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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
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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
TA683.2 N55 2004 TA683.2 H365 2005
TA683.2 M39 2009 TA683.2 R48 2008
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.
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Conduct of Course
Design Projects 20 %
Midterm Exam 40 %
Final Exam 40 %
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
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
Structural Design ConceptStructural Design Concept
Structural Engineering
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Structural Design Concept
�������� Stability
�������� Safety
�������� Serviceability
��������Economy
��������Environment
LIFE-CYCLE OF STRUCTURE
Lesscompetitors
Traditionalactivities
New Design
Construction
Maintenance / Repairs /Renovation
Removal / Failure
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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
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
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�)�+�+��
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 ′
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
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
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
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
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
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)
รอยแตกราวเนื่องจากการดัด
รอยแตกราวเนือ่งจากการเฉือน
คาน
แรงดึง
แรงอัด
รอยแตกราว
บริเวณเกิดแรงดึงสูงสุด
บริเวณเกิดแรงอัดสูงสุด
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
ระนาบของรอยราวที่เปนไปได ซึ่งตองตั้งฉากกบัทิศทางของแรงดึง
ลูกศรแสดงทิศทางของแรงดึงในเนื้อคอนกรีต
รอยแตกราวเนื่องจากการดัดตัวของคาน
เหลก็เสนรับแรงดึง
รอยแตกราวเนือ่งจากการเฉือน
เหล็กปลอกท่ีใชรับแรงเฉือน
รอยแตกหลังคาน
รอยแตกใตทองคาน
ฐานรากทรุดรอยแตกใตทองคาน
รอยแตกหลังคาน
การแตกราวที่ผนัง เปนอาการของการทรุดตัวโครงสราง
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
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
Building CodesBuilding Codes
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Minimum requirements to protect the public
- �.�.�. ����� ����� 2522
- �����""��#ก���$� �����
- $%��""��#
Design Loads
Dead Loads - stationary loads of constant magnitude
Live Loads - moving loads or loads that vary in magnitude
Caused by the weight of structure
Include both the load bearing and non-load
bearing elements in a structure
Generally can be estimated with reasonable
certainty
�&�����ก����ก���� (Dead Load)
�&�����ก��/�ก0�/�������12
���������� kg/m3
���ก�������� �ก 2,400���ก��� ��� 2,320��� 500-1,200�� �ก 7,850
�������������� kg/m2
ก������� ���!" 14ก������� #�$%�&����' 50�� �ก��( ��, * ก�� 5
���������� 10-30���� 5����ก!""#$�"% 180-360����ก!""#$(�)"� 100-200
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
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))
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������� �!"�#$"��%&������� '#"�(#)*�'#�ก+�(kg/m2)
(1) ����
(2) ก� ����������� ก���
(3) �����ก���� ������� � ��� ���� ! ������"#
(4) ����%&" �'ก%&"���()���ก���� ��)�� ����ก ���%�# %�*���� +,��-��.,���������
(5) �! �ก� 0 �
(6) (ก) ���2-)�3 �4" ,������%&" �'ก%&"���()������ก��2-)�3#�"-����� "-����� ������� %�*�������
(,) �����&� �� +� )4������- ,����)�� ����ก ���%�#�! �ก� %�*0 �
30
100
150
200
250
300
300
�&�����ก����ก3����6&������ (�0�)ก�ก������ ��� 6 (�.�. 2527) �.�.. ���������� �.�. 2522
������� �!"�#$"��%&������� '#"�(#)*�'#�ก+�(kg/m2)
(7) (ก) ��� �������- � ��7�*)�# ���#���� 8��������7�*)�# �����4 � �����( �����#���������#�����9�������ก:��&� �3 �������9�ก�� � �3
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(,) �����&� �� +� )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 #�� $"��+�ก 10 ��$�
���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
��� �$�� �.�.. ���������� �.�. 2522
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50(8) �� ���&#'��'&�ก����������'�'/0���� ��"����$�
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 ��
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
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
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)
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
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
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
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
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
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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
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ACI Approximated Coefficients for Moments and ShearsACI Approximated Coefficients for Moments and Shears
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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
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
•• Continuous beamContinuous beam
Location of ReinforcementLocation of Reinforcement
Location of ReinforcementLocation of Reinforcement
• Continuous beam with 2 spans
Behavior of Beam under Load
Working Stress Condition
sε
cf f ′<cε
T = As fs
C
w
L
cε 2.0r cf f f ′< =
cf f ′<Elastic Bending (Plain Concrete)
cε
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
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
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stirrup���+ก���ก
Durability and Fire protection
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
×= = ≈
′ ′
Effective Depth (d) : Distance from compression face to centroid of steel
d
d
compression face
b
kd
cε
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
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
cε
sε
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:
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
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
�������� 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
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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
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)
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
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 (RB9@0.20 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
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
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
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
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
−′ =
′−
���� � ��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
− − ×= = =
′− × −
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′ ′ ′ ′= + + + − +
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
cε
sε
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 ???
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.
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
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
Behavior of Concrete Beam under increasing load
As
h
b
As
d
cε
sε
ctεfs
fc
fct
Before crack
As
After crack
cε
sε fs
fc
Working Stress State
cuε
sε fs
fc
Strength Limit State
Overload
Ultimate strain
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
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
1β
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
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
��$�'"�� ก����$%��&��������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
��$�'"�� ก����$%��&��������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
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
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
=φ
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 = − −
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
=φ
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 ��.
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 ��.
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
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
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)
� 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
����������� ������������
L L
wL2/8
wL2/14 wL2/14d
d d
L L<< L
������ �����������������ก��ก��
d
d d
������ ������������ !"���#���� �$�
Small -M
Critical sectionat face of supports
��%����ก��ก��$����#��#��ก���$�&��!��ก
- ����ก���� � � �������������
- ����ก���ก�����������������
#�� ก#������� :
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
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
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
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
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
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
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′ = =
�� �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
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
xε
′ ′ = −
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′ ′=
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��������$�����#��ก��(��:
222
25.4 10014.4 cm
( ) 4.0(50 6)n
sy
MA
f d d
×= = =
′− −
��������ก�����$��������: 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′ ′ ′= → = = =
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� ����������
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 �!�����
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
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
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:
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"� �"����(���)$��ก*� 0.85 c E
sy
f b tA
f
′≤
("��%� ��� a +��("���ก 0.85
s y
c E
A fa t
f b= ≤
′
ก�����"� +$�$�(, ( / 2) ( / 2)nM T d a C d a= − = −
n 1 2M C (d a / 2) C (d t / 2)= − + −
2
c w
T Ca
0.85 f b−
=′
ก������ 2 : a > t ������"� �"����"��(�� T � ������������ 2 %������ ������ A1 "� �"���� C1
�����ก��� A2 "� �"���� C2
t
bE
As
bw
aA1A2/2 A2/2
Asw
bw
aA1t
bE
Asf
bw
A2/2 A2/2
1 c wC 0.85 f b a′= 2 c E wC 0.85 f (b b ) t′= −
s y 1 2T A f C C= = +
��������� 4.1 �������#ก�����"� �"���� Mn !�����"��(�� T �$�����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$.
→→→→ ��+�)� → ������ bE = 200 %�.
