建筑力学计算题

2024-12-01 01:47:31
推荐回答(1个)
回答1:

1.1.2 混凝土强度等级:C20 fc = 9.55N/mm� ft = 1.1N/mm�
1.1.3 钢筋强度设计值 fy = 300N/mm� Es = 200000N/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)] = 0.8/[1+300/(200000*0.0033)] = 0.550
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*9.55*200)]^0.5 = 226mm
1.2.3 相对受压区高度 ξ = x / ho = 226/460 = 0.492 ≤ ξb = 0.550
1.2.4 纵向受拉钢筋 As = α1 * fc * b * x / fy = 1*9.55*200*226/300 = 1442mm�
1.2.5 配筋率 ρ = As / (b * ho) = 1442/(200*460) = 1.57%
最小配筋率 ρmin = Max{0.20%, 0.45ft/fy} = Max{0.20%, 0.17%} = 0.20%

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