A wood column is 5.5 in. square in cross section and 12 ft long. Its modulus of elasticity is 1.3 × 106 psi. How much will the column shorten under a compressive load of 36,000 lb?
delta L/L = F/area / E
delta L = 144 in * 36*10^3 lb /5.5 in^2/1.3*10^6
= 725 * 10^-3
= .724 inches
(scary! probably fails under buckling but we were not asked) )
Thank you so much!
To calculate the amount the wood column will shorten under a compressive load, we need to use Hooke's Law, which states that the change in length of an object is directly proportional to the force applied to it.
Hooke's Law can be expressed as:
ΔL = (F * L) / (A * E)
where:
ΔL is the change in length
F is the compressive force applied (in pounds)
L is the original length of the column (in feet)
A is the cross-sectional area of the column (in square inches)
E is the modulus of elasticity of the material (in psi)
Now, let's calculate the values needed to plug into the equation:
Given:
F = 36,000 lb (compressive force applied)
L = 12 ft (original length of the column)
A = (5.5 in)^2 (cross-sectional area of the column)
E = 1.3 × 10^6 psi (modulus of elasticity)
First, let's convert the length from feet to inches:
L = 12 ft * 12 in/ft = 144 in
Next, let's calculate the cross-sectional area of the column:
A = (5.5 in)^2 = 30.25 in^2
Now, we can plug these values into the equation:
ΔL = (36,000 lb * 144 in) / (30.25 in^2 * 1.3 × 10^6 psi)
Calculating this expression will give us the change in length of the wood column under the given compressive load.