4–15. The post is made of Douglas fir and has a diameter of 100 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 4 kN>m at y = 0 to w = 12 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.

The image of the Douglas fir post is missing, so i have attached it.

-0.00025918 mm

Explanation:

We are given the points it passes through as;

(y1, w1) = (0, 4)

(y2, w2) = (2, 12)

Thus, the slope is;

K = (w2 - w1)/(y2 - y1)

K = (12 - 4)/(2 - 0)

K = 8/2

K = 4

Now, substituting one of the points into the slope equation, we have;

w - w1 = k(y - y1)

w - 4 = 4(y - 0)

w = 4y + 4

For us to now find the total force the distributed load applies to the body, we will integrate to give;

W(y) = ∫w•dy

Thus;

W(y) = ∫(4y + 4)dy

W(y) = 2y² + 4y

So, the value of this force which is at y = 2 m will be;

W(2) = 2(2²) + 4(2)

W(2) = 16 KN

Now, at equilibrium, we sum up the forces in the vertical y-direction to give;

W - 20 + F = 0

F = 20 - W

F = 20 - 16

F = 4 KN

If we cut through the body to get a free body diagram and sum forces in y-direction to zero, the displacement of the top of the post A with respect to its bottom B is;

δ_B/A = [(L, 0∫)(W - 20)dy]/AE

Where A is πr²

We are given diameter = 100 mm

So, radius = 100/2 = 50 mm = 0.05 m

Also, from tables the Young's modulus of Douglas fir is 13.1 GPa or 13.1 × 10^(9) Pa

Thus;

δ_B/A = [(2, 0∫)(2y² + 4y - 20)dy]/(π × 0.05² × 13.1 × 10^(9))

δ_B/A = [((2 × 2³/3) + (2 × 2²) - 20(2)) - 0] /(0.102887 × 10^(9))

δ_B/A = (16/3 + 8 - 40)/(0.102887 × 10^(9))

δ_B/A = -0.00000025918 m = -0.00025918 mm