Question

A composite rod is 1000 mm long, its two ends are 40 $m{m}^2$ and 30 $m{m}^2$ in area and length are 300 mm and 200 mm respectively. The middle portion of the rod is 20 $m{m}^2$ in area and 500 mm long. If the rod is subjected to an axial tensile load of 1000 N, find its total elongation (in mm). (E = 200 GPa).

• A. 0.165
• B. 0.111
• C. 0.196
• D. none of the above

Answer 1

The correct option is C 0.196

$k=\frac{EA}{L}$

$k_1=\frac{E{A}_1}{L_1}$,$k_2=\frac{E{A}_2}{L_2}$,$k_3=\frac{E{A}_3}{L_3}$

$k_1=E\times \frac{40}{300}\times {10}^{-3}$

$k_2=E\times \frac{30}{200}\times {10}^{-3}$

$k_3=E\times \frac{20}{500}\times {10}^{-3}$

$k_1=\frac{4E}{3}\times {10}^{-4}$

$k_1=\frac{3E}{3}\times {10}^{-4}$

$k_1=\frac{2E}{3}\times {10}^{-4}$

$k_1,k_2,k_3$ are in series.

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}$

$=\frac{3\times {10}^4}{4E}$+$\frac{2\times {10}^4}{3E}$+$\frac{5\times {10}^4}{2E}$

$=\frac{{10}^4}{E}(\frac{3}{4}+\frac{2}{3}+\frac{5}{2})$

$=\frac{{10}^4}{E}\times \frac{47}{12}$

$k_{eq}=\frac{12E}{47}\times {10}^{-4}$

$F=1000N$

$1000=\frac{12E}{47}\times {10}^{-4}\Delta l$

$\Delta l=\frac{1000\times 47}{12E}\times {10}^4$

$\Delta l=\frac{1000\times 47\times {10}^4}{12\times 200\times {10}^9}$

$=19.6\times {10}^{-5}$

$=0.196mm$