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Soln06075
Course: CIVE 1150, Spring 2008School: Toledo
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Complete COSMOS: 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 with ABC and add joints alphabetically), with r = 3, m = 13, n = 8 so r + m = 16 = 2n so completely constrained and determinate ! Structure (c): Simple truss...
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Complete COSMOS: 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 with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8
so
r + m = 16 = 2n
so completely constrained and determinate !
Structure (c):
Simple truss r with = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by any equilibrium equation.
structure is improperly constrained !
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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Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 136.Free-Body Diagram:Fx = 0: Fy = 0: Fy = 0:M O = 0:Bx + C x = 0, or Bx = - C x(1) (2) (3)Ay + By + C y = 0 Az - P = 0, or Az = P = 40.0 lbrOA A + rOB B +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 137.Free-Body Diagram:Fx = 0: Fy = 0: Fz = 0:M O = 0:Bx + C x = 0, or Bx = - C x(1) (2) (3)Ay + By + C y = 0Az - P = 0, or Az = P = 60.0 NrOA A + rOB B +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 128.Free-Body Diagram:Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net moment about this point is zero. (a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 129.Free-Body Diagram:Express the tension in terms of its rectangular components: uuuu r DG = - ( 4.8 in.) j - ( 9 in.) kTDC = TDGuuuu r DG = TDG DG- 4.8 j - 9
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 24.L, mm1 2 3 4 5 75 150x , mmy , mmxL, mm 22812.5 0 45 000 0 5625.0 53 437yL, mm 20 11 250 037.5 0 95.492 0 47.7460 75 0(150 )= 471.2475-112.5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 94. FBD AC:M c = 0:( 0.6 m ) Ax - ( 0.06 m )(170 N ) = 0,A x = 17.00 N-17 N + 15 (170 N ) - Cx = 0, 17 Cx = 133 NFx = 0:FBD CE:Fx = 0:133 N -15 (170 N )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 3.A, mm 21x , mmxA, mm3 729 000 2 460 375 3 189 3751 90 270 = 12 150 2 1 135 270 = 18 225 22 ( 90 ) = 60 3 90 + 1 (135) = 135 3230375xA 3189375 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 88.(a)Let R be the single equivalent force. ThenR = (120 N ) kM B : - a (120 N ) = - ( 0.165 m )( 90 N ) cos15 + ( 0.201 m )( 90 N ) sin15 a = 0.080516 mR = 12
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 82.F= F =Fx2 + Fy2 + Fz2( 450 N )2 + ( 600 N )2 + ( -1800 N )2F = 1950 N !cos x =Fx 450 N = F 1950 N x = 76.7 !cos y =Fy F=600 N 1950 N y = 72.1 !
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 144.Free-Body Diagram:Express forces in terms of their rectangular components: uuur BG - 40 i + 74 j - 32 k 37 16 20 TBG = TBG j- k = TBG = TBG - i + 2 2 2 BG 45
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 145.Free-Body Diagram:Express forces in terms of their rectangular components: uuur BG - 40 i + 74 j - 32 k 37 16 20 = TBG = TBG - i + TBG = TBG j- k 2 2 2 BG 45
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 146.Free-Body Diagram:Express forces in terms of their rectangular components:W1 = W2 = - ( 30 lb ) j uuuu r ( x - 6) i + y j - 6 k BH T=T =T BH ( x - 6 )2 + y 2 +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 70.Structure (a): Non-simple truss with r = 3, m = 16, N = 10, so 2N > m + r partially constrained ! Structure (b): Non-simple truss with r = 3, m = 15, N = 10, so 2N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 59.The mass of the lamp shade is given bym = V = Atwhere A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 87.(a) FBD AC:Note: CE is a two-force memberM A = 0: 2 FCE = 0, 120 N m - ( 0.4 m ) 2 FCE = 150 2 NE x = 150.0 N E y = 150.0 N Fx = 0:(b) FBD CE:!! !
