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BabaBlast [244]
3 years ago
6

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.

Engineering
1 answer:
ehidna [41]3 years ago
5 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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What is the next measurement after 2' -6" on the architect's scale?
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3 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

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:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a
Vaselesa [24]

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4}  }  

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

5 0
3 years ago
4. The outer end of a control arm is attached to the steering knuckle through a
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