大偏心受压构件计算
一( 钢筋混凝土偏心受压计算。
Me,e,e1( ; ; e,i0a0N
he取20mm与mm中的最大值( a30
0.5fAc,,(,1.0)1N l0,,1.15,0.01()(,1.0)2h
2l1,,01 ; ,,,,,,,12ehi,,1400h0
e,0.3h, 当时,为大偏心受压( i0
e,0.3h, 当时,为小偏心受压( i0
2(大偏心受压
'',fbx,fA,fAN=; cyyys1
'''x,,,,,fbxh,,fAh,a Ne=; 1c0ys0s2
he= ,e,,ais2
',,,,2a适用条件:,x bs
''2a2a 当x<时,取x=; ss
'',fbx,fA,fAN= ; cysys1
'',,e,fAh,a N ; ys0s
''h,,e,,e,,a is2
二(
'AA已知,求 SS
1(非对称配筋。
,,,,1方法同上1 判别是大偏压或小偏压
''',,Ne,fAh,ays0s,,,2, ; s2,fbh1c0
,,1,1,2, s
'2a,x,,h 若; 则: sb0
'',fb,h,fA,N1c0ysA, Sfy
''x,2a时;取x,2a; 若 则: ss
'Ne A,S,,fh,a'y0s
2(对称配筋。
,,,,1方法同上1 判别是大偏压或小偏压
'',fbx,fA,fA N=; cyyys1
'''x,,,,,fbxh,,fAh,a Ne=; 1c0ys0s2
he= ,e,,ais2
Nx,h时;为大偏心受压。, x,,2,b0,fb1c
'Ne'‘ 若 x,aA,A,2时;则sSS',,fh,ays0
xNe,fbxh,,,,''c102x,aA,A, 若 2时;则sSS'',,fh,ays0