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- Title: Soln06054
- Type: Notes
- School: Toledo
- Course: CIVE 1150
- Term: Spring
Coursehero >> Ohio >> Toledo >> CIVE 1150
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
Path: Toledo >> CIVE >> 1150 Spring, 2008
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Complete COSMOS: Online Solutions Manual Organization System Chapter 6, Solution 54. FBD Truss: Fx = 0: By symmetry, Ax = 0 A y = M y = 7.5 kips FBD Section IJNM: M J = 0: ( 9 ft )( 7.5 kips - 0.5 kips ) - ( 4.5 ft )(1 kip ) FGI = 11.4858 kips. FGI = 11.49 kips T ! 4 - ( 5.25 ft ) FGI = 0, 17 M G = 0: (12 ft )( 7.5 kip - 0.5 kips ) - ( 7.5 ft )(1 kip ) - ( 3 ft )( 3 kips ) ( + 4.5 ft ) FHJ = 0 FHJ = 15 kips C By inspection of joint H, Fy = 0: FFH = 15.00 kips C ! 1 (11.4858 kips ) + 7.5 kips - 4.5 kips 17 - 3 FGJ = 0 13 FGJ = 6.95 kips T ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
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Soln06055
Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 55. FBD Truss: Fx = 0: A x = 0 By symmetry: A y = v y = 4.5 kips FBD Joint K: Fx = 0: 1 ( FIK - FKN ) = 0 2 1 FIK - 1 kip = 0 2 2 FIK = FKN FIK = Fy = 0: 2 ki...
Soln06056Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 56. FBD Truss: Fx = 0: M N = 0: Ax = 0 ( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft ...
Soln06057Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 57. FBD Truss: Fx = 0: M N = 0: Ax = 0 ( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft ...
Soln06044Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 44. FBD Truss: Fx = 0: Ax = 0 By load symmetry: A y = L y = 840 N M E = 0: 2 24 (1.75 m ) FBD + ( 2 m )( 240 N ) 3 25 - ( 4 m )( 840 N - 120 N ) = 0 FBD Section ABC...
Soln06045Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 45. FBD Truss: Fx = 0: Ax = 0 By load symmetry A y = L y = 840 N slope: GI = FBD Section JKL: 0.4 2 = 3 15 3.5 7 IK = 3 = 4 24 2.3 23 IJ = = 3 30 Resolve FIK and F...
Soln06041Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 41. (a) To check for simple truss, start with ABDE and add three members at a time which meet at a single new joint, successively adding joints G, F, H and C. This is ...
Soln06102Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 102. FBD Stool: w 1 N = ( 56 kg ) 9.81 = 274.68 N 2 2 kg M A = 0: ( 0.350 m ) By - ( 0.150 m )( 274.68 N ) = 0, By = 117.72 N FBD BE: M E = 0: (1.25 m )(117.72...
Soln06103Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 103. FBD BC: Note that only one of the forces B y , By exists. M C = 0: M C B y or B - ( 0.15 m ) y ( ) 1 FFG = 0 2 2 FFG = M B or B y 0.15 m C y ( ) (1...
Soln06105Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 105. Member FBDs: II: M C = 0: M A = 0: ( 7.2 ft ) Bx - ( 24 ft ) By - (12 ft )(14 kips ) = 0 (12.8 ft ) Bx + ( 32 ft ) By - ( 20 ft )( 21 kips ) = 0 Solving: Bx = 2...
Soln06106Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 106. \\ FBD Frame: M A = 0: (12 in.) Dx - ( 28 in.)( 411 lb ) = 0 D x = 959 lb FBD DF: Note that BF and CE are two-force members. Fx = 0: M D = 0: 959 lb + (12 in...
Soln06107Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 107. FBD Frame: M A = 0: (12 in.) Dx - ( 34.5 in.)( 274 lb ) = 0 Dx = 787.75 lb FBD DF: M D = 0: Fx = 0: (12 in.) 3 8 FCE + ( 34.5 in.) FBF - 274 lb = 0 5...
