TUTORIAL

ELASTICITY


1.An iron rod 4.00m long and 0.500cm²in cross section stretches 1.00mm when amass of 225 Kg is hung from its lower end.compute young’s modulus for the iron.

σ = F/A = (225 kg) (9.8m/s²) / (0.500 X 10⁻⁴ m²)
=4.41×10⁷ Pa
ε = ∆s/S∘
= 1.00 x 10 ⁻³m / 4.00 m
=2.5×10⁻⁴
Y= stress/strain
= σ/ε
=4.41x10⁷ / 2.5x10⁻⁴
=1.764×10¹¹
=176 Gpa


2. A load of 50 kg is applied to the lower end of steel rod 80 cm long and 0.60 cm in diameter, How much will the rod strectch?

m=50 kg×9.81 j²=3×10⁻³m ℓ∘ = 80cm
d=0.6cm÷2 A=πj² = 80×10⁻²
=0.3 =πj²(3×10⁻³)² = 0.8
=2.83×10⁻⁵m y=190kPa
Aℓo=Fℓo / Y∆
= (4.90 x 0.8)/(190x 10⁹)(2.83x 10⁵)
=392.4 / 5.377 x 10⁸
=72.97×10⁻⁷
=73μm


3. A platform is suspended by four wires at its corner. The wires are 3.0m long and lave diameter of 2.0mm. Young’s modulus for material is 180GPa.How far will the platform drop (due to elongation of the wires)if a50kg load is placed at the center of the platform?

Y=180GPa
=180×10⁹ Pa
D = 2mm÷2
=1×10×⁻³
Y = Fℓo / YA
ℓo=3m A=πj²
F=50×9.81 = (3.14×10⁻⁵)m
=490.5N
Aℓo=490.5N (3m²) / (180)(3.14 x 10 ˉ⁶)
=1471.5Nm/5.652x 10⁻⁴
=2.6×10⁶m
=2.6/4
=0.65mm

4. Two parallel and opposite forces,each 4000N,aree aplied tangentialy to the upper and lower faces of a cubical metal block 25cm on a side.Find the angel of shear and the displament of the upper surface relative to the lower surface.The shear modulus for the metl is 80GPa.

tanθ=2 x 10ˉ⁷ / 25 x 10⁻²
= 8.0×10⁻⁷ rad
Displament=4000 / (2.5 x 10ˉ²)
80GPa = 64000/∑
∑=64000/80G
=8×10⁻⁷(25×10⁻²)
=2×10⁻⁷

5.The bolk moduls of water is 2.1 Gpa. Compute the volume comtraction of 100ml of water when subjected to a pressure of 1.5 Mpa.

B= 2.1GPa
Vo= 100mL / 1000
=0.1 mL
∆p = 1.5Mpa
∆v = (∆p)(Vo) / B
=(1.5 Mpa)(0.1mL) / 2.1 GPa)
=150000 / 2.1GPa
= -71.43×10³
= -0.0714mL

6. The compressibility of water is 5.0×10⁻¹⁰m²/N,find the decrease in volume of 100ml of water subjected to a pressure of 15mPa.

Bulk strains = ∆Vo / Vo
AVo=B×VoA
=15×10⁶Pa(5.0×10⁻¹Nm⁻²)
=7.5×10⁻³
=0.75mL