工字钢的受力计算

2024-12-21 04:02:31
推荐回答(2个)
回答1:

依据规范: GB50017-2003《钢结构设计规范》、GB50011-2001《建筑抗震设计规范》
计算方式: 计算构件承载力设计值
构件参数:
抗震调整系数 RE: 0.75
热轧普通工字钢: I25a
钢材牌号: Q235
钢材强度折减系数: 1.00
腹板厚度: tw = 8.00 mm
毛截面面积: A = 48.51cm2
截面惯性矩: Ix = 5017.00cm4
半截面面积矩: Sx = 230.70cm3
回转半径: ix = 10.17cm
iy = 2.40cm
截面模量: Wx = 401.40cm3
Wy = 48.40cm3
截面模量折减系数: 0.95
净截面模量: Wnx = 381.33cm3
Wny = 45.98cm3
受压翼缘自由长度: l1 = 2.00m
截面塑性发展系数: x = 1.05
y = 1.05

二、构件承载力
构件截面的最大厚度为 13.00mm, 根据表3.4.1-1, f = 215.00N/mm2, fv = 125.00N/mm2
根据GB/T 700-1988及GB/T 1591-1994, fy =235.00N/mm2
1. 弯曲正应力控制的单向弯矩设计值
Mx1 = 1.00× f × Wnx × x = 1.00 × 215.00 × 381.33 × 103 × 10-6 × 1.05 = 86.09kN•m
2. 只承受与腹板平行的剪力时, 可承受的剪力设计值
Vmax = 1.00× fv × Ix × twSx = 1.00 × 125.00 × 5017.00 × 104× 8.00 230.70 × 103 × 103 = 217.47 kN
3. 整体稳定控制的单向弯矩承载力设计值(绕x-x轴)
简支梁I25a, 钢号Q235, 受压翼缘自由长度l1为2.00m,
跨中无侧向支承, 集中荷载作用在上翼缘
查表 B.2, 并插值计算, 得轧制普通工字钢简支梁的b为2.400
b > 0.6, 根据(B.1-2)式, 得b' = min(1.0 , 1.07 - 0.282b ) = 0.953
整体稳定控制的单向弯矩承载力设计值(绕x-x轴):
Mx2 = 1.00× f × b × Wx/1000. = 1.00 × 215.00 × 0.953 × 401.40 /1000. = 82.20 kN•m

综上, 若该构件只承受与腹板平行的剪力时, 可承受的剪力设计值为217.47kN
Mx1 > Mx2, 整体稳定起控制作用, 构件受弯承载设计值为 Mx2 = 82.20kN•m

回答2:

什么型号的工字型钢,分析受外力还是内部应力(正应力还是切应力)?重物在什么位置?
工字钢受的外载荷主要有重物的拉力,牛腿的支持力,其大小由理论力学求,很简单。至于内力,主要是正应力,先算梁的最大弯矩,又知道25#钢的截面几何尺寸,最大正应力也可求

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