(2) ���������� a -*� ��)��./� a ≤≤≤≤ t
T = fy As = 4.0(40.52) = 162 (��
1623.97
0.85 0.85 0.24 200c E
Ta
f b= = =
′ × ×2$. < 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≥ = =
(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 ′= −
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 ]
(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
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= − = − =
(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$.2
ขอสอบภย
ขอที่ : 50
คานรูปตวัทีโดดๆ มีปกคานกวาง = 80 ซม. หนา = 8 ซม. ตัวคานกวาง = 25 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As = 7.0 ซม.2 ที่ความลึกประสิทธิผล d = 40 ซม. ถาใช fc = 45 กก./ซม.2 และ fs= 1200 กก./ซม.2 จะพบวาตําแหนงแนวแกนสะเทินอยูใตปกคาน ดังนั้น หากสมมุติใหตําแหนงของแรงอัดที่ไดจากคอนกรีตอยูที่กึ่งกลางความหนาของปกคาน จงประมาณคาโมเมนตตานทานปลอดภัยของคานนี้
80 cm
25 cm
As = 7.0 cm2
d = 40 cm
8 cm
jd = 40 – 4 = 36 cm
�ก�%���*�����.(���ก��� ������"� �"��������"��(����4 cm
kd = 12 cm
1 2 c w c E w
1 2kd tC C f b kd f (b b )t
2 2kd−
+ = + −
1
1C 45 25 12 6,750 kg
2= × × × =
2
2 12 8C 45 (80 25) 8
2 12× −
= × × − ××
= 13,200 kg
T = 7.0×1200 = 8,400 kg < C1+C2 � Control
M = 8,400×36/100 = 3,024 kg-m
ขอสอบภย
ขอที่ : 51
คานรูปตวัทีโดดๆ มีปกคานกวาง = 80 ซม. หนา = 8 ซม. ตัวคานกวาง = 25 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As ทีค่วามลึกประสิทธิผล d = 40 ซม. ถาใช fc = 45 กก./ซม.2 และ fs = 1200 กก./ซม.2 จงประมาณคา min As ที่ตองใชตามมาตรฐานกาํหนด
80 cm
25 cm
min As
d = 40 cm
8 cm
= 5.83 cm2
s Wy
14min A b d
f=
1425 40
2400= × ×
ขอสอบภย
ขอที่ : 57
คานรูปตวัทีโดดๆ มีความกวางประสทิธิผลของปกคาน = 120 ซม. หนา = 8 ซม. ตัวคานกวาง = 30 ซม. เสริมเหล็กรับแรงดึงอยางเดียว As = 48.24 ซม.2 ที่ความลึกประสิทธิผล d = 50 ซม. เพื่อรับโมเมนตประลัย (Mu) ชนิดบวก = 50 ตัน-เมตร ถาใช fc’ = 200 กก./ซม.2 และ fy = 3000 กก./ซม.2 จงใชวิธี USD ประมาณคา As ที่ตองใช
120 cm
30 cm
As = ?
d = 50 cm
8 cm ��)�� a = t = 8 cm
C = 0.85 f’c bE t
= 0.85×200×120×8/1,000 = 163.2 ton
Mn = C(d-a/2) = 163.2(50-8/2)/100 = 75 t-m
Mu/φ = 50/0.9 = 55.6 t-m < Mn a < t
�*����� ���(��%��� ����$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
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
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
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
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= −
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 =
�������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 : ��"�������� #$��* ��&ก'� ก
SR24 : f v = 1,200 ksc
SD30 : f v = 1,500 ksc
SD40 : f v = 1,700 ksc
WSD
Example 1 : Shear design by WSD
b = 30 cm, d = 45 cm
f’c = 240 ksc, fy = 4,000 ksc
@ critical section V = 15 ton
dbf29.0V cc ′=
000,1/453024029.0 ××=
= 6.07 ton
dbf32.1V cmax ′= ton 61.27=
ton 63.16dbf795.0 c =′
cs VVV −=
ton 93.807.615 =−=
s
vvV
dfAs = cm 45.13
93.8457.157.1
=××
=
USE Stirrup DB10 @ 13 cm
��� ��ก dbf795.0V c′≤ → smax = 45/2 = 22.5 cm ≤ 60 cm
b0015.0A
s vmax = cm 9.34
300015.057.1
=×
=
Simple formula: 0.53c cV f b d′=
Shear strength with axial load:
Compression : 0.53 1 0.0071 kguc c w
g
NV f b d
A
′= +
Tension : 0.53 1 0.0029 kguc c w
g
NV f b d
A
′= +
0.50 176 0.93uc c c
u
V dV f b d f b d
M
ρ ′ ′= + ≤
Design for Shear (Design for Shear ( SDSDMM)) SDM
ก�������������ก�� :
Design for Shear (Design for Shear ( SDSDMM)) SDM
ก������ �* ���!��#��&ก������� � : Vn = Vc + Vs
ก����������$��#�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=
$�ก�� ก��� ก������ ���&ก'� ก����� ก�� : Vs = Vu / φφφφ - Vc
�����"���&ก'� ก :s
yv
V
dfAs =
ACI318: 11.4.6 ACI318: 11.4.6 –– Minimum Shear ReinforcementMinimum Shear Reinforcement SDM
- h ≤ 2.5 tf
- h ≤ 1/2 bw
11.4.6.1 – A minimum area of shear reinforcement, Av,min , shall be provided
in all reinforced concrete flexural members where Vu ≥ 0.5 φφφφ Vc, except in
members:
• Footings and solid slabs
• Concrete joist construction
• Beams with h ≤ 25 cm
• Beam integral with slabs with h ≤ 60 cm and
bw
tfh
h
b
ACI318: 11.4.6 ACI318: 11.4.6 –– Minimum Shear ReinforcementMinimum Shear Reinforcement SDM
11.4.6.3 – Where shear reinforcement is required, Av,min shall be computed
by
ycmin,v f
sbf2.0A ′=
but shall not be less than 3.5bs/fy �����#!������ ksc 306fc <′
��� ���� ��ก���ก��ก���� !b5.3
fA
bf2.0
fAs yv
c
yvmax ≤
′=
dbf1.1V cs ′≤• smax = d/2 ≤ 60 cm
dbf1.2Vdbf1.1 csc ′≤<′• smax = d/4 ≤ 30 cm
dbf1.2V cs ′>• ������� �"���!
(ก) ก����ก������กก���� �$����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.
