE

bw

As

t

εc = 0.003

εs > εy

d d - a/2

T = As fy

C

ก������ 1 : a ≤≤≤≤ t ก�"�����#��� $���ก� ���%��� ����$&��&��ก���� bE ��ก d

[ ] 0.85 c E s yC T f b a A f′= =

������� �'ก"� �"����(���)$��ก*� 0.85 c E

sy

f b tA

f

′≤

("��%� ��� a +��("���ก 0.85

s y

c E

A fa t

f b= ≤

ก�����"� +$�$�(, ( / 2) ( / 2)nM T d a C d a= − = −

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n 1 2M C (d a / 2) C (d t / 2)= − + −

2

c w

T Ca

0.85 f b−

=′

ก������ 2 : a > t ������"� �"����"��(�� T � ������������ 2 %������ ������ A1 "� �"���� C1

�����ก��� A2 "� �"���� C2

t

bE

As

bw

aA1A2/2 A2/2

Asw

bw

aA1t

bE

Asf

bw

A2/2 A2/2

1 c wC 0.85 f b a′= 2 c E wC 0.85 f (b b ) t′= −

s y 1 2T A f C C= = +

��������� 4.1 �������#ก�����"� �"���� Mn !�����"��(�� T �$�����ก�������%��� ����!�����.�"��

���$�/������ 8 �$(""��� ���"� ������� 4 �$(" ก�� �� �����"����ก"�( f’ c = 240 กก./2$.2

!��� �'ก fy = 4,000 กก./2$.2

bE

30 cm

63.5 cm

12 cm

As= 40.52 cm2

������ (1) �����������ก������ ���!"#���$%&�'�($���$����� )*#�

L/4 = 800/4 = 200 2$.,

bw + 16t = 30 + 16(12) = 222 2$.

"���"� ������� = 400 2$.

→→→→ ��+�)� → ������ bE = 200 %�.

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(2) ���������� a -*� ��)��./� a ≤≤≤≤ t

T = fy As = 4.0(40.52) = 162 (��

1623.97

0.85 0.85 0.24 200c E

Ta

f b= = =

′ × ×2$. < 12 2$. OK

(3) �������ก��"����$�$-�'�$�1*�*

( / 2) 162(63.5 3.97 / 2)nM T d a= − = −

= 9,965 (��-2$. = 99.7 (��-�$(" Ans

��������� 4.2 �������#ก�����"� �"����!�����"��(�� T ������ ก�� �� �����"�.����ก"�(

f’ c= 240 กก./2$.2 !��� �'ก fy = 4,000 กก./2$.2

80 cm

20 cm

N.A.

A1A2/2 A2/2

91 cm

As=85cm2

εcu=0.003

εs > εy

a

0.85f’c

C1

C2

T

40 cm

������ (1) �*��"#����$ก�����"�%*�9*&�!�����

2$. ������./���� bE = 80 2$.

2$. OK

4 4(40) 160E wb b≤ = =

2 40 / 2 20wt b≥ = =

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(2) ���������� a -*� ��)��./� a ≤≤≤≤ t

T = fy As = 4.0(85) = 340 (��

34020.8

0.85 0.85 0.24 80c E

Ta

f b= = =

′ × ×2$. > 20 2$. NG

������(�����ก������"� �"����!�����ก"�(��ก����%��%���

C1 + C2 = 0.85(0.24)(40)a + 0.85(0.24)(80-40)(20) = T = 340 (��

8.16a + 163.2 = 340

C1 = 8.16(21.7) = 177 (��, C2 = 163 (��

(3) �������ก��"����$�$-�'�$�1*�*

a = 21.7 %�.

21.7 20177 91 163 91

2 2nM = − + −

= 27,406 (��-2$. = 274 (��-�$(" Ans

Maximum and Minimum SteelMaximum and Minimum Steel s EA b dρ=

ก������ 1 : a ≤≤≤≤ t ก�"�����#��� $���ก� ���%��� ����$&��&��ก���� bE ��ก d

�"*$�#� �'ก�%"*$���%:���%$�;� Wb b

E

bb

ρ ρ=

10.85 6,1206,120

cb

y y

ff fβ

ρ ′

= +

ก������ 2 : a > t ������"� �"�"��(�� T $���ก��� A2 "� �"���� C2 = 0.85f’c(bE – bw)t ��*�$!��

���������(���$��"*$�#� �'ก�%"*$ Asf ��*�$!�����

( )0.85 c E Wsf

y

f b b tA

f

′ −=

�"*$�#� �'ก�%"*$���%:���%$�;� ( )Wb b f

E

bb

ρ ρ ρ= +

( )0.85f c E Wy W

tf b b

f b dρ ′= −

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x

w

Vu

Mu

w

x

Transverse reinforcement:

� �'ก�%"*$$�ก���%;� 0.75sb

E

Ab d

ρ ρ= ≤

� �'ก�%"*$�������%;� 0.8 14csW

W y y

fAb d f f

ρ′

= ≥ ≥

%�� "� � �'ก�%"*$"� +$�$�(,� ��� ���(��(�� T �����ก���"� �"����

min

1.6 14c

y y

f

f fρ

′≥ ≥

��������� 4.3 �������# ����",�2'�(,� �'ก$�ก%;�����������%;������$. �!����� ���(����.�"��

ก�� �� f’ c = 240 กก./2$.2 ��� fy = 4,000 กก./2$.2

80 cm

20 cm

N.A.

A1A2/2 A2/2

91 cm

As

εcu=0.003

εs > εy

ab

0.85f’c

C1

C2

T

40 cm

xb

������ (1) �(��� ����ก�%���*� xb ��� ab ���%:���%$�;����$��"���

6,120 6,120(91)55.0 cm

6,120 6,120 4,000by

dx

f= = =

+ +

ab = β1 xb = 0.85 × 55.0 = 46.8 cm > [ t = 20 cm ]

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(2) �*��"#��"*$�#� �'ก�%"*$ Asb .�%:���%$�;����$��"���

��������ก ab > t ���(���� ���"�������� 2 %���

C = C1 + C2 = 0.85 f’ c bw ab + 0.85 f’ c (bE - bw) t

= 0.85(0.24)(40)(46.8) + 0.85(0.24)(80-40)(20) = 545 (��

�"*$�#� �'ก�%"*$���%:���%$�;� = 136.3 2$.2545.0

4.0sby

CA

f= =

(3) �*��"#��"*$�#� �'ก$�ก���%;�

max As = 0.75 Asb = 0.75(136.3) = 102.2 2$.2

(4) �*��"#��"*$�#� �'ก�������%;�

20.8 0.8 240(40)(91)11.3 cm

4,000c w

y

f b d

f

′= =

214 14(40)(91) 12.7 cm

4,000wy

b df

= = ��+�)� min As

TT--Beam Design ProcedureBeam Design Procedure

If Mn < Mu / f then a > t NG STEP 4.2

STEP 1: Compute ultimate moment Mu

STEP 2: Determine effective width bE

STEP 3: Assume a = t and Compute Mn

C = 0.85 f’c bE t and Mn = C (d - t/2)

If Mn ≥ Mu / f then a ≤ t OK STEP 4.1

tbE

a

a

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STEP 4.1: Design as section width bE or…

( )/ /

required / 2 0.9

= ≈−

φ φu us

y y

M MA

d a f d f

STEP 4.2:Separate compression to C1 and C2

( ) ( )

( ) ( )

1 2

2

/ / 2 / 2

0.85 / 2 / 2

u

c w

M C d a C d t

f b a d a C d t

φ = − + −

′= − + −

- Required As = (C1 + C2) / fy

( )2 0.85 c E wC f b b t′= −

1- Compute 0.85 c wC f b a′=

- Solve quadratic equation for a

a

a

bE

Assume d – a/2 ≈ 0.9 d

��������� 4.4 ��ก� � �'ก�%"*$"� �"����!����� ���(���� �����"� +$�$�(,��� ��ก ""�;ก�����

50 (��-�$(" +$�$�(,��� ��ก ""�;ก�" 100 (��-�$(" f’ c= 240 กก./2$.2, fy = 4,000 กก./2$.2

������ (1) �����#ก�����"� +$�$�(,������(���ก�"

Mu = 1.4(50) + 1.7(100) = 240 (��-�$("

������(���ก�"!�� (��-�$("240266.7

0.9u

n

MM

φ= = =

(2) �*��"#������� a $�กก��� t "��)$�+��%$$;(*. � a = t

(��

(��-�$("

��������กก�����+$�$�(,���(���ก�"$�กก��� 264.4 (��-�$(" ��������� a (���$�กก��� t

0.85 0.85(0.24)(80)(20) 326.4c EC f b t′= = =

( / 2) 326.4(91 20 / 2) /100 264.4nM C d t= − = − =

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(3) �����#�"*$�#� �'ก As +��� ��������"� �"��������%��%��� (A1 ��� A2)

1 20.85 0.85 ( / 2)2n c c

aM f A d f A d t

′ ′= − + −

20266.7(100) 0.85(0.24)(40 ) 91 0.85(0.24)(40)(20)(91 )

2 2

aa = − + −

24.08 742.5 13448 0a a− + = a = 20.4 2$.

��ก(���������� 4.3 max As = 102.2 2$.2, min As = 12.7 2$.2

min As < As < max As OK

C1 = 0.85(0.24)(40)(20.4) = 166.5 (��

C2 = 0.85(0.24)(80-40)(20) = 163.2 (��

As = (166.5+163.2)/4.0 = 82.4 2$.2

ขอสอบภย

ขอที่ : 50

คานรูปตวัทีโดดๆ มีปกคานกวาง = 80 ซม. หนา = 8 ซม. ตัวคานกวาง = 25 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As = 7.0 ซม.2 ที่ความลึกประสิทธิผล d = 40 ซม. ถาใช fc = 45 กก./ซม.2 และ fs= 1200 กก./ซม.2 จะพบวาตําแหนงแนวแกนสะเทินอยูใตปกคาน ดังนั้น หากสมมุติใหตําแหนงของแรงอัดที่ไดจากคอนกรีตอยูที่กึ่งกลางความหนาของปกคาน จงประมาณคาโมเมนตตานทานปลอดภัยของคานนี้

80 cm

25 cm

As = 7.0 cm2

d = 40 cm

8 cm

jd = 40 – 4 = 36 cm

�ก�%���*�����.(���ก��� ������"� �"��������"��(����4 cm

kd = 12 cm

1 2 c w c E w

1 2kd tC C f b kd f (b b )t

2 2kd−

+ = + −

1

1C 45 25 12 6,750 kg

2= × × × =

2

2 12 8C 45 (80 25) 8

2 12× −

= × × − ××

= 13,200 kg

T = 7.0×1200 = 8,400 kg < C1+C2 � Control

M = 8,400×36/100 = 3,024 kg-m

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ขอสอบภย

ขอที่ : 51

คานรูปตวัทีโดดๆ มีปกคานกวาง = 80 ซม. หนา = 8 ซม. ตัวคานกวาง = 25 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As ทีค่วามลึกประสิทธิผล d = 40 ซม. ถาใช fc = 45 กก./ซม.2 และ fs = 1200 กก./ซม.2 จงประมาณคา min As ที่ตองใชตามมาตรฐานกาํหนด

80 cm

25 cm

min As

d = 40 cm

8 cm

= 5.83 cm2

s Wy

14min A b d

f=

1425 40

2400= × ×

ขอสอบภย

ขอที่ : 57

คานรูปตวัทีโดดๆ มีความกวางประสทิธิผลของปกคาน = 120 ซม. หนา = 8 ซม. ตัวคานกวาง = 30 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As = 48.24 ซม.2 ที่ความลึกประสิทธิผล d = 50 ซม. เพื่อรับโมเมนตประลัย (Mu) ชนิดบวก = 50 ตัน-เมตร ถาใช fc’ = 200 กก./ซม.2 และ fy = 3000 กก./ซม.2 จงใชวิธี USD ประมาณคา As ที่ตองใช

120 cm

30 cm

As = ?

d = 50 cm

8 cm ��)�� a = t = 8 cm

C = 0.85 f’c bE t

= 0.85×200×120×8/1,000 = 163.2 ton

Mn = C(d-a/2) = 163.2(50-8/2)/100 = 75 t-m

Mu/φ = 50/0.9 = 55.6 t-m < Mn a < t

�*����� ���(��%��� ����$ก���� bE : un 2

MR

bd=φ

5

2

50 100.9 120 50

×=

× ×= 18.5 ksc

n

y

2mR11 1

m f

ρ = − −

= 0.0065 As = 0.0065x120x50 = 39.3 cm2

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Shear in Beams 1

� Shear Failure

� Shear Strength of Concrete Section

� Design for Shear (WSD)

� Design for Shear (SDM)

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

A

A

M

V

Shear and Diagonal Tension

A

A

VQIt

ν =Shear stress

V

M

Mcf

I=Bending stress

ν

f

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45o

Web-shear crack

Pure shear at neutral axis:

ν

ν ν

ν45o

ν

ν

(max)tf ν= ν

tf90o

Shear Flexure Effects

ν

ν

ν

ν

tf tf νν

(max)tfmaxα

tf

ν

tf

( ),tf ν

( )0, ν−

(max)tf

max2α

Below neutral axis: Combination of shear stress and tensile stress

Principal stress:

22

(max) 2 2t t

t

f ff ν

= ± +

ft (max)

ft (min)

ft (max) > fr

Crackdirection

Typical cracking due to principal tensionTypical cracking due to principal tension

Shear crackFlexure crack

Shear Flexurecrack

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Shear Stresses

bshear stressdistribution

maxmax

V QbI

ν =approximate V

bdν =

Shear Capacity Mechanism

Vc

carried bycompression

Va

carried byfriction aggregateinterlocking

Vs carried bydowel action

(shear)Vc = concrete

resistance

Vs = (shear) steelresistance

Total resistance = Concrete + Steel resistanceTotal resistance = Concrete + Steel resistance

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Shear Strength of Shear Strength of ConcreteConcrete

Shear strength: cc

w

Vv

b d=

from experiment 20.50 176 0.93 kg/cmuc c w c

u

V dv f f

Mρ′ ′= + ≤

vc

0.50 176 uc c w

u

V dv f

Mρ′= +

0.93 cf ′

/w u uV d Mρ

1.0u

u

V d

M≤

sw

w

A

b dρ =

Design for Shear (WSD)Design for Shear (WSD)

Shear strength of concrete

0.265 91.4 0.464c c c

V dV f b d f b d

M

ρ ′ ′= + ≤

Simple formula: 0.29c cV f b d′=

Shear strength from concrete & steel:c sV V V= +

Required shear strength from steel:s cV V V= −

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Shear Strength Provided by StirrupShear Strength Provided by Stirrup

d

d

s s s /n d s=Number of stirrup

v ss v s

A f dV A f n

s= =

Shear strength provided by stirrup

Av = 2As

Shear Design RequirementsShear Design Requirements

Max. shear strength:max 1.32 cV f b d′=

Max. stirrup spacing:max / 2 60 cms d≤ ≤

Ifmax0.795 / 4 30 cmcV f b d s d′> → ≤ ≤

WSD

Minimum stirrup: sb0015.0Amin v =

orb0015.0

As v

max =

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�������ก���ก��� ��ก������������

Step 1 ���������� � V ������������ก��������� d ��ก������� ���

Step 2 ������ก������ �� �ก��� dbf29.0V cc ′=

�� V < Vc �!��"�� �ก���#�ก������ �#�กก�"������ ก��

→ $%���&ก'� ก�� ����!��(�����"�#�ก���!��)→ ��*����������

Step 3 ������ก������ �#�ก���!�� dbf32.1V cmax ′=

�� V > Vmax �!��"��������#�*���+#",��, → ,��#*����������

Step 4 ������ก������ ������ ก����ก��&ก'� ก cs VVV −=

�����"���&ก'� กs

vvV

dfAs =

WSD

Step 5 �����"���&ก'� ก#�ก���!��b0015.0

As v

max =

�� dbf795.0V c′≤ → smax = d/2 ≤ 60 cm

�� dbf32.1Vdbf795.0 cc ′≤<′ → smax = d/4 ≤ 30 cm

single closed loop stirrup has 2 legs

('� ก�����#�! *�)

Av = 2 As : ,�-������&ก'� ก$�������� �

fv : ��"�������� #$��* ��&ก'� ก

SR24 : f v = 1,200 ksc

SD30 : f v = 1,500 ksc

SD40 : f v = 1,700 ksc

WSD

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Example 1 : Shear design by WSD

b = 30 cm, d = 45 cm

f’c = 240 ksc, fy = 4,000 ksc

@ critical section V = 15 ton

dbf29.0V cc ′=

000,1/453024029.0 ××=

= 6.07 ton

dbf32.1V cmax ′= ton 61.27=

ton 63.16dbf795.0 c =′

cs VVV −=

ton 93.807.615 =−=

s

vvV

dfAs = cm 45.13

93.8457.157.1

=××

=

USE Stirrup DB10 @ 13 cm

��� ��ก dbf795.0V c′≤ → smax = 45/2 = 22.5 cm ≤ 60 cm

b0015.0A

s vmax = cm 9.34

300015.057.1

=

Simple formula: 0.53c cV f b d′=

Shear strength with axial load:

Compression : 0.53 1 0.0071 kguc c w

g

NV f b d

A

′= +

Tension : 0.53 1 0.0029 kguc c w

g

NV f b d

A

′= +

0.50 176 0.93uc c c

u

V dV f b d f b d

M

ρ ′ ′= + ≤

Design for Shear (Design for Shear ( SDSDMM)) SDM

ก�������������ก�� :

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Design for Shear (Design for Shear ( SDSDMM)) SDM

ก������ �* ���!��#��&ก������� � : Vn = Vc + Vs

ก����������$��#�ก������ � : Vn ≥ Vu / φφφφ , φφφφ = 0.85 for shear

Av = 2 As

d

s s

d

������'� ก$�������� � : n = d / s

ก������ ���ก��&ก'� ก :

nfAV yvs =s

dfA yv=

$�ก�� ก��� ก������ ���&ก'� ก����� ก�� : Vs = Vu / φφφφ - Vc

�����"���&ก'� ก :s

yv

V

dfAs =

ACI318: 11.4.6 ACI318: 11.4.6 –– Minimum Shear ReinforcementMinimum Shear Reinforcement SDM

- h ≤ 2.5 tf

- h ≤ 1/2 bw

11.4.6.1 – A minimum area of shear reinforcement, Av,min , shall be provided

in all reinforced concrete flexural members where Vu ≥ 0.5 φφφφ Vc, except in

members:

• Footings and solid slabs

• Concrete joist construction

• Beams with h ≤ 25 cm

• Beam integral with slabs with h ≤ 60 cm and

bw

tfh

h

b

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ACI318: 11.4.6 ACI318: 11.4.6 –– Minimum Shear ReinforcementMinimum Shear Reinforcement SDM

11.4.6.3 – Where shear reinforcement is required, Av,min shall be computed

by

ycmin,v f

sbf2.0A ′=

but shall not be less than 3.5bs/fy �����#!������ ksc 306fc <′

��� ���� ��ก���ก��ก���� !b5.3

fA

bf2.0

fAs yv

c

yvmax ≤

′=

dbf1.1V cs ′≤• smax = d/2 ≤ 60 cm

dbf1.2Vdbf1.1 csc ′≤<′• smax = d/4 ≤ 30 cm

dbf1.2V cs ′>• ������� �"���!

(ก) ก����ก������กก���� �$����d d

d d

(*) ���/0#����� �!������ก�� ก���

���������#$"#�ก���ก�����������#$"#�ก���ก��ACI 11.1.3.1 – For nonprestressed members, sections located less than a distanced from face of support shall be permitted to be designed for Vu computed at a

distance d.

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�"���!%�ก&���� ����ก��ก������ �"���!%�ก&���� ����ก��ก������

Vu

�������ก��

(ก) �������������ก��������

Vu

dd

Vu

(�) ���������-���

Vu

�������ก��

(�) ������������ ���!�

Vu

d

�������ก��

(�) ���"#$%#�������กก�&"���'(����)ก��"#$������

�������ก���ก��� ��ก������������

Step 1 ���������� � Vu ������������ก��������� d ��ก������� ���

Step 2 ������ก������ �� �ก��� dbf53.0V cc ′=

�� Vn < Vc �!��"�� �ก���#�ก������ �#�กก�"������ ก��

→ $%���&ก'� ก�� ����!��(�����"�#�ก���!��)→ ��*����������

Step 4 ������ก������ �#�ก���!�� dbf1.2V cmax,s ′=

�� Vs > Vs, max �!��"��������#�*���+#",��, → ,��#*����������

Step 3 ������ก������ ������ ก����ก��&ก'� ก cs VVV −=

SDM

ก������ ������ ก�� Vn = Vu / φφφφ

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Step 6 �����"���&ก'� ก#�ก���!��b5.3

fA

bf2.0

fAs yv

c

yvmax ≤

′=

�� dbf1.1V cs ′≤ → smax = d/2 ≤ 60 cm

�� dbf1.2Vdbf1.1 csc ′≤<′ → smax = d/4 ≤ 30 cm

single closed loop stirrup has 2 legs

('� ก�����#�! *�)

Av = 2 As : ,�-������&ก'� ก$�������� �

�����"���&ก'� ก����� ก��s

yv

V

dfAs =Step 5

Variation of Shear Capacity

Mid spanSupport

d critical sectionwuL/2

φ Vc φ Vc/2

wu

φ Vn

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��%�������� 6.1 ก�����&ก'� ก������� �$����%"������ ก�����ก����� �ก���= 280 ksc $%���&ก'� ก DB10 ก������&ก!��# fy = 4,000 ksc

PL = 5 tonsPD = 2 tons

PL = 5 tonsPD = 2 tons

wL = 3 t/mwD = 2 t/m

2.5 m 2.5 m4.0 m

A

A

Section A-A40 cm

d =

53

cm

cf ′

1. ����%*�����������

wu = 1.4(2) + 1.7(3) = 7.9 t/m

Pu = 1.4(2) + 1.7(5) = 11.3 ton

!������/0#����� � Vu :

46.85 ton

27.1 ton

15.8 ton

-15.8 ton

-27.1 ton

-46.85 ton

2.5 m4 m

2.5 m

wu = 7.9 t/mVu

d = 53 cm

!##������#ก���!� 30 1#.

Vu/φ ������� d = (46.85 – 7.9(0.15+0.53))/0.85 = 48.80 ton

2. ����%*ก�������������ก��

c cV 0.53 f bd′= 0.53 280 40 53 /1,000= × × = 18.80 ton

3. ����%*ก��������������"��ก�-�ก� ��ก���ก

Vs = Vn – Vc = 48.80 – 18.80 = 30.00 ton

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4. ����%*ก���������� Vs ��ก���� ! %�� �"���!�����!�/���/� ��0��?

= 74.50 tons,max cV 2.1 f bd′= 2.1 280 40 53 /1,000= × ×

[ Vs = 30.00 ton ] < Vs, max �"���!�����!�/���/�

c1.1 f bd 1.1 280 40 53 /1,000′ = × × = 39.02 ton > Vs

smax = d/2 = 53/2 = 26.5 cm < 60 cm

5. ����%*��� ���� ��ก���ก����"��ก�

� $%���&ก'� ก DB10 '� ก'3�(! *�) Av = 2(0.785) = 1.57 cm2

v y

s

A f ds

V=

1.57 4.0 5330.00× ×

= = 11 cm

!������#$" DB10 @ 0.11 �. ������ d -�ก3�%- !����

2.5 m2.5 m

[email protected] [email protected]

3.7 m

[email protected]

6. ��ก��� ��ก���ก���ก���$�%� 27.1 ton

15.8 ton

-15.8 ton-27.1 ton

4 mVu/φ = 15.8/0.85 = 18.6 ton

Vu/φ < [ Vc = 18.8 ton ] Use min. stirrup

v ymax

A fs

3.5b=

1.57 4,00045 cm

3.5 40×

= =×

d/2 = 53/2 = 26.5 cm < 60 cm !������#$" DB10 @ 0.25 �. ���ก���$�%�

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� Shear design summary

� More detail shear design

� Shear span

� Deep beam

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Shear in Beams 2

Shear Design SummaryShear Design Summary

WSD SDM

Shear: V = VDL + VLL Shear: Vu = 1.4 VDL + 1.7 VLL

Vn = Vu / φ

Steel: Vs = Vn - VcSteel: Vs = V - Vc

Spacing: s = Av fy d / VsSpacing: s = Av fs d / Vs

Min. Stirrup: smax = Av fy / 3.5 bMin. Stirrup: smax = Av / 0.0015 b

Chk. light shear: ′≤s cV 1.1 f bd

Concrete: ′=c cV 0.53 f b dConcrete: ′=c cV 0.29 f b d

Chk. light shear: ′≤ cV 0.795 f bd

smax ≤ d/2 ≤ 60 cmsmax ≤ d/2 ≤ 60 cm

Chk. heavy shear: ′≤s cV 2.1 f bdChk. heavy shear: ′≤ cV 1.32 f bd

smax ≤ d/4 ≤ 30 cmsmax ≤ d/4 ≤ 30 cm

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���������ก �������������������ก������������ก �������������������ก���

Max. shear @ midspan

Luu

w LV

8=

LL full spanDL full span

(ก) �����ก�� ����������ก�������������

LL half spanDL full span

(�) �����ก�� ������������������ก��������������

(�) Shear force envelop

uu

w LV

2=

Luu

w LV

8=

Max. shear @ ends

uu

w LV

2=

EXAMPLE 6-2 More Detailed Design of Vertical Stirrups SDM

The simple beam supports a uniformly distributed service dead load of 2 t/m, includingits own weight, and a uniformly distributed service live load of 2.5 t/m. Design verticalstirrups for this beam. The concrete strength is 250 ksc, the yield strength of the flexural reinforcement is 4,000 ksc.

Shear2_11

DL = 2 t/mLL = 2.5 t/m

L = 10 m30 cm

d = 64 cm

wu = 1.4(2) + 1.7(2.5) = 7.05 t/m

wLu = 1.7(2.5) = 4.25 t/m

wuL/2 = 7.05(10)/2 = 32.25 ton

wLuL/8 = 4.25(10)/8 = 5.31 ton

VVuu//φφφφφφφφ Diagram :Diagram :

32.25/0.85 = 37.94 ton

5.31/0.85 = 6.25 ton

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Support Midspan

37.94 t

Vu/φ

6.25 t

Shear2_12

Vu / φ at d = 37.94 – (0.84/5)(37.94 – 6.25) = 32.62 ton

c cShear strength of concrete V 0.53 f bd 0.53 250 (30)(64) /1,000 16.09 ton′= = =

Required Vs

8.05 t

16.09 t

Vc0.5Vc

Is the cross section large enough?Is the cross section large enough?

n,max c cV V 2.1 f bd 16.09 2.1 250 (30)(64) /1,000 79.84 32.62 ton ′= + = + = > OK

84 cm

Critical section32.62 t

c c

max

V 1.1 f bd 16.09 1.1 250 (30)(64) /1,000 55.6 32.62 ton

s d/ 2 60 cm

′+ = + = >

⇒ ≤ ≤

assume column width = 0.40 cm

Shear2_13

Minimum stirrup : (ACI 11.5.6.3)Minimum stirrup : (ACI 11.5.6.3) USE RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2400 ksc

v,min cy

bsA 0.2 f

f′=

Rearranging gives

(ACI Eq. 11-13)

v ymax

c

A f 1.27(2,400)s 32 cm

0.2 f b 0.2 250 (30)= = =

but not less than v ymax

A f 1.27(2,400)s 29 cm

3.5b 3.5(30)= = =

Use smax = 29 cm < [d/2 = 64/2 = 32 cm] < 60 cm

Compute stirrup resuired at d from support

v y

u c

A f d 1.27(2.4)(64)s 11.8 cm

V / V 32.62 16.09= = =

φ − −

Use [email protected]. Change spacing to s = 15 cm where this is acceptable, and thento the maximum spacing of 29 cm.

