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While trying to solve this question and using the flexure formulas of bending for composite beams I obtained a lower value of moment for steel than for aluminum.How is that possible?

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  • $\begingroup$ Showing your calculations would be helpful $\endgroup$ – Fred May 8 at 16:46
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The difference in stiffness is important.

First let's consider the aluminum W8x40 section without any steel plates: It can be loaded until the aluminum reaches a stress of 12 ksi.

Then let's consider what happens if we bolt some very thin steel plates unto the flanges of the aluminum section. Then we can increase the load until the steel plates reach a stress of 18 ksi. As the steel is 3 times stiffer than aluminum, the aluminum will have a stress that is 3 times lower than the steel plates. (I'm assuming we can disregard the slip in the bolts as well as a linear-elastic stress distribution — a plastic stress distribution would be a different calculation.) That results in a maximum stress in aluminum of 6ksi, so we get a total capacity of only half of what we had before we added the steel plates.

If the steel plates aren't extremely thin, they may increase the capacity, but the point is that the steel plates may fail well before we reach the capacity of the aluminum.

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  • $\begingroup$ This is a good point, and one I hadn't thought of before. I actually spent a few minutes going "adding stronger material makes the section weaker? Wut?!" But yeah, it can. In this particular case, though, the plates do strengthen the section. $\endgroup$ – Wasabi May 12 at 1:26
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Let's approximately calculate the S of composite beam, knowing that aluminum section I is 146in^4.

$ I_{COMP}= I_{alum}+ 2*A_{steel} *4.25^2 = 146+(8*0.25*2)4.25^2=146+72.25=218.25in^4 $

The S of aluminum section is 35.5in^3 and the S of steel 72.25/4.5= 16.05in^3

At yield point the steel plates can take a moment of

$ M_s= \sigma*S=18*16.05=288.9 k.in $

But the aluminum beam has not fully loaded at this stage, it has strained the same as steel but stressed only 1/3 of 18ksi so it can take only 18/3=6ksi

$ M_a= 6*35.5=213k.in $

Now we compare the difference adding the steel plates made with the original aluminum section.

$ M_{alumalone}= 12*35.5=426k.in \\ which\ is\ less\ than \ 288.9+213=501.9$

Edit

After a comment by @Wassabi I corrected the stress of Aluminum and recalculated the beam. The concept is the same but the numerical answer is changed.

So we gain roughly 20% of additional strength. But from here on out the advantage of adding steel plates will increase rapidly by adding to the thickness of the plates.

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  • $\begingroup$ -1. This doesn't take into consideration the effect of the difference in elastic moduli between the steel plates and aluminum beam. $\endgroup$ – Wasabi May 12 at 1:23
  • $\begingroup$ @Wasabi, It does. The question says assume Es= 3*Ea $\endgroup$ – kamran May 12 at 1:26
  • $\begingroup$ But I don't see where you take that effect into consideration. $\endgroup$ – Wasabi May 12 at 1:28
  • $\begingroup$ @Wasabi, when I dived the 12k of aluminum by 3 to get 4 k stress. $\endgroup$ – kamran May 12 at 1:30
  • $\begingroup$ Ah, but that's not how it works, I believe. You need elongation continuity at the interface. So you actually need to calculate the steel's stress at the interface (around 17ksi) and divide that by 3. So the aluminum is actually at around 5.7ksi. $\endgroup$ – Wasabi May 12 at 1:40
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The big problem here is that the beam and plates are of different materials. More specifically, that they have different moduli of elasticity.

We are accustomed to seeing the following bending stress profile:

enter image description here

This obviously doesn't work for composite sections, though. Since the different parts are meant to work monolithically, the different materials must have the same elongation at the interface. However, since the materials have different elastic moduli (and that $\sigma = E\epsilon$), equal elongations mean different stresses at the interface. So we know there'll be a stress discontinuity at the interface.

From the standard bending elongation equation (trivially obtained from the more common bending stress equation)

$$\epsilon = \dfrac{My}{EI}$$

we also know that the slopes of the composite-beam elongation and stress diagrams will be different in each material since they have different elastic moduli.

However, we can't yet determine what the slopes will actually be. It may seem clear that they'll just be $\dfrac{M}{EI}$ and $\dfrac{M}{I}$, but that won't work: using those equations with each part's $EI$ will find the results if it were working alone (well, if the beam were alone or if the plates were isolated from the beam). Instead, we'd need to find how much moment $M$ each part will absorb.

