Statics and Mechanics of Materials 2nd Edition

ISBN 13
978-0073398167
ISBN 10
0073398160
Authors
David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf
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151 documents found.
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978-0073398167 Chapter 10 Solution Manual Part 1
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. PROBLEM 10.1
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978-0073398167 Chapter 10 Solution Manual Part 2
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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978-0073398167 Chapter 10 Solution Manual Part 3
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. PROBLEM 10.21 A torque of magnitude 100 N mT= is applied to shaft AB of the gear train shown. Knowing that the diameters of the three solid shafts
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978-0073398167 Chapter 10 Solution Manual Part 4
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. consent of McGraw-Hill Education.
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978-0073398167 Chapter 10 Solution Manual Part 5
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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978-0073398167 Chapter 10 Solution Manual Part 6
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Gear B. 0:BM= 0/B
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978-0073398167 Chapter 10 Solution Manual Part 7
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. PROBLEM
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978-0073398167 Chapter 11 Solution Manual Part 1
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. For rectangle: 12I bh=
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978-0073398167 Chapter 11 Solution Manual Part 10
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. SOLUTION Continued
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978-0073398167 Chapter 11 Solution Manual Part 11
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. SOLUTION 346411(0.120 m)(0.06 m)
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978-0073398167 Chapter 11 Solution Manual Part 12
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. The maximum stress occurs
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978-0073398167 Chapter 11 Solution Manual Part 2
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. consent of McGraw-Hill Education.
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978-0073398167 Chapter 11 Solution Manual Part 3
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 600 32.5 19.5 10
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978-0073398167 Chapter 11 Solution Manual Part 4
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Use wood as the
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978-0073398167 Chapter 11 Solution Manual Part 5
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. consent of McGraw-Hill Education.
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978-0073398167 Chapter 11 Solution Manual Part 6
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. PROBLEM
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978-0073398167 Chapter 11 Solution Manual Part 7
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. (a) Centric loading: 4
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978-0073398167 Chapter 11 Solution Manual Part 8
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 11 Solution Manual Part 9
Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Locate centroid. 2, inA
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978-0073398167 Chapter 12 Solution Manual Part 1
PROBLEM 12.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION 2 consent of McGraw-Hill Education. PROBLEM 12.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine
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978-0073398167 Chapter 12 Solution Manual Part 10
PROBLEM 12.76 (Continued) Shape 3(in )S W27 84 213 W24 68 154 Lightest W-shaped section: W24 68 W21 101 227 W18 76 146 W16 77 134 W14 145 232 consent of McGraw-Hill Education. PROBLEM 12.76 (Continued) Shape 3(in )S W27 84 213 W24 68 154 Lightest W-shaped section: W24 68
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978-0073398167 Chapter 12 Solution Manual Part 11
PROBLEM 12.84 Knowing that beam AB is in equilibrium under the loading shown, draw the shear and bending-moment diagrams and determine the maximum normal stress due to bending. SOLUTION A to C: 0 1.2 ftx << 020203250 1 50 41.6671.241.667 50(41.667 50)0 20.833 500 (20.833 50 )6.944 25xAxAxxwxdV wx
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978-0073398167 Chapter 12 Solution Manual Part 2
PROBLEM 12.8 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. SOLUTION A to C: 0 4 ftx<< 0 : 2 0 2 kipsyF Vx V x = = = 0
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978-0073398167 Chapter 12 Solution Manual Part 3
PROBLEM 12.18 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C. SOLUTION PROBLEM 12.18 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C. SOLUTION
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978-0073398167 Chapter 12 Solution Manual Part 4
PROBLEM 12.26 Solve Prob. 12.25, assuming that 480 NP= and 320 N.Q= PROBLEM 12.25 Knowing that 480 N,PQ = = determine (a) the distance a for which the absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due
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978-0073398167 Chapter 12 Solution Manual Part 5
PROBLEM 12.36 Using the method of Sec. 12.2, solve Prob. 12.8 PROBLEM 12.8 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. SOLUTION PROBLEM 12.36 Using the method of Sec.
