Explanation Given information: EI is constant. Assumptions: Consider that the right side force as positive and left side as negative in horizontal components of forces. Consider that the upward force as positive and downward as negative in vertical components of forces. Consider that the clockwise moment as negative and anticlockwise moment as positive. Virtual-work equation for the beam as follows: 1 ⋅ θ = ∫ 0 L m θ M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of θ , θ is slope caused by the real loads on the beam, m θ is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit moment, M is internal moment in the beam, expressed as a function of x and caused by the real loads, E is modulus of elasticity of a member, and I is moment of inertia of the cross-sectional area about the neutral axis. Internal moment in the beam ( M ): Determine the reactions: Show the free-body diagram of the entire beam as in Figure 1. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 w × 2 L × 2 L 2 − B y ( L ) = 0 B y ( L ) = w × 2 L × 2 L 2 B y = 2 w L Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − w × 2 L = 0 (2) Substitute 2 w L for B y in Equation (2). A y + 2 w L − 2 w L = 0 A y = 0 Region 0 ≤ x < L : Show the free body diagram of the beam for the region 0 ≤ x < L as in Figure 2. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 M 1 + w × x 1 × x 1 2 = 0 M 1 = − w x 1 2 2 Region L < x ≤ 2 L : Show the free body diagram of the beam for the region L < x ≤ 2 L as in Figure 3. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 2 = 0 − M 2 + A y ( x 2 ) − w × x 2 × x 2 2 = 0 (3) Substitute 0 for A y in Equation (3). − M 2 + 0 ( x 2 ) − w ( x 2 ) 2 2 = 0 M 2 = − w x 2 2 2 Internal virtual moment in the beam ( m ): Apply a unit moment at point B and calculate the moments. Determine the reactions: Show the free-body diagram of the entire beam as in Figure 4. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( L ) − 1 = 0 B y = 1 L Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 14.7, Problem 14.122P

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