I'm struggling to identify which equation is used within the first red square.

Could someone please help identify which equation was used as well as why?




The calculation for $e$ that you've boxed is essentially arrived at through geometry. We simply want to find the distance from the centroid where total stress is zero. Drawing the axial, bending, and total stress along the cross-section makes this clear.

stress distribution

$$\begin{align} \sigma_{axial} &= \frac{P}{A} \\ \sigma_{flexure} &= \frac{My}{I} \\ \sigma_{total} &= \sigma_{axial} + \sigma_{flexure} \end{align}$$

So, we can readily calculate $e$ based on the slope of the flexural stress. I tend to phrase it in my head as "at what distance $e$ from the centroid does the flexural stress perfectly counteract the axial stress?"

Find $e$ such that

$$\sigma_{axial} = \frac{2\sigma_{flexure}}{d}e$$

Where $d$ is the depth of the section and $\sigma_{flexure}$ refers to the stress at the extreme fiber (distance $d/2$ from the centroid). Rearranging that expression, and substituting, we see the equation used in the given solution.

$$e = \frac{\sigma_{axial} \cdot \frac{d}{2}}{\sigma_{flexure}} = \frac{\sigma_{axial} \cdot I}{M}$$

It's worth emphasizing, that from the standpoint of developing understanding, memorizing their particular expression for $e$ is not all that useful. It's more important to understand what's happening - that we're simply adding axial and flexural stress, and finding the location where total stress is zero.

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  • $\begingroup$ Also, do you know why the force M is multiplied by 100? as shown in the solutions above. 72*1000*100 $\endgroup$ – Red Queen10101 Jun 21 '19 at 12:14
  • 2
    $\begingroup$ @RedQueen10101 - The moment produced by eccentric load P, is equal to P*eccentricity. The eccentricity is 100mm (distance from point of application to centroid of cross section). So, M = P*100 kN-mm. The 1000 multiplier is converting kN to N (Pascals have units N/m^2, so 1 MPa = 1 N/mm^2) $\endgroup$ – CableStay Jun 21 '19 at 12:28

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