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978-0073398167 Chapter 12 Solution Manual Part 6
PROBLEM 12.44 Using the method of Sec. 12.2, solve Prob. 12.17. PROBLEM 12.17 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C. SOLUTION PROBLEM 12.44 Using the method of Sec. 12.2, solve Prob. 12.17. PROBLEM 12.17 For
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978-0073398167 Chapter 12 Solution Manual Part 7
PROBLEM 12.53 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending. SOLUTION 0 : (2)(1) (3)(4)(2) 4 05.5 kNCMBB = +== 0 : (5)(2) (3)(4)(2) 4 08.5 kNBMCC= + == Shear: A to C: 2 kNV= :C
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978-0073398167 Chapter 12 Solution Manual Part 8
PROBLEM 12.60 For the beam and loading shown, design the cross section of the beam, knowing that the grade of timber used has an allowable normal stress of 1750 psi. SOLUTION PROBLEM 12.60 For the beam and loading shown, design the cross section of the beam, knowing that
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978-0073398167 Chapter 12 Solution Manual Part 9
PROBLEM 12.69 Knowing that the allowable normal stress for the steel used is 24 ksi, consent of McGraw-Hill Education. PROBLEM 12.69 Knowing that the allowable normal stress for the steel used is 24 ksi, consent of McGraw-Hill Education. PROBLEM 12.69 Knowing that the allowable normal stress for
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978-0073398167 Chapter 13 Solution Manual Part 1
PROBLEM 13.1 Three boards, each of 1.5 3.5-in. rectangular cross section, are nailed together to form a beam that is subjected to a vertical shear of 250 lb. Knowing that the spacing between each pair of nails is 2.5 in., determine the shearing force in each nail. SOLUTION 3
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978-0073398167 Chapter 13 Solution Manual Part 2
PROBLEM 13.10 For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a. SOLUTION By symmetry, .ABRR= 0: 10 10 010 kipsy ABABF RRRR= + == = From the shear diagram, 10 kips at
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978-0073398167 Chapter 13 Solution Manual Part 3
PROBLEM 13.20 For the beam and loading shown, determine the largest shearing stress in section n-n. SOLUTION 25 kipsABRR= = At section n-n, 25 kipsV= Locate centroid and calculate moment of inertia. Part 2(in )A (in.)y 3(in )Ay d(in.) 24(in )Ad 4(in )I 4.875 6.875 33.52 2.244 24.55 0.23
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978-0073398167 Chapter 13 Solution Manual Part 4
PROBLEM 13.30 (a) point a, (b) point b. SOLUTION 3 3 646411(80)(80) (56)(68) 1.9460 10 mm12 121.946 10 mI=== (a) 11 2 233632(56)(6)(37) (2)(12)(40)(20) 31.632 10 mm31.632 10 m(2)(12) 24 mm 0.024 maaQ Ay Ayt= +=+== = = = 3666(150 10 )(31.632 10 ) 101.6 10 Pa (1.946
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978-0073398167 Chapter 13 Solution Manual Part 5
PROBLEM 13.40 A beam consists of five planks of 1.5 6-in. cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt
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978-0073398167 Chapter 13 Solution Manual Part 6
PROBLEM 13.49 A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 629 10 psi for the steel and 610.6 10 psi for the
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978-0073398167 Chapter 13 Solution Manual Part 7
PROBLEM 13.59 Solve Prob. 13.58, assuming that the beam is subjected to a horizontal shear V. PROBLEM 13.58 An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stress 9 ksi,= determine
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978-0073398167 Chapter 14 Solution Manual Part 1
PROBLEM 14.1 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 14.2. SOLUTION SOLUTION SOLUTION Stresses
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978-0073398167 Chapter 14 Solution Manual Part 2
PROBLEM 14.11 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress. SOLUTION 18 ksixs=12 ksiys= 8 ksixy= (a) 18 12tan 2 1.8752 (2)(8)xysxy ss + = == 2 61.93s=
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978-0073398167 Chapter 14 Solution Manual Part 3
PROBLEM 14.21 A 400-lb vertical force is applied at D to a gear attached to the solid l-in. diameter shaft AB. Determine the principal stresses and the maximum shearing stress at point H located as shown on top of the shaft. SOLUTION Equivalent force-couple system at center of
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978-0073398167 Chapter 14 Solution Manual Part 4
PROBLEM 14.30 Solve Prob. 14.14, using Mohrs circle. PROBLEM 14.13 through 14.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise. SOLUTION ave8 ksi,12 ksi,6 ksi2 ksi2xyxyxysssss== = += = Plotted points for
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978-0073398167 Chapter 14 Solution Manual Part 5
the maximum shearing stress at point K. SOLUTION 102 51 mm 45 mm 22oo iodr rrt= = = = = ( )4 4 6 4 644.1855 10 mm 4.1855 10 m2oiJ rr= = = 6412.0927 10 m 2IJ= = Force-couple system at center of tube in the plane containing
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978-0073398167 Chapter 14 Solution Manual Part 6
PROBLEM 14.48 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown. SOLUTION PROBLEM 14.48 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition
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978-0073398167 Chapter 14 Solution Manual Part 7
