Answer:
[tex]\mathbf{\dfrac{F}{A}= -\dfrac{( \alpha_A + \alpha _B)\Delta T}{\bigg ( \dfrac{1}{Y_A}+ \dfrac{1}{Y_B}\bigg )} }[/tex]
Explanation:
From the information given.
The formula for Youngs modulus(Y) can be illustrated as follows:
[tex]Y = \dfrac{F/A}{\Delta L /L}[/tex]
where;
F = force
A = area
ΔL / L = the fraction of the change that occurred in the length.
If we make ΔL / L the subject of the formula, the above expression can be re-written as:
[tex]\dfrac{\Delta L}{L } = \dfrac{1}{Y}\dfrac{F}{A}[/tex]
However, the total fractional change that occurred in the length can be computed as:
[tex]\dfrac{\Delta L_{compressive \ stress}}{L} = \dfrac{\Delta L_A}{L_A} + \dfrac{\Delta {L_B}}{L_B}[/tex]
where;
[tex]L = L_A = L_B[/tex]
The area of both rods is said to be equal as well.
Thus;
[tex]\dfrac{\Delta L_{compressive \ stress}}{L} = \dfrac{1}{Y_1} \dfrac{F}{A}+ \dfrac{1}{Y_2}\dfrac{F}{A}[/tex]
[tex]\dfrac{\Delta L_{compressive \ stress}}{L} =\bigg( \dfrac{1}{Y_1} + \dfrac{1}{Y_2} \bigg ) \dfrac{F}{A}[/tex] ----- Let this be equation (1)
Recall that:
[tex]\dfrac{\Delta L }{L }= \alpha \Delta T[/tex]
in which
ΔL = change in length
L = the original length of rod before extension
∝ = coefficient of linear expansion
ΔT = temperature change
Similarly:
The total fractional change in the length is:
[tex]\Delta L _{thermal} = \Delta L_A +\Delta L_B[/tex]
[tex]\Delta L _{thermal} = \alpha_AL \Delta T+\alpha_BL \Delta T[/tex]
[tex]\dfrac{\Delta L _{thermal}}{L} =( \alpha_A+ \alpha _B) \Delta T \ \ \ ---(2)[/tex]
The sum of both the thermal and compressive stress of the combined rod is equal to the fractional change.
Mathematically:
[tex]\Delta L_{net} = \Delta L_{thermal} + \Delta L_{compressive \ stress}[/tex]
where;
[tex]\Delta L_{net} =0[/tex]
∴
[tex]0 = \Delta L_{thermal} + \Delta L_{compressive \ stress}[/tex]
[tex]\Delta L_{compressive \ stress} =- \Delta L_{thermal}[/tex]
From equation (1) & (2)
[tex]\bigg( \dfrac{1}{Y_1} + \dfrac{1}{Y_2} \bigg ) \dfrac{F}{A}= - (\alpha_A + \alpha_B) \Delta T[/tex]
Making the stress [tex]\dfrac{F}{A}[/tex] the subject, we get;
[tex]\mathbf{\dfrac{F}{A}= -\dfrac{( \alpha_A + \alpha _B)\Delta T}{\bigg ( \dfrac{1}{Y_A}+ \dfrac{1}{Y_B}\bigg )} }[/tex]
