ELASTICITY (TUTORIAL)

1. An iron rod 4.00m long 0.500cm in cross section strenches 1.00 mm when a mass of 225kg is hung from its lower end. Compute young’s modulus for the iron.


L.= 4.00 m
y= FL.
A∆L

225x9.81 = 2207.25 N
∆L.= 1x10-³m

F= mg
F= 225x9.81
= 2207.25 N

Y= 2207.25 (4.00)
[0.5x(10-²)² m²] [1x10-³] m
= 1.7658 x10¹¹
= 176.58 x10¹¹x10-²
= 176 x10⁹ Pa
= 176 GPa


2. A load of 50 kg is applied to the lower end of a steel 80cm long and 0.6cm in diameter. How much will the road stretch? Y= 190GPa for steel.

m = 50kg x 9.81
d = 0.6cm ÷ 2
= 0.3
J² 3X10-³ m

A = πJ²
=π (3x10-³)²
=2.83 x 10-⁵

L. = 80cm
=80x10-²
=0.8
Y = 190 GPa

AL. = FL.
YA
=(490.5) (0.8)
(190x10⁹) (2.83x10⁵)
= 392.4
5.377x10⁸
=72.97x10-⁷
=73x10-⁶
=73µm


3. A platform is suspended by four wires at its cover. The wires are 3.0m long and have a diameter of 2.0mm. Young’s modulus for the material is 180GPa. Howfar will the platform drop ( due to elongation of the wires) if a 50kg load is placed at the center of platform?

Y= 180 GPa
=180x10² Pa
D= 2mm ÷ 2
=1x10-³ m

Y= FL. AL. = FL.
A∆L YA

L. = 3m F= 50x9.81
=490.5 N
A= πJ² = (3.14x10-⁶) m
AL. = 490.5x3m²
(180) (3.14x10-⁶)
=1471.5 Nm
5.652x10-⁴
=2.6x10⁶ m
=2.6
4
=0.65 mm


4. Two parallel and apposite forces, each 4000N, are applied tangentially to the upper and lower faces of a cubical metal black 25cm on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus for the metal is 80GPa.

Tan Q =(25×10 ̄ 7/ 25× 10 ̄ 2) = 8.0 × 10 ̄ 7 rad.
Displament = 4000 / (2.8 × 10 ̄ ²)²
80 Gpa = 64000 / Ʃ
Ʃ = 64000 / 80 G
= 80 × 10 ̄ 7 (25 × 10 ̄ ² )
= 2 × 10 ̄ 7.


5. The bulk modulus of water is 2.1 GPa. Compute the volume control contraction of 100mL of water when subjected to a pressure of 1.5 MPa.

D = 2.1 Gpa
Vᴏ = 100 ml / 1000 =0.1 L
A = 1.5 mpa
AV = ( Aᴏ) ( Vᴏ) / B = ( 15 mpa ) ( 0.1 L) / 2.1 Gpa
= 15 000 /2.1 71.43 × 10 ³ = 0.0714 ml.


6. The compressibility of water is 5.0 x 10 . find the decreas in volume of 100mL of water when subjected to a pressure of 15 mPa.

Bulk strains = ∆Vᴏ / Vᴏ
AVᴏ = B × Nᴏ
= 15 × 10 6pa ( 15.0 × 10 ̄ 1 NM² )
= 7.5 × 10 ̄ ³
= 0.75 ml