Explanation Given information: The uniformly districted load act at ‘ a ’ distance from the left of the beam is w The flexural rigidity of the beam is E I . The length of the beam is L . The length of segment AC of the beam is a . Calculation: Show the free body diagram of the beam as in Figure 1. Refer Figure 1. Calculate the support reactions A ( R A ) : Calculate the vertical forces by applying the Equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 R A − ( w × a ) = 0 R A − w a = 0 R A = w a 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 M A − ( w × a × a 2 ) = 0 M A = w a 2 2 Show the free body diagram of the moment function of the beam’s cut segment 1 as in Figure 2. Refer Figure 2. Calculate the moment function with respect of the x 1 distance M ( x 1 ) : M ( x 1 ) = − w x 1 2 2 + w a x 1 − w a 2 2 Show the free body diagram of the moment function of the beam’s cut segment 2 as in Figure 3. Refer Figure 3. Calculate the moment function with respect of the x 2 distance M ( x 2 ) : M ( x 2 ) = 0 Elastic Curve and slope is as follows: E I d 2 υ d x 2 = M ( x ) (1) For the coordinate x 1 , Substitute ( − w x 1 2 2 + w a x 1 − w a 2 2 ) for M ( x ) in Equation (1). E I d 2 υ 1 d x 1 2 = − w x 1 2 2 + w a x 1 − w a 2 2 Integrate both sides of the Equation. E I d υ 1 d x 1 = − w x 1 3 6 + w a x 1 2 2 − w a 2 x 1 2 + C 1 (2) Integrate both sides of the Equation (2). E I υ 1 = − w x 1 4 24 + w a x 1 3 6 − w a 2 x 1 2 4 + C 1 x 1 + C 2 (3) For the coordinate x 2 , Substitute 0 for M ( x ) in Equation (1), E I d 2 υ 2 d x 2 2 = 0 Integrate both sides of the Equation. E I d υ 2 d x 2 = C 3 (4) Integrate both sides of the Equation (4). E I υ 2 = C 3 x 2 + C 4 (5) Boundary conditions are as follows: d υ 1 d x 1 = 0 at x 1 = 0 υ 1 = 0 at x 1 = 0 Apply the first boundary condition in Equation (2). E I ( 0 ) = − w ( 0 ) 3 6 + w a ( 0 ) 2 2 − w a 2 ( 0 ) 2 + C 1 C 1 = 0 Apply the second boundary condition in Equation (3). E I ( 0 ) = − w ( 0 ) 4 24 + w a ( 0 ) 3 6 − w a 2 ( 0 ) 2 4 + C 1 ( 0 ) + C 2 C 2 = 0 Consider x 1 = a and x 2 = a Continuity condition is as follows: d υ 1 d x 1 = d υ 2 d x 2 − w x 1 3 6 + w a x 1 2 2 − w a 2 x 1 2 + C 1 = C 3 − C 1 + C 3 = − w x 1 3 6 + w a x 1 2 2 − w a 2 x 1 2 Substitute a for x 1 and 0 for C 1

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

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Chapter 16.2, Problem 29P

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Find the Equation of the elastic curve using coordinates x1 and x2(υ1 and υ2).

Find the slope at B(θB).

Find the deflection at B(υB).

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Chapter 16.2, Problem 28P

Chapter 16.2, Problem 30P

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Find the Equation of the elastic curve using coordinates x 1 and x 2 ( υ 1 and υ 2 ) . Find the slope at B ( θ B ) . Find the deflection at B ( υ B ) .

Solution Summary: The author calculates the Equation of the elastic curve using coordinates x_1 and

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Videos

Find the Equation of the elastic curve using coordinates x1 and x2(υ1 and υ2).

Find the slope at B(θB).

Find the deflection at B(υB).

Want to see the full answer?

Chapter 16 Solutions

Find the Equation of the elastic curve using coordinates x 1 and x 2 ( υ 1 and υ 2 ) . Find the slope at B ( θ B ) . Find the deflection at B ( υ B ) .

Solution Summary: The author calculates the Equation of the elastic curve using coordinates x_1 and

Concept explainers

Videos

Find the Equation of the elastic curve using coordinates x1 and x2(υ1 and υ2). Find the slope at B(θB). Find the deflection at B(υB).

Want to see the full answer?

Chapter 16 Solutions

Find the Equation of the elastic curve using coordinates x1 and x2(υ1 and υ2).

Find the slope at B(θB).

Find the deflection at B(υB).