Soln06027Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 27. P6.14 Starting with ABC, add, in order, joints E, D, F, G, H simple truss P6.15 Starting with DEF, add, in order, G, C, B, A, I, H, J simple truss P6.23 Starti...
Soln06028Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 28. P6.21 Starting with ABC, add, in order, joints E, D, F, H, J, K, L, I, G, N, P, Q, R, O, M, S, T simple truss P6.25 Starting with ABC, add, in order, joints D, E, ...
Soln07055Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 55. M A = 0: a Dy - ( 5 ft )( 500 lb ) - (10 ft )( 500 lb ) = 0 Dy = 7500 lb ft a Since there are no distributed loads, M is piecewise linear, so only points C and D...
Soln05080Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 80. First, determine the force on the dam face without the silt. Have Pw = 1 1 Apw = A ( gh ) 2 2 1 ( 6 m )(1 m ) 103 kg/m 3 9.81 m/s 2 ( 6 m ) 2 = ( )( ) ...
Soln05002Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 2. A,in 2 1 2 10 8 = 80 x ,in. 5 13 y ,in. 4 xA,in 3 400 702 yA,in 3 320 216 1 9 12 = 54 2 134 4 1102 536 Then and X = Y = xA 1102 = A 134 yA 1102 = A ...
Soln02097Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 97. Have TAB = ( 760 lb )( sin 50 cos 40i - cos 50j + sin 50 sin 40k ) TAC = TAC ( - cos 45 sin 25i - sin 45 j + cos 45 cos 25k ) (a) R A = TAB + TAC ( RA ) x = 0 ...
Soln05015Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 15. A, mm 2 1 2 3 18 240 -1920 x , mm y , mm xA, mm3 72 960 yA, mm3 218 880 -4 - 56 - 41.441 12 54 107520 168 731 -134171 -103 680 - 4071.5 - 41.441 186 731...
Soln07091Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 91. FBD Cable: M A = 0: 4aE y - 3a (1.2 kips ) - 2a ( 0.8 kips ) - a ( 0.4 kips ) = 0 E y = 1.4 kips Fy = 0: Ay - ( 0.4 + 0.8 + 1.2 kips ) + 1.4 kips = 0 A y = 1.0...
Soln07058Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 58. (a) FBD Beam: M C = 0: LAy - M 0 = 0 Ay = Fy = 0: - Ay + C = 0 M0 L M0 L C= M0 dV = w = 0 . at A, and remains constant L dx Moment Diag: M starts at zer...
Soln06091Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 91. \\ First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. M C = 0: rT1 - rT2 = 0 T1 = T2 (a) Replace eac...
Soln02110Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 110. See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below. - 0.61905 TAB + 0.111111TAC + 0.21212 TAD = 0 - 0.761...
Soln05040Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 40. Note that y1 = 0 at x = a, or 0 = 2b 1 - ka 2 , i.e. k = ( ) 1 a2 Also, note that the slope of y2 is - 3b and y2 = 0 at x = 2a. Therefore a y2 = 3b ( 2a - ...
Soln05004Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 4. A,in 2 1 x ,in. 2 ( 21) = 14 3 21 + 1 (13) = 27.5 2 40 - y ,in. 1 ( 24 ) = 32 3 xA,in 3 3528 yA,in 3 8064 1 ( 21)( 24 ) = 252 2 2 (13)( 40 ) = 520 772 20 1...
Soln04080Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 80. Free-Body Diagram: Note that the member is a three-force body. In the free-body diagram, E is the intersection between the lines of action of the three forces. Fro...
Soln07080Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 80. (b) (a) Distributed load Shear: x w = w0 4 - 1 L dV = -w, and V ( 0 ) = 0, so dx V = V = 0 x - wdx = - wo L 1 - x/L 0 x Lwd L 0 x/L 2 x x ...
Soln07073Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 73. (a) Note: C x omitted to avoid clutter. M C = 0: - ( 0.5 m )( 2 kN/m )( 2 m ) + (1.5 m )(1 kN ) - ( 3 m )( 4.5 kN/m )(1 m ) + ( 3.5 m ) By = 0 B y = 4.0 kN Fy =...
