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Soln09131

Course: CIVE 1150, Spring 2008
School: Toledo
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Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 131. from Consider shell to be formed by removing hemisphere of radius r hemisphere of radius r + t. For hemisphere: I = 2 2 mr 5 Area = 1 4 r 2 = 2 r 2 2 ( ) m = V = 1 4 3 2 r = r 3 2 3 3 I = 2 2 4 3 2 r 5 r r = 53 15 Thus For hemispherical shell: I = 4 4 5 ( r + t ) - r 5 = r 5 + 5r 4t + 10r 3t 2 + ... - r 5 15...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 9, Solution 131. from Consider shell to be formed by removing hemisphere of radius r hemisphere of radius r + t. For hemisphere: I = 2 2 mr 5 Area = 1 4 r 2 = 2 r 2 2 ( ) m = V = 1 4 3 2 r = r 3 2 3 3 I = 2 2 4 3 2 r 5 r r = 53 15 Thus For hemispherical shell: I = 4 4 5 ( r + t ) - r 5 = r 5 + 5r 4t + 10r 3t 2 + ... - r 5 15 15 4 4t r 3 Neglect terms with powers of t > 1, I = Mass of shell: I = 2 r 2t m = V = tA = t 2 r 2 = 2 r 2t ( ) ( )2r 3 2 = 2 2 mr 3 I = 2 2 mr 3 Radius of gyration: k2 = I 2 = r2 m 3 k = 0.816r 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 129. Mass of cylindrical ring:m = V= = 4 g 2 2 d 2 - d1 t g4 4 (d2 2- d12 t)Now treat the wheel as a series of 4 concentric rings. (Note - the stee
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 128.h x+h a 2For line BC =-=Alsoh ( a - 2x ) a1 m = V = t ah 2 =1 tah 2(a) HavedI x ==1 2 dm + dm 12 221 2 dm 3where Thendm =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 127.1First notedy = dxThen dy 1+ dx 22 2 2 -x a 3 - x 3 2 2 2 - = 1 + x 3 a3 - x3 -1 3 a 3 = xFor the element shown2 dy dm = md
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 125.For the element shown:ab y y dm = b a dy = 2 y 2dy h h h For thin plate:dI y = dI x + dI z= =1 y 1 y 1 2 2 2 b dm + a dm = 2 b + a y dm 3
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 134.Have (a) anddB =4 - d A = ( 0.33333 - d A ) ft 122 I AA = I GG + md A 2 I BB = I GG + md BThen2 2 I BB - I AA = m d B - d A()2 2 = m ( 0.33333 - d
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 130.First note for the cylindrical ring shown thatm = V = t and, using Figure 9.28, that(d 42 2- d12 =)42 t d 2 - d12()I AA1 d 1 d = m2
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 126.r 2 = y 2 + z 2 = kx :Atx = h,a2 x hr = a;a 2 = kh;k=a2 hThus, r 2 =Have dm = r 2dx = Then m =ha2 xdx hh 0 dm = h 0 xdxa2=Now1 a
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 52.Channel:A = 3780 mm 2 I x = 32.6 106 mm 4I y = 1.14 106 mm 4NowIx = 2 Ix( )C + ( I x )plate= 2 32.6 106 mm 4 +()= 65.2 106 + 2.25d 106 mm 4A
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COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 51.Shape Data:Fig. 9.13AS10 35:A = 10.3 in 2 I x = 147 in 4I y = 8.36 in 4C10 20:A = 5.88 in 2 I x = 78.9 in 4I y = 2.81 in 4Combined section:A = AS
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 48.Locate centroid:y1 =4 (12 in.) 3A1 =2(12 in )2=16in.= 72 in 2y2 = 4 in.A2 = - (12 in.)( 8 in.)= - 96 in 2Theny = yi Ai i 16 2 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 47.Locate centroid:A1 = (16 )( 2.3) = 36.8 in 2x1 = 4 in. x2 = 4 in. x3 = 2 in.y1 = -1.15 in. y2 = 3.2 in. y3 = 1.6 in.A2 =1 (12 )( 9.6 ) = 57.6 in 2 2 1 A3 =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 44.Locate centroid:x1 = 20 mm y1 = 45 mm A1 = ( 40 mm )( 90 mm )= 3600 mm 2x2 = 50 mm y2 = 51 mm A2 == 720 mm 21 ( 48 mm )( 30 mm ) 2Thenx ==xi Ai i(
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 43.Locate centroid:x1 = 8 in. y1 = 3 in. A1 = (16 )( 6 ) = 96 in 2 x2 = -1.5 in. y2 = 2 in. A2 = ( 3)( 4 ) = 12 in 2 x3 = 14 in. y3 = - 5.25 in. A3 = ( 4 )(10.5 ) =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 40.HaveJ A = J C + Ad122 J B = JC + A d2 + a2() ( )(a) Then orJ B = 3 J A; d1 = d 2 = 2.5 in.3 J C + Ad12 = J C + A d12 + a 2()a2 = 2JC + 2d12 A
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 39.HaveJ A = J C + Ad122 J B = JC + A d2 + a2d1 = 2aThenJ A - JB( = A ( 3a) )( )a = 1.500 in.22 - d2Substituting2 256 in 4 - 190 in 4 = 24 in 2
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 36.HaveI x = ( I x )1 - ( I x )2 - ( I x )33 1 = ( 3a )( 2a ) - a 4 - a 4 12 8 8 = 2 - - a4 = 2 - a4 8 8 4 or I x = 1.215a 4 AlsoIy = I
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 35.HaveI x = ( I x )1 + ( I x )2 + ( I x )33 3 1 1 = ( 2a )( 4a ) + ( a )( 3a ) 3 3 2 2 4a 4a + a 4 - a 2 + a 2 3a + 4 3 4 3 16
