# A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and...

## Question:

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at {eq}250^oC, {/eq} if the original lengths are at {eq}40.0^oC? {/eq} Is there thermal stress developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of {eq}brass = 2.0 \times10^{-5} K^{-1} ,steel =1.2 \times 10^{-5} K^{-1}. {/eq}

## Thermal Expansion:

Thermal expansion is defined as the capability to alter the dimensions of a particular object due to the difference in temperature. The thermal expansion of an object is given in terms of the coefficient of linear expansion. Its value is fully dependent on the variation in temperature.

Given data

• The length of the brass rod is: {eq}{l_b} = 50\,{\rm{cm}} = 0.5\,{\rm{m}}{/eq}
• The diameter of the brass rod is: {eq}{d_b} = 3\,{\rm{mm}} = 3 \times {10^{ - 3}}\,{\rm{m}}{/eq}
• The length of the steel rod is: {eq}{l_s} = {l_b} = 0.5\,{\rm{m}}{/eq}
• The diameter of the steel rod is: {eq}{d_s} = {d_b} = 3 \times {10^{ - 3}}\,{\rm{m}} {/eq}
• The change in the temperature across the length is: {eq}\Delta T = 250^\circ \,{\rm{C}} - 40^\circ \,{\rm{C}} = 210\,{\rm{K}}{/eq}
• The coefficient of linear expansion of brass is: {eq}{\alpha _b} = 2.0 \times {10^{ - 5}}\,{{\rm{K}}^{{\rm{ - 1}}}}{/eq}
• The coefficient of linear expansion of steel is: {eq}{\alpha _s} = 1.2 \times {10^{ - 5}}\,{{\rm{K}}^{{\rm{ - 1}}}}{/eq}

The expression for calculating the change in length of brass rod can be obtained from the expression of coefficient of linear expansion for brass is given as,

{eq}\begin{align*} {\alpha _b} &= \dfrac{{{\rm{Strain}}}}{{\Delta T}}\\ {\alpha _b}\Delta T& = \dfrac{{\Delta {l_b}}}{{{l_b}}}\\ \Delta {l_b}& = {\alpha _b}\Delta T{l_b}\, \end{align*}{/eq}

Substituting the value in the above expression,

{eq}\begin{align*} \Delta {l_b}& = 2.0 \times {10^{ - 5}} \times 210 \times 0.5\\ \Delta {l_b}& = 2.1 \times {10^{ - 3}}\,{\rm{m}} \end{align*} {/eq}

The expression for calculating the change in length of steel rod can be obtained from the expression of coefficient of linear expansion for steel is given as,

{eq}\begin{align*} {\alpha _s} &= \dfrac{{{\rm{Strain}}}}{{\Delta T}}\\ {\alpha _s} &= \dfrac{{\dfrac{{\Delta {l_s}}}{{{l_s}}}}}{{\Delta T}}\\ {\alpha _s}\Delta T &= \dfrac{{\Delta {l_s}}}{{{l_s}}}\\ \Delta {l_s} &= {\alpha _s}\Delta T{l_s} \end{align*} {/eq}

Substituting the value in the above expression,

{eq}\begin{align*} \Delta {l_s} &= 1.2 \times {10^{ - 5}} \times 210 \times 0.5\\ \Delta {l_s} &= 1.26 \times {10^{ - 3}}\,{\rm{m}} \end{align*} {/eq}

Now, the expression to calculate the total change in length of brass rod and steel rod combined is,

{eq}\Delta l = \Delta {l_s} + \Delta {l_b} {/eq}

Substituting the value obtained in the above expression,

{eq}\begin{align*} \Delta l &= 2.1 \times {10^{ - 3}} + 1.26 \times {10^{ - 3}}\\ \Delta l &= 3.36 \times {10^{ - 3}}\,{\rm{m}}\\ \Delta l& = 3.36\,{\rm{mm}} \end{align*} {/eq}

Thus, the total change in lengths is {eq}3.36\,{\rm{mm}}{/eq} and the no stress will develop at the point of contact of both the rods because both the rods expand from the extreme end.