Question

A homogenous, linearly elastic rod AB is connected to a linearly elastic spring BC in between the fixed supports at A and C, as shown in the figure. The cross-sectional area, modulus of elasticity, and the coefficient of thermal expansion of the rod AB are 500 mm\(^2\), \(60\times10^3\) MPa, and \(12\times10^{-6}\) per \(^{\circ}\)C, respectively. The stiffness (k) of spring BC is 2500 N/mm.

(Figure not to scale)
The internal force (in kN) that will develop in the spring BC when the temperature of rod AB is increased by \(100^{\circ}\)C is (rounded off to one decimal place).

Hint:

The rod and spring act like two springs in series between fixed supports; the internal force equals the equivalent stiffness of the pair times the rod's free thermal expansion.

Answer 1

Correct Answer: 7.2

Step 1: Set up the problem as a statically indeterminate system.
The rod AB and spring BC sit between two fixed points A and C, so this is a one degree statically indeterminate system: there is one unknown internal axial force that keeps the total length between A and C unchanged when the rod is heated.

Step 2: Release the redundant support and find the free displacement.
Imagine temporarily removing the fixed support at C, so point C (and the free end of the spring) becomes free to move. Heating rod AB by $\Delta T = 100^\circ$C then lets it expand freely by
\[ \delta_T = \alpha L \Delta T = (12\times10^{-6})(3000)(100) = 3.6\text{ mm} \]
Since the spring carries no force in this released state, point C would move outward by exactly this same $3.6$ mm, because the spring does not stretch or compress on its own.

Step 3: Apply the redundant force $F$ back at C to restore the actual fixed condition.
In reality, C cannot move, so an internal axial force $F$ must develop to pull point C back by $3.6$ mm. This force $F$ acts equally on the rod (compressing it) and on the spring (compressing it), since they are connected in series with no branch point in between.
The displacement caused by $F$ alone, with the temperature effect switched off, is the sum of the rod's elastic shortening and the spring's compression:
\[ \delta_F = \frac{FL}{AE} + \frac{F}{k} \]
where $\dfrac{L}{AE} = \dfrac{3000}{500\times60000} = 0.0001\text{ mm/N}$ and $\dfrac{1}{k} = \dfrac{1}{2500} = 0.0004\text{ mm/N}$.

Step 4: Write the compatibility equation.
For point C to end up back at its original fixed position, the displacement due to $F$ must exactly cancel the free thermal displacement:
\[ \delta_F = \delta_T \]
\[ F(0.0001+0.0004) = 3.6 \]
\[ F(0.0005) = 3.6 \]
\[ F = \frac{3.6}{0.0005} = 7200\text{ N} = 7.2\text{ kN} \]

Final Answer:
This redundant force $F$ is exactly the internal force carried by spring BC, since it is connected in series with the rod.
\[ \boxed{F_{BC} \approx 7.2\text{ kN}} \]