工程力學練習冊答案(下)3_機械工業(yè)出版社

1、習題解答習題解答P69 38- -1T2 T4 T3 T1 ABCD0.2kNm1.0kNm0.4kNm0.6kNm0.2kNm0.4kNmMn圖圖Mn圖圖Mnmax = 1.0 kNmMnmax = 0.6 kNm基線基線P69 38- -2T0 xtn 530.5NmMn圖圖T0 = nNk= 530.5 Nm m = 0 tl = T0t = 132.6 Nm/ /m m = 0 Mn(x) = t xMn = t l= 530.5 NmtP70 38- -3T0 T0 A = T0= 1 kNmIP = 32p pd4t tr r = IPMnr r= 61.3610- -8 m4=

2、20.37 MPat tmax = 2t tr r= 40.74 MPa或或WP = 16p pd3= 24.5410- -6 m3t tmax = WPMn= 40.75 MPa: 1 ?P70 38- -40.25kNm0.5kNm0.5kNmMn圖圖= 16p pd3= 24.5410- -6 m3= 20.37 MPat tmax = WPMnmax t t 滿足強度條件滿足強度條件TB TC TD TA t tmax - - t t t t 100 5P71 39- -14.5kNm4.5kNmMn圖圖1.5kNmWP= 16p pd3= 227.3 cm3= 19.80 MPat

3、tmax = WPMnmax IP= 32p pd4= 1193 cm4q qmax = GIPMnmaxp p180 = 0.27 / /m 滿足強度與剛度條件滿足強度與剛度條件Mn1l1 + Mn2l2 + Mn3l3 = GIPp p180 = 0.162 = 2.8310- -3 radT1 T2 T3 T4 剛度校核剛度校核S SP71 39- -2Mn = T0= = WP = 16p pd3t tmax = WPMn= 16Mnp pd3 3d p pt t 16Mnj j = GIPMnlp p180 = Gp p2d4180 32 MnlIP = 32p pd4 4d Gp

4、p2 j j 180 32 Mnl取取 p p180 d = 20 mmP72 39- -3T0 T0 圓軸圓軸:WP = 16p pd3= 196.3 cm3t tmax = WPMn= 70 MPaMn = WPt tmax= 13.74 kNm螺栓螺栓:FC n = 2D0FS = 2T0nD0FC = t t = ASFS t t AS = 4p pd2n p pd12D0t t 8T0np pd12D0 8T0= = 7.3令令取取 n = 8P72 39- -4ABCT0 TA TB Mn圖圖3233T0133T0IP1 = 32p pd14IP2 = 32p pd24= IP1161Mn1 = TAMn2 = TA T0 = j jAC + j jCB= + Mn1a GIP1Mn22a GIP233TA