1-1 Determine the resultant internal normal force acting on the cross section through point A in each column In (a), segment BC weighs 180 lb/ft and segment CD weighs 250 io/ft In (6), the column has a mass of 200 kg/m (a) 5 Hip +T ZF, =0; F-L0-3-3-18-5=0 R F,=13.8kip Ans hi | (2/ø)60) iF 2/6 *ip il i “ip |Ì sKïp LAL 2 022 4@)= né hạ (b) gkn one Cu“ +fEE =0; F-45—-4.5~5.89—6~6—§=0 b, F,=34.9kN Ans 4k + AVE, * S.gqka Fa
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 21-2 Determine the resultant internal torque acting on the cross sections through points C and D The support bearings at A and B allow free turning of the shaft =M, = 0; Te - 250 =0 Te = 250N-m Ans Te =M, = 0; Ih =0 Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 31-3 Determine the resultant internal torque acting on the
cross sections through points B and C A 600 the ` ụ AL 350 thft 3 C tt 500 Ib-ft wo XS 50 (6 e© >> 5M, =0; 1% +350-500= S09 b-$¢ Tp = 1501b-ft Ans ot + EM,=0; Tc-500=0 £ Te =500 lb-ft Ans OS ®-Ƒt x From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 4*1-4 Determine the resultant internal normal and shear force in the member st (a) section a—a and (6) section b-b, each of which passes through point A The 500-Ib load is applied along the centroida! axis of the member 500 tb 500 tb (a) SEF, =0; N,-500=0 2 Nz = 500 Ib Ans 500" | Me +125F=0; W=0 Ans I (b) Wr =0; N,~500cos 30° =0 N, =433lb Ans 500 \y a — x» 30 N, +⁄EF,=0; W-500sin30?=0 W=2501b- Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 51-5 Determine the resultant internal loadings acting on the cross section through point D of member AB 175A 30mm ' 50 mm 134.25 2/479 I15H W Np !3i28w My Segment AD - SER =0; Np+131.25=0; Np =-I31N Ans +12ER,=0; W+175=0; W=-175N Ans EM =0; Mp + 175(0.05) = 0; Mp =-8.75 N-m Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 61-6 - The beam Á ¡s pin supported at A and supported by a cable 8C Determine the resultant internal loadings act- ing on the cross section al point D 6=tan" B =36.87° -/10 ¢ = tan '($)-3687 = 14.47 5 Member AB : (+ >M, = 0; Fac sin 14.47°(10) — 1200(6) = 0 Fac = 2881.46 lb Segment BD : elk, = 0; ~Np — 2881.46 cos 14.47° — 1200 cos 36.87° = 0 Np = ~ 3750 tb = —3.75 kip Ans AEF, = 0; Vp + 2881.46 sin 14.47° — 1200 sin 36.87° = 0 % =0 Ans G EMp = 0; — 2881.46 sin 14.47°(6.25) ~ 1200 sin 36.87°(6.25) — Mp = 0 Mp = 0 Ans BX 1200 ‘6 Notice that member AB is the two - force member ; therefore the shear force and moment are zero Ạ Yy
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 71-7 Solve Prob 1-6 for the resultant internal loadings act- ing at point E `? ah |—aa—+——su—— 8=tan ! B = 36.87° @ = tan (= )- 36.879 = 14.47° Member AB : C+ EM, = 0; — Fac sin 14.47°(10) - 1200(6) = 0 Ze, N, £ ng Fgc = 2881.46 Ib Segment BE : + LF, = 0; — Ng — 2881.46 cos 14.47° ~ 1200 cos 36.87° = 0 Ne = ~— 3750 lb = -3.75 kip Ans X+ ER = 0; Ve + 2881.46 sin 14.47° — 1200 sin 36.87° = 0 =0 ‘Ans Œ IMs = 0; 2881.46 sin 14.47°(3) — 1200 sin 36.87°(3) — Me = 0 Mẹ = 0 Ans Notice that member AB is the two - force member ; therefore the shear force and moment are zero
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 8*1-8 The beam AB is fixed to the wall and has a uniform
weight of 80 Ib/ft If the trolley supports a load of 1500
