1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²) simply supported slab shown in Fig. Solution: 2 m 2 m هنا الاسناد بسيط، لذلك سيتشكل خط خضوع واحد بالمنتصف ( البلاطة متناظرة) = We [5.0x (2x1.5) < (8/2)] x2 We = [158] W=m LO W [mx0x1.5] x2 [2 for two slabs] 8=0*2 -> 0 = 8/2 :. W;= [m × 8/2 × 1.5] <2 = [1.5m 6] :: We = Wi 15 6 = 1.5 m 6 m = 10 kN.m 8/2] -8=1.0 1.5 m E E L 8/2 δ 28 0 = L/2 L

Principles of Foundation Engineering (MindTap Course List)
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Chapter17: Retaining Walls
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Problem 17.5P
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1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²)
simply supported slab shown in Fig.
Solution:
2 m
2 m
هنا الاسناد بسيط، لذلك سيتشكل خط خضوع واحد
بالمنتصف ( البلاطة متناظرة)
=
We [5.0x (2x1.5) < (8/2)] x2
We = [158]
W=m LO
W [mx0x1.5] x2 [2 for two slabs]
8=0*2 -> 0 = 8/2
:. W;= [m × 8/2 × 1.5] <2
= [1.5m 6]
:: We = Wi
15 6 = 1.5 m 6
m = 10 kN.m
8/2]
-8=1.0
1.5 m
E
E
L
8/2
δ
28
0 =
L/2
L
Transcribed Image Text:1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²) simply supported slab shown in Fig. Solution: 2 m 2 m هنا الاسناد بسيط، لذلك سيتشكل خط خضوع واحد بالمنتصف ( البلاطة متناظرة) = We [5.0x (2x1.5) < (8/2)] x2 We = [158] W=m LO W [mx0x1.5] x2 [2 for two slabs] 8=0*2 -> 0 = 8/2 :. W;= [m × 8/2 × 1.5] <2 = [1.5m 6] :: We = Wi 15 6 = 1.5 m 6 m = 10 kN.m 8/2] -8=1.0 1.5 m E E L 8/2 δ 28 0 = L/2 L
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