Answer:
Total BF for the interior wall is 7.50BD
Explanation:
Given Data:
· Size of stud = 2” x 6”
· Height of Wall = 8 ft
· Top plates = 2
· Bottom Plate = 1
BF stands for board feet in lumber/wood terminology. It is the unit of volume.
1 BF (Board feet) = 1 ft x 1 ft x 1 inch
Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.
Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft
Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD
Total BF for the interior wall is 7.50BD
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:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
For pipe 2
Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV
head loss (h)
Now putting the all values
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
