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题目
某受均布荷载矩形截面简支梁,b*h=300*700(mm^2) Mmax=300KN·m ,Vmax=320KN,砼C35,纵向受力筋HRB400,箍筋HPB300,as=70mm(纵向受力筋布两排),求纵筋及箍筋数量
在考试 来个简洁点的可以吗

提问时间:2020-12-17

答案
1.正截面受弯配筋计算 
ξb = β1 / [1 + fy / (Es·εcu)]= 0.8/[1+360/(200000*0.0033)] = 0.518 
x = h0 - [h02 - 2M / (α1·fc·b)]0.5 = 630-(6302-2*300000000/1/16.72/300)0.5 
= 103mm ≤ ξb·h0 = 0.518*630 = 326mm 
As = α1·fc·b·x / fy = 1*16.72*300*103/360 = 1441mm2 
ξ = x / h0 = 103/630 = 0.164 ≤ 0.518 
配筋率 ρ = As / (b·h0) = 1441/(300*630) = 0.76% 
最小配筋率 ρmin = Max{0.20%,0.45ft/fy} = Max{0.20%,0.20%} = 0.20% 
As,min = b·h·ρmin = 420mm2 
取6C18,As =1527mm2
2 斜截面承载力计算 
0.7·ft·b·h0 = 0.7*1575*0.3*0.63 = 208.3kN < V = 320.0kN 
当 V > 0.7·ft·b·h0、500 < h ≤ 800mm 构造要求:
最小配箍面积 Asv,min = (0.24·ft / fyv)·b·s = (0.24*1.575/270)*300*200 = 84mm2 
箍筋最小直径 Dmin = 6mm,箍筋最大间距 smax = 250mm 
一般受弯构件,其斜截面受剪承载力按下列公式计算:
V ≤ αcv·ft·b·h0 + fyv·Asv/s·h0 
Asv = (V - 0.7·ft·b·h0)·s / (fyv·h0) = (320000-0.7*1.575*300*630)*200/(270*630) 
= 131mm2 
矩形截面受弯构件,其受剪截面应符合下式条件:hw = h0 = 630mm,
当 hw/b ≤ 4 时,V ≤ 0.25·βc·fc·b·h0 
Rv = 0.25·βc·fc·b·h0 = 0.25*1*16720*0.3*0.63 
= 790.0kN ≥ V = 320.0kN,满足要求.
Asv = 131mm2,箍筋取A6@250 
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