一、材料：
40
fc =19.1MPa 1.00×19.1=19.10MPa
ftk =2.39MPa Ec =MPa
β1 =0.80εcu=ft = 1.71MPa
混凝土达到90%设计强度时开始对预应力筋进行张拉，则 f′cu =36.0MPa
2. 预应力钢筋：采用MPa 有粘结预应力MPa Es =#######mm 2
1395MPa
1320Mpa 390Mpa kN
3. 非预应力普通钢筋：采用Ⅲ级钢（HRB400）Es =Mpa
4. 本部分梁抗震等级：二、有粘结预应力框架梁配筋计算
1.计算截面：Mk=965
2.53kN·m
A. 截面参数计算
b×h =800×2200hf′=130bf′=2360
截面面积 A=b×h+(bf′-b)×hf′ =mm 2
截面中和轴距梁顶面距离 e1=[b×h +(bf′-b)×hf′ ]/2A＝mm 截面中和轴距梁底面距离 e2 = h - e1 =mm
截面惯性距 I = b×h×[h /12+(h/2-e1) ]+(bf′-b)×hf′×[hf′ /12+(e1-hf′/2) ]
=mm 4
截面抵抗距 W = I / e2 =mm 3
B. 截面受拉区配置预应力筋数量计算
993.1Mpa
139.00fptk =抗压强度设计值 f ′py =二级
跨中截面7.50E+08有粘结预应力框架梁配筋计算书
360 2.0E+050.003301860钢绞线
MPa
183.4875% 3.25E+041.50
预应力筋抗拉强度标准值 fptk =单束预应力筋截面面积 Ap 0=取混凝土拉应力限制系数αct= 1.50
1206.9单束预应力筋拉力设计值 Npy 0 = fpy ×Ap 0 =截面抵抗距塑性影响系数基本值 γm =9.05E+111. 混凝土：混凝土强度等级采用 C 1.96E+061860fy=fy ′=取预应力筋张拉控制应力σcon =预应力筋抗拉强度设计值 fpy =
1.16（当h>1600时，取h=1600）
则αct γftk =MPa mm
mm kN·m
5870.7kN 取计算截面处受拉区预应力筋有效应力σpe =1116.0MPa
计算截面处受拉区单束预应力筋有效拉力 Npe 0 = σpe ×Ap 0 =
kN 梁内受拉区预应力筋数量 np=Npe/Npe 0 =37.8
取48束mm 2
受拉区预应力筋有效拉力 N pe = Ap ×σpe =
kN C. 强度验算： | Md | =kN·m 受拉区预应力筋拉力设计值 Npy = np×Npy 0 =
kN
取预应力强度比 λ =则梁中受拉区配置的普通钢筋截面积 As =（Npy / λ─Npy ）/fy =mm 2
受拉区配置普通钢筋 As = mm 2Ns = As×fy = kN 计算时考虑受压钢筋 As ′ =mm 2Ns′=As′×fy′= kN
受拉区纵向普通钢筋合力点至截面受拉边缘的距离 a s =mm 受压区纵向普通钢筋合力点至截面受压边缘的距离 a s ′=
mm 受拉区非预应力筋重心距梁中和轴的距离 ys = e2-as =
mm 受压区非预应力筋重心距梁中和轴的距离 ys′= e1-as′ =
mm 梁截面有效高度 ho = h-(Npy×a p +Ns×a s )/(Npy+Ns) =
2200-193=
2007mm Np(1)=1.1×σpe×Ap =kN （"(1)"表示仅考虑第一批预应力损失，下同）
epn(1) = yp =σpc(1) = Np(1)/A + Np(1)×epn(1)×yp / I - M2×yp / I =
MPa ρ = (Ap+As)/A =
σL 5 = (35+280×σpc(1)/f′cu)/(1+15×ρ) =
MPa Np =σpe×Ap-σL5×
As =kN 取考虑次弯距的调整系数 β=
956.975
155.12247446.0
965.3受拉区预应力筋有效预加力 Npe= [(|Mk|/w)─ftk]/(1/A+yp/w) =
