A simple beam AB supports a uniform load of intensity q = 1.5 k/ft and a concentrated load P = 5 k (Fig. 9-40). The load P acts at the midpoint C of the beam. The beam has length L = 8.0 ft, modulus of elasticity E = 30 × 10^6 psi, and moment of inertia I = 75.0 in.^4 Determine the downward

Question

A simple beam AB supports a uniform load of intensity q = 1.5 k/ft and a concentrated load P = 5 k (Fig. 9-40). The load P acts at the midpoint C of the beam. The beam has length L = 8.0 ft, modulus of elasticity E = 30 × [latex]10^{6}[/latex] psi, and moment of inertia I = 75.0 [latex]in.^{4}[/latex]
Determine the downward deflection [latex]\delta_{C}[/latex] at the midpoint of the beam by the following methods: (1) Obtain the strain energy of the beam and use Castigliano’s theorem, and (2) use the modified form of Castigliano’s theorem (differentiation under the integral sign).

Accepted Answer

Method (1). Because the beam and its loading are symmetrical about the midpoint, the strain energy for the entire beam is equal to twice the strain energy for the left-hand half of the beam. Therefore, we need to analyze only the left-hand half of the beam.
The reaction at the left-hand support A (Figs. 9-40 and 9-41) is

[latex]R_{A}=\frac{P}{2}+\frac{qL}{2}[/latex]

and therefore the bending moment M is

[latex]M=R_{A}x-\frac{qx^{2}}{2}=\frac{Px}{2}+\frac{qLx}{2}-\frac{qx^{2}}{2}[/latex] (x)

in which x is measured from support A.
The strain energy of the entire beam (from Eq. 9-80a) is

[latex]U=\int{\frac{M^{2}dx}{2EI}}=2\int_{0}^{{L}/{2}}{\frac{1}{2EI}\left(\frac{Px}{2}+\frac{qLx}{2}-\frac{qx^{2}}{2} ight)}^{2}dx[/latex]

After squaring the term in parentheses and performing a lengthy integration, we find

[latex]U=\frac{P^{2}L^{3}}{96EI}+\frac{5PqL^{4}}{384EI}+\frac{q^{2}L^{5}}{240EI}[/latex]

Since the deflection at the midpoint C (Fig. 9-40) corresponds to the load P, we can find the deflection by using Castigliano’s theorem (Eq. 9-87):

[latex]\delta_{i}=\frac{\partial U}{\partial P_{i}}[/latex] (9-87)

[latex]\delta_{C}=\frac{\partial U}{\partial P}=\frac{\partial}{\partial P}\left(\frac{P^{2}L^{3}}{96EI}+\frac{5PqL^{4}}{384EI}+\frac{q^{2}L^{5}}{240EI} ight)=\frac{PL^{3}}{48EI}+\frac{5qL^{4}}{384EI}[/latex] (y)

Method (2). By using the modified form of Castigliano’s theorem (Eq. 9-88), we avoid the lengthy integration for finding the strain energy. The bending moment in the left-hand half of the beam has already been determined (see Eq. x), and its partial derivative with respect to the load P is

[latex]\delta_{i}=\frac{\partial}{\partial P_{i}}\int{\frac{M^{2}dx}{2EI}}=\int{\left(\frac{M}{EI} ight)\left(\frac{\partial M}{\partial P_{i}} ight)dx}[/latex] (9-88)

[latex]\frac{\partial M}{\partial P}=\frac{x}{2}[/latex]

Therefore, the modified Castigliano’s theorem becomes

[latex]\delta_{C}=\int{\left(\frac{M}{EI} ight)\left(\frac{\partial M}{\partial P} ight)dx}[/latex]

[latex]=2\int_{0}^{{L}/{2}}{\frac{1}{EI}\left(\frac{Px}{2}+\frac{qLx}{2}-\frac{qx^{2}}{2} ight)\left(\frac{x}{2} ight)dx}=\frac{PL^{3}}{48EI}+\frac{5qL^{4}}{384EI}[/latex] (z)

which agrees with the earlier result (Eq. y) but requires a much simpler integration.
Numerical solution. Now that we have an expression for the deflection at point C, we can substitute numerical values, as follows:

[latex]\delta_{C}=\frac{PL^{3}}{48EI}+\frac{5qL^{4}}{384EI}[/latex]

[latex]=\frac{(5 k)(96 in.)^{3}}{48(30 imes10^{6} psi)(75.0 in.^{4})}+\frac{5(1.5 k/ft)(1/12 ft/in.)(96 in.)^{4}}{384(30 imes10^{6} psi )(75.0 in.^{4})}[/latex]

= 0.0410 in. + 0.0614 in. = 0.1024 in.

Note that numerical values cannot be substituted until after the partial derivative is obtained. If numerical values are substituted prematurely, either in the expression for the bending moment or the expression for the strain energy, it may be impossible to take the derivative.

Question 9.17: A simple beam AB supports a uniform load of intensity q = 1....

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