Properties of Matter - Elasticity: Important Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

1. A copper wire of 3m length and 1mm diameter is subjected to a tension of 5N. Calculate the elongation produced in the wire if the young's modulus of elasticity of copper is 120 GPa

Given data

 F=5N;

L = 3m;

Y = 120 × 10⁹ Pa;

A = π (0.5 × 10^‒3)² m²

Solution

Formula: Young's Modulus

Y = FL / Al

 l = FL / AY

 = [ 5×3 ] / [ π(0.5×10^‒3)²×120×10⁹ ]

= 1.5 / 94200

 = 1.59 × 10^‒4m

= 15.9 × 10^‒3m

 ∴ The elongation produced in the wire = 15.9 mm

2. A circular and a square cantilever are made of same material and have equal area of cross‒section and length. Find the ratio of their depressions, for a given load.

Solution

Formula

Depression for a cantilever

Ratio of depressions = π/3

3. A cantilever of length 50 cm fixed at one end is depressed by 20mm at the loaded end. Calculate the depression at a distance of 40cm from the fixed end.

Given data

 l = 50 × 10^‒2m

 x = 40 × 10^‒2m

Depression at a distance of 50 cm → y = 20 × 10^‒3m

Depression at a distance of 40 cm → y = ?

Solution

Formula:

(i) Depression at a distance of 50 cm i.e., at (x=1) is y = [ W/YI_g] × [ l³/3]

Formula:

(ii) Depression at a distance of 40 cm from the fixed end is y' =

Depression of the centilever at a distance of 40cm from the fixed end is 19.84mm

ADDITIONAL SOLVED PROBLEMS

4. Calculate the density of lead under a pressure 2×10⁸ N/m². Density of lead = 11.4×10³ kg/m³. Bulk Modulus of elasticity = 8×10⁹ Nm².

Solution

 K = ‒ PV / dV                     …………..(1)

where dV is the change in volume

Also, we know mass (M) = Vρ

Differentiating, Vdρ+ ρdV= 0

 dρ/ρ = ‒dV/V                       …………….(2)

Substituting eqn.(2) in eqn.(1), we have

 K = P / dρ/ρ

 dρ = Pρ / K

 given

P = 2 × 10⁸ Nm^‒2

ρ = 11.4 × 10³ Kgm^‒3

K = 8 × 10⁹ Nm^‒2

dρ = Pρ / K

= [ 2 × 10⁸ ×  11.4 × 10³  ] / [8 × 10⁹]

= 0.285×10³ Kgm^‒3

Density under applied pressure = ρ + dρ

 = 11.4×10³ + 0.285×10³

= 11.685×10³ Kg m^‒3

Density of lead under a pressure of 2×10⁸ Nm^‒2 = 11.685×10³ Kgm³

5. The Modulus of rigidity and poisson's ratio of the material of a wire are 2.87×10¹⁰ Nm^‒2 and 0.379 respectively. Find the value of young's modulus of the material of the wire.

Solution

We have

 σ = (Y/2n) ‒ 1

Y/2n = 1+ σ

Y = 2n(1+ σ)

= 2×2.87×10¹⁰ × (1+0.379)

= 7.915×10¹⁰ Nm^‒2

Youngs modulus of the wire = 7.915×10¹⁰ Nm^‒2

6) Calculate the Poisson's ratio for the material, given Y=12.25×10¹⁰ Nm^‒2 and n=4.55 × 10¹⁰ Nm^‒2

Solution

We have

σ = Y/2n ‒ 1

σ = ( 12.25×10¹⁰  / 2(4.55×10¹⁰) ) ‒ 1

σ = 1.34615‒1

σ = 0.34615

Poisson's ratio σ = 0.34615

7. Calculate the Young's modulus in the cantilever depression method used. The length of cantilever beam is 1m which is suspended with a load of 150 gm. The depression is found to be 4 cm. The thickness of the beam is 5 mm and breadth of the beam is 3 cm.

Solution

= [ 4×9.8×1³ × 150×10^‒3 ] / [ 3×10^‒2 × (5×10^‒3)³ × 4×10^‒2 ]

= [ 5.88 / 1.5 × 10^‒10  ] × 1

Youngs modulus = 3.92 × 10¹⁰ Nm^‒2

ASSIGNMENT PROBLEMS

1. A steel rod of length 100 cm is fixed at one end and 5Kg load is suspended at the other end. Calculate the depression if the Young's Modulus is 8×10¹⁰ Pascals. Diameter is 1.0cm. (Ans. 4.16cm)

2. A bar, one metre long with square cross section (side of square being 5mm) is supported horizontally at its ends and is loaded at the middle point. It is depressed by 1.96 mm by a load of 100gm. Calculate the Young's Modulus of the material of the bar. (Ans.200 GPa)

3. A steel rod of length 5 m is fixed rigidly between two supports. The coefficient of linear expansion of steel = 12×10^‒6/°C. Calculate the stress in the rod for an increase in temperature of 40°C. The Young's modulus of elasticity of steel 2×10¹¹Nm^‒2 (Ans:Stress=9.4×10⁷Nm^‒2)

4. Calculate the Poisson's ratio for the material, given Y=12.25 × 10¹⁰Nm^‒2 and n=4.55×10¹⁰Nm^‒2 (Ans: 0.347)

5. Calculate the Young's modulus of a material, whose bulk modulus K=2.5×10¹⁰Nm^‒2 and rigidity modulus n=4.5×10¹⁰Nm^‒2 (Ans: Y=8.4375×10¹⁰Nm^‒2)

6. Calculate Poisson's ratio of a material. Given rigidity modulus n=3.9×10¹⁰Nm^‒2 and Bulk modulus K=7.9×10¹⁰Nm^‒2 (Ans: 0.288)

7. In an experiment, a bar of length 1.5m is clamped horizontally at one end and a load of 0.1 k attached at its free end. Calculate the depression at the loaded end if Y=9.78 x 10¹⁰Nm^‒2 and the bar is of breadth 0.024m and thickness 0.005m. (Ans: 0.0451m)