Assignment 10

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Assignment 10 Key solutionsChapter 6, Problem 86. Determine the components of the reactions atAand Ewhen a counterclockwise couple of magnitude 192 lb in. isapplied to the frame (a) at B, (b) at D. Chapter 6, Solution 86. (a)FBD AC:(b)FBD CE:Note: CE is a two-force member ( ) ( )1 10: 8in. 2in. 192lb in. 02 2A CE CEM F F _ _ + , ,19.2 2lb,CEF 19.20lbx E t19.20lby E t0: 19.2lb=0,x xF A 19.20lbx A t 0: 19.2lb=0,y yF A 19.20lby A tNote: AC is a two-force member ( )1 40: 3in. 192lb in. =017 17E AE AEM F F _ + + ,12.8 17lb,AEF 4,17x AEA F 51.2lbx A t1,17y AEA F 12.80lby A t0: 51.2lb=0,x xF E 51.2lbx E t 0: 12.80lb=0,y yF E 12.80lby EtChapter 6, Problem 89. The 120-N load can be moved along the line of action shown and can be applied atA,D, or E. Determine the components of the reactions at B and F when the 120-N load is applied (a) at A, (b) at D, (c) at E.Chapter 6, Solution 89. (a)FBD ACF:(b)FBD BCE:Note: BC is a two-force member ( ) ( ) ( )10: 0.1m 120 N 0.3m 0,26F BCM F _ ,40 26 N,BCF 200Nx B t40.0Ny B t( )50: 40 26 N 0,26x xF F 200Nx Ft ( )10: 120 N 40 26 N 0,26y yF F 160.0Ny F tNote ACF is a two-force member ( ) ( ) ( )10: 0.4 m 120 N 0.3m 0,5B CFM F _ ,160 5 N,CFF 320Nx Ft160.0Ny F t( )20: 160 5 N 0,5x xF B 320Nx Bt ( )10: 160 5 N 120 N =05y yF B 40.0Ny B t(c) Moving the 120 N force from D to E does not affect the reactions. The answers are the same as in part (b). tChapter 6, Problem 95. Knowing that the pulley has a radius of 75 mm, determine the components of the reactions at A and B. Chapter 6, Solution 95. FBD Frame & pulley:FBD AD: ( ) ( ) ( ) 0: 0.6m 0.075m 240N 0,B yM A 30.0Ny At 0: 30.0N 240N 0,y yF B + 270Ny B t0: 0,x x x x xF A B A B + ( ) ( )( ) 0: 0.200m 0.075m 240ND xM A ( ) ( ) 0.30m 30N 0 + 45.0Nx A tFrom above ,x xA B 45.0Nx BtChapter 6, Problem 100. For the frame and loading shown, determine the components of the forces acting on member ABC at B and C.Chapter 6, Solution 100. FBD Frame:FBD DE:FBD ABC: ( ) ( )( ) ( ) ( ) 0: 6in. 13.5in. 48lb 16.5in. 20lb 0A xM D 163lbxD 0: A 20lb 48lb+163lb=0x xF + +231lbxA 0: 0 y yF A ( ) ( ) ( ) 0: 19.5in. 163lb 6in. 0E xM B 529.75lbxB ( ) ( ) ( ) 0: 4.5in. 7.5in. 231lb 0, 385lbC y yM B B On ABC: 530lbx B,385lby Bt0: 231lb 529.75lb 0,x xF C + 298.75lbxC 0: 385lb 0,y yF C 385lbyC On : 299 lbxABC C ,385lby CtChapter 6, Problem 106. Knowing that P = 411 lb and Q = 0, determine for the frame and loading shown (a) the reaction at D, (b) the force in member BF.Chapter 6, Solution 106. \FBD Frame: ( ) ( ) ( ) 0: 12in. 28in. 411lb =0A xM D 