Question

An engineer wants to design a structure in which the difference in length between a steel beam and an aluminum beam remains at 0.550 m regardless of temperature, for ordinary temperatures. What must the length of the steel beam be?

Answer 1

Answer:

The  length of the steel beam  is 1.058 m

Explanation:

coefficient of linear expansivity of Aluminum, $\alpha _{AL}$ = 25 x 10⁻⁶ ⁰C⁻¹

coefficient of linear expansivity of steel, $\alpha _{st}$ =  12 x 10⁻⁶ ⁰C⁻¹

Change in length of aluminum; ΔL$_{AL}$ = $L_{Al}$*$\alpha _{AL}$*ΔT

Change in length of steel; ΔL$_{st}$ = $L{st_i}$*$\alpha _{st}$*ΔT

difference in length of Aluminum and steel;

$L_{st_i}$- $L_{Al_i}$ = 0.55 m, for this difference to remain constant, then ΔL$_{AL}$ = ΔL$_{st}$

From the equation above,  $L_{st_i}$   = 0.55 +$L_{Al_i}$

Since,  ΔL$_{AL}$ = ΔL$_{st}$, then  $L_{Al}$*$\alpha _{AL}$*ΔT = $L{st_i}$*$\alpha _{st}$*ΔT

At constant temperature, the equation becomes;

$L_{st_i}$*$\alpha _{st}$ = $L_{Al}$*$\alpha _{AL}$

Recall; $L_{st_i} = 0.55 + L_{Al_i}$

$(0.55 +L_{Al_i})\alpha_{st} = L_{Al_i}\alpha _{Al}\\\\0.55\alpha_{st} + L_{Al_i}\alpha_{st} = L_{Al_i}\alpha _{Al}\\\\ L_{Ali}\alpha _{Al} - L_{Al_i}\alpha_{st} = 0.55\alpha_{st}\\\\L_{Al_i} (\alpha _{Al} -\alpha_{st}) = 0.55\alpha_{st}\\\\L_{Al_i} =\frac{0.55\alpha_{st}}{\alpha _{st} -\alpha_{Al}} = \frac{0.55X12X10^{-6}}{(25X10^{-6}) -(12X10^{-6})}\\\\L_{Al_i} = \frac{0.55X12X10^{-6}}{13X10^{-6}} =0.508 m$

To calculate the length of the steel beam;

$L_{st_i} = 0.55 + L_{Al_i}$

$L_{st_i} = 0.55 + 0.508 = 1.058 m$

Therefore, the  length of the steel beam  is 1.058 m