Budynas ch03

# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 3 3-1 1 R C R A R B R D C A B W D 1 2 3 R B R A W R B R C R A 2 1 W R A R Bx R Bx R By R By R B 2 1 1 Scale of corner magnified W A B (e) (f) (d) W A R A R B B 1 2 W A R A R B B 1 1 2 (a) (b) (c)

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Chapter 3 15 3-2 (a) R A = 2 sin 60 = 1 . 732 kN Ans. R B = 2 sin 30 = 1 kN Ans. (b) S = 0 . 6 m α = tan 1 0 . 6 0 . 4 + 0 . 6 = 30 . 96 R A sin 135 = 800 sin 30 . 96 R A = 1100 N Ans. R O sin 14 . 04 = 800 sin 30 . 96 R O = 377 N Ans. (c) R O = 1 . 2 tan 30 = 2 . 078 kN Ans. R A = 1 . 2 sin 30 = 2 . 4 kN Ans. (d) Step 1: Find R A and R E h = 4 . 5 tan 30 = 7 . 794 m ± + ± M A = 0 9 R E 7 . 794(400 cos 30) 4 . 5(400 sin 30) = 0 R E = 400 N Ans . ± F x = 0 R Ax + 400 cos 30 = 0 R =− 346 . 4N ± F y = 0 R Ay + 400 400 sin 30 = 0 R Ay 200 N R A = ² 346 . 4 2 + 200 2 = 400 N Ans. D C h B y E x A 4.5 m 9 m 400 N 3 4 2 30 ° 60 ° R Ay R A R Ax R E 1.2 kN 60 ° R A R O 60 ° 90 ° 30 ° 1.2 kN R A R O 45 ± ² 30.96 ± ³ 14.04 ± 135 ° 30.96 ° 30.96 ° 800 N R A R O O 0.4 m 45 ° 800 N ± 0.6 m A s R A R O B 60 ° 90 ° 30 ° 2 kN R A R B 2 1 2 kN 60 ° 30 ° R A R B
16 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Step 2: Find components of R C on link 4 and R D ± + ± M C = 0 400(4 . 5) (7 . 794 1 . 9) R D = 0 R D = 305 . 4N Ans. ± F x = 0 ( R Cx ) 4 = 305 . ± F y = 0 ( R Cy ) 4 =− 400 N Step 3: Find components of R C on link 2 ± F x = 0 ( R ) 2 + 305 . 4 346 . 4 = 0 ( R ) 2 = 41 N ± F y = 0 ( R ) 2 = 200 N 3-3 (a) ± + ± M 0 = 0 18(60) + 14 R 2 + 8(30) 4(40) = 0 R 2 = 71 . 43 lbf ± F y = 0: R 1 40 + 30 + 71 . 43 60 = 0 R 1 1 . 43 lbf M 1 1 . 43(4) 5 . 72 lbf · in M 2 5 . 72 41 . 43(4) 171 . 44 lbf · in M 3 171 . 44 11 . 43(6) 240 lbf · in M 4 240 + 60(4) = 0 checks! 4" 4" 6" 4" ± 1.43 ± 41.43 ± 11.43 60 40 lbf 60 lbf 30 lbf x x x O AB C D y R 1 R 2 M 1 M 2 M 3 M 4 O V (lbf) M (lbf • in) O C C D B A B D E 305.4 N 346.4 N 305.4 N 41 N 400 N 200 N 400 N 200 N 400 N Pin C 30 ° 305.4 N 400 N 400 N 200 N 41 N 305.4 N 200 N 346.4 N 305.4 N ( R Cx ) 2 ( R Cy ) 2 C B A 2 400 N 4 R D ( R Cx ) 4 ( R Cy ) 4 D C E Ans.

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Chapter 3 17 (b) ± F y = 0 R 0 = 2 + 4(0 . 150) = 2 . 6 kN ± M 0 = 0 M 0 = 2000(0 . 2) + 4000(0 . 150)(0 . 425) = 655 N · m M 1 =− 655 + 2600(0 . 2) 135 N · m M 2 135 + 600(0 . 150) 45 N · m M 3 45 + 1 2 600(0 . 150) = 0 checks! (c) ± M 0 = 0: 10 R 2 6(1000) = 0 R 2 = 600 lbf ± F y = 0: R 1 1000 + 600 = 0 R 1 = 400 lbf M 1 = 400(6) = 2400 lbf · ft M 2 = 2400 600(4) = 0 checks! (d) ± + ± M C = 0 10 R 1 + 2(2000) + 8(1000) = 0 R 1 = 1200 lbf ± F y = 0: 1200 1000 2000 + R 2 = 0 R 2 = 1800 lbf M 1 = 1200(2) = 2400 lbf · ft M 2 = 2400 + 200(6) = 3600 lbf · ft M 3 = 3600 1800(2) = 0 checks! 2000 lbf 1000 lbf R 1 O O M 1 M 2 M 3 R 2 6 ft 2 ft 2 ft AB C y M 1200 ± 1800 200 x x x 6 ft 4 ft A O O O B ± 600 M 1 M 2 V (lbf) 1000 lbf y R 1 R 2 400 M (lbf •ft) x x x V (kN) 150 mm 200 mm 150 mm 2.6 ± 655 M (N •m) 0.6 M 1 M 2 M 3 2 kN 4 kN/m y A O O O O BC R O M O x x x
18 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (e) ± + ± M B = 0 7 R 1 + 3(400) 3(800) = 0 R 1 =− 171 . 4 lbf ± F y = 0: 171 . 4 400 + R 2 800 = 0 R 2 = 1371 . 4 lbf M 1 171 . 4(4) 685 . 7 lbf · ft M 2 685 . 7 571 . 4(3) 2400 lbf · ft M 3 2400 + 800(3) = 0 checks! (f) Break at A R 1 = V A = 1 2 40(8) = 160 lbf ± + ± M D = 0 12(160) 10 R 2 + 320(5) = 0 R 2 = 352 lbf ± F y = 0 160 + 352 320 + R 3 = 0 R 3 = 128 lbf M 1 = 1 2 160(4) = 320 lbf · in M 2 = 320 1 2 160(4) = 0 checks! (hinge) M 3 = 0 160(2) 320 lbf · in M 4 320 + 192(5) = 640 lbf · in M 5 = 640 128(5) = 0 checks!

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## This document was uploaded on 01/24/2008.

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Budynas ch03 - Chapter 3 3-1 W 2 A RA 1 RB B A RA 1 1 RB W 2 B(a 1 RD 3 RA RB 2 D(b 1 RC C A B W(c W 1 RC RB RA RB 2 RA W(d(e A RA 1 2 W RBx B RBx

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