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### Soln09112

Course: CIVE 1150, Spring 2008
School: Toledo
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Word Count: 161

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Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 112. From Problem 9.111 have, I x I y - I xy = constant Now consider the following two cases Case 1: Case 2: Then or From Figure 9.13B: Now With then Finally I x = I x , I x = I max , I y = I y , I y = I min , I xy = I xy I xy = 0 2 I x I y - I xy = I max I min I xy = I x I y - I max I min I x = 453 103 mm 4 I ave = I y =...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 112. From Problem 9.111 have, I x I y - I xy = constant Now consider the following two cases Case 1: Case 2: Then or From Figure 9.13B: Now With then Finally I x = I x , I x = I max , I y = I y , I y = I min , I xy = I xy I xy = 0 2 I x I y - I xy = I max I min I xy = I x I y - I max I min I x = 453 103 mm 4 I ave = I y = 166 103 mm 4 1 1 Ix + I y = I max + I min 2 2 ( ) ( ) I max = 524 103 4 I mm min = ( 453 + 166 - 524 ) 103 mm 4 = 95.0 103 mm 4 I xy = ( 453)(166 ) - ( 524 )( 95.0 ) 103 mm 4 = 159.43 103 mm 4 The two roots corresponding to the following orientations of the cross section: (a) (a) I xy = -159.4 103 mm 4 (b) I xy = 159.4 103 mm 4 (b) 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 9, Solution 111.First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R are the same for all orientations of the coordina
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 109.First assumeIx &gt; I y(Note: Assuming I x &lt; I y is not consistent with the requirement that the axis corresponding to the I xy obtained by rotating the x axis t
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 108.HaveI ave =1 1 Ix + I y = 640 in 4 + 280 in 4 = 460 in 4 2 2()()1 1 Ix - I y = 640 in 4 - 280 in 4 = 180 in 4 2 2()()Also have Letting the
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 105.Given:I x = 0.166 106 mm 4 , I y = 0.453 106 mm 4 and I xy &lt; 0Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 104.From Prob. 9.43 and 9.77I x = 2703.7 in 4Define Points: ThenI y = 4581.0 in 4x ( 2703.7, -1635.18 ) in 4 y ( 4581.0, 1635.18 ) in 4I xy = -1635.18 in 4I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 101.From Problems 9.74 and 9.83I x = 0.166 106 mm 4 ,Define points NowI y = 0.453 106 mm 4 ,andI xy = -0.1596 106 mm 4Y ( 0.453, 0.1596 ) 106 mm 4X ( 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 100.From Problems 9.73 and 9.81I x = 162.86 106 mm 4I y = 325.72 106 mm 4 I xy = 138.24 106 mm 4Define points NowX (162.86,138.24 ) 106 mm 4Y ( 325.72, -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 97.8From Problem 9.79: Problem 9.67:Ix = I xy =a4Iy =2a41 4 a 2The Mohr's circle is defined by the diameter XY, where1 X a4, a4 2 8Now anda
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 96.HaveI x = 9.45 in 4I y = 2.58 in 4From Problem 9.78 Now2I xy = 2.8125 in 4I ave = Ix + I y 2= 6.015 in 4andR= Ix - I y + I xy 2 ( )2= 4.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 93.From Problems 9.73 and 9.81I xy = 138.24 106 mm 4I x = 51.84 106 mm 4= 162.86 106 mm 4I y = 103.68 106 mm 4= 325.72 106 mm 4NowI ave =1 Ix + I y 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 92.From the solution to Problem 9.72: Problem 9.80:I xy = 501.1875 in 4I x = 865.6875 in 4I y = 4758.75 in 4Now1 I x + I y = 2812.21875 in 4 2()1 I x -
