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HW4-1-2 Solutions

# HW4-1-2 Solutions - ENGR 111-804,5,6 Fall 2009 HW 4-1 4-2...

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ENGR 111-804,5,6 Fall, 2009 HW 4-1, 4-2 Solutions T. Pollock 1 4-1: 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 ( ) + 0 = Ay 41m ( ) 1.00N ( ) 11m ( ) 1.00N ( ) + 60m = Ay 0.867N = Σ Fy = 0

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ENGR 111-804,5,6 Fall, 2009 HW 4-1, 4-2 Solutions T. Pollock 2 Ay Ey + 1.00N - 1.00N - 0 = 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 = 0 Fab sin 49.185° ( ) 0.867N + 0 = Fab 0.867 N sin 49.185° ( ) - = Fab 1.146 - N = Fab = 1.146N (C) Σ Fx = 0 Fac Fab cos 49.185° ( ) + 0 = Fac 1.146 - N ( ) cos 49.185° ( ) + 0 = 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 = 0 1.00 - N 1.146N sin 49.185° ( ) + Fbc sin 63.435° ( ) - 0 = Fbc 1.0 N 1.146 N sin 49.185 ° ( ) - sin 63.435 ° ( ) - = Fbc 0.148 - N = Fbc = 0.148N (C) Σ Fx = 0 Fbd 1.146N cos 49.185 ° ( ) + 1.148 - N ( ) cos 63.435 ° ( ) + 0 = Fbd 1.148 N cos 63.435 ° ( ) 1.146 N cos 49.185 ° ( ) - =
ENGR 111-804,5,6 Fall, 2009

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HW4-1-2 Solutions - ENGR 111-804,5,6 Fall 2009 HW 4-1 4-2...

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