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never [62]
3 years ago
7

The post is made of Douglas fir and has a diameter of 100 mm. if it is subjected to the load of20 kN and the soil provides a fri

ctional resistance that is distributed along its length and varies linearly from IV= 4 kN/m at y = 0 to IV =12 kN/m at y = 2 m. determine the force F at its bottom needed for equilibrium. Also, what is the displacement or the top of the post A with respect to its bottom B? Neglect the weight of the post.

Engineering
1 answer:
Solnce55 [7]3 years ago
5 0

Answer:

Explanation:

Kindly go through the attachment for the step by step approach to the solution of this question.

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Hãy trình bày các bộ phận chính trong một bộ điều khiển điện tử (ECU) dùng trên ô tô. Cho biết công dụng của từng thành phần.
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3 years ago
Assume you have four fins, each with a mass of 8.0 grams. What is the total weight of these four fins? (Hint: watch your units!)
soldier1979 [14.2K]
The answer is 32.0 grams
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3 years ago
B)
Triss [41]

Answer:

2.5 is the required details

8 0
3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
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