Mechanics of solids:Axial load and mechanical properties of materials

MECHANICS OF SOLIDS-SOLVED NUMERICALS

Numericals on - Axial load and mechanical properties of materials

Q-1 Determine the dilatation (bulk modulus) e and the change in volume of the 200 mm length of the rod shown if (a) the rod is made of steel with E = 200 GPa and v = 0.30, (b) the rod is made of aluminum with E = 70 GPa and v = 0.35.

Sol 1:

a). ANS:

WE HAVE :

E = 200 GPa = 2 x 10^⁵ N/m㎡

L = 200 mm

u = 1/m = 0.3

Stress, σ =P/A = 46 x 10^³ x 4 / 3.14 x (22)^²

= 121.01 N/m㎡

Also, strain (e) = stress (σ) / E

= 121.01 / 2 x10^⁵

e = 6.05 x 10^^-4

Also, Bulk modulus = K= E / 3(1-2/m)

= 2 x10^⁵ / 3(1 – 0.6)

= 2 x 10^⁵ / 1.2 = 1.6 x 10^⁵ N/m㎡

Also , dl = l x e = 0.121

Then, Strain of dia. = 0.121 / 200 = 0.000605

d d = 0.000605 x 22

= 0.01331.

Then , dv = ( d^² x dl + 2 l d d d) π/4

= {( 22)² x 0.121 + 2 x 200 x 22 x 0.01331) π/4}

= 551.95 / 4 = 137. 988 m㎥

b). ans:

E = 70 GPa = 70 x 10⁹ x10^-6 N/m㎡

= 70 x 10³ N/m㎡

L = 200 mm

u = 1/m = 0.35

Stress, σ =P/A = 46 x 10³ x 4 / 3.14 x (22)²

= 121.01 N/m㎡

Also, strain (e) = stress (σ) / E

= 121.01 / 70 x10^³

e = 1.72 x 10^^-3

also, bulk modulus = K= E / 3(1-2/m)

= E/ 3(1 – 0.7)

= E/ 0.9

= 7 x 10⁴

= 77777.77 = 7.77 x 10^⁴ N/m㎡

Also , dl = l x e = 200 x 1.72 x 10^^-3 = 0.334 mm

Then, Strain of dia. = 0.334 / 200 = 1.72 x 10^^-3

d d = 1.72 x 10^^-3 x 22

= 0.03784.

Then , dv = ( d^² x dl + 2 l d d d) π/4

= 1046.1 / 4 = 261.525 m㎥

Q-2 A standard tension test is used to determine the properties of an experimental plastic. The test specimen is a 15-mm-diameter rod and it is subjected to a 3.5 kN tensile force. Knowing that an elongation of 11 mm and a decrease in diameter of 0.62 mm are observed in a 120-mm gage length, determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio of the material.

Sol2:

D=15mm

P=3.5KN=3.5 x 10^³N

δl =11mm

Change in diametr=0.62 mm

Gauge length=120 mm

To find : E=?

C=?

POISSON’S RATIO=?

We know that,

σ =P/A

σ =3.5X10^³x 4/3.14 x 225

σ =19.81 N/m㎡

Longitudinal strain = e =δl/l

e =11/120

e =.091

E=σ/e

E=19.81/.091

E=217.62 N/m㎡

Lateral strain= change in lateral dimension / original lateral dimension

= 0.62/15

= 0.041

Poisson’s ratio =1/m=lateral strain/longitudional strain

=.041/.091

=0.45

Let C be the modulus of rigidity

E=2C(1+1/m)

217.32=2C(1+0.45)

Then , C = 108.66/ 1.45 = 74.93 N/m㎡

Q-3 Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when P = 25 kN, (b) the corresponding strain-energy density of portions AB and BC of the rod.

Sol 3:

A_ab= 3.14/4 x (20)^² = 314.16 m㎡ = 3.14 x10^^-6 m.

A_bc= .14/4 x (16)^² = 201.06 m㎡ = 201.06 x10^^-6 m.

Also, p = 25 x 10^³ N.

