- Text
- Determine,
- Shear,
- Portion,
- Solutions,
- Pearson,
- Saddle,
- Copyright,
- Currently,
- Reproduced,
- Diameter,
- Strain,
- Properties

01 Solutions 46060 5/6/10 2:43 PM Page 69 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–1**03**. Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal **stress** for member BC is s allow = 29 ksi and the allowable shear **stress** for the pins is t allow = 10 ksi. 1.5 in. C Referring to the FBD of member AB,Fig.a, a+©M A = 0; 2(8)(4) - F BC sin 60° (8) = 0 F BC = 9.238 kip : + ©F x = 0; 9.238 cos 60° - A x = 0 A x = 4.619 kip B 60 8 ft A + c ©F y = 0; 9.238 sin 60° - 2(8) + A y = 0 A y = 8.00 kip Thus, the force acting on pin A is 2 kip/ft F A = 2 A x 2 + A y 2 = 2 4.619 2 + 8.00 2 = 9.238 kip Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. V A = F A = 9.238 kip V B = F BC 2 = 9.238 2 = 4.619 kip For member BC s allow = F BC A BC ; 29 = 9.238 1.5(t) t = 0.2124 in. Use t = 1 4 in. Ans. For pin A, t allow = V A ; 10 = 9.238 p d A A = 1.085 in. A 4 d2 A For pin B, Use d A = 1 1 8 in t allow = V B ; 10 = 4.619 p A B 4 d B 2 d B = 0.7669 in Use d B = 13 16 in Ans. Ans. 69

01 Solutions 46060 5/6/10 2:43 PM Page 70 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *1–104. Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame. 150 lb/ft Segment AD: : + ©F x = 0; N D - 1.2 = 0; N D = 1.20 kip Ans. 4 ft 1.5 ft A E D B +T©F y = 0; V D + 0.225 + 0.4 = 0; V D = -0.625 kip a +©M D = 0; M D + 0.225(0.75) + 0.4(1.5) = 0 M D = -0.769 kip # ft Ans. Ans. C 2.5 ft 3 ft 5 ft Segment CE: Q+©F x = 0; N E + 2.0 = 0; N E = -2.00 kip Ans. R+©F y = 0; V E = 0 Ans. a+©M E = 0; M E = 0 Ans. •1–105. The pulley is held fixed to the 20-mm-diameter shaft using a key that fits within a groove cut into the pulley and shaft. If the suspended load has a mass of 50 kg, determine the average shear **stress** in the key along section a–a. The key is 5 mm by 5 mm square and 12 mm long. a 75 mm a a +©M O = 0; F (10) - 490.5 (75) = 0 F = 3678.75 N t avg = V A = 3678.75 = 61.3 MPa (0.005)(0.012) Ans. 70

- Page 1 and 2:
FM_TOC 46060 6/22/10 11:26 AM Page

- Page 3 and 4:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 5 and 6:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 7 and 8:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 9 and 10:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 11 and 12:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 13 and 14:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 15 and 16:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 17 and 18:
01 Solutions 46060 5/6/10 2:43 PM P

- Page 19 and 20: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 21 and 22: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 23 and 24: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 25 and 26: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 27 and 28: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 29 and 30: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 31 and 32: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 33 and 34: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 35 and 36: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 37 and 38: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 39 and 40: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 41 and 42: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 43 and 44: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 45 and 46: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 47 and 48: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 49 and 50: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 51 and 52: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 53 and 54: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 55 and 56: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 57 and 58: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 59 and 60: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 61 and 62: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 63 and 64: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 65 and 66: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 67 and 68: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 69: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 73 and 74: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 75 and 76: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 77 and 78: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 79 and 80: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 81 and 82: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 83 and 84: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 85 and 86: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 87 and 88: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 89 and 90: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 91 and 92: 02 Solutions 46060 5/6/10 1:45 PM P
- Page 93 and 94: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 95 and 96: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 97 and 98: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 99 and 100: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 101 and 102: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 103 and 104: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 105 and 106: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 107 and 108: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 109 and 110: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 111 and 112: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 113 and 114: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 115 and 116: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 117 and 118: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 119 and 120: 03 Solutions 46060 5/7/10 8:45 AM P
- Page 121 and 122:
03 Solutions 46060 5/7/10 8:45 AM P