Calculate the deflection at the mid-span point of the beam of Ex. 15.18 by assuming a deflected shape function of the form

$v=v_{1} \sin \frac{\pi x}{L}+v_{3} \sin \frac{3 \pi x}{L}$

in which $v_{1}$ and $v_{3}$ are unknown displacement parameters. Note:

$\int_{0}^{L} \sin ^{2}\left(\frac{n \pi x}{L}\right) d x=\frac{L}{2} \int_{0}^{L} \sin \left(\frac{m \pi x}{L}\right) \sin \left(\frac{n \pi x}{L}\right) d x=0$

Ans. $0.02078 WL ^{3} / EI$ .

Question

Calculate the deflection at the mid-span point of the beam of Ex. 15.18 by assuming a deflected shape function of the form

[latex]v=v_{1} \sin \frac{\pi x}{L}+v_{3} \sin \frac{3 \pi x}{L}[/latex]

in which [latex]v_{1}[/latex] and [latex]v_{3}[/latex] are unknown displacement parameters. Note:

[latex]\int_{0}^{L} \sin ^{2}\left(\frac{n \pi x}{L}\right) d x=\frac{L}{2} \int_{0}^{L} \sin \left(\frac{m \pi x}{L}\right) \sin \left(\frac{n \pi x}{L}\right) d x=0[/latex]

Ans. [latex]0.02078 WL ^{3} / EI[/latex] .

Accepted Answer

This problem may be solved using either complementary energy or potential energy. The strain energy U due to the bending of the beam is given by Eq. (9.21),

[latex]U=\int_{L} \frac{M^{2}}{2 E I_{z}} d x[/latex] (9.21)

i.e.

[latex]U=\int_{0}^{L}\left(M^{2} / 2 E I ight) d x[/latex]

and the potential energy of the applied load is, from Ex. 15.18

[latex]V=-W v_{ B }[/latex]

The total potential energy is then

[latex]TPE =U+V=\int_{0}^{L}\left(\frac{M^{2}}{2 E I} ight) d x-W v_{ B }[/latex]

which, from Eq. (13.3)

[latex]\frac{ d ^{2} v}{ d x^{2}}=\frac{M}{E I}[/latex] (13.3)

may be written

[latex]TPE =\frac{E I}{2} \int_{0}^{L}\left(\frac{ d ^{2} v}{ d x^{2}} ight)^{2} d x-W v_{ B }[/latex]

Substituting for [latex]d ^{2} v / d x^{2}[/latex] from the given deflected shape function

[latex]TPE =\frac{E I}{2} \int_{0}^{L}\left[\left(\frac{\pi^{2} v_{1}}{L^{2}} ight) \sin \left(\frac{\pi x}{L} ight)+\left(\frac{9 \pi^{2} v_{3}}{L^{2}} ight) \sin \left(\frac{3 \pi x}{L} ight) ight]^{2} d x-W v_{ B }[/latex]

or

[latex]TPE =\left(\frac{\pi^{4} E I}{2 L^{4}} ight)\left[v_{1}^{2} \int_{0}^{L} \sin ^{2}\left(\frac{\pi x}{L} ight) d x+18 v_{1} v_{3} \int_{0}^{L} \sin \left(\frac{\pi x}{L} ight) \sin \left(\frac{3 \pi x}{L} ight) d x ight.[/latex]

[latex]\left.+81 v_{3}^{2} \int_{0}^{L} \sin ^{2}\left(\frac{3 \pi x}{L} ight) d x ight][/latex]

Using the values of the integrals given

[latex]TPE =\left(\frac{\pi^{4} E I}{4 L^{3}} ight)\left(v_{1}^{2}+81 v_{3}^{2} ight)-W v_{ B }[/latex]

From the given deflected shape function, at mid-span when x=L/2

[latex]v_{ B }=v_{1}-v_{3}[/latex]

so that

[latex]TPE =\left(\frac{\pi^{4} E I}{4 L^{3}} ight)\left(v_{1}^{2}+81 v_{3}^{2} ight)-W\left(v_{1}-v_{3} ight)[/latex]

Then, from the principle of the stationary value of the total potential energy

[latex]\frac{\partial( TPE )}{\partial v_{1}}=\left(\frac{\pi^{4} E I}{2 L^{3}} ight) v_{1}-W=0[/latex]

[latex]\frac{\partial( TPE )}{\partial v_{3}}=\left(\frac{81 \pi^{4} E I}{2 L^{3}} ight) v_{3}+W=0[/latex]

Solving gives

[latex]v_{1}=\frac{2 W L^{3}}{\pi^{4} E I}, \quad v_{3}=\frac{-2 W L^{3}}{81 \pi^{4} E I}[/latex]

so that

[latex]v_{ B }=\left(\frac{2 W L^{3}}{\pi^{4} E I} ight)\left(1+\frac{1}{81} ight)[/latex]

i.e.

[latex]v_{ B }=\frac{0.02078 WL ^{3}}{E I}[/latex]

Note that Eq. (v) of Ex. 13.5

[latex]v_{ ext {mid-span }}=-\frac{W L^{3}}{48 E I}[/latex] Eq. (v) of Ex. 13.5

gives [latex]v_{ B }=0.02083 W L^{3} / E I[/latex] .

Question 15.P.22: Calculate the deflection at the mid-span point of the beam o...

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