\\this is the qeustion i am facing

First, I thought the neutral axis would just be the y bar which can be obtained from the centroid of the cross-section but is different. I have tried letting the net moment at the wall = 0 and the reaction force I got is 1kn * 1m. But I am kinda lost on how to continue.


2 Answers 2


The stresses from bending and axial should be added up by superposition. This moves the location of zero stress (neutral axis). I have included a figure to demonstrate the principle, and provided the calculations again, please note the axial stress is 40 kN/mm^2

$$ \sigma_{bending}= m/S =\frac{1kN*1m}{(50\times 100^2)/6} =6000N/500mm^2 = 12N/mm^2$$

$$ \sigma_{axial}= 200kN/(50*100)mm^2 = 40Nmm^2 $$

$$ \sigma_{max} = \sigma_{bending} + \sigma_{axial} = 12N/mm^2 + 40N/mm^2 = 52 N/mm^2$$ $$ \sigma_{min} = -\sigma_{bending} + \sigma_{axial} = -12N/mm^2 + 40N/mm^2 = 28 N/mm^2$$

We can see from the scaled image, as well as the math, that the axial stress overcomes the compression stress due to bending. There is no location on the cross section where the stress is zero. Therefore the neutral axis is below the beam. We can determine this distance by similar triangles (or equation of a line, or whatever you prefer).

$$\frac{52-28}{100} =\frac{52}{D_{NA}+D} -> (D_{NA} + D) = \frac{52*100}{52-28} = 216.67mm $$

So the total depth from the bottom surface of the beam to the neutral axis is the difference of the depth and the calculated dimension. $D_{NA} = (216.67 - 100)mm = 116.67mm$ scaled view showing the stress diagram

  • 1
    $\begingroup$ If you prefer a more visual approach, you can calculate the angle at the top right corner of the diagram here by looking at the "bending" triangle, $\tan(\theta)=\frac{50}{12}$, then, knowing that the combined stress is $52\text{N/mm}^2$, you can calculate the height from the top surface as $(D+D_{NA})=52\tan(\theta)=52\times\frac{50}{12}=\frac{650}{3}\text{mm}$ $\endgroup$ Apr 9, 2019 at 19:39

The stress is the superposition of two stresses: tensile due to bending moment and tensile due to 200kN tension. $$ \sigma_{bending}= m/S =\frac{1kN.1m}{(50\times 100^2)/6} =6kN.1000mm/500000mm^2 =6000N/500mm^2 = 12N/mm^2$$

$$ \sigma_{tension}= 200kMN/100^2mm^2 = 20Nmm^2 $$

So maximum tension is 20 +12= 32Nmm^2 on the upper face of the beam at support. And neutral axis can be calculated by finding the slope of stress surface vertically and locating where it turns to zero.

Slope of stress surface from the bottom face to top face is

$ [(20+12) -(20-12)]/100_{mm height}= 0.24Nm\space$

and Neutral axis is at the height from the bottom face of $$\frac{8_N}{0.24} =33.33mm \space or\space\frac{H}{3} $$

  • $\begingroup$ Downvoted because numerous mathematical/typographical errors, such that the final answer is incorrect. The tensile force is 200x larger than the bending force so it's expected that the Neutral axis is further than H/3 away... Don't have time to correct myself now, will happily convert to upvote when fixed. What's a kilo-mega-newton? $\endgroup$ Apr 9, 2019 at 19:25
  • $\begingroup$ @Jonathan R Swift, my apologies for my typos and numerical errors. The answer by ShadowMan is correct, so no point in doing double work. The idea was correct, but I made the errors because I used my mobile app with haste. We all seek the same think answering a question: correct theory and clean answer. $\endgroup$
    – kamran
    Apr 10, 2019 at 0:54

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