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Eng.ALI ALQARNI Page 1
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One way Solid Slab formula:
π π =ππ’ 0.9
1000 β π2β 106 = ___πππ , π = π β 20β
ππ2
= ___ππ
π =ππ¦
0.85 β ππ= __ , π =
1
π 1β 1β
2 β π π β π
ππ¦ = __
π΄π = π β 1000 β π = __ ππ2 , π΄π πππ = 0.0018 β 1000 β π = __ ππ2
ππππ₯ = min π΄ππ΄π β 1000 , 300ππ , 2 β π = __ ππ
πππ ππππ₯π’πππ β π’π π β ππ@ππππ₯
π΄π πππ = 0.0018 β 1000 β π = ___ ππ2
ππππ₯ = min π΄ππ΄π β 1000 , 300ππ , 4 β π = __ππ
πππ π ππππππππ β π’π π β ππ@ππππ₯
Beam formula:-
π π =ππ’ 0.9
1000 β π2β 106 = ___πππ
π = 600β 40β ππ βππ2
= ___ ππ πππ πππ¦ππ
π = 600β 40β ππ β 25βπ
2= ___ ππ π‘π€π πππ¦ππ
π =ππ¦
0.85 β ππ= ___ , π =
1
π 1β 1β
2 β π π β π
ππ¦ = ___
π΄π = π β π β π = __ ππ2 ,
π΄π πππ (1) =1.4
420β ππ€ β π = __ππ2
π΄π πππ (2) = ππ
4 β 420β ππ€ β π = ___ ππ2
π΄π πππ = max π΄π πππ 1 ,π΄π πππ 2 = __ππ2
πππππ π΄π > π΄π πππ ππ
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Eng.ALI ALQARNI Page 5
π =π΄π
π΄π = ππππ ππ’ππππ
πππππ ππππ = 80 + 60 + π β 1 ππ + 25 > π ππ
ππ πππ
= π½1 600
600 + ππ¦ >
π
π β΄ ππ =ππ¦
ππ πππΆπΏππ‘
= 0.375 π½1 >π
ππ‘ β΄ ππππ πππ β ππππ‘ππππππ π πππ‘πππ
β ππ = 0.75 β ππ
6β ππ€ β π β 10β3 = ___ππ
ππ =ππ’ β β ππ
0.75= __ππ
π΄π£ = 2 β πππ
2
4= ___ ππ2 (π‘π€π πππ)
π =π΄π£ β ππ¦ β π
ππ β 10β3___ππ
ππ ππ β€ 2ππ
ππππ₯ = min(600 ,π
2 )
ππ 2ππ < ππ β€ 4ππ
ππππ₯ = min 300 ,π
4
ππ ππ > 4ππ
ππππ π‘π ππππππ π π‘ππ π πππ‘πππ
ππ ππ’ <1
2β ππ
ππ ππππ πππ π ππππ ππππππππππππ‘
ππ β ππ > ππ’ >1
2β ππ
ππππ π‘π ππππ£πππ πππππ’ππ
ππ£ πππ = max{ 1
16 ππ β ππ€ β
π
ππ¦ , (
ππ€ β π
3ππ¦ )}
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Example 1:
The figure shown below shows the architectural plan for 3-story building. The slab will be designed to carry super imposed dead load (2.5KN/m2) and Live load (3KN/ m2) by compressive strength 25 Mpa and steel yield strength 420 Mpa.
Given data:
The one way solid slab is used.
Weight of Wall =15Kn/m
The slab thickness= 130mm
The beams dimensions = (300,600) mm
The columns dimensions= (300,500) mm
B
7m
7m
Strip strip
3m 3m 3m 3m A
C
1-m slab strip
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Eng.ALI ALQARNI Page 7
Determine the following:
1- Check if Slab thickness meets SBC304.
2- Design the slab for Max +Ve moment.
3- Design the slab for Max -Ve moment.
4- Check the slab for Shear reinforcement.
5- Draw the details and show all value.
6- Design the beam (A-B) for flexure and shear.
7- Calculate the axial load acting on column C.
