The Maximum induced bending stress within the beam.

Given that:


The beam depth is 300mm, the width is 150mm. The beam is an I section beam with a 5mm wall thickness.

enter image description here

My calculations:

$$\begin{align} I_{total} &= ( I_1 + A_1 \cdot d_1^2) + ( I_2 + A_2 \cdot d_2^2) + ( I_3 + A_3 \cdot d_3^2) \\ I_3 = I_1 &= \frac{1}{12}bh^3 = \frac{1}{12} \cdot 150 \cdot 5^3 = 1.562 \cdot10^3 \\ I_2 &= \frac{1}{12}bh^3 = 150 \cdot 290^3 = 1.016 \cdot 10^7 \\ A_3 &= A_1 = bh = 150 \cdot 5 = 750\text{ mm}^2 \\ A2 &= bh = 5 \cdot 290 = 1,450\text{ mm}^2 \\ I_{total} &= (1.562 \cdot 10^3 + 750^2 \cdot 150) \\ &+ (1.016 \cdot 10^7 + 1,450^2 \cdot 0) \\ &+ (1.562\cdot10^3 + 750^2 \cdot 150) \\ &= 42.797.499\text{ mm}^4 \\ &= 4.28\cdot10^{7}\text{ mm}^4 \\ \sigma_{bend,max} &= \frac{Mc}{I} \\ M &= 48\text{ kNm} = 48\cdot 10^6\text{ Nm} \\ c &= 150\text{ mm} = 0.150\text{ m} \\ I &= 4.28\cdot 10^{7}\text{ mm}^4 = 4.28 \cdot 10^4\text{ m}^4 \\ \therefore \sigma_{bend,max} &= \frac{48,000 \cdot 0.1475}{4.28 \cdot 10^4} \\ &= 168.23\text{ MPa} \end{align}$$

The maximum induced bending stress within the beam is 168.23;MPa

These are my calculations. Are these correct?

I am not sure on the 147.5. As seen different ways to calculate.

Also not sure if i have done the calculations correct.

  • 1
    $\begingroup$ Check my work questions are technically off-topic. Consider improving your question by adding a sketch, explaining why you did what you did, and where you're uncertain. At a glance, there are issues with the I_total number crunching, with the conversion of I_total from mm^4 to m^4, and with correctly selecting "c" to be for the extreme fiber. $\endgroup$
    – CableStay
    May 29, 2020 at 16:24
  • $\begingroup$ I added a sketch, and explained where i think i have gone wrong. Where number crunching, and conversion from mm to m have i done wrong? Also correctly selecting C? $\endgroup$ May 29, 2020 at 19:02
  • $\begingroup$ Later I'll see everyting but if I'm not wrong c should be the point where you want to analize the stress. Usualy would be c = 150mm (the outermost fiber on the bending axis) $\endgroup$
    – Leafk
    May 29, 2020 at 19:49

2 Answers 2


Two concepts are in play here:

  1. Calculation of moment of inertia for a composite section.
  2. Calculation of elastic beam stresses.

First, the calculation of $I_{total}$ for the wide flange section.

Your equation for $I_{total}$ via the parallel axis theorem is correct, but the execution went awry.

$$I_{total} = \sum (I + A \cdot d^2)$$

Since the section is symmetric, the centroid is at mid-height and the $Ad^2$ term will be zero for the web component. Thus:

$$I_{total} = 2*(I_{flange} + A_{flange} \cdot {d_{flange}}^2) + I_{web}$$

Since we're calculating moment of inertia about a horizontal line, let b = the width and h = the height of each component. Let d = the vertical distance from the composite centroid to the centroid of the component. Note that this is a totally distinct choice and concept from the distance used to calculate maximum stress.

$$I_{flange} = \frac{1}{12} bh^3 = 1,562 \ mm^4$$ $$A_{flange} = bh = 750 \ mm^2$$ $$d_{flange} = 147.5 \ mm$$ $$I_{web} = \frac{1}{12} bh^3 = 1.016 \cdot 10^7 \ mm^4$$

$$I_{total} = 4.2797 \cdot 10^7 mm^4$$

In your numerical calculation of $I_{total}$ it looks as though you squared the A terms instead of the d terms. It also looks as though you measured d to the outer edge of the flange instead of to the centroid. Note also that to convert from $mm^4$ to $m^4$, divide by $1000^4$.

An easy way to check your results is to use one of the many online moment of inertia calculators. You can find ones for an array of common cross sections, including wide flange sections.

