Question

A horizontal bar, fixed at one end (x=0), has a length of 1 m, and cross-sectional area of 100 $m{m}^2$. Its elastic modulus varies along its length as given by E(x) = 100 $e^{-x}$ GPa, where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10kN is applied at the free end (x =1). The axial displacement of the free end is mm.

• A. 1.718

Answer 1

The correct option is A 1.718
Change in length of small strip,

$\delta =\frac{P_{x-x}\left(dx\right)}{A_{x-x}{E}_{xx}}$

Total change in length of bar,

${\delta }_{total}=\underset{o}{\overset{L}{\int }}\delta =\underset{o}{\overset{L}{\int }}\frac{\left(P_{x-x}\right)\left(dx\right)}{\left(A_{x-x}\right)E_{x-x}}$

$P_{x-x}$ = P = Constant
$A_{x-x}$ = A = Constant
$E_{x-x}=100e^{-x}GPa$

${\delta }_{total}=\frac{P}{A}\underset{o}{\overset{L}{\int }}\frac{dx}{100e^{-x}}$

$=\frac{P}{{10}^9\times 100A}\underset{o}{\overset{L}{\int }}\frac{dx}{e^{-x}}$

$=\frac{10000}{\frac{100}{{10}^6}\times 100\times {10}^9}\underset{o}{\overset{1}{\int }}\frac{dx}{e^{-x}}$

$=\frac{1}{1000}\left[1.7183\right]m$

Hence, ${\delta }_{total}$ = 1.7183 mm