Compute Vu/φφφφ where s can be increased to 15 cm.

v yuc

A f dV 1.27(2.4)(64)V 16.09 29.1 ton

s 15= + = + =

φ

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Shear2_14

Support Midspan

37.94 t

Vu/φ

6.25 t8.05 t

Vc = 16.09 t

0.5Vc

84 cm

Critical section32.62 t

500 cm

29.1 t

x

37.94 29.1x 500 140 cm from support

37.94 6.25−

= × =−

Change s to 29 cm, compute Vu/φ

v yuc

A f dV 1.27(2.4)(64)V 16.09 22.82 ton

s 29= + = + =

φ37.94 22.82

x 500 239 cm from support37.94 6.25

−= × =

s=15 cm @x = 140 cm

s=29 cm @x = 239 cm

Support Midspan500 cm

20 cm1 cm

RB9 @ 0.11 m : 20+1+11@11 = 142 cm > 140 cm OK

11@11 cm

RB9 @ 0.15 m : 142+7@15 = 247 cm > 239 cm OK

RB9 @ 0.29 m : 247 + 8@29 = 479 cm

7@15 cm 8@29 cm

[email protected] [email protected] [email protected]

Shear2_15

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Shear Span (a = M /V )

P Pa a

Distance a over which the shear is constant

M = VaMomentDiagram +

V = +P

V = -P

ShearDiagram +

-

Shear2_16

Crack Pattern in Several Lengths of BeamSpan

Mark (m) a/d

1 0.90 1.0

2 1.15 1.5

3 1.45 2.0

4 1.70 2.5

5 1.95 3.0

6 2.35 4.0

7/1 3.10 5.0

8/1 3.60 6.0

10/1 4.70 8.0

9/1 5.80 7.0

Shear2_17

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Variation in Shear Strength with a/d for rectangular beams

a/d0 1 2 3 4 5 6 7

Fai

lure

mom

ent =

Va

Deepbeams

Shear-tension andshear-compressionfailures

Diagonal tensionfailures

Flexuralfailures

Inclined crackingstrength, Vc

Flexural momentstrengthShear-compression

strength

Shear2_18

DEEP BEAMDEEP BEAM

Brunswick Building. Note the deep concrete beams at the top of the ground columns.These 168-ft beams, supported on four columns and loaded by closely spaced fascia columns above, are 2 floors deep. Shear stresses and failure mechanisms were studied on a small concrete model. (Chicago, Illinois)

Shear2_19

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Shear2_20

Deep Beams

When shear span a = M /V to depth ratio < 2

Mechanism:

Compressivestruts

If unreinforced,large cracks may openat lower midspan.

Use both horizontaland vertical mayprevent cracks

Deep beams are structural elements loaded as beams in which a significant amount of the load is transferred to the supports by a compression thrust joining the load and the reaction.

Shear2_21

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Definition of Deep BeamDefinition of Deep Beam

ACI 10.7.1 – Deep beams are members loaded on one face and supported on the opposite face so that compression struts can develop between the loads and thesupports, and have either:

(a) clear spans, Ln, equal to or less than four times the overall member depth; or

(b) regions with concentrated loads within twice the member depth from the faceof the support.

Ln

h Ln / h ≤ 4

h

P

xx < 2 h

Shear2_22

Basic Shear Strength: φVn ≥ Vu

where Vn = Vc + Vs

Location for Computing Factored Shear:

(a) Simply Supported Beams

(Critical section located at distance z from face of support)

- z = 0.15Ln ≥ d for uniform loading

- z = 0.50a ≥ d for concentrated loading

(b) Continuous Beams

Critical section located at face of support

Design Criteria for Shear in Deep BeamsDesign Criteria for Shear in Deep Beams

Limitation on Nominal Shear Strength

n,max cV 2.7 f bd′=

Shear2_23

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c cV 0.53 f bd′=Simplified method:

u uc c c

u u

M V dV 3.5 2.5 0.50 f 176 bd 1.6 f bd

V d M

′ ′= − + ρ ≤

u

u

Mwhere 1.0 3.5 2.5 2.5

V d≤ − ≤

Shear Strength of Concrete, Vc

If some minor unsightly cracking is not tolerated, the designer can use

Shear Reinforcement, Vs

v n vh ns y

v h

A 1 L / d A 11 L / dV f d

s 12 s 12

+ − = +

����� Av = �������� ก��������������� (��.2), Avh = �������� ก�������������� (��.2)

sv = ������ ���� ก!��ก���� (��.), sh = ������ �"����� ก��� (��.)Shear2_24

Minimum Shear Reinforcement

v

dmaximum s 30 cm

5≤ ≤

h

dmaximum s 30 cm

5≤ ≤

and

vh hminimum A 0.0015 b s=

v vminimum A 0.0025 b s=

Shear2_25

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��������� 5.6 ��ก������ ก�����������#$ ����% �&�'��(��'�����$ ���กก��$ #��)*("����$ ���ก+&,� �)� 60 ���+������)*(+�&�'�ก', �% � 3.6 ���� % �ก', � 35 ��. ���%' �� 1ก!��#2324�d = 90 ��. +&, f�

c= 280 กก./��.2 ��� fy = 4,000 กก./��.2

����� (a) �2) �; '� �!<�% ��1ก����=��#$ ����% ���

Ln/h = 360/100 = 3.6 < 4.0 �������!"���� #ก

35 cm

d =

90 c

m

4DB36

40 cm 40 cm3.6 m

5 cm 5 cm35@10 = 3.5 m

60 t

1.20 m

60 t

1.20 m

Shear2_26h

= 1

00 c

m

(b) ��, ��('2ก>�#$ ������$ ���กก��$ �!<�)*( ?(�+&,&�'�% ������ a = 1.20 �.

0.50a = 0.5(1.20) = 0.60 ���� < [d = 0.90 ����]

��, ��('2ก>���@�� 0.60 �. ) ก42'"��)*(������

(c) ก$ ��������"��% �?(�=������ ก��������������, ��ก'2ก>� ��$ �� �กก��$ �!<�)*(!�����%��

1.7 LL = 1.7(60) = 102 ���

�������$ ���ก%���1���,����������ก����$ ���กก��$ �!<�)*( ?(�+&,'23�����(���, ��('2ก>�

102(60)0.67

102(90)u

u

MV d

= =

��'%@;#$ ����% ��1ก%�� 3.5 2.5 3.5 2.5(0.67) 1.83 2.5 u

u

MV d

− = − = < OK

1.83 0.50 176 uc c

u

V dv f

Mωρ

′= +

4(10.18)0.0129

35(90)wρ = =

Shear2_27

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( )176 0.01291.83 0.50 280

0.67cv

= +

= 1.83[8.37 + 3.39] = 21.5 กก./��.2

2Upper limit: 1.6 1.6 280 26.77 kg/cmc cv f ′= = =

ก$ ��������������"��%��ก�� Vc = vc bw d = 21.5(35)(90)/1,000 = 67.8 ���

(d) ก�� ����������'��� ��ก(�

102Required 120 ton

0.85u

n

VV

φ= = =

������) ก Vn ��,��ก � > Vc (120 > 67.8) (�������,��ก ���� ก�#�2������������

����)*

n,max cV 2.7 f bd 2.7 280(35)(90) /1,000′= =

= 142 ��� > 120 ��� OK

Shear2_28

(e) ก���+��*�� ,ก�����������

2

1 / 11 /12 12

v n vh n s

y

A L d A L d Vs s f d

+ − + =

#$ ���� Ln/d = 4 : Vs = 120 – 67.8 = 52.2 ���

b = 35 ��. ��� fy = 4,000 กก./��.2

v vh

v h

A A5 7 52.20.145

s 12 s 12 4.0(90)

+ = =

min Av = 0.0025 b sv max sv = d/5 =18 ��.

min Avh = 0.0015 b sh max sh = d/5 = 18 ��.

���+&, DB12 ' �+���'���+������(, �������� �ก�� sh = 18 ��.

min Avh = 0.0015(35)(18) = 0.945 ��2

%� ������+�, Avh = 2(1.13) = 2.26 ��2 > 0.945 ��2 OK

Shear2_29

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vA 5 2.26 70.145

s 12 18 12

+ =

��%� Avh ��+�#�ก �

[ ]120.145 0.0732 0.172

5= − =vA

s

#$ ������� ก�@ก���� DB12: Av = 2(1.13) = 2.26 ��2

�,��ก � s = 2.26/0.172 = 13.1 ��. < [d/5 = 18 ��.] OK

(������+&, DB12 �!<���� ก�@ก����*ก���� 12 ��. ���(���&�'�% �

Shear2_30

35 cm

90 c

m

4DB36

40 cm 40 cm3.6 m

30@12 = 3.6 m

[email protected]

[email protected]

min Av = 0.0025(35)(18) = 1.58 ��2 < [Av = 2.26 ��.2] OK

��������� 6.1 #$ ����% �(���#(�+��@! �2) �; ��$ ���ก���*ก%�������$ ���ก���*ก)�� ก#*() ก������ก������ ก�����������?(�+&,��� ก!��ก+���'(2�� #��*�2+�,���� #�'����'� ���$ ���ก)������$ ���ก%��+&,� ��� ก�� 1.5 f’ c = 240 กก./��.2 ��� fy = 4,000 กก./��.2

8.0 m c/c

4DB20

8DB20

7.7 m clear

65 cm

35 cm

6 cm

57 cm

4DB20

8DB20

a) ��')#��'� ��� ก(1��� กก'� ����+�,����=��

6,120 6,120 (57)34.5 cm

6,120 6,120 4,000by

dx

f= = =

+ +

max 0.75 0.75(34.5) 25.9 cmbx x= = =

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#$ ���� x = 25.9 ��.

25.9 6(0.003) 0.0023 0.0020 '

25.9s y s yf fε ε−

′ = = > = → =

A’s=12.56 cm2

max As=51.9 cm2

εcu=0.003

εs

max Cc = 0.85bβ1(max x)

= 0.85(0.24)(35)(0.85)(25.9)

= 157.2 ���

Cs = A’sfy = 12.56(4.0) = 50.2 ���

max T = max C = 157.2 + 50.2 = 207.4 ���

2 2max 207.4max 51.9 cm 49.28 cm

4.0sy

TA

f= = = > OK

Real As = 49.28 cm2

(b) � ก$ ���(�( Mn �����$ ���ก���*ก+&,� � ?(�#��*�2+�,��� ก��(%� ก

10.85 c s y s yf b x A f A fβ′ ′+ =

0.85(0.24)(35)(0.85x) + 12.56(4.0) = 49.28 (4.0)

x = 24.2 �.�.

24.2 6.00.003 0.0023

24.2s yε ε−′ = = > ������+**)��.������� ,ก�� ���ก

Cc = 0.85f ‘cbβ1x = 0.85(0.24)(35)(0.85)(24.2) = 146.9 ���

Cs = A’s fy = 12.56 (4.0) = 50.2 ���

T = As fy = 49.28(4.0) = 197.1 ���

157 (0.85) (24.2) 46.7 cm

2 2a

d − = − =

Mn = 146.9(46.7)/100 + 50.2(57-6)/100 = 94.2 ���-����

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21(8) 0.90(94.2) 84.8

8u u nM w Mφ= = = = ���-����

) ก?)�Bก$ ��( wL = 1.5wD (������ wu = 1.4wD + 1.7(1.5wD)

wu = 10.6 ���/����

��$ ���ก���*ก%��+&,� � wD = 10.6/(1.4+2.55) = 2.7 ���/����

��$ ���ก)�+&,� � wL = 1.5(2.7) = 4.0 ���/����

(c) ��ก����� ,ก�+��*�����������

6.8 t

max. shear envelope

6.8 t

LC of support

+

-

Midspan

8.0 m

42.4 tSHD with DL+LLon entire span

42.4 t

Max. shear at support:

10.6(8)42.4 ton

2 2u

u

w LV = = =

Max. shear at midspan whenhalf LL on span:

10.6(8)6.8 ton

8 8u

u

w LV = = =

Critical section from face of support d = 57 cm, support width = 30 cm

Therefore compute Vu at 57+30/2 = 72 cm

(42.4 6.8)42.4 72 36.0 ton

4(100)uV−

= − × =

( )Shear strength of concrete 0.53

0.85 0.53 240 35 57 /1,000 13.9 ton

c c wV f b dφ φ ′=

= × × × × =

C of support Midspan

Face of support

42.4 t

6.8 t

L

d

Critical section36.0 t

72 cm13.9 t

φVc0.5φVc

Required φVs

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Required φVs = Vu - φVc = 36.0 - 13.9 = 22.1 ton

Min φVs = 0.85(3.5)(35)(57)/1,000 = 5.9 ton

Max (for / 2) 0.85 1.1 240 35 57 /1,000 28.9 tonsV s dφ = = × × × =

Since 5.9 ton < Required φVs < 28.9 ton, max s = d/2

0.85 2 0.78 4.0 57USE DB10 stirrup: 13.7 cm

22.1v y

s

A f ds

V

φ

φ× × × ×

= = =

@ Critical section

USE s = 13 cm from z = 0 to 57 cm from face of support

0.85 2 0.78 4.0 5723.2 ton

13v y

s

A f dV

s

φφ

× × × ×= = =

From z = 57 cm, set φVn = Vu

22.157 (400 72)

36.0 6.8sV

zφ−

= + −−

������ 5.1 %' �#�����3B���'� ������� ����ก$ ���#$ ������� ก!��ก+���'(2��

s (cm) φφφφVs (ton) z (cm)

13.7

15

20

25

28.5 (d/2)

51.2 (NG)

22.1 (Max)

20.1

15.1

12.1

10.6

5.9 (Min)

0 to 57

79

135

169

186

238

30 cm 1 cm

6@13cm 4@15cm 2@20cm 8@25cm

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Shear Strength of Members under Combined Bending and Axial Load

Axial Compression

where Nu = Factored axial compressive load

Ag = Gross area of the concrete section

0.53 (1 )140

uc c w

g

NV f b d

A′= +Simplified method:

(0.5 176 ) 0.93 'uc c w w c w

u

V dV f b d f b d

Mρ′= + ≤More detailed equation:

Replace Mu with Mm , where 48m u u

h dM M N

− = −

d - a/2

CA

a/2

h

h/2

Nu

Mu

T

[ΣMA=0]2 2 2u u

a h aT d M N

− = − −

7d/8 d/8

(upper limit) 0.93 135

uc c w

g

NV f b d

A′= +

Axial Tension 0.53(1 )35

uc c w

g

NV f b d

A′= +

0.53

35.2�����( ( Nu = + ), กก./��.2 ���(1� ( Nu = - ), กก./��.2

( ) cguc fANv ′+= 35153.0

cfv ′/

( )gucc ANfv 35193.0 +′=

0.50 176

48

w uc c

m

m u u

V dv f

M

h dM M N

ρ′= +

− = −

( ) cguc fANv ′+= 140153.0

56.3

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Strength Vc - Continuous Beams

0.53c c wV f b d′=Simplified method:

0.50 176 0.93uc c w w c w

u

V dV f b d f b d

′ ′= + ≤

More detailed procedure:

Strength Vs - Continuous Beams

v ys

A f dV

s=

min Av = 0.0015 bws where s ≤ d / 5 ≤ 45 cm

min Avh = 0.0025 bws2 where s2 ≤ d / 3 ≤ 45 cm

Minimum Shear Reinforcement:

Limitation on Nominal Shear Strength

Nominal stress vn = Vn / (φ bwd)

max 2.1 for 2nn c

Lv f

d′≤ <

max 0.18 10 for 2n nn c

L Lv f

d d ′= + ≥

2.1 max 2.7c n cf v f′ ′≤ ≤

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Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

� ���������

� ��� ��ก���ก����������

� ก����������������������

� ก����ก�������������

Design of Slabs 1

���������

One-way slab Two-way slabOne-way slab

Flat plate slab Flat slab Grid slab

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��� ��ก���ก����������

� �������ก������ � (DL) = 2,400 × t

� �������ก�� (LL) �������� ก�����

� �������ก�������� (SDL) : ������, ����ก����, ���

� �������ก� ���!ก " �#�$ �� � ����!"", %&�, ��'�, ()��*

��� ��ก������������

1 m1 m t

��� ��ก������ !��

��������� ��� ��ก (ก.ก./�.2)���+��,���"�!��" 1,658 t

���-.����/��������� 2.5 -�. 55

������������ 2.5 -�. 80

�������� ���� ��� �)�.(��� 2,645 t) 55

���������ก���� (��� 2,800 t) 60

��������+��, 2,400 t

����������� 2,550 t

��������ก���� 1/2” 15

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t

SL

Simple supportson two longedges only

ก����������������������

S S

L

S S

���������� : ������ (L) > ����������������� (S)

การแอนตัวเกิดขึ้นบนดานสั้น

�������������������� == ก�����(�ก�����(� 11 ��*�����*���

���� ����+����������� 4 �,��

L

SL ≥≥≥≥ 2 S

���� ����+�������� 2 �,������ก��

SS

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1.0 m

ก����ก��� : ����������� ��ก������������

ก����ก�������������ก����ก�������������

LS

����ก���� ���ก : �������

����ก���� ก������ : ������

1 m

t

�!"����#�

S

Sn

�!"�������

+��� ���������������+��� ���������������

������������������� 10-15 !�. �����#����ก�$� 2-3 !�.

��'����($���

��� � ��������������� �������

��� � ��������� ��ก�� ���!��������

L / 20Ln

�)����������*����($��L / 24Ln

�)����������+��*���L / 28Ln

�����L / 10Ln

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���"ก���#�$%��ก� ก����&��� ก���$���� �$������'(�# (���"ก���#�ก� ���)

Spacing ≤≤≤≤ 3 t ≤≤≤≤ 45 cm

Main Steel (short direction):As ≥ ∅ 6 mm

Max. Spacing ≤ 3 t ≤ 45 cm

Min. Spacing ≥ f main steel ≥ 4/3 max agg. ≥ 2.5 cm

RB24 (fy = 2,400 ksc) . . . . . . . . . . . . . . 0.0025

DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . 0.0020

DB40 (fy = 4,000 ksc) . . . . . . . . . . . . . . 0.0018

DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . 0.0018 4,0000.0014

yf×

$�#������"ก���#�+ ��� �������$�#������"ก���#�+ ��� �������

�,���+����)-ก�+��� As �����.$���ก�$�.,���( Ag : As/Ag 1 m

t

Ag = b ×××× t = 100 t

&���,����� 9.1 ��ก/00��.���($�� S1 ������,01��,ก0��.#ก2'��� 300 ก.ก./�.2

1��,ก*���,+(#�����.��ก,0 50 ก.ก./�.2 ก1��(����/��.$����2�� f5c = 210 ก.ก./!�.2

/)� fy = 2,400 ก.ก./!�.2

2.7 m

S1

S1 S2

#1��2� 1) ��� � ���������

+1���,0����������*����($�� tmin = L/24

�,����)(�������� fy = 2,400 ksc ��� 2,400

0.4 0.747,000

+ =

min270

t 0.74 8.3 cm24

= × =

����ก��� ���� t = 10 cm

1��,ก��

WSD

= 0.1 x 2,400 = 240 kg/m2

1��,ก�,+(#��;�� = 50 kg/m2

1��,ก�� = 300 kg/m2

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WSD1��,ก��� = 240 + 50 + 300 = 590 kg/m2

<����=.$�����,(��ก>��1���<(�2'����+,����+�.?�@2����� ก.10 (,�$

�. �#(����,0A��2:

�. ก)��'�����:

�. �#(����,0A���ก:

2M 590 2.7 / 9 477.9 kg-m− = × =

2M 590 2.7 /14 307.2 kg-m+ = × =

2M 590 2.7 / 24 179.2 kg-m− = × =

fs = 0.5x2,400 = 1,200 ksc fc = 0.45x210 = 94.5 ksc

134n 9

210= ≈

s

c

1 1k 0.293

f 1,20011

9 94.5n f

= = =++

×

c1 1

R f k j 94.5 0.293 0.902 12.49 ksc2 2

= = × × × =

j = 1 – 0.293/3 = 0.902

�1�������������=.$�2'�2ก����ก/00 :

WSD

d = 10 - 0.45 - 2 = 7.55 cm

����)�ก*������,(<(�+��#�����2'���)-ก RB9 �.�. �����#�� 2 !�.

<����=���.�<(���ก�$� :Mc = R b d2 = 12.49 × 100 × 7.5522 = 71,196 kg-cm

= 712.0 kg-m > M .$���ก��.1� OK

�1�����������)-ก�+���.$�����ก�� : ss

MA

f jd=

&2��� ,� M As ���"ก���#�

�#(����,0A��2 -477.9 5.85 RB9 @ 0.10

ก)��'����� 307.2 3.76 RB9 @ 0.16

�#(����,0A���ก -179.2 2.19 RB9 @ 0.29

RB9 : As = 0.636 cm2

Spacing = 0.636×100/As

��)-ก�+���ก,���� : As,min = 0.0025 × 100 × 10 = 2.5 cm2 USE RB9 @ 0.25 m

RB9 @ 0.25

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���"ก���#������ : 2'���)-ก�+���ก,���� USE RB9 @ 0.25 m

0.7 �. 0.9 �.

2.7 �.

RB9 @ 0.25 �.

RB9 @ 0.16 �.

RB9 @ 0.25 �.RB9 @ 0.10 �.

10 !�.

WSD

&���,����� 9.1 ��ก/00��.���($�� S1 ������,01��,ก0��.#ก2'��� 300 ก.ก./�.2

1��,ก*���,+(#�����.��ก,0 50 ก.ก./�.2 ก1��(����/��.$����2�� f5c = 210 ก.ก./!�.2

/)� fy = 2,400 ก.ก./!�.2

2.7 m

S1

S1 S2

#1��2� 1) ��� � ���������

+1���,0����������*����($�� tmin = L/24

�,����)(�������� fy = 2,400 ksc ��� 2,400

0.4 0.747,000

+ =

min270

t 0.74 8.3 cm24

= × =

����ก��� ���� t = 10 cm

1��,ก�� = 0.1 x 2,400 = 240 kg/m2

1��,ก�,+(#��;�� = 50 kg/m2

1��,ก�� = 300 kg/m2

SDM

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1��,ก���),� wu = 1.4×(240+50) + 1.7×300 = 916 kg/m2

��������)-ก�+�����ก.$�+#( (����� ก.5):

SDM

<����=.$�����,(��ก>��1���<(�2'����+,����+�.?�@2����� ก.10 (,�$

�. �#(����,0A��2:

�. ก)��'�����:

�. �#(����,0A���ก:

2uM 916 2.7 / 9 742.0 kg-m− = × =

2uM 916 2.7 /14 477.0 kg-m+ = × =

2uM 916 2.7 / 24 278.2 kg-m− = × =

ρmax = 0.0341

d = 10 - 0.45 - 2 = 7.55 cm

����)�ก*������,(<(�+��#�����2'���)-ก RB9 �.�. �����#�� 2 !�.

�1�����������)-ก�+���.$�����ก�� : un 2

MR

bd=φ

s c n

y c

A 0.85 f 2R1 1

bd f 0.85 f

′ρ = = − − ′

RB9 : As = 0.636 cm2

Spacing = 0.636×100/As

��)-ก�+���ก,���� : As,min = 0.0025 × 100 × 10 = 2.5 cm2 USE RB9 @ 0.25 m SDM

&2��� ,� Mu As ���"ก���#�

�#(����,0A��2 -742.0 4.75 RB9 @ 0.13

ก)��'����� 477.0 3.01 RB9 @ 0.21

�#(����,0A���ก -278.2 1.73 RB9 @ 0.36RB9 @ 0.25

���"ก���#������ : 2'���)-ก�+���ก,���� USE RB9 @ 0.25 m

0.7 �. 0.9 �.

2.7 �.

RB9 @ 0.25 �.

RB9 @ 0.16 �.

RB9 @ 0.25 �.RB9 @ 0.10 �.

10 !�.

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Exterior span

Bottom bars

Top bars atexterior beams

Top bars atexterior beams

Interior span

Temperature bars

(a) Straight top and bottom bars

Exterior span

Bottom bars

Bent bar Bent bars

Interior span

Temperature bars

(b) Alternate straight and bent bars

ก�����#����"ก+ ��� �������ก�����#����"ก+ ��� �������

������� �� ��� ������� �� ���

RB9 @ 0.10 �.�. �+�����+�

RB9 @ 0.20 �+������KL

$�#������"ก���#���,�ก� (ก���3,�, $���3,�)

RB9 @ 0.10 m As = 6.36 cm2

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31L

41L

81L

L1

Temp. steel

4.0 �.

.13 �.

1.0 �. 1.3 �.

[email protected]

[email protected] [email protected]

4.0 �.

.13 �.

1.0 �. 1.3 �[email protected]) ��������������

[email protected] [email protected] �������AB

���� ���� -.ก�����

Floor Plan

����������������������������

1/14 1/16 1/16

1/24

1/12 1/12 1/12

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&���,����� 9.2 ������ก/00��.���($��+1���,01��,ก�� 500 ก.ก./�2 ก1��(����/��.$����2�� f’c = 210 ksc /)� fy = 2,400 ksc

SDM

AA

3 @ 12 m = 36 m

Section A-A

Ln = 3.7 m Ln = 3.7 m Ln = 3.7 m

0.4 + 2,400/7,000 = 0.74

1) Minimum depth :

Min. h = 0.74×370/24 = 11.4 cm

USE h = 12 cm

Slab weight = 0.12×2,400 = 288 kg/m2

Assume beam + super DL :

Service DL = 350 kg/m2

Service LL = 500 kg/m2

2) Factored Load :

wu = 1.4×350 + 1.7×500 = 1,340 kg/m2

3) Max. Moment :

Mu = 1,340 × 3.72 / 12 = 1,529 kg-m (Interior negative moment)

Max. reinforcement ratio (from Table ก.5) : ρρρρmax = 0.0341

USE RB9 with 2 cm covering: d = 12-2-0.45 = 9.55 cm

'c n

'y c

0.85 f 2R1 1 0.0082

f 0.85 f

ρ = − − =

2un 2 2

M 1,529 100R 18.63 kg/cm

bd 0.9 100 9.55×

= = =φ × ×

< ρρρρmax = 0.0341 OK

Required As = ρbd = 0.0082 × 100 × 9.55 = 7.83 cm2/m

RB9 : As = 0.636 cm2 → s = 0.636×100/7.83 = 8.12 cm

Select [email protected] : As = 0.636×100/8 = 7.95 cm2/m > Required As OK

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Select [email protected] : As = 0.636×100/20 = 3.18 cm2/m

Temp. steel = 0.0025 × 100 × 12 = 3.00 < 7.95 cm2/m OK

������� �����ก�����#����"ก+ ��� �������

RB9 @ 0.20 . �� ก���ก������

RB9 @ 0.08 .

�������������

RB9 @ 0.16 .

�������

RB9 @ 0.08 .

������������� + �������

12 ��.

0.95 . 1.25 .

3.7 .

0.55 . 0.95 .

������������

1 m

S

1 m S

2w SM

2=

/�����0������+���ก�,�� 1 ����

t ≥≥≥≥ S / 10

for deflection control

S

t

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�������� ก�""���,�#�+����.,� ��ก���� �+��4.0 m

S1S2

1.5 m

5 m

0.50

0.10

0.20 3.80 0.20

[email protected] ) ��������������[email protected] �������AB

[email protected] �������AB

0.951.30

0.55

0.95

1.5 m

t

Min. h = 150/10 = 15 cm

USE h = 15 cm

DL = 0.15×2,400 = 360 kg/m2

LL = 200 kg/m2

wu = 1.4×360 + 1.7×200

= 844 kg/m2

Mu = 844 × 1.52 / 2 = 949.5 kg-m (per 1 m width)

USE DB10 with 2 cm covering: d = 15-2-0.5 = 12.5 cm

< ρρρρmin = 0.0035 Use ρρρρmin

un 2

MR

bd=φ 2

949.5 1006.75 ksc

0.9 100 12.5×

= =× ×

c n

y

0.85 f 2R1 1

f 0.85 f

′ρ = − − ′

= 0.0017

Required As = ρbd = 0.0035 × 100 × 12.5 = 4.38 cm2/m

DB10 : As = 0.785 cm2 → s = 0.785×100/4.38 = 17.9 cm

Use [email protected] : As = 0.785×100/17 = 4.62 cm2/m > Required As OK

Use [email protected] : As = 0.785×100/29 = 2.71 cm2/m

Temp. steel = 0.0018×100×15 = 2.70 cm2/m

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0.50

0.10

0.20 3.80 0.20

[email protected] ) ��������������[email protected] �������AB

[email protected] �������AB

0.951.30

0.55

0.95

1.5 m

t

4.0 m1.5 m

0.15 0.10

[email protected]

[email protected]#

[email protected] [email protected] �������AB

[email protected] ) ��������������

ขอสอบภย

ขอที่ : 122

แผนพื้นทางเดียว รับโมเมนตดัดประลัย 1500 กก.-ม. กําหนด fc’ = 280 ksc; fy = 2400 ksc และถาใชปริมาณเหล็กเสริมเหล็กเสริมที่มีอัตราสวนเหล็กเสรมิรับแรงดึงตอหนาตัดประสิทธิผลสูงสุดตามมาตรฐาน ว.ส.ท. จงตรวจสอบหาคา d ที่ต่ําที่สุดที่สามารถออกแบบได (วิธี SDM)

05097.040006120

612085.0

240028085.0

b =

+××

×=ρ

0382.005097.075.0max =×=ρ

ksc 08.742807.1

24000382.0124000382.0 =

××

−×=

ρ−ρ=

c

yymax,n f7.1

f1fR

2nu dbRM φ=

bRM

dn

u

φ=

10008.749.01001500××

×= = 4.74 cm

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ขอสอบภย

ขอที่ : 182

แผนพื้นตอเนื่องมีระยะศนูยถึงศูนยของที่รองรับ = 4.00 เมตร ตองรับน้ําหนักบรรทุกจรแบบแผสม่ําเสมอใชงานเทากับ 500 กก./ม.2 ถาที่รองรับสามารถรับโมเมนตดัดไดเทากับ wL2/24 จงใชวิธี WSD หาขนาดและระยะเรียงของเหล็กเสริมที่ “ประหยัด” ตรงกลางชวงพื้น สมมุติพื้นหนา 20 ซม. เสริมเหล็กรับแรงดึงอยางเดียวที่ระยะ d = 15 ซม. fc’ = 150 กก./ซม.2 และ fy = 3000 กก./ซม.2 ตําแหนงแกนสะเทิน kd = 5 ซม.

