Question 11.34: The beam is made from a plate that has a constant thickness ...

The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress \sigma_{\text {allow }} throughout its length.

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Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a.
Moment Function: The distributed load as a function of x is

\frac{w}{x}=\frac{w_{0}}{L / 2} \quad w=\frac{2 w_{0}}{L} x

The free-body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free-body diagram,

\curvearrowleft +\Sigma M_{o}=0 ; \quad M+\frac{1}{2}\left[\frac{2 w_{0}}{L} x\right]{x}\left(\frac{x}{3}\right)-\frac{1}{4} w_{0} L x=0

 

M=\frac{w_{0}}{12 L}\left(3 L^{2} x-4 x^{3}\right)

Section Properties: At position x, the height of the beam’s cross section is h. Thus

I=\frac{1}{12} b h^{3}

Then

S=\frac{I}{c}=\frac{\frac{1}{12} b h^{3}}{h / 2}=\frac{1}{6} b h^{2}

Bending Stress: The maximum bending stress \sigma_{\max } as a function of x can be obtained by applying the flexure formula.

\sigma_{\max }=\frac{M}{S}=\frac{\frac{w_{0}}{12 L}\left(3 L^{2} x-4 x^{3}\right)}{\frac{1}{6} b h^{2}}=\frac{w_{0}}{2 b h^{2} L}\left(3 L^{2} x-4 x^{3}\right)\quad(1)

At x=\frac{L}{2}, h=h_{0}. From Eq. (1),

\sigma_{\max }=\frac{w_{0} L^{2}}{2 b h_{0}^{2}}\quad(2)

Equating Eqs. (1) and (2),

\frac{w_{0}}{2 b h^{2} L}\left(3 L^{2} x-4 x^{3}\right)=\frac{w_{0} L^{2}}{2 b h_{0}^{2}}

 

h=\frac{h_{0}}{L^{3 / 2}}\left(3 L^{2} x-4 x^{3}\right)^{1 / 2}
2

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