Design-of-critical-two-way-slab.docx
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Design-of-critical-two-way-slab.docx as PDF for free.
More details
- Words: 932
- Pages: 8
Loading documents preview...
Design of Critical Two-way Slab
Scale 1:100 mts
S 3m = 0.75 > 0.5, therefore TWO WAY SLAB. L 4m
For checking the type of slab: m = = GIVEN:
Short side, A: 3 m Long Side, B: 4 m f’c= 28 MPa fy= 420 MPa Φ: 0.9, from Table 421.2.1, NSCP 2015 β: 0.85f’c<30 MPa Concrete Unit Weight: 23.5 kN/m3 = 24 kN/m3 Cover: 20 mm, from Table 420.6.1.3.1, NSCP 2015 Use 12 mmϕ rebars Minimum thickness, h=
=
ln , from Table 408.3.1.1, NSCP 2015 33
4000 mm = 121.2121212 mm 33
Therefore, minimum thickness: h = 125 mm For Effective Depth: dA= h – cover – half of rebar dA=125 – 20 – 6 = 99 mm
dB= h – cover – rebar – half of rebar dB=125 – 20 – 12 - 6 = 87 mm Dead loads: from Table 204-1, NSCP 2015 Slab weight = (24 kN/m3)(0.125m) Floor Fill (Lightweight concrete per mm) Floor Finish (Ceramic or quarry tile (20 mm) on 25 mm mortar bed) Ceiling Load (plaster on tile or concrete) Electrical Load Miscellaneous Load DL Live load: (Call centers and business processing offices) From Table 201-1, NSCP 2015
LL
= 3.00 kPa = 0.015 kPa = 1.10 kPa = 0.24 kPa = 0.023 kPa = 1.00 kPa = 5.378 kPa =2 .9 kPa
Factored Load: WuDL =1.2 DL =1.2(5.378) WuDL = 6.4536 kPa WuLL WuLL Wu Wu
=1.6 LL =1.6(2.9) = 4.64 kPa =1.2 DL + 1.6 LL =1.2(5.378) + 1.6(2.9) = 11.0936 kPa
Solving Moments using Coefficient Method based on 1963 ACI Code Using CASE 4 (2 edges continuous, 2 edges discontinuous) of Moment Coefficients: CA=0.076 CB=0.024 SOLVING FOR MOMENTS: A. Short Span, Middle Strip At edge:
Ratio=
Short span 3 = =0.75 Long span 4
CADL=0.043 CBDL=0.013
CALL=0.052 CBLL=0.016
MA=(Wu) (CA) (LShort Span)2 =11.0936(0.076)(3)2 MA=7.5880 kN.m At mid-span: For dead load MADL = (WuDL) (CADL)(LShort Span)2 = 6.4536(0.043)(3)2 MADL = 2.4975 kN.m For Live load MALL=(WuLL) (CALL) (LShort Span)2 = 4.64(0.052)(3)2 MALL=2.1715 kN.m MA=MADL+MLL= 4.6690 kN.m
B. Long Span, Middle Strip At edge: MB= (Wu) (CB) (LLong Span)2 =11.0936(0.024)(4)2 MB= 4.2599 kN.m At mid-span: For dead load MBDL=(WuDL) (CBDL) (B2) = 6.4536 (0.013)(4)2 MBDL=1.3423 kN.m For Live load MBLL=(WuLL) (CBLL) (B2) = 4.64(0.016)(4)2 MBLL = 1.1878 kN.m MB=MBDL+MBLL= 2.5301 kN.m
SOLVING FOR REINFORCEMENTS: A. Short Span, Middle Strip At edge:
MA
7.5880 x 106 Rn= = = 0.8602296818 MPa ∅ b d 2 0.9∗1000∗992 0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.95795 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00209 1.4 1.4 = fy 420 ρmin =¿ 0.00333
ρmin =
0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd A s=0.00333∗1000∗99=330 mm2
Therefore use ρmin¿0.0033 for As
Using 12 mm bars:
No .of bars=
330 =2.918 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =342.719 mm 330 Spacing=
Smax= 2h Smax=2*125=250mm
Since Smax < S required : Use 3-12mm reinforcing bars @ 250mm spacing
At Mid-span:
Rn=
MA ∅ bd
= 2
4. 6690 x 106 = 0.529311068 MPa 0.9∗1000∗992
0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.52931 ρac t= [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00127
ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ρmax =0.75 [ ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmin > ρact ¿ ρmax Therefore use ρ¿ 0.0033for As
A s=ρbd A s=0.0033∗1000∗99=330 mm2 Using 12 mm bars:
No .of bars=
300 =2.918 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =342.752 mm 300 Spacing=
Smax=2h Smax=2*125=250mm Since Smax < S required : Use 3-12mm reinforcing bars @ 250mm spacing B. Long Span, Middle Strip At edge:
Rn=
MB ∅ bd
= 2
4.2599 x 106 = 0.6253431394 MPa 0.9∗1000∗872
0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.6253431394 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00151
ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd
Therefore use ρmin¿0.00333 for As
A s=0.00333∗1000∗87=290 mm2 Using 12 mm bars:
No .of bars=
290 =2.5641 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =390 mm 290 Spacing=
Smax=2h Smax=2*125=250mm
Smax < S required Use 3-12mm reinforcing bars @ 250mm spacing At Mid-span:
MA
2.5301 x 106 Rn= = = 0.3714126334 MPa ∅ b d 2 0.9∗1000∗872 0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.3714126334 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.000891 ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd A s=0.0033∗1000∗87=290 mm2
Therefore use ρmin¿0.0033 for As
Using 12 mm bars:
No .of bars=
290 =2.5642 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =390 mm 290 Spacing=
Smax=2h Smax=2*125=250mm
Smax < S required
Use 3-12mm reinforcing bars @250mm spacing
Scale 1:100 mts
S 3m = 0.75 > 0.5, therefore TWO WAY SLAB. L 4m
For checking the type of slab: m = = GIVEN:
Short side, A: 3 m Long Side, B: 4 m f’c= 28 MPa fy= 420 MPa Φ: 0.9, from Table 421.2.1, NSCP 2015 β: 0.85f’c<30 MPa Concrete Unit Weight: 23.5 kN/m3 = 24 kN/m3 Cover: 20 mm, from Table 420.6.1.3.1, NSCP 2015 Use 12 mmϕ rebars Minimum thickness, h=
=
ln , from Table 408.3.1.1, NSCP 2015 33
4000 mm = 121.2121212 mm 33
Therefore, minimum thickness: h = 125 mm For Effective Depth: dA= h – cover – half of rebar dA=125 – 20 – 6 = 99 mm
dB= h – cover – rebar – half of rebar dB=125 – 20 – 12 - 6 = 87 mm Dead loads: from Table 204-1, NSCP 2015 Slab weight = (24 kN/m3)(0.125m) Floor Fill (Lightweight concrete per mm) Floor Finish (Ceramic or quarry tile (20 mm) on 25 mm mortar bed) Ceiling Load (plaster on tile or concrete) Electrical Load Miscellaneous Load DL Live load: (Call centers and business processing offices) From Table 201-1, NSCP 2015
LL
= 3.00 kPa = 0.015 kPa = 1.10 kPa = 0.24 kPa = 0.023 kPa = 1.00 kPa = 5.378 kPa =2 .9 kPa
Factored Load: WuDL =1.2 DL =1.2(5.378) WuDL = 6.4536 kPa WuLL WuLL Wu Wu
=1.6 LL =1.6(2.9) = 4.64 kPa =1.2 DL + 1.6 LL =1.2(5.378) + 1.6(2.9) = 11.0936 kPa
Solving Moments using Coefficient Method based on 1963 ACI Code Using CASE 4 (2 edges continuous, 2 edges discontinuous) of Moment Coefficients: CA=0.076 CB=0.024 SOLVING FOR MOMENTS: A. Short Span, Middle Strip At edge:
Ratio=
Short span 3 = =0.75 Long span 4
CADL=0.043 CBDL=0.013
CALL=0.052 CBLL=0.016
