Question 9.17: A simple beam AB supports a uniform load of intensity q = 1....
A simple beam AB supports a uniform load of intensity q = 1.5 kip/ft and a concentrated load P = 5 kips (Fig. 9-40). The load P acts at the midpoint C of the beam. The beam has a 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 deflection \delta_{C} at the midpoint of the beam by the following methods: (a) Obtain the strain energy of the beam and use Castigliano’s theorem. (b) Use the modified form of Castigliano’s theorem (differentiation under the integral sign).
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Strain energy and Castigliano’s theorem.
1, 2. Conceptualize, Categorize: 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, only the left-hand half of the beam must be analyzed.
The reaction at the left-hand support A (Figs. 9-40 and 9-41) is
Therefore, the bending moment M is
M=R_{A} x-\frac{q x^{2}}{2}=\frac{P x}{2}+\frac{q L x}{2}-\frac{q x^{2}}{2} (a)
in which x is measured from support A.
The strain energy of the entire beam [from Eq. (9-95a)] is
U=\int \frac{M^{2} d x}{2 E I}=2 \int_{0}^{L / 2} \frac{1}{2 E I}\left(\frac{P x}{2}+\frac{q L x}{2}-\frac{q x^{2}}{2}\right)^{2} d xSquare the term in parentheses and perform a lengthy integration to find
U=\frac{P^{2} L^{3}}{96 E I}+\frac{5 P q L^{4}}{384 E I}+\frac{q^{2} L^{5}}{240 E I}3. Analyze: Since the deflection at the midpoint C (Fig. 9-40) corresponds to the load P, find the deflection by using Castigliano’s theorem (Eq. 9-107) as
\delta_{C}=\frac{\partial U}{\partial P}=\frac{\partial}{\partial P}\left(\frac{P^{2} L^{3}}{96 E I}+\frac{5 P q L^{4}}{384 E I}+\frac{q^{2} L^{5}}{240 E I}\right)=\frac{P L^{3}}{48 E I}+\frac{5 q L^{4}}{384 E I} (b)
Part (b): Modified Castigliano’s theorem.
1, 2. Conceptualize, Categorize: Use the modified form of Castigliano’s theorem [Eq. (9-118)] to 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. (a)], and its partial derivative with respect to the load P is
\delta_{i}=\frac{\partial}{\partial P_{i}} \int \frac{M^{2} d x}{2 E I}=\int\left(\frac{M}{E I}\right)\left(\frac{\partial M}{\partial P_{i}}\right) d x (9-118)
\frac{\partial M}{\partial P}=\frac{x}{2}3. Analyze: Therefore, the modified Castigliano’s theorem becomes
\delta_{C}=\int\left(\frac{M}{E I}\right)\left(\frac{\partial M}{\partial P}\right) d x= 2 \int_{0}^{L / 2} \frac{1}{E I}\left(\frac{P x}{2}+\frac{q L x}{2}-\frac{q x^{2}}{2}\right)\left(\frac{x}{2}\right) d x=\frac{P L^{3}}{48 E I}+\frac{5 q L^{4}}{384 E I} (c)
which agrees with the earlier result [Eq. (b)] but requires a much simpler integration.
Numerical solution: Now using this expression for the deflection at point C, substitute numerical values, as
= \frac{(5 kips )(96 in .)^{3}}{48\left(30 \times 10^{6} psi \right)\left(75.0 in ^{4}\right)}+\frac{5(1.5 kip / ft )(1 / 12 ft / in .)(96 in .)^{4}}{383\left(30 \times 10^{6} psi \right)\left(75.0 in ^{4}\right)}
= 0.0410 \text { in. }+0.0614 \text { in. }=0.1024 \text { in. }
4. Finalize: 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.
