已知矩形截面梁b×h=250×600,m=210kn.m 砼c40,钢筋hrb400, 环境一类
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解决时间 2021-02-13 02:05
- 提问者网友:溺爱和你
- 2021-02-12 10:22
已知矩形截面梁b×h=250×600,m=210kn.m 砼c40,钢筋hrb400, 环境一类
最佳答案
- 五星知识达人网友:大漠
- 2021-02-12 11:12
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:C40 fc = 19.11N/mm ft = 1.71N/mm
1.1.3 钢筋强度设计值 fy = 360N/mm Es = 200000N/mm
1.1.4 由弯矩设计值 M 求配筋面积 As,弯矩 M = 210kN·m
1.1.5 截面尺寸 b×h = 250*600mm ho = h - as = 600-40 = 560mm
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+360/(200000*0.0033)] = 0.518
1.2.2 受压区高度 x = ho - [ho ^ 2 - 2 * M / (α1 * fc * b)] ^ 0.5
= 560-[560^2-2*210000000/(1*19.11*250)]^0.5 = 85mm
1.2.3 相对受压区高度 ξ = x / ho = 85/560 = 0.152 ≤ ξb = 0.518
1.2.4 纵向受拉钢筋 As = α1 * fc * b * x / fy = 1*19.11*250*85/360 = 1127mm
1.2.5 配筋率 ρ = As / (b * ho) = 1127/(250*560) = 0.81%
最小配筋率 ρmin = Max{0.20%, 0.45ft/fy} = Max{0.20%, 0.21%} = 0.21%
1.1.1 工程名称:工程一
1.1.2 混凝土强度等级:C40 fc = 19.11N/mm ft = 1.71N/mm
1.1.3 钢筋强度设计值 fy = 360N/mm Es = 200000N/mm
1.1.4 由弯矩设计值 M 求配筋面积 As,弯矩 M = 210kN·m
1.1.5 截面尺寸 b×h = 250*600mm ho = h - as = 600-40 = 560mm
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+360/(200000*0.0033)] = 0.518
1.2.2 受压区高度 x = ho - [ho ^ 2 - 2 * M / (α1 * fc * b)] ^ 0.5
= 560-[560^2-2*210000000/(1*19.11*250)]^0.5 = 85mm
1.2.3 相对受压区高度 ξ = x / ho = 85/560 = 0.152 ≤ ξb = 0.518
1.2.4 纵向受拉钢筋 As = α1 * fc * b * x / fy = 1*19.11*250*85/360 = 1127mm
1.2.5 配筋率 ρ = As / (b * ho) = 1127/(250*560) = 0.81%
最小配筋率 ρmin = Max{0.20%, 0.45ft/fy} = Max{0.20%, 0.21%} = 0.21%
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