Chapter 7

Q. 7.7

For a thin elastic beam with rectangular cross section (b × h) and loading that causes bending moment M(x), derive Equation 7.22, the expression for the normal stress σxx(x, z) in a beam, starting from the fact that, as we have shown, the normal stress varies linearly through the thickness.

Given: Rectangular beam in bending.
Find: Normal stress starting from an assumption of a linear distribution.
Assume: Hooke’s law applies; no axial loading on the beam.

Equation 7.22 ( \sigma_{x x}(x, z)=\frac{M_y(x)}{I_y} z. )


Verified Solution

We begin with a general statement of the fact that the normal stress varies linearly through the thickness:

\sigma_{x x}(x, z)=f(x) \cdot z ,

where f (x) is an (as yet) unknown function of x. Then we can integrate this stress over the area of the cross section to find an expression for the axial resultant on the beam, which is zero:

\int_{-h / 2}^{h / 2} \sigma_{x x}(x, z) b \mathrm{~d} z=f(x) \cdot b \int_{-h / 2}^{h / 2} z \mathrm{~d} z=0 .

Now, let us take the moment (about the y-axis) of the axial normal stress acting on a thin strip of cross-sectional area b dz at height z:

d M_y=\left(\sigma_{x x} b \mathrm{~d} z\right) z

And if we sum this moment for all strips through the thickness, we obtain a positive internal moment, M(x):

M(x)=\int_{-h / 2}^{h / 2} \mathrm{~d} M_y=\int_{-h / 2}^{h / 2} z\left(\sigma_{x x} b \mathrm{~d} z\right)=f(x) \int_{-h / 2}^{h / 2} z^2 b \mathrm{~d} z=f(x) I .

We recognize the integral as I, the second moment of area for the cross section about the centroid. Solving for f (x), we obtain

f(x)=\frac{M(x)}{I} .

And we can now write the equation for the normal stress as we have in Equation 7.22:

\sigma_{x x}(x, z)=f(x) \cdot z=\frac{M(x) z}{I} .