4
$\begingroup$

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:

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

eg: https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams

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?

eg: https://en.wikipedia.org/wiki/Newton_(unit)#Examples

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

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Wasabi ♦
    Jul 10, 2017 at 13:34
  • $\begingroup$ Ah, right you are, an error on my part. Thanks! $\endgroup$
    – suprjami
    Jul 10, 2017 at 13:38

2 Answers 2

Reset to default
4
$\begingroup$

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 ($L^2 \cdot F/L^3=F/L$).

Improve this answer
$\endgroup$
0
$\begingroup$

q is going to be the distributed load along the span of the beam. If you are looking at just the self weight, you need to calculate that weight (based on density and cross sectional area) as a distributed force and apply that as q.

Important stuff to understand and learn, but you might be able to use this beam deflection calculator to check your answer.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.