Solution:
ππππ =π
28=
3000
28= 107.14 ππ
ππππ =π
24=
3000
24= 125 ππ
ππππ = max(107.14,125) = 125 ππ
π > ππππ β ππ
π·πΏ = ππ€π π€πππππ‘ + π π’ππππππππ ππ
π·πΏ = 0.13 β 25 + 2.5 = 5.75 ππ π2
πΏπΏ = 3 ππ π2
ππ’ = 1.4π·πΏ + 1.7πΏπΏ
ππ’ = 1.4 5.75 + 1.7 3 = 13.15ππ π2
Lu (m) 3m 3m 3m 3m
Ln (m) 2.7m 2.7m 2.7m 2.7m
Cm 1/12 1/14 1/12 1/12 1/16 1/12 1/12 1/16 1/12 1/12 1/14 1/12
Wu(kn/m2) 13.15 13.15 13.15 13.15 13.15 13.15 13.15 13.15 13.15 13.15 13.15 13.15
Mu(kn.m) 7.99 6.85 7.99 7.99 5.99 7.99 7.99 5.99 7.99 7.99 6.85 7.99
D (mm) 104 104 104 104 104 104 104 104 104 104 104 104
Rn (Mpa) 0.821 0.704 0.821 0.821 0.615 0.821 0.821 0.615 0.821 0.821 0.704 0.821
Ξ‘ 0.002 0.0017 0.002 0.002 0.0015 0.002 0.002 0.0015 0.002 0.002 0.0017 0.002
As (mm2) 208 176.8 208 208 156 208 208 156 208 208 176.8 208
3m
m
3m
m
3m
m
3m
m
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As min(mm2) 234 234 234 234 234 234 234 234 234 234 234 234
As (mm2) 234 234 234 234 234 234 234 234 234 234 234 234
S max (mm) 260 260 260 260 260 260 260 260 260 260 260 260
As rinkashge (mm2)
234 234 234 234 234 234 234 234 234 234 234 234
S max (mm) 300 300 300 300 300 300 300 300 300 300 300 300
Cv 1 1.15 1 1 1 1 1 1
Vu (kn) 17.75 20.42 17.75 17.75 17.75 17.75 20.42 17.75
ΟVc (kn) 65 65 65 65 65 65 65 65
ππππππ ππ πππππππππππ
πΏπ’ = 3π , πΏπ = 3 β 0.15β 0.15 = 2.7π
πΆπ =1
12 , ππ’ =
1
12β 13.15 β 2.72 = 7.99 ππ.π
π = 130β 20β12
2= 104ππ , π π =
7.99 0.9
1000 β 1042β 106 = 0.821 πππ
π =420
0.85 β 25= 19.765 , π =
1
19.765 1β 1 β
2 β 0.985 β 19.765
420 = 0.002
π΄π = 0.002 β 1000 β 104 = 208 ππ2 , π΄π πππ = 0.0018 β 1000 β 130 = 234 ππ2
ππππ₯ = min 113.04
234β 1000 = 483.1 ππ, 300ππ, 2 β 130 = 260 = 260ππ
πππ ππππ₯π’πππ β π’π π β 12@260 ππ
π΄π π ππππππππ = 0.0018 β 1000 β 130 = 234 ππ2
ππππ₯ = min 113.04
234β 1000 = 483.1ππ, 300ππ, 4 β 130 = 520 = 300ππ
πππ π ππππππππ β π’π π β 12@300 ππ
πΆπ£ = 1.15 , ππ’ =1.15 β 13.15 β 2.7
2= 20.42 ππ
β ππ = 0.75 β 25
6β 1000 β 104 β 10β3 = 65 ππ
β ππ > ππ’ β π ππππ ππππππππππππ‘ ππ πππ‘ ππππππ
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Design of beam (A-B) :-
Load acting on the beam (A-B)
1- Dead load:
π·πΏ = 5.75 β 1.5 + 1.5 + 15 + 0.47 β 0.3 β 25 = 35.775ππ π
2- Live load
πΏπΏ = 3 β 1.5 + 1.5 = 9ππ π
ππ’ = 1.4 35.775 + 1.7 9 = 65.385 ππ π
B
7m
7m
Strip strip
3m 3m 3m 3m A
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Eng.ALI ALQARNI Page 11
ππππππ ππ πππππππππππ
πΏπ’ = 7π , πΏπ = 7 β 0.15β 0.15 = 6.7π
πΆπ =1
9 , ππ’ =
1
9β 65.385 β 6.72 = 326.13 ππ.π
π = 600β 40β 10β20
2= 540 ππ πππ πππ¦ππ
π π =326.13 0.9
300 β 5402β 106 = 4.14 πππ
π =420
