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Review 2 - Review 2 5-93 If the maximum intensity of the...

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Unformatted text preview: Review 2 5-93. If the maximum intensity of the distributed load n acting 011 the beam is w = 4 kN/m, determine the reactions at the pin A and roller B. II M. 6 kl LXE=0: AK=0 Ans (+IMA=0; ~12(1.5)—6(4)+B,(6)=0; 31:75:»; An, ”213:0; A, +7—1s=o; A, =11kN Ans 5-94. Determine the normal reaction at the rollerA and horizontal and vertical components at pin B for equilibrium of the member. [OkN Equation of Equilibrium : The annual auction N‘ an heobtained directly by summing moms about point B. (+ m‘ =0; 10(o.a+ 1.2003 «mama -N, (1.2+ 1.200: 60") = o NA = 8.00 icN Ans ’0 “’ 1.25:0; 8,—6cos30°=0 B,=5.20kN Ans mu; :0; n,+s.oo-6smso°-io-o a, =5.00kN Ans 5-97. The uniform ladder rests along the wall of a building at A and on the roof at B. If the ladder has a weight of 25 lb and the surfaces at A and B are assumed smooth, determine the angle 9 for equilibrium. x W's. (+234. = o; —-R4(185ine)+ 25(9cose) = o $m=o; m—Rssinw =0 +TZF} =0; Racos40°—25=0 Solving, R5 = 32.641b R4 = 20.93 lb m 9 = [fl] 20.9808) 0 = 30.8” An: 5-99. A vertical force of 80 lb acts on the crankshaft. Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, 2 components of force at the smooth journal bearing A and the thrust bearing 8.11m bearings are properly aligned and exert only force reactions on the shaft. m,=o; P(8)-80(10)=0 P=1001b Ans zu=; H,(28)—80(14)=o 19,:wa Ans EM: == 0; - B,(28) - 100(10) = 0 B, = -3§.7 lb Ans m4; 0; A,+(-35.7)-100=0 A, = 136 "J A. 25:0; By=0 AIIS “=0: A¢+40—80=0 Al=401b Ans Negative sign indicates that B. acts in the opposite seme to that shewn on the FBD. 6—127. The structure is subjected to the loadings shown. Member AB is supported by a ball-and-socket at A and a smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C. From FBD(a) 2M, =0; My, =0 £M=0; —MB,+800=0 MB,=800N-m 2M: =0; By(3)_3x(2)=0 (1) 2Fl =0; Az=0 Ans 214} :0; —A,+Bx =0 (2) 2F, =0; —A,+B, =0 (3) From FBD(b) EM, = 0; B, (1.5)+800—250 cos 45°(5.5) =0 B, = 114.85 N From Eq.(1) 114.85(3)—B, (2) = 0 B, = 172.27 N From Eq. (2) A, = 172 N Ans From Eq.(3) A, = 115 N Ans 2F, =0; C, +250 cos 60°-172.27 =0 C, =47.3 N Ans 2]“ =0; 250 cos 45°;— 114.85—Q, :0 C; =61.9N ' Ans EF;=0; 250cos 60°—Q=0 Q=125N Am EM, = 0; MC, - l72.27(1.5)+ 250 cos 60°(5.5) = 0 MC,=—429N. m Ans 224:0; Mez=0 Ans Negative sign indicates that M; , acts in the. opposite sense to that shown on FBD. *6—128. Determine the resultant forces at pins B and C on member ABC of the four-member frame. Prob. 6—128 5 FEE 4 A Fbpfn " 3155;?(2) = 0 F}. ——> T (,‘f‘ZMF = O. 4 F1 F0 Hm = 0; —150(7)(3.5) + 3%(5) — FCD(7) = 0 . 15mm, 211 5’1? A ' —————— “l 5,5 = 153111; = 1.53 kip Ans " 4T 3 ., Fm = 350 lb Ans 6-130. Determine the force in each member of the truss and state if the members are in tension or compression. 20 kN “my =0; -20(1.5) — 10(4.5) + E,(6) = o = 3M + ‘ _ A: —)T ¢ & T 425:, Are A 9040* low E 3 3 +T2F, =0; Ay —2o-10+12.5=0 W5“ L5“ L5," 15m A” = 17.5kN Joint A: 4 +T21v; =0; 175-?” =0 «9 :4» F” FB = 21.88 = 21.9kN (C) Ans 0 49F”. r 3 .5»: $21; = 0; F16 - g(21.88) = 0 ’7 50 = 13.125 = 13.1 kN (T) Ans Joint B: $21; = ; «Em + 201.88) = o 1 .._ Em = 13.1 kN (C) Ans iii F‘c—“ x 4 2m n +T2F, =0; 3(21-88) - Fsa = 0 k F” Fm = 17.51;»: (T) Ans Joint G: . ”E 4 +T2F, =0; 17.5—20+ EFGC=0 Vac Fac = 3.125: 3.12kN (T) Ans 3 (2“,: :[email protected] x $23 = ; 39125) + E”, — 13.125: 0 “if. Fm, = 11.2kN (T) Ans .Cont'd 6 - 130 .Cont'd Joint C: +1321; =0; + all”; = 0, Joint D: 3215 = 0; +TEI'; =0; Joint F: + ~92}; = 0, +T2F} =0; 4 4 3 CF — 3(3.125) = 0 Fa. = 3.12m (C) Ans 1%”??ch —ac ., s 3 .1125an3 3 3 F“ 13.12 — 3(3125) — 3(3125) — FCD = 0 FUD = 9.375 = 9.38 kN (C) Ans 3 ‘9’ 9375—3013”) = o I 1.315401 ® a ‘5 13,). = 15.63 = 15.6kN (C) Ans '35 l: F175 4 F9: 3(15‘63) — 13,. = 0 For = 12.5 kN (T) Ans 3 f 38.125) — 11.25 + F)"; = O 3.125%” 1251‘») :_:_T® . ,_ _ 11.: Far 15” — 9.38 kN (T) Ans 5m 1, 10k» 4 12.5 — 10 — 5(3'125) = 0 Check! 6-134. Determine the horizontaland vertical components of force that pins A and B exen on the two-member frame. Set 1’ = 500 N. MunberAC: “EMA = 0; -600(0.75) — C, (1.5605 60°) + C, (1.5 sin 60") = 0 Mamba CB : (32% = o. -c,(1) — 6,0) + 500(1) = o Solving. c, = 402.6N C, = 97.4N MemberAC: imp, = o; —A, +600sin60° —402.6 = o A. =117N Ans +T2F, =0; A,—60000560°—97.4=0 A, =397N Ans l>z,=o-, 402.5—soo+a,=o B,=97.4N Ans +TZF; =0; —B,+9'7.4=0 B, = 97.4N A‘ns ...
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