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 90.From Problems 9.78 and 9.84I xy = 2.81 in 4I x = 9.45 in 4I y = 2.58 in 4Then1 I x + I y = 6.015 in 4 2()1 I x - I y = 3.435 in 4 2()Equation (9
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 121.At Then Nowx = a, y = b:y = b 2 x a2b = ka 2ork =b a2dm = r 2 dx( ) b = 2 x 2 dx a Then2m = b2 a 4 x dx a4 0a1 b2 = 4 x5 5 a 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 117.Mass = m = V = tA m I mass = tI area = I area A 1 A = bh 2(a)Axis AA: 1 b 3 1 3 I AA, area = 2 h = hb 12 2 48 m m 1 3 1 I AA, mass = I AA, area
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 113.Mass = m = tAI mass = tI area =Area = A =m I area A1 2 a 2I AA, area = I DD, area =I AA, mass = I DD, mass =1 4 1 4 a = a 2 4 8m m 1 4 1 2 I AA,
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 106.From Figure 9.13I x = 9.45 in 4I y = 2.58 in 4 I xy = - 2.81 in 4X ( 9.45, - 2.81) in 4 Y ( 2.58, 2.81) in 4From Problem 9.78The Mohr's circle is defined b
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 102.From Problems 9.75 and 9.82I x = 0.70134 106 mm 4 ,NowI y = 7.728 106 mm 4 ,I xy = 1.5732 106 mm 4I ave =1 1 I x + I y = ( 0.70134 + 7.728 ) 106 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 98.From the solution to Problem 9.72:I xy = 501.1875 in 4From the solution to Problem 9.80:I x = 865.6875 in 4I y = 4758.75 in 41 I x + I y = 2812.21875 in 4 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 94.From Problems 9.75 and 9.82I x = 0.70134 106 mm 4I y = 7.728 106 mm 4 I xy = 1.5732 106 mm 4NowI ave =1 I x + I y = 4.2147 106 mm 4 2()andR=
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 86.First note that cable tension is uniform throughout, henceFSP1 = FSP2k1x1 = k2 x2x2 =k1 6 lb/in. x1 = x1 k2 3 lb/in.x2 = 2 x1Now, with C midway between
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 87.Stretch of Springs = AB - rs = 2 ( r cos ) - r s = r ( 2cos - 1)Potential Energy:V = V =1 2 ks - Wr sin 2 2W = mg1 2 2 kr ( 2 cos - 1) - Wr sin 2 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 91.Consider a small clockwise rotation of the plate about its center. ThenV = 2VP + 4VSPwherea VP = P cos 2 = 1 ( Pa cos ) 2 1 2 kySP 2andVSP =Now
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 122.r1 = 2 r2a = L m = VNowor Now 2 2 = ( 2L )( 2r2 ) - ( L )( r2 ) 3 3 7 = Lr22 3 3 m = 7 Lr22 r -r r = 2 1 x + r1 L x = r2 2 - L dI z = dI z + x 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 118.From Prob. 9.117: 1 I AA = mb 2 24 1 I BB = mh 2 18Note that AA and BB are centroidal axes. 1 mb 2 + md 2 Hence I DD = I AA + md 2 = 24I DD = I EE = I BB + md
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 114.Mass = m = V = tAI mass = tI area =m I area AArea = A = r22 - r12 = r22 - r12()I AA, area =4r24 -4r14 =4(r4 2- r14)(a) I AA
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 110.Consider the regular pentagon shown, with centroidal axes x and y. Because the y axis is an axis of symmetry, it follows that I xy = 0. SinceI xy = 0,the x an
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 107.From Figure 9.13B: Have NowI x = 7.20 106 mm 4 ,I xy = I xyI y = 2.64 106 mm 4I xy = I xy + x yA and I xy = 0( )1 + ( I xy )2 ,wherex1 =102 - 25.3
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 103.From the solution to Problem 9.71I xy = 300 103 mm 4NowI x = ( I x )1 - ( I x )2 - ( I x )33 3 2 1 1 = (120 mm )( 80 mm ) - 2 ( 50 mm )( 20 mm ) + 1000 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 99.From the solution to Problem 9.76 NowI xy = 567 in 4I x = ( I x )1 - ( I x )2 - ( I x )3 , where( I x ) 2 = ( I x )3=4(15 in.)4 - 2 ( 9 in.)( 6 in.)
