Question

A composite tube is made by striking a thin steel tube on a brass tube. If $A_s$ and $A_B$ are the respective sectional areas of the steel and brass tubes and $Y_s$ and $Y_B$ their Young's moduli, then find the Young's modulus of single tube of the same length and total sectional area, which would behave in the same fashion as that of the composite tube.

Answer 1

Method 1 : Let the composite tube be subject to an axial (tensile) force F and $\delta l$ be the corresponding elongation. As discussed in the theory section, the stress borne by steel tube

${\rho }_{steel}=\frac{F{Y}_s}{A_s{Y}_s+A_B{Y}_B}$

But we know $Y_s=\frac{{\sigma }_s}{{\epsilon }_s}=\frac{{\sigma }_s}{\left(\delta l/L\right)}$

$\delta ı=\frac{{\sigma }_{s}L}{Y_s}=\frac{FL}{(A_s{Y}_s+A_B{Y}_B)}$

If Y be the required Young's modulus of the tube which behaves in the same fashion as that of the composite tube, then, corresponding to the same external force F, the deflection $\delta l$ should be the same, i.e.,

$\delta l=\frac{FL}{(A_s+A_B)Y}$

Comparing Eqs. (iii) and (iv), we have

$Y=\frac{A_s{Y}_s+A_B{Y}_B}{A_s+A_B}$

Method 2 : As discussed in the theory section, we can compare composite rod

System with spring combination. We can right equivalent force contact of spring as

$K_{eq}=\frac{I}{L}(A_s{Y}_s+A_B{Y}_B)$

If replace the composite rods into a single rod of length L and area (As + AB)

Here $K_{eq}=\frac{(A_S+A_B)Y}{L}$

Comparing Eqs. (v) and (Vi) we have equivalent Young's modulus as

$Y=\frac{A_S{Y}_S+A_B{Y}_B}{(A_S+A_B)}$