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 31.\By inspection of joint C:FBC = 0 FBE = 0Then, by inspection of joint B: Then, by inspection of joint E:FDE = 0 FFH = 0 and FHI = 0By inspection of joint
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 71.The distributed load can be represented in terms of resultants: R1 = ( 8 m )( 300 N/m ) = 2400 NR2 =1 ( 8 - a ) m ( 2400 N/m ) = 1200 ( 8 - a ) N 21 - 8 Ay
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 142.Free-Body Diagram:There is only one unknown of interest and, therefore only one equation is needed:M AB = 0Geometry: = tan -1 1.05 m = 16.2602 3.5 m
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 49.Based on M x = ( P cos ) ( 0.225 m ) sin - ( P sin ) ( 0.225 m ) cos (1) (2) (3)M y = - ( P cos )( 0.125 m ) M z = - ( P sin )( 0.125 m )ByEquation
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 118.From the solutions to Problems 2.111 and 2.112, haveTBE + TCF + TDG = 0.2 65-TBE sin 45 + TCF sin 30 - TDG sin15 = 0(2)(3)TBE cos 45 + TCF cos 30 - TDG co
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 119.d AB = d AC =( - 30 ft )2 + ( 24 ft )2 + ( 32 ft )2= 50 ft = 38 ft( - 30 ft )2 + ( 20 ft )2 + ( -12 ft )2TAB = TAB AB =TAB - ( 30 ft ) i + ( 24 ft ) j
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 12.FBD Truss:Fx = 0:Ax = 0By symmetry: A y = B y = 4.90 kNandFAB = FEG , FAC = FFG , FBC = FEF FBD = FDE , FCD = FDFJoint FBDs: Joint A:Fx = 0: Fy = 0:5
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 82.FBD Frame:M A = 0:( 0.2 m ) Bx - ( 0.36 m ) 2 180 N = 0 2 Bx = 162 2 N, Fx = 0: -162 2 N + Ax - Ax = 252 2 N, Fy = 0: By -B y = 90 2 N,B x = 229 N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 97.Free-Body Diagram:Fx = 0:M D = 0:Dx = 0( - 7 in.) i C + ( 2 in.) i + ( 3 in.) k ( 530 lb ) j + ( -192 lb ) k + ( -3 in.) i + ( 6 in.) j ( -96 lb )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 43.Free-Body Diagram:Geometry: Equation of the slot: y =x2 4 2x dy = 1.20000 = slope of slot at C = 4 dx C ( x = 2.4 in.) It follows for the angles tha
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 81.FBD Ring:(a)M A = 0: FBC =(8 in.) ( FBC cos 20 - 6 lb ) = 0FBC = 6.39 lb C 6 lb cos 20 = 0, cos 20 Ax = 6 lb6 lb , cos 20 Ax -(b)Fx = 0:Fy = 0:Ay -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 41.For y2 ThenNow and foratx = a, y = b : a = kb2y2 = b 1/2 x aork =a b2xEL = x 0 x a y b x1/2 x1/2 : yEL = 2 = , dA = y2dx = b dx 2 2 2 a aFora 1 b
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 77.(a)Fx = (180 lb ) cos 35 sin 20 = 50.430 lbFx = 50.4 lb !Fy = - (180 lb ) sin 35 = -103.244 lb Fy = -103.2 lb ! Fz = (180 lb ) cos 35 cos 20 = 138.555 lb Fz =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 125.FBD half-span:sB = 65 m,hB = 30 mw = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m2 2 yB = c 2 + sB()( c + hB )2c=2 = c 2 + sB2 22 2 ( 65 m ) - ( 30 m )
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 92.FBD Pipe 2:Fx = 0:Fy = 0:NG - ND -3 ( 220 lb ) = 0, 5 4 ( 220 lb ) = 0, 5NG = 132 lb N D = 176 lbFBD Pipe 1: = 90 - 2 tan -1Fx = 0:3 = 16.2602 43 (
