Mechanics Of Materials 7th Edition Solution Manual

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1. Mechanics of Materials 7th Edition Beer Solutions Manual Full clear download( no error formatting) at: beer-solutions-manual/ CCHHAAPPTTEERR 22. PROBLEM 2.1 A nylon thread is subjected to a 8.5-N tension force.

Knowing that E 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread. SOLUTION (a) Strain:  L 1.1 0.011 100 Stress: E (3.3 109 )(0.011) 36.3 106 Pa P A Area: A P  8.5 234.16 109 m2 36.3 106 Diameter: d  4A  (4)(234.16 109 ) 546 106 m d 0.546 mm  (b) Stress: 36.3 MPa. 4 PROBLEM 2.2 A 4.8-ft-long steel wire of 1 -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: PL; AE d2 A  4 Area: (b) Stress: Area: (0.25 in.)2 2 2 A 4.9087 10 in 4 (750 lb)(4.8 ft 12 in./ft) (4.9087102 in2 )(29106 psi) 3.0347 102 in.

P A (750lb) (4.9087102 in2 ) 0.0303 in. 1.52790 104 psi 15.28 ksi. PROBLEM 2.3 An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.

It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire. E 200 GPa, SOLUTION (a) PL, or P AE (b) AE L with A 1 d2 1 (0.005)2 19.6350 106 m2 4 4 (0.045 m)( m2 )(200109 N/m2 ) P 9817.5 N 18 m P 9817.5 N 6 50010 Pa P 9.82 kN  500 MPa A 19.635010 6 m 2.

PROBLEM 2.4 Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load is applied, determine (a) the stress in the rod, (b) the factor of safety. SOLUTION (a) L L0 250.28 mm 250 mm 0.28 mm  L0 0.28 mm 250 mm 1.11643 104 E (73 109 Pa)(1.11643 104 ) 8.1760107 Pa 81.8 MPa  (b) F.S. U  140 MPa 81.760 MPa 1.71233 F.S. 1.712. PROBLEM 2.5 An aluminum pipe must not stretch more than 0.05 in.

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When it is subjected to a tensile load. Knowing that E 10.1 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips. SOLUTION (a) PL AE Thus, (b) P A L EA P E  (10.1 106 )(0.05) 14 103 L 36.1 in.

Thus, A P  127.5 103  14 103 A 9.11 in2. PROBLEM 2.6 A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN. Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the smallest diameter rod that should be used, (b) the corresponding maximum length of the rod. SOLUTION (a) P; A d2 A  4 Substituting, we have P d2  d 4P 4   4(4 103 N) d  (180 106 Pa) (b) E;  L d 5.3192 103 m d 5.32 mm Substituting, we have E  L L E  (105 109 Pa)(3 103 m) L  (180 106 Pa) L 1.750m.

PROBLEM 2.7 A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. When a 2-kip tensile load is applied to it. Knowing that E 29106 psi, determine (a) the smallest diameter rod that should be used, (b) the corresponding normal stress caused by the load. SOLUTION (a) PL: 0.04 in.  (2000 lb)(5.512 in.) 6 AE A(2910 psi) A 1 d2 0.113793 in2 4 d 0.38063 in. (b) P  A 2000 lb 0.113793 in 2 17575.8 psi 17.58 ksi.

PROBLEM 2.8 A cast-iron tube is used to support a compressive load. Knowing that E 10 106 psi and that the maximum allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.

SOLUTION (a) 0.00025 L 100 E; E  L  L (10 106 psi)(0.00025) 2.50 103 psi 2.50 ksi (b) P;  A A P  1600 lb 0.64 in2 2.50 103 psi A d2 d2  4 o i d2 d2 4A i o  2 d2 (2.0 in.)2 4(0.64 in ) 3.1851 in2 i  di 1.78469 in. T 1 (d d ) 1 (2.0 in.

Mechanics Of Materials 7th Edition Solution Manual

1.78469 in.) 2 o i 2 t 0.107655 in. T 0.1077. PROBLEM 2.9 A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when the rod is subjected to a 10-kN axial load.

Knowing that the rod. E 200 GPa, determine the required diameter of SOLUTION L 4 m 3 103 m, 150 106 Pa E 200 109 Pa, P 10 103 N Stress: P A 3 A P  10 10 N 66.667 106 m2 66.667 mm2 150 106 Pa Deformation: PL AE 3 A PL E The larger value of A governs: (10 10 )(4) 66.667 106 m2 66.667 mm2 (200 109 )(3 103 ) A 66.667 mm2 A d2 d 4A  4(66.667) d 9.21 mm 4. PROBLEM 2.10 A nylon thread is to be subjected to a 10-N tension. Knowing that E 3.2 GPa, that the maximum allowable normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the required diameter of the thread.

