When there is one rectangular beam like below image,

enter image description here

I would like to know how the beam is deflected according to thicknesses of top plate and bottom plate.

My concerning cross section patterns are like below.

enter image description here

The conditions are as below.

Thicknesses of both side plates are same.

Type 1 : thickness of top < thickness of bottom

Type 2 : thickness of top = thickness of bottom

Type 3 : thickness of top > thickness of bottom

Which type is less deformed? and Why?


The deflection of a simply-supported beam under a concentrated load at midspan, regardless of the cross-section, is equal to:

$$\delta = \dfrac{PL^3}{48EI}$$

So the only thing that changes in your cases is the inertia, $I$. If Type 3 is the same as Type 1, only inverted, then their deflection should be the same.

If the sum of the top and bottom thicknesses is equal in all the cases (for instance, Type 1 is $1 + 3$, Type 2 is $2 + 2$, and Type 3 is $3 + 1$), then the inertia of Type 2 will be greater than that of Type 1 and 3. To prove this, we need to dive into the equations.

Using the following dimensions:

enter image description here

we can calculate all the properties. Type 2 is merely the case where $t_1=t_2$. The centers of gravity of these cross-sections are:

$$\begin{align} \overline{y} &= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{h_2}{2}\right)}{h_1b_1-h_2b_2} \\ &= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)\right)}{h_1b_1-h_2b_2} \\ &= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\frac{h_1}{2} - h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\ &= \frac{\frac{h_1}{2}\left(h_1b_1 - h_2b_2\right)}{h_1b_1-h_2b_2} - \frac{h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\ &= \frac{h_1}{2} - \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)} \end{align}$$

For Type 2, this simplifies to $\frac{h_1}{2}$, as expected, since $t_1=t_2$, zeroing the second term.

Now, the inertia of these sections is equal to:

$$I = \frac{b_1h_1^3}{12} + b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - \frac{b_2h_2^3}{12} - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2$$

For Type 2, this simplifies to

$$I_2 = \frac{b_1h_1^3}{12} - \frac{b_2h_2^3}{12}$$

So, to see that Type 2 will always have a larger inertia, we need to find the difference between $I_{1,3}$ and $I_2$:

$$\begin{align} \Delta I &= I_{1,3} - I_2 \\ &= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2 \\ &= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)-\overline{y}\right)^2 \\ &= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(\frac{1}{2}\left(h_1+t_1-t_2\right)-\overline{y}\right)^2 \\ &= b_1h_1\left(\frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 - b_2h_2\left(\frac{t_1-t_2}{2} + \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 \\ &= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(1 + \frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2\right) \\ &= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(\frac{h_1b_1}{h_1b_1-h_2b_2}\right)^2\right) \\ &= \left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2\left(b_1h_1\left(h_2b_2\right)^2 - b_2h_2\left(h_1b_1\right)^2\right) \\ \Delta I &= \underbrace{\left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2}_{\geq0}\underbrace{b_1h_1b_2h_2}_{>0}\underbrace{\left(h_2b_2- h_1b_1\right)}_{<0} \\ \end{align}$$

Since $\Delta I$ is therefore the product of three values, two of which are positive (or zero) and one of which is negative (since $h_1$ and $b_1$ are greater than $h_2$ and $b_2$, respectively), $\Delta I \leq 0$. This therefore means that $I_{1,3}$ (the inertia of Types 1 and 3) will always be lesser than $I_2$ (the inertia of Type 2). Therefore, the deflections under Type 2 will be smaller than those of Types 1 and 3.

All this above assumes that the material is isotropic and has the same behavior under both tension and compression. Something like reinforced concrete, where the concrete under tension is useless since it'll just crack, Type 3 is probably the best, since it puts the most concrete in the compression zone and leaves the steel to resist the tension.

This answer was initially incorrect, stating that Types 1 and 3 would have a larger inertia. I assumed the parallel axis component would not be so significant, and we all know what happens when one assumes. After seeing @kamran's answer, I double-checked my work and realized I was sorely mistaken. I apologize for the inconvenience.

  • $\begingroup$ One question : neutral surface of type 1 is the same with that of type3? $\endgroup$
    – KKS
    Jun 10 '16 at 13:58
  • $\begingroup$ @김갑수: Neutral line is the same distance from the same face. If they are mirrored, then, if the neutral line is at a distance $y$ from the thick (bottom) face of Type 1, then it will be at a distance $y$ from the thick (top) face of Type 3, which is equivalent to saying it will be at $h-y$ from the thin (bottom) face of Type 3, where $h$ is the section's total height. $\endgroup$
    – Wasabi
    Jun 10 '16 at 14:03
  • $\begingroup$ In my understand, you mean, in classical theory, type1 and type 3 will be samely deflected with some assumption. But in real world, type 3 will prevent compression, effectively, so type 3 would be less deformed. Is it right understanding?? $\endgroup$
    – KKS
    Jun 10 '16 at 14:22
  • $\begingroup$ @김갑수: It depends on the section material. If it's steel, which behaves almost identically under tension and compression, then the deflection will be the same. If it's reinforced concrete, then having more concrete under compression is better, since the concrete under tension will be ignored, since we assume that only the steel reinforcement resists tension. $\endgroup$
    – Wasabi
    Jun 10 '16 at 14:27

The deflection in type 2 is the least. Type one and type 3 will deform more because of the fact that neutral axis if the beam will move nearer to top or bottom.
When this happens the new strain on the narrower flange if the beam will be bigger resulting in more elongation at that flange. Meaning more deflection of the beam.
One crude way of looking at it is the new neutral axis will consume some of the web of the beam from the thick side. Hence the beam bends more to stress the left over cross section more to make up for its reduced area. It can easily be seen in drawing of beams cross section stress/ strain digram which i tried to draw on my phone by finger! cross section if the beam.

  • $\begingroup$ But yes, the neutral axis definitely shifts toward the thick flange. (Y = Sigma Yi * Ai/ sigma Ai) Meaning the centroid lies on Z axis and the Y from the bottom face of lower flange is the sum of all rectangular areas x their distances from y=0 divided by total cross section area. Meaning the Y is dragged up and down by the thicker area. if we take a look at U.S. railroad profiles which is similar to this beam, if we stick the two sides to one stem, we see non of them has the neutral axis passing mid height. icrr.net/rails.htm $\endgroup$
    – kamran
    Jun 14 '16 at 8:38

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