# Determining the ratio of bending moments in parts of a composite beam

Consider a composite beam formed by sandwiching a wood beam between two steel plates, whose cross section is shown. Say the beam is loaded by some arbitrary loading, due to which at any cross section of this composite beam the bending moment is $$M$$. The bending moment $$M$$ at that section will be distributed between wood and steel. Let them be $$M_w$$ and $$M_s$$, such that

$$M= M_s + M_w$$

is there any way I can determine in what ratio will the bending moment be distributed between wood and steel, i.e.

$$\frac{M_s}{M_w}$$

Edit: I was actually trying to mathematically justify that a composite beam like the one in figure can carry more bending moment than a wooden beam of same cross section and area. I need to get your opinion on its correctness.

Let the allowable stresses in wood and steel be $$\sigma_{a,w}$$ and $$\sigma_{a,s}$$

The maximum bending moment that wood will be able to carry will be,

$$(M_{max})_w = \sigma_{a,w} \,Z_w$$

where $$Z_w$$ is the section modulus of the wood portion.

Similarly,

The maximum bending moment that steel will be able to carry will be,

$$(M_{max})_s = \sigma_{a,s} \,Z_s$$

where $$Z_s$$ is the section modulus of the steel portion.

At first I thought I can simply sum these values to obtain the maximum bending moment that the composite beam can carry, but after some thought concluded that can't be the case. So,

If the ratio of the bending moments in each portion at any cross section if the composite beam were to be loaded is,

$$\frac{M_s}{M_w} = \frac{E_s I_s}{E_w I_w}$$

then the maximum beding moment which the composite beam can carry will be,

$$M_{max,composite} = (M_{max,})_w + \frac{E_s I_s}{E_w I_w}(M_{max,})_w$$

$$M_{max,composite} = (M_{max,})_w (1+ \frac{E_s I_s}{E_w I_w})$$

$$M_{max,composite} = \sigma_{a,w} \,Z_w(1+ \frac{E_s I_s}{E_w I_w}) = M_2$$

if the entire beam were to made up of wood then the maximum bending moment that the beam would have carried would be,

$$M_{max,homogeneous} = \sigma_{a,w} \,Z = M_1$$

where $$Z$$ is the section modulus of the beam entirely made of wood (having the same cross sectional area and shape as that of composite beam)

Taking the ratio,

$$\frac{M_2}{M_1} = \frac{Z_w}{Z} (1+ \frac{E_s I_s}{E_w I_w})$$

Am I correct up to this point? If yes, how should I proceed further to show that $$M_2 > M_1$$

• In practice, I think you neglect the wood since it's meant to act as a shear web and keep the caps separated rather than carry any tensile or compressive load. This is especially true with something like carbon and foam where the stiffness of one is vastly higher than the other. Feb 14, 2022 at 23:40

TL;DR: In the elastic range it should be :

$$\frac{M_e}{M_i} =\frac{I_e E_e}{I_i E_i}$$

where:

• $$I$$ is the second moment of area
• $$E$$ is the youngs modulus
• index $$e$$ refers to the external material
• index $$i$$ refers to the internal material

During bending of a beam from two different materials the stress and the strain distribution is as in the following image. I.e.: the bending strain is a linear function with the depth, and then (in the elastic region) because $$\sigma= E\cdot\epsilon$$, you get this discontinuous distribution (in this case there are only two material, but there another one.

The total bending moment carried out by the section is $$M_t = \int_A y \sigma \delta A$$

however because strain has a linear relationship with the variable y, ($$\epsilon = k\cdot y + m$$), and $$\sigma= E\cdot\epsilon$$:

$$M_t = \int_A y E(k\cdot y + m )d A$$ $$M_t = \int_A E (k\cdot y^2 + m\cdot y )dA$$ $$M_t = \int_A kE\cdot y^2 dA + \int_A mE\cdot y dA$$ $$M_t = k\int_A E\cdot y^2 dA + m\int_A E\cdot y dA$$

For the symmetric case, of your example $$\int_A E\cdot y dA=0$$ about the neutral line. sot the total bending moment is given by (k is just a constant):

$$M_t = k\int_A E\cdot y^2 dA$$

Assuming that:

• b is the breadth of the beam then $$dA=b\cdot dy$$
• the thickness of wood is $$2*t_2$$, and the thickness of each steel plate is $$t_s$$ then

$$M_t = k\int_y E\cdot y^2 \cdot b\cdot dy$$ $$M_t = k\cdot b\int_{-tw-ts}^{ tw+ts} E\cdot y^2 \cdot dy$$

