Answer:
550kN
Explanation:
We are required to find the maximum allowable load that will satisfy geotechnical ULS requirements
P' = pu/FOS ---1
Pu = design load
FOS = factor of safety
We get the design of safety
Pu = 0.75p
= 0.75 x 2200
= 1650kN
Factor of safety = 3
Design load = 1650
P' = 1650/3
= 550kN
So we have the maximum allowable load to be 550kN