SOLUTION Stress criterion: 40 MPa 40 106 Pa P 10 N P: A A P  10 N 250 109 m2 40 106 Pa A d2: d 2 A 2 250 109 564.19 106 m Elongation criterion: 4 1% 0.01 L PL: AE  d 0.564 mm P/E 10 N/3.2 109 Pa A  312.5 109 m2 /L 0.01 9 d 2 A 2 312.5 10 630.78 106 m2  The required diameter is the larger value: d 0.631 mm d 0.631 mm.  PROBLEM 2.11 A block of 10-in. Length and 1.8 × 1.6-in. Cross section is to support a centric compressive load P.

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The material to be used is a bronze for which E 14 × 106 psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at most 0.12% of its original length. SOLUTION Considering allowable stress, Cross-sectional area: 18 ksi or 18103 psi A (1.8 in.)(1.6 in.) 2.880 in2 P  A Considering allowable deformation,  L P A (18103 psi)(2.880 in2 ) 5.1840104 lb or 51.840 kips 0.12% or 0.0012 in. PL AE P AE  L  P (2.880 in2 )(14 106 psi)(0.0012 in.) 4.8384104 lb or 48.384 kips The smaller value for P resulting from the required deformation criteria governs. 48.4 kips. PROBLEM 2.12 A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing that E 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.

SOLUTION 180 106 Pa P 40 103 N E 105 109 Pa 2.5 103 m (a) PL L AE E E (105109 )(2.5103 ) L 1.45833 m (b)  P A 180106 3 L 1.458 m A P  40 10 222.22 106 m2 222.22 mm2 180 106 A a2 a  A 222.22 a 14.91 mm. P 130 kips PROBLEM 2.13 Rod BD is made of steel (E 29106 psi) and is used to brace the axially A 72 in.

D B compressed member ABC. The maximum force that can be developed in member BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the smallest- diameter rod that can be used for member BD. SOLUTION FBD 0.02P (0.02)(130) 2.6 kips 2.6103 lb Considering stress, 18 ksi 18103 psi FBD A FBD 2.6 0.14444 in2 A 18 Considering deformation, (0.001)(144) 0.144 in.

FBD LBD F L (2.6103 )(54) A BD BD 0.03362 in2 Larger area governs. AE E A 0.14444 in2 (29106 )(0.144) A d2 d  4A  (4)(0.14444) d 0.429 in.

4. BC B 2.5 m P PROBLEM 2.14 The 4-mm-diameter cable BC is made of a steel with E 200 GPa. Knowing 3.5 m A C 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 shown. 4.0 m SOLUTION Use bar AB as a free body. LBC  62 42 7.2111 m M 0: 3.5P (6) 4 F 0 A 7.2111 BC  P 0.9509FBC Considering allowable stress, 190106 Pa A d2 (0.004)2 12.566106 m2 4 4 FBC  A FBC A (190106 )(12.566106 ) 2.388103 N Considering allowable elongation, 6103 m 6 9 3 FBC LBC  AE FBC AE LBC (12.56610 )(20010 )(610 ) 2.091103 N 7.2111 Smaller value governs.

F 2.091103 N P 0.9509FBC (0.9509)(2.091103 ) 1.988103 N P 1.988 kN. 2 BC  1.25-in.

Diameter d A PROBLEM 2.15 A single axial load of magnitude P = 15 kips is applied at end C of the steel rod ABC. Knowing that E = 30 × 106 psi, determine the diameter d of portion 4 ft B C 3 ft P BC for which the deflection of point C will be 0.05 in. SOLUTION PLi PL  PL C AiEi AE AB AE BC LAB 4 ft 48 in.; LBC 3 ft 36 in. Substituting, we have AAB d2 (1.25 in.)2 1.22718 in 4 4 15 103 lb  48 in. 30 106 psi 1.22718 in2  A 0.59127 in2  ABC ABC  or d  d2 4 4ABC 4(0.59127 in2 ) d  d 0.86766 in. PROBLEM 2.16 A 250-mm-long aluminum tube (E 70 GPa) of 36-mm outer 36 mm 28 mm diameter and 28-mm inner diameter can be closed at both ends by means of single-threaded screw-on covers of 1.5-mm pitch. With one cover screwed on tight, a solid brass rod (E 105 GPa) of 25 mm 250 mm 25-mm diameter is placed inside the tube and the second cover is screwed on.

Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.

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Mechanics Of Materials 7th Edition Solution Manual Chapter 1

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