Notice that the value of E is depended on the height so its not constant through y. We can break the integral:

$$M_t = k\cdot b \left(\int_{-t_w-t_s}^{ -t_w} E_s\cdot y^2 \cdot dy + \int_{-t_w}^{t_w} E_w\cdot y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} E_s\cdot y^2 \cdot dy \right)$$

In those integration ranges E is constant so:

$$M_t = k\cdot b \left(E_s\cdot \int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy } + E_s\cdot \int_{t_w}^{ t_w+t_s} y^2 \cdot dy \right)$$

$$M_t = k\cdot b \left(E_s\cdot \int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + E_s\cdot \int_{t_w}^{ t_w+t_s} y^2 \cdot dy + \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy }\right)$$

$$M_t = k\cdot b \cdot E_s\cdot \left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right) + k\cdot b \cdot \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy }$$

On the right hand side:

• the first term $$k\cdot b \cdot E_s\cdot \left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right)$$ is the moment generated/carried by steel $$M_s$$,
• the second term $$\left(k\cdot b \cdot \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy } \right)$$ is the moment generated by the wood $$M_w$$

$$M_t = k\cdot b \cdot E_s\cdot \left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right) + k\cdot b \cdot \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy }$$

On the right hand side:

• the first term $$k\cdot b \cdot E_s\cdot \left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right)$$ is the moment generated/carried by steel $$M_s$$,
• the second term $$\left(k\cdot b \cdot \color{green}{E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy } \right)$$ is the moment generated by the wood $$M_w$$

So the total $$M_t = M_s + M_w$$. To get the fraction:

$$\frac{M_s}{M_w}= \frac{k\cdot b \cdot E_s\cdot \left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right) }{k\cdot b \cdot E_w\cdot \int_{-t_w}^{t_w} y^2 \cdot dy }$$

$$\frac{M_s}{M_w}= \frac{E_s}{E_w}\cdot \frac{\left(\int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy\right) }{ \int_{-t_w}^{t_w} y^2 \cdot dy }$$

however the latter fraction can be simplified because:

• for steel: $$I_s = \int_{-t_w-t_s}^{ -t_w} y^2 \cdot dy + \int_{t_w}^{ t_w+t_s} y^2 \cdot dy$$
• for wood: $$I_w = \int_{-t_w}^{t_w} y^2 \cdot dy$$

which yields:

$$\frac{M_s}{M_w}= \frac{E_s}{E_w}\cdot \frac{I_s}{ I_w }$$

• Can we use the moment-curvature relationship, developed for homogeneous beams for composite beams, as you do? Feb 15, 2022 at 5:19
• I am not using the moment curvature, I am using the strain compatibility across the cross-section.
– NMech
Feb 15, 2022 at 8:19

You should transform the composite section into a cross-section of homogeneous material using the concept of strain compatibility.

For the sandwiched scetion, the corresponding strains at where the two (steel & wood) materials meet are,

$$\epsilon_s = \dfrac{f_s}{E_s}$$ and $$\epsilon_w = \dfrac{f_w}{E_w}$$

Due to strain compatibility, $$\epsilon_s = \epsilon_w$$ ---> $$f_s = \dfrac{E_s}{E_w} f_w = nf_w$$

That is, $$\dfrac{P}{A_s}= n\dfrac{P}{A_w}$$

Then, let's hold $$d$$ as constant,

$$\dfrac{P}{b_s d} = \dfrac{nP}{b_w d}$$, thus, $$b_w = nb_s$$

From here, you should adjust the width of the materials, calculate the moment of inertia, and get the resulting stress distribution on the homogeneous cross-section. From which, you can get the moment carried by each element.

You can multiply the width of steel,

$$B$$, by

$$n=\frac{E_{steel}}{E_{wood}}$$

Now the second area moment of the composite beam Ic is the I of the section nBs *h subtracted by the I of the two missing sides on the wood.

' $$I_{c} = \frac{ nB *h^3}{ 12} - \frac{B(n-1)*(h-2t)^3}{12}$$.

and $$I_{wood} =\frac{B* (h-2t)^3}{12}$$

t = Thickness of the steel flange

$$I_{steel} =I_{c}-I{wood}$$

We can calculate the stress at any point distanced C

$$\sigma= MC/I_c$$

except if $$C>(h-t)/2$$ we multiply $$\sigma$$ by n.