Tuesday, January 12, 2010

Elasticity.

tutorial.

1. An iron rod 4.00m long an 5.00cm^2 in cross section stretches 1.00mm. When a mass of 22.5kg is hung from its lower end. Compute Young’s Modulus from the iron.

Solution :

σ= F/A
= (225kg) (9.8m/s^2) / (0.500 x 10^ -4)
= 4.41 x 10^7 Pa

ε = Δƪ / ƪ˳
= (1.00 x 10^ -3 m) / (4.00m)
= 2.5 x 10^ -4

Y = σ / ε
= (4.41 x 10^7 Pa) / (2.5 x 10^ -4)
= 1.764 x 10^11
= 176 GPa

2. A load of 50 kg is applied to the lower end of a steel rod 8cm long and 0.60cm in diameter. How much will the rod sterch Y = 190 GPa for stell?

Solution :

Formula : Y = F ƪ˳ / AΔ ƪ A ƪ = F ƪ˳ / Y. A

m = 50 kg x 9.81
d = 0.6 cm / 2
= 0.3
j^2 = 3 x 10^ -3m
A = п j^2
= п (3 x 10^ -3)^2
= 2.83 x 10^ -5

ƪ˳ = 80 cm
= 80 x 10^ -2
= 0.8
Y = 190 GPa
Aƪ˳ = F ƪ˳ / YA
= (490.5) (0.8) / (190 x10^9) (2.83 x 10^5)
= 392.4 / 5.377 x 10^8
= 72.97 x 10^ -7
= 73 x 10^ -6
= 73μm

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

Solution :

ƪ˳ = 3m
F = 50 x 9.81
= 490.5N
A = пj^2
= (3.14 x 10^ -6)m
A ƪ˳ = 490.5Nm / (5.652 x 10^ -4)
= 2.6 x 10^6
= 0.65m

4. Two parallel and opposite force. Each 400N are applied tangentially to the upper and lower forces of a cubical matel block 25cm on a side. Fine the angle of shear and the displacement of the upper surface relative to the surface. The shear modulus for the metal is 80GPa.

Solution :
F = 400N
h = 25cm
A = 0.0625m^2
S = 80 x 10^4 p
tanθ = (2 x 10^ -7) / (25 x 106 -2)
= 8.0 x 10^ -7 rad
x = FAh / SA
x = (4000) (0.25 x 10^ -3) / (8.0 x 10^6Pa) (0.0625)
= 2 x 10^ -7m

5. The bulk modulus of water is 2.1GPa. Compute the volume contraction of 100ml of water when subjected to a pressure of 1.5mPa?

Solution :
Av˳ = - Pv˳ / B
B = - Pv˳ / Δ v˳
B = 2.1GPa
P = 1.5MPa
= 1.5 x 10^6Pa
v˳ = (100ml) / (1000)
Δ v˳ = - Pv˳ / B
= (1.5 x 10^6) (0.1) / (2.1 x 10^9)Pa
= (-7.14 x 10^ -5mm) x (1000)
= -0.0714ml

6. The compressibility of water is 5.0 x 10^ -10m^2/N. Find the decrease in volume of water of 100ml when subjected to a pressure of15mPa?

Solution :

A v˳ = B x v˳
= (15 x 10^6) (5.0 x 10^ -10Nm^ -2)
= 7.5 x 10^ -3) x(100)
= 0.75ml

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