MA=(Wu) (CA) (LShort Span)2 =11.0936(0.076)(3)2 MA=7.5880 kN.m At mid-span: For dead load MADL = (WuDL) (CADL)(LShort Span)2 = 6.4536(0.043)(3)2 MADL = 2.4975 kN.m For Live load MALL=(WuLL) (CALL) (LShort Span)2 = 4.64(0.052)(3)2 MALL=2.1715 kN.m MA=MADL+MLL= 4.6690 kN.m
B. Long Span, Middle Strip At edge: MB= (Wu) (CB) (LLong Span)2 =11.0936(0.024)(4)2 MB= 4.2599 kN.m At mid-span: For dead load MBDL=(WuDL) (CBDL) (B2) = 6.4536 (0.013)(4)2 MBDL=1.3423 kN.m For Live load MBLL=(WuLL) (CBLL) (B2) = 4.64(0.016)(4)2 MBLL = 1.1878 kN.m MB=MBDL+MBLL= 2.5301 kN.m
SOLVING FOR REINFORCEMENTS: A. Short Span, Middle Strip At edge:
MA
7.5880 x 106 Rn= = = 0.8602296818 MPa ∅ b d 2 0.9∗1000∗992 0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.95795 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00209 1.4 1.4 = fy 420 ρmin =¿ 0.00333
ρmin =
0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd A s=0.00333∗1000∗99=330 mm2
Therefore use ρmin¿0.0033 for As
Using 12 mm bars:
No .of bars=
330 =2.918 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =342.719 mm 330 Spacing=
Smax= 2h Smax=2*125=250mm
Since Smax < S required : Use 3-12mm reinforcing bars @ 250mm spacing
At Mid-span:
Rn=
MA ∅ bd
= 2
4. 6690 x 106 = 0.529311068 MPa 0.9∗1000∗992
0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.52931 ρac t= [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00127
ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ρmax =0.75 [ ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmin > ρact ¿ ρmax Therefore use ρ¿ 0.0033for As
A s=ρbd A s=0.0033∗1000∗99=330 mm2 Using 12 mm bars:
No .of bars=
300 =2.918 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =342.752 mm 300 Spacing=
Smax=2h Smax=2*125=250mm Since Smax < S required : Use 3-12mm reinforcing bars @ 250mm spacing B. Long Span, Middle Strip At edge:
Rn=
MB ∅ bd
= 2
4.2599 x 106 = 0.6253431394 MPa 0.9∗1000∗872
0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.6253431394 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.00151
ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd
Therefore use ρmin¿0.00333 for As
A s=0.00333∗1000∗87=290 mm2 Using 12 mm bars:
No .of bars=
290 =2.5641 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =390 mm 290 Spacing=
Smax=2h Smax=2*125=250mm
Smax < S required Use 3-12mm reinforcing bars @ 250mm spacing At Mid-span:
MA
2.5301 x 106 Rn= = = 0.3714126334 MPa ∅ b d 2 0.9∗1000∗872 0.85 f ' c 2 Rn [1− 1− ] fy 0.85 f ' c 0.85∗28 2∗0.3714126334 ρact = [1− 1− ] 420 0.85∗28
√
ρact =
√
ρact =0.000891 ρmin =
1.4 1.4 = fy 420
ρmin = 0.00333 0.85 f ' cβ 600 ( )] fy 600+ fy 0.75∗0.85∗28∗0.85 600 ρmax = ( )¿ 420 600+420 ρmax =0 .02125 ρmax =0.75 [
ρmin > ρact ¿ ρmax A s=ρbd A s=0.0033∗1000∗87=290 mm2
Therefore use ρmin¿0.0033 for As
Using 12 mm bars:
No .of bars=
290 =2.5642 , say 3 pcs. π∗122 4
A ¯¿ ¿ As π∗122 ∗1000 4 Spacing= =390 mm 290 Spacing=
Smax=2h Smax=2*125=250mm
Smax < S required
Use 3-12mm reinforcing bars @250mm spacing
More Documents from "Bevelyn Manalo"
Design-of-critical-two-way-slab.docx
January 2021 0
Preboard - Set A
January 2021 1
2019-civil-law-bar-examinations-q-and-suggested-as.pdf
February 2021 0
1001 Solved Problems In Electrical Engineering
February 2021 0
