Explanation Given information: EI is constant. Method adopted: Castigliano’s theorem. 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. Castigliano’s theorem to determine the displacement for the beam as follows: Δ = ∫ 0 L M ( ∂ M ∂ P ) d x E I (1) Here, Δ is displacement of the point caused by the real loads acting on the beam, P is an external force of variable magnitude applied to the beam at the point and in the direction of Δ , M is internal moment in the beam, expressed as a function of x and caused by both the force P and the actual loads on the beam, 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 ): Apply an external force P at the point where the displacement is to be determined. 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 ) + P ( 2 L ) = 0 B y ( L ) = w × 2 L × 2 L 2 + 2 P L B y = 2 w L + 2 P 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 − P = 0 (2) Substitute ( 2 w L + 2 P ) for B y in Equation (2). A y + 2 w L − 2 w L + 2 P − P = 0 A y = − P 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 + P ( x 1 ) = 0 M 1 = − w x 1 2 2 − P x 1 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 – P for A y in Equation (3). − M 2 − P ( x 2 ) − w ( x 2 ) 2 2 = 0 M 2 = − w x 2 2 2 − P x 2 Calculation of vertical displacement: Tabulate the calculated values as in Table 1

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Chapter 14.10, Problem 14.144P

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