lb, determine the resultant internal loadings acting on the cross sections through points C and D 208-10 r3 zxrnn F) Segment BC : CEE =0; Nc=0 Ans +T†1£E =0, W- 20-15 =0 i Ne | 1500 tb a Ye = 3.50 kip ake ee se¿.0X?P (+EM =0; ~Me ~ 2(2.5) —1.5 (15) = 0 Mc = —47.5kip-ft Ans Segment BD : CEF,=0; Np =0 Ans +TER =0; W-024=0 Np <8, 0-083) = 0-24 RIP VY = 0.240 kip Ans (+ EMp = 0; ~My ~ 0.24(1.5) = 0 Mp = -0.360 kip - ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 91-9 Determine the resultant internal loadings acting on the cross section at point C The cooling unit has a total
weight of 52 kapand a center of gravity at G From FBD (a) G2IM,=0; %(6)-523)=0; 7% =26kip From FBD (b) ŒEMo=0:; — Tesin 30°(6)-26(6)=0; Te = 52 kip From FBD (c) +
>» TF, =0; —NĐc ~ 52 cos 30° = 0; Ác =-45.0kip Ans
+TXR=0; V+52sin309-26=0, Y=0 Ans
EMc=0; — 52cos 30°(0.2)+52 sin 30°(3)~ 26(3)~ Mc = 0 Mc = 9.00 kip ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 101-10 Determine the resultant internal loadings acting on the cross se¢tions through points D and E of the frame Member AG : ( IM, = 0; (+ EM = 0: * > EE =0, For point D : EF, = 0: +TIF =0; Œ EM =0: For point E : * ¬kÈEF, = 0; +TFE, = & EM = 0; si (3) — 75(4)(5) ~ 150 cos30 (7) = 0; Fy¢ = 1003.89 Ib A, (3) — 15(4)(2) - 150cos30(4) = 0; A, = 373.201b Fae 75 C4) Ib A, ~ 3(1003.89) + 150sin30 = 0; A, = 527.33 ib Np + 527.33 = 0 Np = ~—5271b Ans -373.20 - Vp = 0 YW = —373 1b Ans Mp + 373.201) = 0 Mp = —3731b-ft Ans f~“~—~~ i_—-—-~¬ N Ms yt ' 150 sin30 - Ne= 0 E Ibe , - Ng = 7501 Ans Bit a W — 715(3) — 150cos 30 = 0 isolt lý = 355 Ib Ans — Mg — 75(3)(1.5) ~ 150 cos 30 (3) = 0; Mẹ = —7271b-ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 111-11 Determine the resultant internal loadings acting on the cross sections through points £ and G of the frame Member AG : (tÈM = 0; For point F : teh Fe = 0; \ LF, = 0; (+ EM = 0: For point Œ : + cÈF =0, +TIF, =0: ¢ *2⁄Mẹ = 0; 1501b 4 ° 5 far) — 300(5) - 150 cos 30 (7) = 0 Fer = 1003.9 lb Far 7z(4)z 32ao !6 s -} - VY =0 Ans Ax Fix! : od | Np ~ 1003.9 = 0 3C 1271127 15016 Ne = 1004 Ib Ans Fag =/003 9 (6 Mr= 0 Ans Nc - 150sin30 = 0 Ne = 75.01b Ans VW ~ 75(1) ~ 150 cos 30 = 0- Vj = 205 Ib Ans ~Mg ~ 75(1)(0.5) — 150 cos 30 (1) = 0 Mg = ~1671b-ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 12*1-12 Determine the resultant internal loadings acting on {a) section a-a and (b) section b-b Each section is located through the centroid, point C \+ ZK =0; Nc +5.091 sin 45° =0 Nc = —3.60 kip Ans ALE =0; Y+5.091 cos 45°-2.4=0 Y =-1.20 kip Ans G >Mc =0; — Mc —2.4(2) + 5.091 cos 45°(4) = 0 Mc = 9.60 kip - ft Ans (b) CER =0; Ne+24cos 45°=0 Nc =-1.70kip — Ans +TER =0; %+5.091-2.4sin 45°=0 Ye=-3.39kip Ans (+E Mc =0; —Mc~2.4(2)+5.091cos 45°(4) =0 Mc = 9.60 kíp - ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 131-13 Determine the resultant internal loadings acting on the cross section through point C in the beam The load D :has a mass of 300 kg and is being hoisted by the motor M
