工程力学习题答案第7章 应力状态和强度理论.docx

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1、第7章应力状态和强度理论 习题答案7-2 F = 94.2kN o7-3 %. = 100 MPa ,工m=50 MPa。7-4 (a) (Ja = 35 MPa, ra = 60.6 MPa。(b) oa - 70 MPa, ra = 0o(f =62.5 MPa,工 =21.6 MPa。(c) o。= -12.5 MPa,工a = 65 MPa。7-5 (a) ox - 57 MPa,*= -7 MPa, g= -1920r,t= 32 MPa。(b) rr. = 57 MPa，%= -7 MPa，= -1920r,t= 32 MPa。(c) ox - 25 MPa,g= -25 MPa,

2、 () = -45 , rmax = 25 MPa。I，v111 a A=11.2MPa, %=71.2MPa, 为=37。59, r n =41.2MPa o(d) o. = 4.7MPa, % = 84.7MPa, a() = 1317 i = 44.7MPao 。 1JUIII ciXcTj =37MPa3 = 27MPa,& = 1920rniax = 32MPa o7-6% = 10.66MPa,cr3 = -0.06MPa,a = 4.73 7-7 (1) ox- 150 MPa, o) = 75 MPa, rmax = 75 MPa。(2) (7a = 131 MPa,工“=32

3、.5MPa。7-8,= 26MPa2 = ，% - -96MPa a0 = 27.57-9 4皿=80MPa, rmqx = 120MPa。7-10 (a)- 50MPa, o? = 50MPa, o. = -50MPa, rnvix = 50MPa。14。111 d zvcr. =52.2MPa,= 50MPa, o. - -42.2MPa, rmqx = 47.2MPa o14J111 cl z7, = 130MPa,= 30MPa, % = 30MPa,工 =80MPa。1 4D111 d zv7-11 M- 3 kN-m o7-12 o x =80 MPa， 7 = 0 oxy7-13 Ah = 1.46x 10 3 mm。7-14Me2Elbh初TV。7-15 cr12 = 26.8MPa (7t , crrM = 25.8MPa 7t,安全。7-16 %=24.3MPa,= 26.6MPa。7-17 (q) crr3 = J a2 + 4r2 (rr4 = a/o-2 + 3r2S) crr3 = cr + r % =o2 + 3r27-18 crr3 =43.3MPa7.19 F = 2kN , Me = 2kN - mr4 = 26.9MPa o v17-20按第三强度理论p =L2MPa；按第四强度理论 =L38MPa。