Soln07084Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 84. (a) FBD section AC: M C = 0: ( 395 - 215 ) N m + ( 0.1 m )(1000 N/m )( 0.2 m ) - ( 0.2 m )VA = 0, VA = 1000 N M B = 0: 395 N m + ( 0.45 m )(1 kN/m )( 0.4 m ) -...
Soln07048Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 48. FBD Angle: Fy = 0: T - C y = 0 Cy = T M C = ( 0.3 ft ) T M C = 0: M C - ( 0.3 ft ) T = 0, By symmetry, Dy = T and M D = ( 0.3 ft ) T (a) Beam AB: Fy = 0: 2T ...
Soln07111Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 111. Cable profile: Eqn: y= w 2 x 2T0 xB - x A = 45 in. at A: at B: 4 in. = 1 in. = w ( xB - 45 in.)2 2T0 w 2 xB 2T0 =4 (1) (2) or ( xB - 45 in.)2 2 xB 2 xB + ...
Soln07083Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 83. M A = 0: LBy - V =- w0 L + 20 L1 w0 L = 0 54 By = w0 L 20 x1 0 w0 3 w 4 x dx1 = 0 3 5x1 - L4 3 1 20L L or V = ( ) w0 4 5 L - x ) - L4 3 ( 20 L ...
Soln10101Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 10, Solution 101. From sketch y A = 4 yC Thus, (a) Virtual Work: y A = 4 yC U = 0: P= 1 F 4 P y A - F yC = 0 F = 300 N: P= 1 ( 300 N ) = 75 N 4 P = 75.0 N (b) Free body...
Soln05027Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 27. The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity of the wire must lie directly under B. Also, because the wire is ...
Soln04057Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 57. Free-Body Diagram: First note T = tension in spring = ks where s = elongation of spring = AB ( ) - ( AB ) = 90 90 = 2l sin - 2l sin 2 2 1 = 2l...
Soln03046Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 46. Have rE = ( 0.90 m ) i + (1.50 m ) j TDE = TDE DE DE - ( 2.30 m ) i + (1.50 m ) j - ( 2.25 m ) k = (1349 N ) 2 2 2 ( - 2.30 ) + (1.50 ) + ( - 2.25) m = - ( ...
Soln06024Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 24. FBD Truss: Fx = 0: M A = 0: Ax = 0 (12 m ) (M y - 1 kN) - (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0 M y = 5.05 kN Fy = 0: Ay - 2(1 kN) - 5(1.5 kN) + M...
Soln03058Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 58. (a) For edge OA to be perpendicular to edge BC, uuu uuu r r OA BC = 0 where From triangle OBC ( OA) x ( OA) z = a 2 a 1 a = 2 3 2 3 = ( OA ) x tan 30 = a...
Soln03055Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 55. Have M AD = AD rE/ A FEF ( ) where AD = AD = AD AD ( 7.2 m ) i + ( 0.9 m ) j ( 7.2 m )2 + ( 0.9 m )2 = 0.99228 i + 0.124035 j rE/ A = ( 2.1 m ) i - ( ...
Soln03052Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 52. First note: TBA = TBA BA BA = ( 70 lb ) ( 4 ) i + 1.5 - ( LBC + 1) j + ( - 6 ) k + 1) + ( - 6 ) 2 2 2 ( 4 )2 + 1.5 - ( LBC 52 + ( 0.5 - LBC ) = ( 70 l...
Soln05022Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 22. L, in. 1 2 3 4 5 19 15 4 10 8 56 x , in. 9.5 14.5 10 5 0 y , in. 0 6 10 8 4 xL, in 2 180.5 217.5 40 50 0 488 y , in 2 0 90 40 80 32 242 X = Then and xL 488...
Soln06126Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 126. FBD AB: M A = 0: ( 6 in.) FBC cos 30 - 660 lb in. = 0, FBC = 127.017 lb T FBD Piston: 180 lb - FCE cos = 0, 180 lb C cos Fy = 0: FCE = FBD Joint C: Fy = 0:...
Soln06063Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 63. FBD Section ACFBD: M D = 0: ( 4.3 m ) 27 FFI - ( 2.7 m )( 40 kN ) = 0 793 FFI = 26.196 kN, 27 ( 26.196 kN - FDG ) = 0 793 FFI = 26.2 kN C ! Fy = 0: FDG = 26...
