材料力学练习册答案
第一章 绪论及基本概念
1.C 2.C 3.A 4.D 5.D 6.D 7.02
3==a x M
, 20
max
2
3qa M
M
x =
== 8.
011=-N , P Q =-11, 211Pa M =-, 022=-N , P Q =-22, 2
22Pl
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=ατ α=45o,MPa 50=ασ,MPa 50=ατ α=60o,MPa 25=ασ,MPa 3.43=ατ 3.1
212ln )(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?3 8.P =236.7kN ,d ≥0.208m ,∴取d =21cm 9.(1)45=θo
(2)E
a
Dy ][4σ=?
10.E =70GPa , μ = 0.32 11.3100.2-?=P ε
12.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-=σ 14.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σ 16.MPa 100=螺栓σ,MPa 50-=铜套σ 17.[P ]=12.24kN
18.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=τ
2.MPa 132=τ,MPa C 176=σ,MPa 140=σ 3.n =10只(每边5只) 4.n = 4 5.d =12 mm
6.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 MPa 3.1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 , -58.30 6.10.66 MPa , -0.06 MPa , 4.730 7.(1) - 48.2 MPa , 10.2 MPa (2) 110 MPa , 0 MPa , - 48.8 MPa 8.(1) 2.13 MPa , 24.3 MPa ;
(2) 84.9 MPa , 0 MPa , -5 MPa 9. 略
10.6.021-==σσMPa , 103-=σMPa 11.略
12.64510390-?=οε 13.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 ?o (4)最大剪应力变小 4.
51.0=实
空
A A 5.d =111mm 6.强度满足
7.AC 段:MPa 4.49max =τ,77.1max =θo/m DB 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 ?o 9. n = 8只 10.d = 82.7 mm
11.m kN ?=37.10max τ , 58.3=?o 12.(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= -2P
x =0, M=0;x =a, M=Pa ;x =2a, M=3Pa 2.(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 / 2 x =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=qa
x =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=P x =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 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=qa x =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/2 x =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=0 5.(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 =σMPa 3. (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 σMPa 6. (1) []18.4=q kN/m. (2) 4.19=D mm , []7.15max =q kN/m 7. 6.12max =τMPa , 6.8=a τMPa 8. 158max =σMPa , 9.24max =τMPa 9. 88.6max =σMPa , 75.0max =τMPa 10. (1) 右图所示的放置形式合理。 (2) []27.7=F kN 11. 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 =4.2MPa 12. 略 第八章 梁的变形 1.略 2.(a) )3(62 a l EI Pa f B --= ,EI Pa B 22-=θ (b) )2(a l EI ma f B -- = , EI ma B - =θ 3.相对误差为:2 max 31?? ? ??l v 4.EI Pl f C 25633 = 5. (a) )16163(4822a al l EI Pa f --= , )31624(4822l al a EI P -+=θ (b) )65(242a l EI qal f += , )125(242 a l EI ql +-=θ (c) EI qa f 2454-= , EI qa 43 -=θ (d) )43(24323l l a a EI qa f -+- = , )44(24323l l a a EI q -+-=θ 6. )()(32122123 2 131l l EI l Pl I l I l E P f +-+-= , )2 (21222121l l EI Pl EI Pl +--=θ 7. lEI x l Px x v 3)()(2 2-= 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。 3.(1)略 (2)[τ] = 0.577[σ] 4.(1)[P] = 9.8 kN (2)[P] = 2.07 kN 5.MPa r 983=σ,MPa r 9.844=σ,∴强度满足。 6.MPa r 1754=σ 第十一章 组合变形 1.略 2.№.40C 工字钢 3.(1)MPa 98.9max =σ,MPa 98.9min =σ(压) (2)MPa 7.10max =σ,MPa 7.10min =σ(压) 4.MPa 5.62max =σ,mm f 74.5max = 5. 8倍 6.(1)略 ; (2)P = 1837.5 kN , e = 17.9 mm 7.№.16工字钢 8.m a 89.31=,m a 5.32= 9.略 10.MPa 5.23=σ,MPa 3.18=τ ,MPa 7.21max =τ 11. d ≥5.97 cm 12.MPa r 7.1973=σ,强度不满足。 13.N P 800max = 14.A 点:MPa 293max =σ D 点:MPa r 1.723=σ,MPa r 5.674=σ ∴ 强度满足。 第十二章 压杆稳定 1.(a ),(c ) 2.kN F cr 25401=,kN F cr 47102=,kN F cr 48303= 3.(1)kN F cr 355= (2)b / h = 0.5 4.(1)kN F cr 8.16= (2)kN F 4.50max = 5.kN F 6.18max = 6.[ P ] = 24.5 kN 7.[ F ] = 26.6 kN 8.[ P ] = 1948 kN 9.[ P ] = 378 kN 10.d = 180 mm 11.梁:MPa 163max =σ 柱:MPa 6.79=σ,MPa cr 2.83][=σ ∴ 结构安全。 12.75.1=n ,08.4=w n 第十三章 动荷载 1. 均安全 2. 48.4MPa 3. 154.7MPa 4. ????? ??? ????+++++)8316(2113242 32432 23Gd l a Ebh a Ed l P H l a d P πππ 5. ??? ? ??? ?+++3 2)(48114gPl EI gl v W Pl 6. 两梁的最大应力之比 2 1=AB CD σσ 两梁的吸收能量之比4 1 =AB CD U U 附录Ⅰ 平面图形的几何性质 1. 形心距离上边95mm ,I y = 2.08?108 mm 4 ,I z = 2.05?108 mm 4 2. I y = 5.90?1010 mm 4 , I z = 1.97?1010 mm 4 3. b =143mm 4. I z2 = 0.224bh 3 附录Ⅱ 平面应力条件下的应变分析 1.6max 10800-?=ε,6min 10200-?-=ε,ο60max =α 2.6max 10800-?=ε,0min =ε,ο7.31max -=α,6max 10800-?=γ 3.6max 10281-?=ε, 6min 10181-?-=ε, ο75max -=α,6max 10800-?=γ 4.6110190-?=ε, 621045-?-=ε, 631085-?-=ε 5.99.12=φu KN ?m /m 3