The simply supported tapered rectangular beam with constant width b supports the concentrated forces P. Determine the absolute maximum normal stress developed in the beam and specify its location.

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Accepted Answer

Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a.
Moment Function: Considering the moment equilibrium of the free-body diagrams of the beam's cut segment shown in Figs. b and c. For region AB,
[latex]\curvearrowleft +\Sigma M_{O}=0\quadM-P x=0 \quad M=P x[/latex]

For region BC,

[latex]\curvearrowleft +\Sigma M_{O}=0\quad M-P\left(\frac{L}{3} ight)=0 \quad M=\frac{P L}{3}[/latex]

Section Properties: Referring to the geometry shown in Fig. d,

[latex]\frac{h-h_{0}}{x}=\frac{h_{0}}{L / 2} ; \quad h=\frac{h_{0}}{L}(2 x+L)[/latex]

At position x, the height of the beam's cross section is h. Thus

[latex]I=\frac{1}{12} b h^{3}=\frac{1}{12} b\left[\frac{h_{0}}{L}(2 x+L) ight]^{3}=\frac{b h_{0}{ }^{3}}{12 L^{3}}(2 x+L)^{3}[/latex]

Then

[latex]S=\frac{I}{c}=\frac{\frac{b h_{0}^{3}}{12 L^{3}}(2 x+L)^{3}}{\frac{h_{0}}{2 L}(2 x+L)}=\frac{b h_{0}^{2}}{6 L^{2}}(2 x+L)^{2}[/latex]

Bending Stress: Since the moment in region BC is constant and the beam size at this region is larger than that of region AB, the maximum moment will not occur at this region. For region AB, the flexure formula gives

[latex]\sigma_{\max }=\frac{M}{S}=\frac{P x}{\frac{b h_{0}^{2}}{6 L^{2}}(2 x+L)^{2}}[/latex]

[latex]\sigma_{\max }=\frac{6 P L^{2}}{b h_{0}^{2}}\left[\frac{x}{(2 x+L)^{2}} ight]\quad(1)[/latex]

In order to have absolute maximum bending stress,

[latex]\frac{d \sigma_{\max }}{d x}=0[/latex].

[latex]\frac{d \sigma_{\max }}{d x}=\frac{6 P L^{2}}{b h_{0}^{2}}\left[\frac{(2 x+L)^{2}(1)-x(2)(2 x+L)(2)}{(2 x+L)^{4}} ight]=0[/latex]

[latex]\frac{6 P L^{2}}{b h_{0}^{2}}\left[\frac{(L-2 x)}{(2 x+L)^{3}} ight]=0[/latex]

Since [latex]\frac{6 P L^{2}}{b h_{0}^{2}} eq 0[/latex], then

[latex]L-2 x=0 \quad x=\frac{L}{2}[/latex]

Since [latex]x=\frac{L}{2}>\frac{L}{3}[/latex], the solution is not valid. Therefore, the absolute maximum bending stress must occur at

[latex]x=\frac{L}{3} ext { and, by symmetry, } x=\frac{2 L}{3}[/latex]

Substituting [latex]x=\frac{L}{3}[/latex] into Eq. (1).

[latex]\sigma_{abs max }=\frac{18 P L}{25 b h_{0}^{2}}[/latex]

Question 11.33: The simply supported tapered rectangular beam with constant ...

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