Explanation Given information: The wide-flange beam acts as a cantilever. It is installed at an angle θ with the vertical. The W 10 × 15 beam is made of an A992 steel. The applied load is P = 1.5 kip . The Length of the beam is L = 12 ft . The vertical angle is θ = 10 ° . Calculation: The Modulus of elasticity of A992 steel is E = 29 , 000 ksi . The moment of inertia for the W 10 × 15 beam is I x = 68.9 in 4 and I y = 2.89 in 4 . Sketch the components as shown in Figure 1. Refer to Figure 1. Resolve the moment into components in terms of θ as shown below. Along x direction, x max = ( P sin θ ) L 3 3 E I y (1) Here, P is the applied load, θ is the angle, L is the length of the beam, E is the modulus of elasticity, and I y is the moment of inertia about y axis. Along y direction, y max = ( P cos θ ) L 3 3 E I x (2) Here, I x is the moment of inertia about x axis. Find the ratio of the deflection in x direction to the deflection in y direction at A as shown below. x max y max = ( P sin θ ) L 3 3 E I y ( P cos θ ) L 3 3 E I x = ( P sin θ ) L 3 3 E I y × 3 E I x ( P cos θ ) L 3 = I x sin θ I y cos θ = I x I y tan θ Hence, the ratio of the deflection at A is I x I y tan θ _

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 12.5, Problem 12.97P

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The ratio of the deflection in x direction to the deflection in y direction at A and numerically the deflection in x direction and the deflection in y direction.

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Chapter 12.5, Problem 12.96P

Chapter 12.5, Problem 12.98P

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