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