29Closedsectshearflows

29Closedsectshearflows - AAE 352 Lecture 29 Shear flow in...

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AAE 352 1 AAE 352 Lecture 29 Shear flow in closed sections
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AAE 352 2 16,000 lb. 3A R A R 0.081” 0.032” 0.040” 3A R A R 40” 10” A four spar example-compute the shear flows in a thin-wall cross-section q o q 2 q 1 q 3 0.040”
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AAE 352 3 Loads in the stringers depend on their cross- sectional areas and location with respect to the neutral axis ( 29 ( 29 ( 29 ( 29 2 2 2 2 4 3 5 3 5 5 5 200 zz R R R R R I A A A A A in = + + + = 16,000 200 i i i zz R VQ P a Q a I A = = ( 29 ( 29 3 * 5 R Q A = - 16,000 lb. 40” 5” ( 29 3 *5 R Q A = ( 29 ( 29 * 5 R Q A = - ( 29 ( 29 * 5 R Q A =
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AAE 352 4 16,000 lb. 40” 5” Relationships for the net loads in the stringers 80 i i R P Q a A = 15 R Q A = 15 R Q A = - 5 R Q A = 5 R Q A = - ( 29 80 15 front R R P A a A = ± P i P i P i P i ( 29 Re 80 5 ar R R P A a A = ±
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AAE 352 5 16,000 lb. 40” 5” Computing net loads in the stringers 1200 front P a = ± P i P i P i P i Re 400 ar P a = ±
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AAE 352 6 The net loads in the stringers 16,000 lb. 1200 P a = 400 P a = 1200 P a = 400 P a =
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AAE 352 7 Define the unknown shear flows 1200 P a = 400 P a = 1200 P a = 400 P a = q o q 2 q 1 q 3
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This note was uploaded on 01/18/2012 for the course AAE 352 taught by Professor Chen during the Fall '08 term at Purdue.

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29Closedsectshearflows - AAE 352 Lecture 29 Shear flow in...

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