HW4-1-2 Solutions

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

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Unformatted text preview: 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: atan 22 19 = 49.185 = atan 22 11 = 63.435 = 180 - - = 67.38 = Reactions at the supports: Fx = 0 Ex 0.00N = Me = 0 60 m ( )- Ay ( ) 60m 19m- ( ) 1.00N ( ) + 60m 49m- ( ) 1.00N ( ) + = Ay 41m ( ) 1.00N ( ) 11m ( ) 1.00N ( ) + 60m = Ay 0.867N = Fy = 0 ENGR 111-804,5,6 Fall, 2009 HW 4-1, 4-2 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 Fy = Fab sin 49.185 ( ) 0.867N + = Fab 0.867 N sin 49.185 ( )- = Fab 1.146- N = Fab = 1.146N (C) 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 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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HW4-1-2 Solutions - ENGR 111-804,5,6 Fall, 2009 HW 4-1, 4-2...

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