# Question 7.7: For a thin elastic beam with rectangular cross section (b × ......

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.

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

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.

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:

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

Question: 7.9

## An axial load is applied to a solid circular bar that contains an offset in order to fit in a tight space in a machine. Compute the maximum tensile and compressive normal stresses at section a. Given: Magnitude of axial force and bent bar geometry. Find: Maximum normal stresses at specified ...

Although we have learned about bars subject to axi...
Question: 7.8

## 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: ...

The first of the equilibrium equations represents ...
Question: 7.6

## The beam shown is subjected to a distributed load. For the cross section at x = 0.6 m, determine the average shear stress (a) at the neutral axis and (b) at z = 0.02 m. Given: Dimensions of and loading on simply supported beam. Find: Shear stress at two locations along height of cross section at x ...

First we need to consult our FBD and find the reac...
Question: 7.5

## An I-beam is made by gluing five wood planks together, as shown. At a given axial position, the beam is subjected to a shear force V = 6000 lb. (a) What is the average shear stress at the neutral axis z = 0? (b) What are the magnitudes of the average shear stresses acting on each glued joint? ...

We obtained a formula for shear stress at a given ...
Question: 7.4

## A steel T-beam is used in an inverted position to span 400 mm. If, due to the application of the three forces as shown in the figure, the longitudinal strain gage at A (3 mm down from top of beam and at the x location shown) registers a compressive strain of −50 × 10^−5, how large are the applied ...

The gage at A, in the upper portion of the cross s...
Question: 7.3

## Find the centroid and the second moment of area about the horizontal (y) axis of the cross section shown. All dimensions are in millimeters. If a beam is constructed with the cross section shown from steel whose maximum allowable tensile stress is 400 MPa, what is the maximum bending moment that ...

The symmetry of the cross section shown suggests t...
Question: 7.2