Soln07114Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 114. FBD Cable: M B = 0: LACy + aT0 - M B loads = 0 (Where M B loads includes all applied loads) (1) x M C = 0: xACy - h - a T0 - M C left = 0 L (2) FBD AC: ...
Soln06110Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 110. Member FBDs: M A = 0: ( 0.3 m ) FBF - ( 0.9 m ) 4 FDG = 0 5 M E = 0: Solving: 4 - ( 0.3 m ) FBF + ( 0.6 m ) FDG - ( 0.9 m )( 0.8 kN ) = 0 5 FBE = - 3 kN, FBF ...
Soln06111Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 111. Member FBDs: M A = 0: ( 0.6 m ) 4 4 FCH - ( 0.3 m ) FBG = 0 5 5 M E = 0: Solving: 4 4 - ( 0.9 m ) FCH + ( 0.6 m ) FBG - ( 0.9 m )( 0.8 kN ) = 0 5 5 FBG = 6 kN...
Soln06099Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 99. FBD Frame: M A = 0: ( 0.3 m ) Dx = - ( 0.4 m ) ( 900 N ) = 0, Ax = 1.200 kN, Dx = 1200 N D x = 1.200 kN Fx = 0: 1.200 kN - Ax = 0, Fy = 0: A x = 1.200 kN A y =...
Soln02051Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 51. Free-Body Diagram: Fx = 0: FC + ( 2.3 kN ) sin15 - ( 2.1 kN ) cos15 = 0 or FC = 1.433 kN Fy = 0: FD - ( 2.3 kN ) cos15 + ( 2.1 kN ) sin15 = 0 or FD = 1.678 kN ...
Soln07132Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 132. w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m L = 48 m, Tmax 1800 N ( ) Tmax = wyB yB = Tmax w yB 1800 N = 61.162 m 29.43 N/m x yB = c cosh B , c Solving numeri...
Soln02071Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 71. Free-Body Diagram: Pulley C Fx = 0: TACB ( cos 30 - cos 50 ) - P cos 50 = 0 or P = 0.34730TACB (1) Fy = 0: TACB ( sin 30 + sin 50 ) + P sin 50 - 2000 N = 0 or ...
Soln05008Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 8. A, mm 2 1 2 35 343 x , mm y , mm xA, mm3 2 250 006 yA, mm3 0 63.662 31.831 0 - 4417.9 30925.1 - 31.831 -140 626 2 109 380 140 626.2 140 626.2 Then and ...
Soln06165Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 165. Member FBDs: FBD I: M C = 0: R ( FBD sin 30 ) - R (1 - cos 30 ) ( 200 N ) - R ( 100 N ) = 0 FBD = 253.6 N Fx = 0: (b) FBD = 254 N T - C x + ( 253.6 N ) c...
Soln07014Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 14. FBD AB: B = 0: LP - hAx = 0, Ax = L P h Geometry: y = kx 2 , at B: h = kL2 , so k = ha 2 L2 h hx 2 , y= 2 L2 L at J: y J = ka 2 = slope = dy = 2kx, dx a...
Soln06122Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 122. FBD ABC: = sin -1 M C = 0: ( 45 mm ) sin 25 - 10 mm 100 mm = 5.1739 ( 70 mm )(110 N ) - ( 45 mm ) sin 25 FBD cos 5.1739 + ( 45 mm ) cos 25 FBD sin 5.1739...
Soln06033Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 33. By inspection of joint C: FBC = 0 ! By inspection of joint M : FLM = 0 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russe...
Soln06120Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 120. FBD Stamp D: Fy = 0: E - FBD cos 20 = 0, E = FBD cos 20 M A = 0: FBD ABC: ( 0.2 m )( sin 30)( FBD cos 20) + ( 0.2 m )( cos 30 )( FBD sin 20 ) - ( 0.2 m ) sin 3...
