Explanation Given information: The beam is made of wooden. The modulus of elasticity of the wooden ( E ) is 1.6 × 10 3 ksi . The width of the beam ( b ) is 6 in . The depth of the beam ( d ) is 12 in . The uniformly varying load act 9 ft distance from left end of the beam ( w ) is 0.8 kip/ft . The point load act at a right end of the beam ( P ) is 1.5 kip . The moment of inertia of the beam is I The length of the beam ( L ) is 18 ft . The length of segment AB and BC ( L A B and L B C ) is 9 ft . Calculation: Show the free body diagram of the beam as in Figure 1. Refer to Figure 1. Calculate the support reactions A and B ( A y and B y ) : Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 A y + B y − 1.5 − ( 0.8 × 9 × 1 2 ) = 0 A y + B y − 5.1 = 0 A y + B y = 5.1 (1) Calculate the moment about point A by applying the equation of equilibrium: Sum of moments about point A is equal to 0. ∑ M A = 0 ( B y × 9 ) − ( 1.5 × 18 ) − ( 1 2 × 0.8 × 9 × 2 × 9 3 ) = 0 9 B y − 27 − 21.6 = 0 9 B y − 48.6 = 0 9 B y = 48.6 B y = 48.6 9 B y = 5.4 kip Substitute 5.4 kip for B y in Equation (1). A y + 5.4 = 5.1 A y = 5.1 − 5.4 A y = − 0.3 kip A y = 0.3 kip ↓ Calculate the moment function with respect of the x distance M ( x ) : M ( x ) = [ − 0.3 ( x − 0 ) − 1 6 × 1.6 18 ( x − 0 ) 3 − ( − 5.4 ) ( x − 9 ) − ( − 0.8 2 ) ( x − 9 ) 2 − 1 6 ( − 0.8 9 ) ( x − 9 ) 3 ] = [ − 0.3 x − 0.0148 x 3 + 5.4 ( x − 9 ) + 0.4 ( x − 9 ) 2 + 0.0148 ( x − 9 ) 3 ] Elastic Curve and slope is as follows: E I d 2 υ d x 2 = M ( x ) (2) Substitute [ − 0.3 x − 0.0148 x 3 + 5.4 ( x − 9 ) + 0.4 ( x − 9 ) 2 + 0.0148 ( x − 9 ) 3 ] for M ( x ) in Equation (2). E I d 2 υ d x 2 = − 0.3 x − 0.0148 x 3 + 5.4 ( x − 9 ) + 0.4 ( x − 9 ) 2 + 0.0148 ( x − 9 ) 3 (3) Integrate both sides of the Equation (3). E I d υ d x = [ − 0.3 x 2 2 − 0.0148 x 4 4 + 5.4 ( x − 9 ) 2 2 + 0.4 ( x − 9 ) 3 3 + 0.0148 ( x − 9 ) 4 4 + C 1 ] = [ − 0.15 x 2 − 0.003704 x 4 + 2.7 ( x − 9 ) 2 + 0.13330 ( x − 9 ) 3 + 0.003704 ( x − 9 ) 4 + C 1 ] (4) Integrate both sides of the Equation (4). E I υ = [ − 0.15 x 3 3 − 0.003704 x 5 5 + 2.7 ( x − 9 ) 3 3 + 0.13330 ( x − 9 ) 4 4 + 0.003704 ( x − 9 ) 5 5 + C 1 x + C 2 ] = [ − 0

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Author: HIBBELER

Publisher: PEARSON

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Chapter 12.3, Problem 12.48P

To determine

The equation of the elastic curve of the beam (υ) .

The deflection at the end C of the beam (υC) .

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Chapter 12.3, Problem 12.47P

Chapter 12.3, Problem 12.49P

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The equation of the elastic curve of the beam ( υ ) . The deflection at the end C of the beam ( υ C ) .

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The equation of the elastic curve of the beam (υ) . The deflection at the end C of the beam (υC) .

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Chapter 12 Solutions

The equation of the elastic curve of the beam (υ) .

The deflection at the end C of the beam (υC) .