�"���!%�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��
→ $%���&ก'� ก�� ����!��(�����"�#�ก���!��)→ ��*����������
Step 4 ������ก������ �#�ก���!�� dbf1.2V cmax,s ′=
�� Vs > Vs, max �!��"��������#�*���+#",��, → ,��#*����������
Step 3 ������ก������ ������ ก����ก��&ก'� ก cs VVV −=
SDM
ก������ ������ ก�� Vn = Vu / φφφφ
Step 6 �����"���&ก'� ก#�ก���!��b5.3
fA
bf2.0
fAs yv
c
yvmax ≤
′=
�� dbf1.1V cs ′≤ → smax = d/2 ≤ 60 cm
�� dbf1.2Vdbf1.1 csc ′≤<′ → smax = d/4 ≤ 30 cm
single closed loop stirrup has 2 legs
('� ก�����#�! *�)
Av = 2 As : ,�-������&ก'� ก$�������� �
�����"���&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
��%�������� 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
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
DB10@0.11m DB10@0.11m
3.7 m
DB10@0.25m
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���$�%�
� 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
���������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
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 RB9@0.11m. 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
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
RB9@0.11 RB9@0.15 RB9@0.29
Shear2_15
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
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
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
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
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
��������� 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
( )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
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
DB12@0.18
DB12@0.12
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= = =
#$ ���� 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 ���-����
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
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
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
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
Mρ
′ ′= + ≤
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′ ′≤ ≤
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
��� ��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
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
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
�)����������*����($��L / 24Ln
�)����������+��*���L / 28Ln
�����L / 10Ln
���"ก���#�$%��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����������*����($�� tmin = L/24
�,����)(�������� fy = 2,400 ksc ��� 2,400
0.4 0.747,000
+ =
min270
t 0.74 8.3 cm24
= × =
����ก��� ���� t = 10 cm
1��,ก��
WSD
= 0.1 x 2,400 = 240 kg/m2
1��,ก�,+(#��;�� = 50 kg/m2
1��,ก�� = 300 kg/m2
WSD1��,ก��� = 240 + 50 + 300 = 590 kg/m2
<����=.$�����,(��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
���"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����������*����($�� tmin = L/24
�,����)(�������� fy = 2,400 ksc ��� 2,400
0.4 0.747,000
+ =
min270
t 0.74 8.3 cm24
= × =
����ก��� ���� t = 10 cm
1��,ก�� = 0.1 x 2,400 = 240 kg/m2
1��,ก�,+(#��;�� = 50 kg/m2
1��,ก�� = 300 kg/m2
SDM
1��,ก���),� wu = 1.4×(240+50) + 1.7×300 = 916 kg/m2
��������)-ก�+�����ก.$�+#( (����� 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 !�.
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,�, $���3,�)
RB9 @ 0.10 m As = 6.36 cm2
31L
41L
81L
L1
Temp. steel
4.0 �.
.13 �.
1.0 �. 1.3 �.
RB9@0.10
RB9@0.18 RB9@0.07
4.0 �.
.13 �.
1.0 �. 1.3 �.RB9@0.07) ��������������
RB9@0.18 RB9@0.14 �������AB
���� ���� -.ก�����
Floor Plan
����������������������������
1/14 1/16 1/16
1/24
1/12 1/12 1/12
&���,����� 9.2 ������ก/00��.���($��+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 RB9@0.08 : As = 0.636×100/8 = 7.95 cm2/m > Required As OK
Select RB9@0.20 : 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
�������� ก�""���,�#�+����.,� ��ก���� �+��4.0 m
S1S2
1.5 m
5 m
0.50
0.10
0.20 3.80 0.20
DB10@0.20 ) ��������������DB10@0.40 �������AB
DB10@0.40 �������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 DB10@0.17 : As = 0.785×100/17 = 4.62 cm2/m > Required As OK
Use DB10@0.20 : As = 0.785×100/29 = 2.71 cm2/m
Temp. steel = 0.0018×100×15 = 2.70 cm2/m
0.50
0.10
0.20 3.80 0.20
DB10@0.20 ) ��������������DB10@0.40 �������AB
DB10@0.40 �������AB
0.951.30
0.55
0.95
1.5 m
t
4.0 m1.5 m
0.15 0.10
DB10@0.17
DB10@0.20#
DB10@0.20 DB10@0.34 �������AB
DB10@0.17 ) ��������������
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
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
×= =
×
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
t
SL
Simple supportson all four edges
ก�����������������
ก ���������ก����������������� � �ก ������� ��� !��ก����������� ����
�� ����� S �"#�� �� � L
ก����������������
$#��$ �% �&�������'(����ก�� � b = 1 ��� ����������� ������ ����� S �"#
�� �� � L ������ก�������������ก��������������
L
S
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
�� ��!"����� �#��$!�!���
����������ก��� !��"����������ก��� !��"
��#� �� ( !�������������) ��&�! !������������� �
��&�! !������������� ��&�! !������������� ��&�! !�������������
����!"����� �#��$!�!��� ( 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
����!"����� �#��$!�!��� ( 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
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���,������������
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)
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)
0.50
0.10
0.20 4.80 0.20
DB10@0.20 ��� ������������DB10@0.40 ��������2
DB10@0.40 ��������2
���������
0.701.20
1.201.60
0.50
0.10
0.20 3.80 0.20
DB10@0.20 ��� ������������DB10@0.40 ��������2
DB10@0.40 ��������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�������*���">������������
ขอสอบภย
ขอที ่: 131
พื้น S1 ขนาด 5x5 เมตร หนา 12 ซม. เหล็กเสริมโมเมนตบวก(เหล็กลาง)กลางแผนพืน้กําหนดใหเทากับ RB12@0.15# 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) �������
�����-.������-.�
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����-.�
���������� (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
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
������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 ���
�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��"�?�
� ������������ �������������� �
� ��� ��������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����
���������������
������������
� ����ก���ก�����������
� ������ก������ � �� (DL) = ����� � + ������ �
� ������ก�� (LL) "��������ก�����
� ������ก��#�$%&'�� (SDL) : ������, ����ก����, ��
� ������ก#���%��ก�����) ��� ������ �, '���, �������
� ����ก����������
��������� � ����ก (ก.ก./ .2)%&�����%�������� 1,658 t
��*+ ���,��������� 2.5 *�. 55
����������� 2.5 *�. 80
��%&�����������������+(��� 2,645 t) 55
��%&����ก���� (��� 2,800 t) 60
��%&������� 2,400 t
��%&���%&� 2,550 t
��%&%� ก���� 1/2” 15
�����#������ก�$������%�& �����
������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����$ก�'���� wu = 1.4(280+180)+1.7(300) = 1154 กก./�.2
�����<ก d = 10 – 2 – 0.45 = 7.55 *�.
������ก����� � 0.10(2400) = 280 กก./�.22 215 2525
+
=� ���,��ก��������� �+��� Mu = 1154(2.0)2/8 = 577 กก.-�.
%����? ��;ก #��� As = 0.0048(100)(7.55) = 3.62 *�.2/����ก���� 1 ���
#�+�,&��-./��
��� @��%������������ก���� 1 ��� Vu = wL/2 = 1154(2.0)/2 = 1154 กก.
ก����������� @��������ก�+�:
un 2
MR
bd=
φ2
2
577 10011.25 kg/cm
0.90 100 7.55×
= =× ×
c n
y c
0.85 f 2R1 1 0.0048
f 0.85 f
′ρ = − − = ′
ρρρρmin < ρρρρ < ρρρρmax OK
-0/�ก1�$-�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�$�
�����#������'�
��������ก
����������
��������
L
w
L
w
w w
H
L = Projected length
w
3�����3�������#������'�
�������%���B��������� ����#����#�+���%����#�ก
�4�'�%�
�*�,��&ก��� (*�.) ≤ 20 ≤ 19
�&ก��� (*�.) ≥ 22 ≥ 24
����#&� H (�.) ≤ 3.0 ≤ 4.0
������� L (�.) ≤ 4.0 ≤ 4.0
����ก���� (�.) ≥ 0.9 ≥ 1.5
��&ก��� � (*�.) 2 – 2.5 2 – 2.5
ก�5 -�0�������-�6�%�
��� %C�������� �+������������ ���� �<�ก����กD?�=���#����
t
0.025
�&ก��� 25-30 *�.
��&ก��� �
�&ก���15-18 *�.
t
�����������
t0.025
�&ก���
��&ก��� �
�&ก���
�����������
����3��#/ ������&0�3��3� ������
����'���� 8.4 ����ก������ ����������� 2.50 ��� ���������������������
��ก ������������ก�� 300 กก./��.�. �����กก���� 1.0 ��� ������ �ก���� 25 *�.
#����ก����� 15 *�. f’c= 240 กก./��.*�. fy = 4,000 กก./��.*�.
��*��� ����� ���� 250/20 = 12.5 *�.
�����<ก d = 12 – 2 – 1.2/2 = 9.4 *�.
������ก������ � = 0.5(0.15)(2400) = 180 กก./�.2
������ก����$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������ @��
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
��������� �
�����&��'/��+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./ �.
0.25 m
1.50 m
262 kg/m
21(262)(1.5) 295 kg-m
2uM = =
=� ���,��ก�+�#$�"�������:
22 2
295(100)6.53 kg/cm
0.90(10)(22.4)u
n
MR
bdφ= = =
#�+�,-�02ก-��� :
ρmin = 14/fy = 14/4000 = 0.0035
1
61200.85 0.0262
6120c
by y
f
f fρ β
′= = +
ρmax = 0.75ρb = 0.75(0.0262) = 0.0197
min
0.85 21 1 0.0017
0.85c n
y c
f R
f fρ ρ
′= − − = < ′
USE ρmin
%����? ��;ก #��� As = 0.0035(10)(22.4) = 0.784 *�.2/������ � 1 ���
��ก"�� ��;ก #��� 1 DB12 �.�. (As = 1.13 *�.2)
#�+�,&��-./��:
��� @��%����� Vu = wL = 262(1.5) = 393 กก.
ก����������� @��������ก�+�:
φVc = 0.85(0.53)(√ 240)(10)(22.4) = 1563 กก. > 2Vu OK
0.10
1.50 0.40
0.800.15
��;ก�����<� RB9@0.20
��;ก #������ก 1DB12
��;ก�<������� �RB9 @ 0.20
��;ก�����<����1RB9 �$ก�$�
0.25
0.10
0.10
0.150.25
��;ก #������ก 1DB12
&���'0�-��'�ก�-��� -�02ก�����'/��+ก%����ก�&#�
Assignment: +���ก&�������1����3$�0�
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
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��#����!�%$�����&� �ก�����+�#� ,������ #������ก�ก��-������%�ก����ก�����.,������ก���
����������ก2
4s s s
dA f f
π= =
�������� �� d Luπ=
�������� �� = ����������ก2
4 s
dd Lu f
ππ =
���������������� �����ก�����4
sd fL
u=
L
T = As fs
Anchorage Bond
u = ������������� ��#� *��!������ก����� (กก./2�.2)
WSD
Bond Stress Based on Simple Cracked Analysis
jd
T + dT
V
C + dC
T
V
C
dxO
[ ΣΣΣΣMO = 0 ]
T T + dT
u
1
[ ΣΣΣΣFx = 0 ]
u Σo dx = dT 2
21
Elastic CrackedSection Equation
Flexural Bond
jd(dT) V(dx)=
dT Vdx jd
=
WSD
ΣΣΣΣo = *����!��������)������ก����������� (2�.)
djV
uoΣ
=
ขอสอบภย
ขอที่ : 129
เหล็กเสริม DB12 SD30 ฝงในเน้ือคอนกรีตลึก 50 cm กําหนดให หนวยแรงยึดเหน่ียวที่ยอมให 11 ksc เมื่อออกแบบการรับแรงดึงโดยวิธีหนวยแรงใชงาน เหล็กเสริมจะรับแรงดึงสูงสุดทีย่อมใหไดเทาไหร
ขอสอบภย
ขอที่ : 130
จงหาหนวยแรงยดึเหน่ียวท่ีเกิดของจุดตอระหวางคานกับเสา เมื่อคานรับแรงเฉือนที่จุดตอ V=6555 kg มีคา jd=39.735 cm มีเหล็กเสริมที่พิจารณาในการคํานวณแรงยึดเหน่ียวคือ 4 เสน ขนาด RB15
= 8.75 ksc
Control
2
s
dT f
4π
=
21.21,500
4π×
= ×
= 1,696 kg
T dLu= π
1.2 50 11= π× × ×
= 2,073 kg
o
Vu
jd=Σ
65554 1.5 39.735
=× π× ×
WSD���������� ����� �.�.�. 1007-34
djV
uoΣ
=
�����5�������)���� 2 �������ก�������)����!���ก�)*����)����� ������������������ก�%�ก����#� �������6�����%�ก
�&( �,7��ก��ก����8!������ก����� ��(�ก��,����ก��( ��%�ก��������� ����ก�������)����#� �������6 %$��������������$�$� � L #� �&���&�
L
TL
Tu
oΣ=
���� ������)����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
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
=
��������������� ��!��"��#���� ������$������
Actual Distribution of Flexural Bond Stress
CL
Actual T
MT
jd=
Actual u
o
Vu
jd=Σ
�����%�&�'����()�!���������������������� ����'��"���������� �!�ก��������#��*���'������'��
Bond Stress in a Pull-out Test
T
fs = T/A
Bar stress = fs Bond stress = u
����ก��#��)ก���������� ���$���������ก������$���ก��� ���������&������ก���
,> 1950 ����%�ก��+�%$���ก��#��)����#�
%�กก��#��)%$&)���������������� �������.����#� ��������������ก����
ก��������� ���� !ก"�##�#
uT
�����ก�$#��������$�������#���$���������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���
Radial componentof bearing pressure
Circumferentialtensile stresses
ก��ก�'$�������$�กก������� �%#�ก��&
���5,�$ก�)!�������#�@��@��%�ก����ก����#������ก����������ก�$%����ก.,���
���ก�������)
Cylindrical zones of
circumferential tension��������%$�������$�$#��
����+�%��������$�$�����$���������ก������$�$�$!�)#� �&���&�
Minimum Bar Covering and Spacing
cb
Minimum bar covering ( cb )
cs
Minimum bar spacing ( 2cs )
2cs
Splitting of concrete along reinforcement
(�!!��� ��+�#�ก���#ก������� splitting
�$�$�����$���������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���
�$�$�=����$�$�$��������ก�����
����ก,��ก
SDM����ก)� *� �.�.�. 1008-38
�������� �� ld = �������� ��&(+�A�� ldb ×××× ����8:,��)���
'������������ �����ก ������������
�������� ��&(+�A�� b ydb
c
0.06 A f
f=
′ℓ ������)����ก!��.���ก�� 36 �.�.
����8:,��)���%$!+�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=
+ ′
ℓ
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 =
(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��������$���� ��#��,?�)�����%����8�������������ก�������:����
���� �� ��ก������� (cb + Ktr) / db = 1.5
DB20 and smaller DB25 and larger
Case A:(1) covering = db
clear c-c = db
min. stirrup
(2) covering = db
clear c-c = 2db
(A - 1) (A - 2)
bc
ety df
f15.0
′
ψψb
c
ety df
f19.0
′
ψψ
Case B: others
(B - 1) (B - 2)
bc
ety df
f23.0
′
ψψb
c
ety df
f28.0
′
ψψ
SDM&����%��(�������
������) f’c = 240 กก./2�.2 ��$ 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
��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. '����$'������������3��&*��6#����������� y t e sd b
b trc
b
0.28 fd
c Kfd
ψ ψ ψ=
+ ′
ℓ
cb �(����#� ����ก����$�����
(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�.
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
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
��������������������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����$����������ก���� (D)
�.�.�. 6 ��. 1� 25 ��.28 ��. 1� 36 ��.44 ��. 1� 57 ��.
6 db
8 db
10 db
ACI DB10 – DB25
DB28 – DB36
DB40 – DB60
8 db
10 db
12 db
ก������� !� �����"����ก���ก �.�.�.
Tdb D
6db
db DT
12db
Tdb
6db
135o
"������ ���ก���ก���6 ��. 1� 16 ��.
"������ ���ก���ก���20 ��. 1� 25 ��.
"������ ���ก���ก���6 ��. 1� 25 ��.
�������������� ���ก����������� SDM�.�.�.
�����������"������ ���ก��������������',���45�ก��������������#$��%�
$�ก���������������&%ก������&%����� � �� ����!� ����ก� �� �*+,��กก� ���� 8 db
��- 20 '�.
LdLdh
Combined actions: - Bond along straight length
- Anchorage provided by hook
Tdb
Critical sectionfull bar tension
Tdb
Ldh
12db
4db
4db ≥ 6 cm
�������������� ���ก����������� SDM�.�.�.
������������������"������ ���ก�����������',� fy = 4,000 กก./'�.2
�.�.�.
ACI 318-08
bhb
c
320 d
f=
′ℓ
e yhb b
c
0.075 fd
f
ψ=
′ℓ
����&%����� ����"8��ก��/9�����+���+�
���� ก�������ก����ก��� *+, fy �+� �� ��$�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 '�.
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) $-/9�ก��1 �� *����-�� �� ���ก "��8��/��-�-
*�� /9�ก�� ���ก���!� ก#� 36 ��.
ก��� �*�� ���ก "��*+,����� ���ก�� /��/9��������*+,����ก������� �- "��
��- �#,��-�-���$����� "��/������ "���������- 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�
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 ���*+,$4���������9 �� +,����-�����#"�-
��������,�
���ก "�#� 2 "��
���ก "�#� 1 "��
0
d ���� 12db d ���� 12db
Required moment Mu
ก����� ���ก ���� %&'�������������$��� %&��
���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=
ℓ
$-!�� ��������������ก*+,"4*+,���/�����
nd
u
MVφ
=ℓ
ก����� ���ก ���� %&'�������������$��� %&��
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
���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
�������� 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
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
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
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
DB10@0.20m
40 cm
60 cm
Section B-B
2DB25
4DB25
DB10@0.20m
3DB28
� 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
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
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��
(�) 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= × ×
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
�������� 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
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
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� $ก
1 1t
x yA 2A
s+ =
ℓ
....(44-19)
���$,�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
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)
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=
� 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
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Solid Hollow
0 1 2 3Percent of torsional reinforcement
0
Tn
Tcr solid section
Tcr hollow section
Torsional Strength of Reinforced Concrete SDM
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�� ��$�ก�� �����ก����"% �"ก�����ก�� ���������������ก�����
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
�� ��$� �����!�����ก��� cp cpcr c
cp
2A 0.75AT 1.06 f 2
3 p
′= × ×
2cp
cr ccp
AT 1.06 f
p
′=
������ #��)��� ก������+�� 4/TT cru φ≤
���#'���ก�� �����ก���� φ = 0.85
( )2cp
ccp
A0.265 f
p
′≤ φ
ตัวอยางที ่7.1 คานยื่นรับน้ําหนักบรรทกุประลัย 3 ตันที่มุมหนาตัดหางจาก
ศูนยกลางหนาตัด 15 ซม. ตรวจสอบดูวาจําเปนตองคิดผลของการบิดในการออกแบบ
หรือไม กําหนด f’c = 240 กก./ซม.2
��������'������� pcp = 2(60+30) = 180 /�.
-+$���������� 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
+
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 ′=
Torsional Crack
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����ก���ก�9�
A0h = -+$�������
T
45o
Torsional crack
Txo
yo
����ก���ก�����ก���
θV1V2
V3
V4
����ก���������
�)��#��ก������� ����� ���
Space Truss Analogy
ก�����*+�� q ,�)�� ���"�&'ก��� ��ก��(��� �*+�� V1 V2 V3 ��� V4 �������
����� #� &�ก�������� ,��'�
! "��#����������������$%�: 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=
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
=φ
-+$��������ก��������������� ,���-����%$��-+�������ก���� yvth
y
fAA p
s f
=
ℓ
ℓ
���������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 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 �.�.
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
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< ������&�����!&��!�
Torsional reinforcement
Combined shear & torsion stirrup
t u
0 yv
A Ts 2 A f
=φ
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
=
ℓ
ℓ
= 0.0575 × 148 = 8.51 cm2
yvc t,min cp h
y y
f1.33 f AA A p
f s f
′= −
ℓ
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
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)
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�����
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What is COLUMN ?What is COLUMN ?- vertical member ?
- axial compression ?
- carrying floor load ?
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)
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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
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
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.
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
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
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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bmin = 20 cm
ρρρρg = Ast / Ag 0.01 ≤≤≤≤ ρρρρg ≤≤≤≤ 0.08
4DB12 6DB12
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f2
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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
[ 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
��*�����+��()���� �.�. . 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
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
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
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 − +
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]
4π
= × × × × × − × + × ×
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
�"�"�����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 ]
4π
= × × × × − ρ + × ρ
�"�"���������"����&��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
4π
= × = 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
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
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
��������������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
���������������� 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)(
)( −+=
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
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
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 −′′+′=
��������������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′=
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��������!�����"���:
����#$��%&�": e = Mn / Pn
Small e →→→→ fs < fy when εc = εcu = 0.003 (compression failure)
Large e →→→→ fs = fy when εc = εcu = 0.003 (tension failure)
Pn
ePn
ePn
ePn
e
Small Eccentricity Large Eccentricity
Tension & Compression Failure
Pn
Mn
P0
e=
0
esm
all
Compressionfailure range
eebb : Balance failure: Balance failure
e large Tension failure range
e = ∞
�&�'��()��� ��*� (Interaction diagram) SDM
���#����(� e *�3���'�'��#)����(� � Pn ��� Mn
��� ����(����(����, ��'��ก�����(� e �(��3 ก,��!�" ก��+1 �'�(&���)����*��+"
(Mn, Pn)
�"��&�'��$'�
e =M n
/P n
b b be M /P=
ก���������������� (Balanced failure) SDM
cb
d
d′
yε
s′εcuε
�� �/���ก��������4)��#�,ก�����������)�2)��0����ก εεεεy ��" �ก��� ก�'�2�ก ����ก$'�#(��ก�����#�����0� εεεεcu = 0.003
cub
cu y
c dε
=ε + ε y
6,120d
6,120 f=
+
b 1 ba c= β
bs y
b
c df 6,120 f
c
′−′ = ≤
b c b s s s yP 0.85 f a b A f A f′ ′ ′= + −
bb c b s s s y
ah h hM 0.85 f a b A f d A f d
2 2 2 2 ′ ′ ′ ′= − + − + −
Case 1: e < eb
εcu
cbεyMb
fs < fy
Case 2: e > e b
ก����+����ก����������,-� ebSDM
Compression Failure Tension Failure
M < Mb
εy
εcu
cb
Mb
M > Mb
c > cb εs < εy fs > fyc < cb εs > εy
SDM����+��ก���%����ก%����������
�� 25 x 40 4�. ����#�,ก�� 4DB28 ���$'� 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 ��
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�.
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�����*��
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)
��#�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
=
′ =
=
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
������������ ( h ) WSD
*#"2� �(������( ��(����#�(����%��(����%+1������� ����� ��� ก��กก� '����( !+'%
h
������%!�"��
�����ก��%2)��+:#����
h
������%!�"��
�����ก��%2)���%
h
������%�'��� ����
�����ก��%2)�$" ���
ขอสอบภย
ขอที่ : 233
เสาปลอกเดี่ยวขนาด 30 x 30 ซม. อยูในเฟรมที่เซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 6.0 ม. ใหใชวิธี WSD ประมาณคาตวัคูณลดคา R
= 1.32 – 0 .006 × 66.7 = 0.92
7.66303.0
600rh
=×
= 60 < h / r ≤≤≤≤ 100 ��ก�����!�/0��1ก����" R
R = 1.32 – 0.006 h / r ≤ 1.0
ขอสอบภย
ขอที่ : 234
เสาปลอกเดี่ยวขนาด 40 x 40 ซม. อยูในเฟรมที่เซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 6.0 ม. ใหใชวิธี WSD ประมาณคาตวัคูณลดคา R
50403.0
600rh
=×
= h / r < 60 R = 1.00 .��������ก�����
ขอสอบภย
ขอที่ : 235
เสาปลอกเดี่ยวสี่เหลี่ยมจตุรัส อยูในเฟรมท่ีเซไมได เสาน้ีจะโกงสองทาง ความยาวปราศจากการค้ํายันของเสาคือ 8.0 ม. ใหใชวิธี WSD ประมาณขนาดอยางนอยของเสาตนน้ีที่จะถือวาเปนเสาสั้น
h / r = 60 60b3.0
800rh
=×
= b = 44.4 cm Use 50 x 50 4�.
�/ก������ก%�����%����������� ( R ) WSD
���#�����$'��'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
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< −
ℓ
���#�������$'��'ก��4 !�(�" ������� �������������
uk22
r<
ℓ
ขอสอบภย
ขอที่ : 239
เสาปลอกเดี่ยวขนาด 50 x 50 ซม. อยูในโครงเฟรมทีเ่ซได ถาพบวาคา effective length factor เทากับ 1.5 ดังน้ัน ชวงความยาวเสาปราศจากการค้ํายันควรเปนเทาใดตามวิธี USD จึงจะเปนเสาสั้น
����% :
ขอสอบภย
ขอที่ : 240
เสาปลอกเกลียวขนาดเสนผาศูนยกลาง 50 ซม. อยูในโครงเฟรมที่เซได ถาพบวาคา effective length factor เทากับ 1.5 ดังน้ัน ชวงความยาวเสาปราศจากการค้ํายันควรเปนเทาใดตามวิธี USD จึงจะเปนเสาสั้น
r = 0.25 x 50 = 12.5 4�.
r = 0.3 x 50 = 15 4�.
uk22
r<
ℓ u1.522
15×
=ℓ
= 2.20 mu 220 cm=ℓ
u1.522
12.5×
=ℓ����% : uk
22r
<ℓ
u 183 cm=ℓ = 1.83 m
� ���������ก
� ��� ���ก�������������ก
� ��ก ������ ก���ก�����������
� ��ก ��� �
� ��ก ������ �!
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Reinforced Concrete DesignReinforced Concrete Design
Design of Footing 1Design of Footing 1
"�� ��.�.$�%& '�! ����
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�����ก���������� ������������������ก��������������� � ��������������������� �
�����!���"����# �$ก �ก��� ��% ���� �����ก����&�%�� �ก$ก �ก�����ก�����
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ก����&�%��$���)ก��$����# �(��$ก �ก����&�
%������%ก%���ก�� (differential settlement)
*+��',���(���������$ก ��� $������
Wall
Property line
�����������ก
-� ���������ก',�+��ก����ก./,���� � ��,$����������ก������
�����ก������� (wall footing)
0���
�����ก
w�������ก�0�� ���$� �
%� ��
�����ก
�ก������������ก���$���� ��������ก#!�����������'�ก0���
�����ก��������������� (spread footing)
�ก������������ก������ ��������ก#!�������� 1���'�ก$����+��%��
P
�����ก���� (combined footing)
(-�����������ก'�ก$�����%������ �กก��� *+������(ก��$���ก��
P
P
A B
Rectangular, PA = PB
A B
Rectangular, PA < PB
Property line
A B
Rectangular, PA < PB
Property line
A B
Strap or Cantilever
Property line
�����ก������������
�����ก!��� ��������0��(mat footing)
Pile cap
PilesWeak soil
Bearing stratum
�����ก��$��$�2 (pile cap)
�����ก��������
���ก����� �ก����ก��������ก
�����ก��ก��%$�� $��2ก*+��0 ���ก��%�� 0��� �%���$��� �������(-��,�,�&�
��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
� �$�����
P
HeaveHeave
� �����
������������#$��%!&�������
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��,������������ ����$ก ��+�� ก���������������ก���� ������ (�� qa = 8.0 %��/%�. .
0.4 m
1.1 m 30x30 cm column
P = 50 ton !�(��� � &% �������ก�����ก 10%
����������������ก���%���ก��:
0.810.150
A d'req×
= = 6.88 %�. .
(-������ก���� 2.7 × 2.7 . (A = 7.29 %�. .)
���(-������ก��� 0.4 . �������ก�����ก: W = 0.4×2.72×2.4 = 7.0 %��
������������+������� ����� �: q = (50 + 7) / 2.72 = 7.8 %��/%�. . < qa OK
������������+���&�; ���� �: qn = 50 / 2.72 = 6.9 %��/%�. .
�������� 12.1 ก�������ก���� ���� �.�.�. ����������� �.�. 2522
)*�+���� ก�& ���ก��(� �/�..$.)
*�����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≤
���������������#$�!���+�+���$��()��*�����
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
� !��/��� 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������������ $�����
a = 0.90 – 0.40 = 0.50 .
min 2
P 6Mq
BL BL= −
2
80 6 80 0.151.2 1.8 1.2 1.8
× ×= −
× ×
= 37.0 – 18.5 = 18.5 %��/%�. .
qmax = 37.0 + 18.5 = 55.5 %��/%�. .
max
2Pq
3ab= 2 80
3 0.5 1.2×
=× ×
= 88.9 %��/%�. .
1-m slice on whichdesign is based
Wall
Footing
wUniformly loaded wall
w
Bending deformation
�����ก������� (Wall footing)
Concrete column,
pedestal or wall
Critical section
Column with steel
base plate
s
s/2
Critical section
���������ก,�������������*���
����%��� ก5%�������������%� ������ก�����
����%��� ก5%����ก+��ก����,����������,1���6ก���ก�����ก��� �
����%��� ก5%����ก+��ก����,��������%� ����,����0��$��2ก���(%�$��
Masonry wall
b/2 b/2
b/4
Factored wall load = wu t/m
Factored soil pressure, qu = (wu )/L
Required L = (wDL+wLL)/qa
qa = Allowable soil pressure, t/m2
221 1
( )2 2 8u u u
L bM q q L b
− = = −
2u u
L bV q d
− = −
Min t = 15 cm for footing on soil, 30 cm for footing on piles
Min As = (14 / fy ) (100 cm) d
�����*�-+����.��� ������ก�������
b
d
wu = 1.4wDL+1.7wLL
d
L
qu
� !��/��� 13.3 ��ก��������ก���0���$��������������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�����$7����������%����ก��%: c cV 0.29 f bd′=
30.29 240 100 22 /10= × ×
= 9.88 %�� > V OK
WSD��ก�����&7ก���$ �"$�$���� �
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
−
−= 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
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��$7����,�& (punching shear)
���� �!�ก6���� �"$�$���� �
���� �!�ก6���� �ก�5���
Footing Type
One-way
Two-way
Square Footing Rectangular Footing
L
B
s (typ.)
L
L
s (typ.)
L
B
B/2 B/2
As2As2 As1
AsL
AsB
s (t
yp.)
1
12
21
2
s sL
sL ss
A A
A AA
LB
β
β
= +
−=
=
ก��ก�+/����-0ก��������ก�����
c1 + d
c2c2 + d
c1
d/2
b0
����.������!�� (����.���!+-%)%� ��$'�,�,�&�����ก�������$����$!:���!!P�� � ����%��� ก5%(-����$��������!����'�ก%� ����ก �$!:��,�, d / 2
P
ก�& ��5����*&�:
SDM
WSD dbf53.0V 0cc ′=
dbf06.1V 0cc ′=
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�����$7�����ก��%:
= 75.7 %�� > V OK
dbf53.0V 0cc ′= 30.53 240 288 32 /10= × ×
�!'���ก�5���%��� ���� �!�ก6�*�* d = 32 8$. '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 #
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
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 #
� �����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���$ก��ก��!�����ก��%���&'� ������ก
'�� 1/2 < P2/P1 < 1
��������ก ������!(���!�
P1 P2
� ���
����
��
P P
��!�� 2 ��������ก��ก��
P1 P2
� ���
����
��
��!������������)��� P1 < P2
� ���
����
��
'�� P2/P1 < 1/2 ���������ก��������ก����������ก�� !*��!(�������!
ก��ก��������������ก����
C
s
B
nm
L/2 L/2
P1 P2Rn m
q
���*���� c ���*����"+& R ���
�#�����ก P1 *�� P2 :
RsP
PPsP
n 1
21
1 =+
=
��ก�$� 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
+ −=
+
= −
+ =
ก���ก��������������ก����
L
B
Side view
�!���ก����������ก,�����%���&ก��*����"+&����#�����ก���$ก ��� �*��������������
��ก�����*��*/� !���� !� �����ก��!ก��)ก��*����������-
ก�����������ก�������ก����
ก��� ��!���0ก!�#��������� *������ �#� !�#�������*���������� �1#�ก��*/�2�!�
)!�!��&���
h
q
P2P1
V
M
ก�����������ก������
P2
q
ก�����������ก������
���������%�� �#� �#�����ก�'�����!���ก��!����ก���������"�#�����ก�����ก��
( �!ก ���� ���!��� ก d/2 ��*��������
B
Side viewc + d
d/2d/2
c
Punching shearperimeter
�� ����ก�� ก!�����ก���� �"���������� ���!�� ���������ก�� 5 �!�� )��!� ����������������������ก� ������ก 40 7!. �#����� �*��*�ก��������(�� 10 ���/��.!. ก����� f<c = 240 กก./7!.2 *�� fy = 4,000 กก./7!.2
A B
40 x 40 cm
45 x 45 cm
D = 50 tonL = 25 ton
D = 80 tonL = 40 ton
5.0 m
40 cm
R
C.G.
x
(75 120) 120(5)
3.1 m
x
x
+ =
=
��#$�%� 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
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
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����� $� As,min
����ก������0ก 12 DB20 (As = 37.68 7!.2)
3cV 0.85 1.06 240 388 52 /10φ = × × ×
6. ����� ก���./ ��0�4 "������� �*������� ���*����� qnu = 12.66 ���/��.!.
��� A : b0 = 4(40+52) = 368 7!.
Vu = 112.5 – 12.66×0.922 = 101.8 ���40+52 7!.
Pu
qnu
3cV 0.85 1.06 240 368 52 /10φ = × × ×
= 267 ��� > Vu OK
��� B : b0 = 4(45+52) = 388 7!.
Vu = 180 – 12.66×0.972 = 168.1 ���
= 282 ��� > Vu OK
7. ����� ก���./ � �� ��ก*/�2�!�*���E��� Vu,max = 113.2 ���
ก������E������(��ก��: 3cV 0.85 0.53 240 330 52 /10φ = × × ×
= 119.8 ��� > Vu OK
8. ก!����ก����������� "��������ก/� � �*���������ก!������� d/2
40x40cm
45x45cm
d/2 = 52/2 = 26 cm
A B 3.3 m
20+40+26 = 86 cm 26+45+26
= 97 cm7.0 m
3.30 m
0.60 m
PA = 112.5 ton
1.45 m0.40 m
3.30 m
0.60 m
1.425 m0.45 m
PB = 180 ton
��� A : be = 20+40+26 = 86 7!., wu = 112.5 / 3.3 = 34.1 ���/!.
Mu = 34.1×1.452/2 = 35.9 ���-!.5
2
35.9 100.9 86 52
×=
× ×2u
ndb
MR
φ= = 17.2 กก./7!.2
′
−−′
=ρc
n
y
c
f85.0R2
11f
f85.0= 0.0045
����ก������0ก 7DB20
(As = 21.98 7!.2)dbAs ρ= = 20.1 7!.2= 0.0045×86×52
��� B : be = 26+45+26 = 97 7!., wu = 180 / 3.3 = 54.5 ���/!.
Mu = 54.5×1.4252/2 = 55.3 ���-!.5
2
55.3 100.9 97 52
×=
× ×2u
ndb
MR
φ= = 23.4 กก./7!.2
′−−
′=ρ
c
n
y
c
f85.0R2
11f
f85.0= 0.0062
����ก������0ก 10DB20
(As = 31.4 7!.2)dbAs ρ= = 31.3 7!.2= 0.0062×97×52
As = 0.0018(100)(60) = 10.8 cm2
9. ����ก������+�����ก��!�ก�+��USE DB20 @ 0.15
(As = 12.56 cm2/m)
0.40 m
7.0 m
0.60 m
5.0 m0.40 m 0.45 m
0.86 m7DB20
0.97 m10DB20
10DB28
12DB20
A B
DB20 @ 0.15 m
DB20 @ 0.15 m
�����ก��������� (pile cap)
Pile cap
PilesWeak soil
Bearing stratum
�!���ก�����*�ก��������!,!��"��"� ������ก�������
��ก*��*/� ������ ���0!��ก�� ��'����#�����ก���$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�
���������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���$ก���!�������#�������! ,!��ก�� 2 ���/��.!.
