A uniform simply supported beam, span L, carries a distributed loading which varies according to a parabolic law across the span. The load intensity is zero at both ends of the beam and wo at its midpoint. The loading is normal to a principal axis of the beam cross section and the relevant flexural

Question

A uniform simply supported beam, span L, carries a distributed loading which varies according to a parabolic law across the span. The load intensity is zero at both ends of the beam and [latex]w_{ o }[/latex] at its midpoint. The loading is normal to a principal axis of the beam cross section and the relevant flexural rigidity is EI. Assuming that the deflected shape of the beam can be represented by the series

[latex]v=\sum_{ i =1}^{\infty} a_{ i } \sin \frac{ i \pi x}{L}[/latex]

find the coefficients [latex]a_{ i }[/latex] and the deflection at the mid-span of the beam using only the first term in this series.
Ans. [latex]a_{ i }=32 w_{ o } L^{4} / E I \pi^{7} i ^{7} ext { (i odd), } w_{ o } L^{4} / 94.4 E I[/latex].

Accepted Answer

The problem is solved in a similar manner to P.15.20. Therefore Eq. (iii) of S.15.20 applies,

[latex]C=\int_{0}^{L} \frac{E I}{2}\left(\frac{ d ^{2} v}{ d x^{2}} ight)^{2} d x-\int_{0}^{L} w v d x[/latex] Eq. (iii) of S.15.20

that is

[latex]C=\int_{L} \frac{E I}{2}\left(\frac{ d ^{2} v}{ d x^{2}} ight)^{2} d x-\int_{L} w v d x[/latex] (i)

in which

[latex]v=\sum_{i=1}^{\infty} a_{i} \sin \frac{i \pi x}{L}[/latex] (ii)

and w may be expressed as a function of x in the form [latex]w=4 w_{0} x(L-x) / L^{2}[/latex] which satisfies the boundary conditions of w = 0 at x = 0 and x = L and [latex]w=w_{0}[/latex] at x = L/2.
Substituting in Eq. (i)

[latex]C=\frac{E I}{2} \int_{0}^{L} \sum_{i=1}^{\infty} a_{i}^{2} \frac{i^{4} \pi^{4}}{L^{4}} \sin ^{2} \frac{i \pi x}{L} d x-\frac{4 w_{0}}{L^{2}} \int_{0}^{L} x(L-x) \sum_{i=1}^{\infty} a_{i} \sin \frac{i \pi x}{L} d x[/latex] (iii)

Now

[latex]\int_{0}^{L} \sin ^{2} \frac{i \pi x}{L} d x=\int_{0}^{L} \frac{1}{2}\left(1-\cos \frac{i 2 \pi x}{L} ight) d x=\left[\frac{x}{2}-\frac{L}{i 2 \pi} \sin \frac{i 2 \pi x}{L} ight]_{0}^{L}=\frac{L}{2}[/latex]

[latex]\int_{0}^{L} L x \sin \frac{i \pi x}{L} d x=L\left[-\frac{x L}{i \pi} \cos \frac{i \pi x}{L}+\int \frac{L}{i \pi} \cos \frac{i \pi x}{L} d x ight]_{0}^{L}=\frac{-L^{3}}{i \pi} \cos i \pi[/latex]

[latex]\int_{0}^{L} x^{2} \sin \frac{i \pi x}{L} d x=\left[-\frac{x^{2} L}{i \pi} \cos \frac{i \pi x}{L}+\int \frac{L}{i \pi} \cos \frac{i \pi x}{L} 2 x d x ight]_{0}^{L}=-\frac{L^{3}}{i \pi} \cos i \pi+\frac{2 L^{3}}{i^{3} \pi^{3}}(\cos i \pi-1)[/latex]

Substituting these expressions in Eq. (iii) gives

[latex]C=\sum_{i=1}^{\infty} \frac{E I a_{i}^{2} i^{4} \pi^{4}}{4 L^{3}}-\frac{4 w_{0}}{L^{3}} \sum_{i=1}^{\infty} \frac{2 a_{i} L^{3}}{i^{3} \pi^{3}}(1-\cos i \pi)[/latex] (iv)

The value of [latex](1-\cos i \pi)[/latex] is zero when i is even and 2 when i is odd. Therefore Eq. (iv) may be written

[latex]C=\frac{E I a_{i}^{2} i^{4} \pi^{4}}{4 L^{3}}-\frac{16 w_{0} a_{i} L}{i^{3} \pi^{3}}[/latex] i is odd

From the principle of the stationary value of the total complementary energy

[latex]\frac{\partial C}{\partial a_{i}}=\frac{E I i^{4} \pi^{4} a_{i}}{2 L^{3}}-\frac{16 w_{0} L}{i^{3} \pi^{3}}=0[/latex]

which gives

[latex]a_{i}=\frac{32 w_{0} L^{4}}{E I i^{7} \pi^{7}}[/latex]

Then

[latex]v=\sum_{i=1}^{\infty} \frac{32 w_{0} L^{4}}{E I i^{7} \pi^{7}} \sin \frac{\pi x}{L}[/latex] i is odd

At mid-span where x = L/2 and using the first term only in this expression

[latex]v_{ ext {mid }- ext { span }}=\frac{w_{0} L^{4}}{94.4 E I}[/latex]

Question 15.P.21: A uniform simply supported beam, span L, carries a distribut...

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