เสริมเหล็ก 12 มม. @ 20 ซม.

�������ก��� + �������ก�� = 0.2×2400 + 500 = 980 กก./�.2

(�����/���ก'��� � M+ = 980××××42/14 = 1120 กก.-�.

��ก kd = 5 -�. jd = 15 – 5/3 = 13.33 -�.

ss

MA

f jd= 5.60 1�.2

��'Eก 12 ��. (As = 1.13 -�.2) s = 100×1.13/5.60 = 20.18 -�.

1120 1001500 13.33

×= =

×

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Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

� Design of Two-Way Slab

� Moment Coefficient Method

� Load Transfer from Two-Way Slab

� Bar Detailing

� Design Example

� Slab on Ground

Design of Slabs 2

G

G

B B

B

B

B B

L

S

(ก) ������� ������ (�) ���������� �

����������� ��������

������� ������������ �� � L � ����ก�������� ����� ����� S

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t

SL

Simple supportson all four edges

ก�����������������

ก ���������ก����������������� � �ก ������� ��� !��ก����������� ����

�� ����� S �"#�� �� � L

ก����������������

$#��$ �% �&�������'(����ก�� � b = 1 ��� ����������� ������ ����� S �"#

�� �� � L ������ก�������������ก��������������

L

S

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ก���ก���������� ( L < 2S )

L

SS

������������ ������ tmin :

Perimeter 2 (L S)10 cm

180 180+

= ≤

d�� �� � d�� ����

1∅ ����ก���������ก���������������

As ≥ RB 9 ≥ Temp. steel

Max. Spacing ≤ 3 t ≤ 45 cm

Min. Spacing ≥ ∅ main steel ≥ 4/3 max agg. ≥ 2.5 cm

S/4

S/4

S/2

L/4 L/2 L/4

���ก" �

�����

�����

���ก" � ����� �����

-Ms

-Ms

+Ms

+ML

-ML -ML

Middle strip moment: MM = CwS2

Column strip moment: MC = 2MM/3

�� ��!"����� �#��$!�!���

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����������ก��� !��"����������ก��� !��"

��#� �� ( !�������������) ��&�! !������������� �

��&�! !������������� ��&�! !������������� ��&�! !�������������

����!"����� �#��$!�!��� ( C )

��ก()�����

*������*����

�� �� �)��� m

1.0 0.9 0.8 0.7 0.6 0.5

����3���+���,"�-�� ���������

-�� �����������+���,��ก��ก" �.���

0.033-

0.025

0.040-

0.030

0.048-

0.036

0.055-

0.041

0.063-

0.047

0.083-

0.062

0.033-

0.025

����8!�������9������ ��+���,"�-�� ���������

-�� �����������+���,��ก��ก" �.���

0.0410.0210.031

0.0480.0240.036

0.0550.0270.041

0.0620.0310.047

0.0690.0350.052

0.0850.0420.064

0.0410.0210.031

����8!�������9������+���,"�-�� ���������

-�� �����������+���,��ก��ก" �.���

0.0490.0250.037

0.0570.0280.043

0.0640.0320.048

0.0710.0360.054

0.0780.0390.059

0.0900.0450.068

0.0490.0250.037

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����!"����� �#��$!�!��� ( C )

��ก()�����

*������*����

�� �� �)��� m

1.0 0.9 0.8 0.7 0.6 0.5

����8!�������9��!���+���,"�-�� ���������

-�� �����������+���,��ก��ก" �.���

0.0580.0290.044

0.0660.0330.050

0.0740.0370.056

0.0820.0410.062

0.0900.0450.068

0.0980.0490.074

0.0580.0290.044

����8!�������9�� 9���+���,"�-�� ���������

-�� �����������+���,��ก��ก" �.���

-0.0330.050

-0.0380.057

-0.0430.064

-0.0470.072

-0.0530.080

-0.0550.083

-0.0330.050

ก��'!� �(����ก)�ก�������������ก��'!� �(����ก)�ก�������������

L

S

w wS/3

�� ���ก�� ��� ���� =3

w S

45o

w wS/3(3-m2)/2

�� ���ก�� ��� �� � −

=

233 2

w S m

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ก�����������ก����������ก�����������ก����������

Sn

Sn / 7 Sn / 4

Sn / 4 Sn / 3

�*+ ���������

Ln

Ln / 7 Ln / 4

Ln / 4 Ln / 3

�*+ ������ ��

����ก����

����ก

L/5

L/5

L = �� ����� ��������

ก�����������ก���,������������ก�����������ก���,������������

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Example: Design two-way slab as shown below to carry the live load 300-kg/m2

fc’ = 240 kg/cm2, fy = 2,400 kg/cm2

3.80

4.00

4.805.00

Floor plan

0.10

0.50

0.20 0.20

Cross section

Min h = 2(400+500)/180 = 10 cm

DL = 0.10(2,400) = 240 kg/m2

wu = 1.4(240)+1.7(300) = 846 kg/m2

m = 4.00/5.00 = 0.8

max

0.85(240)(0.85) 6,1200.75

4,000 6,120 4,000

0.0197

ρ = +

=

Short span

Moment coeff. C

-M(�����������) +M -M(��������)

0.032 0.048 0.064

Max. M = C w S 2 = 0.064 × 846 × 4.02 = 866 kg-m/1 m width

d = 10 - 2(covering) - 0.5(half of DB10) = 7.5 cm

As = 0.0045(100)(7.5) = 3.36 cm2

As,min = 0.0018(100)(10) = 1.8 cm2 > As OK

��&�! !������������� (m = 0.8)

2un 2 2

M 86,600R 17.11 kg/cm

bd 0.9 100 7.5= = =φ × ×

c n

y c

0.85f 2R1 1 0.0045

f 0.85f

′ρ = − − = ′

≤ ρmax = 0.0197 OK

Select short span reinforcement DB10 @ 0.20 m (As = 3.90 cm2)

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Max. M = C w S 2 = 0.049 × 846 × 4.02 = 663 kg-m/1 m width

d = 10 - 2(covering) - 1.5(half of DB10) = 6.5 cm

��&�! !������������� (m = 0.8)

Long span

Moment coeff. C

-M(�����������) +M -M(��������)

0.025 0.037 0.049

c n

y c

0.85 f 2R1 1 0.0046

f 0.85 f

′ρ = − − = ′

≤ ρmax = 0.0197 OK

As = 0.0045(100)(6.5) = 2.97 cm2

As,min = 0.0018(100)(10) = 1.8 cm2 > As OK

Select short span reinforcement DB10 @ 0.20 m (As = 3.90 cm2)

un 2

MR

bd=φ 2

66,3000.9 100 6.5

=× ×

= 17.44 ksc

���:���ก;��������<��������ก� �

����/����/"�� Vu = wuS/4 = (846)(4.0)/4

= 846 ก.ก./.

ก "���������/������ก��� φVc = 0.85(0.53) (100)(7.5)

= 5,234 ก.ก./. OK

240

���=ก����!ก�����

As,min = 0.0018(100)(10) = 1.8 cm2

Select temp. steel reinforcement DB10 @ 0.30 m (As = 2.60 cm2)

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0.50

0.10

0.20 4.80 0.20

[email protected] ��� ������������[email protected] ��������2

[email protected] ��������2

���������

0.701.20

1.201.60

0.50

0.10

0.20 3.80 0.20

[email protected] ��� ������������[email protected] ��������2

[email protected] ��������2

�����������

0.951.30

0.55

0.95

*���">����?������

$#� 3!�ก "������&������"�"� �"#ก ��������$#�������

.����'4�B X B

ACI !��� �.�.�. ก !��3!�� .����'4�+�����&"���ก "������&��������� :

(1) .����'4���6�3��������������������ก" ������ก�� +��3.��!"7ก������� ก��3��������.����'4�

(2) 3��������������������ก" ������ก�� �'4�.���ก�� �������ก�� 1/8 ����� ก�� ������ �� �����)

(3) 3��������������ก��$ ก����� !�������"#���ก" �!������ � ���'4�.������ +���!"7ก������� �! ��'3����"#���� ��������ก�� 1/4 ����!"7ก����3��������) "��!)���=ก����!� 9��8"�������!���9!8��� 9�������*���">������������

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ขอสอบภย

ขอที ่: 131

พื้น S1 ขนาด 5x5 เมตร หนา 12 ซม. เหล็กเสริมโมเมนตบวก(เหล็กลาง)กลางแผนพืน้กําหนดใหเทากับ [email protected]# ถาตองการเปดชองโลงกลางแผนพื้น ขนาด 0.80x0.80 เมตร ตองเสริมเหล็กทดแทนอยางนอยเทาไร?

RB12 � As = 1.13 cm2

$ ���������! ��' = 80/15 = 5.33 ����

�������!"7ก��! ��' = 1.13×5.33 = 6.02 8.2

�����!"7ก�������������� � = 6.02/2 = 3.01 8.2

����! 2 DB16 (As = 4.02 @!.2) �������

�����-.������-.�

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ก����ก/00�����-.�

-- ������������������������ 1010 ++ 22 5�5�..

-- �7����ก�� 8���(� �ก��59��7����ก�� 8���(� �ก��59�

-- ����8�����ก�� &�!���'*ก/��/������8�����ก�� &�!���'*ก/��/��

���� EpoxyEpoxy ก��59����-.�ก��59����-.�

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���������� (Slab-On-Ground)

����3���

����3���ก

�� �.���������� 5-10 8.GB GB

3-5 8. ≤ t/22-2.5 8.

�� �.���������� 5-10 8.

5-10

8.

5 8.

45o

�� �.���������� 5-10 8. GB

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PM

PM

PM

PM

Soil

���������� (Slab-On-Ground)

Soil modeled as springs in the solution of beam on elastic foundation

������������ (Joint Spacing)

0 5 10 15 20 30

Slab thickness, cm

0

3

6

9

Max

.joi

nt s

paci

ng,m

Range ofmax. spacing

spacing

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������ก�ก���� ����ก����� ��ก���� ��ก��������� ��ก. 737-2531 ��" fy

= 5,000 ก.ก./'�.2

1.4810.9430.943∅ 6 ��.× 6 ��., 30 '�.× 30 '�.

1.7761.1311.131∅ 6 ��.× 6 ��., 25 '�.× 25 '�.

2.2201.4141.414∅ 6 ��.× 6 ��., 20 '�.× 20 '�.

0.6580.4190.419∅ 4 ��.× 4 ��., 30 '�.× 30 '�.

0.7900.5030.503∅ 4 ��.× 4 ��., 25 '�.× 25 '�.

0.9880.6290.629∅ 4 ��.× 4 ��., 20 '�.× 20 '�.

1.3170.8380.838∅ 4 ��.× 4 ��., 15 '�.× 15 '�.

���:���+ ���� ����:���

�(����ก(กก./ �.�.)

��������� ��( �.5�. / �.)∅∅∅∅ :������, :��� �/ก��

�/ก���������ก�(����)�*+ �/ก���������ก�(����)�*+ (Wire mesh)(Wire mesh)

��ก/00����ก�����D� ��E���ก/00����ก�����D� ��E� SubgradeSubgrade DragDrag

W

T

L

s s

FL WT A f

2= =

ss

FL WA

2f=

, �- T ��"�ก.�/�01ก�������2�3����ก��.� As

= 456���"����ก��.�������ก��� 1 ��� ('�.2)W = �6�����ก456���456���" (ก.ก./��.)L = ����3��456����ก�� (���)F = ���9��.�:.;���3���� (1.5 ก<�=���>���1�)fs

= ���3, ��"3�����>� ����ก��.� (ก.ก./�.'�.)

ก<��?� Wire mesh fy = 5,000 กก./�.'�. ,� fs ��"3������4�3 1,700 ก.ก./�.'�. �����6�

sy y

FL(1.4 W) FL WA

2f 1.43 f= =��E�ก(���� :

9.��<����ก������ก��� 1 ���

/�=����, �- T ������ก��� 1 ���

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�6�����ก456� W = 0.12(2,400)

= 288 ก.ก./��.

sy

FL W 1.5 6.0 288A

1.43 f 1.43 5,000× ×

= =×

��5�ก�?� WWR ∅ 4 �.�.× 4 �.�., 30 '�.× 30 '�. (As = 0.419 �.'�./���) �

��� !����� 9.7 / ��ก,@@456����ก���� @��.�2�3ก������3��3�����ก�@ 6.0 ��� 2�3�?� Wire mesh ��ก���� ��ก fy = 5,000 ก.ก./'�.2

��E��(� ��5�ก�������2�3�?�ก�A =�� 12 '�.

0 5 10 15 20 30

Slab thickness, cm

0

3

6

9

Max

.joi

nt s

paci

ng,m

Range ofmax. spacing

= 0.363 �.'�./���

9.��<����ก��"�?�

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� ������������ �������������� �

� ��� ��������ก�������������������� �

� ��� ��������������

� ��� ���������ก������ก�����

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Design of Stairs

������������ก���

�����������������

�����������ก�����ก����

���������������

������������

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� ����ก���ก�����������

� ������ก������ � �� (DL) = ����� � + ������ �

� ������ก�� (LL) "��������ก�����

� ������ก��#�$%&'�� (SDL) : ������, ����ก����, ��

� ������ก#���%��ก�����) ��� ������ �, '���, �������

� ����ก����������

��������� � ����ก (ก.ก./ .2)%&�����%�������� 1,658 t

��*+ ���,��������� 2.5 *�. 55

����������� 2.5 *�. 80

��%&�����������������+(��� 2,645 t) 55

��%&����ก���� (��� 2,800 t) 60

��%&������� 2,400 t

��%&���%&� 2,550 t

��%&%� ก���� 1/2” 15

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�����#������ก�$������%�& �����

������ก��� ����������� 1 ���

L = ��'���%�& �����

�����#������ก�$������%�& �����

����'���� 1 ��ก������ �ก���� 2.0 �. ������ �ก���� 25 *�.

#����ก 15 *�. ������������ก�� 300 กก./�.2 ก����� f’c = 240 ksc

��� fy = 2,400 ksc

��*��� ��กก�����������ก�+���� ��;ก #��� ρmax = 0.0389

"������� ���� 200/20 = 10 *�.

������ก������ � 0.5(0.15)(2400) = 180 กก./�.2

������ก����$ก�� LL = 300 กก./�.2

������ก����$ก�'���� wu = 1.4(280+180)+1.7(300) = 1154 กก./�.2

�����<ก d = 10 – 2 – 0.45 = 7.55 *�.

������ก����� � 0.10(2400) = 280 กก./�.22 215 2525

+

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=� ���,��ก��������� �+��� Mu = 1154(2.0)2/8 = 577 กก.-�.

%����? ��;ก #��� As = 0.0048(100)(7.55) = 3.62 *�.2/����ก���� 1 ���

#�+�,&��-./��

��� @��%������������ก���� 1 ��� Vu = wL/2 = 1154(2.0)/2 = 1154 กก.

ก����������� @��������ก�+�:

un 2

MR

bd=

φ2

2

577 10011.25 kg/cm

0.90 100 7.55×

= =× ×

c n

y c

0.85 f 2R1 1 0.0048

f 0.85 f

′ρ = − − = ′

ρρρρmin < ρρρρ < ρρρρmax OK

-0/�ก1�$-�02ก-��� RB9 @ 0.15 m (As = 4.24 cm2 / m)

cV 0.85 0.53 240 100 7.55φ = × × × = 5269 กก. > 2 VuNo need forshear reinf.

0.25

0.150.10

φ 9 ��. @ 0.20( ��;ก�<������� �)

1 φ 9 ��. �$ก�$�( ��;ก�<������� �)

φ 9 ��. @ 0.20

( ��;ก�����<�)

φ 9 ��. @ 0.15

( ��;ก #��� �ก)

0.10

2.00

��������� �

φ 9 ��. @ 0.15

φ 9 ��. @ 0.20

&���'0�-��'�ก�-��� -�02ก�����#����ก�$�

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�����#������'�

��������ก

����������

��������

L

w

L

w

w w

H

L = Projected length

w

3�����3�������#������'�

�������%���B��������� ����#����#�+���%����#�ก

�4�'�%�

�*�,��&ก��� (*�.) ≤ 20 ≤ 19

�&ก��� (*�.) ≥ 22 ≥ 24

����#&� H (�.) ≤ 3.0 ≤ 4.0

������� L (�.) ≤ 4.0 ≤ 4.0

����ก���� (�.) ≥ 0.9 ≥ 1.5

��&ก��� � (*�.) 2 – 2.5 2 – 2.5

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ก�5 -�0�������-�6�%�

��� %C�������� �+������������ ���� �<�ก����กD?�=���#����

t

0.025

�&ก��� 25-30 *�.

��&ก��� �

�&ก���15-18 *�.

t

�����������

t0.025

�&ก���

��&ก��� �

�&ก���

�����������

����3��#/ ������&0�3��3� ������

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����'���� 8.4 ����ก������ ����������� 2.50 ��� ���������������������

��ก ������������ก�� 300 กก./��.�. �����กก���� 1.0 ��� ������ �ก���� 25 *�.

#����ก����� 15 *�. f’c= 240 กก./��.*�. fy = 4,000 กก./��.*�.

��*��� ����� ���� 250/20 = 12.5 *�.

�����<ก d = 12 – 2 – 1.2/2 = 9.4 *�.

������ก������ � = 0.5(0.15)(2400) = 180 กก./�.2

������ก����$ก�� = 300 กก./�.2

������ก����$ก��� wu = 1.4(336+180)+1.7(300) = 1,232 กก./�.2

-0/�ก1�$%� �� 12 9 .

������ก����� � = 0.12(2400) = 336 กก./�.22 215 2525

+

=� ���,��� Mu = 1,232 × 2.52 / 8 = 962.5 กก.-�.

%����? ��;ก #��� As = 0.0035(100)(9.4) = 3.29 *�.2/����ก���� 1 ���

��ก"�� ��;ก #��� DB12 . . @ 0.20 . (As = 5.65 *�.2/�.)

#�+�,&��-./��

��� @��%������������ก���� 1 ��� Vu = wL/2 = 1,232(2.5)/2 = 1,540 กก.

ก����������� @��������ก�+�:

un 2

MR

bd=

φ 2

962.5 1000.90 100 9.4

×=

× ×= 12.10 ก.ก./*�.2

c n

y c

0.85f 2R1 1

f 0.85 f

′ρ = − − ′

= 0.0031 < [ ρmin = 0.0035 ] Use ρρρρmin

cV 0.85 0.53 240 100 9.4φ = × × × = 6,560 กก.

������ก φVc > 2Vu ������ #��� ��;ก������ @��

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DB12 �.�. @ 0.20 �.

��;ก #��� �ก

φ9 �.�. �$ก�$�

��;ก�<����

φ9 �.�. @ 0.20 �.

��;ก�<����

2.50

1.05

0.150.25

φ9 �.�. @ 0.20 �.

��;ก�����<�

0.12

&���'0�-��'�ก�-��� -�02ก�����#���'�

%�& �����3�������#������'�

UP

��������� ����������� #��+�ก�������#&�����������

1st Floor

�����ก

2nd Floor

��������� �

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�����&��'/��+ก%����ก�&#�

Cantilever beamLoad area

Main steel

Deflected shape

Load

0.25

0.10

0.10

0.15

0.25

����'���� 8.5 ��ก������ ������ก������ก�����ก���� 1.50 ��� ������ก�� 300 กก./��.�.

�+����"������� 2.5 ��� ������ �ก���� 25 *�. #����ก����� 15 *�. f’ c= 240 กก./��.*�.

fy = 4,000 กก./��.*�.

��*��� ��ก����� 10 *�. �����?������� %C�

������� 10x25 *�. ���"��&%

�����<ก��� d = 25-2-1.2/2 = 22.4 *�.

������ก������ ���<����� = (0.15 + 0.25)(0.10)(2,400) = 96 ก.ก./ �.

������ก���������� ���<����� = 0.25(300) = 75 ก.ก./ �.

������ก�'�%����� wu = 1.4(96)+1.7(75) = 262 ก.ก./ �.

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0.25 m

1.50 m

262 kg/m

21(262)(1.5) 295 kg-m

2uM = =

=� ���,��ก�+�#$�"�������:

22 2

295(100)6.53 kg/cm

0.90(10)(22.4)u

n

MR

bdφ= = =

#�+�,-�02ก-��� :

ρmin = 14/fy = 14/4000 = 0.0035

1

61200.85 0.0262

6120c

by y

f

f fρ β

′= = +

ρmax = 0.75ρb = 0.75(0.0262) = 0.0197

min

0.85 21 1 0.0017

0.85c n

y c

f R

f fρ ρ

′= − − = < ′

USE ρmin

%����? ��;ก #��� As = 0.0035(10)(22.4) = 0.784 *�.2/������ � 1 ���

��ก"�� ��;ก #��� 1 DB12 �.�. (As = 1.13 *�.2)

#�+�,&��-./��:

��� @��%����� Vu = wL = 262(1.5) = 393 กก.

ก����������� @��������ก�+�:

φVc = 0.85(0.53)(√ 240)(10)(22.4) = 1563 กก. > 2Vu OK

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0.10

1.50 0.40

0.800.15

��;ก�����<� [email protected]

��;ก #������ก 1DB12

��;ก�<������� �RB9 @ 0.20

��;ก�����<����1RB9 �$ก�$�

0.25

0.10

0.10

0.150.25

��;ก #������ก 1DB12

&���'0�-��'�ก�-��� -�02ก�����'/��+ก%����ก�&#�

Assignment: +���ก&�������1����3$�0�

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Bond, Anchorage, and Development Length

� ������������� ����������ก������������ก

� ก���������� ����������������������

� ก��� �����������ก!������

� �������� �� (Development Lengths)

� �������� ��������ก����� �.�.#. ��$ ACI

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Bond Stresses in Beam

Concrete

Reinforcing bar

PEnd slip

Greased or lubricated

Free slip εs ≠ εc

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Bond forces acting on concrete

Bond forces acting on steel

Bond Stresses in Beam

Round bar reinforcing

j d

Mmax

Little or no bond

Tied-arch action in a beam with little or no bond

���%$��&'��ก������(�� Tied arch ������!�������������ก�����)����� ��$����ก�������)����

��( ��%�กก���������� ��!������กก��*������)��������� ก��#����!�%$�����&� �ก�����+�#� ,������ #������ก�ก��-������%�ก����ก�����.,������ก���

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����������ก2

4s s s

dA f f

π= =

�������� �� d Luπ=

�������� �� = ����������ก2

4 s

dd Lu f

ππ =

���������������� �����ก�����4

sd fL

u=

L

T = As fs

Anchorage Bond

u = ������������� ��#� *��!������ก����� (กก./2�.2)

WSD

Bond Stress Based on Simple Cracked Analysis

jd

T + dT

V

C + dC

T

V

C

dxO

[ ΣΣΣΣMO = 0 ]

T T + dT

u

1

[ ΣΣΣΣFx = 0 ]

u Σo dx = dT 2

21

Elastic CrackedSection Equation

Flexural Bond

jd(dT) V(dx)=

dT Vdx jd

=

WSD

ΣΣΣΣo = *����!��������)������ก����������� (2�.)

djV

uoΣ

=

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ขอสอบภย

ขอที่ : 129

เหล็กเสริม DB12 SD30 ฝงในเน้ือคอนกรีตลึก 50 cm กําหนดให หนวยแรงยึดเหน่ียวที่ยอมให 11 ksc เมื่อออกแบบการรับแรงดึงโดยวิธีหนวยแรงใชงาน เหล็กเสริมจะรับแรงดึงสูงสุดทีย่อมใหไดเทาไหร

ขอสอบภย

ขอที่ : 130

จงหาหนวยแรงยดึเหน่ียวท่ีเกิดของจุดตอระหวางคานกับเสา เมื่อคานรับแรงเฉือนที่จุดตอ V=6555 kg มีคา jd=39.735 cm มีเหล็กเสริมที่พิจารณาในการคํานวณแรงยึดเหน่ียวคือ 4 เสน ขนาด RB15

= 8.75 ksc

Control

2

s

dT f

=

21.21,500

4π×

= ×

= 1,696 kg

T dLu= π

1.2 50 11= π× × ×

= 2,073 kg

o

Vu

jd=Σ

65554 1.5 39.735

=× π× ×

WSD���������� ����� �.�.�. 1007-34

djV

uoΣ

=

�����5�������)���� 2 �������ก�������)����!���ก�)*����)����� ������������������ก�%�ก����#� �������6�����%�ก

�&( �,7��ก��ก����8!������ก����� ��(�ก��,����ก��( ��%�ก��������� ����ก�������)����#� �������6 %$��������������$�$� � L #� �&���&�

L

TL

Tu

oΣ=

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���� ������)����ก!��ก��*������)������ก������� ��� ����.���ก�� 11 กก./2�.2

���� ����ก)�����-�����ก�����������#� �����ก�����8��������ก��+��ก�� 30 2�.

WSD���������� ����� �.�.�. 1007-34

��������������#� �����:.� ���������.���ก�����#� ก����.�����.,��+

���� �����������ก ������������:

����ก)� ��(�.���ก�� 25 กก./2�.2D

f29.2 c′

��(�.���ก�� 35 กก./2�.2����ก�( ���ก%�ก����ก)�D

f23.3 c′

���� �����������ก ����������� ��(�.���ก�� 28 กก./2�.2cf72.1 ′

ขอสอบภย

ขอที่ : 27

ถาไมทํา “ของอมาตรฐาน” ระยะที่ตองฝงเหล็กกลมผิวเรียบ (RB 15มม.) จากหนาตัดวิกฤต (critical section) มีคาประมาณเทาใด กําหนดให หนวยแรงยึดเหน่ียวที่ยอมให u = 11 กก./ตร.ซม. (สูตรคํานวณ L = dbfs/4u

RB15 � db = 1.5 cm = 40.9 2�.u4fd

L sb=11412005.1××

=

ขอสอบภย

ขอที่ : 72

ถาระยะฝงยึดของเหล็กเสริมรับแรงดึง(ที่ไมใชเหล็กบน) ถูกจํากัดใหไมเกินกวา 80 ซม. จงใชวิธี WSD หาขนาดโตสุดของเหล็กขอออย (SD30) ที่สามารถนํามาใชได กําหนด fc’ = 150 กก./ซม.2

= 74.3 2�.

u4fd

L sb=

13.14415008.2

L××

=

b

c

d

f23.3u

′=

bb d6.39

d15023.3

==

sb f

Lu4d =

150080d/6.394 b ××

= = 2.91 cm

��(�ก����ก DB28 : ksc 13.148.215023.3

u == < 35 ksc OK

< 80 2�. OK

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ขอสอบภย

ขอที่ : 73

จงประมาณระยะฝงยึดจากหนาตัดวิกฤตถึงตําแหนงที่ตองเริ่มงอเหล็กเสริมเพ่ือทําเปน สําหรับเหล็ก DB25 (As = 4.91 ซม.2) ที่รับแรงดึง ซึ่งวิธี WSD กําหนดวา “ของอมาตรฐาน” มีกําลังรับแรงดึงไดเทากับ 700 กก./ซม.2 กําหนดให fc’ = 200 กก./ซม.2, fy = 3000 กก./ซม.2 และหนวยแรงยดึเหน่ียวที่ยอมใหของ DB25 = 13 กก./ตร.ซม.