PROBLEM 14.58 The bulk storage tank shown in Photo 14.3 has an outer diameter of 3.3 m and a wall thickness of 18 mm. At a time when the internal pressure of the tank is 1.5 MPa, determine the maximum normal stress and the maximum shearing stress in the
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978-0073398167 Chapter 14 Solution Manual Part 8
PROBLEM 14.67 (Continued) ave 1( ) 35.69 MPa 2xy = += 229.40 MPa2xy xy R = += max ave 45.1MPaR = += max 45.1MPa = max(in-plane) 9.40 MPaR = = max (in-plane)9.40 MPa= consent of McGraw-Hill Education. PROBLEM 14.67 (Continued) ave 1( ) 35.69 MPa 2xy = += 229.40
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978-0073398167 Chapter 14 Solution Manual Part 9
PROBLEM 14.74 For the state of stress shown, determine the range of values offor which the magnitude of the shearing stress is equal to or less than 8 ksi. SOLUTION ave222216 ksi, 06 ksi1( ) 8 ksi 22( 8) (6) 10 ksi2(2)(6) tan 2 0.75162 36.87018.435xyxyxyxy xy xypxy pbR====
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978-0073398167 Chapter 15 Solution Manual Part 1
PROBLEM 15.1 For the loading shown, determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free end. SOLUTION 0: ( ) 0 J M M PL x = = ()M
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978-0073398167 Chapter 15 Solution Manual Part 2
PROBLEM 15.9 (Continued) 36 4 64 PROBLEM 15.9 (Continued) 36 4 64 PROBLEM 15.9 (Continued) 3 6 4 64 Data: 60 10 N,P= 26.9 10 mm 26.9 10 mI == 9200 10 PaE= 625.38 10 N mEI = 2mL= (a) 326(60 10 )(2)(16)(5.38 10 )A= 32.79 10 radA= (b) 33 3 6(60 10
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978-0073398167 Chapter 15 Solution Manual Part 3
PROLEM 15.14 (Continued) 33311[ , 0] Eq. (4): ( ) 066Px L y bL L L a C LL= = += 32 32131 1 11()6 6 66PPC C L a bL b bLLL == = Make xa= in Eq. (2). 33 233 2111 ( )666 6CP P ba b
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978-0073398167 Chapter 15 Solution Manual Part 4
PROBLEM 15.21 (Continued) 10, 0 0 0 0dyxCdx= = ++ = 10C= [ ]20, 0000 0xy C= = +++ = 20C= ,2L dy dyxdx dx = = 22311 110028 28AA AAML RL ML RL C++=++ 30C= ,2Lx yy= = 23 23 4 11 1100 008 48 8 48AA AAML RL
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978-0073398167 Chapter 15 Solution Manual Part 5
PROBLEM 15.27 For the beam and loading shown, determine (a) the deflection at point C, (b) the slope at end A. PROBLEM 15.27 For the beam and loading shown, determine (a) the deflection at point C, (b) the slope at end A. PROBLEM 15.27
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978-0073398167 Chapter 15 Solution Manual Part 6
PROBLEM 15.36 (Continued) Slope at B. 315 6.98 10 2148B= = 36.98 10 radB= Deflection at B. 328.125 13.09 10 ft 2148By= = 0.1571 in.By= Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. PROBLEM 15.36 (Continued) Slope at B. 315
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978-0073398167 Chapter 15 Solution Manual Part 7
29 10 psi.E= 29 10 psi.E= 29 10 psi.E= PROBLEM 15.45 The two beams shown have the same cross section and are joined by a hinge at C. For the loading shown, determine (a) the slope at point A, (b) the deflection at point
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978-0073398167 Chapter 15 Solution Manual Part 8
PROBLEM 15.53 Knowing that beam AE is an S200 27.4 rolled shape and that 17.5 kN, 2.5 m, 0.8 mP La= = = and 200 GPa,E= determine (a) the equation of the elastic curve for portion BD, (b) the deflection at the center C of
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978-0073398167 Chapter 15 Solution Manual Part 9
PROBLEM 15.60 For the uniform beam shown, determine the reaction at each of the three supports. SOLUTION Consider CR as redundant and replace loading system by I and II. Loading I: (Case 4 of Appendix C.) 348CCRLyEI = Loading II: (Case 7 of Appendix C.) 32 02062 216CMLL yLEILMLEI
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978-0073398167 Chapter 16 Solution Manual Part 1
PROBLEM 16.1 Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the critical load cr.P SOLUTION PROBLEM 16.1 Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the critical load
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978-0073398167 Chapter 16 Solution Manual Part 2
PROBLEM 16.11 Determine the radius of the round strut so that the round and square struts have the same cross-sectional area and compute the critical load for each. Use 200 GPa.E= SOLUTION For square strut, 2225 625 mmA= = 4 3 4 941(25) 32.552 10 mm 32.552 10 m
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978-0073398167 Chapter 16 Solution Manual Part 3
SOLUTION Buckling in xz-plane. 3,12 edb L LI= = 2 23cr 1 22 23cr 112 () 12 ()( . .) 12 eEI EdbPLL PEdb FS PPL = == = Buckling in yz-plane. 32, 12 ebd L LI= = 2 23cr 2 22 23cr 222 () 12(2
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978-0073398167 Chapter 16 Solution Manual Part 4
PROBLEM 16.30 A sawn lumber column with a 7.5 5.5-in. cross section has an 18-ft effective length. Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is 1200 psiCs= and that the adjusted modulus is 3470 10 psi,E= determine the
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978-0073398167 Chapter 16 Solution Manual Part 5
SOLUTION Continued Determine allP for various values of x. x u v PC all(lb)P 1.0 0.73620 0.22240 0.17087 12,920 1.2 0.78513 0.32026 0.24092 26,227 1.1 0.75955 0.26910 0.20473 18,729 1.09 0.75712 0.26423 0.20124 18,075 1.089 0.75687 0.26374 0.20089 18,011 all18 kips18,000 lbP= = For all 18,000 lb 1.089Px= =