with constant velocity —Í, Ä1_ Đ— 4 —+ Q ì) 2 +“ /m “9a, 5.880 KH 300/98!) 2.993 KN 3 = 2.83 KH M 298 7 o.lm Po Ne FS 2ø 2.943 KN €- EF =0; No+2.943=0; Nạ=-294kN Ans +TIF =0; Y-2.943=0; %=2.94KN Ans GEM =0; ~Mc—2.943(0.6)+ 2.943(0.1) = 0 Mức =—1.47 kN-m Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 141-14 Determine the resuitant internal loadings acting ơn the cross section through point E of the beam in Prob 1-13 + ZF, = 0; Ne + 2943 = 0 Ng = —2.94kKN Ans Me +†EEQ =0, ~2943-W =0 W = —2.94kN Ans 2242 8l Œ IM; = 0; M,; + 2943 (1) = 0 Mr; = ~2.94kN-m Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 151-15 The 800-†b load is being hoisted at a constant speed using the motor M, which has a weight of 90 lb Determine the resultant internal loadings acting on the cross section through point B in the beam The beam has a weight of 40 Ib/ft and is fixed to the wall at A 0.49 Hạ Hạ & lo, 2 ta oe tl? i SIR =0; -M-04=0 Np =—0.4 kip Ans +T†šFE =0; Y%-08-01650 % = 0.960 kíp Ans ¢ LMyp = 0; — Mẹ — 016(2) ~ 0.8(4.25) + 0.4(1.5) = 0 Mg = —3.12 kip: ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 16*1-16 Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob 1- } 5S, M 4 D C B i ăn a 4R— For point C : © ER =0; Ne+04=0; Neo = ~O4kip +TIF =0; W-08-004(7)=0; Y= 108 kip Œ Mc = 0; ~ Mc — 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 Mẹ =— 6.18 kíp -ft Ans Ans Ans 0.09 kip _ FHF Al A ~~ > Mp] > % 0-04 14) Kip For point D : obKip 7ft | F257 «IF =0; Np = 0 Ans +TxzE = 0; Vb - 0.09 - 0.04(14) - 0.8= 0; = 1.45 kip Ans G ZMp = 0; ~ Mp - 0,09(4) — 0.04(14)(7) — 0.8 14.25) = 0 Mp =~-15.7 kip * $4 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 171-17 Determine the resultant internal loadings acting on the cross section at point B SER =0; Ny=0 Ans +T ER =0; We -5(48)112) =0 Vp = 288 Ib Ans ¢ IMp=0; —-Mp- 2(4812)4) =0 Ans My = -~1152 Ib-ft= — 1.15 kíp - ft
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 18
1-18 The beam supports the distributed load shown Determine the resultant internal loadings acting on the cross section through point C Assume the reactions at the sup- ports A and 8 are vertical Ans 4 IR =0; Nc=0 +L2F=0; %4+054+15-3.75=0 Y= 175 kN Ans & >Mc =0; Mc +0.5(1) + 1.5(1.5)— 3.75(3) =0 0-503) san Ä(0:383)(4) =0 5 kzJ Mc=8.50kN-m Ans es Ae ba a/Me © 853 kenya, 3-14 ˆ# % mn
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 19
1-19 Determine the resultant internal loadings acting on the cross section through point D in Prob 1-18 3 ER =0; Np =0 Ans +TEE =0, 375-3~-2-l =0 Vp=-125kN Ans (+ EMp =0; Mo+2(20+3(3)~3.75 (6) = 0 Mp = 9.50kN-m Ans Š,2€tA) zy 24) _ 1 eeT ANE, a zh t 6x my No 2.75 kN Vv >
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 20#1-20 The scrving tray 7 used on an airplane is supported on each side by an arm The tray is pin
connected to the arm at A, and at B there is a smooth
pin (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown WN Nn jan v fj ‘an 2 (Stas Vial số Ne “ 4 # + *SAP mm ứ € M, / Ne x ft LE, = 0; Nc + 9 cos 30° + 12cos30° = 0; No = -18.2N Ans 4 2 WIE = Ye - 9sin 30° — 12 sin 30° = 0, % = 10.5N Ans H 2 (+2 Mc ~Me ~ 9(0.5 cos 60°+ 0.115) - 12(0.5 cos 60° + 0.265) = 0 Mc = —946N-m Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 211-21 The metal stud punch is subjected to a force of 120 N on the handle Determine the magnitude of the reactive
force at the pin A and in the short link BC Also,
determine the internal resultant loadings acting on the cross section passing through the handle arm at D (tiM = 0; Foc cos 30°(50) — 120(500) = 0 Fạc = 1385.6N = 1.39 kN Ans +TER =0; A, — 1385.6 - 1200s 30° = 0 A, = 1489.56N C EE=0 A-120sin309=0; Á,= 60N F, = 1489.56? + 60? = 1490N = 1.49kN Ans Segment : WLR = 0; Np - 120=0 Np = 120N=Q.12KN Ans YEE, = 0; =0 Ans GEMp=0, Mp - 120(0.3) = 0 Mp = 36.0N-m Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 221-22 Solve Prob 1-21 for the resultant internal loadings
