Answer:
45.873 KN/mL is the ultimate bearing capacity of the foundation.
Explanation:
C' = 10.0 kN/m², Ф' = 25, r = 18 kN/m²
ration of Df/B ≤ 1 , Df ≤ B (1 ≤ 2.5m)
It is a shallow foundation
∴ Ultimate bearing Capacity
qu = C' + r. + Br
NФ = tan²(45 + Ф/2)
= tan²(45 - 25/2)
= 0.41
Nq = NФ×e^(πtanФ') = 0.41e^(πtan25)
= 1.756
Nr = 1.8tanФ(Nq - 1) = 1.8tan25(1.756 - 1)
= 0.634
Nc = 1.621
putting the above values in equation.
qu = 45.873 KN/mL
Answer:
minimum sight distance = 699 ft
Explanation:
given data
road lane = 4 divided road
median width = 12 ft
grade road = 5%
solution
we take here time gap factor for minor road vehicle when enter to major road from table
time gap = 8.1 sec
and for median width of 12 ft
time gap = 8.2 + 0.7 ( 1 + )
time gap = 9.5 second
so minimum sight distance will be
minimum sight distance = 1.47 × design speed × time gap
minimum sight distance = 1.47 × 50 × 9.5
minimum sight distance = 699 ft
The answer would be letter A
Complete Question
The complete question is shown on the first uploaded image
Answer:
a)
b)
Explanation:
The explanation is shown on the second and third uploaded image