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978-0073398167 Chapter 16 Solution Manual Part 6
PROBLEM 16.47 Solve Prob. 16.46, assuming that the effective length of the column is decreased to 20 ft. PROBLEM 16.46 A square structural tube having the cross section shown is used as a column of 26-ft effective length to carry a centric load of 65 kips. Knowing that the
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978-0073398167 Chapter 16 Solution Manual Part 7
PROBLEM 16.54 Member AB consists of a single C130 10.4 steel channel of length 2.5 m. Knowing that the pins at A and B pass through the centroid of the cross section of the channel, determine the factor of safety for the load shown with respect to
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978-0073398167 Chapter 2 Solution Manual Part 1
PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of
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978-0073398167 Chapter 2 Solution Manual Part 10
consent of McGraw-Hill Education. consent of McGraw-Hill Education. consent of McGraw-Hill Education. PROBLEM 2.86 Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb cylinder is suspended
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978-0073398167 Chapter 2 Solution Manual Part 11
PROBLEM 2.92 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 0,Q= find the value of P for which the tension in cable AD is 305 N. SOLUTION 0: 0A AB AC AD = + + +=F TTTP where
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978-0073398167 Chapter 2 Solution Manual Part 12
SOLUTION Continued Finally, 22 2(0.4 m) (1.6 m) (0.86 m)( 0.4 m) (1.6 m) ( 0.86 m) 1.86 m[ (0.4 m) (1.6 m) (0.86 m) ]1.86 m( 0.2151 0.8602 0.4624 )AE AE AEAEAE AEAEAEAETTAETT=+= + + == ==+=+ ij kT ij kT ijk SOLUTION Continued Finally, 22 2(0.4 m) (1.6
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978-0073398167 Chapter 2 Solution Manual Part 13
PROBLEM 2.107 Knowing that a = 40, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos20 56.382 lb(60 lb)sin 20 20.521lbxyFF= == = 80-lb Force: (80 lb)cos60 40.000 lb(80 lb)sin60 69.282 lbxyFF= == = 120-lb Force: (120 lb)cos30 103.923 lb(120 lb)sin30 60.000 lbxyFF= == =
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978-0073398167 Chapter 2 Solution Manual Part 2
PROBLEM 2.11 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N. SOLUTION Using the law of cosines: 222(1600
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978-0073398167 Chapter 2 Solution Manual Part 3
PROBLEM 2.21 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION 22(650 mm) (720 mm)970 mmBC = + = (a) 650970xPP= or 970650970325
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978-0073398167 Chapter 2 Solution Manual Part 4
PROBLEM 2.31 For the post loaded as shown, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. SOLUTION 960 24 4(500 N) (200 N)1460 25 548 640 N
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978-0073398167 Chapter 2 Solution Manual Part 5
PROBLEM 2.41 A sailor is being rescued using a boatswains chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that = 30 and = 10 and that the combined weight of
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978-0073398167 Chapter 2 Solution Manual Part 6
PROBLEM 2.51 A 600-lb crate is supported by several rope- and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.)
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978-0073398167 Chapter 2 Solution Manual Part 7
PROBLEM 2.61 Solve Problem 2.60, assuming that point A is located 15 north of west and that the barrel of the gun forms an angle of 25 with the horizontal. PROBLEM 2.60 A gun is aimed at a point A located 35 east of north. Knowing that the barrel
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978-0073398167 Chapter 2 Solution Manual Part 8
PROBLEM 2.71 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB:
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978-0073398167 Chapter 2 Solution Manual Part 9
PROBLEM 2.80 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION Free-Body Diagram at A: The forces applied at A are: , , , andAB
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978-0073398167 Chapter 3 Solution Manual Part 1
PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that = 25, determine the moment of the force about Point B by resolving the force into horizontal and vertical components. SOLUTION Free-Body Diagram of
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978-0073398167 Chapter 3 Solution Manual Part 10
PROBLEM 3.87 A machine component is subjected to the forces shown, each of which is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at A. SOLUTION For equivalence :BCD A ++=FF F F R ( ) ( ) ( )
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978-0073398167 Chapter 3 Solution Manual Part 11
PROBLEM 3.96 Three children are standing on a 5 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if
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978-0073398167 Chapter 3 Solution Manual Part 12
PROBLEM 3.106 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular. SOLUTION First note 22 2(12) (12) ( 6) 18
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978-0073398167 Chapter 3 Solution Manual Part 2