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 31.Area: A = 2 (10 mm )( 40 mm ) + ( 90 mm )(10 mm ) = 1700 mm 2 Part :I x = I x + Ad 2 =1 (10 mm )( 40 mm )3 + (10 mm )( 40 mm )( 25 mm )2 12= 303.3 103 mm
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 28.h x b 2By observationy =b y 2hor Now And Thenx= b dA = xdy = 2hdI x = y 2dA = y dy b 3 y dy 2hhI x = dI x = 2 0b 3 y dy 2hb y4 = h 4From
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 27.Have:A = 2 04 0a cos 2rdrd= 04 a 2 cos 2 2 dsin 4 4 =a + 8 0 2=Have:2 8a2J O = r 2dA= 2 04 0a cos 2r 2 ( rdrd )1 = 04 a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 24.First note: x = r 2 - y2JP = Ix + I yr r 2 - y2I x = y 2dA = 2 r 02y 2dxdy= 2 r y 2 r 2 - y 2 dy2rLet Then Thus Nowy = r sin dy = r cos d ;y
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 142.FBD wheel:M E = 0: or- M E + ( 7.5 in.)(T2 - T1 ) = 0 M E = ( 7.5 in.)(T2 - T1 )FBD lever:M C = 0: or Impending slipping: or So( 4 in.)(T1 + T2 ) - (16 i
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 138.FBD pin C:FAB = P sin10 = 0.173648P FBC = P cos10 = 0.98481P Fy = 0:FBD block A:N A - W - FAB sin 30 = 0 N A = W + 0.173648P sin 30 = W + 0.086824Por Fx = 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 137.FBD Collar:Stretch of spring x = AB - a =a -a cos a 1 - a = (1.5 kN/m )( 0.5 m ) - 1 Fs = k cos cos 1 = ( 0.75 kN ) - 1 = ( 750 N )( sec - 1)
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 134. FBDs Top block:(a) Note: With the cable, motion must impend at both contact surfaces. Fy = 0: Impending slip: Fx = 0: N1 - 40 lb = 0 N1 = 40 lbF1 = s N1 = 0.4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 8, Solution 133. FBD block (impending motion to the right) s = tan -1 s = tan -1 ( 0.25 ) = 14.036P W = sin s sin ( - s )sin ( - s ) =W sin s PW = mg(a) ( 30 kg ) 9
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 14.First note: or Have-a At x = a : b = k 1 - e a k =b 1 - e-1I y = x 2dAb= = a 1- e 0 0-1 -x 1- e a 2 x d ydx-x 0 x 1 - e-1 1 - e a
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 9.x2 y2 + 2 =1 a2 bx = a 1-y2 b2dA = xdy dI x = y 2dA = y 2 xdyI x = dI x = - b xy 2dy = a - b y 2 1 -Set: y = b sin bby2 dy b2dy = b cos dI x
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 2.Atx = a, y = b :b = ka 25ork =b a25 y =b a5 2x25orx=a b2 5y52dI y =1 3 x dy 3=1 a3 6 y 5 dy 3 b6 51 a3 b 6 y 5 dy 3 b6 0
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 124.For the element shown:ab y y dm = dV = a b dy = 2 y 2dy h h h dIx =Parallel-axis theorem1 y 1 b 2 2 ab 2 1 ab3 4 y 2 y dy = y dy b dm =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 123.At Then Nowx = 2a2a = k ( 2a )3ork =y=1 3 x 4a 21 4a 2dm = r 2dx()2 6 1 = 2 x3 dx = x dx 16a 4 4a Thenm==16a 442a 6 a x
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 120.First locate the centroid CX A = xA:or1 X 2a 2 + a 2 = a 2a 2 + 2a + 2a a 2 3 ()( )( )X =14 a 9Z A = zA:or (a) Have1 1 Z 2a 2 + a 2 =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 119.mass = m = V = tA1 = t ( 2a )( a ) + ( 2a )( a ) 2 First note= 3 ta 2AlsoI mass = tI area=(a) Nowm I area 3a 2I x, area = ( I x )1, area + (
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 116.Locate centroid:y1 = y2 =Thenb 2 3 b 2y =A1 = 3ab A2 = abyi Ai ib 3 ( 3ab ) + b ( ab ) 2 = 2 3ab + ab=Uniform thickness: (a)3 b 4 3 m 4 m2 = 1 m 4
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 115.First notemass = m = V = tA= t4(r2 2- r12)AlsoI mass = tI area=4( )m r22 - r12)I area(a) Using Figure 9.12, ThenI AA, area =
Toledo - CIVE - 1150
COSMOS: Complete Online Solutions Manual Organization SystemChapter 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 FinallyI x = I
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 > I y(Note: Assuming I x < 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 < 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
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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
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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
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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
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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
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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
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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
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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