ψ = 1.1 - 0.65 * ftk / (ρte * σsq) (混凝土规范式 7.1.2-2)
σsq = Mq / (0.87 * ho * As) (混凝土规范式 7.1.4-3)
σsq = 2.82*10^6/(0.87* 76.* 251.) = 169.497N/mm2
矩形截面,Ate=0.5*b*h=0.5*1000*100.= 50000.mm2
ρte = As / Ate (混凝土规范式 7.1.2-4)
ρte = 251./ 50000.= 0.005
当 ρte <0.01 时,取ρte = 0.01
ψ = 1.1 - 0.65* 2.01/( 0.01* 169.50) = 0.331
ωmax = αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) (混凝土规范式 7.1.2-1)
ωmax =1.9*0.331*169.497/200000.*(1.9*20.+0.08* 8.00/0.01000) = 0.054mm,满足规范要求!
④、下端支座跨中裂缝:
裂缝间纵向受拉钢筋应变不均匀系数 ψ, 按下列公式计算:
ψ = 1.1 - 0.65 * ftk / (ρte * σsq) (混凝土规范式 7.1.2-2)
σsq = Mq / (0.87 * ho * As) (混凝土规范式 7.1.4-3)
σsq = 3.91*10^6/(0.87* 76.* 503.) = 117.698N/mm2
矩形截面,Ate=0.5*b*h=0.5*1000*100.= 50000.mm2
ρte = As / Ate (混凝土规范式 7.1.2-4)
ρte = 503./ 50000.= 0.010
ψ = 1.1 - 0.65* 2.01/( 0.01* 117.70) = -0.002
当 ψ<0.2 时,取ψ = 0.2
ωmax = αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) (混凝土规范式 7.1.2-1)
ωmax =1.9*0.200*117.698/200000.*(1.9*20.+0.08* 8.00/0.01005) = 0.023mm,满足规范要求!
⑤、右端支座跨中裂缝:
裂缝间纵向受拉钢筋应变不均匀系数 ψ, 按下列公式计算:
ψ = 1.1 - 0.65 * ftk / (ρte * σsq) (混凝土规范式 7.1.2-2)
σsq = Mq / (0.87 * ho * As) (混凝土规范式 7.1.4-3)
σsq = 2.82*10^6/(0.87* 76.* 251.) = 169.497N/mm2
矩形截面,Ate=0.5*b*h=0.5*1000*100.= 50000.mm2
ρte = As / Ate (混凝土规范式 7.1.2-4)
ρte = 251./ 50000.= 0.005
当 ρte <0.01 时,取ρte = 0.01
ψ = 1.1 - 0.65* 2.01/( 0.01* 169.50) = 0.331
ωmax = αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) (混凝土规范式 7.1.2-1)
ωmax =1.9*0.331*169.497/200000.*(1.9*20.+0.08* 8.00/0.01000) = 0.054mm,满足规范要求!
⑥、上端支座跨中裂缝:
裂缝间纵向受拉钢筋应变不均匀系数 ψ, 按下列公式计算:
ψ = 1.1 - 0.65 * ftk / (ρte * σsq) (混凝土规范式 7.1.2-2)
σsq = Mq / (0.87 * ho * As) (混凝土规范式 7.1.4-3)
σsq = 3.91*10^6/(0.87* 76.* 503.) = 117.698N/mm2
矩形截面,Ate=0.5*b*h=0.5*1000*100.= 50000.mm2
ρte = As / Ate (混凝土规范式 7.1.2-4)
ρte = 503./ 50000.= 0.010
ψ = 1.1 - 0.65* 2.01/( 0.01* 117.70) = -0.002
当 ψ<0.2 时,取ψ = 0.2
ωmax = αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) (混凝土规范式 7.1.2-1)
ωmax =1.9*0.200*117.698/200000.*(1.9*20.+0.08* 8.00/0.01005) = 0.023mm,满足规范要求! 苏州某物流有限公司配套用房结构设计+CAD图纸(6):http://www.youerw.com/gongcheng/lunwen_51015.html