The answer is choice C
Explanation:
As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .
So it will be useful for the construction crews to connect the pipes to the sewer lines before the foundation is poured.
But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,
After this process is over, the city inspector comes to check whether the foundation has been laid down as per the code of construction.
Only after that the rest of the construction activity follows through.
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:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
Answer:
Hello Adam here! (UWU)
Explanation:
The advantages of 3D modeling for designers is not limited to productivity and coordination, it is an excellent communication tool for both the designer and end user. 3D models can help spark important conversations during the design phase and potentially avoid costly construction mishaps.
Happy to Help! (>.O)
This question is incomplete, the complete question is;
Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.
Use Fy=50 ksi and assume Cb=1.0 (if needed).
Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft
Explanation:
Given that;
section W 21 x 73 steel beam;
now from the steel table table for this section;
Zx = Sx = 151 in³
also given that; fy = 50 ksi and Cb = 1.0
QMn = 0.9 × Fy × Zx
so we substitute
QMn = 0.9 × 50 × 151
QMn = 6795 k-inch
we know that;
12inch equals 1 foot
so
QMn = 6795 k-inch / 12
QMn = 566.25 f-ft
Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft
