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Question 7.8: Having confirmed the normal stress equation in Example 7.7, ......

Having confirmed the normal stress equation in Example 7.7, derive the shear stress distribution in a rectangular beam (b × h) subject to loading that causes bending moment M(x) and shear force V(x), starting from the plane stress elasticity equations of equilibrium from Chapter 4:

\begin{aligned} & \frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \sigma_{x z}}{\partial z}=0, \\ & \frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{z z}}{\partial z}=0. \end{aligned}

Given: Rectangular beam in bending.
Find: Normal stress starting from plane stress elasticity equations.
Assume: Hooke’s law applies; no axial loading on the beam.

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The first of the equilibrium equations represents equilibrium in the x-direction. We can rearrange the equation and then substitute the known expression for σ_{xx} and taking the partial derivative with respect to x, recognizing that neither z nor I is a function of x:

\frac{\partial \sigma_{x z}}{\partial z}=-\frac{\partial \sigma_{x x}}{\partial x}=-\frac{\partial(M(x) z / I)}{\partial x}=-\frac{z}{I} \frac{\mathrm{d} M}{\mathrm{~d} x} .

We can now integrate through the beam thickness, from the z location of interest to the top of the beam at h/2, to obtain

\begin{aligned} \int_z^{h / 2} \frac{\partial \sigma_{x z}(x, z)}{\partial z} \mathrm{~d} z & =\int_z^{h / 2}-\frac{z}{I} \frac{\mathrm{d} M(x)}{\mathrm{d} x} \mathrm{~d} z=-\frac{1}{I} \frac{\mathrm{d} M(x)}{\mathrm{d} x} \int_z^{h / 2} z \mathrm{~d} z, \\ \sigma_{x z}\left(x, \frac{h}{2}\right)-\sigma_{x z}(x, z) & =-\frac{1}{2 I} \frac{\mathrm{d} M(x)}{\mathrm{d} x}\left(\frac{h^2}{4}-z^2\right), \\ \sigma_{x, z}(x, z) & =\frac{1}{2 I} \frac{\mathrm{d} M(x)}{\mathrm{d} x}\left(\frac{h^2}{4}-z^2\right). \end{aligned}

We have used the fact that the shear stress must be zero on the top surface of the beam as there is no layer above it to provide a shear force to the surface. Now instead of just stating that dM(x)/dx = V(x), we can show this by writing V(x) as the integral of the shear stress over the cross section:

V(x)=\int_{-h / 2}^{h / 2} \sigma_{x z}(x, z) b \mathrm{~d} z .

Then, by substituting the previous result into this equation we find that:

V(x)=\frac{1}{2 I} \frac{\mathrm{d} M(x)}{\mathrm{d} x} \int_{-h / 2}^{h / 2}\left(\frac{h^2}{4}-z^2\right) b \mathrm{~d} z=\left.\frac{1}{2\left(b h^3 / 12\right)} \frac{\mathrm{d} M(x)}{\mathrm{d} x} b\left(\frac{h^2}{4} z-\frac{z^3}{3}\right)\right|_{-h / 2} ^{h / 2}=\frac{\mathrm{d} M(x)}{\mathrm{d} x} .

We can now write an explicit expression that shows how a beam carries a resultant shear force V(x) with a spatially varying stress distribution on a section:

\sigma_{x z}(x, z)=\frac{1}{2 I} V(x)\left(\frac{h^2}{4}-z^2\right) .

Now that we have this expression, we can use it to determine maximum value of shear stress on the section. We can see that we have a parabolic distribution of shear stress, with a maximum at z = 0, the neutral axis, and at that level

\sigma_{x z, \max }(x)=\frac{3 V(x)}{2 b h}=1.5 \frac{V(x)}{A} .

The maximum shear stress is 1.5 times the average. Remember that these results are only for a rectangular cross section.

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