SOLUTION Free-Body Diagram of Rod AB: (a) 221350 N (450) (600) 750 mmF AC= = += 450 600cos 0.6 sin 0.8750 750= = = = cos sin(1350 N)0.6 (1350 N)0.8 (810 N) (1080N)FF== += +F i+ j ijijF /(0.45 m) (0.24 m) AB =r ij /( 0.45 0.24
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978-0073398167 Chapter 3 Solution Manual Part 3
PROBLEM 3.21 In Prob. 3.15, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.15 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line
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978-0073398167 Chapter 3 Solution Manual Part 4
SOLUTION (a) Using Eq. (3.30): 2 222 2211 12 24(11) ( 12) (24) 29 in.31 12 24(31) ( 12) (24) 41 in.CACACBCB=+= + + ==+= + + =ijkijk cos ( )( ) (11 12 24 ) (31 12 24 )(29)(41)(11)(31) ( 12)( 12) (24)(24)(29)(41)0.89235CA CBCA CB=+ +=+ +==ijk ijk
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978-0073398167 Chapter 3 Solution Manual Part 5
PROBLEM 3.41 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force AR exerted on the davit at A must not exceed 279 lbft in absolute value. Determine the largest allowable
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978-0073398167 Chapter 3 Solution Manual Part 6
PROBLEM 3.49 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance
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978-0073398167 Chapter 3 Solution Manual Part 7
SOLUTION Continued 123(3.84 2.88 ) 2.40 ( 1.92 1.6 )(1.92 N m) (1.44 N m) (1.28 N m)MM M M= + + = + + = + + j k j ikijk 222( 1.92) (1.44) (1.28) 2.72 N mM=+ + = 2.72 N mM= cos 1.92/2.72cos 1.44/2.72cos 1.28/2.72xyz= == 134.9
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978-0073398167 Chapter 3 Solution Manual Part 8
PROBLEM 3.68 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna. SOLUTION We have 2
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978-0073398167 Chapter 3 Solution Manual Part 9
PROBLEM 3.77 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the
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978-0073398167 Chapter 4 Solution Manual Part 1
PROBLEM 4.1 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. SOLUTION Free-Body Diagram: (a) Reaction at A: 0: 0xxFA= = 0: (15 lb)(28 in.) (20 lb)(22 in.) (35 lb)(14 in.)(20 lb)(6 in.) (6 in.) 0ByMA= + ++
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978-0073398167 Chapter 4 Solution Manual Part 10
SOLUTION Continued Setting the coefficients of the unit vectors equal to zero: ( ) : 450 2.2283 0 CE F+ =k 201.95 lb CE F= or 202 lb CE F= : 1.11417(201.95) ( ) 0 y 225.00 lb ftyAM= : 600 0xAM+ =i 600 lb ftxAM= ( )0: 0.55709
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978-0073398167 Chapter 4 Solution Manual Part 11
PROBLEM 4.78 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when P = 80 lb. SOLUTION Assume equilibrium: 0: (50 lb)sin 30 (80 lb)cos 40 0xFF = + = 36.284 lbF= + 0: (50 lb)cos 30 (80 lb)sin 40
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978-0073398167 Chapter 4 Solution Manual Part 12
SOLUTION Continued 0: (32 in.) (12 in.) (24 in.) 0ABM PW N= + = 8 3 6 0 0.250.3PPW P W+ = = tip( 0.25 OK)P WP= < 0.25(120 lb)P= or 30.0 lb=P (c) Casters at B free, so 0BF= Impending slip: A sAFN= 0: 0x A A sAF
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978-0073398167 Chapter 4 Solution Manual Part 13
PROBLEM 4.95 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each of weight W. Knowing that 80 = and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the largest value of P
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978-0073398167 Chapter 4 Solution Manual Part 14
belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION We replace BT and BT by their resultant ( 180 N)j and CT and CT by their resultant
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978-0073398167 Chapter 4 Solution Manual Part 2
PROBLEM 4.11 The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb. SOLUTION Free-Body Diagram: 0: (5 in.) (7.5
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978-0073398167 Chapter 4 Solution Manual Part 3
PROBLEM 4.20 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F.
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978-0073398167 Chapter 4 Solution Manual Part 4
SOLUTION Since CD is a two-force member, the force it exerts on member ABD is directed along DC. Free-Body Diagram of ABD: (Three-Force member) The reaction at B must pass through E, where D and the 250-N load intersect. Triangle CFD: 60tan 0.610030.964aa= == Triangle EAD: 200tan 120 mm120
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978-0073398167 Chapter 4 Solution Manual Part 5
PROBLEM 4.36 Determine the reactions at A and B when a = 150 mm. SOLUTION Free-Body Diagram: Force triangle 80 mm 80 mmtan 150 mm 28.072a= == 320 Nsin 28.072A= 680 N= A28.1 320 Ntan 28.072B= 600 N= B consent of McGraw-Hill Education. PROBLEM 4.36 Determine the reactions
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978-0073398167 Chapter 4 Solution Manual Part 6
SOLUTION Free-Body Diagram: (Three-Force body) 160 mmtan 300 mm 28.07aa== SOLUTION Free-Body Diagram: (Three-Force body) 160 mmtan 300 mm 28.07aa== SOLUTION Free-Body Diagram: (Three-Force body) 160 mm tan 300 mm 28.07 a a = = PROBLEM 4.45 Solve Problem 4.44, assuming that the 170-N force applied at
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978-0073398167 Chapter 4 Solution Manual Part 7
PROBLEM 4.53 A 4 8-ft sheet of plywood weighing 40 lb has been temporarily SOLUTION Free-Body Diagram: We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained ( 0).xF= / //0: ( ) ( 40 lb) 0A