Soln07045Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 45. FBD CE: Fx = 0: Fy = 0: M C = 0: Cy = 0 C y - 4 kN = 0 C y = 4 kN M C - ( 0.5 m )( 4 kN ) = 0 M C = 2 kN m Fx = 0: Ax = 0 Ay - 4 kN - 2 kN - 1 k = 0 Ay = 7 ...
Soln07041Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 41. (a) FBD Beam: Fy = 0: ( 4 m )( w) - ( 2 m )(12 kN/m ) = 0 w = 6 kN/m Along AC: Fy = 0: - x ( 6 kN/m ) - V = 0, V = - ( 6 kN/m ) x V = -6 kN at C ( x = 1 m ) ...
Soln06029Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 29. Fx = 0: Fx = 0 FFG = 0 FGH = 0 FIJ = 0 FHI = 0 FEI = 0 Then, by inspection of joint F, Then, by inspection of joint G, By inspection of joint J, Then, by inspect...
Soln06118Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 118. FBD ABC: M C = 0: 0.045 m + ( 0.30 m ) sin 30 ( 400 N ) sin 30 + 0.030 m + ( 0.30 m ) cos 30 ( 400 N ) cos 30 12 5 - ( 0.03 m ) FBD - ( 0.045 m ) F...
Soln07069Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 69. (a) M F = 0: (18 ft )(1 kip/ft )( 6 ft ) - (12 ft ) Dy + ( 4.5 ft )( 2 kips/ft )( 9 ft ) + ( 3 ft )( 33 kips ) = 0 D y = 24 kips Fy = 0: 24 kips + Fy + 33 kips...
Soln06117Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 117. (a) Member FBDs: FBD I: FBD II: M A = 0: aF1 - 2aP = 0 M B = 0: - aF2 = 0 F1 = 2P; F2 = 0 Fy = 0: Ay - P = 0 Ay = P Fx = 0: Bx + F1 = 0, Bx = - F1 = -2P B x...
Soln04152Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 152. Free-Body Diagram: From free-body diagram of beam Fx = 0: Bx = 0 so that B = By Fy = 0: A + B - (100 + 200 + 300 ) N = 0 or A + B = 600 N the other support re...
Soln05014Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 5, Solution 14. First, by symmetry X = 90 mm A, mm 2 1 2 3 y , mm yA, mm3 1 296 000 (180 )(120 ) = 21 600 - - 60 4 ( 90 )(120 ) = - 8482.3 ( 90 )(120 ) = - 8482.3 4635.4 ...
Soln06116Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 116. Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces...
Soln07004Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 4. FBD Frame: Note: AC is a two-force member, resolve FAC at C: M E = 0: ( 0.25 m + 0.25 m ) 1 FAC - ( 0.15 m )( 320 N ) = 0 2 FAC = 96 2 N FBD sect. AB: Fx = 0:...
Soln04158Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 158. Free-Body Diagram: Since Slope of ED is 45 45 yED = xED = a, slope of HC is Also and DE = 2a a 1 DH = HE = DE = 2 2 a/ 2 = R a 2R For triangles DHC and EHC...
Soln04159Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 159. Free-Body Diagram: First note W = mg = (17 kg ) 9.81 m/s 2 = 166.77 N ( ) h= From free-body diagram of plywood sheet (1.2 )2 - (1.125)2 = 0.41758 m (1.125...
Soln04150Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 150. Free-Body Diagram: Express forces in terms of their rectangular components: uuur FG 18 i - 6 j - 9 k T TFG = TFG = TFG = FG ( 6 i - 2 j - 3 k ) 2 2 2 FG 7 (18) + ...
Soln06101Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 101. FBD Frame: M E = 0: ( 6 in.)(84 lb ) - ( 42 in.) Ay = 0 Ay = 12.00 lb FBD CF: (I) FBD ABC: (II) I: II: M D = 0: M B = 0: ( 3.5 in.) Cx + ( 7 in.) C y - ( 5 ...
Soln06075Path: Toledo >> CIVE >> 1150 Spring, 2008
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 75. Structure (a): Rigid truss with r = 3, m = 14, n = 8 so r + m = 17 > 2n = 16 so completely constrained but indeterminate ! Structure (b): Simple truss (start wi...