A composite tube is made by strinking a thin steel tube on a nras 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 be have in the same facition as that of the composite tube.

Explanation

Method1: Let the composite tube be subjected to an axial (tensile) force `F` and `deltal` be the corresponding elongation.
As discussed in theory, the stress borne by steel tube
`rho_("steel")=(FY_(S))/(A_(S)Y_(S)+A_(B)Y_(B))`
But we know `Y_(S)=(sigma_(S))/(epsilon_(S))=(sigma_(S))/(deltal//L)`
`deltal=(sigma_(s)L)/(Y_(S))=(FL)/((A_(S)Y_(S)+A_(B)Y_(B))`.........iii
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 `deltal` should be the same i.e,
`deltal=(3L)/((A_(S)+A_(B))Y)` ..........iv
comparing eqn iii and iv
we have
`Y=(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 write equivalent force contact of spring is
`k_(eq)=1/L(A_(S)Y_(S)+A_(B)Y_(B))` ........v
If replace the composite rods into a single rod of length `L` and area `(A_(S)+S_(B))`
Here `k_(eq)=((A_(S)+A+B)Y)/L`......vi
Comparing eqn v and vi we ave equivalent Young's modulus as
`Y=(A_(S)Y_(S)+A_(B)Y_(B))/((A_(S)+A_(B)))`