同济大学 材料力学 习题解答3(练习册P71-P74等).pptVIP

* * 习题解答(三) P71 39-1 4.5kN·m 4.5kN·m Mn图 1.5kN·m │Mn│max = 4.5 kN·m WP = 16 pd3 = 227.3 cm3 = 19.80 MPa tmax = WP │Mn│max ＜ [t ] IP = 32 pd4 = 1193 cm4 │q│max = G·IP │Mn│max p 180 × = 0.27 o/m ＜ [q ] 满足强度与刚度条件 Mn1l1 + Mn2l2 + Mn3l3 = G·IP p 180 × j14 = j34 + j23 + j12 = - 0.162 o = - 2.83×10-3 rad T1 T2 T3 T4 刚度校核 扭转角 S P71 39-2 Mn = T0 = 10.505 N·m = 9550 n Nk WP = 16 pd3 tmax = WP Mn = 16Mn pd3 ≤ [t ] 3 d ≥ p[t ] 16Mn = 9.63 mm j = G·IP Mnl p 180 × = Gp2d4 180 × 32 Mnl IP = 32 pd4 ≤ [j ] = 1° = 19.8 mm 4 d ≥ Gp2 [j ] 180 × 32 Mnl 取 或 q = G·IP Mn p 180 × ≤ [q ] = 2m 1° Nk = 0.33 kW d = 20 mm P72 39-3 T0 T0 圆轴： WP = 16 pd3 = 196.3 cm3 tmax = WP Mn = 70 MPa Mn = WPtmax T0 = = 13.74 kN·m 螺栓： FC × × n = T0 2 D0 FS = 2T0 nD0 FC = t = AS FS ≤ [t ] AS = 4 pd2 n ≥ pd12D0[t ] 8T0 npd12D0 8T0 = = 7.3 令 取 FC FC FC FC FC FC FC FC T0 FC FC n = 8 P72 39-4 A B C T0 TA TB Mn图 32 33 T0 1 33 T0 IP1 = 32 pd14 IP2 = 32 pd24 = IP1 16 1 Mn1 = TA Mn2 = TA - T0 jAB = jAC + jCB = + Mn1a GIP1 Mn22a GIP2 33TA - 32T0 = a GIP1 = 0 令 ∴ TA = T0 33 32 ?m = 0 TB = T0 - TA = T0 33 1 = T0 33 32 反力：受力图 = - T0 33 1 P110 58-1(b) z y O 200 200 100 100 zC yC C 负面积法 A1 = 800 cm2 A1 A2 A3 A2 = - 25p cm2 A3 = - 56.25p cm2 y1 = 20 cm y2 = 30 cm y3 = 10 cm = 21.8 cm yC = A1y1 + A2y2 + A3y3 A1 + A2 + A3 对称性 zC = 10 cm IzC = S(IzCi+Aibi2 ) i = 76396 cm4 IzC1 1 IzC2 2 IzC3 3 12 1 = bh3 64 p = d14 64 p = d24 组合法 yC (106667+2592)cm4 -(491+5281)cm4 -(2485+24606)cm4 错的不少 P111 58-3 R z1 z2 zC R O IzC = Iz1 - pR2×yC2 2 1 yC = 3p 4R yC Iz2 = IzC + pR2×( R + yC )2 2 1 3 4 =( p + )R4 8 5 = 3.30R4 P282： Iz = IzC+Ab2 平行移轴公式 附录A 第四项 P111 58-4 z y 20 C1 C2 Iz1 = Iy1 = 179.51 cm4 A1 = 19.26 cm2 z0 = 2.84 cm z0 Iz = 2Iz1 = 359 cm4 Iy =