建筑力学47页计算题的第2小题怎么做
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发布时间:2024-04-01 15:13
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热心网友
时间:2024-04-03 15:23
550
1.2.4 纵向受拉钢筋 As = α1 * fc * b * x Ǘ.1.2 混凝土强度等级.20%?
1.2.5 配筋率 ρ = As /.55*200*226/300 = 1442mm;[1+300/(200000*0.0033)] = 0; fy = 1*9.1N/mm:C20 fc = 9; (b * ho) = 1442/(200*460) = 1.57%
最小配筋率 ρmin = Max{0?
1.1.3 钢筋强度设计值 fy = 300N/.550
1.2.2 受压区高度 x = ho - [ho ^ 2 - 2 * M / (α1 * fc * b)] ^ 0.5
= 460-[460^2-2*150000000/(1*9;mm? Es = 200000N/, 0.17%} = 0;mm?
1.1.4 由弯矩设计值 M 求配筋面积 As,弯矩 M = 150kN·m
1.1.5 截面尺寸 b×h = 200*500mm ho = h - as = 500-40 = 460mm
1.2 计算结果:
1.2.1 相对界限受压区高度 ξb
ξb = β1 / [1 + fy / (Es * εcu)] = 0.20%.8/.55*200)]^0.5 = 226mm
1.2.3 相对受压区高度 ξ = x / ho = 226/460 = 0.492 ≤ ξb = 0, 0.45ft/fy} = Max{0.55N/mm? ft = 1
热心网友
时间:2024-04-03 15:21
太多了吧!!!!