A four-lane northbound section of interstate has rigid pavement and was designed with an 8-inch slab, 90% reliability, a 700 lb/
in2 concrete modulus of rupture, a 5 million lb/in2 modulus of elasticity, a 3.0 load transfer coefficient, and an overall standard deviation of 0.3. The initial PSI is 4.6 and the TSI is 2.5. The pavement was conservatively designed (assuming the upper limit of the W18 design lane load) to last 20 years, and the CBR is 25 with a drainage coefficient of 1.0. A design mistake was made that ignored 1000 total northbound (daily) passes of trucks with 22-kip single and 30-kip tandem axles. What slab thickness should have been used?
1 answer:
You might be interested in
Answer:
1113kN
Explanation:
The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:
Inside Diameter = Outside Diameter - Thickness
Inside Diameter = 61cm - 0.9cm = 60.1cm
Converting this diameter to meters, we have:
This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:
The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:
Now, water density is given by the equation , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:
With the water mass we can find the weight of water:
Finally we find the total weight add up the weight of the water and the weight of the pipe,so:
Converting this total weight to kN, we have:
