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: Find the free thermal expansion of rod AB.
If rod AB were not restrained by anything, it would simply expand along its length when heated. The free expansion is given by
\[ \delta_T = \alpha L \Delta T \]
where \(\alpha = 12\times10^{-6}\) per \(^{\circ}\)C, \(L = 3\ \text{m} = 3000\ \text{mm}\) (the length of rod AB), and \(\Delta T = 100^{\circ}\)C.
\[ \delta_T = 12\times10^{-6}\times3000\times100 = 3.6\ \text{mm} \]

Step 2: Find the axial stiffness of rod AB.
The rod behaves like an axial spring with stiffness
\[ k_{rod} = \frac{AE}{L} \]
where \(A = 500\ \text{mm}^2\) and \(E = 60\times10^3\ \text{MPa} = 60000\ \text{N/mm}^2\).
\[ k_{rod} = \frac{500\times60000}{3000} = 10000\ \text{N/mm} \]

Step 3: Treat the rod and spring as two springs in series between the fixed supports.
Since A and C are both fixed, the total distance between them cannot change. As rod AB tries to expand by \(\delta_T\), this expansion has to be absorbed jointly by an elastic shortening of the rod and a compression of the spring, one after the other, exactly like two springs connected in series.
The equivalent stiffness of two springs in series is
\[ k_{eq} = \frac{k_{rod}\,k_{spring}}{k_{rod}+k_{spring}} \]
\[ k_{eq} = \frac{10000\times2500}{10000+2500} = \frac{25000000}{12500} = 2000\ \text{N/mm} \]

Step 4: Find the internal force that develops.
The free thermal expansion \(\delta_T\) is what would happen with no restraint at all; because both ends are fixed, this entire amount is absorbed by the elastic system, and the internal force that develops equals the equivalent stiffness times this free expansion:
\[ F = k_{eq}\,\delta_T = 2000\times3.6 = 7200\ \text{N} = 7.2\ \text{kN} \]
This same force \(F\) passes through the rod and then into the spring, since they are connected end to end in series, so the internal force developed in spring BC is also \(F\).

Final Answer:
\[ \boxed{F_{BC} \approx 7.2\ \text{kN}} \]