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1、习题4-1 解: 各杆的扭矩图4-2 解（1）扭矩图（2）M9.55 N - 9.55 x 卫-2.388KN1A 200M9.55 x 0.4775KN - m2200M-M - 2M - 0.955KN - m3421 和 3 轮换位：（M ）- 0.4775 + 0.955 -n max4-3解兀d 454兀J61.36cm 4P3232WP2JP-d-24.54cm3-m1.4325KN - mTATmax-兰=32.59MPaJP-也- 必10- 40.75MPaW24.54 X10-6P4-4 解(d 2 一d 2) = d24 外 内 4=-X 52(一 0.52)d 2 = 5
2、2外d = 5.77cmd = 2.89cm外内兀兀W =一d 3(l-a4)= 一x5.773 x(l-0.54)= 35.36cm3P 16 外 16M lx l03T = n =max W35.36 X10-6P=28.28MPa4-5 解N15000M = 9.55= 9.55 x = 573KN - mnn250W = D 3(1-(上)4)=叟 X 553(1-(巴)4)= 29775.96cm3 P 16D 1650M 573 x1000T=n =max W29775.96 x 10-6P=19.24MPa t4-6 解N7.5M = 9.55= 9.55 x = 0.71625
3、KN - mnn100Mn 4.50cm116 MnD 3(1-0.54) 2TD = 4.599cm24-7 解M = 2AS8t = 2 x x152 x10-4 x 0.25 x10-2 x 56 x 106 = 4948N - m n4M l M l(P = =n GJ2G r 35n4948 x 4.62x 80x109 x x (15 2)3 x 10-6 x 0.25 x 10-2= 4. 29 x 1 0-2 rad = 2.464-8 解 d5.6J = d 4(1-()4)= x104(1-(一)4)= 885.20cm4 p 32 外 d3210max2Jn=177.04
4、cm31325177.04 x 10-6=7.48 MPaMlW = nAB GJP1325 x 280 x10-380x109 x 885.2x10-8=5.23x10-4(rad)4-9 解M = 9.55 = 9.55 x 出=17.5KN - m n n 60兀兀J =D4 =x154 = 4970.10cm 4p 3232W = Mp = 662.68cm3p D=26.4MPa t M17.5x105T = n =max W662.68 x 10-6pM175000 = l = 4.4x 10-3rad /m = 0.252 /m 0GJ80 x 105 x 49 70 x 10-
5、8p4-10 解(1) M = 9.55 = 9.55 x 巴=0.53K - mnn1802)m = nTa=2泸=13.25 m 加i. 乂时=24.14 MPa t16 x 3514.4 “ c厂BC 段 t - 34.95MPa t max兀 x 0.083(3) A轮与B轮对换受力合理，强度也就够了。4-13 解兀M -tW - 80x106 xx 63 x10-6 - 3.393KN- mA max P16M M x -10.149KN-m d 3 0.0865m 8.65cmCD A 623 兀T3 兀 x 80 x 106申AD 5M x 0.3A兀80 x109 xx 64
6、x 10-832M x 0.25CD兀80x109 xx 8.654 x10-832-1.00x1020.577x102-0.423x102rad -0.2424-14 解(1) 实心轴脚标用 1，空心轴脚标用2，则=TWP1WP2maxD =-= 1.0667D AFDC 不再保持平面。1 -0.541D2 = 1.0217D1=1.022 倍D24(1 - 0.54)2D412)=TWP1WP2max* =-= 2.1333D131-0.54D = 1.287D192P1D24(1-0.54)=2.57倍4-15 解(1)D 41(2)工X = 0平面ACDF上，应力分布以x轴为对称轴，截
7、面对称，应力反对称故满足I x = 0。 端面ADC与DEF形状面积相同，而应力各处对应大小相等方向相反故满足Iz = 0。 由于 ABC 及 DEF 两端面剪应力形成的 M 是大小相等方向相反的，故满足XIM = 0。x平面ACDF应力的合力形成的对Y轴的力矩等于端面DEF上的水平合力对y轴的 力矩，故IM y = 0满足。所有应力对x轴不形成力矩，故IM = 0，自然满足。z（AFDC 变形图）4-16 解Wp1WP2=8d3(t i)maxM + M1 2WP1（T 2）max=WP21)i max=(T 2)2 maxM + M1 2WP1=M_=WP2M1WP1=监=WP2即竺+1
8、= 8 M 24-17 解M = 9.55 N = 9.55 x = 1.061KN mn27按强度条件计算：I16MId=3；FX=316 x 1.061x 1031= 5.13cm兀 x 40x IO6按刚度条件计算：180x 兀32 x 1.061 x 103 x 180.=5.27 cm兀 2 x 80 x IO9取 d = 53mm。轴：Mn1 = WP冗 冗d 3 k=x 43 x10-6 x 60x106 = 753.98N161 16-x63 xlO-616x45x106 =1531.5N -mQj=t l tM M 1n =n2-r02t 2-r02n套筒：M = -D-(1
9、 -a4)t 1=n2 16 As t销钉：M = Q - d = At d = -d t d = -1.3x90x106 x4x10-2 = 477.84Nn3整个纵街面上的合剪力为t -1 -1，由n个柳钉承受每个柳钉的剪力Q.x434.可以传替477.84N-m转矩。4-19 解dd ddcos45。y bdcos2 45。yg=ad adbd / cos 45。bd24-20 解, y = 2g 4. t= G y = 2G gmax4-D3M =t W = 2Gg(1 -a 4)n max P45。 16(1-=2x80x109 x 425 x10-6 x - x0.4316=743
10、.77x103N-m=743.77KN-mN = Mn-n = 743.77 x12 = 9345.8KW9.559.554-21 解 纵截面上的剪应力4-22 解(1) M = ml 一 mx (0 x 1)nml7Trrm(2) d申xf xd9xmxdxGJPfx0mx , dx =GJP1 mx22 GJP(0 x1)或9x=J xd9=J xml 一 mxGJpdx =GJp(lxx2(Q x l)4-23 解= ml-21 一 丄 ml 2=0=ml=ml4-24 解9外 =9内95 = 1.396(d -1 )38.5=M= 20202.396=8.347KN - m=11.653KN - m内管：&347 x1Q3外管：叫外4-25 解2ASt8.52x1Q-4 x Q.5 x1Q-2147.2MPa11.653x1Q32ASt9.52x1Q-4xQ.5x1Q-211.653x1Q3x18Q164.4MPa2.48 / m