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978-0073398167 Chapter 4 Solution Manual Part 8
SOLUTION Continued Coefficient of j: 240 240 0 13 17BDTT+ = 17 17 (520) 680 lb 13 13BD BDTT T= = = 0: 320 0AD AE BFC= + + + =F TTT j Coefficient of i: 20 8(520) (680) 052 17xC + += 200 320 0 120 lbxxCC+ +=
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978-0073398167 Chapter 4 Solution Manual Part 9
SOLUTION Continued ( )( ) ( )( ): 0.74278 3 ft 300 lb 1.5 ft 0CEF=k 201.94 lbCEF= or 202 lbCEF= ( ) ( ) ( ) ( ) ( ): 4 ft 0.55709 201.94 lb 4 ft 0.37139 201.94 lb 3 ft 0xA +=j 56.250 lbxA= ( ) (
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978-0073398167 Chapter 5 Solution Manual Part 1
PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION Area 1: Rectangle 72 mm by 45 mm. Area 2: Triangle
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978-0073398167 Chapter 5 Solution Manual Part 10
SOLUTION Continued Then 2 22 22 (1.25312 in. ) 144 in. 0.054678 ftsA= =Finally Number of gallons 400 0.054678= 21.87 gallons= Order 22 gallons consent of McGraw-Hill Education. 22 (1.25312 in. ) 144 in. 0.054678 ftsA= =Finally Number of gallons 400 0.054678= 21.87 gallons= Order 22 gallons consent of
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978-0073398167 Chapter 5 Solution Manual Part 2
PROBLEM 5.11 SOLUTION First note that symmetry implies 0X= 2,mA ,my 3,myA 1 44.5 3 18 3 = 1.2 21.6 2 2(1.8) 5.08942= 0.76394 3.8880 12.9106 17.7120 Then 3217.7120 m12.9106 myAYA = = or 1.372 mY= consent of McGraw-Hill Education. PROBLEM 5.11 SOLUTION First note that symmetry implies
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978-0073398167 Chapter 5 Solution Manual Part 3
SOLUTION SOLUTION SOLUTION PROBLEM 5.21 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion BCD of the wire is horizontal. First note that for equilibrium, the center of gravity of the wire
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978-0073398167 Chapter 5 Solution Manual Part 4
PROBLEM 5.30 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies 0x= For the element (EL) shown 2 (Figure 5.8B) EL r ydA rdr== Then ( )2211222 2122rrrr rA dA rdr r r= = = = and ( )2211 3 33212 12(
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978-0073398167 Chapter 5 Solution Manual Part 5
PROBLEM 5.37 SOLUTION From the solution of Problem 5.1, we have 2332632.5 mm111,172.5 mm63,787.5 mmAxAyA=== Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x-axis: area3Volume 2 22 (63,787.5 mm )y A yA= = = or 33Volume 401 10 mm= ( )line line22 33 4422Area 2 2
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978-0073398167 Chapter 5 Solution Manual Part 6
PROBLEM 5.47 Determine the volume and total surface area of the bushing shown. SOLUTION PROBLEM 5.47 Determine the volume and total surface area of the bushing shown. SOLUTION PROBLEM 5.47 Determine the volume and total surface area of the bushing shown. SOLUTION Volume: The
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978-0073398167 Chapter 5 Solution Manual Part 7
PROBLEM 5.55 Determine the reactions at the beam supports for the given loading. SOLUTION IIIIII(200 lb/ft)(15 ft)3000 lb1(200 lb/ft)(6 ft) 2600 lbRRRR==== 0: (3000 lb)(1.5 ft) (600 lb)(9 ft 2ft) (15 ft) 0AMB= + + = 740 lbB= 740 lb=B 0: 740 lb 3000 lb 600 lb 0yFA= +
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978-0073398167 Chapter 5 Solution Manual Part 8
PROBLEM 5.62 For the machine element shown, locate the z coordinate of the center of gravity. SOLUTION For half-cylindrical hole, III0.95 in.4(0.95)1.5 3 1.097 in.ry== = For half-cylindrical plate, IV1.5 in.4(1.5)5.25 5.887 in.3rz==+= 3, inV , in.y , in.z 4, inyV 4, inzV PROBLEM 5.62 For the machine
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978-0073398167 Chapter 5 Solution Manual Part 9
PROBLEM 5.72 A brass collar, of length 2.5 in., is mounted on an aluminum rod of length 4 in. Locate the center of gravity of the composite body. (Specific weights: brass = 0.306 lb/in3, aluminum = 0.101 lb/in3) SOLUTION Aluminum rod: 32(0.101lb/in ) (1.6 in.) (4 in.)40.81229 lbWV== =
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978-0073398167 Chapter 6 Solution Manual Part 1
PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss 0: 0 0 x xx FC= = =C 0: (1.92 kN)(3 m) (4.5 m) 0 By MC= + = 1.28 kN
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978-0073398167 Chapter 6 Solution Manual Part 10
PROBLEM 6.66 Solve Problem 6.65 assuming that pipe 1 is removed and that only pipe 2 is supported by the frames. PROBLEM 6.65 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft by a small frame like that shown. Knowing that the combined weight of
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978-0073398167 Chapter 6 Solution Manual Part 11
SOLUTION Free body: Entire Frame We have Horizontal force exerted on block. ( )4 in. sin15 1.03528 in.1.03528 in.sin = 9.93596 in.BF= ==4cos15 6cos9.9359 9.7737 in.AD AG FD=+= + = ( )0: (100 lb)(10 in.)cos 15 ( )sin (9.7737 in.) 0A BDMF= =572.77 lbBDF= ( )( ) cos (572.77
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978-0073398167 Chapter 6 Solution Manual Part 12
SOLUTION SOLUTION SOLUTION Free body: Piston: 0 : ( ) 0 ( ) = y BC yBC yF FPFP= = Since BC is two-force member: ()() 75 , () ()75 250 250 ( ) =0.3BC yBC x BC x BC y BC xFFFF FP= =
