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hw13_sol

# hw13_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-76 A wire is stretched tight between two pylons, one of which is shown in Fig. P7-76. The 1.50-kN force is parallel to the x-y plane and makes an angle of 20° with the y-axis. The supports are equivalent to a ball-and- socket joint at B and short links at A and C. Determine the support reactions and the force in each member of this space truss. SOLUTION From a free-body diagram for joint D: Z? = TAD + Tao + T + F CD D A A _ O _ O _ 0 _ 0 - TAD(cos 45 1 cos 45 E) TBD E + TCD(cos 45 J cos 45 E) A + 1.5(cos 700 1 - cos 200 j) = (0.7071TAD + 0.5130) 1 + (0.7071TCD - 1.4095) J (-o.7o71TA — TB — 0.7071TCD) E = O D D Solving yields: TAD — -0.7254 RN 9 0.725 kN (C) Ans. TBD -0.8966 RN 2 0.897 kN (C) Ans. TCD 1.9934 RN 2 1.993 kN (T) Ans. ~—N ‘7 ~ — \ .4 !--_ DD ) From a free-body diagram for joint A: ‘ \ 1’ 5* U \ SE = TAB + TAC + TAD + A at \ N A 0A 0A -TAB 1 + TAC(—cos 45 1 + cos 45 J) (-0.7254J(-cos 45° 3 + cos 45° E) + A2 E [-TAB — 0.7071TAC+ 0.5129) 1 + (0.7071TAC) J + (-0.5129 + A2) E = O ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-76 (Continued) Solving yields: A2 = 0.5129 kN K a 0.513 E kN Ans. = 0 Ans. 0.5129 kN Q 0.513 kN (T) Ans. From a free-body diagram for joint C: EF=TAC+TBC+TCD+C A = 0 - TBC J + (1.9934)(-cos 45° 3 + cos (Cx) 1 + (-1.4095 - TBC) J + (—1.4095 Solving yields: = 0 = 1.4095 kN C a 1.410 E kN Ans. TBC = -1.4095 kN H 1.410 kN (C) Ans. From a free-body diagram for joint B: 2F=TAB+TBC+TBD+B = (+0.5129) i + (-1.4095) J + (-0.8966) R + 13x 3 + By 3 + Bz E (+0.5129 + Bx) i + (-1.4095 + By) 3 + (-0.8966 + Bz) E = U Solving yields: = -0.5129 kN 1.4095 kN 0.8966 kN B a -0.513 E + 1.410 3 + 0.897 E kN ENGINEERING MECHANICS - STATICS, 2nd. - ' F. RILEY AND L. D. STURGES 7-80 The space truss shown in Fig. P7-80 is supported by a ball-and-socket joint at A and by short links at B and C. Determine the support reactions and the forces in members AB. BC, BD. and BE of the truss. SOLUTION From a free-body diagram 'for the complete truss: A (2.75 E) x (B 3 + B 3) + (1 i) x (c 3) x y Y + 2 3) x (500 3 - 800 E) + (2 3) x (-300 i) .758y - 1600) 3 + (2.7513x + 800) 3 + (cy + 1100) E = 0 -290.9 N -58l.8 N = -291 i - 582 3 N A -1100 N -1100 J 1Axi + Ay3 + AZE) - 300 i + 500 3 - 800 E A - 290.9 3 — 581.8 3 - 1100 J (Ax - 59019) i + (Ay — 1181.8) 3 + (Az - 800) E = 6 590.9 N I Ay = 1181.8 N 800 N K = 591 i + 1182 3 + 800 E N ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. 7-80 (Continued) From a free-body for joint C: From a free-body diagram for joint E: = TDE - 300 = 0 300 lb = 300 lb (T) TBE sin ¢E = 0 —TAE - TBE cos ¢E From a free-body diagram for joint B: 1 . [------—---—--—-]TBD - 290. Y ..-)2 1112 +12)2 + (-2.13 1031.1 lb 2 1030 lb (IT) Ans. ’——_§______—__-——_—___E TBD + (-2.75) (1) +12)2 ——72'—”— (1031.1) /_____ +1—2.~75)2 (1)2 +12)2 -800 1b = 800 1b (C) Ans‘ F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—87* A pin—connected system of levers and bars is used as a toggle for a press as shown in Fig. P7—87. Determine the force F exerted on the Ican at A when a force ? = 100 lb is applied to the lever at G. SOLUTION From a free—body diagram for the lever: + C EMF =100(30) — FDE(8) = 0 FDE = 375 lb From a free—body diagram for the pin at D: + —» ZFX = —F cos 67° — F cos 78° - 375 = 0 BD CD . 