Now the calculation of maximum bending stress

When we're assuming that plane sections remain plane and that the section remains elastic (an Euler-Bernoulli beam), the equation for pure bending stress is as you noted.

$$\sigma = \frac{My}{I}$$

The derivation of this equation should be available in any introductory mechanics of materials textbook. Wikipedia also contains a discussion of Euler-Bernoulli bending theory.

In this equation, $y$ is the distance from the neutral axis (our centroid). At the neutral axis, bending stress is zero. The maximum bending stress must therefore occur at the cross section height that is farthest from the neutral axis - at the extreme fiber. For the beam you sketched, the extreme fiber is located at the outer edge of the flange, $150 \ mm$ from the centroid. Because we are so often interested in the maximum bending stress, this particular distance of $y$ is assigned its own variable name - typically, $c$.

In design of steel beams, we often talk about the "elastic section modulus" which simply lumps together the equations you've looked at here, giving engineers a quick equation to calculate the maximum bending stress in the elastic section.

$$Elastic \ Section \ Modulus, \ S = \frac{I}{c}$$

This gives us a quick way to calculate the yield Moment, $M_y$.

$$M_y = \sigma_y \cdot S$$

Where $\sigma_y$ is the yield stress of the material.

Some notes on real-world steel beams

As Leafk noted, wide flange beams are usually designed to optimize the use of material, which typically results in flanges being thicker than webs. And while it's beyond the scope of this question it's worth noting that there are several other beam design considerations beyond pure bending stress.

  • $\begingroup$ +1, I'd also mention how the inertia can be most easily calculated as the inertia of the bounding rectangle minus the "void rectangle". $\endgroup$
    – Wasabi
    Jun 3, 2020 at 16:23

You are on the right track but your equation is wrong on some places and there's some misinterpretation.
First error your drawing doesn't show that the midle piece thickness is 5mm (for this type of beam usualy the web and flange thickness is different).

Just note some numbers aren't fully in cientific notation. I've done this to stick with your precision level and to facilitate readability
Sorry, I use dot to separate thousands and comma for decimal (the oposite way), and I'll stick with my method.

$I_1$ = $I_3$ = $1.562mm^4 \enspace {\color{green} {(ok)}}$
$I_2$ = $(b*h³)/12 = (5*290³)/12 = 1.016*10^4mm^4 \enspace {\color{green} {(ok)}}$
$A_1$ = $A_3$ = $750mm² \enspace {\color{green} {(ok)}}$
$A_2$ = $1.450mm² \enspace {\color{green} {(ok)}}$
$I_{total}$ = $(1.562+750⋅147.5²)+(1.016⋅10^4+1450⋅0²)+(1.562+750⋅147.5²) \enspace {\color{red} {(wrong)}}$
$I_{total}$ = $42.797.499mm^4 = 4,28*10^7mm^4$ (here you got $mm^4$)

Usualy I stick with the same unit system when I can't find errors. So, converting new units.
$M$ = $48kN*m$ = $48*10^6N*mm \enspace {\color{blue} {(edit)}}$ Good thing with conversion is that errors are found more easily.
$c$ = $150mm$
$\sigma_{bend,máx}$ = $48*10^6*150/4,28*10^7$
$\sigma_{bend,máx}$ = $168,23 MPa$ on the external fibers in the bending axis

My analysis is only for what your equation shows.
In a real case you should note that you can find other points of failure other than the bending stress too.

Edit.: Corrected Moment value thanks to ingenørd tip.

  • 1
    $\begingroup$ That's not quite right, 48kNm = 48*10⁶ Nmm. $\endgroup$
    – ingenørd
    May 30, 2020 at 6:23
  • $\begingroup$ @ingenørd are you sure. I've stoped on this some time when doing the answer too. If i work with 1000mm arm the force will be 48N. And with 1m arm will be 48kN (ok I'm wrong here). But the correct should be 48*10^3 Nmm right? $\endgroup$
    – Leafk
    May 30, 2020 at 14:39
  • $\begingroup$ The answer should be $*10^6$ (when editing I read more carefully) I just can't find where I got wrong in my reasoning. $\endgroup$
    – Leafk
    May 30, 2020 at 14:48
  • $\begingroup$ Thank you for helping. Can i ask why C is 150? I thought it should be 147.5? $\endgroup$ Jun 1, 2020 at 9:54
  • $\begingroup$ Kyle C should be the point of maximum stress you're checking. With this geometry isn't too aparent. But think how the calculation would work if you're analising without the flanges (only the retangular beam) $\endgroup$
    – Leafk
    Jun 1, 2020 at 10:24

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