Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
Answer:
Jesus is always the answer
Answer:
14.36((14MPa) approximately
Explanation:
In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.
Please check attachment for complete solution and step by step explanation
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
