Question

A stepped rod of length 2 m is fixed at both ends A and C. The area of cross-section of AB is 200 mm$^{2}$ and that of BC is 100 mm$^{2}$. A force $F$ is applied at section B causing a displacement of 0.1 mm in the direction of the force. Young’s modulus of the rod is 200 GPa. The applied force $F$ in N is _________ (round off to the nearest integer).

Hint:

When a rod is fixed at both ends and loaded at an intermediate point, the stiffness at the load point is the sum of segment stiffnesses.

Answer 1

Correct Answer: 5990

Convert areas to m$^{2}$: \[ A_{AB} = 200 \times 10^{-6}, \qquad A_{BC} = 100 \times 10^{-6} \] Lengths: \[ L_{AB}=1~\text{m}, \qquad L_{BC}=1~\text{m} \] Total stiffness at point B for two axial springs in series from fixed ends: \[ k_{AB} = \frac{EA_{AB}}{L_{AB}}, \qquad k_{BC} = \frac{EA_{BC}}{L_{BC}} \] \[ k_{AB} = \frac{200\times10^{9}(200\times10^{-6})}{1} = 40\times10^{6}~\text{N/m} \] \[ k_{BC} = \frac{200\times10^{9}(100\times10^{-6})}{1} = 20\times10^{6}~\text{N/m} \] Total stiffness seen at B: \[ k = k_{AB} + k_{BC} = 60 \times 10^{6}~\text{N/m} \] Displacement: \[ \delta = 0.1~\text{mm} = 1\times10^{-4}~\text{m} \] Force: \[ F = k\delta = (60\times10^{6})(1\times10^{-4}) \] \[ F = 6000~\text{N} \] \[ \boxed{6000} \]