(2)求受拉钢筋应变不均匀系数v 由题知A。=1005mm2,b=1000m,k=200-15-8=17m 1005 =0.01005 0.5bh0.5×1000×200 38.25×10 sk 247.16MPa 0.87A4h0.87×177×1005 0.65f从 0.65×2.01 =0.574 0.01005×247.16 3)求短期刚度B E.2×10 E.3X104667（2）求受拉钢筋应变不均匀系数ψ 2 0 1005mm , 1000mm, 200 15 8 117mm 1005 0.01005 0.5 0.5 1000 200 s te b h A bh  = = = − − = = = =   由题知As 6 0 38.25 10 247.16MPa 0.87 0.87 177 1005 0.65 0.65 2.01 1.1 1.1 0.01005 247.16 s sk s tk te sk M A h f      = = =    = − = −  ＝0.574 （3）求短期刚度Bs 5 4 2 10 6.67 3 10 s E c E E   = = =  目录