acting dn the cross section passing through the handle
arm at E and at a cross section of the short link BC 120N, Member : G EM, = 0; — Faccos 30°(50) ~ 120(500) = 0 Foc = 1385.6N = 1.3856KN it: : Segmen' An yi = 0 Ne= 0 WER =0 Y-120=0, = 120N Ans (Œ EM =0 Me - 120(04)= 0, Mg = 48.0N-m Ans [3856 Ka) =¬ *Ít < Short link : Ans + ZF, = 0; v=0 +T EF, = 0; 1.3856 -N=0,; N= 139KN Ans Ans GIM,;=0; M=0
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,
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Trang 231-23 The pipe has a mass of 12 kg/m If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B Neglect the weight of the wrench CD 2F=0; (Ng): =0 Ans ZF, =0; (%), =0 Ans ZF =0; (Vp), —60+60-(0.2)(12)(9.81) — (0.4)(12)(9.81) = 0 (W), = 70.6 N Ans ÈM, =0; Œ§);+60(0.4)—60(0.4)— (0.4)(12)(9.81)(0.2) = 0 (Ty); = 9.42N-m Ans XM, =0; (My), + (0.2)(12)(9.81)(0.1) + (0.4)(12)(9.81)(0.2) — 60(0.3) = 0 (Mạ), = 6.23N-m Ans =M,=0; (Mp), =0 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 24*1~24 The main beam AB supports the load on the wing of the airplatie The loads consist of the wheel reaction of 35.000 Ib at C, the 1200-lb weight of fuel in the tank of the wing, having a center of gravity at D, and the 400-Ib weight of the wing, having a center of gravity at E If it is fixed to the fuselage at A, determine the resultant internal loadings on the beam at this point Assume that the wing does not transfer any of the loads to the fuselage, except through the beam ZF, =0; (YU) = 0 Ans ZF, = 0; (Nyy = 0 Ans ZF =0; (CÚ), — 1200 — 400 + 35000 = 0 (YU) = ~33.4 kip Ans =M, = 0; (Ma)s — 1200(6) + 35000(10) - 400(12) = 0 (My), = 338 kip- ft Ans IM, = 0: (Œá3% + 1200(1.5) - 4001) = 0 (T%)y = — 140 kíp -ft Ans =M, = 0; (M,), = 0 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 251-25 Determine the resultant internal loadings acting on the cross section through point 8 of the signpost The post is fixed to the ground and a uniform pressure of 7 Ib/ft? acts perpendicular to the face of the sign ⁄ 75° 7(⁄4z/2s 4 Ve) Oy ey (uly tà) (No) ¢ * tor LF, =0; (Ve), — 105 = 0; (CÚ), = 105 Ib Ans ZF =0; (%) =0 Ans > FE, = 0; (Neh =0 Ans =M,=0; (Mg), =0 Ans EM,=0;, (Mp), -105(7.5)=0; (Mg)y =7881b-ft Ans = M, =0; (Ts); — 105(0.5) = 0; (Tg), = 52.5 Ib- ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 261-26 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft Determine the resultant internal loadings acting on the cross section located at point C The 300-N forces act in the —z direction and the 500-N forces act in the +x direction The journal bearings at A and B exert only x and z components of force on the shaft EF =0; (%),+1000-750=0; (%), =-250N , Ans ZF =0; (Ne), =0 Ans EF=0; (%),+240=0; (%), =-240N Ans IM, =0; (Mc), + 2400.45) = 0; (Mc), =-108 N-m Ans 2M, =0; = (Te)y =0 Ans EM, =0; (Mc), - 1000(0.2)+750(0.45)=0; (Mc), =-138N-m Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 271-27 A hand crank that is used in a press has the dimen- sions shown Determine the resultant internal loadings act- ing on the cross section at A if a vertical force of 50 Ib is applied to the handle as shown, Assume the crank is fixed to the shaft at Ư, S01 LF, =0; (Y)x =0 Ans LF, =0; (Na)y +50 sin 30° = 0; (Na)y = —25 Ib Ans LF=0; (),-50cos30°=0, (4), =43.3 1b Ans
