# Round beam deflection under its own weight

I'm trying to calculate how much simply-supported round beams of several materials and diameters deflect under their own weight plus an additional weight in the middle.

I know the material densities so I can calculate weight of the entire beam, and I know the material Young's Modulus so I can calculate Moment of Inertia.

I've got the deflection due to an added weight calculated correctly, but I'm stuck on deflection due to the shaft's own weight.

Several places have given the formula:

$$\delta C = \frac {5 q L^4} {384 E I}$$

But I am not sure what q represents.

Wikipedia describes that as "Uniform load on the beam (force per unit length)" and I see other sites mention it in Newtons.

Is that gravity's acceleration (9.80665 m/sec^2) times the beam's own weight in kilograms to get Newtons of downward force?

For example, a 102g beam would exert 1N of downward force, so q would be 1?

• A small correction: you don't need the Young's Modulus ($E$) to calculate the moment of inertia ($I$). The product of both is usually called the beam's stiffness, but they are each independent of the other. – Wasabi Jul 10 '17 at 13:34
• Ah, right you are, an error on my part. Thanks! – suprjami Jul 10 '17 at 13:38

No, as the Wikipedia article states, a uniform load's unit is in "force per unit length" ($F/L$), which means it is a load applied along a given distance.
So, if you have a 102 g beam, it weighs a total of 1 N. However, that unit is merely "force" ($F$), so we still need to get it into "per unit length" ($1/L$). So, if your beam is 2 m long, then $q = \frac{0.102\cdot9.8}{2} = 0.5\text{ N/m.}$
Another way of calculating this would be by multiplying the beam's cross-sectional area (unit is $L^2$) by its specific weight ($F/L^3$), which gives you the beam's linear weight ($L^2 \cdot F/L^3=F/L$).