Explanation Displacement: Virtual-work equation for the entire beam is as follows: 1 ⋅ Δ = ∫ 0 L m M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of Δ , Δ is displacement caused by the real loads on the beam, m is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit load, 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. Slope: Virtual-work equation for the entire beam is as follows: 1 ⋅ θ = ∫ 0 L m θ M E I d x (2) 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. Internal moment in the beam ( M ): Determine the reactions: Entire beam: 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 B y ( L ) − P ( L 2 ) = 0 (3) 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 − P = 0 (4) Along the horizontal direction: Determine the horizontal reaction at point B by resolving the horizontal component of forces. ∑ F x = 0 B x = 0 Show the calculation of values as follows: Conclusion: Solve Equation (3). B y ( L ) = P ( L 2 ) B y = P 2 Substitute P 2 for B y in Equation (4). A y + P 2 − P = 0 A y = P 2 Region 0 ≤ x < L 2 : Show the free body diagram of the beam for the region 0 ≤ x < L 2 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 − A y ( x 1 ) = 0 (5) Substitute P 2 for A y Equation (5). M 1 − P 2 ( x 1 ) = 0 M 1 = P x 1 2 Region L 2 < x ≤ L : Show the free body diagram of the beam for the region L 2 < x ≤ 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 ( L 2 + x 2 ) + P ( x 2 ) = 0 (6) Substitute P 2 for A y in Equation (6). M 2 − P 2 ( L 2 + x 2 ) + P ( x 2 ) = 0 M 2 − P L 4 − P x 2 2 + P x 2 = 0 M 2 = P L 4 − P x 2 2 Determination of displacement at point C : Internal virtual moment in the beam ( m ): Apply a unit load at point C and calculate the moments. Determine the reactions: Entire beam: 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 ( L 2 ) = 0 (7) 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 − 1 = 0 (8) Along the horizontal direction: Determine the horizontal reaction at point B by resolving the horizontal component of forces. ∑ F x = 0 B x = 0 Show the calculation of values as follows: Conclusion: Solve Equation (7). B y ( L ) − 1 ( L 2 ) B y = 1 2 Substitute 1 2 for B y in Equation (8). A y + 1 2 − 1 = 0 A y = 1 2 Region 0 ≤ x < L 2 : Show the free body diagram of the beam for the region 0 ≤ x < L 2 as in Figure 5. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 m 1 − A y ( x 1 ) = 0 (9) Substitute 1 2 for A y in Equation (9). m 1 − 1 2 ( x 1 ) = 0 m 1 = x 1 2 Region L 2 < x ≤ L : Show the free body diagram of the beam for the region L 2 < x ≤ L as in Figure 6. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 2 = 0 m 2 − A y ( L 2 + x 2 ) + 1 ( x 2 ) = 0 (10) Substitute 1 2 for A y in Equation (10). m 2 − 1 2 ( L 2 + x 2 ) + 1 ( x 2 ) = 0 m 2 − L 4 − x 2 2 + x 2 = 0 m 2 = L 4 − x 2 2 Calculation of displacement: Substitute the calculated moment values as follows: 1 ⋅ Δ C = ∫ 0 L / 2 m 1 M 1 E I d x 1 + ∫ 0 L / 2 m 2 M 2 E I d x 2 (11) Substitute x 1 2 for m 1 , P x 1 2 for M 1 , ( L 4 − x 2 2 ) for m 2 , and ( P L 4 − P x 2 2 ) for M 2 in Equation (11)

11th Edition

ISBN: 9780137605514

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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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 14.7, Problem 105P

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The displacement at point C and the slope at B.

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Chapter 14.7, Problem 104P

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The displacement at point C and the slope at B .

Solution Summary: The author explains the virtual-work equation for the entire beam, which is as follows: displacement, slope, internal moment, and inertia.

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The displacement at point C and the slope at B.

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