0.85 β 25= 19.765 , π =
1
19.765 1β 1 β
2 β 4.41 β 19.765
420 = 0.0111
π΄π = 0.0111 β 300 β 540 = 1798.2 ππ2 ,
Lu (m) 7m 7m
Ln(m) 6.7m 6.7m
Wu(kn/m) 65.385
Cm 1/16 1/14 1/9 1/9 1/14 1/16
Mu (kn.m) 210 326.13 326.13 210
d(mm) 540 (one layer) 517.5(two layer) 517.5(two layer) 540 (one layer)
Rn(Mpa) 2.667 4.51 4.51 2.667
Ξ‘ 0.00681 0.01221 0.01221 0.00681
As(mm2) 1103.2 1896 1896 1103.2
As min (mm2) 540 517.5 517.5 540
n 3.51β4 6.03β7 6.03β7 3.51β4
Fs=Fy Ok Ok Ok Ok
TCL Ok Ok Ok Ok
b min (mm) 275 275 275 275
Use 4β 20 7β 20 7β 20 4β 20
Cv 1 1.15 1.15 1
Vu (kn) 219.04 252 252 219.04
ΟVc (kn) 101.25 97.03 97.03 101.25
Vs (kn) 157.05 206.67 206.67 157.05
Av (mm2) 157 157 157 157
S (mm) 226.72 165.1 165.1 226.72
S max (mm) 270 258.75 258.75 270
Use β 10@270 β 10@160 β 10@160 β 10@270
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Eng.ALI ALQARNI Page 12
π΄π πππ =1.4
420β 300 β 540 = 540 ππ2
π΄π πππ = 25
4 β 420β 300 β 540 = 483 ππ2
π΄π πππ = max 540,483 = 540ππ2
π =1798.2
314= 5.72 β 6
ππππ = 80 + 60 + 5 20 + 25 = 365ππ > π = 300 β΄ π’π π π‘π€π πππ¦ππ
π = 600β 40β 10β 20β25
2= 517.5 ππ π‘π€π πππ¦ππ
π π =326.65 0.9
300 β 517.52β 106 = 4.51 πππ
π =420
0.85 β 25= 19.765 , π =
1
19.765 1β 1β
2 β 4.51 β 19.765
420 = 0.01221
π΄π = 0.01224 β 300 β 517.5 = 1896 ππ2 ,
π΄π πππ =1.4
420β 300 β 517.5 = 517 ππ2
π΄π πππ = 25
4 β 420β 300 β 517.5 = 462.1 ππ2
π΄π πππ = max 517,462.1 = 517ππ2
π =1900.3
314= 6.03 β 7
π =7 β 314 β 420
0.85 β 25 β 300= 144.81 ππ
πππ
= 0.5 >π
π= 0.28 β΄ ππ =ππ¦
πππΆπΏππ‘
= 0.31875 >π
ππ‘= 0.268 β΄ ππππ πππ β ππππ‘ππππππ π πππ‘πππ
πππ ππππ₯π’ππ π’π π 7β 20
πΆπ£ = 1.15 , ππ’ =1.15 β 65.385 β 6.7
2= 252 ππ
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Eng.ALI ALQARNI Page 13
β ππ = 0.75 β 25
6β 300 β 517.5 β 10β3 = 97.03ππ
ππ =252.3β 97.03
0.75= 206.67ππ
π΄π£ = 2 β π102
4= 157 ππ2 (π‘π€π πππ)
π =157 β 420 β 517.5
206.67β 10β3 = 165.1 ππ
ππ β€ 2ππ
ππππ₯ = πππ 300,517.5
2= 258.75 = 258.75 ππ
π’π π β 10@160ππ
Axial load on column C:-
Beam s in x- direction:
π·πΏ = 0.3 β 0.47 β 25 + 15 β3
2 + 0.3 β 0.47 β 25 + 15 β
3
2 = 55.575 ππ
πΏπΏ = 0 + 0 = 0 ππ
Beams in y- direction:
π·πΏ = 0.3 β 0.47 β 25 + (15) + 5.75 β 3 β7
2 + 0.3 β 0.47 β 25 + (15) + 5.75 β 3 β
7
2
= 250.425 ππ
πΏπΏ = 3 β 3 β7
2 + 3 β 3 β
7
2 = 63 ππ
ππ’ = 1.4 55.575 + 250.425 + 1.7 63 = 388.5 ππ
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Example 2:
Given data:-
ππ = 30 πππ ,ππ¦ = 420 πππ
π π’πππ πππππ ππ = 2.5ππ/π2 , πΏπΏ = 3ππ/π2
π€πππππ‘ ππ π€πππ = 12ππ/π
ππ = 170ππ , ππππ π ππ§π = 700,400 , ππππ’πππ π ππ§π (500,400)
Required:-
Design the one way solid slab for flexure and shear.