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 95.From Problems 9.74 and 9.83I x = 0.166 106 mm 4I y = 0.453 106 mm 4 I xy = -0.1596 106 mm 4NowI ave =R=1 I x + I y = 0.3095 106 mm 4 2()and Ix
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 91.8From Problem 9.79:Ix =a4Iy =Problem 9.67:2a4I xy =1 4 a 2The Mohr's circle is defined by the diameter XY, where1 X a4, a4 8 2 Nowand
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 89.From Problems 9.74 and 9.83 I x = 0.166 106 mm 4I y = 0.453 106 mm 4 I xy = -0.1596 106 mm 4Then1 I x + I y = 0.3095 106 mm 4 2()1 I x - I y = -0.143
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 86.From the solutions to Problem 9.72 and 9.80I xy = 501.1875 in 41 I x + I y = 2812.21875 in 4 2()1 I x - I y = -1946.53125 in 4 2()Then Equation (9.2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 88.Have Thenl AD = 2r sin y A = - LAD sin ( 90 - ) - 45 = - 2r sin sin ( 45 - )Also for springs = l AB - r= 2r cos - r = r ( 2 cos - 1)Potential Ener
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 10, Solution 89.HavexC = d sin yB = h cosPotential Energy:1 2 V = 2 kxC + WyB 2 = kd 2 sin 2 + Wh cosThendV = 2kd 2 sin cos - Wh sin d = kd 2 sin 2 - Wh sin a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 84.From Problem 9.78I xy = 2.8125 in 4From Figure 9.13I x = 9.45 in 4 ,NowI y = 2.58 in 41 I x + I y = 6.015 in 4 2()1 I x - I y = 3.435 in 4 2()
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 85.8From Problem 9.79: Problem 9.67:Ix =a4Iy =2a4I xy =2I xy1 4 a 21 2 a4 2 a4 -Now, Equation (9.25):tan 2 m = -Ix - I y=-82a4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 83.From Problem 9.74I xy = -0.1596 106 mm 4From Figure 9.13I x = 0.166 106 mm 4I y = 0.453 106 mm 4Now1 I x + I y = 0.3095 106 mm 4 2()1 I x - I y
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 82.From Problem 9.75 Now where andI xy = 1.5732 106 mm 4I x = ( I x )1 + ( I x )2 + ( I x )3( I x )1 = ( I x )2 = ( I x )3=1 (150 mm )(12 mm )3 = 21 600 mm 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 79.From Figure 9.12:Ix ==16( 2a )( a )38a4Iy ==From Problem 9.67: First note16( 2a )3 ( a )2a4 1 4 a 2I xy =1 1 5 I x + I y = a4 +
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 78.Have For each rectangleI xy = I xy( )1 + ( I xy )2I xy = 0 (symmetry)I xy = I xy + x yAThenand I xy = x yA = ( -0.75 in.)( -1.5 in.) ( 3 in.)( 0.5 in
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 75.Have Now symmetry implies and for the other rectangles ThusI xy = I xy( )1 + ( I xy )2 + ( I xy )3( I xy )1 = 0I xy = I xy + x y A where I xy = 0 (symmetry)
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 74.Have For each rectangle ThusI xy = I xy( )1 + ( I xy )2and I xy = 0 (symmetry)I xy = I xy + Ax y I xy = x y AA, mm 21 2x , mmy , mmAx y , mm 476
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 72.Note: Orientation of A3 corresponding to a 180 rotation of the axes. Equation 9.20 then yieldsI xy = I xySymmetry implies Using Sample Problem 9.6 and Similarl
