Question

Consider two straight bars of uniform cross section made of the same material. Bar 1 has an axial length of and a square cross section with side length . Bar 2 has an axial length of and a round cross section with diameter. When subjected to axial tension, bar 1 elongates by . What will be the elongation of bar 2, , if it is compressed axially with a load of ?

Answer 1

Given data:
1st rod: L₁, T₁, x₁, a => A₁=a²;
2nd rod: L₂, T₂, x₂, d => A₂= πd²/4;
Hook’s Law for rods:
σ=Eε
T/A = E•x/L,
1st rod: T₁/a²= E•x₁/L₁,
2nd rod: 4T₂/π•d²= E•x₂/L₂,
x₂ = 4•x₁•L₂•a²•T₂ /L₁•T₁•π•d²=….

Answer 2

step
1) Derive a formula for the young's modulus E with the knowns of the first bar. It'll look something like

(F_1 / A_1) / (∆_1 / l_1 )
Notice that all of these are known. Don't solve yet, do all your analysis so you can isolate the physical meaning from the numerical minutia and that way understand it better.

2) Derive another formula for the young's modulus E, but now using the data of the second bar, you'll get the elongation ∆_2 as an unknown.

3) Since the 2 bars share the material and the young's modulus is a material property, you can set them equal to one another.

4) You know have a large equation with only one unknown. Solve algebraically.

5) Plug in the numbers and solve numerically.