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同济大学材料力学练习册答案

材料力学练习册答案第一章 绪论及基本概念1.C2.C3.A4.D5.D6.D7.023==a x M , 20max 23qa M M x === 8.011=-N , P Q =-11, 211Pa M =-, 022=-N , P Q =-22, 222Pl M =-, Pa M n =-22第二章 轴向拉伸与压缩1.略2.α=30º,MPa 75=ασ,MPa 3.43=ατα=45º,MPa 50=ασ,MPa 50=ατα=60º,MPa 25=ασ,MPa 3.43=ατ3.1212ln )(b b b b Et Pl l -=∆ 4.mm Ay 365.1=∆(↓)5.2576.0m A =上,2665.0m A =下,mm Ay 24.2=∆(↓)6.kN N AB 2.19=,n ≥38.2 ,∴n =39(根)7.kN N AB 75=,27.468mm A ≥,∴选2∠40⨯40⨯38.P =236.7kN ,d ≥0.208m ,∴取d =21cm9.(1)45=θº(2)Ea Dy ][4σ=∆ 10.E =70GPa , μ = 0.3211.3100.2-⨯=P ε12.kN N 6.381=,kN N 14.322=MPa CE 5.96=σ<[σ] ,MPa BD 161=σ<[σ]13.kN N 4.351=,kN N 94.82=,kN N 74.73-=()MPa 1771=σ,()MPa 8.292=σ,()MPa 4.193-=σ14.P N N N 278.0321===,P N N 417.054==15.kN N 60=钢(压),kN N 240=混(压),MPa 4.15-=钢σ,MPa 54.1=混σ16.MPa 100=螺栓σ,MPa 50-=铜套σ17.[P ]=12.24kN18.q =1.55MPa , MPa 5.77=钢筒σ,MPa 4.18-=铜套σ第三章 剪切1.MPa b 67.6=τ2.MPa 132=τ,MPa C 176=σ,MPa 140=σ3.n =10只(每边5只)4.n = 45.d =12 mm6.a = 60 mm , b =12 mm , d = 40 mm第四章 应力应变状态分析1.略2.(a) 130.6 MPa , -35 MPa ; -450 ; 140 MPa , 0 MPa , 450 ; 70 MPa (b) 34.8 MPa , 11.7 MPa ; 59.80 , -21.20 ; 37 MPa , -27 MPa , 109.30 ; 32 MPa (c) 5 MPa , 25MPa ; 900 , 56.30 ; 57 MPa , -7 MPa , -19.30 ; 32 MPa3.1点: 0 MPa , 0 MPa , -120MPa ; 2点: 36 MPa , 0 MPa , -36MPa ; 3点: 70.3 MPa , 0 MPa , -10.3MPa ; 4点: 120 MPa , 0 MPa , 0MPa 。

4.略5.(a) 19.14 MPa , -9.14 MPa , 31.70(b) 1.18 MPa , -21.8 MPa , -58.306.10.66 MPa , -0.06 MPa , 4.7307.(1) - 48.2 MPa , 10.2 MPa(2) 110 MPa , 0 MPa , - 48.8 MPa8.(1) 2.13 MPa , 24.3 MPa ;(2) 84.9 MPa , 0 MPa , -5 MPa9. 略10.6.021-==σσMPa , 103-=σMPa11.略12.64510390-⨯= ε13.m = 125.7 KN •m第五章 扭转1.略2.略3.(1)略(2)DC 段:MPa 41.2max =τ,CB 段:MPa 83.4max =τ,BA 段:MPa 1.12max =τ (3)646.0=DA ϕº(4)最大剪应力变小4.51.0=实空A A 5.d =111mm6.强度满足7.AC 段:MPa 4.49max =τ,77.1max =θº/mDB 段:MPa 2.21max =τ,434.0max =θº/m∴强度和刚度满足。

