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: 9780137605521

Author: HIBBELER

Publisher: RENT PEARS

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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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