Design the beam (A-D) for flexure and shear.
Compute the axial load on column C.
B
4.4m 4.4m 4.4m 4.4m
C
9 m
9 m
9 m
A
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Eng.ALI ALQARNI Page 15
ππππ =π
28=
4400
28= 157.14 ππ
ππππ =π
24=
4400
24= 183.33 ππ
ππππ = max(157.14 , 183.33) = 183.33 ππ
π < ππππ β πππ‘ π ππ‘ππ ππππ πππππππ‘πππ ππππ’ππππππ‘
π·πΏ = ππ€π π€πππππ‘ + π π’ππππππππ ππ
π·πΏ = 0.17 β 25 + 2.5 = 6.75 ππ π2
πΏπΏ = 3 ππ π2
ππ’ = 1.4π·πΏ + 1.7πΏπΏ
ππ’ = 1.4 6.75 + 1.7 3 = 14.55ππ π2
Lu (m) 4.4m 4.4m 4.4m 4.4m
Ln (m) 4m 4m 4m 4m
Cm 1/24 1/14 1/10 1/11 1/16 1/11 1/11 1/16 1/11 1/10 1/14 1/24
Wu(kn/m2) 14.55 14.55 14.55 14.55 14.55 14.55
Mu(kn.m) 16.63 23.28 23.28 16.63
d (mm) 144 144 144 144
Rn (Mpa) 0.891 1.25 1.25 0.891
Ξ‘ 0.00216 0.0031 0.0031 0.00216
As (mm2) 311 446.4 446.4 311
As min(mm2) 306 306 306 306
As (mm2) 311 446.9 446.9 311
S max (mm) 300 253 253 300
As rinkashge (mm2)
306 306 306 306
S max (mm) 300 300 300 300
Cv 1.15 1.15
Vu (kn) 33.465 33.465
ΟVc (kn) 98.59 98.59
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Eng.ALI ALQARNI Page 16
ππππππ ππ πππππππππππ
πΏπ’ = 4.4π , πΏπ = 4.4β 0.2β 0.2 = 4π
πΆπ =1
10 , ππ’ =
1
10β 14.55 β 42 = 23.28ππ.π
π = 170β 20β12
2= 144ππ , π π =
23.28 0.9
1000 β 1442β 106 = 1.25 πππ ,
π =420
0.85 β 30= 16.471 , π =
1
16.471 1β 1 β
2 β 1.25 β 16.471
420 = 0.0031
π΄π = 0.0031 β 1000 β 144 = 446.4 ππ2 , π΄π πππ = 0.0018 β 1000 β 170 = 306 ππ2
ππππ₯ = min 113.04
446.4β 1000 = 253.22ππ, 300ππ, 2 β 170 = 340 = 253.22ππ
πππ ππππ₯π’πππ β π’π π β 12@250 ππ
π΄π π ππππππππ = 0.0018 β 1000 β 170 = 306 ππ2
ππππ₯ = min 113.04
306β 1000 = 452.88ππ, 300ππ, 4 β 170 = 680 = 300ππ
πππ π ππππππππ β π’π π β 12@300 ππ
πΆπ£ = 1.15 , ππ’ =1.15 β 14.55 β 4
2= 33.465 ππ
β ππ = 0.75 β 30
6β 1000 β 144 β 10β3 = 98.59 ππ
β ππ > ππ’ β π ππππ ππππππππππππ‘ ππ πππ‘ ππππππ
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Eng.ALI ALQARNI Page 17
Load acting on the beam (A-B)
1- Dead load:
π·πΏ = 6.75 β 2.2 + 2.2 + 12 + 0.53 β 0.4 β 25 = 47ππ π
2- Live load
πΏπΏ = 3 β 2.2 + 2.2 = 13.2ππ π
