# What are the units used in beam bending equations? Do they matter?

I was just wondering whether or not the standard beam bending equations worked for all units. E.g. imperial and metric etc.

From what I'm guessing it shouldn't really, since once we plug everything in the units should cancel out properly.

• units are always valid if measured correctly... it's people that fail to properly track conversions through their uses. do a sanity check between units on left and right sides of equations...
– Abel
Jul 1 '21 at 12:55

## 3 Answers

Yes, the equations are valid for both imperial and metric systems. The most important thing is to use consistent units of a system throughout the calculations. Consistency of units matters as most mistakes are caused by mixing the imperial and metric units without conducting correct unit conversions (such as 1 in = 25.4 mm, 1MPa = 145 psi), or forgetting to multiply/divide the unit factors within the same unit system (such as 1' = 12", 1000N = 1 kN).

Typical Units of (Imperial), [Metric] systems:

$$P$$ - (lbs, kips), [N, kN]

$$w$$ - (plf, klf), [N/m, kN/m]

$$L, x$$ - (in, ft), [mm, m]

$$E$$ - (psi, ksi), [Pa, MPa]

$$I$$ - (in^4), [mm^4]

$$\delta$$ - (in), [mm]

Edit:

To confirm that the equations are valid for both systems, here is the deflection using the Imperial units and those parameters in the example provided by NMech.

$$P = 4 kN = 0.9 kips$$

$$L = 2 m = 78.74 in$$

$$E = 200 GPa = 29,000 ksi$$

$$I = 4 cm^4 = 0.0961 in^4$$

$$\delta = \frac {0.9 kips * (78.74 in)^3}{3*29000*0.0961} = 52.55 in$$

Metric:

$$\delta = \frac {4000*2^3}{3*200*10^9*4*10^{-8}} = 1.33 m = 52.49 in$$, check!

## SI

In the SI, you are correct that the idea is that you don't need to modify the equations. You can use any units in the equation and perform the conversions in the calculation. Typically, what one (usually a beginner) would do is convert to the basic SI units and then perform the calculation.

For example a steel (E=200GPA) cantilever beam with L = 2[m], and $$I=4 cm^4$$, with a load of 4 kN would have the following calculation for $$\delta_{max}$$:

$$\delta_{max} = \frac{PL^3}{3EI}$$ $$\delta_{max}= \frac{4 [kN] \cdot (2 [m])^3}{3\cdot 200 [GPa]\cdot 4[cm^4]}$$ $$\delta_{max} = \frac{4\cdot 10^3 [N] \cdot (2 [m])^3}{3\cdot 200\cdot 10^9 [Pa]\cdot 4\cdot 10^{-8}[m^4]}$$

The result of the above calculation would be in the main unit of length in the SI (i.e. m). You can derive it, but you can also depend on it -if you've done the calculation correctly). This is important, because as you can see you have force units, pressure units and length units in the final line, but you can still depend on the result being in the basic SI unit.

## Imperial

(Although I don't use it professionally) in the Imperial things are more tricky. For example, Modulus is usually given in ksi, and moment of area is given in $$in^4$$, however the length can be in feet or other units.

You can use the above approach (i.e. convert all quantities to a single unit length, force unit), however -at least to my understanding- there is no standard imperial unit for length for example, so you need to perform the additional step of deriving the units of the result.

# What's an equation?

In life there are two types of equations:

• theoretical equations are obtained from first principles: make some assumptions and then play around with variables until you get a useful equation
• empirical equations are obtained by experiment and then finding an equation that adequately describes the observed behavior.

Theoretical equations are always unit-agnostic. This is because the coefficients originated from your mathematical manipulations (probably integration or derivation, etc) and don't actually represent any physical quantity. For this reason, dimensional analysis must, by definition, always work out for theoretical equations. So if you have a theoretical equation $$y = 3x^2$$, you can be sure that those coefficients popped out from some numerical manipulation, and that therefore the unit of $$y$$ will be equal to the square of $$x$$'s unit (so if $$x$$ is a length $$[L]$$, $$y$$ is an area $$[L^2]$$, for example).

Empirical equations are another matter entirely and are usually unit-system-dependent. This is because the equations are effectively improvised with no mathematical backing to them, so the coefficients (including exponents) which are adopted may need to be entirely different. So much so that dimensional analysis will more often than not fail for empirical equations. This means you may be dealing with an empirical $$y = 3x^2$$ where $$x$$ is a length $$[L]$$ and $$y$$ is a mass $$[M]$$, which of course makes no sense analytically.

Indeed, it only makes sense if you know that the 3 coefficient is actually 3 kg/m2: therefore, this equation only works if you use meters for $$x$$ and expect kilograms for $$y$$.

If you know the coefficient's units, however, you could convert this equation to other units of measurement pretty easily: $$3\text{ kg/m}^2 = 3000 \text{ g/m}^2 = 0.3 \text{ g/cm}^2 \approx 0.048 \text{ g/in}^2$$. But still, each equation (using 3, 3000, 0.3, 0.048 or any other coefficient you can think of) will only be valid for the specific units used to define that coefficient. And more complex cases can make this far more complicated to figure out.

# And what about bending equations?

Well, now answering your question is very straightforward: the beam bending equations are 100% theoretical. Euler and Bernoulli made assumptions (parallel sections remain parallel, etc), played around with variables and -- poof! -- out came the fundamental beam equation, from which we can then derive the deflections for different boundary conditions.

Being theoretical equations, we can be sure that they are, by definition, unit-agnostic.