Proving the advantage of a composite beam

Question

The textbook that I'm following for studying bending stresses in beams, states that - "Engineers purposely design beams in this manner (composite manner) in order to develop a more efficient means for supporting loads"

I do get an intuition that for same cross sectional area and shape a composite beam will be able to take more bending moment than a homogeneous beam, but I was trying to prove it mathematically. Consider for instance, a wooden beam with some given cross sectional area and length-

Let's say the maximum bending moment this beam can carry is $M_1$

If I were to attach two steel plates at the top and bottom by keeping the height and width the same

then let $M_2$ be the maximum bending moment that this composite beam can carry

How can I prove $M_2 > M_1$?

This question is followed by a question that I asked earlier - Determining the ratio of bending moments in parts of a composite beam

Answer 1

I think the easiest way is to argue the following. Assume:

• a beam of uniform cross-section
• with breath $b$
• two layers 1 and 2

The fraction of the contribution of each layer is:

$$\frac{M_1}{M_2} = \frac{I_1 \cdot E_1}{I_2\cdot E_2}$$

then the fraction of $M_1$ to the total of this beam with two different materials would be equal to:

$$\frac{M_1}{M_t} = \frac{I_1 \cdot E_1}{I_1 \cdot E_1 + I_2\cdot E_2}$$

or :

$$M_t = \frac{I_1 \cdot E_1 + I_2\cdot E_2}{I_1 \cdot E_1}\cdot M_1$$

Now consider two cases:

• A) both sections are from the same material 1 and $$M_{T,A} = \frac{I_1 \cdot E_1 + I_2\cdot \color{red}{E_1}}{I_1 \cdot E_1}\cdot M_1$$

(NOTICE above that $I_1\neq I_2$ but $E_1 = E_2$)

• B) sections are from the different material 1 and material 2 $$M_{T,B} = \frac{I_1 \cdot E_1 + I_2\cdot E_2}{I_1 \cdot E_1}\cdot M_1$$

Now, by taking the ratio of $M_{T,A}$ over $M_{T,B}$:

$$\frac{M_{T,A}}{M_{t,B}} = \frac{\frac{I_1 \cdot E_1 + I_2\cdot \color{red}{E_1}}{I_1 \cdot E_1}\cdot M_1 } {\frac{I_1 \cdot E_1 + I_2\cdot E_2}{I_1 \cdot E_1}\cdot M_1}$$

Simplyfiying:

$$\frac{M_{t,A}}{M_{t,B}} = \frac{I_1 \cdot E_1 + I_2\cdot \color{red}{E_1}} {I_1 \cdot E_1 + I_2\cdot \color{blue}{E_2}}$$

From here you can see that the value of $E_2$ dominates which section carries more moment. If $E_1>E_2$ then $M_{t,A}>M_{t,B}$, and vice versa.

Answer 2

I find it a bit easier to explore this in terms of stiffness.

The strength of a cross-section to bending is

$$M_R = \dfrac{f_yI}{c} = \dfrac{\epsilon_yEI}{c}$$

where $f_y$ is the yield stress, which is itself equal to $\epsilon_yE$ (yield strain times elastic modulus); $I$ is the moment of inertia; and $c$ is the largest distance to the centroid (half the height, in your case).

Now, we can further simplify this by defining stiffness as $K = EI$.

$$M_R = \dfrac{K\epsilon_y}{c}$$

So, the bending moment for a section is linearly proportional to its stiffness $K$, its yield strain $\epsilon_y$, and inversely proportional to $c$. (Obviously, $c$ and $K$ are related since both are a function of the section height)

Now let's focus our attention on $K$. It is a simple product of two independent variables: $E$, which describes the material, and $I$, which describes the geometry. However, $K$ doesn't really differentiate between them: if we have two beams A and B, where A is made of a material that's twice as hard as B, but B's inertia is twice that of A, then they will have the exact same bending strength.

This means we can play around with the actual values of $E$ and $I$ that we use in our beam if we want. Specifically, we can use this to transform a composite beam into an equivalent uniform one.

In your case, you've replaced some wooden parts of the beam with steel. You can then change that steel back into "wood" as long as you modify the moment of inertia of those parts accordingly. This is done by "widening" those segments by $\dfrac{E_s}{E_w}$. Equivalently, you could pretend the whole beam is steel and then "thin" the wooden segment by $\dfrac{E_w}{E_s}$.

That is, a rectangular beam made of wood and steel plates above and below is identical to a wooden I-shape beam, as below (forgive the sloppiest mspaint job of all time):