����#�+��� Asfs = 4.91x1500 = 7365 kg

�������� ��#� ����ก�� = 7365 – 700x4.91 = 3928 kg

�$�$� �� : πx2.5x13xL = 3928 L = 38.5 cm

Steel Force and Bond Stress

T

M

T

MCracked concrete segment

u stresses on bar

������ก���<�����ก���%$�ก��������!+�2 �����ก%$.����ก�)���ก���

%�.����������������� ��#� ������� u = 0

steel tension TdjM

T = ���� T ������ก������������ก#� �=#� �������

bond stress u

dxdT1

uoΣ

=

��������������� ��!��"��#���� ������$������

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Actual Distribution of Flexural Bond Stress

CL

Actual T

MT

jd=

Actual u

o

Vu

jd=Σ

�����%�&�'����()�!���������������������� ����'��"���������� �!�ก��������#��*���'������'��

Bond Stress in a Pull-out Test

T

fs = T/A

Bar stress = fs Bond stress = u

����ก��#��)ก���������� ���$���������ก������$���ก��� ���������&������ก���

,> 1950 ����%�ก��+�%$���ก��#��)����#�

%�กก��#��)%$&)���������������� �������.����#� ��������������ก����

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ก��������� ���� !ก"�##�#

uT

�����ก�$#��������$�������#���$���������ก��$���ก���

RT

ก��� ������ก�#� !������

����กก��+������������ ��ก�)���ก�����ก����ก�$#��������$�������#��

2 �%$����������������( �����ก����-8ก�����( ����ก����%$��!�����

T

u

��ก�:�!������ก ��� ��ก%�กก����ก�$#��������$�������#������

!������%$#������ก�ก��� ������ก��������,?�ก�����ก�$#��������ก���

R R

V-notch failure

Side-split failure

R

Vertical cracking of bottom cover

��$�กก������� �%#�ก��&

R R

R R

*�!��ก��� ��#������ก����,?�ก����������ก���

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Radial componentof bearing pressure

Circumferentialtensile stresses

ก��ก�'$�������$�กก������� �%#�ก��&

���5,�$ก�)!�������#�@��@��%�ก����ก����#������ก����������ก�$%����ก.,���

���ก�������)

Cylindrical zones of

circumferential tension��������%$�������$�$#��

����+�%��������$�$�����$���������ก������$�$�$!�)#� �&���&�

Minimum Bar Covering and Spacing

cb

Minimum bar covering ( cb )

cs

Minimum bar spacing ( 2cs )

2cs

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Splitting of concrete along reinforcement

(�!!��� ��+�#�ก���#ก������� splitting

�$�$�����$���������ก�������(��$�$�=�����ก��� -����.���&���&���%�ก�ก����ก����

ก�������)����!�����ก���

���%�กก��� ��-������ก��%#������ก�ก����ก����

ld

T = Ab fy

T = 0

%��(������� (Development Length)

��( ��%�ก������������� �������.����������������������ก����� ����A�� ACI ��$ �.�.#. ��, %%=)��%��,�� ������������!��'�������������#�

'������������ ld �(����������+�#� �=#� ��������������ก������&� �!+�%�ก@8��5-�

ก��������ก fy

(�!!��� ��+�#�'������������ ld ก�������)����!�����ก���

�$�$�=����$�$�$��������ก�����

����ก,��ก

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SDM����ก)� *� �.�.�. 1008-38

�������� �� ld = �������� ��&(+�A�� ldb ×××× ����8:,��)���

'������������ �����ก ������������

�������� ��&(+�A�� b ydb

c

0.06 A f

f=

′ℓ ������)����ก!��.���ก�� 36 �.�.

����8:,��)���%$!+�ก�)�$�$�=�� �$�$��������ก����� ��$����ก�����#��!���

'������������ �����ก �����������

b ydb

c

0.075d f

f=

′ℓ�������� ��&(+�A�� �������.������ก��� 0.0043 db fy

�.�.. 1008-38 = ACI 1989 �������=����ก����+���+���� ACI 1995 %���ก���,�� ���,������

SDM����ก)� *� ACI318-08

'������������ �����ก ������������

��#� �#���ก� ��ก�)ก����)�� (cb + Ktr)/db %$���.�.���ก�� 2.5 ��$ trtr

40 AK

sn=

��( � n �(�%���������!������ก�������ก��� ������$��)ก����ก����

��%������ Ktr = 0 ������)ก����ก�))���������������%$������ก�����#��!���

Atr = &(+�#� ����ก�����#��!���#�+���<�����$�$ s

s = �$�$��������ก�����#��!�����ก#� �=<����������� ld

�������� �� �������.������ก��� 30 2�.y t e sd b

b trc

b

0.28 fd

c Kfd

ψ ψ ψ=

+ ′

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cb �(����#� ����ก����$�����

(ก) �$�$%�ก@8��5ก�������ก�����-�*�����ก���#� �ก��#� �=

( ) �� ��� �!���$�$�����$���������ก�����

Definition of Definition of AAtrtrPotential planeof splitting

Atr

SDM

2 cb cb

cb

e 1.5ψ =����ก��������$�$�=������ก��� 3 db ��(��$�$�����$���������ก��������ก��� 6 db

SDM

����ก������'����-��ก.� eψ

����ก��������(�)��&���ก2� ก�:��( �#�+��� e 1.2ψ =

e 1.0ψ =����ก�����.�����(�)��&���ก2�

�����.�ก����*��8:!�� ψt ψe %$����.���ก�� 1.7

����8:,��)���������)�������� ��!������ก!��������)���� :

����ก�����)� �,B�����ก������������������ก�����8��)(+��������กก��� 30 2�.

t 1.3ψ =

#�����������ก�����tψ

����ก��������������( � t 1.0ψ =

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(As required) / (As provided)����8:,��)���������)����ก������ก�� :

SDM

��%��������� ��.���( �����ก����������5�������)�������ก��ก���#�

����ก��%�กก��������$�5 �ก����ก�:�#� ก���,��� ��(�ก��� ��������)

fy ���#� ����ก�����C&�$ ��(�����ก�����#� -8ก��ก�))<�����!��ก����

!����( ���*����.��

sψ �������ก�����

s 0.8ψ =����ก����!�� 20 �.�. ��$���กก���

s 1.0ψ =����ก����!�� 25 �.�. ��$��D�ก���

SDM�,&�%)���-%��(�������#������

�#�ก������8��������$���� ��#��,?�)�����%����8�������������ก�������:����

���� �� ��ก������� (cb + Ktr) / db = 1.5

DB20 and smaller DB25 and larger

Case A:(1) covering = db

clear c-c = db

min. stirrup

(2) covering = db

clear c-c = 2db

(A - 1) (A - 2)

bc

ety df

f15.0

ψψb

c

ety df

f19.0

ψψ

Case B: others

(B - 1) (B - 2)

bc

ety df

f23.0

ψψb

c

ety df

f28.0

ψψ

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SDM&����%��(�������

������) f’c = 240 กก./2�.2 ��$ fy = 4,000 กก./2�.2

����ก����'������������ (.�.)

Case A Case B

DB10DB12DB16DB20DB25DB28DB32DB36

DB40

38.746.562.077.5123137157177

196

49.158.978.598.1181202231260

289

#������ ก��'����$'������������ �����ก�����#�����������

39 c

m

45 c

m

1.5 m

A 3DB25

Ld

B

Wall

Construction joint

Construction joint 40 cm

DB16 at 30 cm O.C.

����( ���ก%�ก*������ก������������ก)� 3DB25 �����:�$�$� ������#� �=!������ก�������*��� ก���� f’c = 240 กก./2�.2 ��$ fy=4,000 กก./2�.2

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��1��� ���#��#�+���������(� ����8����������� ��$�8��������$���� �������A�� ACI

1. '����$�����������ก��3����3������ก(�ก

�$�$�=�����!��� = 4 + 1.6 = 5.6 2�. (2.24db)

�$�$��������ก���� = (40 – 2(4+1.6) – 3×2.5) / 2 = 10.65 2�. (4.26db)

��ก�:���+.��������ก,��ก ���������ก DB16 ������ �<����*���#�+����!���

��( ��%�ก�$�$�=����กก��� db ��$�$�$��������ก������กก��� 2db ��$����ก���� DB25 ����+��,B�ก�:� (A-2)

= 159.4 cm

2. '����$'������������

y t ed b

c

0.19 fd

f

ψ ψ=

′ℓ

0.19 4,000 1.3 1.02.5

240

× × ×= ×

����ก)�

����+������������ ���!��.,��*��� 1.60 ������(�%�ก����� = 123 x 1.3

= 159.9 cm

3. '����$'������������3��&*��6#����������� y t e sd b

b trc

b

0.28 fd

c Kfd

ψ ψ ψ=

+ ′

cb �(����#� ����ก����$�����

(ก) �$�$%�ก@8��5ก�������ก�����-�*�����ก���#� �ก��#� �=

( ) �� ��� �!���$�$�����$���������ก�����

�$�$�=�����!��� = 4 + 1.6 + 2.5/2 = 6.85 2�.

40 2 6.850.5

2− × =

= 6.58 2�. cb = 6.58 .�.

trtr

40 AK

sn= ��( � s �(��$�$��������ก�����#��!������$�$� ��

Atr �(�&(+�#� ����ก,��ก���$��)��ก����

= DB16 #�+����!��� = 2 × 2.01 = 4.02 2�.2

n = %���������ก����#� � �� = 3

= 30 2�.

40 4.0230 3×

= 1.79 2�.

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USE 2.5b tr

b

c K 6.58 1.79d 2.5+ +

= = 3.35 > 2.5

y t e sd b

b trc

b

0.28 fd

c Kfd

ψ ψ ψ=

+ ′

ℓ0.28 4,000 1.3

2.5240 2.5

× ×= ×

×= 94.0 2�.

39 c

m

45 c

m

1.5 m

A 3DB25

Ld = 95 cmB

Wall

Construction joint

Construction joint

USE 95 cm

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Bond, Anchorage, and Development Length

� �������������� ���ก���������������

� �������������� ���ก�����ก�� ���ก��

� �������������������������

� Bar Cutoff and Bend Points in Beams

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Part 2

�������������� ���ก���������������

b ydc

c

0.075 d f

f=

′ℓ������������������ �� ����!� ����ก� � 0.0043 db fy

SDM�.�.�.

�����������"������ ���ก����������������#$��%�$�ก������������������

�&%ก������&%����� � �� ����!� ����ก� � 20 '�.

����&%����� ������������"������ ���ก���������������:

���� ����ก�����ก� ก� �*+,����ก��$�กก���# ���-�. (As *+,����ก��)/(As *+, "�#�$�#�)

���� ����ก���ก�ก���� ��� ≥ 6 ��. ��- �-�- ก�+�� ≤ 10 '�. 0.75

���� ����ก���ก������ ��� ≥ 12 ��. ��- �-�-� �� ≤ 10 '�. 0.75

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��������������������ก���������������ก� ��� ก�� SDM

�.�.�.

�������������� ���ก�� �- "��*+,�����ก�� ���ก�� ��,�����������������

���� ���� �#,��+ก�����- 20 "������ ���ก 3 "�������ก�� ���ก��

���� ���� �#,��+ก�����- 33 "������ ���ก 4 "�������ก�� ���ก��

/�ก���#$��%�������� ��0/��1��� � ���ก*+,�����ก�� ���ก�� ��� "���� ���ก "��

+��*+,�+��� "��2 �3&��.ก���*+,����$�ก����*+,������ *+�� * �',��+$43&��.ก���

���ก��

ก����������ก������� !� �"#����

LdLdh ก��������� ���ก "�#� ��,� �#,�ก������ ��,�����

��� ���ก*+,����ก������-�-�����+!� �+����

ก��*�������������$-�+��� 90o ��- 180o

Tdb D

4 db ≥ 6 cm180o Hook

Tdb D

12 db90o Hook

� �����ก����$����������ก���� (D)

�.�.�. 6 ��. 1� 25 ��.28 ��. 1� 36 ��.44 ��. 1� 57 ��.

6 db

8 db

10 db

ACI DB10 – DB25

DB28 – DB36

DB40 – DB60

8 db

10 db

12 db

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ก������� !� �����"����ก���ก �.�.�.

Tdb D

6db

db DT

12db

Tdb

6db

135o

"������ ���ก���ก���6 ��. 1� 16 ��.

"������ ���ก���ก���20 ��. 1� 25 ��.

"������ ���ก���ก���6 ��. 1� 25 ��.

�������������� ���ก����������� SDM�.�.�.

�����������"������ ���ก��������������',���45�ก��������������#$��%�

$�ก���������������&%ก������&%����� � �� ����!� ����ก� �� �*+,��กก� ���� 8 db

��- 20 '�.

LdLdh

Combined actions: - Bond along straight length

- Anchorage provided by hook

Tdb

Critical sectionfull bar tension

Tdb

Ldh

12db

4db

4db ≥ 6 cm

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�������������� ���ก����������� SDM�.�.�.

������������������"������ ���ก�����������',� fy = 4,000 กก./'�.2

�.�.�.

ACI 318-08

bhb

c

320 d

f=

′ℓ

e yhb b

c

0.075 fd

f

ψ=

′ℓ

����&%����� ����"8��ก��/9�����+���+�

���� ก�������ก����ก��� *+, fy �+� �� ��$�ก 4,000 กก./'�.2 fy / 4,000

���� ����ก��� ≤≤≤≤ DB36 ���ก�+��4��������� ≥ 6 '�. ��-

"��������:�ก*+,�+�-�-�4�� ������ ≥ 5 '�.0.7

���� ����ก�����ก� ก� �*+,����ก��$�กก���# ���-�. (As *+,����ก��)/(As *+, "�#�$�#�)

�������������� ���ก����������� SDM�.�.�.

���� ����ก��� ≤≤≤≤ DB36 ����8��/����ก�+�-�-� �����

����������� ≤ 3 db

0.8

���ก�+��4������������ ≥ 6 '�.

���ก�+��4��������:�ก ≥ 5 '�.

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dh

2400L 65.3 39.2 cm

4000= × =

320 2.565.3 cm

150

×= =b

hb

c

320dL

f=

ขอสอบภย

ขอที่ : 73

จงประมาณระยะฝงยึดจากหนาตัดวิกฤตถึงตําแหนงที่ตองเริ่มงอเหล็กเสริมเพ่ือทําเปน สําหรับเหล็ก DB25 (As = 4.91 ซม.2) ที่รับแรงดึง ซึ่งวิธี WSD กําหนดวา “ของอมาตรฐาน” มีกําลังรับแรงดึงไดเทากับ 700 กก./ซม.2 กําหนดให fc’ = 200 กก./ซม.2, fy = 3000 กก./ซม.2 และหนวยแรงยึดเหน่ียวที่ยอมใหของ DB25 = 13 กก./ตร.ซม.

����*����� Asfs = 4.91x1500 = 7365 kg

ขอสอบภย

ขอที่ : 74

จงใชวิธี USD ประมาณระยะฝงยึดจากหนาตัดวิกฤตถึงตําแหนงโคงงอเหล็กเสริมเมื่อทาํเปน “ของอมาตรฐาน” สําหรับเหล็ก RB25 (As = 4.91 ซม.2) ที่รับแรงดึง กําหนดให fc’ = 150 กก./ซม.2, fy = 2400 กก./ซม.2 และให modification factor = 1.0

���� ��+,��*+,����ก�� = 7365 – 700x4.91 = 3928 kg

�-�-���� : πx2.5x13xL = 3928 L = 38.5 cm

ก����� ���ก ���� (Splicing)

ก��� � ���ก "�#�/������- "���ก ก#���� ��0/����ก �"���� ',���$*��!������#<+

!��ก ก��� �*�� ก��� � 9#�ก� ก��� � 9�,�� ��-ก��� �������ก*��

��������ก���� ��������

ก��� �*�� ���ก���2#�!� "��2�"/���� ���ก "��������& � ��ก�� ≤ 1/5 �-�-*��

��-!� ก#� 15 '�.

ก� %���" (lap splicing) $-/9�ก��1 �� *����-�� �� ���ก "��8��/��-�-

*�� /9�ก�� ���ก���!� ก#� 36 ��.

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ก��� �*�� ���ก "��*+,����� ���ก�� /��/9��������*+,����ก������� �- "��

��- �#,��-�-���$����� "��/������ "���������- 2-3 "�� �#,��-�-*��

20% " �����- 4 "�� �#,��-�-*�� 33% ����� � ���ก*�������# �% +��ก��

ก� %��&��ก� ���� �*��ก���������������������!��� ������*+,"4 1.25 fy

��� ���ก "��

ก� %��&'��� ���� � 9�,������ �9���������+ก������� ������*+,"4 1.25 fy ���

���ก "��

ก� %�����ก������������"#���� ก����9����4%8�� A ��- B �� *����+�

�-�-*������!� ����ก� � 30 '�.

ก��� �9����4%8�� A d1.0 ℓ

ก��� �9����4%8�� B d1.3 ℓ

ก����� ���ก ��������� (Bar Cutoff)

�2�8&�#5� ���.�

�����ก����

���ก "�#������ก����ก�� "�#� ���ก/����� � ��,��

���ก$-1&ก "�#� ���!�/����ก�+���# �%*+,������� /����� � ��,��$- "�#� ���ก� ��

*+,ก���9 �� ��,����5� ���.��ก ��- "�#� ���ก��*+,$4������ ��,����5� ���.��

�('�������)��������ก����"����� *)+,ก��$�- "���.���/�% ���ก�

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Moment capacity of beam:2s y

aM A f d

= −

5MLd

Ld

Ld

2 bars 4 bars 5 bars

Moment capacity Mn

Required moment Mu / φ

CL

������������ก������ ��!� ���"�������

M

2M

3MMmax

�#��������� ���ก ���������$��� %&��

������ ���4 ���ก*��*=>?+

/�ก����� ���ก "�#�������� ������,� ���ก ��$4*+,!� ����������!� ����-�- d

���� 12 db 5�/9�� �*+,��กก� � �ก ���*+,$4���������9 �� +,����-�����#"�-

��������,�

���ก "�#� 2 "��

���ก "�#� 1 "��

0

d ���� 12db d ���� 12db

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Required moment Mu

ก����� ���ก ���� %&'�������������$��� %&��

���9 �� +,������������ก���*4ก�2 "�,�� "��

�+�2�8&�#5� ���.���/��&�

aℓ L

ก�����5� ���.$- �#,� ��� "�����$�ก3&��.*+,

�������$��+ก����� ���*+,8��/��-�- dℓ

dℓ

Moment capacity φφφφMn

��$$-�+9 ��*+,��������ก��5� ���.��ก ก#�

ก�����5� ���. ��$*��/�� ก#ก���#���# :��-*+,

$�ก���� ��+,�� (local bond failure)

����������*��/������9�����ก�����5� ���.

!� ����ก� � "��"��2�" O-A ��/��&�

O

Mu

AB

φφφφMn

����9�����ก�����5� ���. n dM /= φ ℓ

����9�������������ก��5� ���. uu

dMV

dx=

����������9�����ก�����5� ���.����*+,"4���

nu

d

MV

φ=

$-!�� ��������������ก*+,"4*+,���/�����

nd

u

MVφ

=ℓ

ก����� ���ก ���� %&'�������������$��� %&��

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SDMก��������� ���ก �������!� ���"��ก �.�.�.

� ������,� ���ก "�#��� ������ 1 /� 3 ��� ���ก���5� ���.��ก/����9 �� +,��

�� ���!�/����������!� ����ก� � 15 '�.

� ������,� ���ก "�#��� ������ 1 /� 4 ��� ���ก���5� ���.��ก/����� � ��,�� �� ���!�/����������!� ����ก� � 15 '�.

15 cm

+ AsAs/4

15 cm

+ AsAs/3

SDMก��������� ���ก �������!� ���"��ก �.�.�.

� *+,$4���������9 �� +,����-*+,$4�ก��� ���ก "�#����5� ���.��ก�����+���*+,

$��ก� ��,�/��������������+� �!� ก#�

$4���������9 �� +,��

au

nd V

M3.1 ℓℓ +≤

$4�ก������� � ��,��

au

nd V

Mℓℓ +≤

5�*+, Mn = ก�����5� ���.������%5�"��4�#/�� ���ก "�#�*�����*+,�������+�� �����1� fy

Vu = ��� :�����-���*+,������

*+,$4���������9 �� +,������-�-���*+, ��$43&��.ก���*+,������aℓ

*+,$4�ก������� � ��,������-�-���*+, ��$4�ก��� d ����

12db 5�/9�� �*+,��กก� �aℓ

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���ก "�� a ���ก "�� b

SDMก����� ���ก �������!� ���"��ก �.�.�.

���*$����"����� &%��������

dMax. ℓ

aℓun V/M3.1

CL

CL

�-�-����*�����

�������#ก=�"������ ���ก "�� a*+,������ ���4 ���ก*��*=>?+"������ ���ก "�� b

SDMก����� ���ก �������!� ���"��ก �.�.�.

���*$����ก��"����� %�� '���

dMax. ℓ

un V/M

$4�ก���

da =ℓ ���� 12db

���ก "�� a�������#ก=�"������ ���ก "�� a

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�������� ก�����'���ก��������� ���ก���!� ���"��ก

aℓ

= 15 cm

L = 5 m

wu = 12 t/m

2DB40

40 cm

d = 54 cm

1. ��� �.���������������� DB40

(A - 2)ก�%+ y t ed b

c

0.19 fd

f

ψ ψ=

′ℓ

0.19 4,000 1.0 1.04.0

240

× × ×= × = 196 '�.

2. �*��"���������������� DB40

= 12.3 '�.

$4���������9 �� +,��

au

nd V

M3.1 ℓℓ +≤

[ C = T ] s y

c

A fa

0.85 f b=

2 12.57 4,0000.85 240 40× ×

=× ×

n s y

aM A f d

2 = −

12.32 12.57 4.0 54 /100

2 = × × × −

= 48.1 ���- ���

*+,$4������ uu

w LV

2=

12 52×

= = 30 ���

na

u

M 1.3 48.1 1001.3 15

V 30× ×

+ = +ℓ = 223 '�.

����������� '�. ����ก� � 223 '�. ������ DB40 "����1/9�!�d 196=ℓ

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As

SDMก����� ���ก �������!� ���"�� �.�.�.

$4�ก���

� ������,� ���ก "�#��� ������ 1 /� 3 ��� ���ก���5� ���.�� ��$4�ก���

!� ����ก� � d ���� 12db ���� 1/16 ����-�-9 ��� ����� 5�/9�� �*+,��กก� �

dℓ

��,� ���ก As/3 ����-�-!� ����ก� �d ���� 12db ���� Ln/16

Bar Cutoff requirements of the ACI Code

Moment capacityof bars O

Moment capacity

of bars M

Fac

e of

sup

port

Inflection pointfor +As

Inflection point for -As

+ M

- M

Bars N

Bars Od or 12db

Ld

C of spanL

Ld

15 cm for at least1/4 of +As

Bars M

Bars L

Ld

d or 12db

Ld

Greatest of d , 12db or Ln/16for at least 1/3 of -As

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Standard Cutoff and Bend Points for Bars

For approximately equal spans with uniformly distributed loads

15 cm

L1

L1/4 L1/3

0 cm

15 cm

L1/8

L2

L2/8

L2/3 L2/3

L1/7 L1/4 L2/4 L2/4

L1/3 L2/3

����%�� 6.3 $��#$��%����������� ���ก����- ���ก� ��������� � ��,��9 ����ก"4���"��������ก�2 ��-���*�������� wu = 8.0 ��� ก���� f’c = 280 กก./'�.2, fy = 4,000 กก./'�.2,b = 40 '�., h = 60 '�. ��-���ก�+��4�� 4 '�.

wu

Exterior column Interior column

Ln = 7.6 m

1. ��ก#""����ก�����"���� 4#�)#��5'� �"'6�� �

a. -&�ก������)�4#""�)��.�%����� 4#�)#��5'�

Interior face ofexterior support

-Mu = wuLn2/16 = 8(7.6)2/16 = -28.88 t-m

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Mid span positive +Mu = wuLn2/14 = 8(7.6)2/14 = 33.01 t-m

Exterior face of firstinterior support

-Mu = wuLn2/10 = 8(7.6)2/10 = -46.21 t-m

Exterior face of firstinterior support

Vu = 1.15wuLn/2 = 1.15(8)(7.6)/2 = 34.96 t-m

b. (�*�.�����ก�����"���� 4��� 5�/9����ก�+��4�� 4 '�. ���ก���ก DB10 ��- ���ก "�#����5� ���.� DB25 ���� DB28 � � d ≈ 60-4-1.0-1.4 ≈ 53.6 '�.

Mu As required Bars As provided

- 28.88 t-m

+ 33.01 t-m

- 46.21 t-m

15.97 cm2

18.44 cm2

26.76 cm2

4DB25

4DB25

2DB25+3DB28

19.63 cm2

19.63 cm2

28.29 cm2

A

A

B

B

4DB25 2DB25+3DB28

4DB25

40 cm

60 cm

Section A-A

4DB25

4DB25

[email protected]

40 cm

60 cm

Section B-B

2DB25

4DB25

[email protected]

3DB28

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� Torsional effects

� Torsion in plain concrete

� Torsion Design by WSD

� Cracking Torque Tcr

� Torsion Design by USD

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Torsion 1Torsion 1

T

Torsional effects in reinforced concrete

T

T

mt

Torsion at a cantilever slab

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T

T

mt

Torsion at an edge beam

A B

A B

Stiff edge beam

A B

Flexible edge beam

Torsion in plain concrete members

T

T

x

yτmax

Rectangular section: max 2

Tx y

τα

=

y/x

α

1.0 1.5 2.0 3.0 5.0 ∞

0.208 0.219 0.246 0.267 0.290 1/3

WSD

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WSD���ก����ก���ก��� ���� ��� �.�.�. 1007-34

(ก) ������������� ��������� ������ L ������ T �����������������������ก�� !"���ก��#��������������"

t 2

3.5Tx y

τ =Σ

����$ x ��$�����"� � % y ������� $���� ���������&!����%ก$� !"�����'����������"�

(������������������� L ���$��� T ���������'������� �����$�)�� x ��$�����"� � % y ��$�������

����ก���� $��*ก������+���ก��������� ��� $� ΣΣΣΣx2y ,%��$�-���ก�� 3 ������������*ก ���$-���ก�� 1/12 $�+������ )���+��������$�ก���

6402 ก���� ������ ���

t

≤ 3 t or 1/12 L

��$��.��� ���������� ,�ก�%�������������/�$�-���ก���ก�01ก������"

ก������������������������ max c1.32 f ′τ =

(�) ���������� ���������ก ������������$�����������-�� �������������� t ��$� -����$�ก��� x/4 (�������$����� x/4 ���-����$�ก��� x/10 ������� ��� Σx2y

���� 4t/x

t

x

() ���������� ���������ก ��������������$�ก��� x/10

���������)���+��234&!����'����$����� ��

��ก��������$$ก��� ���(�$����������������#��ก�� !"����%�% d ,�ก $� $����$����

6403 ���ก�������������� ������

(ก) ก�������������/�$�,�กก������.��$���������ก��

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(�) ก�������������������/�$�� %����������ก�� ����������������"�$�� ������ก��

t v 2

3.5T Vbdx y

τ + τ = +Σ

���� �� 5ก� $ก �ก��"���!�� � t

c v

A Ts 2A f

=

���� �� 5ก������������� %�#� lc s

T zA

2A f=

Ac y0 h

x0

b

����$ Ac ��$���"$���ก��$�ก��6������ 5ก� $ก

z ��$�%�%�%������� 5ก����������)���/ ��

o o2x 2 yz

4+

=fv ��$���������!����$���� $��� 5ก� $ก

fs ��$���������!����$���� $��� 5ก����������

ก������������������������

cmax f65.1 ′=τ

6404 ก�!��"!��#ก�������

����!��"!��#ก�������(�����������/�$����ก�� !"� τt ���$ τt + τv c0.29 f ′>

ขอสอบภย

ขอที่ : 79

คานกลวงขนาด 30x40 ซม. ผนังดานขางหนา 10 ซม. ผนังดานบนและลางหนา 12.5 ซม. ถาคานนี้รับโมเมนตบิดเพียงอยางเดียว จงใชวิธี WSD ประมาณคาโมเมนตบิดใชงานสูงสุดที่ไดจากคอนกรีตเพียงอยางเดียว กําหนดให fc’ = 150 กก./ซม.2

= 365 kg-m

t 2

3.5Tx y

τ =Σ c0.29 f ′≤ 2

c cT 0.29 f x y / 3.5′= Σ

2cT 0.29 150 30 40 / 3.5 /100= × ×

ขอสอบภย

ขอที่ : 80

คานกลวงขนาด 30x40 ซม. ผนังดานขางหนา 10 ซม. ผนังดานบนและลางหนา 12.5 ซม. ถาคานนี้รับโมเมนตบิดเพียงอยางเดียว ตามวิธี WSD เม่ือเสริมเหล็กรับแรงบิดแลว คาโมเมนตบิดใชงานสูงสุดที่หนาตัดนี้จะรับได=? กําหนดให fc’ = 150 กก./ซม.2

= 1663 kg-m

t 2

3.5Tx y

τ =Σ c1.32 f ′≤ 2

max cT 1.32 f x y / 3.5′= Σ

2maxT 1.32 150 30 40 / 3.5 /100= × ×

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1.0 m

40 cm

w kg/m per 1 m width

20 cm

10 cm

�� ����� $$ก������+����������� 4.0 �����.��$�$����.�"�ก����+������� 1.0 ���� ��� 10 &�. �"�����ก����#ก,� 100 กก./�.2 ก����� f’c = 150 กก./��.&�. fy =

3000 กก./��.&�. (�� 5ก������) fy = 2400 กก./��.&�. (�� 5ก� $ก)

T

T

mt �"�����ก,� + �"�����ก.�"� :

w = 100 + 0.1x2400 = 340 kg/m

mt = (1/2)(340)(1.0)2 = 170 kg-m

� �$ก������� 20x40 cm � d = 35 cm

d

4.0 m

T = 170(4.0/2 – 0.35) = 281 kg-m

1.0 m

40 cm

20 cm

10 cm

30 cm

t 2

3.5Tx y

τ =Σ

��������������ก�� !"�

t 2 2

3.5 281 10020 40 10 30

× ×τ =

× + ×

= 5.18 kg/cm2

�"�����ก��� = 0.2x0.4x2400 = 192 kg/m

�"�����ก��� = 340 + 192 = 532 kg/m

����/�$����%�% d = 35 cm ,�ก,#��$����

V = 532(4.0/2 – 0.35) = 878 kg

���������/�$�τv = 878/(20x35) = 1.25 kg/cm2

���������/�$����τt + τv = 5.18 + 1.25 = 6.43 kg/cm2

���������/�$����$����

1.65 150 20.2= > 6.43 ksc OK

���������/�$��$�ก��

0.29 150 3.55= < 6.43 ksc NG

����!��"!��#ก$��ก����!%&��

t

c v

A Ts 2A f

=281 100

2 15 35 1200×

=× × ×

= 0.0223 cm2/cm

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�������� 5ก� $ก��$���#� :