U = E P^²L/2EA , here E = sigma

= (25 x 10^³)^² x 1.2 / 2 x (200 x 10^⁹ ) x 3.14 x10^^-6

.+ (25 x 10^³)^² x 0.8 / 2 x (200 x 10^⁹) x 201.06 x10^^-6

a). U = 5.968 + 6.213 = 12.18 Nm = 12.18 J.

b). STRESS_ab= 25 x 10^³/ 3.14 x10^^-6 = 79.58 x 10^⁶ Pa.

U_ab = (STRESS_ab)^² /2E = (79.58 x 10^⁶)^² / 2 x 200 x 10^⁹.

= 15.83 x 10^³ = 15.83 kJ/㎥.

c). STRESS_bc= 25 x 10^³/ 201.06 x 10^^-6 = 124.28x 10^⁶ Pa.

U_bc= (STRESS_bc)^² /2E = (124.28 x 10^⁶)^² / 2 x 200 x 10^⁹.

= 38.6 x 10^³ = 38.6 kJ/㎥.

Q-4 Two bars of length L and of same material are subjected to the axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for a length L/3 and a diameter 2d for the remaining length. Compare strain energies of two bars?

Sol 4: For first bar:

Diameter = 2d

Stress = p /3.14 x (2d)^².

= p/ 3.14 x (d)^²

Here, π = 3.14

Now, Strain energy stored = (stress)^²/ 2E x volume of bar.

= 1/ 2E x (p)^²/ (3.14)^² x (d)^⁴ x 3.14(2d)^² l/4

= (p)^²l/ 2E x 3.14(d)^².

For second bar:

Then, stress in the part AB = p/ 3.14/4 x (d)^²

= 4p/ 3.14 x d^².

Also, stress in the part BC = p/ 3.14/4 x (2d)^²

= p/ 3.14 x d^².

Now, strain energy stored = strain energy stored by AB + strain energy stored BC.

= (stress₁)^²/ 2E x volume of AB + (stress₂)^²/ 2E x volume of BC.

= 16p^²/(3.14)^² (d)^⁴ x l/2E x 3.14 d^²/4 x l/3

+ p^²/(3.14)^² (d)^⁴ x l/2E x 3.14 (2d)^²/4 x2 l/3

= 2/3 x(3.14)^² l /2E x 3.14 d^² + 1/3 x(3.14)^² l /2E x 3.14 d^²

= (3.14)² l / E x 3.14 d^².

Therefore, Ratio of strain energies of two bars = p^² l / 2E 3.14 d^² x E 3.14 / p^² l

= 1/2

Q-5 A steel specimen 1.5 c㎡ in cross- section, stretches 0.005 cm over a 5 cm gauge length under an axial load of 30 KN. Calculate strain energy stored in the specimen at this point. If the load at the elastic limit for the specimen is 50 KN, calculate the elongation at elastic limit and proof resilience.

Sol5 :

A = 1.5 c㎡ = 150 m㎡.

dl = 0.005 cm = 0.05 mm.

l = 5 cm = 50 mm

p = 30 kN= 30 x 1000 = 30000 N

Load at elastic limit = 50 KN = 50000 N

Now, Strain energy stored = work done = 1/2 p (dl)

= 1/2 x 30000 x 0.005 = 750 Nmm

= 0.75Nm

We have , elongation due to 30 kN = 0.05 mm

Then , elongation due to 50 kN = 50/30 (0.005) = 10/ 6 (0.005)

= 5/ 3 (0.005)

= 1/12 mm = elongation at elastic limit.

Proof resilence = work done = 1/2 x (load at elastic limit) x ( extension )

= 1/2 x 50000 x 1/12

= 2083.3 Nmm.

= 2.083 Nm

= 2.083 J

Q-6 Abar is 3 m long and 60 mm diameter. It is subjected to a tensile load of 200 kN. Find the stress and the elongation when the load is applied gradually. What would be the maximum stress and the maximum elongation if the load has been suddenly applied? Take E_s= 2x10^⁵ N/m㎡.

Sol6:

GIVEN:

P=200KN=200 x 10^³N

L=3m=3 x 10^³mm

D=60mm

E_s=2 x 10^⁵N/mm²

To find:

GRADUALLY STRESS?

ELONGATION?

We know that:

Gradually stress =σ= P/A

σ =200x10^³x4/3.14x60x60