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 88.! " EB = ( 36 in.) i - ( 45 in.) j + ( 48 in.) kEB =( 36 in.)2 + ( - 45 in.)2 + ( 48 in.)2! " EB EB= 75 in.TEB = TEBEB = TEB TEB =60 lb ( 36 in.) i - ( 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 89.! " BA = ( 4 m ) i + ( 20 m ) j - ( 5 m ) kBA =F = F BA( 4 m )2 + ( 20 m )2 + ( - 5 m )2= 21 m! " BA 2100 N ( 4 m ) i + ( 20 m ) j - ( 5 m ) k = F = 21
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 90.! " DA = ( 4 m ) i + ( 20 m ) j + (14.8 m ) kDA =F = F DA( 4 m )2 + ( 20 m )2 + (14.8 m )2= 25.2 m! " DA 1260 N ( 4 m ) i + ( 20 m ) j + (14.8 m ) k = F
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 138.The forces applied at A are:TAB , TAC , TAD and Pwhere P = Pj . To express the other forces in terms of the unit vectors i, j, k, we write uuu r AB = - ( 0.72
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 72.Structure (a): Simple truss (start with ABC and add joints alphabetical to complete truss), withr = 4, m = 13, n = 8sor + m = 17 > 2n = 16Structure is comple
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 140.Free-Body Diagram:Express the forces in terms of rectangular components:W = - ( mg ) j = - ( 3 kg ) 9.81 m/s 2 j = - ( 29.43 N ) j()N B = N B ( 0.8j + 0.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 96.First note For small values of and :A = a sin = b sin a = b= V = P ( a + b ) cos - 2Q ( a + b ) cos a = ( a + b ) P cos - 2Q cos b a dV a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 136.The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Fig. 5.8A
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 26.Have where andM C = rA/C FBA rA/C = ( 0.96 m ) i - ( 0.12 m ) j + ( 0.72 m ) k FBA = BA FBA - ( 0.1 m ) i + (1.8 m ) j - ( 0.6 m ) k ( 228 N ) = ( 0.1)2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 119. FBD Blade:Fy = 0:3 kN -12 FBD = 0: FBD = 3.25 kN C 13FBD ABC:M C = 0: 0.045 m + ( 0.3 m ) sin 30 P sin 30 + 0.030 m + ( 0.3 m ) cos 30 P cos 30 -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 62.(a) and (b)M B = 0:( 0.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) - ( 9.6 ft ) AyA y = 12.625 kips=0Shear Diag: dV V is piecewise co
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 52.Free-Body Diagram:Fx = 0:- FB cos15 + 2.4 kN + (1.9 kN ) sin15 = 0orFB = 2.9938 kN FB = 2.99 kNFy = 0:FD - (1.9 kN ) cos15 + ( 2.9938 kN ) sin15 = 0FD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 78.(a)Fx = (180 lb ) cos 30 cos 25 = 141.279 lbFx = 141.3 lb !Fy = - (180 lb ) sin 30 = - 90.000 lb Fy = - 90.0 lb ! Fz = (180 lb ) cos 30 sin 25 = 65.880 lb Fz
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 103.Given:l AB = 3.6 in. lBC = 1.6 in. lCD = 1.2 in.lDE = 1.6 in.lEF = 1.6 in. lFG = 4.8 in.Assume y A : yC =(1.6 in.) y = 4 y ( 3.6 in.) A 9 A4 9 yD
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 33.For the element (EL) shownAtx = a, y = h: h = k1a3ork1 =h a3ora = k 2 h3Hence, on line 1k2 =a h3y =and on line 2h 3 x a3 h 1/3 x a1/3y =Th
