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. )

Step-by-Step

Learn more on how do we answer questions.

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) zAnd 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} .

Question: 7.9

Although we have learned about bars subject to axi...

Question: 7.8

The first of the equilibrium equations represents ...

Question: 7.6

First we need to consult our FBD and find the reac...

Question: 7.5

We obtained a formula for shear stress at a given ...

Question: 7.4

The gage at A, in the upper portion of the cross s...

Question: 7.3

The symmetry of the cross section shown suggests t...

Question: 7.2

In each case, we will first find the external reac...

Question: 7.1

Our strategy is to find the reactions at the suppo...