0 . 0 -FBI) s1n 67 + FCD sin 78 -639.5 lb +T2F 0 y F BD F -601.8 lb II CD From a free-body diagram for the piston at B: T : — , ' o : + ZFY Ay 6395s1n 67 0 R3 Ay = 588.7 lb = 588.7 lb T Force on the can: F g 589 lb l L: Ans. £27 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES ?—91* Determine all forces acting on member ABCD of the frame shown in Fig! P?-910 SOLUTION From a free-body diagram for the complete frame: lttn + C EMA = F(12} - 75(24} = 0 F = 150.0 1b = 150.0 lb T A - 75 = 0 X ?5.0 lb = 75.0 lb-9 4. l 5? tﬁm 3- ll 4%; A + 150 = 0 y A = —150.0 1b = 150.0 lb ¢ 2 2 {(Ax} + (AV) 2 2' {(75.01 + {—150.01 = 167.71 1b _ '1 -15000 _ _ 0 6A - tan 75.0 - 53.43 + _._) [‘1 I11 ‘4 II :1:- ll K a 167.7 's 63.4° Member BE is a two~force member; therefore, the line of action of force 3 is known as shown on the free-body diagram for member ABCD: 433 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES T-Ql {Continued} + c EMC 2 -B cos 45° (6) ~ 75413) - T5i6} = o B = -424.3 lb B = 424 lb 5 45.o° Ans. 0x + (-424.3) cos 45° + 75 — 75 = 0 300 1b = 300 lb -é + T EFY = 0V — {~424.3) sin 45° + 150 = 0 -150.0 lb = -150.0 lb ¢ /{Cx}2 + (Cylz = /{30012 + (-150.0)2 = 335.41 lb _ -1 -150 ~ _ o 3C — tan —§55 — 26.57 C = 335 lb 1 26.6” 43‘! »ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—103* The fold—down chair of Fig. P7-103 weighs 25 lb and has its center of gravity at G. Determine all forces acting on member ABC. SOLUTION From a free-body diagram for the chair: + C EMA = 25(3)_- E(21) = o E = 3 571 1b = 3.571 lb +— A - 3.571 = 0 X 3.571 lb = 3 5 = 25 lb = 25 lb T l \$- = /<A )2 + (A )2 x Y /(3.571)2 + (25)2 = 25.25 15 _ -1 25 ex ‘ tan 3.571 K = 25.3 1b a 81.9° = 81.87° Member ED is a two—force member; therefore, the line of action of force E is known as shown on the free-body diagram for member ABC: + C ZMC = B cos 39.81° (9) - 3.571(21) 0 B = 10.846 lb E = 10.85 lb 8 @ 39.8° = c + 3.571 — 10.846 cos 39.810 = o c = 4.761 lb X X = Cy + 25 — 10.846 sin 39.810 = 0 Cy = -18.056 1b‘ = ﬂex)2 + (cy)2 = /(4.761)2 + (—18.056)2 18.673 15 _ -1 -18 056 _ _. o _ o - tan "ZT761 _ 75.23 C _ 18.67 lb 8 75.2 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-105 Determine all forces acting on member DEF of the frame shown in Fig. P7-105. SOLUTION For the distributed load: §t14o)(5 - 2 tan 30°) = 269.2 lb §I5 - 2 tan 30°} = 2.564 ft From a free-body diagram for the complete frame: + C EMA = Fxtzl - 269.2(2.564} = 0 FX = 345.1 lb = 345.1 lb -* H. Members BE and CD are two~foroe members; therefore, their lines of action are known as shown on the free-body diagram for member DEF: + —% 2Fx = 345.1 + E cos 75° - D cos 600 = 0 + T ZFY = FY + E sin 75° + 0 sin 60° = o + C EMF 2 E sin 75° (2) + 9 sin 60° (5) = 0 Solving yields: = 319.5 lb D a 320 1b 5 60.0° = -?15.1 lb E u 716 lb r r5.o° 415.0 lb /(Fx 2 + try)2 = /(345.112 + (415.0)2 = 539.7 lb F a 540 lb 2 50.3° ...
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