22 14. 91 mma 3. 5 m 4. 0 m 2. 5 m B AC PROBLEM 2. The 4-mm-diameter cable BC is made of a steel with E200 GPa. 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 shown. SOLUTION 22 6 4 7. 2111 m BC L Use bar AB as a free body. 0: 3. 5 (6) 0 7. 0. ABC MP F PF Considering allowable stress, 190 10 Pa 22 62 663 (0. 004) 12. 566 10 m (190 10)(12. 566 10) 2. 388 10 N Ad F FA Considering allowable elongation, 610 m 693 (12. 566 10)(200 10)(6 10) 3 2. 091 10 N BC BC FL AE F AE L Smaller value governs. FBC2. 091 10 N 33 PF 0. 9509 BC (0. 9509)(2. 091 10) 1. 988 10 N 1. 988 kNP 1. 25-in. diameter 4 ft 3 ft C PROBLEM 2. A single axial load of magnitude P = 15 kips is applied at end C of the steel rod ABC. Knowing that E = 30 × 10 psi, determine the diameter d of portion BC for which the deflection of point C will be 0.
This is proprietary material solely for authorized instructor use. PROBLEM 2. 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 E200 GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire. SOLUTION (a), or PLAE P AEL with 112262 (0. 005) 19. 6350 10 m 44 Ad 62 9 2 (0. 045 m)(19. 6350 10 m)(200 10 N/m) 9817. 5 N 18 m P P9. 82 kN (b) 62 9817. 5 N 500 10 Pa 19. 6350 10 m P A 500 MPa PROBLEM 2. An aluminum pipe must not stretch more than 0. 05 in. when it is subjected to a tensile load. Knowing that E10. 1 10 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, 3 (1 0. 1 1 0) ( 0. 0 5) 14 10 EA E L P L36.
C C H H A A P P T T E E R R 2 2 Mechanics of Materials 7th Edition Beer Solutions Manual Full Download: Full download all chapters instantly please go to Solutions Manual, Test Bank site: PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. PROBLEM 2. A 4. 8-ft-long steel wire of 1 4 -in. -diameter is subjected to a 750-lb tensile load. Knowing that E = 29 × 10 6 psi, determine (a) the elongation of the wire, (b) the corresponding normal stress. SOLUTION (a) Deformation: 2; PLd A AE Area: 2 (0. 25 in. ) 22 4. 9087 10 in A 22 6 (750 lb)(4. 8 ft 12 in. /ft) (4. 9087 10 in)(29 10 psi) 2 3. 0347 10 in. 0. 0303 in. (b) Stress: P A Area: 22 (750 lb) (4. 9087 10 in) 4 1. 52790 10 psi 15. 28 ksi PROPRIETARY MATERIAL.
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05 in. SOLUTION i ii AB BC PL PL PL AE AE AE 4 ft 48 in. ; 3 ft 36 in. AB BC LL 22 (1. 2 5 i n. ) 2 1. 22718 in AB Substituting, we have 15 10 lb 48 in. 36 in. 0. 05 in. 30 10 psi 1. 22718 in BC ABC0. 59127 in 4 BC 4 BC A or d 4(0. 59127 in) 0. 86766 in. d 0. 868 d in. P 5 350 lb AB CD 1 in. 1 in. 1. 2 in. 0. 4 in. 1. 6 in. PROBLEM 2. The specimen shown has been cut from a -in. -thick sheet of vinyl (E = 0. 45 × 10 psi) and is subjected to a 350-lb tensile load. Determine (a) the total deformation of the specimen, (b) the deformation of its central portion BC. SOLUTION 3 (350 lb)(1. 9778 10 in. (0. 45 10 psi)(1 in. )(0. ) AB PL EA 3 (350 lb)(2 in. ) 15. 5556 10 in. (0. 45 10 psi)(0. 4 in. ) 4. 9778 10 in. CD AB (a) Total deformation: ABBCCD 3 25. 511 10 in. 25. 5 10 in. (b) Deformation of portion BC: 15. 56 10 in. 375 mm 1 mm DA PROBLEM 2. The brass tube AB E( 105 GPa) has a cross-sectional area of 140 mm and is fitted with a plug at A.
1 in. (b) P A Thus, 127. 5 10 14 10 P A 2 A9. 11 in PROBLEM 2. 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 d Substituting, we have 4 4 4(4 10 N) (1 8 0 1 0 P a) 5. 3192 10 m PP d d d5. 32 mm (b); E L Substituting, we have E EL L 93 (105 10 Pa) (3 10 m) (1 8 0 1 0 P a) L 1. 750 mL PROBLEM 2. A cast-iron tube is used to support a compressive load. Knowing that E10 10 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. L 100 ; E L E L 6 (10 10 psi)(0.
00025) 2. 50 10 psi 2. 50 ksi 1600 lb;0. 64in 2. 50 10 psi P P A A 22 4 oi A dd 224 io 22 4(0. 64 in) 2 (2. 0 in. ) 3. 1851 in i 1. 78469 in. d 11 () (2. 1. 78469 in. ) oi t dd 0. 107655 in. t t0. 1077 PROBLEM 2. 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 E200 GPa, determine the required diameter of the rod. SOLUTION L4m 36 93 310m, 15010Pa EP200 10 Pa, 10 10 N Stress: 62 2 10 10 N 66. 667 10 m 66. 667 mm 150 10 Pa P Deformation: (1 0 1 0) ( 4) 66. 667 mm (200 10)(3 10) PL AE PL A E The larger value of A governs: A66. 667 mm 2 44 (66. 667) 4 A Add 9. 21 mmd PROBLEM 2. A block of 10-in. length and 1. 8 × 1. 6-in. cross section is to support a centric compressive load P. The material to be used is a bronze for which E 14 × 10 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.
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The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder (E72 GPa) with a cross-sectional area of 250 mm 2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1 mm. Determine the force P that must be applied to the cylinder. SOLUTION Shortening of brass tube AB: 262 9 96 375 1 376 mm 0. 376 m 140 mm 140 10 m 105 10 Pa (0. 376) 25. 578 10 (105 10)(140 10) AB AB LA PL P EA Lengthening of aluminum cylinder CD: 262 9 0. 375 m 250 mm 250 10 m 72 10 Pa (0. 375) 20. 833 10 (72 10)(250 10) CD CD CD CD CD CD L AE PL P P EA Total deflection: AABCD where A0. 001 m 99 0. 001 (25. 578 10 20. 833 10) P P21. 547 10 N 21. 5 kNP