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 10.Free-Body Diagram:For Qmin , TD = 0 M B = 0:( 7.5 kN )( 0.5 m ) - Qmin ( 3 m ) = 0Qmin = 1.250 kNFor Qmax , TB = 0 M D = 0:( 7.5 kN )( 2.75 m ) - Qmax ( 0.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 148.FBD Whole:M A = 0: ( 5.4 ft ) Bx - ( 7.8 ft )( 90 lb ) = 0B x = 130 lbFBD BE with pulleys and cord:M E = 0: ( 5.4 ft )(130 lb ) - ( 7.2 ft ) By+ ( 4.8 ft
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 65.From the solution of Problem 3.60: DI = 0.97014 i - 0.24254 jTEG = ( 2.4 lb ) i - ( 7.2 lb ) j - ( 23.4 lb ) k TEG = 24.6 lbM DI = - 224.74 lb ftOnly the pe
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 68FBD Section ABDC: Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be slack, FCF = 0Fx = 0:M D = 0:20
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 64.At Collar A .Have For stretched lengthL = ABFs = k ( L - LAB ) AB(12 in.)2 + (16 in.)2L = 20 in. ABFor unstretched lengthLAB = 12 2 in.ThenFs = 4 l
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 63.TBC must be perpendicular to FAC to be as small as possible.Free-Body Diagram: CForce Triangle is a Right Triangle(a) We observe: (b)TBC = ( 400 lb ) sin 60
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 74.(a)Fx = - ( 810 lb ) cos 45 sin 25 = - 242.06 lbFx = -242 lb !Fy = - ( 810 lb ) sin 45 = - 572.76 lb Fy = - 573 lb ! Fz = (810 lb ) cos 45 cos 25 = 519.09 lb
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 69.(a)M = Fd12 N m = F ( 0.45 m ) or F = 26.7 N (b)M = Fd12 N m = F ( 0.24 m ) or F = 50.0 N (c)M = FdWhere d =( 0.45 m )2 + ( 0.24 m )2= 0.51 m12 N
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 97.(a) FBD Entire Machine:M A = 0:(145 in.) (2B) - ( 35 in.)(18 kips )- (170 in.)( 50 kips ) = 0 2B = 62.966 kips, B = 31.5 kips!Fy = 0:2 A + 62.966 kips - 1
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 50.For steel/steel contact, s1 = tan -1 s1 = tan -1 ( 0.3) = 16.6992For steel/concrete interface, s2 = tan -1 s2 = tan -1 ( 0.6 ) = 30.964FBD Plate CD:Fy = 0
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 131.l = total length yB = c cosh or xB , c L + c = c cosh 1 = cosh L L - 2c c L = 4.933 c L/2 cSolving numerically, sB = c sinh So xB cl L = c sinh 2 2cl = 10 ft
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 62.FBD Section above a-a:M G = 0:( 5.9 m ) 27 FIK - ( 2.7 m )( 40 kN ) - ( 5.4 m )( 40 kN ) = 0 793 FIK = 57.275 kN,Fy = 0: 27 ( 57.275 kN - FGJ ) = 0 793
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 46.First note that by symmetry y = 0. Using the area element shown in the figure,xEL =2 2 r cos = R cos 2 cos 3 3dA =1 2 1 r d = R 2 cos 2 2 d 2 2 1 1 4 4
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 131.FBD BD: = 90 - 25 = 65M B = 0: 2 ( 0.2 m ) cos 25 D sin 65 -15 N m = 0 D = 45.654 NFBD AC:M A = 0:M A = ( 0.2 m )( 45.654 N ) = 0M A = 9.13 N m!
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 61.\FBD Truss:Fx = 0: A x = 0By symmetry Ay = Ny = 1.6 kipsFBD Joint J:Fx = 0: - FGJ - FHJ = 0, FHJ = - FGJBy inspection of joint I, FIJ = 0.6 kip CFBD Sec
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 130.Free-Body Diagram:Express forces and moments in terms of rectangular components:FDE = FDE- 40 i - 70 j + 40k( - 40 )2+ ( - 70 ) + ( 40 )22=FDE