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ValentinkaMS [17]
2 years ago
8

What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A

. individual footings B. strip footings C. mat foundations D. end-bearing piles E. friction piles
Engineering
1 answer:
ikadub [295]2 years ago
8 0

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

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A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is
Talja [164]

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × \pi × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × \pi × 0.2 × 1500 ( 4.5³ - 3.5³ )

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3 0
3 years ago
The _______ is a function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different
scoundrel [369]

Answer:

B: Directional Transfer Function

Explanation:

The function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different frequencies that arrive at each ear from different locations in space is called Directional Transfer Function.

8 0
1 year ago
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

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Pressure = Stress* Area

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Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 =  370.758 + 16^2

d^2 =   626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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