2M, =0; (My) — 50 cos 30°(7) = 0; (Mg), = 303 Ib - in Ans 2M, =0,; (T4)y +50 cos 30°93)=0; ˆ (7)y =—l130lb-m Ans =M,=0; (MM), +50sin 30°(3)=0; (My), =—75 Ib- in Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 28*1-28 Determine the resultant internal loadings acting on the cross section of the frame at points F and G The con- tact at E is smooth Member DEF : G = Mp = 0; Ng (5)~80 9) = 0 Ng = 144lb Member BCE : G =Mp=0; Fac Da) 144 sin 30° (6) = 0 Fac = 1801b + > ZF, =0; B, + 180 (=) 144 cos 30° = 0 B, = 16.708 Ib 144dh — 2°“ 4 4 +TEP,=0; -B,+180)-144sin302=0 sy „cÍư¿ B, = 7201b N For point F : : Ye ALF, =0; Ne =0 Ans nr Rey yedo yr = 0; W.-80 = 0; Wẹ = 80Ib Ans ( EMc=0; M;-80(2) =0; Mp = 160Ib-ft Ans For pọnt G : y * - =," = 4 > IF, =0; 16.708-NG=0: Ne=167lb Ams l¿¬oƒ4ụ N h I, 6 +T EE, =0; W-720=0; Wz=7201b Ans St m, 72.015 ¢ = Mg = 0; 72 (1.5)—Mg = 0, Mg = 108 lb-ft Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 291-29 The bolt shank is subjected to a tension of 80 Ib Determine the resultant internal loadings acting on the cross section at point C Segment AC : 3 EE=0; Ne+80=0; Ne =-801b Ans +TER=0; k=0 Ans (+ >Mc =0; Mc + 80(6) = 0; Mc =-4801b-in Ans bin te 80 áo From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 301-30 Determine the resultant internal loadings acting on the cross section at points B and C of the curved member From FBD (a) JAZ F, =0; 400cos 30° +300 cos 60° — % = 0 = 496 Ib Ans \tZB-=0; Nz +400 sin 30° - 300 sin 60° = 0 Na = 59.80 = 59.8 Ib Ans ¢ =Mo=0; 300(2)—59.80(2)- Mp = 0 Mg = 480 Ib- ft Ans From FBD (b) PtXF,=0; 400 cos 45° +300 cos 45°- Ne = 0 Ne = 495 Ib Ans NGEK-=0; -% +400 sin 45°- 300 sin 45° = 0 Ye = 70.7 Ib Ans Œ >Mọạ =0; 300(2)+495(2)— Mc =0 Mc =1590lb-ft= 1.59kip-ft Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 311-31 The curved rod AD of radius has a weight per iength of w, If it lieé in the vertical plane, determine the resultant internal loadings acting on the cross section through point B Hint: The distance from the centroid C of segment AS to point O is OC = [2r sin (@2)]/6 wtÈ5=®0 NĐg +wrỠ cos@ = 0 Ng =—wr@ cosØ Ans 5 =0; —W-wré sind =0 Ve = —wr sin® Ans 2r sin (8/2
(+E Mo =0; — wr (cos 2g 2 +(Ng)r +My = 0
My =—Ngrem wr? 2 sin (8/2) cos (6/2) Mg = wr" (@ cos@-sin@ ) Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 32#1-32 The curved rod 4D of radius r has a weight per
length of w If it lies in the horizontal plane, determine
the resultant internal loadings acting on the cross section through point B Hint: The distance from the centroid C of segment AB to point O is CO = 0.9745r 2 ay II e ~ 2= 0; W=0785wr Ans LF, = 0; M =0 Ans ™ = il 2 he = r(0.09968r) =0; Te = 0.0783 wr? Ans EM, =0; Mg + : (0.3729 r) = Ú; Mp = - 0.293wr? Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 331-33 A differential element taken from a curved bar is shown in the figure Show that dN/dé=V, dV/d@ = —N, dM/d@ = —T, and dT/de = M T†đdf the dạ dạ Men SẼ + Ven TC — (N + ÁN) em T + (+ đ9) in = 0 2 re = 0; ae do Man SỐ - ven 22 + we any sin! + (V+ dv) cos = = 0 (2) = 2 2 2 2 LM, = 0; Z 46 đỡ Tem + Man © - (+ 4D ome S + (M + aah sin = 0 (3) do nM i dạ rain? - Moos te + (T + áT) sa SẼ + (M + độ oi CC = 0 (4 de # 2 2 2 2 +4 è dé , đg — độ đỡ Ínce —¬ —=—.œs— “ Ì Since 5 is small, then sin > 3 ms dvd8 Eq (1) becomes Vd@ - dN + > =0 Neglecting the second order term, Vd ~ dN =0 Nov QED đã dNdg lđØ + đV+ ——— = 0 Bq.(2) becomes Nd@ + 3 Neglecting the second order term, Nd@ + dv=o0 He-Ne độ ide Eg.(3) becomes Md@ ~ dT+ = =0 Neglecting the second order term, Md@ - dT = 0 Tom Qev 46 Bạ (4) boenmex T40 + di « “TT =0 Neglecting the second order term, Td@ + dM = 0 “ser F7 QED
