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hw7-sol - Problem 10.47 Each bar has a 500 m m2...

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Problem 10.47 Each bar has a 500 m m 2 cross- sectional area and modulus of elasticity E = 72 GPa . If a 160-kN downward force is applied at A , what is the resulting displacement of point A ? Free Body Diagram: Solution: The symmetry of the truss dictates that ALL of the displacement will be vertical. We see that the axial loads in the two angled members will be equal. In other words, δ B = δ C . Summing vertical forces where the load is applied: [1] Σ F y = 0 = 160 , 000 N + P A + 2 P B (sin60 ) The vertical displacement for each of the two angled members is: v angle = δ angle (1 / sin60 ) Using the relationship between displacements for the straight member and the two angled members: δ A = δ B ( 1 sin 60 ) = δ C ( 1 sin 60 ) P A (0 . 3 m) AE = P B ( 0 . 3 m sin 60 ) AE ( 1 sin 60 ) [2] P A = 1 . 333 P B = 1 . 333 P C Substituting Equation [2] into Equation [1]: 0 = 160 , 000 N + 1 . 333 P B + 2 P B (sin60 ) P B = P C = 52 , 201 N From Equation [2] we get the axial load in member A : P A = 1 . 333(52 , 201 N) = 69 , 584 N The displacement of point A is: ANS: δ A = P A L A AE = (69 , 584 N)(0 . 3 m) ( 500 × 10 6 m 2 )( 72 × 10 9 N / m 2 ) = 0 . 580 mm
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Problem 14.9 Model the ladder rung as a simply sup- ported (pin-supported) beam and assume that the 200- lb load exerted by the person’s shoe is uniformly dis- tributed. Determine the internal forces an moment at A . Solution: The distributed load due to the weight exerted by the foot is: w 0 = (200 lb) / (4 in) = 50 lb / in Draw the FBD for the ladder rung and determine the reactions at the left and right ends. Σ M L = 0 = [(50lb / in)(4in)](10in)+ R y (15in) R y = 133 . 3lb Σ F y = 0 = [(50lb / in)(4in)]+ R y + L Y = 200lb+133 . 3lb+ L y L y = 66 . 67lb Since no forces are exerted in the x -direction: L x = R x = 0 Cut the beam through point A and draw the FBD: Σ M A = 0 = (66 . 67lb)(10in)+[(50lb / in)(2in)](1in)+ M A ANS: M A = 567 in lb Σ F y = 0 = L y (50lb / in)(2in)+ V A = 66 . 67lb 100lb+ V A ANS: V A = 33 . 3 lb Σ F x = 0 = P A ANS: P A = 0
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Problem 14.28 (a) Determine the internal forces and moment as functions of x . (b) Draw the shear force and bending moment diagrams.
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