The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr
1 answer:
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Answer:
γ =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
Explanation:
<u><em>General Considerations</em></u>
The design of the yard will affect the natural surface and subsurface drainage pattern of a watershed or individual hillslope. Yard drainage design has as its basic objective the reduction or elimination of energy generated by flowing water. The destructive power of flowing water increases exponentially as its velocity increases. Therefore, water must not be allowed to develop sufficient volume or velocity so as to cause excessive wear along ditches, below culverts, or along exposed running surfaces, cuts, or fills.
A yard drainage system must satisfy two main criteria if it is to be effective throughout its design life:
1. It must allow for a minimum of disturbance of the natural drainage pattern.
2.It must drain surface and subsurface water away from the roadway and dissipate it in a way that prevents excessive collection of water in unstable areas and subsequent downstream erosion
The diagram below ilustrate diffrent sturcture of yard to be consider before planing to utiliza rainwater
