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4020-F2011-HW7Solutions

4020-F2011-HW7Solutions - MSE 4020 HW7-SOLUTION Problem 1...

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1 MSE 4020 HW7-SOLUTION 11/10/2011 Problem 1 Force applied on the shank is, F SN = ! D 2 4 2K ( ) = ! D 2 K 2 ! D "#" K = F SN = ! D 2 K 2 $ # = D 2 $ # = # = D 2 / 1 n = D " n 2
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2 Problem 2 First, divide the whole system into blocks A, B, C and D (see sketch above), and analyze the their relative displacement q (with different indices) between contacting blocks. Angle α is to be determined. W = F t q A = 4 l AB ! q AB + l BD ! q BD ( ) ! K From triangular geometry: q AB sin30 ° = q A sin45 ° ! q AB = 1 2 q A 1/ 2 = 2 2 q A q BD sin75 ° = q AB sin " 2 ! q BD = sin75 ° # q AB sin " 2 = 2 2 q A # sin75 ° sin " 2 l AB # sin75 ° = 2.5t ! l AB = 2.5t sin75 ° l BD # sin " 2 = 0.5t ! l BD = 0.5t sin " 2
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3 ! F t = 4 q A 2.5t sin75 ° " 2 2 q A + 2 2 " 0.5t sin # 2 " sin75 ° sin # 2 q A $ % & & & & ' ( ) ) ) ) " K = 2 2 t 2.5 sin75 ° + 0.5 " sin75 ° sin 2 # 2 $ % & & & & ' ( ) ) ) ) " K Also from geometry, tan30 ° = 2t l ! l = 2t tan30 ° = 2 3 t ! 2.5t tan75 ° + 0.5t tan " 2 = 2 3 t ! 0.5 tan " 2 = 2 3 # 2.5 tan75 ° = 2.72 ! tan " 2 = 0.5 2.72 = 0.184 ! " 2 = 10.8 ° ! " = 21.6 °
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4 Problem 3 ! " z = D #" 4L = D # 2 $ 4L = D # $ 2L = 3 # $ 2 # 12 = 0.433 % ! f = 2 3 #! " z = 2 3 # 0.433 = 0.453 & f = 28000 + 220000 # ! f ( ) 1/2 = 28000 + 220000 # 0.453 ( ) 1/2 = 176000 psi & f = 3 & " z % & " z = & f 3 Twist moment M = $ D # t ( ) area !"# #& " z # D / 2 = $ D # t ( ) # & f 3 # D / 2 = $ D 2 # t 2 ' ( ) * + ,# & f 3 = $# 3 2 # 1 16 2 ' ( ) ) ) ) * + , , , , # 176000 3 = 89700 in # lb
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5 Problem 4 The method to solve this problem is called slab (see the sketch above) approach, F x 1 ! = 0 " # 11 + d # 11 ( ) $ D + dD ( ) % # 11 $ D + 2 μ P cos & $ cos &$ dx 1 + 2P $ sin & cos & $ dx 1 = 0 " # 11 $ dD + D $ d # 11 + d # 11 $ dD cancelled out ! " # $ # + 2P μ + tan & ( ) $ dx 1 = 0 From the sketch dD dx 1 = 2tan & " dx 1 = dD 2 $ cot & " # 11 $ dD + D $ d # 11 + P μ + tan & ( ) $ dD $ cot & = 0 " D $ d # 11 + # 11 + P μ $ cot & + 1 ( ) ' ( ) * $ dD = 0 " d # 11 # 11 + P + + 1 ( ) = % dD D ,where + = μ $ cot & # 22 = % P " # 11 % # 22 = # 0 " # 11 + P = # 0 " P = # 0 % # 11 " d # 11 # 11 + # 0 % # 11 ( ) $ + + 1 ( ) = % dD D " d # 11 %+$# 11 + + + 1 ( ) $# 0 = % dD D " d # 11 +$# 11 % + + 1 ( ) $# 0 = dD D
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6 ! dD D D i D " = d # 11 $%# 11 & $ + 1 ( ) %# 0 0 # 11 " ! ln D D i = 1 $ ln $%# 11 & $ + 1 ( ) %# 0 ( ) 0 # 11 ! $%# 11 & $ + 1 ( ) %# 0 & $ + 1 ( ) %# 0 = D D i ' ( ) ) * + , , $ ! $%# 11 & $ + 1 ( ) %# 0 = & $ + 1 ( ) %# 0 % D D i ' ( ) ) * + , , $ ! # 11 = # 0 % $ + 1 $ ' ( ) * + , % 1 & D D i ' ( ) ) * + , , $ ' ( ) ) * + , , To find the stress acting at D = D 0 , ! 11 ( ) 0 = ! 0 " # + 1 # $ % & ' ( ) " 1 * D D i $ % & & ' ( ) ) # $ % & & ' ( ) ) # = μ " cot + = 0.05 " cot 5 ° ( ) = 0.5715 F 0 = !
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