8.(1)D =102 mm(2)163.0-=-A D ϕº9. n = 8只10.d = 82.7 mm11.m kN ⋅=37.10max τ , 58.3=ϕº12.(1)[T] = 10.37 kN-m(2)[T] = 0.142 kN-m第六章 梁的内力1.(a )0 ≤ a ≤ x , Q=0; x =2a , Q= -qa ; x =2a +,Q=qa ; x =3a ,Q=0 0≤x ≤a ,M=0; x =2a , M= -qa 2/2; x =3a , M=0(b )x =0, Q= -P ;x =a +, Q= -2P ; x =2a, Q= -2Px =0, M=0;x =a, M=Pa ;x =2a, M=3Pa2.(a )x =0, Q=3qa /2; x =a +, Q=qa/2; x =3a ,Q=-3qa /2;x =4a, Q=-3qa /2 x =0, M=0; x =a, M=1.5qa 2; x =1.5a, M ma x =1.625qa 2;x =3a -, M=0.5qa 2;x =3a +, M=1.5qa 2;x =4a, M=0(b )x =0, Q=0;x =a -, Q=qa ; x =a +, Q= -qa/2; x =2a, Q= -3qa / 2x =0, M=0;x =a, M=qa 2/2; x =2a, M=-qa 2/2(c )x =0, Q=0;x =a, Q= -qa ; x =2a -, Q= -qa ; x =2a +, Q=qa ; x =3a, Q=qax =0, M=0;x =a, M=-qa 2/2;x =2a -, M=-3/2qa 2;x =2a +, M=qa 2;x =3a, M=0 (d )x =0, Q=16;x =2, Q= -4;x =4-, Q= -4;x =4+,Q= -24;x =5, Q= -24 (单位:kN)x =0, M=0;x =1.6, M=12.8; x =2, M=12; x =3-, M=8;x =3+, M=28;x =4,M=24;x =5, M=0 (单位:kN-m)3.(a ) x =0, Q=P ; x =l /3(左), Q=P ; x =l /3(右), Q=0; x =2/3l (左), Q=0; x =2/3l(右), Q=P ; x =l , Q=Px =0, P(上);x =l /3, P (下); x =2l /3, P (上), M=2 P l /3(逆时针); x =l , P(下)(b )x =0, Q= -3; x =1-, Q= -3; x =1+, Q=4.2; x =5, Q= -3.8; x =6, Q= -3.8 x =0, P= -3(下); x =1, P=7.2(上); 1<x <5, q= -2(下); x =5, M=6(顺时针); x =6, P=3.8(上)(单位:kN/m ,kN ,kN-m)4.(a )x =0, Q=0; x =a, Q= -qa ; x =2a -, Q= -qa ; x =2a +, Q=qa ; x =3a, Q=qax =0, M=qa 2/2; x =0, M=0;x =2a, M=qa 2; x =3a, M=0(b )x =0, Q=P/2; x =l /2, Q=P/2; x =l -, Q=P/2; x =l +, Q= -P/2; x =3/2l , Q= -P/2x =0, M= -5/4P l ; x =l -/2, M= -P l ;x =l +/2, M=0;x =l , M=P l /4;x =3/2l , M=05.(a )x =0, M= -qa 2/8; x =a/2, M=0; x =a, M= -qa 2/8(b )x =0, M=0; x =a, M= -qa 2; x =3/2a, M=0第七章 梁的应力1. I-I 截面:A σ= -7.41MPa , B σ= 4.93MPa , A D σσ-= , C σ=0.II-II 截面: A σ= 9.26MPa , B σ= -6.17MPa , A D σσ-= , C σ=0.2. 250max =σMPa3. (1) No25a 工字钢(2) 两个No22a 槽钢4. (1) []03.2=+M kN·m , []38.5=-M kN·m.(2) 10=A σMPa , 09.4-=M kN·m.5. (1) 24=δmm(2) 7.84max =c σMPa6. (1) []18.4=q kN/m.(2) 4.19=D mm , []7.15max =q kN/m7. 6.12max =τMPa , 6.8=a τMPa8. 158max =σMPa , 9.24max =τMPa9. 88.6max =σMPa , 75.0max =τMPa10. (1) 右图所示的放置形式合理。

(2) []27.7=F kN11. A 截面: 20max =t σMPa,40max -=c σMPa.C 截面: 32max =t σMPa,16max -=c σMPa.A 截面: 当b 上 100=mm 时 , τ上 =6.0MPa当b 下 =25mm 时 , τ下 =4.2MPa12. 略第八章 梁的变形1.略2.(a) )3(62a l EIPa f B --= ,EI Pa B 22-=θ (b) )2(a l EI ma f B --= , EIma B -=θ 3.相对误差为:2max 31⎪⎭⎫ ⎝⎛l v 4.EIPl f C 25633= 5. (a) )16163(4822a al l EI Pa f --= , )31624(4822l al a EIP -+=θ (b) )65(242a l EI qal f += , )125(242a l EIql +-=θ (c) EI qa f 2454-= , EIqa 43-=θ(d) )43(24323l l a a EI qa f -+-= , )44(24323l l a a EIq -+-=θ 6. )()(321221232131l l EI l Pl I l I l E P f +-+-= , )2(21222121l l EI Pl EI Pl +--=θ 7. lEIx l Px x v 3)()(22-= 8. 3/2l a =第十章 强度理论1.(a )MPa r 1003=σ(b )MPa r 1003=σ, ∴ 相同2.MPa r 4.321=σ,MPa r 1.332=σ,强度满足。

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