Q2_solution

# Q2_solution - 1. See ﬁgure below. A beam AB is simply...

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Unformatted text preview: 1. See ﬁgure below. A beam AB is simply supported and subjected to a concentrated force P at the mid-span and a distributed load of intensity q over one—half of the span. Neglect the weight of the beam. (a) Find the reactions at A and B. (10) (b) Find the internal shear force and bending moment at a section of distance 20 inch to the left of B. (15) (c) Draw the shear force and bending moment diagrams for the entire beam AB. Mark the maximum bending moment and its location. (10) P r: lb g :22: 51bit. _ it C“) «roll: zooib (a) *L—‘k— ‘ f C3) : 26?” (lo _. . .7 t7 =- 0 (3) A7! 30 T'léo 'fl%00 (2) Tm»): ,4):70 [[9 and Bj=l7olb 4olb [b ‘00”) L> FBD ‘"~M v ( : t J” ” M“ t (3’) A :79”, 5] :I7o/b /——— 7 (w Vs~7wb (5’) M 5 17:2 x?0—-/00x/o : 24w[[b~a‘n) 2. Determine the range of mass m for which the lOO-kg block is in static equilibrium. Neglect the mass of the pulley. (a) Assume all wheels and pulleys have negligible friction. (15) (b) Determine the range when the coefﬁcient of static friction between the belt and the pulleys is 0.1. Assume that both pulleys do not rotate. (10) Hint: For part I), use the formula for belt friction: T2 = T16”; . (a) F59 N; 2T Q Lower boom-d 0% m a 0 Tl“: 2m} 09w" = 01/511420" ~ﬂsWW2" =57/V 2) m, :1 3142/ C2,) Z“ 0 /@ «(WWW 1mm“: WSWO :Z7'7L (7) + i 2. 1/1/1251 005/00; WSCMZOD'I'MSol/(oszaa ’- ..—v- élz/l/ - é? w; 3/7/37, (2) ﬁg 77 1 mgfasloo (I) t ’90 72 z 7-. 6%‘81 , ﬂso‘l, E‘Tﬁﬁ (I) “ iMPZ M30! 37‘ 7—3 " 72, e ) 1L” (9 L WZY bound: ‘ 0 0 WSW?" 1' T; +75+ﬂSWOJSZ> (D TL 3 7, a”! (I) 7; ‘1: TL eﬂpz- (t) 2) WM =1W 163/ (I) ' o ,g W a) @ bound LUSH/x20 : 72 “(’73 a #5 59);; I m 7}, : (84¢, a U) 73f:_z,€”upz" .ﬁ 3. See ﬁgure below. The ZOO-kg slider A is held in place on the smooth vertical bar by the cable AB. (a) Find the magnitude and direction of the force in the cable. (15) (b) Find the magnitude and direction of the contact force between the slider and the vertical bar. (10) (c) Find the moment about 0 due to the force in the cable exerted on the hook at B. (15) (a) W :. m3:1762//‘/ W :—-[7{Zj (2) w m '" - e (Z) VL 7 “ms v C‘) IA5:*Z&+§£TZ£ % Y“ I 6mg “ ﬂag” =- -0.q/85'V“z, +0.72%} (5) w Y‘ )Msi ‘l'OJI’S’S’v/f /V (S normed #9 +hL VerhcoJ bar) they; B ag/w'lt'énwm M W W 7 0.7lg u) El: :4) £1762. +£7 2.0 => 5g 1 = 45°}; H7612 +1391]: (2-) [fl/Z]:- (g‘I’X/l/ (2) é” :0‘70 v2; -“‘ 0‘7D7vféz) a—l : r JBA (3) wwx :cfj+25)xcmﬂjzlﬂz : & 2 cf 0 3’ 2r ' (3 07 «I762, ~I3o7 ( zo> jﬁyUﬁ V\o ...
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## This note was uploaded on 04/18/2011 for the course E M 306 taught by Professor Rodin during the Spring '08 term at University of Texas.

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Q2_solution - 1. See ﬁgure below. A beam AB is simply...

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