959lbx DFBD DF: Note that BF and CE are two-force members.4 150: 959lb+ 05 17x CE BFF F F ( ) ( )3 80: 12in. 34.5in. 05 17D CE BFM F F _ _ + , ,Solving:357lb,BFF ( ) ( ) ( )801: 22.5in. 357lb 12in. 017E yM D 1 1 ]315lby D , so (a)1009lb D18.18 t(b)357lbTBFF t Chapter 6, Problem 118. The shear shown is used to trim electronic-circuit-board laminates. Knowing thatP = 400 N, determine (a) the vertical component of the force exerted on the shearing blade atD, (b) the reaction at C.Chapter 6, Solution 118. FBD ABC: ( ) ( ) 0: 0.045m+ 0.30msin30 400Nsin30CM1 1 ] ]( ) ( ) 0.030m+ 0.30mcos30 400Ncos301 1 + ] ]( ) ( )12 50.03m 0.045m 013 13BD BDF F _ _ , ,3097.64NBDF ( ) ( )50: 3097.64 N 400 N sin30 013 x xF CFBD Blade:991.39Nx C ( ) ( )120: 3097.64 N 400 N cos30 013 y yF C2512.9Ny C (a) Vertical component at ( )123097.64 N13D==2.86kNt(b)2.70kN C 68.5 tChapter 6, Problem 122. The control rod CEpasses through a horizontal hole in the body of the toggleclampshown. Determine(a) theforceQrequiredtoholdthe clamp in equilibrium, (b) the corresponding force in link BD.Chapter 6, Solution 122. FBD ABC:( )145 mm sin 25 10 mmsin 5.1739100 mm 1 ] ( ) ( ) ( ) 0: 70 mm 110 N 45 mm sin 25 cos5.1739C BDM F1 ]

( ) 45 mm cos 25 sin5.1739 01 + ]BDF504.50 NBDF ( ) ( ) 0: 504.50 N cos5.1739 110 N sin 25 0,x xF C 455.96 NxC FBD CE:0: 455.96 N = 0xF Q 455.96 N, Q (a)456 N Q t(b)540 N TBDF tChapter 7, Problem 1. For the frame and loading of Prob. 6.99, determine the internal forces at a point J located halfway between points B and E.Chapter 7, Solution 1. FBD FRAME:FBD JEHDF: 0:AM ( ) ( ) ( ) 0.3 m 0.4 m 900 N 0xD 1200 Nx D0:xF 1200 N + 0 V1200 N V 0: 0yF F 0:JM ( ) ( ) 0.15 m 1200 N 0 M 180 N m M + Thus, ( ) on JE 0 F t1200 N Vt180.0 N m M t Chapter 7, Problem 5. Determine the internal forces at point J of the structure shown. Chapter 7, Solution 5. FBD Frame:Note:AB is a two-force member, so12 5yxAA (1)( ) ( ) ( ) 0: 15 in. 24 in. 78 lb 0C xM A 124.8 lbx AFrom (1) above,52.0 lby A FBD AJ:0: 124.8 lb + 0xF F 124.8 lb Ft 0: 52 lb 0yF V 52.0 lb Vt( ) ( ) 0: 10 in. 52 lb 0JM M 520 lb in. MtChapter 7, Problem 6. Determine the internal forces at point K of the structure shown.Chapter 7, Solution 6. FBD CD:Note: AB is a two-force member( ) ( )5 120: 18 in. 7.5 in.13 13C AB ABF F _ _ + , ,( ) ( ) 24 in. 78 lb 0 135.2 lbABF ( )120: 135.2 lb 013x xF C 124.8 lbx C( ) ( )50: 135.2 lb 78 lb 013y yF C + FBD CK:26 lby C ( ) ( )12 50: 124.8 lb 26 lb 013 13xF F + + 125.2 lb F22.6 t( ) ( )5 120: 124.8 lb 26 lb 013 13yF V + 24.0 lb V67.4 t( ) ( ) ( ) ( ) 0: 5 in. 124.8 lb 12 in. 26 lb 0KM M 312 lb in. M t