Answer:
t=2.025 inches
Explanation:
Given that
P = 400 Psi
Yield stress ,σ = 80 ksi
Diameter ,d= 45 ft
We know that
1 ft = 12 inches
d= 540 inches
Factor of safety ,K= 3
The required thickness given as

t=thickness


t=2.025 inches
Therefore thickness will be 2.025 inches.
Answer: The total vehicle delay is
39sec/veh
Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.
Effective red time (r) = 25sec
Arrival rate (A) = 900veh/h = 0.25veh/sec
Departure rate (D) = 1800veh/h = 0.5veh/sec
STEP1: FIND THE TRAFFIC INTENSITY (p)
p = A ÷ D
p = 0.25 ÷ 0.5 = 0.5
STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE
The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.
Dt = (A × r^2) ÷ 2(1 - p)
Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)
Dt = 156.25 ÷ 4 = 39.0625
Therefore the total vehicle delay after one cycle is;
Dt = 39
Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are 
The two long are 
The two shors are 
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
![Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3](https://tex.z-dn.net/?f=Q_c%20%3D%20%5B%282%2A122.08%29%2B%282%2A86-375%29%5D%2A14%2A0.833%5C%5CQ_c%3D4864.02ft%5E3)
That in cubic yards is equal to 
Hence, we need order 5% plus that represent with the quantity

Answer:
They don't make a positive change in the world because they have made mistakes that aren't able to be made fixed and there are a lot of engineers who haven't study enough and know the important basis of coming to engineer.