受拉区预应力筋重心距梁中和轴的距离 yp = e2 - ap =
6165.8108.780%截面抵抗距塑性影响系数 γ =(0.7+120/h)×γm =
1.12500.94%1131.9918.10.680σcon =25
Ap=np×Ap0 =14013.025
11781.68807.011512.54241.4
24956.98190.5
4241.47511.4411781.66672.04.17则次弯距 M2 = (β-1) × Mk =受拉区预应力筋重心距梁底面的距离 ap =
epn = (σpe×Ap×yp-σL5×As×ys)÷(σpe×Ap-σL5×As) =mm MPa kN·m MPa
ξb=β1/[1+0.002/εcu+(fpy-σpo)/(Es×εcu)] =砼受压区高度 x=hf′+(Npy+Ns-Ns′-bf′×hf′×α1×fc)/(b×α1×fc)
mm x/ho =< ξb，满足要求
受压区砼重心距梁顶面距离 x1 =
mm 极限弯距Mu=α1×fc×b×x×(ho-x/2)+α1×fc×(bf′-b)×hf′×(ho-hf′/2)+Ns′×(ho-as′)
=
kN·m >Md =kN·m
E.抗震验算
若取砼受压区高度 x = 0.25ho ，则极限弯距 Mu 计算如下：
Mu =α1×fc×b×(0.25ho)×(ho-0.25ho/2)+α1×fc×(bf-b)×hf×(ho-hf/2)+Ns′×(ho-as′)
=kN·m >Md=kN·m 若取砼受压区高度 x = 0.35ho ，则极限弯距 Mu 计算如下：
Mu =α1×fc×b×(0.35ho)×(ho-0.35ho/2)+α1×fc×(bf-b)×hf×(ho-hf/2)+Ns′×(ho-as′)
=kN·m >Md=kN·
m 纵向受拉钢筋按照非预应力钢筋抗拉强度设计值折算的配筋率
ρ =( As + Ap × fpy / fy )/(b × h) =
设计配筋的预应力度
λ=0.675≤0.75 满足二级抗震要求As′/As=1
0.3/(1-λ)=0.92As′/As ≥0.3/(1-λ)满足二级抗震要求
1/3*(fpy×hp/(fy×hs))*Ap=959≤As 满足要求
纵向普通钢筋配筋率 As/(b×h)=≥0.2% 满足要求E. 裂缝宽度验算
αcr = 1.5C =25
mm deq = (n s ×ds 2+n p ×dp 2) / (n s ×υs ×d s +n p ×υp ×d p ) =mm Ate = b × h / 2 =
mm 2ρte=(As+Ap)/Ate=
Npo =σpo×Ap-σL5×As =kN ep = ho - e1 - epn =mm 1164.7 2.06%29180.3927.06491.0
920.6
14013.014013.033488.790.1610.43
8.12
24821.814013.08.80E+050.021093.3
161.4σpo = σpe + σpc × Ep /Ec =σpc = Np/A + Np×epn×yp/ I - M2×yp / I =322.9Mcr = (σpc + γftk)·Wo =
8173.7=
0.67%
e = ep+|Mk+M2|/Npo =
mm
γf ′= (b f ′-b)×h f ′/(b×ho) =z=[0.87-0.12×(1-γf ′)×(ho/e)2]×ho=mm
σsk=[|Mk+M2|-Npo×(z-ep)]/[(Ap+As)×z] =
MPa ψ=1.1-0.65×ftk/(ρte×σsk) =Wmax=αcr×ψ×σsk ×(1.9×C+0.08×deq/ρte)/Es =mm
0.01514630.12631729.10.2
64.1
-hf′/2) ]
Ns′×(ho-as′) )+Ns′×(ho-as′) )+Ns′×(ho-as′)