ππ’ = 1.4 47 + 1.7 13.2 = 88.24 ππ π
Lu (m) 9m 9m 9m
Ln(m) 8.6m 8.6m 8.6m
Wu(kn/m) 88.24
Cm 1/16 1/14 1/10 1/11 1/16 1/11 1/10 1/16 1/11
Mu (kn.m) 466.2 652.6 652.6 466.2
d(mm) 615.5 615.5 615.5 615.5
Rn(Mpa) 3.4 4.8 4.8 3.4
Ξ‘ 0. 009 0.0128 0.0128 0. 009
As(mm2) 2215.8 3151.4 3151.4 2215.8
As min (mm2) 821 821 821 821
N 7.05β8 10.03β11 7.05β8 10.03β11
b min (mm)
Fs =Fy Ok Ok Ok Ok
TCL Ok Ok Ok Ok
Use 8Ρ20 11Ρ20 8Ρ20 11Ρ20
Cv 1.15 1.15
Vu (kn) 436.35 436.35
ΟVc (kn) 168.56 168.56
Vs (kn) 357.1 357.1
Av (mm2) 226.08 226.08
S (mm) 163.7 163.7
S max (mm) 307.75 307.75
Use Π€12@160 Π€12@160
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Eng.ALI ALQARNI Page 18
ππππππ ππ πππππππππππ
πΏπ’ = 9π , πΏπ = 9β 0.2β 0.2 = 8.6π
πΆπ =1
10 , ππ’ =
1
10β 88.24 β 8.62 = 652.6 ππ.π
π = 600β 40β 12β 20β25
2= 615.5 ππ π‘π€π πππ¦ππ
π π =652.6/0.9
400 β 615.52β 106 = 4.8 πππ
π =420
0.85 β 30= 16.471 , π =
1
16.471 1β 1β
2 β 4.8 β 16.471
420 = 0.0128
π΄π = 0.0128 β 400 β 615.5 = 3151.4 ππ2 ,
π΄π πππ =1.4
420β 400 β 615.5 = 821 ππ2
π΄π πππ = 30
4 β 420β 400 β 615 = 803 ππ2
π΄π πππ = max 821,803 = 821ππ2
π =3151.4
314= 10.03 β 11
π =11 β 314 β 420
0.85 β 30 β 400= 142.22 ππ
πππ
= 0.5 >π
π= 0.23 β΄ ππ =ππ¦
πππΆπΏππ‘
= 0.31875 >π
ππ‘= 0.222 β΄ ππππ πππ β ππππ‘ππππππ π πππ‘πππ
πππ ππππ₯π’ππ π’π π 11β 20
πΆπ£ = 1.15 , ππ’ =1.15 β 88.24 β 8.6
2= 436.35 ππ
β ππ = 0.75 β 30
6β 400 β 615.5 β 10β3 = 168.56 ππ
ππ =436.35β 168.56
0.75= 357.1ππ
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Eng.ALI ALQARNI Page 19
π΄π£ = 2 β π122
4= 226.08ππ2 (π‘π€π πππ)
π =226.08 β 420 β 615.5
357.1β 10β3 = 163.7ππ
ππ < 2ππ
ππππ₯ = πππ 300,615.5
2= 307.75 = 300 ππ
π’π π β 12@160ππ
Axial load on column C:-
Beams in x- direction:
π·πΏ = 0.4 β 0.53 β 25 + 12 β4.4
2 + 0.4 β 0.53 β 25 + 12 β
4.4
2 = 76.12 ππ
πΏπΏ = 0 + 0 = 0 ππ
Beams in y- direction:
π·πΏ = 0.4 β 0.53 β 25 + 12 + 6.75 β 4.4 β9
2 + 0.4 β 0.53 β 25 + 12 + 6.75 β 4.4 β
9
2
= 423 ππ
πΏπΏ = 3 β 4.4 β9
2 + 3 β 4.4 β
9
2 = 118.8ππ
ππ’ = 1.4 76.12 + 423 + 1.7 118.8 = 900.728 ππ