= 0.030 < 2(0.0223) OK

� �$ก�� 5ก� $ก RB9 (As = 0.636 cm2)

s = 0.636/0.0223 = 28.5 cm

d/2 = 35/2 = 17.5 cm < 60 cm

!��"!��#ก$��ก RB9 @ 0.17 m

!��#ก!��"��"�� :

z = (15 + 35) / 2 = 25 cm

lc s

T zA

2A f=

281 100 252 15 35 1500

× ×=

× × ×

= 0.446 cm2

)�����1��� :21

M 532 4.0 1064 kg-m8

= × × =

s

1064 100A

1500 0.904 35×

=× ×

�� 5ก���� ��� :

= 2.24 cm2

�� 5ก���� �����"���� :

= 2.24 + 2×0.446 = 3.13 cm2

3 DB12

�� 5ก��������"���� := 2×0.446 = 0.89 cm2

2 DB12

0.40

0.20

2DB12

3DB12

RB9 @ 0.17 m

vAmin 0.0015b

s= 0.0015 20= ×

Cracking Torque

T

T

45o

Torsion cracks

τ

τ

τ

τ

ft max at 45o

TT

Tt

Tb 45o

Bending: Tb = T cos45o

Torsion: Tt = T cos45o

Plain concrete rectangular section in torsion

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a

a

TTb

Tt

45o

τ

τ

ft max at 45o

Concrete crack occurs when

,max 0.80 0.80 2.0 1.6t r c cf f f f′ ′= = × =

fr = Modulus of rupture

Sectional Modulus:2

31 2/( / 2)

12 cos45 6cos45a a a a o o

y x yS I x x

x− −

= = =

,,max 2 2

36cos45cos45 1.6

ob cr o cr

t cr ca a

T Tf T f

S x y x y−

′= = = =

Cracking Torque: ( )2

1.63cr c

x yT f ′=

USD���ก����ก���ก��� ���� ��� �.�.�. 1008-38

4406 ก�����!%&����(ก��������� "ก��

-����$����� $�ก���������$ 4/TT cru φ≤ )yxf13.0( 2c Σ′φ≤

������ �ก�� ��������ก����� φ = 0.85

ก�� ����� $���������� 5ก : Tn = Tc + Ts

$$ก��������������ก�� ����� : Tn ≥ Tu / φ

ก�� ����������)���$�ก�� :2

ut

u

2c

c

TCV4.0

1

yxf21.0T

+

Σ′=

yx

dbC,

2tΣ

=

x1x

y1 y

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USD���ก����ก���ก��� ���� ��� �.�.�. 1008-38

�������� 5ก �ก��"�(�� 5ก� $ก)��$���#�

+=+

sA

2s

As

A tvtv

yfb

5.3≥

Av : 2 legs

At : 1 leg

ก�� ����������)���� 5ก���� :s

fAyxT yt11t

s

α=

y11t

st

fyxT

sA

α=

50.1xy

33.066.01

1t ≤+=α

Ts ����)"�!ก�� 4Tc

USD���ก����ก���ก��� ���� ��� �.�.�. 1008-38

�������� 5ก���������� Al &!��ก�%,��$���)���$��� 5ก� $ก

1 1t

x yA 2A

s+ =

....(44-19)

���$,�ก

u 1 1t

uyu

t

T x y28 x sA 2A

Vf sT3C

+ = − +

ℓ....(44-20)

)������+��������กก��� ��� Al ��������,�ก�ก�� (44-20) -��,����'���$��ก��

�����-��,�กก����� 2 At ���� w

y

3.5b df

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ขอสอบภย

ขอที่ : 79

คานกลวงขนาด 30x40 ซม. ผนังดานขางหนา 10 ซม. ผนังดานบนและลางหนา 12.5 ซม. ถาคานนี้รับโมเมนตบิดเพียงอยางเดียว จงใชวิธี WSD ประมาณคาโมเมนตบิดใชงานสูงสุดที่ไดจากคอนกรีตเพียงอยางเดียว กําหนดให fc’ = 150 กก./ซม.2

= 365 kg-m

t 2

3.5Tx y

τ =Σ c0.29 f ′≤ 2

c cT 0.29 f x y / 3.5′= Σ

2cT 0.29 150 30 40 / 3.5 /100= × ×

ขอสอบภย

ขอที่ : 81

คานกลวงกวาง 30 ซม. ลึก 40 ซม. ผนังดานขางหนา 10 ซม. ผนงัดานบนและลางหนา 12.5 ซม. ถาคานรับโมเมนตบิดอยางเดียว จงใชวิธี USD ประมาณคาโมเมนตบิดจากคอนกรีต กําหนด fc’ = 150 กก./ตร.ซม.

= 787 kg-m

( )2c cT 0.21 f x y′φ = φ Σ

20.85 0.21 150 30 40 /100= × × ×

ขอสอบภย

ขอที่ : 82

คานกลวงกวาง 30 ซม. ลึก 40 ซม. ผนังดานขางหนา 10 ซม. ผนงัดานบนและลางหนา 12.5 ซม. เสริมเหล็กปลอกแบบวงปดขนาด 9 มม. ทุกระยะ 15 ซม. สมมุติให x1 = 24 ซม. y1 = 30 ซม.กําลังครากเหล็กปลอก 2400 กก./ตร.ซม. fc’ = 150 กก./ตร.ซม. ถาคานรับโมเมนตบิดอยางเดียว จงใชวิธ ีWSD ประมาณคาโมเมนตบิดที่ไดจากเหล็กปลอก

t s 1 12A f x yT

s=

2 0.636 1200 24 3015

× × × ×=

= 73267 kg-cm = 733 kg-m

ขอสอบภย

ขอที่ : 83

คานกลวงกวาง 30 ซม. ลึก 40 ซม. ผนังดานขางหนา 10 ซม. ผนงัดานบนและลางหนา 12.5 ซม. เสริมเหล็กปลอกแบบวงปดขนาด 9 มม. ทุกระยะ 15 ซม. สมมุติให x1 = 24 ซม. y1 = 30 ซม.กําลังครากเหล็กปลอก 2400 กก./ตร.ซม. fc’ = 150 กก./ตร.ซม. ถาคานรับโมเมนตบิดอยางเดียว จงใชวิธ ีUSD ประมาณขนาดเหล็กเสริมทางยาว ที่แตละมุม สําหรับโมเมนตบิดประลัยเพียงอยางเดียว

l t 1 1A 2A (x y ) / s= + = 2×0.636×(24+30)/15 = 4.58 cm2

�� 5ก����� %�#� = 4.58/4 = 1.15 cm2 Use DB16 (As=2.01 cm2)

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ขอสอบภย

ขอที่ : 84

คานหนาตัดสี่เหล่ียมผืนผาตัน ขนาด 0.25 x 0.60 เมตร ระยะ d = 50 ซม. รบั Mu = 5000 กก.-เมตร ที่กลางชวงคาน และ Vu = 3750 กก. กับ Tu = 2250 กก.-เมตร ที่หนาตัดวิกฤต สมมุติใช fc’ = 200 กก./ตร.ซม. fy = 3000 กก./ตร.ซม. (สําหรับเหล็กตามยาว) fy = 2400 กก./ตร.ซม. (สําหรับเหล็กปลอกทางขวาง) ถา φTc = 900 กก.-เมตร αt = 1.32 และให x1 = 20 ซม. y1 = 40 ซม. ดังนั้นตองการปริมาณเหล็กปลอก (ขาเดียว) สําหรับโมเมนตบิด At/s เทากับ (สูตร Ts = αtx1y1Atfy/s กก.-ซม.)

= 0.0627 cm2/cm

ขอที่ : 85 ถา φVc = 1550 กก. ตองการเหล็กปลอก (สองขา) สําหรับแรงเฉือน Av/s = ?

s u cT (T T ) /= − φ φ = (2250 – 900)/0.85 = 1558 kg-m

t s

t 1 1 y

A Ts x y f=α

1588 1001.32 20 40 2400

×=

× × ×

s u cV (V V ) /= − φ φ = (3750 – 1550)/0.85 = 2588 kg

sv

y

VAs f d=

25882400 50

= 0.0216 cm2/cm

ขอสอบภย

ขอที่ : 88

คานสี่เหล่ียมตันขนาด 0.30 x 0.50 เมตร fy = 2400 กก./ซม.2 สําหรับเหล็กปลอกทางขวาง T = 1450 กก.-เมตร ถา x1 = 24 ซม.y1 = 42 ซม. ปริมาณเหล็กปลอกขาเดียวสําหรับโมเมนตบิด At/s=?

t s 1 12A f x yT

s=

= 0.060 cm2/cm

t

s 1 1

A Ts 2f x y=

1450 1002 1200 24 42

×=

× × ×

ขอสอบภย

ขอที่ : 89

คานสี่เหล่ียมตันขนาด 0.25 x 0.60 เมตร d = 50 ซม. fc’ = 200 กก./ตร.ซม. เพ่ือตานทาน V = 1875 กก. และ T = 1125 กก.-เมตร จะพบวาหนวยแรงเฉือนรวมมีคา <=> หนวยแรงที่ยอมใหของคอนกรีต?

v

Vbd

τ =1875

1.5 ksc25 50

= =×

����!��"!��#ก�������τt + τv = 1.5 + 10.5 = 12.0 ksc c0.29 f ′>

t 2 2

3.5T 3.5 1125 10010.5 ksc

x y 25 60× ×

τ = = =Σ ×

0.29 200 4.1 ksc=

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� Thin-walled Tube

� Combined Shear & Torsion

� Space Truss Analogy

� Torsion Design by ACI318

� Compatibility Torsion

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Torsion 2Torsion 2

T

�����������ก�� ����� ����������������������ก�� �����������ก�������������ก����������������ก���ก������������� ��

Solid Hollow

0 1 2 3Percent of torsional reinforcement

0

Tn

Tcr solid section

Tcr hollow section

Torsional Strength of Reinforced Concrete SDM

����� �����!�����ก����� ������ก�� "���ก�� ก�� ����������� ������������ก�� ��#����ก��������������������ก��������ก��

�� ��$�ก�� �����ก����"% �"ก�����ก�� ���������������ก�����

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Shear Stress in Thin-walled Tube

Cracking Torque (Tcr)

SDM

#������ ���"�&'ก#������(����)�� � ��������ก�ก� #��ก���

A0

t

Shear flow (q) Shear flow:0

Tq kg/cm

2A=

��#�# �������������'����ก� #����)�� ���

������� �*+��"กก���� τt = q / t :

t0

T2A t

τ =

ก���ก����ก���%$���+�� τ ��#�&% c1.06 f ′ crcr c

0

T1.06 f

2A t′τ = =

�� ��$� �����!�����ก��� ( )cr c 0T 1.06 f 2 A t′=

SDMACI ก���,�����#�#����)�� ���� t = 0.75Acp/pcp ���

-+$����������� A0 = 2Acp/3 ��+��

pcp #+���������'�.���ก�� ������#��ก���

Acp #+�-+$����/%� ������� �� pcp

�� ��$� �����!�����ก��� cp cpcr c

cp

2A 0.75AT 1.06 f 2

3 p

′= × ×

2cp

cr ccp

AT 1.06 f

p

′=

������ #��)��� ก������+�� 4/TT cru φ≤

���#'���ก�� �����ก���� φ = 0.85

( )2cp

ccp

A0.265 f

p

′≤ φ

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ตัวอยางที ่7.1 คานยื่นรับน้ําหนักบรรทกุประลัย 3 ตันที่มุมหนาตัดหางจาก

ศูนยกลางหนาตัด 15 ซม. ตรวจสอบดูวาจําเปนตองคิดผลของการบิดในการออกแบบ

หรือไม กําหนด f’c = 240 กก./ซม.2

��������'������� pcp = 2(60+30) = 180 /�.

-+$���������� Acp = (60)(30) = 1,800 /�.2

������������� ���������� ������

���"ก�� �����!��� φTcr/4 ( )2cp

ccp

A0.265 f

p

′= φ

218000.85 0.265 240

180

= × ×

= 62,812 กก.-/�. = 0.63 ���-����

�����!�������ก��� Tu = (3)(0.15) = 0.45 ���-���� OK< 0.63 ���-����

SDM

3 ton

30 cm

15 cm

60 cm

tA2T

0t =τ

Combined Shear and Torsion

Torsionalstresses

Shearstresses

Hollow section

Torsionalstresses

Shearstresses

Solid section

SDM

������� �*+��"กก��*+��:db

V

wv =τ

������� �*+��"กก����:��+�� t = 0.75Acp/pcp

��� A0 = 2Acp/3

�����������������ก�����:

2h0

hu

w

u

A7.1

pTdb

V+

2

2h0

hu2

w

u

A7.1

pTdb

V

+

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Torsional Geometric Parameters

Gross area A0h = x0 y0

Shear perimeter ph = 2(x0 + y0)

x0 , y0 = Distance from center to

center of stirrup

SDM

y0 h

x0

b

�����������ก�� &�)�� ��������ก��,��,5�#�����,��������� �ก����6� ���&�)�� ��#���� t < A0h/ph ,��������� �*+��"กก������(�

2h0

hu

A7.1

pTtA7.1

T

h0

u

Maximum Shear + Torsion Stress SDM

��������� ������-�� -����"��,��������� �*+������ก�����#�����ก�����"ก��

(a) For solid sections

2h0

hu

w

u

A7.1

pTdb

V+

2

2h0

hu2

w

u

A7.1

pTdb

V

+

′+φ≤ c

w

c f12.2db

V

(b) For hollow sections

′+φ≤ c

w

c f12.2db

V

For reinforced concrete cc f53.0V ′=

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Torsional Crack

ก���ก���"กก��������ก7����(��)������

���6����#��ก����������� �% �������ก�����#�ก

ก�� �����ก��������,�8��"ก����ก���ก�9� ����ก��������� ����)��#��ก������� ����� ���

#��ก�����ก����ก���ก���#�����)����ก���� �� ��$���� ก���ก���"�,5�-+$����A0h /%� &'ก������� ������ก���ก

����ก���ก�9�

A0h = -+$�������

T

45o

Torsional crack

Txo

yo

����ก���ก�����ก���

θV1V2

V3

V4

����ก���������

�)��#��ก������� ����� ���

Space Truss Analogy

ก�����*+�� q ,�)�� ���"�&'ก��� ��ก��(��� �*+�� V1 V2 V3 ��� V4 �������

����� #� &�ก�������� ,��'�

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! "��#����������������$%�: V4

V4

x0

V4y0

s

At fyv

At fyv

At fyv

y0 cot θ �����!���"ก�� �*+�� V4 :

4 04

V xT

2=

�� �*+�� V4 = n At fyv

"������ก,������*+��:

o0 0y cot 45 y

ns s

= =

t yv 04

A f yV

s=

t yv 0 04

A f x yT

2s= 1 2 3T T T= = =

�����!�����$ ��� : n 1 2 3 4T T T T T= + + +

t yv 0 0n

2 A f x yT

s= t yv 0h2A f A

s=

! "��#������������������&��

V4y0

∆N4/2

∆N4/2θθθθ

V4

θθθθ

D4

N4

�� �% ,�����ก���������: ∆N4 = V4 cot θ

t yv 04

A f yN

s∆ = 2N= ∆

t yv 01 3

A f xN N

s∆ = ∆ =

perimeter of stirrup2(xo+yo)

y0

x0�� �% ��$ ���: ∆N = ∆N1 + ∆N2 + ∆N3 + ∆N4

t yvy 0 0

A fA f 2(x y )

s= +

ℓ ℓ

t vy hA f p

s=

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Torsion Design by ACI318-08 SDM

��+�� �����!��� Tu ��#��ก�� φφφφTcr/4 ,����ก������ �����!����-+��,��

φφφφTn ≥≥≥≥ Tu

,�ก�#��� Tn �����!�����$ ���"�&'ก��� ��

����ก���ก�������ก��������� ��,�� Tc = 0

���������ก���ก#��� �� 0 t yvn

2A A fT

s= ��+�� A0 = 0.85A0h

t n

0 yv

A Ts 2A f

= u

0 yv

T2 A f

-+$��������ก��������������� ,���-����%$��-+�������ก���� yvth

y

fAA p

s f

=

���������ก���ก������ ก�������� �*+����� �����!���

v t tvA AATotal 2

s s s+ = +

2 legs Av

1 leg At

SDMMinimum Torsion Reinforcement

���������ก���ก��������6� ( ) wv t c

yv

b sA 2A 0.199 f

f′+ = w

yv

b s3.5

f≥

������ ����ก���ก��#�����ก��#��������ก���� ph/8 ��+� 30 /�.

-+$��������ก�����������������6� yvc t,min cp h

y y

f1.33 f AA A p

f s f

′= −

ℓ ℓ

����� At/s "���� �������ก�� 1.8 bw/fyv

���� ���������ก�������ก����ก��"� �����.�,�����ก���ก ���������� �ก����6� 30 /�.

���� ���������ก������� ������%� �������������6��� ����ก���ก

���� ,5�����ก��������� ∅∅∅∅ ≥≥≥≥ 1/24 ��������� ����ก���ก ≥≥≥≥ 10 �.�.

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Example 14-1: The 8-m span beam carries a cantilever slab 1.5 m. The beam supports a live load of 1.2 t/m along the beam centerline plus 200 kg/m2 over the slab surface. The effective depth of beam is 54 cm, and the distance from the

surface to the stirrup is 4 cm. f’c= 280 kg/cm2, fy = 4,000 kg/cm2

8 m 1.5 m

60 cm

30 cm15 cm

Load from slab:

wu = 1.4(0.15)(2,400)(1.5)+1.7(200)(1.5) = 1,266 kg/m

Eccentricity = 1.5/2 = 0.75 m

Load from beam:

wu = 1.4(0.6)(0.3)(2,400)+1.7(1,200) = 2,645 kg/m

2 2(1,266 2,645)(8.0)25.0 t-m

10 10 1,000u

w LM

+= = =

×

�� �*+��������: uu

w L (1,266 2,645)(8.0)V /1,000 15.6 ton

2 2+

= = =

�����!���������: tuu

w L 1,266 0.75 8.0T /1,000 3.80 t-m

2 2× ×

= = =

�����!���������:

Flexural Design

22 2

25(1,000)(100)31.8 kg/cm

0.9(30)(54)u

n

MR

bdφ= = =

ρmin = 0.0035, ρmax = 0.0229

0.85 21 1 0.0086

0.85c n

y c

f Rf f

ρ ′

= − − = ′ OK

Al,flexure = ρ b d = 0.0086(30)(54) = 13.9 cm2

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Shear Design

Cracking Torque

60cm

15 cm

30 cm

Acp = 30 × 60 = 1,800 cm2 pcp = 2(30 + 60) = 180 cm

( )2cp

cr ccp

AT / 4 0.265 f

p

′φ = φ

21,8000.85 0.265 280

180= × ×

= 67,845 kg-cm = 0.68 t-m < Tu = 3.80 t-m

∴∴∴∴ ����ก������'ก�(� ����)����*� �

sv

y

VAs f d

=

cV 0.53 280 30 54 /1,000= × × = 14.37 ton

Vu / φ = 15.6 / 0.85 = 18.35 ton

Vs = Vu / φ – Vc = 18.35 – 14.37 = 3.98 ton

3.984.0 54

= 0.0184 cm2 / cm / two legs

bwd = (30)(54) = 1,620 cm2

xo = 30 - 2(4) = 22 cm

yo = 60 - 2(4) = 52 cm

Aoh = xoyo = (22)(52) = 1,144 cm2

Ao = 0.85(1,144) = 972.4 cm2

ph = 2(22+52) = 148 cm

Torsional Geometric Parameters

30 cm

60 c

m

Check adequacy of section

���"����'�������������,�8��-�� -���+����?

2

2h0

hu2

w

u

A7.1

pTdb

V

+

′+φ≤ c

w

c f12.2db

V

2 2

2

15.6 3.80 100 14830 54 1.7 1,144

× × + × × ( )0.85 0.53 2.12 280 /1,000+

0.0271 t/cm2 0.0377 t/cm2< ������&�����!&��!�

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Torsional reinforcement

Combined shear & torsion stirrup

t u

0 yv

A Ts 2 A f

3.80 1002 0.85 972.4 4.0

×=

× × ×= 0.0575 cm2 / cm / one leg

v t tvA AATotal 2

s s s+ = +

= 0.0184 + 2 × 0.0575 = 0.1334 cm2 / cm

two legs

��+�ก����ก���ก�9� DB12 (-+$������ � 2(1.13) = 2.26 /�.2)

������ ����ก���ก������ ก� s = 2.26 / 0.1334 = 16.9 cm

������ ����ก���ก�ก����6� smax = ph / 8 = 148 / 8 = 18.5 cm < 30 cm

Use closed stirrup DB12 @ 0.16 m (Av+t/s = 2.26/16 = 0.141 cm2 / cm)

Torsion longitudinal steel

Minimum torsion reinforcement

v t w wc

y y

A b bmin 0.199 f 3.5

s f f+ ′= ≥ f’c > 309 ksc

control

303.5

4,000= × = 0.0263 cm2 / cm < (Av+t/s = 0.141 cm2 / cm) OK

yvth

y

fAA p

s f

=

= 0.0575 × 148 = 8.51 cm2

yvc t,min cp h

y y

f1.33 f AA A p

f s f

′= −

ℓ ℓ

1.33 2801,800 8.51

4,000= × − = 1.51 cm2

t w

yv

A b 300.0575 1.8 1.8 0.0135

s f 4,000 = > = =

OK

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Total Longitudinal Steel

Bending: Al = 13.9 cm2 (Top reinforcement)

Torsion: Al = 8.51 cm2 (Distributed along perimeter)

Provide 4DB16 in the bottom half of beam

60 cm

30 cm

To satisfy 30 cm max. spacing of Al

and add 8.51 – 4(2.01) = 0.47 cm2 to flexural steel

Steel area required at top = 13.9 + 0.47 = 14.37 cm2

USE TOP 3DB25 (As = 14.73 cm2)

Section @ Supports

60 cm

3DB25

2DB25

DB12 @ 0.16 m

30 cm

2DB16

2 2(1,266 2,645)(8.0)17.87 t-m

14 14 1,000u

w LM

+= = =

×At midspan:

Required As = 9.68 cm2 (2DB25)

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Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Design of Column 1Design of Column 1� ก�����������ก��� �ก�����

� ����������

� �����������������������ก�

� ก����ก���������������������� ���

� ก����ก�������������ก����

What is COLUMN ?What is COLUMN ?- vertical member ?

- axial compression ?

- carrying floor load ?

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Pont-du-Gard. Roman aqueduct built in 19 B.C. to carry wateracross the Gardon Valley to Nimes. Spans of the first and secondlevel arches are 53-80 feet. (Near Remoulins, France)

ก�����������ก��� �กก�����������ก��� �ก

�����ก��� �ก!"�#ก����� !�ก!����������ก$���%&�����������'���(���� ��!���� �����#�)*�����

ก����ก���ก����"���(���� ��!"�&+����ก�����ก,-"ก������� �����ก��� �ก

�� � ��ก�����������ก �.��� �.���!" �� %� �(���� �� �.��&�"�� ��/�) �.���

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Tributary AreaTributary Area

When loads are evenly distributed over a surface, it is often possible to assignportions of the load to the various structural elements supporting that surface

by subdividing the total area into tributary areas corresponding to each member.

Half the load of the tablegoes to each lifter.

100 kg/m23 m

6 m

Half the 100 kg/m2 snow load on the cantileveredroof goes to each column.

The tributary area for each column is 3 m x 3 m.

So the load on each column is

100 (3 x 3) = 900 kg

Column load transfer from beams and slabs

1) Tributary area method: Half distance to adjacent columns

y

x

Load on column = area × floor load

y

x

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9 m 12 m 9 m

4.5 m

6 m

6 m

All area must be tributed to columns

C1

C1 : Corner column

C2

C2 : Exterior column

C3

C3 : Exterior column

C4

C4 : Interior column

9 m 12 m 9 m

4.5 m

6 m

6 m

C1

C1 C1

C2C2

C2

C3

C3C3

C4 C4

C4

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2) Beams reaction method:

B1 B2

RB1

RB1 RB2

RB2

Collect loads from adjacent beam ends

C1B1 B2

B3

B4

Load summation on column section for design

Design section

Design section

Design section

ROOF

2nd FLOOR

1st FLOOR

Footing

Ground level

Load on pier column= load on 1st floor column + 1st floor + Column wt.

Load on 1st floor column= load on 2nd floor column + 2nd floor + Column wt.

Load on 2nd floor column= Roof floor + Column wt.

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C1 (A-6)��������������ก��� �ก�����

���������

������

3.50 m

0.3 x 0.3 m

������

���� ����

3.50 m

0.3 x 0.3 m

���� ����

�����ก

1.50 m

0.4 x 0.4 m

RB2 = 5280 kgRB4 = 4800 kgRB19 = 4416 kgT1 = 960 kgCol.Wt. = 756 kg

Floor load = 16212 kg

2B5 = 10764 kg2B4 = 14736 kgCol.Wt. = 756 kg

Floor load = 26256 kg

Cum. load = 42468 kg

2B5 = 10764 kg2B4 = 14736 kgCol.Wt. = 576 kg

Floor load = 26076 kg

Cum. load = 68544 kg

T1

RB2

RB4

RB19

B4

B5

B4

B5

B4

B5

B4

B5

Type of Columns

Tie

Longitudinalsteel

Tied column

Spiral

s = pitch

Spirally reinforced column

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Strength of Short Axially Loaded Columns

Short columns are typical in most building columns.

P0

A A

Section A-A

.001 .002 .003

fy

cf ′

Steel

Concrete

Strain

Str

ess

P0

fyfy

cf ′

Fs = Ast fyFc = (Ag - Ast) cf ′

[ ΣFy = 0 ]

0 ( )st y c g stP A f f A A′= + −

From experiment:

0 0.85 ( )st y c g stP A f f A A′= + −

where

Ag = Gross area of column section

Ast = Longitudinal steel area

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Failures of RC Columns

CrushingBuckling

Pu

0Axial deformation ∆

Initial failure

Axi

al lo

ad Tied columnLightspiral

ACI spiral

Heavy spiral

Column Failure by Axial Load

Pu

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��ก��������������� �.�. . SDMWSD

� ���(��*��� �$��0#��1ก���%���.�ก��� 20 2�. �ก�� ���� �.��#��"���������ก

�*�%�������������*.�� 24.�!"� ����� ���(� �.���%���.�ก��� 15 2�.

)*�� �.� ������5ก������*������ � ��%��� ��ก��� 0.01 ��"%���ก�� 0.08 ���

)*�� �.� ������� Ag

��5ก �.�� � ��%����5กก��� 12 ��. ���&��ก�ก������ %��� ��ก��� 6 �� � ���&��ก

���.���� %��� ��ก��� 4 �� �

bmin = 20 cm

ρρρρg = Ast / Ag 0.01 ≤≤≤≤ ρρρρg ≤≤≤≤ 0.08

4DB12 6DB12

������������

(��ก������������24.�$��(��ก������$������������� 7.5 2�.

�������ก�������� ก�����

�"�"!�ก$��(��ก����4�$����ก��������5ก&��ก���.�� ��5ก&��ก�ก���� �*���5ก�#ก���� ��ก�-� �.%������5ก���ก���� � ����4�$����ก�����5ก����� �.��#���ก���

- ��(��: ��5ก�������ก ��5ก�#ก���� 3.0 2�.

- �����: ��5ก&��ก�ก���� ��5ก&��ก���.�� 3.5 2�.

(��ก��� �.���$������*��#ก���7�

- �������5ก����������8�ก��� 16 ��. 5.0 2�.

- �������5ก��������� 16 ��. ��"��5กก��� 4.0 2�.

(��ก��� �.%�����$������*�%���#ก���7�

- �������5ก����������8�ก��� 44 ��. 4.0 2�.

- �������5ก��������� 35 ��. ��"��5กก��� 2.0 2�.

SDMWSD

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�� ก!��ก���"��

����ก���ก� �������������������������� ������������� ��ก��� ����!

– ��5ก&��ก 6 ��. �������5ก�*����� 20 ��. �*���5กก���

SDMWSD

��5ก�*� �ก�� �!"� ���#ก��� ����&��ก���.�����������������

– ��5ก&��ก 9 ��. �������5ก�*����� 25 - 32 ��.

