Elasticity — Question 3

Question 3

Series segments share force while their extensions add

Two rods are joined in series and pulled by $F=100\,\mathrm N$. Segment 1 has $(L_1,A_1,E_1)=(1\,\mathrm m,2\,\mathrm{mm^2},100\,\mathrm{GPa})$; segment 2 has $(L_2,A_2,E_2)=(.5\,\mathrm m,1\,\mathrm{mm^2},200\,\mathrm{GPa})$. Neglect joint deformation.

1. Find each stress and extension, and the total extension.

2. Find the stored elastic energy and explain which quantities are equal across the joint.

Question 3 - Solution

Series segments share force while their extensions add

(a) Derivation and calculation

The same axial force passes through both segments: $$\boxed{\sigma_1=F/A_1=50\ \mathrm{MPa}},\qquad \boxed{\sigma_2=F/A_2=100\ \mathrm{MPa}}.$$ Their extensions are $$\delta_1=\frac{100(1)}{(100\times10^9)(2\times10^{-6})}=.500\ \mathrm{mm},$$ $$\delta_2=\frac{100(.5)}{(200\times10^9)(10^{-6})}=.250\ \mathrm{mm}.$$ Hence $\boxed{\delta_{\mathrm{tot}}=.750\ \mathrm{mm}}$.

(b) Result and interpretation

The stored energy is $$\boxed{U=\tfrac12F\delta_{\mathrm{tot}}=\tfrac12(100)(.000750)=.0375\ \mathrm J}.$$ Force and joint displacement are continuous; segment stresses and extensions differ. Here the strains happen to agree: $$\epsilon_1=\epsilon_2=5.00\times10^{-4},$$ because the two axial rigidities $E_iA_i$ are both $2.00\times10^5$ N.

Physical check

The longer segment stores twice the extension and energy at the same force and axial rigidity.