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assignment10-1

# assignment10-1 - Assignment 10 Key solutions Chapter 6...

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Assignment 10 Key solutions Chapter 6, Problem 86. Determine the components of the reactions at A and E when a counterclockwise couple of magnitude 192 lb in. is applied to the frame ( a ) at B , ( b ) at D . Chapter 6, Solution 86. ( a ) FBD AC : ( b ) FBD CE : Note: CE is a two-force member ( 29 ( 29 1 1 0: 8 in. 2 in. 192 lb in. 0 2 2 A CE CE M F F Σ = - - + = 19.2 2 lb, CE F = 19.20 lb x = E t 19.20 lb y = E t 0: 19.2 lb = 0, x x F A Σ = - 19.20 lb x = A t 0: 19.2 lb = 0, y y F A Σ = - 19.20 lb y = A t Note: AC is a two-force member ( 29 1 4 0: 3 in. 192 lb in. = 0 17 17 E AE AE M F F Σ = + + 12.8 17 lb, AE F = - 4 , 17 x AE A F = 51.2 lb x = A t 1 , 17 y AE A F = 12.80 lb y = A t 0: 51.2 lb = 0, x x F E Σ = - 51.2 lb x = E t 0: 12.80 lb = 0, y y F E Σ = - 12.80 lb y = E t

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Chapter 6, Problem 89. The 120-N load can be moved along the line of action shown and can be applied at A , D , or E . Determine the components of the reactions at B and F when the 120-N load is applied ( a ) at A , ( b ) at D , ( c ) at E . Chapter 6, Solution 89. ( a ) FBD ACF : ( b ) FBD BCE : Note: BC is a two-force member ( 29 ( 29 ( 29 1 0: 0.1 m 120 N 0.3 m 0, 26 F BC M F Σ = - = 40 26 N, BC F = 200 N x = B t 40.0 N y = B t ( 29 5 0: 40 26 N 0, 26 x x F F Σ = - = 200 N x = F t ( 29 1 0: 120 N 40 26 N 0, 26 y y F F Σ = - - = 160.0 N y = F t Note ACF is a two-force member ( 29 ( 29 ( 29 1 0: 0.4 m 120 N 0.3 m 0, 5 B CF M F Σ = - = 160 5 N, CF F = 320 N x = F t 160.0 N y = F t ( 29 2 0: 160 5 N 0, 5 x x
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