– ��5ก&��ก 12 ��. �������5ก�*��8�ก��� 32 ��. �4��%&

�"�"��������5ก&��ก� ��%����กก���(�����%&���

– 16 � �����������5ก�*�

– 48 � �����������5ก&��ก

– (���ก� ��� ���� �.��5ก �.���

�� ก!��ก���"�� SDMWSD

� ��!��� ��5ก�*� �ก�����"�� ��� ��� ���ก�� ���������

��������5ก&��ก ��������5ก&��ก%���ก�� 135o

�������5ก�*� �.%���������5ก&��ก�4� �"�"�����������

��5ก�� �����ก���� � �.�������5ก&��ก�4�� ��%���ก�� 15 2�.

135o max

xx

x ≤≤≤≤ 15 cmxx

x

x

xx

x >>>> 15 cmxx

x

x

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ก��#������� ก!��ก���"�� SDMWSD

6 BARS 8 BARS 10 BARS

12 BARS 14 BARS 16 BARS

��"��#����ก���ก�ก����$�����ก��

Pu

f2

f2

1

��*.����&��ก�ก�������������(��ก���� ���ก&��ก!"ก"� �"��ก (spalled off)

Ag Acore

ก����� �� �������� �.�#8����%&(*�

c g core0.85 f (A A )′ −

���$�!�กก�����������&��ก�ก���� �� ������� ��� �����ก�(��ก���

�� ก����������(��ก����)�.��4��

f c 2f f 4.1f′= +

&����-��5ก&��ก� �����)���)� �.!" �� ก���� �.�#8����%&!�กก��ก"� �"�#ก��������ก���� �.�)�.��4��!�กก��������

c g core 2 core0.85 f (A A ) 4.1f A′ − =

)�!��-�� ���������������������(�4.��&+��������

Core

Spiral

s

hcore

Ab fy

Ab fy

s

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[ ΣFx = 0 ] hcore s f2 = 2 Ab fy

hcore

Ab fy

Ab fy

s

b y2

core

2A ff

h s= 2

12

g b yc

core core

A 4.1(2A f )0.85 f 1

A h s

′ − =

3

������%������ก���ก�ก����

sh4

hA

2core

corebs π

π=ρ

shA4

core

b=

ρs 3

′=ρ 1

A

A

ff42.0

core

g

y

cs

����:�� ACI ��" �.�. . &���(�� 0.42 �&+� 0.45

′=ρ 1

A

A

ff45.0

core

g

y

cs

'��ก���(�ก��()��� ก!��ก�ก���� SDMWSD

����ก���ก�ก��������������������������� ������������� ��ก��� ����!

�����ก��ก���� ���������� �. ������5ก

&��ก�ก����� ��%��� ��ก��� 9 ��.

�"�"���������"����&��ก�ก����� ��%���ก��

7.5 2�. ��"� ��%��� ��ก��� 2.5 2�.

�����������5ก&��ก�ก���� ρρρρs � ��%��� ��ก���

g cs

core yt

A f0.45 1

A f

′ ρ = −

��� �. fyt (*� ก����(��ก�����5ก&��ก�ก���� ���� ��%��� ��ก���

4,000 กก./2�.2

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��*�����+��()���� �.�. . 1007-34

�����ก �.(���-��/�(����� %� �=)�"��*.�%����ก����ก������*.��!�ก(����"�#�

WSD

��� �. fs = ������� �.���� �����5ก������������.������� � �� � ��ก�� 0.40 fy ���� ��%���ก�� 2,100 กก./2�.2

ρρρρg = �����������*�� �.��5ก�������������.������*�� �. ����� = Ast / Ag

������ก�ก���� )ff25.0(AP gscg ρ+′=

������ก� ��� )ff25.0(A85.0P gscg ρ+′=

��� �.�.�. �ก�����ก rrstscg AfAffA225.0P ++′=

��� �. fr = ������� �.���� ����ก���5ก ���� ��%���ก�� 1,200 กก./2�.2 �������5ก ��ก.116-2529 ����(�-/�) Fe24

ก������������ก'������� ก �"���

������ก� ��� )ff25.0(A85.0P gscg ρ+′=0.2 m ×××× 0.2 m

4 DB12

����� f’ c = 240 กก./2�.2 ��" fy = 4,000 กก./2�.2

= 26.5 ton

������ก�ก���� )ff25.0(AP gscg ρ+′=

stg

g

A 4 1.130.0113

A 20 20×

ρ = = =×

0.2 m

6 DB12

2P 0.85(0.25 0.24 20 0.4 4.0 4 1.13)= × × + × × ×

stg 2

g

A 6 1.130.022

A ( / 4) 20×

ρ = = =π ×

= 29.7 ton

2P 0.25 0.24 ( / 4) 20 0.4 4.0 6 1.13= × × π × + × × ×

WSD

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ρg = 0.028

ρg = �" ��

��������ก����ก+�����

������ก� ��� )ff25.0(A85.0P gscg ρ+′=

!���ก����������&��ก���.���)*.���������ก��� �ก�� ��� 80 ��� ก��� f’ c = 240

กก./2�.2 ��" fy = 4,000 กก./2�.2

WSD

��������� 30××××30 '�.

2g80 0.85 30 (0.25 0.24 0.4 4.0 )= × × + × × ρ

��������� 40××××40 '�.

2g80 0.85 40 (0.25 0.24 0.4 4.0 )= × × + × × ρ

�� ��%������� (�)%��(��� ������ ��*�(+�����ก����$���

0.01 ≤≤≤≤ ρρρρg ≤≤≤≤ 0.08 OK

Ast = 0.028 × 302 = 25.2 2�.2

Use 6DB25 (29.45 '�.2)

��5ก�*����� DB25-DB32 �� ��5ก&��ก���� RB9

�"�"�����5ก&��ก : � ���(� �.��� = 30 2�.

16 � ����5ก�*� = 16 × 2.5 = 40 2�.

48 � ����5ก&��ก = 48 × 0.9 = 43 2�.

Use RB9 @ 0.30 m

0.3 m ×××× 0.3 m

6 DB25

RB9 @ 0.30 m

WSD

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2

2

30 2400.45 1

2,40024

= −

��������ก����ก+�����

!���ก����������&��ก�ก�����)*.���������ก��� �ก�� ��� 80 ��� ก��� f’ c = 240

กก./2�.2 ��" fy = 4,000 กก./2�.2

WSD

������ก�ก���� )ff25.0(AP gscg ρ+′=

ρg = 0.033

��������� ∅∅∅∅ 30 '�.

0.01 ≤≤≤≤ ρρρρg ≤≤≤≤ 0.08 OK

Ast = 0.033 × (π / 4) × 302 = 23.5 2�.2

Use 6DB25 (29.45 '�.2)

)0.44.024.025.0(304

80 g2 ρ××+××

π=

�����������5ก&��ก�ก���� g cs

core yt

A f0.45 1

A f

′ ρ = −

= 0.0253

�"�"���������"����&��ก�ก����� ��%���ก�� 7.5 2�. ��"%��� ��ก��� 2.5 2�.

shA4

core

bs =ρ

�����ก��ก���� ���������� �. ������5ก

&��ก�ก����� ��%��� ��ก��� 9 ��.RB9 : Ab = 0.636 cm2

4 0.6360.0253

24s×

= s = 4.19 cm

USE RB9 @ 0.04 m

Diameter = 0.3 m

6 DB25

RB9 @ 0.04 m

WSD

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ขอสอบภย

ขอที่ : 115

เสากลมขนาดเสนผาศูนยกลาง 0.30 ม. เสริมเหลก็ตามยาว 6-DB12 เหล็กปลอก RB9 @ 0.05 ม. จะรับน้ําหนกัไดเทาไร ถา fc’ = 240 กก./ซม.2, fy = 3000 กก./ซม.2 (วิธี WSD)

)ff25.0(AP gscg ρ+′=

2gc 3025.024025.0Af25.0 ×π××=′ kg 412,42=

000,34.013.16fA ss ×××= kg 136,8=

P = 42,412 + 8,136 = 50,548 kg

)ff25.0(A85.0P gscg ρ+′=

ขอสอบภย

ขอที่ : 138

เสากลมขนาดเสนผาศูนยกลาง 0.30 ม. เสริมเหลก็ตามยาว 6-DB12 เหล็กปลอก RB9 @ 0.05 ม. จะรับน้ําหนกัไดเทาไร ถา fc’ = 240 กก./ซม.2, fy = 3000 กก./ซม.2 (วิธี WSD)

303025025.0Af25.0 gc ×××=′ kg 250,56=

000,34.014.34fA ss ×××= kg 072,15=

P = 0.85(56,250 + 15,072) = 60,624 kg

��*�ก���� �.�. . 1008-38 SDM

�����ก��� �ก&�"��� �.ก�" ������� : Pu = 1.4 DL + 1.7 LL

��ก���� ����� ��ก���� : Pn ≥ Pu / φφφφφφφφ = 0.75 ��������&��ก�ก����

φφφφ = 0.70 ��������&��ก���.��

ก��������������������ก�������&��ก�ก���� :

ก��������������������ก�������&��ก���.�� :

n,max c g st y stP 0.80 [0.85 f (A A ) f A ]′φ = φ − +

n,max c g st y stP 0.85 [0.85 f (A A ) f A ]′φ = φ − +

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ก������������ก'������� ก �"���

����� f’ c = 240 กก./2�.2 ��" fy = 4,000 กก./2�.2

= 55.3 ton

stg

g

A 4 1.130.0113

A 20 20×

ρ = = =×

stg 2

g

A 6 1.130.022

A ( / 4) 20×

ρ = = =π ×

= 57.3 ton

n,max c g st y stP 0.85 [0.85 f (A A ) f A ]′φ = φ − +������ก�ก���� 0.2 m

6 DB12

������ก� ���

0.2 m ×××× 0.2 m4 DB12

n,max c g st y stP 0.80 [0.85 f (A A ) f A ]′φ = φ − +

2uP 0.8 0.7 [0.85 0.24 (20 4 1.13) 4.0 4 1.13]= × × × × − × + × ×

2uP 0.85 0.75 [0.85 0.24 ( 20 6 1.13) 4.0 6 1.13]

= × × × × × − × + × ×

SDM

ρg = 0.009

ρg = �" ��

��������ก����ก+�����

!���ก����������&��ก���.���)*.���������ก��� �ก&�"��� 120 ��� ก��� f’ c = 240

กก./2�.2 ��" fy = 4,000 กก./2�.2

��������� 30××××30 '�.

��������� 40××××40 '�. �� ��%������� (�)%��(��� ������

��*�(+�����ก����$���

Use ρρρρg = 0.01

Ast = 0.01 × 302 = 9.00 2�.2

Use 4DB20 (12.56 '�.2)

SDM

������ก� ��� u g c g y gP 0.80 A [0.85 f (1 ) f ]′= φ − ρ + ρ

2g g120 0.8 0.7 40 [0.85 0.24(1 ) 4.0 ]= × × × − ρ + × ρ

2g g120 0.8 0.7 30 [0.85 0.24(1 ) 4.0 ]= × × × − ρ + × ρ

< 0.01

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�"�"�����5ก&��ก : � ���(� �.��� = 30 2�.

16 � ����5ก�*� = 16 × 2.0 = 32 2�.

48 � ����5ก&��ก = 48 × 0.6 = 28.8 2�.

0.3 m ×××× 0.3 m

4 DB20

RB6 @ 0.25 m

SDM��5ก�*����� DB20 �*���5กก����� ��5ก&��ก���� RB6

Use RB9 @ 0.25 m

2

2

30 2400.45 1

2,40024

= −

��������ก����ก+�����

!���ก����������&��ก�ก�����)*.���������ก��� �ก&�"��� 120 ���

ก��� f’ c = 240 กก./2�.2 ��" fy = 4,000 กก./2�.2

ρg = 0.016

��������� ∅∅∅∅ 30 '�.

0.01 ≤≤≤≤ ρρρρg ≤≤≤≤ 0.08 OK

Ast = 0.016 × (π / 4) × 302 = 11.3 2�.2

Use 6DB16 (12.06 '�.2)

�����������5ก&��ก�ก���� g cs

core yt

A f0.45 1

A f

′ ρ = −

= 0.0253

SDM

������ก�ก���� u g c g y gP 0.85 A [0.85 f (1 ) f ]′= φ − ρ + ρ

2g g120 0.85 0.75 30 [0.85 0.24(1 ) 4.0 ]

= × × × × − ρ + × ρ

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�"�"���������"����&��ก�ก����� ��%���ก�� 7.5 2�. ��"%��� ��ก��� 2.5 2�.

shA4

core

bs =ρ

�����ก��ก���� ���������� �. ������5ก

&��ก�ก����� ��%��� ��ก��� 9 ��.RB9 : Ab = 0.636 cm2

4 0.6360.0253

24s×

= s = 4.19 cm

USE RB9 @ 0.04 m

Diameter = 0.3 m

6 DB16

RB9 @ 0.04 m

SDM

ขอสอบภย

ขอที่ : 119

จงหาวาเสาสั้นปลอกเกลียวขนาดเสนผาศูนยกลาง 30 ซม. มีเหล็กเสริมยืน 6-DB20 มม. fc’ = 210 ksc, fy = 3000 ksc รับน้ําหนักประลยัตามแนวแกนไดเทาไร

= 114 ton

ขอสอบภย

ขอที่ : 120

จงคํานวณกําลังรับน้ําหนกัที่สภาวะประลัยของเสาสั้นปลอกเดี่ยวขนาด 40x40 ซม. มีเหล็กเสริมยืน 6-DB20 กําหนด fc’ = 210 กก./ซม.2, fy= 3000 กก./ซม.2

2 2gA 30 707 cm

= × = 2st, A 6 3.14 18.84 cm= × =

u nP P 0.85 0.75[0.85 0.21(707 18.84) 3.0 18.84]= φ = × × − + ×

2uP 0.80 0.70[0.85 0.21(40 18.84) 3.0 18.84]= × × − + ×

= 190 ton

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ขอสอบภย

ขอที่ : 198

เสาสั้นปลอกเดี่ยว เสริมเหล็กยืน As = As’ รับน้ําหนักบรรทุกคงที่ PD = 130 ตนั และน้ําหนักบรรทุกจร PL = 98.5 ตนั กําหนด fc’ = 280 ksc, fy = 4000 ksc จงหาเนื้อที่ของหนาตัดเสาที่เล็กที่สุด วิธี WSD

Ag = 1,358 cm2

เสาสั้นปลอกเดี่ยว เสริมเหล็กยืน As = As’ รับน้ําหนักบรรทุกคงที่ PD = 130 ตนั และน้ําหนักบรรทุกจร PL = 98.5 ตนั กําหนด fc’ = 280 ksc, fy = 4000 ksc จงหาเนื้อที่ของหนาตัดเสาที่เล็กที่สุด วิธี USD

ขอสอบภย

ขอที่ : 199

� ���������5ก �.��� �� ��5ก��ก �.��� ρρρρg = 0.08

g c s g s yP 0.85 A (0.25 f f ), f 0.4 f′= + ρ =

g228.5 0.85 A (0.25 0.28 0.4 4.0 0.08)= × + × ×

ρρρρg = 0.08 Pu = 1.4×130 + 1.7×98.5 = 349.5 ton

u g c g y gP 0.8 A [0.85 f (1 ) f ] , 0.7′= φ − ρ + ρ φ =

g349.5 0.8 0.7 A [0.85 0.28(1 0.08) 4.0 0.08]= × × − + ×

Ag = 1,158 cm 2

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Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Design of Column 2Design of Column 2

� ก����ก������� ���ก�������� ��� (WSD)

� ��� ���������� (SDM)

� ��������������ก����������� (WSD)

� ��������������ก����������� (SDM)

Combined Axial Load and Bending MomentsCombined Axial Load and Bending Moments

Bending moments can occur in columns because:

- Unbalance gravity loads

- Lateral loads: wind, earthquake

wind

earthquake

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��������������ก����������� WSD

��� ��������������ก P ����������� M ������� �!�"#�� ก����� P

��ก��$�������% �&��� ����'������% �&��� e = M / P

P

M

Pe

����� �����(�� �#(�����$'�ก���)%��ก���������ก������������( #(�����$'�� �*#" ���" �!�(ก� 1.0

.+ + ≤bya bx

a bx by

ff f1 0

F F F

��� fa = #(����� ��$'�ก���)%gA

P=

fbx = #(��������$'�ก���)%� ��ก x =x

xx

IcM

fby = #(��������$'�ก���)%� ��ก y =y

yy

I

cM

P

ey

x

y

ex

cx cy

��������������ก����������� WSD

Fa = #(����� ��$'�� �*#" cg fm1340 ′ρ+= )(.

Fb = #(��������$'�� �*#" cf450 ′= .

g

stg A

A=ρ ���

g

y

f850

fm

′=

.

�(������ � ���'� Ix, Iy � �#"������

���� ��ก��%$'�#"�����+��� �#�,ก����

stt A1n2A )( −=

(2 n – 1) A st

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���������������� WSD

Ds = g DD

Circular section:

( )2

4 2 164 8

sx y st

DI I D A n

π= = + −

Ds = g Dt

Square section with circular steel:

( )2

412 1

12 8s

x y st

DI I t A n= = + −

t

���������������� WSD

g b

t

b

Rectangular section:

( ) ( )2

312 1

12 6x st

gbI b t A n= + −

g t( ) ( )2

312 1

12 6y st

gtI bt A n= + −

y

yx x

g b

t

b

Rectangular section:

g t

y

y

x x 4gb

1n2Atb121

I2

st3

x)(

)( −+=

4gt

1n2Abt121

I2

st3

y)(

)( −+=

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P

M = P e

eb

Tension controle > eb

Compression control

e < eb ������ก�ก��:

0.43 0.14b g se m D tρ= +

������ก��� :

( )0.67 0.17b ge m dρ= +

��� Ds = �"�(�&���ก���+� กก�'��

t = �����)ก$�%�#��� �#"����

d = �����)ก+����$-���� �#"����d t

. �/������0� #(�����*#�,ก�������� ก�'��'�(�$(�ก��#(�����$'�� �*#"

������������������� ( eb ) WSD

P

M = PeMsMbMoMa

Pb

PaZone 1

Zone 2

Zone 3

ea

eb

Po

Zone 1 : e < ea ก���+1��������������ก�"�

1 1a s

a o

e MP P

= −

(0.25 ) for spiral column

0.85 (0.25 ) for tied column

g c s g

a

g c s g

A f fP

A f f

ρ

ρ

′ +

= ′ +

, /o a g s bP F A M F I c= =

ก�� �!������ก������ WSD

��� ��������������ก P ����������� M = P e ����(�ก������� �

ก+1 3 �(�����������% �&��� e

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P

M = PeMsMbMoMa

Pb

PaZone 1

Zone 2

Zone 3

ea

eb

Po

Zone 2 : ea < e < eb Compression control

1 or 1a b

o s a b

f fP MP M F F

+ ≤ + ≤

Zone 3 : e > eb Tension control

M and P are proportioned between

(Mb , Pb) and (Mo , 0)

WSD

Mo = ������������� ��������� �������������������������������

Spirally reinforced column: 0.12o st y sM A f D=

Symmetric Tied column: ( )0.40o s yM A f d d ′= −

Unsymmetric Tied column: 0.40o s yM A f jd=

Biaxial Bending: 1.0yx

ox oy

MMM M

+ ≤

WSD

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ขอสอบภย

ขอที่ : 220

เสาปลอกเดี่ยวขนาด 50 x 50 ซม. เสริมเหล็กยืน 6DB25 (Ast = 29.45 ซม.2) โดยท่ี As = As’ ระยะหุมคอนกรีต 5 ซม. ใหใชวิธี WSD ประมาณคาโมเมนตอินเนอรเชียของหนาตัดเสา กําหนด n = 9

ขอสอบภย

ขอที่ : 247

เสาสั้นปลอกสั้นปลอกเดี่ยวขนาด 25 x 25 ซม. เสริมเหล็กยืน 6DB20(Ast = 18.84 ซม.2) โดยท่ี As = As’ ระยะหุมคอนกรีต 4 ซม. ใหใชวิธี WSD ประมาณคากําลังตานแรงอัดใชงาน Pb ที่สภาวะสมดุล สมมุติคาหนวยแรงอัดที่ยอมให = 120 ksc หนวยแรงดัดที่ยอมให = 112.5 ksc ระยะเย้ืองศูนยสมดุล = 8.5 ซม. และโมเมนตอินเนอรเชียของหนาตัด = 55700 ซม.4

= 721,093 cm2

4)gb(

)1n2(Atb121

I2

st3

x −+=

440

)192(45.2950121 2

4 −×+×=

eb

P

M

b ba b

g

P Pf 0.0016P

A 25 25= = =

×

b b bb b

P e c P 8.5 12.5f 0.0019P

I 55700× ×

= = =

a b

a b

f f1

F F+ =

b b0.0016P 0.0019P1

120 112.5+ =

Pb = 33,015 kg = 33 ton

��������������ก����������� SDM

��� ��������*������ Pn ก��$��$'�������% �&��� e

Pn

e

hwidth = b

c

d

d′

sεsε ′

cuε

a

s sA f s sA f′ ′

0.85 cf ′

b

sA sA′

h

��ก���0�� ����*�������#"������ [ ΣΣΣΣ Fy = 0 ]

TCCP scn −+=

sssscn fAfAbaf85.0P −′′+′=

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��������������ก����������� SDM

T CsCcCL

d’

a/2

h/2

d

h

Pn

e

d - h/2

ก������"�$��������� Mn = Pn e #���ก���0������� ��ก&���2(��������ก� �#"���� [ ΣΣΣΣ MC = 0 ]

L

−+

′−+

−=

2h

dTd2h

C2a

2h

CM scn

#�� #���ก���0������� �#�,ก�������)�

( )ddC2a

dCM sc2n ′−+

−=

−+=

2h

dePn

c

d

d ′

sεsε ′

cuε

a

s sA f s sA f′ ′

0.85 cf ′

���"ก�������#� ���"ก��������� ��� ���ก�$ SDM

����ก���������: ssfAT =

26s kg/cm 1004.2E ×=

003.0cu =εc

cdcus

−ε=ε

sss Ef ε=

ccd

Escu−

ε=c

cd120,6

−= yf≤

s s sC A f′ ′=����ก���������:

s cu

c dc

′−′ε = ε

s s s y

c df E 6,120 f

c

′−′ ′= ε = ≤

���ก���: c cC 0.85 f ab′=

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ก%���������������ก� ���ก%����������� SDM

c

d

d ′

sεsε ′

cuε

a

s sA f s sA f′ ′

0.85 cf ′

��� ������ก���������ก���� �� c

s y

d cf 6,120 f

c−

= ≤

s y

c df 6,120 f

c

′−′ = ≤

1a c= β

s sT A f=

s s sC A f′ ′=

c cC 0.85 f ab′=

TCCP scn −+=

−+

′−+

−=

2h

dTd2h

C2a

2h

CM scn

ก�����������������ก�:

ก��������!�����"���:

����#$��%&�": e = Mn / Pn

Small e →→→→ fs < fy when εc = εcu = 0.003 (compression failure)

Large e →→→→ fs = fy when εc = εcu = 0.003 (tension failure)

Pn

ePn

ePn

ePn

e

Small Eccentricity Large Eccentricity

Tension & Compression Failure

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Pn

Mn

P0

e=

0

esm

all

Compressionfailure range

eebb : Balance failure: Balance failure

e large Tension failure range

e = ∞

�&�'��()��� ��*� (Interaction diagram) SDM

���#����(� e *�3���'�'��#)����(� � Pn ��� Mn

��� ����(����(����, ��'��ก�����(� e �(��3 ก,��!�" ก��+1 �'�(&���)����*��+"

(Mn, Pn)

�"��&�'��$'�

e =M n

/P n

b b be M /P=

ก���������������� (Balanced failure) SDM

cb

d

d′

s′εcuε

�� �/���ก��������4)��#�,ก�����������)�2)��0����ก εεεεy ��" �ก��� ก�'�2�ก ����ก$'�#(��ก�����#�����0� εεεεcu = 0.003

cub

cu y

c dε

=ε + ε y

6,120d

6,120 f=

+

b 1 ba c= β

bs y

b

c df 6,120 f

c

′−′ = ≤

b c b s s s yP 0.85 f a b A f A f′ ′ ′= + −

bb c b s s s y

ah h hM 0.85 f a b A f d A f d

2 2 2 2 ′ ′ ′ ′= − + − + −

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Case 1: e < eb

εcu

cbεyMb

fs < fy

Case 2: e > e b

ก����+����ก����������,-� ebSDM

Compression Failure Tension Failure

M < Mb

εy

εcu

cb

Mb

M > Mb

c > cb εs < εy fs > fyc < cb εs > εy

SDM����+��ก���%����ก%����������

�� 25 x 40 4�. ����#�,ก�� 4DB28 ���$'� As = As’ ����� ก�'�#0"� 5 4�. ก��#� f’c = 280 กก./4�.2 ��� fy = 4,000 กก./4�.2

20 cm 20 cm

12.5 cm

12.5 cm

5 cm5 cm

�(�����,� :

by

6,120c d

6,120 f=

+

6,12035

6,120 4,000= ×

+

= 21.2 4�.

b 1 ba c 0.85 21.2= β = × = 18.0 4�.

bs

b

c df 6,120

c

′−′ =

21.2 56,120

21.2−

= = 4,677 > fy sf ′′′′ = 4,000 กก./4�.2

b c b s y s yP 0.85f a b A f A f′ ′= + −

s sA A′ =

0.85 0.28 18.0 25= × × × = 107 ��

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40 cm

5 cm

35 cm

18 cm

h

d

d

a

=

′ =

=

=

SDMb

b c b s y s y

ah h hM 0.85f a b A f d A f d

2 2 2 2 ′ ′ ′= − + − + −

40 18 40 40107 12.32 4.0 5 12.32 4.0 35

2 2 2 2 = − + × − + × −

= 2,656 ��-4�. = 26.6 ��-�.

bb

b

Me

P=

2,656107

= = 24.8 4�.

c < cb = 21.2 4�. e > eb : tension failure

� ��� ก c = 10 4�. �� ���ก+1ก���������������)� ����% fs = fy

Pn

Mn

ebMb, Pb

a 0.85 10= × = 8.5 4�. cC 0.85 0.28 8.5 25= × × × = 50.6 ��

s

10 5f 6,120

10−′ = < fy= 3,060 กก./4�.2 OK

nP 50.6 12.32 3.06 12.32 4.0= + × − × = 39 ��

30 56,120

30−

=

a 0.85 30= ×

2,10239

=

nM 50.6(20 8.5 / 2) 12.32 3.06(20 5) 12.32 4.0(35 20)= − + × − + × − SDM

= 2,102 ��-4�. = 21.0 ��-�.

= 53.9 4�.n

n

Me

P=

c > cb = 21.2 4�. e < eb : compression failure

� ��� ก c = 30 4�. �� ���ก+1ก�������������� �� ����% fs < fy

Pn

Mn

ebMn, Pn

= 25.5 4�. cC 0.85 0.28 25.5 25= × × × = 152 ��

s

d cf 6,120

c−

=35 30

6,12030−

= < fy= 1,020 กก./4�.2 OK

bs

b

c df 6,120

c

′−′ = = 5,100 > fy sf ′′′′ = 4,000 กก./4�.2

nP 152 12.32 4.0 12.32 1.02= + × − × = 189 ��

nM 152(20 25.5 / 2) 12.32 4.0(20 5) 12.32 1.02(35 20)= − + × − + × −

= 2,030 ��-4�. = 20.3 ��-�. e = 2,030/189 = 10.7 4�.

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n

n

Me

P=

MnM0

e =e b

Pb

Pn

P0

e=0

Mb

�&�'��()��� ��*� (Interaction diagram) SDM

M0 = Nominal moment strength

P0 = Nominal axial strength

e = ∞

c g st y st0.85 f (A A ) f A′= − +

0.003

s yε ε=�/������0�

0.003s yε ε<

����0�����

�� ��

0.003

s yε ε>

����0��������)�

n(max) c g st y stP 0.80 [0.85 f (A A ) f A ]′φ = φ − +

φPn(max)

Design curve

c g0.1f A′

�&�'��()��� ��*��%�����ก����ก��� SDM

Pn

Mn

Nominal strength

������ก�ก�� : n(max) c g st y stP 0.85 [0.85 f (A A ) f A ]′φ = φ − +

������ก��� :

φ = 0.75 : ��+� กก�'��φ = 0.70 : ��+� ก�'���

0.70 ≤ φ ≤ 0.90

φ = 0.90 : ก�����*��

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0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10

'n

g c

PA f

φ

'n

g c

MA hf

φ

γ h

h

b

γ = 0.80

ρgm=0

0.5

1.0

1.5

2.0

2.5

3.0

Normalized Interaction Diagram

u

g c

M 30 100A hf 30 50 50 0.24

×=

′ × × ×

u

g c

P 200A f 30 50 0.24

=′ × ×

SDM����+��ก����ก������

�� ก���������� ������������ก Pu = 200 �� ����������� Mu = 30 ��-��� ก��#� f’c = 240 กก./4�.2 ��� fy = 4,000 กก./4�.2

�+� �� � �#"���� 30 x 50 4�.

