6 feet
Explanation: A few extra feet of free fall can significantly increase the arresting force on the employee, possibly to the point of causing injury. Because of this, the free fall distance should be kept at a minimum, and, as required by the standard, in no case greater than 6 feet (1.8 m).
Answer:
b)false
Explanation:
Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.
We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.
So given statement is wrong.
I think it’s manufacturing
This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ = 
/ t₂
t₂ = t₁
t₂ = t₁ /
t₂ = ( / )t₁
t₂ = 
/ 
× t₁
so we substitute
t₂ = 0.0049 / 0.0018 × 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs
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
