Deflections

Last updated January 2, 2024
By Ian Story

Codes limit deflections to a fraction of the span length, assuming fixed supports at each end of the span. But what happens if we use spring supports instead of fixed supports? Where should the deflection be measured from?

This issue becomes even more pronounced with finite element analysis where the supports cannot be easily defined. For the example that prompted this article: I’ve got a grade beam with regularly spaced pin piles acting as spring supports, with perpendicular grade beams adding additional springs at certain points. It’s difficult to classify which of these is the “support” for determining span length, and what happens when there is differential settlement along the length of the beam?

To work this out, I’ve reverse-engineered the code specifications for deflection into a format that is easier to check locally.

Often, when we check deflection limits for beams, we aren’t actually concerned about how far the beam dips. Rather, we want the changes to be subtle enough that they aren’t noticeable and don’t cause cracks in finishes. These criteria are a function of two things:

Slope
Curvature

I’ll approach these issues one at a time.

Slope

Using a simply supported beam with a uniform load as a code basis, let’s work out the relationship between maximum local slope and absolute deflection. Here are the governing equations for this beam type:

$q(x) = w$

$V(x) = wx + V_0$

$M(x)=EI\phi (x)= \frac{wx^2}{2} + V_0x + M_0$

$EI\theta (x)= \frac{wx^3}{6} + \frac{V_0x^2}{2} + M_0x + EI\theta_0$

$EI\Delta (x)= \frac{wx^4}{24} + \frac{V_0x^3}{6} + \frac{M_0x^2}{2} + EI\theta_0x + EI\Delta_0$

Using the following boundary conditions:

$M(0)=0$

$M(L)=0$

$\Delta (0)=0$

$\Delta (L)=0$

We find:

$\theta_0 = \frac{wL^3}{24EI}=\theta_{max}$

From the standard beam formulas, we know that:

$\Delta_{max} = \frac{5wL^4}{384EI}$

The ratio between these two is:

$\frac{\theta_{max}}{\Delta_{max}} = \frac{16}{5L}$

$\theta_{max} = \frac{16\Delta_{max}}{5L}$

For the deflection limit of L/240, this gives a maximum allowed slope of:

$\theta_{max} = \frac{16L/240}{5L} = \frac{1}{75}\text{ radians} = 0.76\text{ degrees}$

Running the numbers for other common allowed deflections gives the following slope limits:

Deflection Limit	Equivalent Slope Limit (approx)
L/120	1.50°
L/180	1.00°
L/240	0.75°
L/360	0.50°

Working backwards, if you have a known slope you can work out the equivalent deflection limit using the following formula:

$\text{L/x}(\theta)=\frac{16}{5\theta}$

Where $\theta$ is in radians

Curvature

To determine the code-intended curvature, let’s consider a simply-supported beam with a single point load at the center as our worst-case scenario. We will assume Euler-Bernoulli bending to eliminate the effects of shear deformation on curvature, which is conservative in this case.

Since curvature is simply $M / EI$ , we know that the maximum curvature is:

$\phi (x)=\frac{PL}{4EI}=\phi_{max}$

and the maximum deflection is

$\Delta_{max} = \frac{PL^3}{48EI}$

The ratio between these two is:

$\frac{\phi_{max}}{\Delta_{max}} = \frac{12}{L^2}$

$\phi_{max} = \frac{12\Delta_{max}}{L^2}$

For the deflection limit of L/240, this gives a maximum allowed slope of:

$\phi_{max} = \frac{12L/240}{L^2} = \frac{1}{20L}$

Unlike slope, the maximum curvature allowed under the building code is dependent on the span length. To simplify this problem, let us assume that the code provisions were derived for a beam with a minimum span of 10 feet. This gives:

$\phi_{max} = \frac{1}{20*120}=4.17*10^{-3}$