51_ch 03 Mechanical Design budynas_SM_ch03

# 51_ch 03 Mechanical Design budynas_SM_ch03 - = tan 1 F y F...

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64 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-75 r i = 0 . 125 in, r o = 0 . 125 + 0 . 1094 = 0 . 2344 in From Table 3-4 for h = 0 . 1094 r c = 0 . 125 + 0 . 1094 / 2 = 0 . 1797 in r n = 0 . 1094 / ln(0 . 2344 / 0 . 125) = 0 . 174 006 in e = r c r n = 0 . 1797 0 . 174 006 = 0 . 005 694 in c i = r n r i = 0 . 174 006 0 . 125 = 0 . 049 006 in c o = r o r n = 0 . 2344 0 . 174 006 = 0 . 060 394 in A = 0 . 75(0 . 1094) = 0 . 082 050 in 2 M = F (4 + h / 2) = 3(4 + 0 . 1094 / 2) = 12 . 16 lbf · in σ i =− 3 0 . 082 05 12 . 16(0 . 0490) 0 . 082 05(0 . 005 694)(0 . 125) =− 10 240 psi Ans. σ o =− 3 0 . 082 05 + 12 . 16(0 . 0604) 0 . 082 05(0 . 005 694)(0 . 2344) = 6670 psi Ans. 3-76 Find the resultant of F 1 and F 2 . F x = F 1 x + F 2 x = 250 cos 60 + 333 cos 0 = 458 lbf F y = F 1 y + F 2 y = 250 sin 60 + 333 sin 0 = 216 . 5 lbf F = (458 2 + 216 . 5 2 ) 1 / 2 = 506 . 6 lbf This is the pin force on the lever which acts in a direction
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Unformatted text preview: = tan 1 F y F x = tan 1 216 . 5 458 = 25 . 3 On the 25 . 3 surface from F 1 F t = 250 cos ( 60 25 . 3 ) = 206 lbf F n = 250 sin(60 25 . 3 ) = 142 lbf r c = 1 + 3 . 5 / 2 = 2 . 75 in A = 2[0 . 8125(0 . 375) + 1 . 25(0 . 375)] = 1 . 546 875 in 2 The denominator of Eq. (3-63), given below, has four additive parts. r n = A ( d A / r ) 25.3 206 507 142 2000 lbf in...
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