Mmax=Q墙顶X-q11X2/2+(q22-q11)X3/6H=44.44449KN·M
(四)配筋计算
混凝土抗压强度fcd=16.7
ho=265
钢筋设计强度fy=360
计算宽度b=1000N/mm2mmN/mm2mmM支座 =fcdbx(h0-x/2)
130230625.00 =
x =
得
As = M支座/(ho-x/2)/fy =1492
Mmax =fcdbx(h0-x/2)
44000000.00 =16700x(265-x/2)
x =10.136 mm≤
解得
跨中As = Mmax/(ho-x/2)/fy =483
(五)裂缝计算
钢筋直径d=18
钢筋间距100
每延米实配钢筋As=2544.69mmmmmm2
-108.53KN·M
184.983N/mm2mm2ξbh0 =0.53×265.00 =mm216700x(265-x/2)31.273 m≤ξbh0 =0.53×265.00 =标准组合下Mk支座=-H2×(8q2+7q1)/120=σsk=Mk支座/(0.87hoAs)=
αcr=2.1
ρte=0.0169646
ftk=2.2
ψ=0.644319395
所以ψ取0.644319395
Es=200000
c=25
deq=18
裂缝宽度Wfk=0.165673631mm<1 且>0.2裂缝满足要求