Explanation Given information: The modulus of elasticity of wood is E = 250 GPa . The cross sectional dimension of the beam is 100 mm × 200 mm . Calculation: Find the moment of inertia of the beam ( I ) as shown below. I = b d 3 12 Here, b is the width of the beam and d is the depth of the beam. Substitute 100 mm for b and 200 mm for d . I = ( 100 mm × 1 m 1 , 000 mm ) ( 200 mm × 1 m 1 , 000 mm ) 3 12 = 6.667 × 10 − 5 m 4 Sketch the Elastic curve of the beam as shown in Figure 1. Refer to Figure 1. Apply the method of superposition as shown below. Find the slope of the beam at A as shown below. θ A = ( θ A ) 1 + ( θ A ) 2 (1) Here, ( θ A ) 1 is the slope due to the load applied at 1.5 m from A and ( θ A ) 2 is the slope due to the load applied at 3 m from A . Find the slope ( θ A ) 1 as shown below. ( θ A ) 1 = P a b ( L + b ) 6 E I L Here, P is the applied load, a is the distance of the application of the load from A , b is the distance of the application of the load from B , L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia. Substitute 3 kN for P , 1.5 m for a , 4.5 m for b , 6 m for L , 6.667 × 10 − 5 m 4 for I , and 10 GPa for E . ( θ A ) 1 = ( 3 kN × 1 , 000 N 1 kN ) × ( 1.5 m ) × ( 4 .5 m ) × ( 6 m + 4.5 m ) 6 × ( 10 GPa × 10 9 N / m 2 1 GPa ) × ( 6.667 × 10 − 5 m 4 ) × ( 6 m ) = 212 , 625 24 , 001 , 200 = 0.008589 Find the slope ( θ A ) 2 as shown below. ( θ A ) 2 = P L 2 16 E I Substitute 3 kN for P , 6 m for L , 6.667 × 10 − 5 m 4 for I , and 10 GPa for E . ( θ A ) 2 = ( 3 kN × 1 , 000 N 1 kN ) × ( 6 m ) 2 16 × ( 10 GPa × 10 9 N / m 2 1 GPa ) × ( 6.667 × 10 − 5 m 4 ) = 108 , 000 10 , 667 , 200 = 0.010124 Substitute 0.008589 for ( θ A ) 1 and 0.010124 for ( θ A ) 2 in Equation (1)

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 12.5, Problem 12.88P

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The slope at A and the deflection at point C.

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The slope at A and the deflection at point C .

The slope at A and the deflection at point C.

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