= 0.56

= 0.17

��ก��/��� (��(� ρρρρg m = 0.65

ρρρρg = 0.65 × 0.85 × 0.24 / 4.0 = 0.033

Ast = 0.033 × 30 × 50 = 49.5 4�.2

USE 8-DB28 (Ast = 49.28 4�.2)

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��#�ก����ก���ก DB10 :

����#(��#�,ก+� ก�" �!�(" �ก�(� : 16 × 2.8 = 44.8 4�.

48 × 1.0 = 48 4�.

30 4�. ���,�USE DB10 @ 0.30 �.

50 4�.

30 4�.#�,ก�� 8-DB28

#�,ก+� ก DB10 @ 0.30 �.

Column strength interaction diagram. A 25 x 40 cm column is reinforced with 4DB28.

Concrete strength f’c = 280 ksc and the steel yield strength fy = 4,000 ksc

20 cm 20 cm

12.5 cm

12.5 cm

5 cm5 cm40 cm

5 cm

35 cm

h

d

d

=

′ =

=

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ก����ก%�������.��!�ก������������ WSD

���#���������� �" �#���(���� ����������4)��!�"��กก��������#��"���9ก� �� R ���#�����ก�����ก( ��"��)����� ก���

����� �ก����ก�������� �����(����������� ��� h / r ���'%

D

h / r < 60

b r = 0.30 b

r = 0.25 D

��&�'!����� r � �#"������ :

60 < h / r ≤≤≤≤ 100

h / r > 100

.��������ก���������������

��ก�����!�/0��1ก����" R

�������"!�������2��'�3�����!ก�� � �*� �3�$�

P�����$�P����=

RM�����$�

M����=R

�/ก������ก%�����%����������� ( R ) WSD

���#�����$'�!�(�'ก��4$���"��"�� +���$�%�� ��)��( ����'#)���0����ก�����#�(��+���$�%�� �

(1)

(1) (2)

R = 1.32 – 0.006 h / r ≤≤≤≤ 1.0

���#�����$'�!�(�'ก��4$���"��"�� +���$�%�� ��)��( ������ก(������"��'���

(2)

R = 1.07 – 0.008 h / r ≤≤≤≤ 1.0

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������������ ( h ) WSD

*#"2� �(������( ��(����#�(����%��(����%+1������� ����� ��� ก��กก� '����( !+'%

h

������%!�"��

�����ก��%2)��+:#����

h

������%!�"��

�����ก��%2)���%

h

������%�'��� ����

�����ก��%2)�$" ���

ขอสอบภย

ขอที่ : 233

เสาปลอกเดี่ยวขนาด 30 x 30 ซม. อยูในเฟรมที่เซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 6.0 ม. ใหใชวิธี WSD ประมาณคาตวัคูณลดคา R

= 1.32 – 0 .006 × 66.7 = 0.92

7.66303.0

600rh

= 60 < h / r ≤≤≤≤ 100 ��ก�����!�/0��1ก����" R

R = 1.32 – 0.006 h / r ≤ 1.0

ขอสอบภย

ขอที่ : 234

เสาปลอกเดี่ยวขนาด 40 x 40 ซม. อยูในเฟรมที่เซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 6.0 ม. ใหใชวิธี WSD ประมาณคาตวัคูณลดคา R

50403.0

600rh

= h / r < 60 R = 1.00 .��������ก�����

ขอสอบภย

ขอที่ : 235

เสาปลอกเดี่ยวสี่เหลี่ยมจตุรัส อยูในเฟรมท่ีเซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 8.0 ม. ใหใชวิธี WSD ประมาณขนาดอยางนอยของเสาตนน้ีที่จะถือวาเปนเสาสั้น

h / r = 60 60b3.0

800rh

= b = 44.4 cm Use 50 x 50 4�.

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�/ก������ก%�����%����������� ( R ) WSD

���#�����$'��'ก��4$���"��"�� +���$�%�� ��)��((3)

(3)

R = 1.07 – 0.008 h’ / r ≤≤≤≤ 1.0

R = 1.18 – 0.009 h’ / r ≤≤≤≤ 1.0

2"�ก������ 2�ก����0�������*����� �( �����

�����(��!#� �������9ก� �� R 'ก�" ��� 10 4)���'�(����'%

��� h’ = �������+����$-���� ���

������(����0*�&� ( h’ ) WSD

*������"��$'�!�(�'ก��4 *#"*�"�������+����$-��� h’ $(�ก��������� ���� h

*������"��$'��'ก��4 *#"*�" h’ 4)������ ��กก� '����( !+'%

+����"��#)��2�ก�)���%� ��� 'ก+���2�ก�)�#�0 :

h’ = 2 h (0.78 + 0.22 r’ ) ≥≥≥≥ 2 h

+���$�%�� ��"��2�ก�)���%� : h’ = h (0.78 + 0.22 r’ ) ≥≥≥≥ h

��+��� ������2�ก�)� : h’ = 2 h

r’ �� �����(���������9�� ����( ��������9�� ���$'�+�����

r’ =Σ (EI/h)���

Σ (EI/L)���

r’ > 25 2� �(�+������'�/���)�#�0

r’ = 0 #�� 1 2� �(�+������'�/���)��(

+ก��*�"�(�;�'��� �+��������(�� r’ = (r’T + r’B) / 2

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ขอสอบภย

ขอที่ : 236

เสาปลอกเดี่ยวขนาด 40 x 40 ซม. อยูในเฟรมแบบ Portal ชวงเด่ียวและช้ันเดียวซึ่งเซได โดยที่ปลายเสาเปนแบบยึดแนน และที่หัวเสายึดกับคานมีคา I/L = 200 ซม.3 ความยาวเสาปราศจากการค้ํายนัคือ 8.0 ม. ใหใชวิธี WSD ประมาณความยาวประสิทธิผลของเสาตนน้ี

( I / L )���

440 /12800

= = 267 4�.3

+����(���)��( r’B = 1

+�����)�ก���� r’T = 267 / 200 = 1.335

r’ = (1 + 1.335)/2 = 1.17

h’ = h (0.78 + 0.22 r’) = 8.0 (0.78 + 0.22×1.17) = 8.30 m

ขอสอบภย

ขอที่ : 238

เสาปลอกเดี่ยวขนาด 40 x 40 ซม. อยูในเฟรมแบบ Portal ชวงเด่ียวและช้ันเดียวซึ่งเซได โดยที่ปลายเสาเปนแบบยึดแนน และที่หัวเสายึดกับคานมีคา I/L = 200 ซม.3 เสาตนน้ีจะโกงสองทาง ความยาวเสาปราศจากการคํ้ายันคือ 8.0 ม. ใหใชวิธี WSD ประมาณคาตัวคูณลดกําลัง R ของเสาตนน้ี

r = 0.3 x 40 = 12 4�.

R = 1.07 – 0.008(h’/r) = 1.07 – 0.008×830/12 = 0.52

&�������-����

M1b

M2b

-

M1b

M2b

+

SDM

��������(�$'�!�(�'ก���)���%� lu � ���$(�ก�������( ��(����#�(����%

*������"��$'�!�(�'ก��4 ����� �������+����$-��� k ≤ 1.0 ���#�������$'��'

ก��4 k > 1.0

��&�'!����� r = 0.30b ���#������'�#�'��� ��� r = 0.25D ���#�����ก��

���#�������$'�!�(�'ก��4 !�(�" ������� �������������

u 1b

2b

k M34 12

r M< −

���#�������$'��'ก��4 !�(�" ������� �������������

uk22

r<

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ขอสอบภย

ขอที่ : 239

เสาปลอกเดี่ยวขนาด 50 x 50 ซม. อยูในโครงเฟรมทีเ่ซได ถาพบวาคา effective length factor เทากับ 1.5 ดังน้ัน ชวงความยาวเสาปราศจากการค้ํายันควรเปนเทาใดตามวิธี USD จึงจะเปนเสาสั้น

����% :

ขอสอบภย

ขอที่ : 240

เสาปลอกเกลียวขนาดเสนผาศูนยกลาง 50 ซม. อยูในโครงเฟรมที่เซได ถาพบวาคา effective length factor เทากับ 1.5 ดังน้ัน ชวงความยาวเสาปราศจากการค้ํายันควรเปนเทาใดตามวิธี USD จึงจะเปนเสาสั้น

r = 0.25 x 50 = 12.5 4�.

r = 0.3 x 50 = 15 4�.

uk22

r<

ℓ u1.522

15×

=ℓ

= 2.20 mu 220 cm=ℓ

u1.522

12.5×

=ℓ����% : uk

22r

<ℓ

u 183 cm=ℓ = 1.83 m

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� ���������ก

� ��� ���ก�������������ก

� ��ก ������ ก���ก�����������

� ��ก ��� �

� ��ก ������ �!

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Design of Footing 1Design of Footing 1

"�� ��.�.$�%& '�! ����

�����ก����������

�����ก���������� ������������������ก��������������� � ��������������������� �

�����!���"����# �$ก �ก��� ��% ���� �����ก����&�%�� �ก$ก �ก�����ก�����

��ก'�ก�������%�����ก���(�������ก �

ก����&�%��$���)ก��$����# �(��$ก �ก����&�

%������%ก%���ก�� (differential settlement)

*+��',���(���������$ก ��� $������

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Wall

Property line

�����������ก

-� ���������ก',�+��ก����ก./,���� � ��,$����������ก������

�����ก������� (wall footing)

0���

�����ก

w�������ก�0�� ���$� �

%� ��

�����ก

�ก������������ก���$���� ��������ก#!�����������'�ก0���

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�����ก��������������� (spread footing)

�ก������������ก������ ��������ก#!�������� 1���'�ก$����+��%��

P

�����ก���� (combined footing)

(-�����������ก'�ก$�����%������ �กก��� *+������(ก��$���ก��

P

P

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A B

Rectangular, PA = PB

A B

Rectangular, PA < PB

Property line

A B

Rectangular, PA < PB

Property line

A B

Strap or Cantilever

Property line

�����ก������������

�����ก!��� ��������0��(mat footing)

Pile cap

PilesWeak soil

Bearing stratum

�����ก��$��$�2 (pile cap)

�����ก��������

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���ก����� �ก����ก��������ก

�����ก��ก��%$�� $��2ก*+��0 ���ก��%�� 0��� �%���$��� �������(-��,�,�&�

��ก��%%����&� 7.5 * .

!.�.�.

�� �+ก��������ก$����$��2ก$�� ����:

%���# �����ก��� 15 * . ������������ก������ � ��,

%���# �����ก��� 30 * . ������������ก�����$��$�2 7.5 * .

15 * .

$��%� ����!ก� ������!����$����� ��' �$� ���$������%����!���$����� '�%&���

*+�� ��������$���ก�� $����(-�(�ก��ก���������% �� ก5%��� $ �%6 ���$7��� ��,

ก��89��+����$��2ก$��

A A A

P

q = ��������ก���

��������ก!��"�� �������ก

�������ก����&ก����0��������ก�����-���� �',$ก ���������ก���

����������ก��%��1���6ก��������ก',� &% (�������� �ก��

�0�ก�,'��� ���$� �$���ก�������������

(��� $!:�'� �����������# �� ���$� �$�����'�ก :

�� �����&����������ก

�� �+ก��������ก(%�0 �� �

-� ����� �: � �$����� ���� � �����

P

� �$�����

P

HeaveHeave

� �����

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������������#$��%!&�������

L

PB

�����ก ��� ก���� B �� ��� L ����������ก����&ก

P ก�,�����ก����������

W

P

qn

�������+������ �(%������ก q = (P + W) / BL

�%�(�ก������/ $ �%6�����,���$7���(������ก',

(-��������+���&�; ���� � (net soil pressure : qn)

qn = P / BL

$�����'�ก�������+�����������ก%������������ก� ��

$���������ก

��',%���# �$ก ������������ (������ � q ≤≤≤≤ qa

�������� : �����ก����������$�����ก���()��*

$��%� ������������ก����&ก(-����%� ����ก������ � 50 %�� �������������ก���$����� '�%&��� '�� '��/����������ก��,������������ ����$ก ��+�� ก���������������ก���� ������ (�� qa = 8.0 %��/%�. .

0.4 m

1.1 m 30x30 cm column

P = 50 ton !�(��� � &% �������ก�����ก 10%

����������������ก���%���ก��:

0.810.150

A d'req×

= = 6.88 %�. .

(-������ก���� 2.7 × 2.7 . (A = 7.29 %�. .)

���(-������ก��� 0.4 . �������ก�����ก: W = 0.4×2.72×2.4 = 7.0 %��

������������+������� ����� �: q = (50 + 7) / 2.72 = 7.8 %��/%�. . < qa OK

������������+���&�; ���� �: qn = 50 / 2.72 = 6.9 %��/%�. .

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�������� 12.1 ก�������ก���� ���� �.�.�. ����������� �.�. 2522

)*�+���� ก�& ���ก��(� �/�..$.)

*�����ก����� �ก�������������������� ���������������ก�����������ก��� ��!�

� ���������� �� ����$%2 ��� 2

� �!��ก��������������� 5

� ����������������� 10

ก�������� ���� 20

� �� ���� 25

� �!������� ����� 30

� �����������# ��!��"�� 100

�����ก����$�����ก���!%ก���$��()��*

eP

qminqmax

e

L

P B

(�ก�/�����������ก����&ก P ก�,���$�����1���6 ���������ก����������ก����&ก��� ก�� $ �%6���

������������+������ �(%������ก q ',ก�,'��%�����$- �$��� ��� �ก����&� ��,��������&������������ก

12/LBI

2/Lc3=

=

IMc

AP

qmin −=

IMc

AP

qmax +=

2LB

M6LB

P−=

2LB

M6LB

P+= aq≤

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���������������#$�!���+�+���$��()��*�����

e1

P

qmin

qmax

e2

P

qmax

0

e3

P

qmax

��/!���� ��)0�&�

����� ���ก$�����������������+������ �$!:���ก���$!:����������+� *+��(���

$!:�'� �# ��� ������0���#!��������ก#�� ��'���(�������ก�� ก����#��

0I

McAP

qmin =−=

$����(��������(%����$!:������������ � ก�����(�� qmin = 0

IceP

AP

=cA

Ie = 12/LBI

LBA

2/Lc

3=

=

=

)2/L(LB12/LB

e3

max = = L / 6

Pemax = L / 6

L/3 L/3 L/3

Middle Third

a

3a

eP

qmaxR

������+�+���$��()��* e > L / 6

������������+������ �(%����',ก�,'��$!:���!�� $����� ��������;6�+�� R

',%��ก���������ก����&ก P

������������+������ � �ก����&���#��'�ก

max

1P R 3a b q

2= = × × ×

max

2Pq

3ab= $ ��� a = L / 2 – e

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� !��/��� 13.1 �����ก���� 1.8 x1.2 . ����������ก����&ก 80 %��ก�,�������,�,$�����1���6(������� 0.15 . '�� '��/�������� �(%������ก ��������/*�����ก������������,�,$�����1���6 0.40 .

0.60 m

0.60 m

0.90 m 0.90 m

e

Load

!�(��� e = 0.15 . < [1.8 / 6 = 0.30 .]

$�����'�ก e = 0.40 . > [1.8/6 = 0.30 .] (-�� ก������������ $�����

a = 0.90 – 0.40 = 0.50 .

min 2

P 6Mq

BL BL= −

2

80 6 80 0.151.2 1.8 1.2 1.8

× ×= −

× ×

= 37.0 – 18.5 = 18.5 %��/%�. .

qmax = 37.0 + 18.5 = 55.5 %��/%�. .

max

2Pq

3ab= 2 80

3 0.5 1.2×

=× ×

= 88.9 %��/%�. .

1-m slice on whichdesign is based

Wall

Footing

wUniformly loaded wall

w

Bending deformation

�����ก������� (Wall footing)

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Concrete column,

pedestal or wall

Critical section

Column with steel

base plate

s

s/2

Critical section

���������ก,�������������*���

����%��� ก5%�������������%� ������ก�����

����%��� ก5%����ก+��ก����,����������,1���6ก���ก�����ก��� �

����%��� ก5%����ก+��ก����,��������%� ����,����0��$��2ก���(%�$��

Masonry wall

b/2 b/2

b/4

Factored wall load = wu t/m

Factored soil pressure, qu = (wu )/L

Required L = (wDL+wLL)/qa

qa = Allowable soil pressure, t/m2

221 1

( )2 2 8u u u

L bM q q L b

− = = −

2u u

L bV q d

− = −

Min t = 15 cm for footing on soil, 30 cm for footing on piles

Min As = (14 / fy ) (100 cm) d

�����*�-+����.��� ������ก�������

b

d

wu = 1.4wDL+1.7wLL

d

L

qu

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� !��/��� 13.3 ��ก��������ก���0���$��������������ก����&ก���� wD = 12 %��/ .��,�������ก����&ก'� wL = 8 %��/ . ����������ก������� ��� 10 %��/%�. . ก����� fKc = 240 กก./* .2 ��, fy = 4,000 กก./* .2

D = 12 t/mL = 8 t/m

25 cm

L

8 cmclear

5 cmtypical

1.50 m

!�(��� � '��/����ก���� 1 $ %�

� &% �����ก��� t = 30 * .

%!$�!��ก�� ����ก:

a

(D L) WL

q+ +

=(12 8) 1.1

10+ ×

=

= 2.2 $ %� USE 2.2 m

�������ก�����ก W = 0.3 × 2.2 × 2.4 = 1.58 %��/ .

�������+������ � q = (12 + 8 + 1.58) / 2.2 = 9.81 < [qa = 10 %��/%�. .] OK

WSD��ก���"��!�(���/!��������

�������+���&�; ���� � qn = (12 + 8) / 2.2 = 9.1 %��/%�. .

�!'���ก�5����� ���� �!�ก6�*�* d 'ก�����$/�

75.5 cm

d = 22 cm

25 cm

9.1 t/m2

30 cm

L bV q d

2− = −

2.2 0.259.1 0.22

2− = −

= 9.1 × 0.755 = 6.87 %��

ก�����$7����������%����ก��%: c cV 0.29 f bd′=

30.29 240 100 22 /10= × ×

= 9.88 %�� > V OK

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WSD��ก�����&7ก���$ �"$�$���� �

97.5 cm25 cm

9.1 t/m2

2

2bL

q21

M

=

2

225.02.2

1.921

××=

= 9.1 × 0.9752 / 2 = 4.33 %��- .

ก���������������%����ก��%:

= 8.10 %��- . > M OK

2cc dbjkf

21

M =

52 10/22100883.0351.010821

×××××=

!� �/$��2ก$�� ���%���ก��:jdf

MA

ss =

22883.0700,11033.4 5

×××

= = 13.1 * .2

!� �/$��2ก$�� ��������&�: Min As = 0.0018×100×30 = 5.4 * .2 < As OK

WSD

$���ก(-�$��2ก DB12 @ 0.20 (As = 5.65 * .2/$ %�)

$���ก(-�$��2ก DB16 @ 0.15 (As = 13.40 * .2/$ %�)

��ก�����&7ก�$�! (-�$����%������ก���%ก����

As = 0.0018×100×30 = 5.4 * .2/$ %�

2.20 m

0.30 m

0.05 m ��ก��%����0.05 m �������������

DB16 @ 0.15 m

DB12 @ 0.20 m

0.25 m

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−= d

2bL

qV uu

−= 22.0

225.02.2

8.13

SDM��ก���"��!�(���/!�ก�& �

�������+���&�; !�,������� � qnu = (1.4 × 12 + 1.7 × 8) / 2.2 = 13.8 %��/%�. .

�!'���ก�5����� ���� �!�ก6�*�* d 'ก�����$/�

75.5 cm

d = 22 cm

25 cm

13.8 t/m2

30 cm

= 13.8 × 0.755 = 10.42 %��

ก�����$7����������%����ก��%:

= 15.35 %�� > V OK

dbf53.0V cc ′φ=φ

310/2210024053.085.0 ×××=

SDM��ก�����&7ก���$ �"$�$���� �

97.5 cm25 cm

13.8 t/m2

2

uu 2bL

q21

M

=

2

225.02.2

8.1321

××=

= 13.8 × 0.9752 / 2 = 6.56 %��- .

2u

ndb

MR

φ=

2

5

221009.0

1056.6

××

×= = 15.06 กก./* .2

−−′

=ρc

n

y

c

f85.0R2

11f

f85.0= 0.0039

!� �/$��2ก$�� ���%���ก��: dbAs ρ= 221000039.0 ××= = 8.58 * .2

!� �/$��2ก$�� ��������&�: Min As = 0.0018×100×30 = 5.4 * .2 < As OK

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SDM

$���ก(-�$��2ก DB12 @ 0.20 (As = 5.65 * .2/$ %�)

$���ก(-�$��2ก DB16 @ 0.20 (As = 10.05 * .2/$ %�)

��ก�����&7ก�$�! (-�$����%������ก���%ก����

As = 0.0018×100×30 = 5.4 * .2/$ %�

2.20 m

0.30 m

0.05 m ��ก��%����0.05 m �������������

DB16 @ 0.20 m

DB12 @ 0.20 m

0.25 m

1d/2 d

d3

21 ก��$7����,�&

2 ก��$7�����(�� 1�������

3 ก��$7�����(�� 1������

1

2

1 $ �%6(�� 1�������

2 $ �%6(�� 1������

�����ก�����������������ก����������ก����&ก'�ก$��%��$����� (�ก����ก���%���� '��/� $ �%6��� ก��$7����� (beam shear) ��,ก��$7����,�& (punching shear)

���� �!�ก6���� �"$�$���� �

���� �!�ก6���� �ก�5���

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Footing Type

One-way

Two-way

Square Footing Rectangular Footing

L

B

s (typ.)

L

L

s (typ.)

L

B

B/2 B/2

As2As2 As1

AsL

AsB

s (t

yp.)

1

12

21

2

s sL

sL ss

A A

A AA

LB

β

β

= +

−=

=

ก��ก�+/����-0ก��������ก�����

c1 + d

c2c2 + d

c1

d/2

b0

����.������!�� (����.���!+-%)%� ��$'�,�,�&�����ก�������$����$!:���!!P�� � ����%��� ก5%(-����$��������!����'�ก%� ����ก �$!:��,�, d / 2

P

ก�& ��5����*&�:

SDM

WSD dbf53.0V 0cc ′=

dbf06.1V 0cc ′=

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aqWLD

A++

=

� !��/�ก��ก�����ก��� �!�� ��&� �$' �� � '���ก��������ก$��������$����� '�%&������$�����$����� '�%&������� 40 * . $�����������������ก����&ก���� 40 %�� ��,�������ก����&ก'� 30 %�� ����������ก������� ��� 10 %��/%�. . ก����� fKc = 240 กก./* .2 ��, fy = 4,000 กก./* .2

D = 40 tL = 30 t

40 cm

b

h

� &% �����ก��� h = 40 * . � d = 32 8$.

!�(��� �����ก�� ����ก:

= 7.7 %�. .10

1.1)3040( ×+=

�&��ก��ก��� 2.8 x 2.8 $. (A = 7.84 �.$.)

�������ก�����ก W = 0.4 × 2.82 × 2.4 = 7.53 %��

�������+������ � q = (40 + 30 + 7.53) / 2.82 = 9.89 < [qa = 10 %��/%�. .] OK

WSD��ก���"��!�(���/!��������

�������+���&�; ���� � qn = (40 + 30) / 2.82 = 8.93 %��/%�. .

ก�5����*&��� ���� �!�ก6�*�* d/2 = 16 8$. 'ก�����$/�

40 cm72 cm

2.8 m

40 cm

72 cm

d / 2 = 16 cm

2.8 m

V = 8.93(2.82 – 0.722) = 65.4 %��

���$7����������%��� ก5%:

$��������! b0 = 4 × 72 = 288 * .

ก�����$7�����ก��%:

= 75.7 %�� > V OK

dbf53.0V 0cc ′= 30.53 240 288 32 /10= × ×

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�!'���ก�5���%��� ���� �!�ก6�*�* d = 32 8$. 'ก�����$/�

V = 8.93 × 0.88 × 2.8 = 22.0 %��

40

2.8 m

2.8 m

32 8888 cm

d = 32 cm

40 cm

8.93 t/m2

40 cm

���$7����������%��� ก5%:

ก�����$7�����ก��%:

= 40.3 %�� > V OK

30.29 240 280 32 /10= × ×c cV 0.29 f bd′=

WSD

2 5c

1M 108 0.351 0.883 280 32 /10

2= × × × × ×

WSD��ก�����&7ก���$ �"$�$���� �

M = 8.93 × 2.8 × 1.22 / 2 = 18.0 %��- .

= 48.0 %��- . > M OK

120 cm40 cm

8.93 t/m2

!� �/$��2ก$�� ���%���ก��:jdf

MA

ss = = 37.5 * .2

!� �/$��2ก$�� ��������&�: Min As = 0.0018×280×40 = 20.2 * .2 < As OK

518.0 101,700 0.883 32

×=

× ×

$���ก(-�$��2ก 19 DB16 #

(As = 38.19 * .2)

40 cm

2.80 m

0.40 m0.05 m ���ก�����0.05 m ��� �� ����

19 DB16 #

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SDM��ก���"��!�(���/!�ก�& �

�������+���&�; !�,������� � qnu = (1.4×40 + 1.7×30) / 2.82 = 13.65 %��/%�. .

ก�5����*&��� ���� �!�ก6�*�* d/2 = 16 8$. 'ก�����$/�

40 cm

72 cm

d / 2 = 16 cm

2.8 m

Vu = 13.65(2.82 – 0.722) = 99.9 %��

���$7���!�,����������%��� ก5%:

$��������! b0 = 4 × 72 = 288 * .

ก�����$7�����ก��%:

= 128.6 %�� > Vu OK

3cV 0.85 1.06 240 288 32 /10φ = × × ×

SDM�!'���ก�5���%��� ���� �!�ก6�*�* d = 32 8$. 'ก�����$/�

Vu = 13.65 × 0.88 × 2.8 = 33.6 %��88 cm

d = 32 cm

40 cm

13.65 t/m2

40 cm

���$7���!�,����������%��� ก5%:

ก�����$7�����ก��%:

= 62.5 %�� > Vu OK

3cV 0.85 0.53 240 280 32 /10φ = × × ×

��ก�����&7ก���$ �"$�$���� �

120 cm40 cm

13.65 t/m2

Mu = 13.65×2.8×1.22/2 = 27.5 %��- .

$ �%6���!�,����������%��� ก5%:

5

n 2

27.5 10R

0.9 280 32×

=× ×

= 10.66 กก./* .2

−−′

=ρc

n

y

c

f85.0R2

11f

f85.0= 0.0027

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0.0027 280 32= × ×

SDM

!� �/$��2ก$�� ���%���ก��: dbAs ρ= = 24.2 * .2

!� �/$��2ก$�� ��������&�: Min As = 0.0018×280×40 = 20.2 * .2< As OK

$���ก(-�$��2ก 13 DB16 #

(As = 26.13 * .2)40 cm

2.80 m

0.40 m0.05 m ���ก�����0.05 m ��� �� ����

13 DB16 #

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� �����ก�������� ��

� �����ก���������

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Design of Footing 2Design of Footing 2

��� ��.��.�� � ���������

�����ก�������� ��

�����ก���� ������ก��ก������� ������ก�������� ��������� ���������ก�� !�"������

�#�����ก���$ก��ก��!�����ก��%���&'� ������ก

'�� 1/2 < P2/P1 < 1

��������ก ������!(���!�

P1 P2

� ���

����

��

P P

��!�� 2 ��������ก��ก��

P1 P2

� ���

����

��

��!������������)��� P1 < P2

� ���

����

��

'�� P2/P1 < 1/2 ���������ก��������ก����������ก�� !*��!(�������!

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ก��ก��������������ก����

C

s

B

nm

L/2 L/2

P1 P2Rn m

q

���*���� c ���*����"+& R ���

�#�����ก P1 *�� P2 :

RsP

PPsP

n 1

21

1 =+

=

��ก�$� c ก�����( �!�� ��ก,-�#�

���������ก�� L/2 :

L = 2 (m + n)

( �!ก ��������ก :

LqWPP

B 21 ++=

Cb1

nm

L

c1 c2

b2

2

1

1 2

1 21

1 2

1 22

1 2

3( )2 3( )

2( )

( 2 )3( )

(2 )3( )

e

b n m Lb L n m

Rb b

q L

L b bc

b b

L b bc

b b

+ −=

− +

+ =

+=

+

+=

+

Cb1

nm

L1 L2

b2

21

1 1 2

1 12

2 2

1 1 2 2

2( )( )

e

e

n m Lb

L L L

R L bb

q L L

RL b L b

q

+ −=

+

= −

+ =

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ก���ก��������������ก����

L

B

Side view

�!���ก����������ก,�����%���&ก��*����"+&����#�����ก���$ก ��� �*��������������

��ก�����*��*/� !���� !� �����ก��!ก��)ก��*����������-

ก�����������ก�������ก����

ก��� ��!���0ก!�#��������� *������ �#� !�#�������*���������� �1#�ก��*/�2�!�

)!�!��&���

h

q

P2P1

V

M

ก�����������ก������

P2

q

ก�����������ก������

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���������%�� �#� �#�����ก�'�����!���ก��!����ก���������"�#�����ก�����ก��

( �!ก ���� ���!��� ก d/2 ��*��������

B

Side viewc + d

d/2d/2

c

Punching shearperimeter

�� ����ก�� ก!�����ก���� �"���������� ���!�� ���������ก�� 5 �!�� )��!� ����������������������ก� ������ก 40 7!. �#����� �*��*�ก��������(�� 10 ���/��.!. ก����� f<c = 240 กก./7!.2 *�� fy = 4,000 กก./7!.2

A B

40 x 40 cm

45 x 45 cm

D = 50 tonL = 25 ton

D = 80 tonL = 40 ton

5.0 m

40 cm

R

C.G.

x

(75 120) 120(5)

3.1 m

x

x

+ =

=

��#$�%� 1. �%�!����!���'#( R :

������ก C.G. '1���������ก����7���

= 3.1 + 0.4 = 3.5 m

( �!�� �����ก L = 2 x 3.5 = 7.0 m

2. �����������ก L :

ก����� C.G. �������ก�� R

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3. ���ก�+�������ก B : *�����������!��� qa = 10 ���/��.!.

