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Unformatted text preview: ENGR 111804,5,6 Fall, 2009 HW 41, 42 Solutions T. Pollock 1 41: The truss drawn to the right is loaded with two external forces, of magnitude P, acting vertically downward. Each member can withstand a force of 75 kN in tension and 66 kN in compression. Using a factor of safety of 2.5, calculate the maximum allowable load P. Solution: Let P = 1.00 N. Truss FBD: Calculate angles: α atan 22 19 = α 49.185 ° ⋅ = β atan 22 11 = β 63.435 ° ⋅ = γ 180° α β = γ 67.38 ° ⋅ = Reactions at the supports: Σ Fx = 0 Ex 0.00N = Σ Me = 0 60 m ⋅ ( ) Ay ( ) ⋅ 60m 19m ( ) 1.00N ( ) ⋅ + 60m 49m ( ) 1.00N ( ) ⋅ + = Ay 41m ( ) 1.00N ( ) ⋅ 11m ( ) 1.00N ( ) ⋅ + 60m = Ay 0.867N = Σ Fy = 0 ENGR 111804,5,6 Fall, 2009 HW 41, 42 Solutions T. Pollock 2 Ay Ey + 1.00N 1.00N = Ey 2.00 N ⋅ Ay = Ey 2.00 N ⋅ 0.867 N = Ey 1.133N = Internal forces in the members, by joint: 0.867 N Fab Fac A 49.185 o Σ Fy = Fab sin 49.185° ( ) ⋅ 0.867N + = Fab 0.867 N ⋅ sin 49.185° ( ) = Fab 1.146 N = Fab = 1.146N (C) Σ Fx = 0 Fac Fab cos 49.185° ( ) ⋅ + = Fac 1.146 N ( ) cos 49.185° ( ) ⋅ + = Fac 1.146 N ( ) cos 49.185° ( ) ⋅ = Fac 0.749N = Fac = 0.749N (T) B 63.435 o 49.185 o 1.00 N 1.146 N Fbd Fbc Σ Fy = 1.00 N 1.146N sin 49.185° ( ) ⋅ + Fbc sin 63.435° ( ) ⋅ = Fbc 1.0 N ⋅ 1.146 N ⋅ sin 49.185 ° ⋅ ( ) ⋅ sin 63.435 ° ⋅ ( ) = Fbc 0.1480....
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This note was uploaded on 04/04/2011 for the course ENGR 111 taught by Professor Walker during the Fall '07 term at Texas A&M.
 Fall '07
 walker

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