Answer:
stress = 50MPa
Explanation:
given data:
Length of strain guage is 5mm
displacement
stress due to displacement in structural steel can be determined by using following relation
where E is young's modulus of elasticity
E for steel is 200 GPa
stress = 50MPa
Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
Answer:
Explanation:
To solve this problem we use the expression for the temperature film
Then, we have to compute the Reynolds number
Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number
but we also now that
but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is
In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
Answer and Explanation:
O decreases linearly with the distance from the generator