LqWPP

Ba

21 ++=

0.71015.1)40802550(

××+++

= = 3.21 m USE 3.3 m

������( �!��������ก 60 7!. ( �!�1ก d = 52 7!.

�#�����ก�����ก W = 0.6 × 3.3 × 7.0 × 2.4 = 33.26 ���

*�������������ก q =0.73.3

26.3340802550×

++++= 9.88 ���/��.!.

< qa OK4. !��-���!���./ �!�0������(1����������

� � A : Pu = 1.4×50 + 1.7×25 = 112.5 ���

� � B : Pu = 1.4×80 + 1.7×40 = 180 ���

*��������-����� :3.30.7

1805.112qnu ×

+= = 12.66 ���/��.!.

�#�����ก*/�-����� :0.7

1805.112wu

+= = 41.79 ���/!.

Column A :Pu = 112.5 ton

Column B :Pu = 180 ton

0.4 m 5.0 m1.6 m

7.0 m

41.79 t/m2

2.29 m

Vu (ton)16.7 t

-95.8 t

113.2 t

-66.8 t

Mu (t-m)error = 6.4 t-m

Mu,max = -106.4 t-m

3.34 t-m

47.0 t-m

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5. ����ก������������������(�� )!�!��&��!�ก� $�ก����� � –Mu = 106.4 ���-�!��

2u

ndb

MR

φ=

2

5

523309.0

104.106

××

×= = 13.6 กก./7!.2

−−′

=ρc

n

y

c

f85.0R2

11f

f85.0= 0.0035

-��!�����0ก� ��!�����ก��: dbAs ρ= = 60.1 7!.2

-��!�����0ก� ��!����� $�: As,min = 0.0018×330×60 = 35.6 7!.2< As OK

����ก������0ก 10 DB28 (As = 61.58 7!.2)

= 0.0035×330×52

������)!�!��&� ก 47 *�� 3.3 ���-!. ������0ก����� $� As,min

����ก������0ก 12 DB20 (As = 37.68 7!.2)

3cV 0.85 1.06 240 388 52 /10φ = × × ×

6. ����� ก���./ ��0�4 "������� �*������� ���*����� qnu = 12.66 ���/��.!.

��� A : b0 = 4(40+52) = 368 7!.

Vu = 112.5 – 12.66×0.922 = 101.8 ���40+52 7!.

Pu

qnu

3cV 0.85 1.06 240 368 52 /10φ = × × ×

= 267 ��� > Vu OK

��� B : b0 = 4(45+52) = 388 7!.

Vu = 180 – 12.66×0.972 = 168.1 ���

= 282 ��� > Vu OK

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7. ����� ก���./ � �� ��ก*/�2�!�*���E��� Vu,max = 113.2 ���

ก������E������(��ก��: 3cV 0.85 0.53 240 330 52 /10φ = × × ×

= 119.8 ��� > Vu OK

8. ก!����ก����������� "��������ก/� � �*���������ก!������� d/2

40x40cm

45x45cm

d/2 = 52/2 = 26 cm

A B 3.3 m

20+40+26 = 86 cm 26+45+26

= 97 cm7.0 m

3.30 m

0.60 m

PA = 112.5 ton

1.45 m0.40 m

3.30 m

0.60 m

1.425 m0.45 m

PB = 180 ton

��� A : be = 20+40+26 = 86 7!., wu = 112.5 / 3.3 = 34.1 ���/!.

Mu = 34.1×1.452/2 = 35.9 ���-!.5

2

35.9 100.9 86 52

×=

× ×2u

ndb

MR

φ= = 17.2 กก./7!.2

−−′

=ρc

n

y

c

f85.0R2

11f

f85.0= 0.0045

����ก������0ก 7DB20

(As = 21.98 7!.2)dbAs ρ= = 20.1 7!.2= 0.0045×86×52

��� B : be = 26+45+26 = 97 7!., wu = 180 / 3.3 = 54.5 ���/!.

Mu = 54.5×1.4252/2 = 55.3 ���-!.5

2

55.3 100.9 97 52

×=

× ×2u

ndb

MR

φ= = 23.4 กก./7!.2

′−−

′=ρ

c

n

y

c

f85.0R2

11f

f85.0= 0.0062

����ก������0ก 10DB20

(As = 31.4 7!.2)dbAs ρ= = 31.3 7!.2= 0.0062×97×52

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As = 0.0018(100)(60) = 10.8 cm2

9. ����ก������+�����ก��!�ก�+��USE DB20 @ 0.15

(As = 12.56 cm2/m)

0.40 m

7.0 m

0.60 m

5.0 m0.40 m 0.45 m

0.86 m7DB20

0.97 m10DB20

10DB28

12DB20

A B

DB20 @ 0.15 m

DB20 @ 0.15 m

�����ก��������� (pile cap)

Pile cap

PilesWeak soil

Bearing stratum

�!���ก�����*�ก��������!,!��"��"� ������ก�������

��ก*��*/� ������ ���0!��ก�� ��'����#�����ก���$ก��

����#����71�������1ก��,-

��������(��������0ก �#�����ก���$ก,!�!�ก �������0!

�#�71���-F���0!��ก ก��'����#�����ก����%��( �!GH�

��� ���/� � ���0!ก����#����)�����

��������(��������I� �#�����ก���$ก!�ก �������0!

�� 71���-F���0!��ก������0!���� ก��'����#�����ก����%��

( �!GH���� ���/� � ���0!ก����#����)����� *��ก�����

*��*�ก���-���� ���0!����#����*�0�

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���������ก (driven pile)*�

�GH��

��/�

��0!

� ���0!��ก����-F� � ���0!,!� � ���0!���0ก *��)�� � ���I������� ���0!(��ก��

�+ �$ ��(�'�ก ( �($!($�2�"� ���0!,�� *��ก����ก�����

� ���0!GJ�*���ก����#���������'����#�����ก,���

�+ ��$� ���-JK�������ก����ก�ก��� �����*��ก�� ��� �����

��� �����ก� ���0!���� ���� *����� ��/�ก������

��(�������(��

�������ก����� ���� �

��ก���,!�!��ก ��* ��/�ก��� ��($� !����������

�#�����ก���$ก���!�������#�������! ,!��ก�� 2 ���/��.!.

��� �*��GH����!��� :

���������( �!�1ก,!��ก�� 7 �!��

��� �*��GH����!��� 600 กก./��.!.

���������( �!�1ก�ก�� 7 �!��

��� �*��GH����!��� 800 + 200L กก./��.!.

)��� L (��( �!�� � ���ก�� 7 �!��

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��������!�" (bored pile)

� ���0!���� �!��' ��'����#�����ก�� ����#������1ก��

,-71��� ���0!��ก�����,-,!�'1�

ก������ ���0!��ก���� �!�ก������ก��ก��*�����

-��!��!�ก��� ��/������(�������(��

Pile cap

BED ROCK

�����ก���������#����ก���$��%&

P

RR R

!!$������ ���0!$ก�������#�����ก���$ก���ก��

aRnP

R ≤=

)��� R = �#�����ก���$ก�� �*�������������

P = �#�����ก���$ก�������#��!�

D = �#�����ก���$ก(��

L = �#�����ก���$ก��

W = �#�����ก�����ก*�����'!����������ก

= D + L + W

n = ���� �� ���0!

Ra = �#�����ก���$ก���!������� ���0!*�������

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�����ก���������#����ก�%'#��$��%&

P

R2R1 R3

� ���0!$ก���������#�����ก���$ก,!����ก�� � ���0!�������)!�!��&���������ก��*�����������#�����ก!�ก�1#�

a2n

n Rd

dMnP

R ≤Σ

±=

)��� R = �#�����ก���$ก�� �*�������������

M = )!�!��&����������ก����ก����!��

dn = ������������ ���0!*���������ก*ก�%���&'� ����ก�$�!� ���0!

P = �#�����ก���$ก�������#��!� = D + L + W

n = ���� �� ���0!

Ra = �#�����ก���$ก���!������� ���0!*�������

M

15 cm

3D

D

3D1.5D1.5D

1.5D

1.5D

3D

3D

ก����ก������ก�������

!��1���������%��� ก!�����ก:

WSD D LR

n+

=

SDM u

1.4D 1.7LR

n+

=

ก%�������������ก:

����������� ���� ���0! 3D

������� ���� ���0!'1���������ก 1.5D

)��� D (������� ���0!

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1.5D

3D

1.5D

1.5D1.5D

2 PILES

1.5D

3D

1.5D3D

1.5D

3 PILES

3D1.5D

3D

1.5D

1.5D1.5D

4 PILES

3D

1.5D

3D

1.5D

1.5D1.5D

6 PILES

3D3D

1.5D

1.5D

1.5D1.5D

5 PILES

D23

D23

1.5D

1.5D

7 PILES

D23

3D

3D

3D

1.5D

3D

1.5D

1.5D1.5D

8 PILES

3D3D3D

1.5D

1.5D

1.5D1.5D

D23

D23D23

ก��!������������������ก

1.5D

3D

1.5D

1.5D1.5D

9 PILES

3D3D

3D

10 PILES

1.5D

1.5D

3D

D33

1.5D 1.5D3D 3D 3D

3D

11 PILES

1.5D

1.5D

3D

D33

1.5D 1.5D3D 3D 3D

3D

1.5D

3D

1.5D

1.5D1.5D

12 PILES

3D3D

3D

3D

ก��!������������������ก

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������� �ก�$� �!��(����� ก)

0.16 x 0.16

0.18 x 0.18

0.22 x 0.22

0.26 x 0.26

0.30 x 0.30

0.35 x 0.35

0.40 x 0.40

15

21

30

43

50

80

100

Section Size(m) Load capacity(ton)

0.18 x 0.18

0.22 x 0.22

0.26 x 0.26

0.30 x 0.30

0.35 x 0.35

0.40 x 0.40

15

22

30

43

57

80

����������0

0.25(0.85 )a c gP f A′=

����('���������ก���������+�����ก;�

-x +x

� ���0!������ก������� �กL� ≥≥≥≥ dp/2 ���(��*��-M�ก������#��!�

<�� ��!��R=0 pd

2Rpd

2

� ���0!������������� �กL� ≥≥≥≥ dp/2 ���(��*��-M�ก������-F�%���&

� ���0!�������� � -dp/2 ≤≤≤≤ x ≤≤≤≤ dp/2 ���(��*��-M�ก������-F� �� � �)�����:

p

1 xR R

2 d ′ = +

)��� x (��������� ���������� �กL�*��%���&ก���� ���0! �-F����!���� ���0!����2����������� �กL� *���-F�� ก�!���� ���0!������ก

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Example 12.7 ����ก*�������ก� ���0!�����"�������#�����ก(�� 100 ���*�� �#�����ก�� 50 ��� � � ������!���$�� ���� 40×40 7!. f’c= 240 กก./7!.2 fy = 4,000 กก./7!.2 *����� ��#�����ก��� γs = 2.0 ���/��.!. �����ก�����1ก 1.50 !. ��ก�����/� ���

��������0!�������� 40 7!. "�#�� A = (π/4)×402 = 1,256 7!.2

a c gP 0.25(0.85 f A )′= = 0.25×0.85×240×1,256/103 = 64 ���

USE ∅∅∅∅ 40 cm bored pile with safe load 50 ton

1.20

0.60

0.60

0.60 1.20 0.60

2.40

2.40

!!$���#�����ก�����ก*�����'! 15%

���� �� ���0! n = 1.15(100+50)/50 = 3.45

USE 4 piles

��#$�%�

3cV 0.85 1.06 240 288 82 /10φ = × × ×

�����������ก��� 40 7!. ( �!�1ก-�� �+�/� d = 32 7!.

�#�����ก�����ก*�����'! W = (0.4×2.4 + 1.1×2.0)(2.4)2

OK= 18.2 ���

< �-��!��, � 22.5 ���

1.10

0.40

0.60 1.20 0.60

2.40

��%���ก���4ก>�0�����������!���0�+�:

SDM u

1.4D 1.7LR

n+

=1.4 100 1.7 50

4× + ×

= = 56.3 ���/���

����� ก���./ ��0�4

40 cm

72 cm

d/2 = 16 cm

Vu = Pu = 1.4×100 + 1.7×50 = 225 ���

b0 = 4×72 = 288 7!.

= 330 ��� OK> Vu

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����� ก���./ � ��

40 cm

80 cm

d = 32 cm

120 cm2Ru

x = 8 ?�.1.20

0.60

0.60

0.60 1.20 0.60

2.40

2.40

0.68

��+�����ก;� ���1ก�+������� :

p

1 xR R

2 d ′ = +

1 856.3

2 40 = + ×

Vu = 2×39.4 = 78.8 ���

= 39.4 ���

3cV 0.85 0.53 240 240 32 /10φ = × × ×

= 53.6 ��� 1�+����ก��ก��������!���./ �NG< Vu

Vs = (78.8 – 53.6)/0.85

= 29.7 ���

Mu = 2×56.3×0.4 = 45.0 ���-�!��

ก!����ก������������(��

40 cm

80 cm

40 cm

120 cm2Ru

5

2

45.0 100.9 240 32

×=

× ×2u

ndb

MR

φ= = 20.4 กก./7!.2

−−′

=ρc

n

y

c

f85.0R2

11f

f85.0= 0.0054

����ก������0ก 14DB20# (As = 43.96 7!.2)

dbAs ρ= = 41.5 7!.2= 0.0054×240×32

����-F����0ก��ก��#����*���E����� �

v ys

A f dV

s=

s = 240/14 = 17.14 7!.

2 3.14 4.0 3217.14

× × ×=

= 46.9 ��� OK> Vs �����ก��

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0.40

0.050.10

14DB20#

DB20 ������

(��ก��������������*���

��0!���� ∅ 0.40 !. ��� �.�.-���2������� 50 ������� � 4 ���

!����0� $�������ก����������0

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� �����ก������ �����������

� ก����ก�����������������

� ก��������� �����

� ���������������������������

� ก���������!�ก�"#�$��ก%���&ก

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

ServiceabilityServiceability

��� ��.��.�� � ��������

∆∆∆∆

w

�����ก������

��ก����ก�%%�����������"� ��ก!�ก!��������'�����(� ������ ������ �

����������������#����)���������(��!�����(ก����%�"#�$��ก%���&ก����

ก����ก�������� ������%��� �ก��!�����

ก�����������"�������"�� �����������)���ก����*ก#�$��

ก����*�����+��

���� �

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Strength Design Method

Crackings

Deflections

More slender members

More service loadproblems

- more accurate assessment of capacity

- higher strength materials

ก�� �ก�����������������ก�����������������, ���� n = Es / Ec ≈ 8-10

���, ����ก$�ก �����ก��� cr f2f ′= ≈ 30 กก.//�.2

$����������$ 1ก����� 2���*���ก������*��ก��ก����ก���:

fs ≈ 8×30 = 240 กก.//�.2 << fy

��������ก�������ก��ก����������� ��!��"�#��ก!���$ก%ก�

ก����ก�!!������� !�(������$�����ก�����

���� 1ก%��� �ก��!������������

�(+*������%���! ���,�� �����������

����4��ก��ก��ก��ก���� ���$ 1ก�����

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����ก��������� w

����ก�������������#���2������ก�� �� Gerely � � Lutz �(+*������2����ก�����ก��*�&� ����������*�����%����5� �����

���ก���&�������� 33s cw 0.011 f d A 10−= β × �.�.

�����* w = ����ก�����ก��*�&� ��������, �.�.

fs = $��������5����$ 1ก����� 2���%�"#�$��ก%���&ก���� = M / (As jd)

= 0.6 fy ��ก�2���*)����ก���#���2�����*��!���

dc = �������ก���$&�!�ก�����%����5�'5�7,���ก ���$ 1ก������'� ����&�

�.�.�.

�.�.�.

β = �����������$��������!�ก�����%����5�'5��ก��������������!�ก

7,���'����$ 1ก�����'5��ก������� = h2 / h1

β = 1.20 �#�$��%���, 1.35 �#�$��%(+"���������

A = (+"���*���ก���$��$&��$ 1ก�����$�5*����

y2y

dc

h1 h2

(+"���*��%����5����������

�ก�������

7,���'����$ 1ก�����

bw

=(+"���*��������� !#�����$ 1ก

w2ybn

=

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����ก��������������������

���'���������"� �.�.

���ก����������������(#�

��ก�7�$�$�+���ก��$&���� 0.016 0.41

��ก�7+"�$�+���������� 0.012 0.30

������������� � ���"#�� 1� 0.007 0.18

�������"#���� $�+���ก����8�ก�$�� �%ก�� 0.006 0.15

��������ก�"��"#� 0.004 0.10

��ก�������������� �.�.�.

ก����ก�%%�������ก#� ����$ 1ก����� fy = 3,000 - 5,600 ksc (%�������������

���$ 1ก�������*�����ก������������,���ก �#��$������ก�����ก 5"�

����9�� ACI $�+� �.�.�. !5�ก#�$���$�#���2��������ก��������� z �#�$��%

��� ��������� β = 1.20 :

33s cw 0.011 f d A 10−= β × 3

s cz f d A=w w

0.011 1.20 0.013= =

×

�����(�: z ≤ 31,000 กก.//�. (w ≤ 0.41 �.�.)

�������ก: z ≤ 26,000 กก.//�. (w ≤ 0.34 �.�.)

'*������������(�: z ≤ 31,000(1.2/1.35) = 28,000 กก.//�. (w ≤ 0.41 �.�.)

'*��������������ก: z ≤ 26,000(1.2/1.35) = 23,000 กก.//�. (w ≤ 0.34 �.�.)

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ACI Provision for Crack ControlACI Provision for Crack Control

Gergely-Lutz euqation was replaced in the 1999 ACI Code.

New ACI provisions on crack control through reinforcement distribution limits

the spacing in RC beam and slab to :

= −

2,80038 2.5 c

s

s cf

but not greater than 30(2,800/fs), where

(ACI Eq.10.4)

fs = calculated stress (ksc) in reinforcement at service load = unfactored momentdevided by the steel area and the internal arm moment, fs = M/(As jd).Alternatively, fs = (2/3) fy may by used; an approximate jd = 0.87d may by used.

cc = clear cover from the nearest surface in tension to the flexural tension reinforcement (cm)

s = center-to-center spacing of flexural tension reinforcement nearest to the extreme

concrete tension face (cm)

SDM

�����+� ����ก��ก� ���&��#�,ก��� ��'*���!$�ก����ก����

30 cm

3DB28

2DB25

5 cm

5 cm

4 cm

fs = 0.6 fy = 0.6×4,000

= 2,400 กก.//�.2

dc = 5 + 1.4 = 6.4 /�.

3 6.16 6.4 2 4.91 13.05y

3 6.16 2 4.91× × + × ×

=× + ×

= 8.71 /�.

w2ybA

n=

2 8.71 305

× ×= = 104.5 /�.2

3s cz f d A= 32,400 6.4 104.5= ×

= 20,988 กก.//�. < 26,000 กก.//�. �#�$��%�������� OK

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As

2

max

0.10( 75) cm / m

/ 6 30 cm

skA d

s d

≥ −

≤ ≤

ก����� ��#�,ก��!����(�%-ก���.%������'*���!$��������

%����2��*��%������� %����������+*�� �$ก��!���$ 1ก������$��*����������������� $�+�����$�5*�����% �����������������*���ก���

As

10/Lbb E ≤≤

ก����� ��#�,ก���� �����&��&�����ก�'*���!$��������

�#�$��%�����* 5ก�ก�� 90 /�. ���������$ 1ก��*������ �� Ask

ก��!����*#����������� d/2 !�ก�#��$����$ 1ก�������%����5�

d/2

dAsk

s

≥90

/�.

Minimum number of bar in one layer

bw4 cmcover

dc2dc

Total tensile area = 2 dc bw

Tensile area per bar: 2 c wd bA

m=

m = number of bars in one layer

( )

3 2 2

33

2 2From

/c w c w

s cs s

d b d bzz f d A m

f m z f

= → = → =

Example: SD40: fy = 4,000 kg/cm2, fs = 0.6(4,000) = 2,400 kg/cm2

covering = 4 cm

stirrup ∅ ≈ 9-10 mmdc = 5 + 0.5 db

( )( )

2

3

2 5 0.5max 2

/ 2,400

b wd bm m

z

+= ⇒ =

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Box BeamFlexure Testing

Failure of BeamFailed Beam

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Deflection of Elastic Sections

1) Excessive deflection Wall

2) Ponding effect of roof

rain

4) Visually offensive sag

Working Stress Design (WSD) Deflection is controlled indirectly by

limiting service load stress result in large member.

Ultimate Stress Design (USD) Members become more slender and/or

smaller sections may result in deflection problems.

3) Misalignment of machine

cracking of partitions

����� ก��������������� ���!"����#����

���+�������

(+"��������� L / 20

��� L / 16

'������ 5ก�����*����#����������" )������#���2�����������

�/�������+���*��%��������

L / 24

L / 18.5

���+���*����%���

L / 28

L / 21

���*��

L / 10

L / 8

�#�$��% fy ��*)������ก�% 4,000 กก.//�.2 �$�,2��� 0.4 + fy / 7,000

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�$��%������������������� �$�����&���'(�

$ ������*)�������%$�+����ก�%������*��������������*������!��ก����������$��!�กก�����������ก�ก�����

�� �&��/���� � � ��+�������' ���0� ' ก��� � ��+�

L / 180����������������!�ก�"#�$��ก%���&ก!�

(+"���*)�������%$�+����ก�%������*��������������*������!��ก����������$��!�กก�����������ก�ก�����

L / 360����������������!�ก�"#�$��ก%���&ก!�

$ ����$�+�(+"���*�����%$�+����ก�%������*��������������*������!��ก����������$��!�กก�����������ก�ก�����

L / 480

�������������"�$����*�ก�� 5"�$ ��!�กก���5����ก�%������*������������ � �����������������ก� �� ���+*��!�ก�"#�$��ก%���&ก�������"�$�� � �������������������+*��!�ก�"#�$��ก!���*�(�*� 5"�

$ ����$�+�(+"���*�����%$�+����ก�%������*��������������*������!�)���ก����������$��!�กก�����������ก�ก�����

L / 240

ก���&���'���� �$�������������*����� L ��%�"#�$��ก%���&ก�%%�����*#����� w !�)�

w

L

45 w L384 EI

∆ =

��������:�!�����*� �������ก!������������+*��!�ก!&������:���+"������ก��

wMa Mb

Ma Mb

2

0

wLM

8=( )

2

0 a b

L5M 3 M M

48 EI∆ = − +

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���.����*�#�$+� (Modulus of Elasticity)

������/��ก���� (Cracking Moment)

���, ���+�$�&�� �����ก��� �#���2)�!�ก�,��: 1.5c c cE 4,270 w f ′=

�#�$��%���ก����"#�$��ก�ก��: c cE 15,100 f ′=

b

hyt

�+�������������*�#��$$��������5������ก���������ก�� r cf 2 f ′=

r gcr

t

f IM

y=

yt = ����!�ก�ก�������'5������%����5�

Ig = ������������������ ��$�������"�$��

�#�$��%$�������*�$ �*���+���������ก��� b ������ 5ก h: t

hy ,

2= 3

g

1I b h

12=

���)#��ก�����ก$�� ������ก�� �ก���

MM

�������ก���ก���

�������������ก���

Ig = gross moment of inertia

Icr = cracked moment

of inertia

M

Mcr

Deflection ∆

∆cr ∆e

Icr

Ig Ie

Ie = effective moment of inertia

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*�#�������#����#�������� �ก���, Icr

$�������ก������ก�%������ก�����%�������$�+��ก�������� �(+"���* ���$ 1ก��%

����5�/5*�!�',ก�� �����������������, ����

As

h

b

d

�#��$����ก������� : s

xb x n A (d x)

2= − 2

s s

bx n A x n A d 0

2+ − =

32

cr s

b xI nA (d x)

3= + −

xN.A.

nAs

n = Es/Ec

$������� �

*�#�������#����#��,�����-�.�, Icr ≤≤≤≤ Ie ≤≤≤≤ Ig

��+*������������ก��������������ก��� ������������������ ��$�����!� � �!�ก

Ig '5� Icr 5"�ก�%������������*��ก���#�

3 3cr cr

e g cr ga a

M MI I 1 I I

M M

= + − ≤

Ie

Ig

Icr

1 2 3 Ma/Mcr

�����* Mcr = ���������ก���t

gr

y

If=

fr = ���, ����ก$�ก cf0.2 ′= �#�$��%���ก����"#�$��ก�ก��

Ig = ������������������ ��$�������"�$��

Ma = ���������ก��*�&��������������*���������

h

b

yt

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Mcr

∆cr

Ig

(Ie)D

∆D

(Ie)D+L

MD

MD+L

∆L

∆D+L

ก�� �$����/�ก�"&����ก����� ���"&����ก/�

ก����+������ก��"�#��ก��� :

Dec

2D

aD )I(ELM

β=∆

�ก�� 5"�� ���� ��#��$$�������ก���)�%������ (Ie)D

ก����+������ก��"�#��ก����� ��"�#��ก�� :LDec

2LD

aLD )I(ELM

+

++ β=∆

ก����+������ก��"�#��ก�� : DLDL ∆−∆=∆ +

2a

1M 0.7 10

8= × ×

�����+���� 10.3 !����!��%ก��������� �����������*����� 10 ���� ก#�$�� f’c = 280 กก.//�.2 � � fy = 4,000 กก.//�.2

8 ton (LL)

10 m

5 mBeam weight700 kg/m(DL)

52 c

m60

cm

40 cm

8DB25, As = 39.27 cm2

� 4��"� ���� 5ก�����*�&� L/16 = 1,000/16 = 62.5 /�. > 60 /�. Ckeck ∆∆∆∆

1. ก����+������ก��"�#��ก!���$ก���

��������������������"�$�� 3g

1I 40 60

12= × × = 720,000 /�.4

������������ก��*�&� = 8.75 ���-����

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40 cm

x

N.A.

nAs

#�����������%�:

f’c = 280 กก.//�.2

cE 15,100 280= = 254,512 กก.//�.26

s

c

E 2.04 10n 8

E 254,512×

= = =

"���0�"��#�+�ก�� �� �:2x

40 8(39.27)(52 x)2

= − x = 21.8 /�.

������/� �����/����#��������ก����: Icr = Iconcrete + Isteel

3 2cr

1I 40 21.8 8(39.27)(52 21.8)

3= × × + − = 424,663 /�.4

rf 2.0 280= = 33.5 กก.//�.2

r gcr

t

f IM

y=

33.5 720,00030 100×

= 8,040 กก.-����

cr

a

M 8,0400.92

M 8,750= =

3cr

a

M0.78

M =

� �

������/� �����/����%� � �4 ��:3 3

cr cre g cr g

a a

M MI I 1 I I

M M

= + − ≤

Ie = 0.78×720,000 + 0.22×424,663 = 655,026 /�.4

� � ��+������ก��"�#��ก!���$ก���:

4

Dc e

5 wL384E I

∆ =45 700 /100 1,000

384 252,671 655,026× ×

=× ×

= 0.55 /�.

10 m

Beam weight700 kg/m(DL)

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2. ก����+������ก��"�#��ก����+��ก�!��"�#��ก��

Ma = 8.75 + 8×10/4

8 ton (LL)

10 m

5 mBeam weight700 kg/m(DL)

= 28.75 ���-����3

cr

a

M0.022

M =

� �cr

a

M 8,0400.28,

M 28,750= =

Ie = 0.022×720,000 + 0.978×424,663 = 431,160 /�.4

4 3

D Lc e c e

5 wL PL384E I 48E I+∆ = +

45 7 1,000384 252671 431160

× ×=

× ×

38,000 1,00048 252671 431160

×+

× ×

= 0.84 + 1.53 = 2.37 /�.

3. ก����+������ก��"�#��ก!���$ก�� ∆L = ∆L+L – ∆D = 2.37 – 0.55 = 1.82 /�.

��� ∆L ��*����$L 1,000

360 360= = = 2.78 /�. > 1.82 /�. OK