FinalSol - §awp1a Common fimt Beam 423M 2:104 Z I 700...

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Unformatted text preview: §awp1a Common fimt Beam 423M 2:104 Z I 700 T002 -* Z (Tou\(7m)a05(lfilfl) ; c; = HBQLK.. 7(1) [40° 5w: —80ws(203(1.z) Cosmo) ) + gogg'nflo)U-Z + erSin(3°3) :Wfl. fa‘J/GZMH .932 Am 1 IX fi/ez- imam“ = 3'4"?” .- a 3 1,, : €44 mums)? f/,,L'L£3_2_ig~fi-._.,.! b b 2%: 409(3)“ 600(4) - 903001» We): 0 7-:- Wow 9001b D J G .111 H 40 MC = Ema-MM) we, (3) :0 / .EéfiMLJ I..-“ A C a: z 400": ”- ‘2 E 411 r are ¢ Boil!) é R, = izos<n(20)-mestfi3u) 4' NW N: 132,363 F:JM:N :, 3052.393} 3-" eqb‘f/b Fx :3 1200:3(203— ZCDSI'MSQ) ‘ F :0 3°“ N 6) I 5&1 74:7 kf-a'. 4% .1. .... I 5 00(2 ) V " x 0‘ X wa :f xzcmdx a K O (-21.12 5L J.— T 1 m z w )L 5(1xcs.s—)-zs)mm.<) (8 (7) w 1(4) 1.4.35.1} x40! 943.1,; +chz) (73(93w _. .- 514x43?- m: F=250N MW"+ quUV‘V‘ [EX : O H MQ$\.‘\{.-}"\+ ..-- I..— 25'0 (4) =1 - 223-0 ' ' MuW+ my-..Zri:__:v3i‘i£i?> ..?T..§.90‘“ C 1’1 7 AV: TU 7210)“ 0'1: M 7}: W025 A Mfis: All) T (43:10 3? ZZfr') ...- _.- L 5'06] ZQ‘MHb + 2 9 a14(2) "" 0 :7 @Unfiflms 1' AX: All/4;: 1:; TE. 7.“ = $685+ 4; ~51) 7: == 3%" (bi +5; +2k) ZMY: Egmm- 3mm :0 I “5.91”. -}.7M1;=o -320(4)+f%z—(6m33 +§twy 312211 +Z,S7Tz=lzao 3m. 5wka .50 '1‘; W- Q93 . : 54:15:,“ «Lu-mama. n—......_..... w..-" .- ,— l6" '6' 12" @Uhk’nwfi‘) ‘3 a, A514," Ex, 5);} Cxfly, 0,409., H 0): @2M0:—5o-:>Lze,)-c,m)=o £9 CY=—Szvng @iFY'» “500+ 321' +0), =0 3"? by = “'22:”: ©2155,: *By+Zz-S‘ ==o fifmfi-zzmb GDZ/VIA: qulb)“§Z5-(BZJ‘CXUB)ZO =3’Cx=*82‘5'lb @2sz D; «SH-so '2‘? 0x582?lb @é Mrs: *8ZS'OM + My!) =0 =7 H: S'TOIL (‘5) £6 : 5'5T3*82S"Bx =0 => BXL 521m @ilfix; “Ax#5zfi'+‘83§=o 217/4xx‘3‘00lb W-.-QL.§FW Ax, +' 225’ J 72’3'50 #7)“! .-.-: 300M: )4) I £505 m 5-212???— 12 :13?an )5) g) r: at: 4- 05,2. ;Z§_TT/b_:/ '73. 900 lb/ft K :— 4(em3 = gees 2%, = 2460 HZL 690000} =10 z @2800: 3 2a.: ,4 + 2300 “seam Afim .T x w l0 zMi M‘I'Zfloo *fimx :0 gm M5 €00): - Moo . 900 r L Moo 2V1 gob—*v—‘iwa'm _ . _ lujm x]: ~900x #3003 ;T gm: NH 2.1m ‘fi’wx + 900(x—83( 5%) €860 WIS-450x“ +frwoxi312w )3) \1 6w «mm—x W ...+.._a‘a____..w. ) “myL ** 280MB 9 o q a K w w fl--——u-~-~—w .X‘ I \Jrzaoo _ .. _..—-m —n_a-..-a.. um..-”— .. ' 4351;, ,___...____H-_.A_I._-,_._ m M you /Q\__ gr; 1%."; ” A - M__+‘._M \ A 8 I Z O 20)‘ Mrw Wax? +800Ux «312m: (-3: Jilmx ’ ‘ _ _ W H i' x =-— 5/8: 8*? 9‘15 m m 652; ......._..4_u.-ur—u-un-.n.u..-._n __‘ A" ___ __I _ __ ‘ “u Maui-II-m Sample Common Final Exam ESM 2104 Closed book and closed notes, but you may use one 81/2" x 11" formula sheet . Sign the following pledge: l have neither given nor received unauthorized aid on this exam. 1. The magnitude of the resultant of the 2 force vectors is most nearly 700 N a. 1200 N 1130N 1091 N 758 N 400 N 451 N a 80-N force P is applied at point C. The moment of the force P = 80 N [bout the point A is most nearly. RN (16.4i—78.1j)Nm (78.1i+16.4j)Nm (49.3i—78.1j)Nm (78.1i+49.3j)Nm —61.7kNm 61 .7kNm —28.9kNm 28.9kNm f.