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978-0073398167 Chapter 6 Solution Manual Part 13
SOLUTION Free body: ( ) ( )0: 15 in. 35 in. 0AxMQ C= = 7 (1) 3xQC= 0: 800 lb 0 or 800 lbyy yFC C= = = Spring force. Unstretched length: 07 in.= Stretched length: 10 in.= 50 lb/in.k= 0()50(10 7) 150 lbF kx k= = =
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978-0073398167 Chapter 6 Solution Manual Part 14
SOLUTION Continued Free body: Joint B: 0: (800 lb)cos16.26 0y BCFF = = 768.0 lb 768 lbBC BCF FC== 0: 3613.8 lb (800 lb) sin16.26 0x BDFF= + + = 3837.8 lbBDF= 3840 lbBDFC= Free body: Joint C: 0: sin32.52 (4114.3 lb)sin32.52 (768 lb)cos16.26 0y CEFF = + = 2742.9
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978-0073398167 Chapter 6 Solution Manual Part 15
SOLUTION Free body: 16-ft length of pipe: (500 lb/ft)(16 ft) 8 kipsW= = Force Triangle 8 kips35 4BD= = 6 kips 10 kipsBD= = Determination of CB = CD. We note that horizontal projection of horizontal projection of =BO OD CD C C C+ 34()55r r CD+= consent of
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978-0073398167 Chapter 6 Solution Manual Part 2
PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: 0: (3 m) (24 kN 8 kN)(1.5 m) (7 kN)(3 m) 0 Dy MF= + = 23.0 kN y =F 0: 0 xx F= =F 0: (7
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978-0073398167 Chapter 6 Solution Manual Part 3
consent of McGraw-Hill Education. consent of McGraw-Hill Education. consent of McGraw-Hill Education. PROBLEM 6.14 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: 0: 0xxF= =F 0: 5 kNFM=
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978-0073398167 Chapter 6 Solution Manual Part 4
PROBLEM 6.19 Determine whether the trusses of Problems 6.21 and 6.24 are simple trusses. SOLUTION Truss of Problem 6.21 Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further thus, this truss is not a
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978-0073398167 Chapter 6 Solution Manual Part 5
PROBLEM 6.29 Determine the force in members DE and DF of the truss shown when P = 20 kips. SOLUTION Reactions: 2.5P= =CK 7.5tan 18 22.62== Member DE: 0: (2.5 )(6 ft) (12 ft) 0A DEM PF= = 1.25DEFP= + For 20 kips,P= 1.25(20) 25 kipsDEF=+=+ 25.0 kipsDEFT= Member DF:
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978-0073398167 Chapter 6 Solution Manual Part 6
PROBLEM 6.39 Determine the force in members AD, CD, and CE of the truss shown. SOLUTION Reactions: 0: 9 kips(8 ft) (45 ft) 5 kips(30 ft) 5 kips(15 ft) 0 kMB= + + = 6.6 kips=B 0: 9 kips 0 9 kipsxxxFK= + = =K 0: 6.6 kips 5
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978-0073398167 Chapter 6 Solution Manual Part 7
PROBLEM 6.47 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): PROBLEM 6.47 Classify each of the structures shown as completely, partially, or
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978-0073398167 Chapter 6 Solution Manual Part 8
PROBLEM 6.55 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its
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978-0073398167 Chapter 6 Solution Manual Part 9
SOLUTION Continued (b) Load applied at D. Free body: Member ACF. ACF is a two-force member. Thus, the reaction at F must be ACF is a two-force member. Thus, the reaction at F must be ACF is a two-force member. Thus, the reaction at F
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978-0073398167 Chapter 7 Solution Manual Part 1
PROBLEM 7.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. SOLUTION 1 21( );xy h h h dA ydxa=+ =221 21212310342113312()3412 4yayxdI x dA x h h h dxahhI h x x dxahhaaha ba h a= = += += +=
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978-0073398167 Chapter 7 Solution Manual Part 2
PROBLEM 9.10 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis. SOLUTION 1:y At 2 , 0:x ay= = 0 sin (2 )c ka= 2 or 2 ak k a = = At ,:xa yh= =
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978-0073398167 Chapter 7 Solution Manual Part 3
of the rectangle shown with respect to the midpoint of its (a) longer sides, (b) shorter sides. SOLUTION 313xdI a dx= 33412(2 )3 33axa aI a dx a a+= = = 22( )ydI x dA x a dx= = 324233aayaa xI a x dx a++ = = = 442233oxyJII
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978-0073398167 Chapter 7 Solution Manual Part 4
PROBLEM 7.25 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis. SOLUTION First note that 123222[(24)(6) (8)(48) (48)(6)] mm(144 384 288) mm816 mmAA A A=++ = ++= ++= Now 123() () ()xx x xII I I=++ where 3
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978-0073398167 Chapter 7 Solution Manual Part 5
PROBLEM 7.35 Determine the moments of inertia xI and yI of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB. SOLUTION First locate centroid C of the area. 2, inA , in.x , in.y 3, inxA 3, inyA 1 5 8 40= 2.5
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978-0073398167 Chapter 7 Solution Manual Part 6
PROBLEM 7.40 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area. SOLUTION Section Area, in2 , in.x 3, inxA 2Then 23: (8.835 in ) 21.375 inX A xA X== or 2.419 in.X= Then 44 4(4.5 in.)
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978-0073398167 Chapter 7 Solution Manual Part 7
SOLUTION Continued Also 2() ()y yL yCII I= + where 2 64 2 264( ) 0.512 10 mm (929 mm )[(127 21.2) mm]10.9109 10 mm()yL yyC yI I AdII=+= + = = Then 64[2(10.9109) 28.0] 10 mmyI= + or 6449.8 10 mmyI= Copyright McGraw-Hill Education. All rights reserved.
