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**The rod of the picture is loaded with the force 2P which is uniformly distributed on Section D where applied. If the material of the rod is homogenous with Elastic Modulus E find the variation in length of the rod. The area of the cross section in part AC is 2A while CD is A.**

__Solution__

We calculate the reaction R at the rigid end connection A from the equilibrium of forces.

ΣP_{X}= 0 → 2P - R = 0 → R = 2P **(1)**

The rod doesn’t have everywhere constant area of cross section so we examine parts AC and CD.

At part AC the cross section with area 2A is loaded with axial force equal to R. So at this part we have equal uniform stress σ_{1} and deformation ε_{1}.

σ_{1} = R / 2A

ε_{1} = Δl_{1} / L

Because stress is equal and the material is homogenous Hook Law is applied for section AC.

σ_{1}= Eε_{1 }→ R / 2A = E ( Δl_{1 }/ L) →Δl_{1 }= (R / 2A) (L / E) **(2)**

At part CD we have axial force 2P so:

σ_{2} = 2P / A

ε_{2} = Δl_{2} / 3L

And Hooks Law:

σ_{2} = Eε_{2 }→ 2P / A = E (Δl_{2} / 3L) → Δl_{2}= (2P / A )(3L / E) **(3)**

From equations (1),(2),(3) we have the total change of length of the rod.

Δl= Δl_{1} + Δl_{2} = (R / 2A)(L / E) + (2P / A)(3L / E)

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