A simply supported beam AB of span L and uniform section carries a distributed load of intensity varying from zero at A to w0/unit length at B according to the law w = 2wox/L (1-x/2L) per unit length. If the deflected shape of the beam is given approximately by the expression v = a1 sin πx/L +a2

Question

A simply supported beam AB of span L and uniform section carries a distributed load of intensity varying from zero at A to [latex]w_{ o } / ext { unit }[/latex] length at B according to the law

[latex]w=\frac{2 w_{0} x}{L}\left(1-\frac{x}{2 L} ight)[/latex]

per unit length. If the deflected shape of the beam is given approximately by the expression

[latex]v=a_{1} \sin \frac{\pi x}{L}+a_{2} \sin \frac{2 \pi x}{L}[/latex]

evaluate the coefficients [latex]a_{1}[/latex] and [latex]a_{2}[/latex] and find the deflection of the beam at mid-span.
Ans. [latex]a_{1}=2 w_{ o } L^{4}\left(\pi^{2}+4 ight) / E I \pi^{7}, a_{2}=-w_{ o } L^{4} / 16 E I \pi^{5}, 0.00918 w_{ o } L^{4} / E I[/latex].

Accepted Answer

The total complementary energy of the system is given by (see Eq. (15.39))

[latex]\delta\left(\sum_{j=1}^{n} \int_{0}^{F_{j}} \delta_{j} d F_{j}-\sum_{k=1}^{r} \Delta_{k} P_{k} ight)=0[/latex] (15.39)

[latex]C=\int_{L} \int_{0}^{M} d heta d M-\int_{L} w v d x[/latex] (i)

From symmetrical bending theory (see Fig. 9.5 and Eq. (9.7))

[latex]M=\frac{E I}{R}[/latex] (9.7)

[latex]\delta heta=\frac{\delta x}{R}=\frac{M}{E I} \delta x[/latex]

Eq. (i) then becomes

[latex]C=\int_{L} \int_{0}^{M} \frac{M}{E I} d x d M-\int_{L} w v d x[/latex]

from which

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

The expression on the right hand side of Eq. (ii) could have been obtained by considering the total potential energy of the system.
Substituting for M in Eq. (ii) from Eq. (13.3)

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

[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] (iii)

Now

[latex]v=a_{1} \sin \frac{\pi x}{L}+a_{2} \sin \frac{2 \pi x}{L}[/latex],

[latex]w=\frac{2 w_{0} x}{L}\left(1-\frac{x}{2 L} ight)[/latex]

so that

[latex]\frac{ d ^{2} v}{ d x^{2}}=-a_{1} \frac{\pi^{2}}{L^{2}} \sin \frac{\pi x}{L}-a_{2} \frac{4 \pi^{2}}{L^{2}} \sin \frac{2 \pi x}{L}[/latex]

Substituting in Eq. (iii) and expanding gives

[latex]C =\frac{E I \pi^{4}}{2 L^{4}} \int_{0}^{L}\left(a_{1}^{2} \sin ^{2} \frac{\pi x}{L}+8 a_{1} a_{2} \sin \frac{\pi x}{L} \sin \frac{2 \pi x}{L}+16 a_{2}^{2} \sin ^{2} \frac{2 \pi x}{L} ight) d x[/latex]

[latex]-\frac{2 w_{0}}{L} \int_{0}^{L}\left(a_{1} x \sin \frac{\pi x}{L}+a_{2} x \frac{\sin 2 \pi x}{L}-\frac{a_{1} x^{2}}{2 L} \sin \frac{\pi x}{L}-\frac{a_{2} x^{2}}{2 L} \sin \frac{2 \pi x}{L} d x ight)[/latex] (iv)

Eq. (iv) may be integrated by a combination of direct integration and integration by parts and gives

[latex]C=\frac{E I \pi^{4}}{2 L^{4}}\left(\frac{a_{1}^{2} L}{2}+8 a_{2}^{2} L ight)-a_{1} w_{0} L\left(\frac{1}{\pi}+\frac{4}{\pi^{3}} ight)+\frac{ a _{2} w_{0} L}{2 \pi}[/latex]

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

[latex]\frac{\partial C}{\partial a_{1}}=0[/latex] and [latex]\frac{\partial C}{\partial a_{2}}=0[/latex]

From the first of these

[latex]0=a_{1} \frac{E I \pi^{4}}{2 L^{3}}-\frac{w_{0} L}{\pi^{3}}\left(\pi^{2}+4 ight)[/latex]

so that

[latex]a_{1}=\frac{2 w_{0} L^{4}}{E I \pi^{7}}\left(\pi^{2}+4 ight)[/latex]

From the second

[latex]0=a_{2} \frac{8 E I \pi^{4}}{L^{3}}+\frac{w_{0} L}{2 \pi}[/latex]

which gives

[latex]a_{2}=-\frac{w_{0} L^{4}}{16 E I \pi^{5}}[/latex]

The deflected shape of the beam is then

[latex]v=\frac{w_{0} L^{4}}{E I}\left[\frac{2}{\pi^{7}}\left(\pi^{2}+4 ight) \sin \frac{\pi x}{L}-\frac{1}{16 \pi^{5}} \sin \frac{2 \pi x}{L} ight][/latex] (v)

At mid-span when x = L/2 the deflection is

[latex]v_{ ext {mid }- ext { span }}=0.00918 \frac{w_{0} L^{4}}{E I}[/latex]

Question 15.P.20: A simply supported beam AB of span L and uniform section car...

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