A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a ma
ximum load of 250 kips, and the planned column support will consist of a 6 foot by 6 foot square footing placed 2 feet below the ground surface (to the bottom of the footing). Subsurface conditions consist of a 15-foot-thick layer of uniform silty sand (unit weight = 122 pcf, soil modulus = 160,000 psf) over stiff clay (unit weight = 118 pcf, soil modulus = 230,000 psf). Groundwater is deep.
a) Sketch the problem (freehand, not to scale) and state any necessary assumptions.
b) Calculate the immediate (elastic) settlement.
a) Sketch the problem (freehand, not to scale) and state any necessary assumptions.
b) Calculate the immediate (elastic) settlement.
1 answer:
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Answer: I am not for sure
Explanation:
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
where;
fracture toughness K = 137 MPa
geometry factor Y = 1
applied stress = ???
crack length a = 2mm = 0.002
∴
Now, the tensile impact obtained is:
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN
Answer:
Total time taken = 0.769 hour
Explanation:
using the velocity method
for sheet flow ;
Tt =
Tt = travel time
n = manning CaH
Pl = 25years
L = how length ( ft )
s = slope
For Location ( 1 )
s = 0.045
L = 1000 ft
n = 0.06 ( from manning's coefficient table )
Tt1 = 0.128 hour
For Location ( 2 )
s = 2.5 %
L= 750
n = 0.13
Tt2 = 0.239 hour
For Location ( 3 )
s = 1.5%
L = 500 ft
n = 0.15
Tt3 = 0.237 hour
For Location (4)
s = 0.5 %
L = 250 ft
n = 0.011
Tt4 = 0.165 hour
hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4
= 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour