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Mongkol JIRAVACHARADET Reinforced Concrete Design Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING กก Design of Slabs 1 One-way slab Two-way slab One-way slab Flat plate slab Flat slab Grid slab

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Page 1: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

Page 2: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

��� ��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

Page 3: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

Page 4: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

Page 5: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

���"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

Page 6: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

Page 7: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

���"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

Page 8: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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 !�.

Page 9: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

Page 10: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

31L

41L

81L

L1

Temp. steel

4.0 �.

.13 �.

1.0 �. 1.3 �.

[email protected]

[email protected] [email protected]

4.0 �.

.13 �.

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

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

���� ���� -.ก�����

Floor Plan

����������������������������

1/14 1/16 1/16

1/24

1/12 1/12 1/12

Page 11: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

SDM

AA

3 @ 12 m = 36 m

Section A-A

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

0.4 + 2,400/7,000 = 0.74

1) Minimum depth :

Min. h = 0.74×370/24 = 11.4 cm

USE h = 12 cm

Slab weight = 0.12×2,400 = 288 kg/m2

Assume beam + super DL :

Service DL = 350 kg/m2

Service LL = 500 kg/m2

2) Factored Load :

wu = 1.4×350 + 1.7×500 = 1,340 kg/m2

3) Max. Moment :

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

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

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

'c n

'y c

0.85 f 2R1 1 0.0082

f 0.85 f

ρ = − − =

2un 2 2

M 1,529 100R 18.63 kg/cm

bd 0.9 100 9.55×

= = =φ × ×

< ρρρρmax = 0.0341 OK

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

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

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

Page 12: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

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

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

RB9 @ 0.20 . �� ก���ก������

RB9 @ 0.08 .

�������������

RB9 @ 0.16 .

�������

RB9 @ 0.08 .

������������� + �������

12 ��.

0.95 . 1.25 .

3.7 .

0.55 . 0.95 .

������������

1 m

S

1 m S

2w SM

2=

/�����0������+���ก�,�� 1 ����

t ≥≥≥≥ S / 10

for deflection control

S

t

Page 13: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

S1S2

1.5 m

5 m

0.50

0.10

0.20 3.80 0.20

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

[email protected] �������AB

0.951.30

0.55

0.95

1.5 m

t

Min. h = 150/10 = 15 cm

USE h = 15 cm

DL = 0.15×2,400 = 360 kg/m2

LL = 200 kg/m2

wu = 1.4×360 + 1.7×200

= 844 kg/m2

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

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

< ρρρρmin = 0.0035 Use ρρρρmin

un 2

MR

bd=φ 2

949.5 1006.75 ksc

0.9 100 12.5×

= =× ×

c n

y

0.85 f 2R1 1

f 0.85 f

′ρ = − − ′

= 0.0017

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

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

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

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

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

Page 14: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

0.50

0.10

0.20 3.80 0.20

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

[email protected] �������AB

0.951.30

0.55

0.95

1.5 m

t

4.0 m1.5 m

0.15 0.10

[email protected]

[email protected]#

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

[email protected] ) ��������������

ขอสอบภย

ขอที่ : 122

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

05097.040006120

612085.0

240028085.0

b =

+××

×=ρ

0382.005097.075.0max =×=ρ

ksc 08.742807.1

24000382.0124000382.0 =

××

−×=

ρ−ρ=

c

yymax,n f7.1

f1fR

2nu dbRM φ=

bRM

dn

u

φ=

10008.749.01001500××

×= = 4.74 cm

Page 15: Design of Slabs 1 - eng.sut.ac.theng.sut.ac.th/ce/oldce/CourseOnline/430431//RC10_Slab1Way.pdf · Reinforced Concrete Design S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY

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

×= =

×