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

01 Solutions 46060 5/6/10 2:43 PM Page 49 © 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. 1–71. Determine the average normal **stress** at section a–a and the average shear **stress** at section b–b in member AB. The cross section is square, 0.5 in. on each side. Consider the FBD of member BC,Fig.a, a+©M C = 0; F AB sin 60°(4) - 150(4)(2) = 0 F AB = 346.41 lb Referring to the FBD in Fig. b, + b©F x¿ = 0; N a - a + 346.41 = 0 N a - a = -346.41 lb C 150 lb/ft 4 ft 60 a b a B Referring to the FBD in Fig. c. + c ©F y = 0; V b - b - 346.41 sin 60° = 0 V b - b = 300 lb b The cross-sectional areas of section a–a and b–b are A a - a = 0.5(0.5) = 0.25 in 2 and 0.5 A b - b = 0.5 a . Thus cos 60° b = 0.5 in2 s a - a = N a - a = 346.41 = 1385.64 psi = 1.39 ksi A a - a 0.25 Ans. A t b - b = V b - b = 300 = 600 psi A b - b 0.5 Ans. 49

01 Solutions 46060 5/6/10 2:43 PM Page 50 © 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. •1–73. Member B is subjected to a compressive force of 3 800 lb. If A and B are both made of wood and are thick, 1 determine to the nearest 4 in. 8 in. the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear **stress** for the segment is t allow = 300 psi. B 800 lb t allow = 300 = 307.7 ( 3 2 ) h h = 2.74 in. A h 13 12 5 Use h = 2 3 4 in. Ans. 1–74. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear **stress** for the key is t allow = 35 MPa. a A d a 20 mm 500 mm a + ©M A = 0; F a - a (20) - 200(500) = 0 200 N F a - a = 5000 N t allow = F a - a A a - a ; 35(10 6 ) = 5000 d(0.025) d = 0.00571 m = 5.71 mm Ans. 1–75. The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear **stress** for the bolts is t fail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5. 30 mm 80 kN 30 mm 350(10 6 ) 2.5 = 140(10 5 ) t allow = 140(10 6 ) = 20(1**03** ) p 4 d2 d = 0.0135 m = 13.5 mm 40 kN 40 kN Ans. 50

- 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: 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 and 70: 01 Solutions 46060 5/6/10 2:43 PM P
- Page 71 and 72: 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