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 341-34 The column is subjected to an axial force of 8 KN at
its top If the cross-sectional area has the dimensions shown
in the figure, determine the average normal stress acting at section a—a Show this distribution of stress acting over the
area’s cross section @ kN 1-82 Min A = (2150110) + (140)(10) = 4400 mm? = 4.4(103) m? Ớ= — “ ———— = 1.82 MPa Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 351-35 The anchor shackle supports a cable force of 600 Ib If the pin has a diameter of 0.25 in., determine the average shear stress in the pin Y.w ée@4\ +T EK =0: V= 300 ib avg = = ee sbi ksi Ans A 40.25? From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 36“1-36 While running the foot of a 150-Ib man is momen- tarily subjected to a force which is 5 times his weight Determine the average normal stress developed in the tibia T of his leg at the mid section a~a The cross section can be assumed circular, having an outer diamcter of 1.75 in and an inner diameter of f in Assume the fibula F does not sup- port a load P = 5(150 Ib) = 750 Ib = —=——————-=463pậi A7} 75Ẻ- 0) PS Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 371-37 The smail block has a thickness of 0.5 in If the stress distribution at the support developed by the load varies as shown, determine the force F applied to the block, and the distance d to where it is applied F= | o dA = volume under load diagram F = 20(1.5)(0.5) + 5 (20)(1.510.5) = 22.5 kip Ans Z0kgt Fd= | x(ơ dA) (22.5) d = (0.75)(20)(1.5)(0.5) + (1.5) 120)0.500.5) (22.5) d= 18.75 d=0.833in Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 381-38 The small block has a thickness of 5 mm If the stress distribution at the support developed by the load varies as shown, determine the force F applied to the block, and the
distance d to where it is applied F= J odA = volume under stress diagram F= 2(006/40410240.00%) +(0.120)(40)(10°)(0.005) + 5 (0.120)(20)(10°}0.005) F=36kN Ans Require Fd =| xodA) 36.0(10")d = $10.06); }(0.06)(40)(10°)(0.005) + (0.06 + 50 120))(0.120)(40)(10°)(0.005) + (0.06 + 0 120)G }(0.120)(20)(10°)(0.005) 36.0(10°)d = 3960 d=0.110= 110mm Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 391-39 The lever is held to the fixed shaft using 2 tapered pin AB, which has a mean diameter of 6 mm If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever g l2mm E Ơ_ ri4 vn Cừ , c => = / A | 250mm | 250mm | E==—L-==—| zao* Zon 20N 20N Gz Mo =0; -F(12)—20(500)=0; F = 833.33N tive == Ok =29.5MPa Ans v %/_6 s2 A š(T8ng) From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 40#1-40 The supporting wheel on a scaffold is held in place on the leg using a 4-mm-diameter pin as shown If the wheel
is subjected to a normal force of 3 KN, determine the average
shear stress developed in the pin Neglect friction between the inner scaffold puller leg and the tube used on the wheel
Ban
+TER=0; 3kN-2V=0; V=LSKN
=—=>———— =119MPa_ Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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