��� �*��GH����!��� :
���������( �!�1ก,!��ก�� 7 �!��
��� �*��GH����!��� 600 กก./��.!.
���������( �!�1ก�ก�� 7 �!��
��� �*��GH����!��� 800 + 200L กก./��.!.
)��� L (��( �!�� � ���ก�� 7 �!��
��������!�" (bored pile)
� ���0!���� �!��' ��'����#�����ก�� ����#������1ก��
,-71��� ���0!��ก�����,-,!�'1�
ก������ ���0!��ก���� �!�ก������ก��ก��*�����
-��!��!�ก��� ��/������(�������(��
Pile cap
BED ROCK
�����ก���������#����ก���$��%&
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!*�������
�����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���$ก���!������� ���0!*�������
M
15 cm
3D
D
3D1.5D1.5D
1.5D
1.5D
3D
3D
ก����ก������ก�������
!��1���������%��� ก!�����ก:
WSD D LR
n+
=
SDM u
1.4D 1.7LR
n+
=
ก%�������������ก:
����������� ���� ���0! 3D
������� ���� ���0!'1���������ก 1.5D
)��� D (������� ���0!
1.5D
3D
1.5D
1.5D1.5D
2 PILES
1.5D
3D
1.5D3D
1.5D
3 PILES
3D1.5D
3D
1.5D
1.5D1.5D
4 PILES
3D
1.5D
3D
1.5D
1.5D1.5D
6 PILES
3D3D
1.5D
1.5D
1.5D1.5D
5 PILES
D23
D23
1.5D
1.5D
7 PILES
D23
3D
3D
3D
1.5D
3D
1.5D
1.5D1.5D
8 PILES
3D3D3D
1.5D
1.5D
1.5D1.5D
D23
D23D23
ก��!������������������ก
1.5D
3D
1.5D
1.5D1.5D
9 PILES
3D3D
3D
10 PILES
1.5D
1.5D
3D
D33
1.5D 1.5D3D 3D 3D
3D
11 PILES
1.5D
1.5D
3D
D33
1.5D 1.5D3D 3D 3D
3D
1.5D
3D
1.5D
1.5D1.5D
12 PILES
3D3D
3D
3D
ก��!������������������ก
������� �ก�$� �!��(����� 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
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
����� 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��
0.40
0.050.10
14DB20#
DB20 ������
(��ก��������������*���
��0!���� ∅ 0.40 !. ��� �.�.-���2������� 50 ������� � 4 ���
!����0� $�������ก����������0
� �����ก������ �����������
� ก����ก�����������������
� ก��������� �����
� ���������������������������
� ก���������!�ก�"#�$��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����*�����+��
���� �
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�����
����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
=
����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 �.�.)
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
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
+= ⇒ =
Box BeamFlexure Testing
Failure of BeamFailed Beam
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
�$��%������������������� �$�����&���'(�
$ ������*)�������%$�+����ก�%������*��������������*������!��ก����������$��!�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 ��%�"#�$��ก%���&ก�%%�����*#����� w !�)�
w
∆
L
45 w L384 EI
∆ =
��������:�!�����*� �������ก!������������+*��!�ก!&������:���+"������ก��
wMa Mb
∆
Ma Mb
2
0
wLM
8=( )
2
0 a b
L5M 3 M M
48 EI∆ = − +
���.����*�#�$+� (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
�+�������������*�#��$$��������5������ก���������ก�� r cf 2 f ′=
r gcr
t
f IM
y=
yt = ����!�ก�ก�������'5������%����5�
Ig = ������������������ ��$�������"�$��
�#�$��%$�������*�$ �*���+���������ก��� b ������ 5ก h: t
hy ,
2= 3
g
1I b h
12=
���)#��ก�����ก$�� ������ก�� �ก���
MM
�������ก���ก���
�������������ก���
Ig = gross moment of inertia
Icr = cracked moment
of inertia
M
Mcr
Deflection ∆
∆cr ∆e
Icr
Ig Ie
Ie = effective moment of inertia
*�#�������#����#�������� �ก���, Icr
$�������ก������ก�%������ก�����%�������$�+��ก�������� �(+"���* ���$ 1ก��%
����5�/5*�!�',ก�� �����������������, ����
As
h
b
d
�#��$����ก������� : s
xb x n A (d x)
2= − 2
s s
bx n A x n A d 0
2+ − =
32
cr s
b xI nA (d x)
3= + −
xN.A.
nAs
n = Es/Ec
$������� �
*�#�������#����#��,�����-�.�, Icr ≤≤≤≤ Ie ≤≤≤≤ Ig
��+*������������ก��������������ก��� ������������������ ��$�����!� � �!�ก
Ig '5� Icr 5"�ก�%������������*��ก���#�
3 3cr cr
e g cr ga a
M MI I 1 I I
M M
= + − ≤
Ie
Ig
Icr
1 2 3 Ma/Mcr
�����* Mcr = ���������ก���t
gr
y
If=
fr = ���, ����ก$�ก cf0.2 ′= �#�$��%���ก����"#�$��ก�ก��
Ig = ������������������ ��$�������"�$��
Ma = ���������ก��*�&��������������*���������
h
b
yt
Mcr
∆cr
Ig
(Ie)D
∆D
(Ie)D+L
MD
MD+L
∆L
∆D+L
ก�� �$����/�ก�"&����ก����� ���"&����ก/�
ก����+������ก��"�#��ก��� :
Dec
2D
aD )I(ELM
β=∆
�ก�� 5"�� ���� ��#��$$�������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���
��������������������"�$�� 3g
1I 40 60
12= × × = 720,000 /�.4
������������ก��*�&� = 8.75 ���-����
40 cm
x
N.A.
nAs
#�����������%�:
f’c = 280 กก.//�.2
cE 15,100 280= = 254,512 กก.//�.26
s
c
E 2.04 10n 8
E 254,512×
= = =
"���0�"��#�+�ก�� �� �:2x
40 8(39.27)(52 x)2
= − x = 21.8 /�.
������/� �����/����#��������ก����: Icr = Iconcrete + Isteel
3 2cr
1I 40 21.8 8(39.27)(52 21.8)
3= × × + − = 424,663 /�.4
rf 2.0 280= = 33.5 กก.//�.2
r gcr
t
f IM
y=
33.5 720,00030 100×
=×
= 8,040 กก.-����
cr
a
M 8,0400.92
M 8,750= =
3cr
a
M0.78
M =
� �
������/� �����/����%� � �4 ��:3 3
cr cre g cr g
a a
M MI I 1 I I
M M
= + − ≤
Ie = 0.78×720,000 + 0.22×424,663 = 655,026 /�.4
� � ��+������ก��"�#��ก!���$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)
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
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