Personally, I find that a hassle, though. Why not replace the steel plates with equivalent aluminum plates, finding the new "all-aluminum beam"'s moment of inertia and do all the math assuming a single elastic modulus? That's way easier.1

So let's do that:

From steel shape tables, we know that the moment of inertia of a W8x40 beam is equal to 146 in4. We now need to calculate the change in moment of inertia due to the steel plates.

However, since the beam and plates have different materials, we also need to take their different elastic moduli into account.

This can be done by considering the fact that bending moment doesn't actually care about moment of inertia $(I)$, nor about elastic modulus $(E)$. Bending moment cares about stiffness ($EI$). If two beams have the same $EI$, they'll behave the same under bending (assuming elastic behavior), even if one is tiny but made of hard material and the other is large but made of plastic. Same $EI$ means same behavior.

We can use this knowledge to our advantage in this case and transform the steel plates into "equivalent aluminum" ones. We do this by setting the steel plates' elastic modulus as equal to aluminum's and then adjusting the plates' moment of inertia so that their final $EI$ is unchanged.

The question then becomes how to change the plate's geometry (and therefore their moment of inertia) to satisfy that condition. Thankfully, we know that moment of inertia is directly proportional to width and to the cube of height. Obviously, the easiest dimension to change is therefore the width: if you double the width, you double the moment of inertia; changing the height would require taking cubed-roots, and ain't nobody got time for that.2

Now, the question tells us to assume that

$$\dfrac{E_{steel}}{E_{alum}} = 3$$

Which means that the steel plates are three times stiffer than similarly-sized aluminum plates. So, for aluminum plates to be as stiff as the steel ones, they'd have to be three times wider. After all

$$\begin{align} (EI)_{plate} &= E_{steel}\cdot I_{plate\ on\ beam} \\ &= E_{steel}\cdot (I_{plate} + A_{plate}d^2) \\ &= 3E_{alum}\cdot \left(\dfrac{bh^3}{12} + bhd^2\right) \\ &= E_{alum} \cdot 3\left(\dfrac{bh^3}{12} + bhd^2\right) \\ &= E_{alum} \cdot \left(\dfrac{(3b)h^3}{12} + (3b)hd^2\right) \end{align}$$

So if we replace our 8x0.25" steel plates with 24x0.25" aluminum ones, the beam's behavior will be identical (we'll get to the difference in allowable bending stress further ahead).

So now let's calculate this "equivalent beam"'s moment of inertia. Using the parallel axis theorem and being thankful that the symmetric reinforcement won't change the neutral axis, we know it'll be

$$\begin{align} I_{equiv\ beam} &= I_{beam} + 2\left(I_{equiv\ plate} + A_{equiv\ plate}\cdot d^2\right) \\ &= I_{beam} + 2\left(\dfrac{(3b)h^3}{12} + (3b)h\cdot d^2\right) \\ &= 146 + 2\left(\dfrac{24\cdot0.25^3}{12} + 24\cdot0.25\cdot\left(\dfrac{8.25 + 0.25}{2}\right)^2\right) \\ &= 362.8\text{ in}^4 \end{align}$$

Now we can use the aluminum's elastic modulus with this equivalent moment of inertia to get the bending stress throughout the beam with the standard equation:

$$\sigma = \dfrac{My}{I_{equiv}}$$

However, there's a hitch: the stress on the plates is actually three times what we calculate with that equation. After all, we're simulating the plates with an area three times greater than it actually is. The actual area of steel will need to absorb that same internal force but with a smaller area and therefore greater stress.

Using this moment of inertia, the allowable stress in the plates is therefore just $f_{y,steel}/3 = 18/3 = 6\text{ ksi}$.

The allowable bending moment of the reinforced beam is therefore:

$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{6\cdot 362.8}{8.25/2+0.25} = 497.6\text{ k.in}$$

The original, unreinforced beam, however, could resist

$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{12\cdot 146}{8.25/2} = 424.7\text{ k.in}$$

The reinforcement therefore increased the beam's bearing capacity by ~17%.


1 All I wrote might make it seem complicated, but this method really is simple. It's basically: (1) multiply the plate's width by $E_{plate}/E_{beam}$; (2) calculate the equivalent beam's moment of inertia; (3) divide the plate's allowable bending stress by $E_{plate}/E_{beam}$; (4) calculate the allowable bending moment. It's effectively the same as the traditional method under the hood, but this one just seems more intuitive for me.

2 Changing the height would also complicate the stress calculation later on, since it depends on the distance to the fiber. Changing the width doesn't affect this at all.

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