Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 =
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/() =
m^2
recall that area =
=
=
=
=
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
The radius of the specimen is 60 mm
<u>Explanation:</u>
Given-
Length, L = 60 mm
Elongated length, l = 10.8 mm
Load, F = 50,000 N
radius, r = ?
We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this
elongation using Equation:
ε = Δl / l₀
ε = 10.8 / 60
ε = 0.18
We know,
σ = F / A
Where A = πr²
According to the stress-strain curve of brass alloy,
σ = 440 MPa
Thus,
Therefore, the radius of the specimen is 60 mm
The answer & explanation for this question is given in the attachment below.
Answer:
-4.5 m/s
Explanation:
Assuming steady and incompressible flow and uniform properties at each section
Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections
At the surface, is negative hence the equation above will be
Making the subject of the formula then
Substituting the given values then