I've been given a problem which I've attempted to solve, and hope some fellow engineers could possibly see if my approach is correct.

We are asked to calculate a column's axial deformation due to an applied load of F = 25 kN and the force due to the column's own weight. We are also given that ρ = 7800 kg/m3, E = 200 GPa and g = 9.81 m/s. It's vertical length is L = 1 m and its radius varies vertically as r = (0.03)/(1+y).

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This is the approach that I took where I first found the force due to the column's weight as a function of y:

$$F(y)=\int_0^y \rho g A(y) \, dy$$ $$F(y)=\int_0^y \rho g \pi \left (\frac{0.03}{1+y} \right )^2\, dy$$ $$F(y)=(0.03)^2 \rho g \pi \int_0^y \frac{dy}{(1+y)^2}$$ $$F(y)=(0.03)^2 \rho g \pi \left (\frac{y}{1+y} \right )$$

I then put this into the axial deformation equation and integrated to get the following:

$$\delta_{w}=\int_0^L \frac{F(y)dy}{A(y)E}$$ $$\delta_{w}=\int_0^1 \frac{(0.03)^2 \pi \rho g \left (\frac{y}{1+y} \right )dy}{\pi \left (\frac{0.03}{1+y} \right )^2 E}$$ $$\delta_{w}=\int_0^1 \frac{(0.03)^2 \pi \rho g y(1+y)^2dy}{(0.03)^2 \pi E (1+y)}$$ $$\delta_{w}=\frac{\rho g}{E}\int_0^1 y(1+y)dy$$ $$\delta_{w}=\frac{5 \rho g}{6 E}$$ $$\delta_{w}=\frac{5 (7800)(9.81)}{6 (200)(10^9)}$$ $$\delta_{w}=3.18825(10^{-7})$$

For the applied load I did:

$$\delta_{F}=\int_0^L \frac{Fdy}{A(y)E}$$ $$\delta_{F}=\frac{F}{E}\int_0^L \frac{dy}{\pi \left (\frac{0.03}{1+y} \right )^2}$$ $$\delta_{F}=\frac{F}{(0.03)^2 \pi E}\int_0^L (1+y)^2 dy$$ $$\delta_{F}=\frac{7 F}{3 \pi (0.03)^2 E}$$ $$\delta_{F}=\frac{7 (25)(10^3)}{3 \pi (0.03)^2 (200)(10^9)}$$ $$\delta_{F}=1.03156(10^{-4})$$

And finally, adding the two deformations I got:

$$\delta_{total}=\delta_{w}+\delta_{F}$$ $$\delta_{total}=-1.0347(10^{-4})$$

Where the negative sign is due to the fact that the forces are compressive. Is this result for the deformation of the beam accurate? My main point of concern is whether I calculated $F(y)$ correctly.


Your integration limits should be y to 1 to get force at height y. For example at y=1m force due to self weight is zero, and deformation is again zero; Again when y=0, self weight is maximum. I will leave the further work outs to you.


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