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978-0073398167 Chapter 7 Solution Manual Part 8
PROBLEM 7.55 Determine the moments of inertia xI and yI of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB. SOLUTION Dimensions in mm First locate centroid C of the area. Symmetry implies 30 mm.Y= 2, mmA , mmx 3, mmxA PROBLEM
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978-0073398167 Chapter 8 Solution Manual Part 1
PROBLEM 8.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that 150d= mm and 230d= mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC. SOLUTION (a) Rod AB 32 2 3 2 32136
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978-0073398167 Chapter 8 Solution Manual Part 2
PROBLEM 8.11 A couple M of magnitude 1500 N m is applied to the crank of an engine. For the position shown, determine (a) the force P required to hold the engine system in equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450-mm2
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978-0073398167 Chapter 8 Solution Manual Part 3
PROBLEM 8.21 For the assembly and loading of Prob. 8.8, determine (a) the average shearing stress in the pin at C, (b) the average bearing stress at C in member BC, (c) the average bearing stress at B in member BC. PROBLEM 8.8 Two horizontal 5-kip forces are applied
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978-0073398167 Chapter 8 Solution Manual Part 4
PROBLEM 8.27 The 1.4-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice shown. Determine the normal and shearing stresses in the glued splice. SOLUTION 202201400 lb 90 60 30(5.0)(3.0) 15 incos (1400)(cos30 )15PAPAs= = = =
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 8 Solution Manual Part 5
SOLUTION Continued (a) For member AB: F.S.U U ABAB ABPAFF= = 362 6(F.S.) (3.2)(17 10 ) 181.333 10 m 300 10ABAB U FA= = = 2181.3 mmABA= (b) For member AC: F.S.U U ACAC ACPAFF= = 362 6(F.S.) (3.2)(20 10 ) 213.33 10 m 300 10ACAC U FA= =
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 8 Solution Manual Part 6
PROBLEM 8.46 Solve Prob. 8.45, assuming that the structure has been redesigned to use 516-in-diameter pins at A and C as well as at B and that no other changes have been made. PROBLEM 8.45 Link AC is made of a steel with a 65-ksi ultimate normal stress and
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 8 Solution Manual Part 7
PROBLEM 8.56 A 58-in.-diameter steel rod AB is fitted to a round hole near end C of the wooden member CD. For the loading shown, determine (a) the maximum average normal stress in the wood, (b) the distance b for which the average shearing stress is 100 psi on
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 1
PROBLEM 9.1 A 4.8-ft-long steel wire of 14-in.-diameter is subjected to a 750-lb tensile load. Knowing that E = 29 106 psi, determine (a) the elongation of the wire, (b) the corresponding normal stress. SOLUTION (a) Deformation: 2;4 PL dAAEd= = Area: 222 (0.25 in.) 4.9087 10 in 4A= =
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 10
PROBLEM 9.79 (Continued) 1() 2bb bbP AA= + ( )1.5bbPA= 6(80 10 Pa)(0.04 m)(0.06 m)(1.5)= P 52.880 10 N= P 288 kNP= consent of McGraw-Hill Education. PROBLEM 9.79 (Continued) 1() 2bb bbP AA= + ( )1.5bbPA= 6(80 10 Pa)(0.04 m)(0.06 m)(1.5)= P 52.880 10 N= P 288 kNP=
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 2
PROBLEM 9.11 The 4-mm-diameter cable BC is made of a steel with 200 GPa.=E Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 3
PROBLEM 9.21 6 SOLUTION (a) 226 5 7.810 ft 93.72 in.= += = ABL Use joint A as a free body. 350: 28 07.81043.74 kip 43.74 10 lb= == = y ABABFFF 36(43.74 10 )(93.72)(29 10 )(0.80)= = AB ABAB AB FLEA 0.1767 in.=AB (b) Use joint B as
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 4
PROBLEM 9.30 (Continued) C to D: 3 3696( 60 10 )(0.100)74.220 101.34735 10 80.841 10= = = ABCARPLEARD to E: 33696100 10100 mm 0.100 m( 100 10 )(0.100)74.220 101.34735 10 134.735 10AADEAPRLRPLEAR== == = = A to E: 963.85837 10 242.424 10=+++= AE AB BC CD DEAR AE(a) 96
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 5
PROBLEM 9.37 The brass shell 6( 11.6 10 / F)=b is fully bonded to the steel core 6( 6.5 10 / F).=s Determine the largest allowable increase in temperature if the stress in the steel core is not to exceed 8 ksi. SOLUTION Let axial force developed in the
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 6
PROBLEM 9.45 Knowing that a 0.02-in. gap exists when the temperature is 75 F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to 11 ksi, (b) the corresponding exact length of the aluminum bar. SOLUTION 33311 ksi 11 10 psi(11 10
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 7
PROBLEM 9.54 A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, v = 0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width of portion AB of the test coupon, (c) in the
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 8
PROBLEM 9.64 An elastomeric bearing (G = 130 psi) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 38 in. when a 5-kip lateral load is applied as shown. Knowing that the maximum allowable shearing stress is
Statics and Mechanics of Materials 2nd Edition
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978-0073398167 Chapter 9 Solution Manual Part 9
PROBLEM 9.74 The brass tube ( 105 GPa)AB E = has a cross-sectional area of 140 mm2 and is fitted with a plug at A. The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder (
Statics and Mechanics of Materials 2nd Edition

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