# Curtain Rod – A Simply Supported Beam

I saw that curtain rods at my house were bent due to the weight of curtains on them and wondered whether a beam analysis can be carried out to verify its deformation under the load.

Here is a picture of the initially straight curtain rod without curtains.

And here is a picture of the bent curtain rod when curtains were hung from it.

I analyzed this scenario as a linear, elastic, small deformation simply supported beam problem with uniformly distributed loads on it to check whether the maximum deformation (sagging) of the beam (curtain rod) at its center matches with theoretical calculations.

Though the problem appears to be a pretty straight forward one, the challenge was to measure or estimate physical quantities (dimensions, weight, material properties) with reasonable accuracy to validate the actual results theoretically.

1st I measured the dimensions. The span length of the beam was easy to measure using an inch tape. It turned out to be $$L=2884$$ mm.

Next I needed the cross-section dimensions with reasonable accuracy (something like $$\frac{1}{10}$$th of an mm). Additionally the OD of the rod was corrugated which prompted me to find out the mean OD to simplify the analysis. Here’s a picture of the cross-section (The ends are slightly worn out here, but still you’ll get the idea.)

To measure the OD I rolled up three rounds of thread on the outer surface of the rod, unrolled it, measured its length and divided by $$3\pi$$. Next using a scale I visually estimated the height of the corrugations to be around 0.8 mm. Using the measured outside OD and approximate height of corrugations I calculated the mean OD of the rod which turned out to be $$OD_{mean}=238/3\pi-0.8=24.45$$ mm.

Here’s a picture showing thread rolled over the rod.

To estimate the ID, I inserted a sheet of rolled paper inside the hole of the rod to match its ID. Then repeated a similar procedure, this time rolling thread over the paper to estimate the ID as $$ID=211/3\pi=22.39$$ mm.

Here’s a picture of the thread rolled over the paper inserted inside the curtain rod hole.

The curtain rod was  made of Aluminium with a brown paint over it. Young’s Modulus assumed for the Aluminium rod was $$E=69$$ GPA.

Next I set out to compute the uniformly distributed load on the curtain rod with the hanging curtains. To accomplish this I weighed the curtains on a weighing machine and figured out its weight as $$wt_{curtains}=3$$ kg.

Here’s a picture showing the curtain weight.

Next I estimated the weight of the Aluminium rod as its volume*density.

$$wt_{Al}=\frac{\pi}{4}*(24.45^2-22.39^2)*2884*(2700*10^{-9})=0.59$$ kg.

Knowing the weights I calculated the udl intensity as

$$w=\frac{wt_{curtains}+wt_{Al}}{L}=\frac{3.59}{2884}=0.001245$$ kg/mm $$=0.0122$$ N/mm

Next I calculated the moment of inertia for the cross-section before moving to the last part of this analysis.

$$I=\frac{\pi(OD_{mean}^4-ID^4)}{64}=5206$$ $$mm^4$$

Now the final part. For the sake of brevity, I wouldn’t do a derivation here, but the maximum central deformation for a simply supported beam with a uniformly distributed load is given by $$\delta=\frac{5wL^4}{384EI}=\frac{5*0.0122*2884^4}{384*69*1000*5206}=30.6$$ mm. When I measured the central deformation of the rod it came out to be 31 mm (I used the shadow of rod on the wall before and after hanging the curtains to measure this). That the result got validated with such high accuracy in spite of the rough method used in the analysis actually amused me as might have amused you.

## 1 thought on “Curtain Rod – A Simply Supported Beam”

• Ramesh S says:

good method and effort by you to amuse you for simply supported calculation.
I am got feeling of practical done and also learned thanks.