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 80.From the solution to Problem 9.72I xy = 501.1875 in 4A2 = A3 = 20.25 in 2First compute the moment of inertiaI x = ( I x )1 + ( I x )2 + ( I x )3with( I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 76.Symmetry implies( I xy )1 = 0Using Sample Problem 9.6 and Equation 9.20, note that the orientation of A2 corresponds to a 1 2 2 90 rotation of the axes; thus I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 71.Have Where and NowI xy = I xy( )1 - ( I xy )2 - ( I xy )30and( I xy )1 = 0I xy = I xy + A x y for areas A2 = ( 50 mm )( 20 mm )= 1000 mm 2x2 = -15 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 81.From Problem 9.73,I xy = 138.24 106 mm 4I x = ( I x )1 + ( I x )24 = 2 (120 mm ) 8 ( I x )1 = ( I x )2= 51.84 106 mm 4Iy = Iy( )1 + ( I y )2( I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 77.Have Where Then Section :I xy = I xy + x yAI xy = 0 for each rectangleI xy = I xy( )1 + ( I xy )2 + ( I xy )3 = x yAx1 = - ( 8.92 in. - 8 in.) = - 0.92 i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 73.Have For each semicircle ThusI xy = I xy( )1 + ( I xy )2andI xy = I xy + x y A I xy = x y AI xy = 0 (symmetry)A, mm 21x , mmy , mmAx y , mm 469.
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 128.Fn = 0:or N - T + (T + T ) sin =0 2N = ( 2T + T ) sin Ft = 0: 2 (T + T ) - T cos - F = 0 2 F = T cosF = s Nor Impending slipping: So 2T cos
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 129.Small belt section: Side view: End view:Fy = 0: Fx = 0: Impending slipping:2N sin - T + (T + T ) sin =0 2 2 2 (T + T ) - T cos - F = 0 2F = s N
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 130.FBD motor and mount: Impending belt slip, cw rotations T2 = T1esin 2T2 = T1e sin18 = 58.356T1( 0.40 )M D = 0:(12 in.)(175 lb ) - (13 in.) T1 - ( 7 in
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 127.FBD wrenchNote: EC =( 0.20 m ) ,sin 75EA = EC - 0.03 m = 75so = 285 = 4.9742 radM E = 0: 0.20 m 0.20 m - 0.03 m F - cos 75 - 0.03 m T = 0 sin 7
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 126.FBD wrench:Note: EC =( 0.2 m ) ,sin 65EA = EC - 0.03 m = 65so = 295 = 5.1487 rad 0.20 m 0.20 m - 0.03 m F - cos 65 - 0.03 m T = 0 sin 65 sin
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 125. FBD pin B:(a) By symmetry: 2 Fy = 0: B - 2 2 T1 = 0 T1 = T2or B= 2T1(1)FBD Drum:For impending rotation : T4 &gt; T2 = T1 &gt; T3 , so T4 = Tmax = 5.6 kN
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 121.Geometry and force notation: = sin -15 in. = 30 = rad, so contact angles are: 10 in. 6C = -6=5 2 = , D = + , 6 2 6 3E =2(a) D and E fixed,
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 122.FBD drum B:M B = 0:( 0.02 m )(TA - T ) - 0.30 N m = 0TA - T = 0.30 N m = 15 N 0.02 mImpending slip: TA = Te s = Te0.40 Solving; T e0.40 - 1 = 15 N()
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 123.FBD drum B:M B = 0:( 0.02 m )(TA - T ) - 0.3 N m = 0TA - T = 15 NImpending slip:TA = Te s B = Te0.40Solving, T e0.40 - 1 = 15 N()T = 5.9676 NIf
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 119.Geometry and force notation: = sin -1r = 30 = , so contact angles are: 2r 6C = D =2+6=2 , 3E = (a) D and E fixed, so slip on these surfaces
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 120.Geometry and force notation: = sin -15 in. = 30 = rad, so contact angles are: 10 in. 6C = -6=5 2 = , D = + , 6 2 6 3E =2(a) All pulleys loc