‘ 'the triangular area shown the area moment of 6” rtia about the line a-a is 162 in“. The area moment nertia about the line b-b is most nearly 1782 in.4 9" 3 1563 in.4 6" 1134 in.4 1 162 in.4 b """""""""""""""""""""""""""""""" """ " b -810 in.4 -1458 in.4 l a force in member F6 is most nearly. 800 lb T 800 lb C 1000 lb T 1000 lb C 1200 lb T 1200 lb C 7999.6 5. The force P = 120 lb is applied to the 200-lb block when it is at rest. The magnitude and direction of the friction force exerted by the surface on the block is most nearly 52.0 lb up the slope 52.0 lb down the slope 39.7 lb up the slope 39.7 lb down the slope 33.0 lb up the slope 33.0 lb down the slope 12.76 lb up the slope 12.76 lb down the slope ensuring!» 6. 3 area moment of inertia about the y axis for the area ween the straight line and the parabola is 4 a ______ u N amend a b0 # CD 3 >4 4 3? V II II 7. a centroid is located at 2 2 WM 900 x = 4.00in., y = 3.50m. u u x = 4.00in., y = 2.94m. 5 5 u x=40mm,y=25mm ¢' 7 x = 4.00in., y = 2.38in. x = 4.00in., y = 2.00m. O 8" 8. r the same area of the previous problem, the area moment inertia about the y-axis is most nearly 229 in.4 493 in.4 578 in.4 848 in.4 e. 1008 in.4 9. 10. ar 11 The moment of the 250-N force about point A is most nearly 9m / ‘_-1_6‘m‘_ F=250N a. (0i+0j—3000k)Nm B / b. (0i—2250j+0k)Nm T ““N‘Df ; i (0i—2250j—3000k)Nm 12m ___ i i E d. (0i+2250j+3000k)Nm ;/ ‘lT‘i/L‘NN x e. (3000i+0j—2250k)Nm /f\\~\4‘/ f. (—3000i + 0j + 2250k)Nm Z c The cylinder is pinned at O and has a radius of 2 in. There is H = 0 22 a cable that goes around the cylinder. The minimum 5 = 0'15 moment M that will allow the cylinder to begin to turn is most “1‘ ' nearly P = 50 1b 180.6 in.-lb 299 in.-lb 354 in.-lb 500 in.-lb 500 in.-lb a. uniform sign weighs 320 lb and is supported by a ball (et joint at A and by two cables. iforce in cable ED is most nearly 142.6 lb 173.5 lb 92.0 lb 76.5 lb 225 lb force in cable CE is most nearly 78.9 lb 235 lb 42.7 lb 356 lb 325 lb x-component of the force in the pin at A is most nearly 320lb 160lb 390lb d. 25.4lb e. 73.8lb Members ABC, BDH, and DOE are rigid bodies. The weight of each member may be neglected. 14. The force on the roller at H is most nearly a. 1050 lb l 550 lb 471 lb 352 lb 293 lb 15 3 force in the pin at C is most nearly 525 lb 825 lb 550 lb 734 lb 978 lb 16 3 force in the pin at D is most nearly 855 lb 825 lb 525 lb 300 lb 225 lb Be fOI long 1/3 of the beam. 17 3 force under the roller at C is most nearly 3200 lb 3000 lb 2800 lb 2400 lb-ft AC is pinned at A and on a roller at C. There is a point moment applied at B, and a distributed 900 lb/ft 1800 lb 0 18 3 largest absolute value of the shear force in the beam is 800 lb 1800 lb 2800 lb 900 lb 4500 lb 19 3 largest absolute value of the bending moment in the beam is 3200 lb-ft 4000 lb-ft 4360 lb-ft 11200 lb-ft 2400 lb-ft 20 3 largest absolute value of the bending moment occurs most nearly at A B C 6.00 ft to the right of A 7.23 ft to the right of A 8.89 ft to the right of A ...
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FinalSol - §awp1a Common fimt Beam 423M 2:104 Z I 700...

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