Question

A concrete cylinder of diameter 150 mm and length 300 mm when subjected
to an axial compressive load of 240 kN resulted in an increase of diameter by 0.127 mm and a
decrease in length of 0.28 mm. Compute the value of Poisson's ratio u=1/m
and modulus of
elasticity E.
​

Answer 1

Answer:

Explanation:

Original Diameter(D)= 150mm

Change in diameter(d)=+0.127 mm

Original Length(L)=300mm

Change in length (l)= -0.28mm

v(Poisson's Ratio)= (d x L)/(D X l)=$\frac{0.127*300}{150*-0.28}$=-0.90

Stress= 240 x 10^3 /3.14 x 75 x 75 =13.58 MPa

Answer 2

A concrete cylinder of diamter of 150 mm and length 300 mm when subjected to an axial compresssive load of 240 kN resulted in an increase of diameter by 0.127 mm and decrease in length of 0.28 mm.

We have to find the value of Poisson's ratio and the modulus of elasticity.

Poisson's ratio : it is the ratio of lateral strain to the longitudinal strain.

i.e., μ =  $\frac{\text{lateral strain}}{\text{longitudinal strain}}$

given, decrease in the length of cylinder, |ΔL| = 0.28mm

so the longitudinal strain = $\frac{\Delta L}{L}$ = $\frac{0.28mm}{300mm}$

similarly given, increase in the diameter of cylinder, |Δd| = 0.127 mm

so the lateral strain = $\frac{\Delta d}{d}=\frac{0.127mm}{150mm}$

now Poisson's ratio, μ = $\frac{0.127mm}{150mm}\times\frac{300mm}{0.28mm}$ = 0.907

Therefore the Poisson's ratio is 0.907.

 modulus of elasticity is given by, E = $\bf\frac{stress}{strain}$

= $\frac{FL}{A\Delta L}$

here, applied compressive load = F = 240 kN = 240000 N

cross sectional area = A = π(75mm)² = 3.14 × 5625 × 10⁻⁶ m²

change in length of cylinder , ΔL = 0.28 mm = 0.00028 m

length of cylindr , L = 300 mm = 0.3 m

now E =  $\frac{240000\times 0.3}{\pi\times5625\times10^{-6}\times0.00028}$ = 14.55 GN/m²

Therefore the modulus of elasticity of cylinder is 14.55 GN/m²