3.3.3. 20米跨门式刚架 风荷载内力计算

44三、在风荷载作用下内力计算对于作用在屋面的风荷载可以分解为水平方向的分力q和竖向分力q。xy（1）在迎风面横梁上风荷载竖向分力（1）在迎风面横梁上风荷载竖向分力q作用下（见图）1H20/63.3330■fh1/60.167kh's610.04990.5973k(3)30.5970.167(30.167)4.1259）=1164.1259）=1164.1259850.167)=0.1338Vq-1.8920kN4.73kNE88VA(1.89104.73)kN14AHHql(1.8920AE4MMql21.89202BD44Mql2[2(1)]1.89202C

17kN3.3330.13384)kN4.21kN0.1338kN=25.29kN[0.5(10.167)0.1338]417.74kN(2)在背风面横梁上风荷载竖向分力q作用下(见图)^4y九=I；H=206=3.333W=f!h=1,6=0.167k=hs=610.0499=0.597y=3+k+屮(3+屮)=3+0.597+0.167x(3+0.167)=4.1259①=1加+呵=16x(8+5XO.167"1338ql8T.ql8T.23X20kN=8—3.08kNV=(—1.23X10+3.08)kN=-9.22kNEH=—H=—ql九①=—(—1.23x20x3.333x0.1338..'4)kN=2.74kNAE4x0.1338kN=16.46kNM=M=竺o=・23x202BD4x0.1338kN=16.46kNM=M=普[a2—(1+屮)0]=—1.23X202x[0.52—(1+0.167)0.1338]/4=—11.54kN在背风面横梁上风荷载竖向分力作用下M图(3)在迎风面柱上风荷载q作用下(见图)G)1X=l.H0=20.''6=3.333W=f/h=16=0.167k=hs=610.0499=0.597y=3+k+屮(3+屮)=3+0.597+0.167x(3+0.167)=4.1259a=1①二丄[6(2+^)+5k]=1[6(2+0.167)+5x0.597]=0.96874y4x4.1259AE2lAE2l0.47x622x20kN=-0.42kNM=M=如(2-①)=M=M=如(2-①)=BD4H=1(1x0.47-4.36)kN=2.1kNA62H=(0.47x6-2.1)kN=0.7kNEM=qh2a2[1-(1+屮)0]=0.47x62x12[1-(1+0.167)x0.9687]=-0.55kN・mc44(4)在背风面柱上风荷载竖向分力q作用下(见图)g)3AEX=l'：H0=206=3.333W=fh=1,6=0.167k=加s=610.0499=0.597y=3+k+屮(3+屮)=3+0.597+0.167x(3+0.167)=4.1259a=1①=丄[6(2+^)+5k]=[6(2+0.167)+5x0.597]=0.96874卩4x4.1259V一V=空一L°4x62kN=-0.94kNAE2l2x20qha—1.04x6xl1H=_(2一①)=(2一―x0.9687)kN=4.73kNE2222H=(1.04x6-4.73)=1.51kNM=(4.73x6-1.04x62..'2)kN=9.66kNDM=1.51x6kN=9.06kNBM=qh^l[1-(1+屮)0]=1.04x62x12[1-(1+0.167)x0.9687]=-1.22kNC4q(5)在迎风面横梁上风荷载水平分力作用下(见图)w2x

W=fh=16=0.167k=hs=6/10.0499=0.597y=3+k+屮(3+屮)=3+0.597+0.167x(3+0.167)=4.1259a=1①=兰(4+3屮)=0.167(4+3x0.167)=0.04558卩8x4.1259V=—V=-qf-(2h+f)=-189x1(2x6+1)kN=0.61kNAE212x20H=f1+①)=1.89x1(1+0.0455)kN=0.98kNa22H=(1.89x1-0.98)=0.91kNEM=qfh[0.5屮-(1+屮)0]C2189x1x62M=0.98x6=5.88kN・mBM=0.91x6=5.46kN2M=0.98x6=5.88kN・mBM=0.91x6=5.46kN・mD6）在背风面横梁上风荷载水平分力作用下（见图）w4x1.89kN/mW=f；h=1/6=0.167k=hs=610.0499=0.597y=3+k+屮(3+屮)=3+0.597+0.167x(3+0.167)=4.1259a=1TOC\o"1-5"\h\z屮0167①=一(4+3屮)=-(4+3x0.167)=0.04558卩8x4.1259V=—V=-qf-(2h+f)=-1.23x1(2x6+1)kN=0.40kNAE2l2x20H=生(+①)=1.23x1(1+0.0455)kN=0.64kNe22H=(1.23x1-0.64)=0.59kNAM=qfh[0.5W—(1+W)①]C22M=0.5