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计算如下:
1 受弯构件: L-1
1.1 基本资料
1.1.1 工程名称: 工程一
1.1.2 混凝土强度等级 C30, fcu,k = 30N/mm?, fc = 14.331N/mm?, ft = 1.433N/mm?
1.1.3 钢筋材料性能: fy = 300N/mm?, Es = 200000N/mm?,
1.1.4 弯矩设计值 M = 280.1kN·m
1.1.5 矩形截面,截面尺寸 b×h = 250×500mm, h0 = 457.5mm
1.2 正截面受弯配筋计算
1.2.1 相对界限受压区高度 ξb = β1 / [1 + fy / (Es·εcu)]
= 0.8/[1+300/(200000*0.0033)] = 0.550
1.2.2 单筋矩形截面或翼缘位于受拉边的T形截面受弯构件受压区高度 x 按下式计算:
x = h0 - [h02 - 2M / (α1·fc·b)]0.5 = 457.5-(457.52-2*280100000/1/14.331/250)0.5
= 227mm ≤ ξb·h0 = 0.550*457.5 = 252mm
1.2.3 As = α1·fc·b·x / fy = 1*14.331*250*227/300 = 2716mm?
1.2.4 相对受压区高度 ξ = x / h0 = 227/457.5 = 0.497 ≤ 0.55
配筋率 ρ = As / (b·h0) = 2716/(250*457.5) = 2.37%
最小配筋率 ρmin = Max{0.20%, 0.45ft/fy} = Max{0.20%, 0.21%} = 0.21%
As,min = b·h·ρmin = 269mm?
上述计算看似没有任何问题,但对于计算配筋量为2716mm?,是需要配置双层钢筋的,对于假设的as也是需要修改的。 |
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