工程力学练习册答案(下)8_机械工业出版社.ppt

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1、习题解答习题解答组合变形组合变形P96P96 51-1Fl bA l h z zy yCBFa aj jb bFy=Fcosj jFz=Fsinj j=9.66 kN=2.59 kN14Mz=Fy2l=7.24 kNm14My=Fz2l=1.94 kNm I Iz z =1212bhbh3 3=1104 cm4 I Iy y =1212hbhb3 3=5625 cm4s s =-y-z IzMzIyMy(1)s s =0 令令tana a=z0y0=Iy MzIz My=tanj jbh2()=0.476a a=25.5WWz z =6 6bhbh2 2=1103 cm3WWy y =6 6h

2、bhb2 2=750 cm3=+WzMzWyMys st max=s sc max=9.83 MPa(2)Fy(2l)348EIzfy=Fz(2l)348EIyfz=5.43 mm=2.59 mmf=fy2+fz 2=6.02 mmtanb b=fyfz=tana a b b=a a=25.5f跨中截面跨中截面跨中截面跨中截面:中性轴中性轴中性轴中性轴 s s s st t maxmaxs s s sc c maxmax中性轴中性轴拉压区拉压区P96P96 51-2F lA lCBF z zy ya a查表查表查表查表:b=20 cmz0=5.69 cmIz=4554.55 cm4 Iy=11

3、80.04 cm4 L L L L 200200 2020z z0 0s st max 在在B点点:危险截面危险截面:跨中跨中Fy=Fz22=F=17.68 kNMy=1 14 4Mz=F Fy y 2 2l l =17.68 kNms s =-y-z IzMzIyMys s =0 令令tana a=z0y0=3.86a a=75.5zB=-z02yB=0=-80.47 mms sB=120.6 MPas sA=-146.2 MPas sc max 在在A点点:=60.95 mmzA=b-z0222=141.4 mmyA=b22中性轴中性轴中性轴中性轴 弯曲中心弯曲中心弯曲中心弯曲中心 形心形

4、心形心形心主惯性平面主惯性平面主惯性平面主惯性平面 P97P97 51-3zyx2FA AB BC CD DlMMz z MMy y FN=2F=20 kN23h2Mz=Fl-2F 12My=Fl+2F b2=7.46 kNm=5.6 kNmA A =bh=bh=72 cm2s s =+y+z IzMzIyMyAFNs sA =2.78+51.81+77.78s sB =2.78+51.81-77.78s sC =2.78-51.81-77.78s sD =2.78-51.81+77.78=132.4 MPa=-23.2 MPa=-126.8 MPa=28.8 MPaWWz z =6 6bhb

5、h2 2=144 cm3WWy y =6 6hbhb2 2=72 cm3 斜弯曲斜弯曲偏心受压偏心受压F FN NF3030s s角点角点角点角点 =+WzMzWyMyAFN-P97P97 51-4FFeez zy y50250横截面为横截面为横截面为横截面为 50502 250 mm50 mm2 2FN=FA=bhM=FeWz=6bh2s s =IzMzAFNmax+-min=(1 )h6ebhF+-e e 左左=Es s s sminmine e 右右=Es s s smanman=(1-)h6eEbhF=(1+)h6eEbhFF=(e e左左 +e e右右 )Ebh2e=h6e e右右

6、+e e左左e e右右 -e e左左=625 kN=25 mm讨论讨论:(1)s smax 0s smin 0 0(e 0 0 0-整个截面整个截面:s s s s 0 0-0 0(e )h6-虎克定律虎克定律偏心受拉偏心受拉P98P98 52-2hbD风压力的合力风压力的合力:底截面底截面:14F=p p pD2 M=F h Mn=F b=314 Nm=235.6 Nms sr3=M2+Mn2W1=h2+b2d3(1-a a4)8pD21-a a4a a 0.3086-d d=(1-a a)d2 2 2.64 mm-取取圆杆内径圆杆内径圆杆内径圆杆内径:54.6 mm 54.6 mm-弯扭组

7、合弯扭组合(圆杆圆杆)WW =3232p p p pd d3 3(1 1-a a a a4 4)-s s 危险截面危险截面d d=2.7 mmP99P99 52-3lllABCDx xy yz zF3F4F1F24.09 kN4.09 kN15.91 kN15.91 kN7.96 kN7.96 kN0.68 kN0.68 kNMMe eMMe eMn 图图0.204 kN0.204 kNmm1 kN1 kNmm1.227 kN1.227 kNmm3 kN3 kNmm1.092 kN1.092 kNmmMy 图图Mz 图图反力反力 内力图内力图危险截面危险截面B:Mz=1.092 kNmMy=3

8、 kNmM M=MMz z2 2+MMy y2 2s sr4=M2+0.75Mn2W1=3.193 kN=3.193 kNmmMMn n=-1 kN 1 kNmm弯扭组合弯扭组合-s s M2+0.75Mn2 s s -M2+0.75Mn2 s s 323p p=69.6 mm取取F4F1内力图内力图=3232p p p pd d3 3WW =WWz z 弯扭组合弯扭组合(圆轴圆轴):d=7.0 cmP99P99 52-4T T M M 4545A At ts sA As s=WMW=32p pd3t t=WPTWP=16p pd3e e 00=Es ss s=Ee e 0 虎克定律虎克定律虎克定律虎克定律:M=Ws s =105 MPa=278 Nm2s ss sa a=+cos2a a (t t)sin2a a2s s2s ss s 4545 =+t t2s ss s-4545 =-t t1Ee e 4545 =(s s4545 -nsns-4545 )广义虎克定律广义虎克定律广义虎克定律广义虎克定律:1-n n2=e e00 +t t 1+n nE1-n n2t t=(e e4545 -e e00)1+n nE=40.4 MPaT=WPt t =214 Nm