What is the required moment of inertia (Iₓ) for a simply supported beam 20' long with a uniformly distributed total service load of 800 plf throughout the full length of the beam, assuming A992 steel and a W-flange member?
Calculation for Required Moment of Inertia
To determine the required moment of inertia (Iₓ) for a simply supported beam with a uniformly distributed load, we will use structural engineering formulas. First, we calculate the maximum bending moment (M) using the formula:
M = wL² / 8
Where:
M = Maximum bending moment
w = Uniform load per unit length
L = Length of the beam
For the given scenario:
w = 800 plf (pounds per linear foot)
L = 20 feet
M = 800 plf * (20 ft)² / 8
M = 40000 ft-lb
Next, we convert this moment to inch-pounds:
M = 40000 ft-lb * 12 in/ft
M = 480000 in-lb
Using the flexural formula:
σ = M / S
Where:
σ = Allowable stress of A992 steel
S = Section modulus
Assuming the bending stress for A992 steel is 36 ksi with a factor of safety of 0.90:
σ = 36 ksi * 0.90
σ = 32.4 ksi
Converting ksi to psi:
σ = 32.4 ksi * 1000 psi/ksi
σ = 32400 psi
Calculate S:
S = 480000 in-lb / 32400 psi
S = 14.81 in³
To find the required moment of inertia:
Iₓ = S * c
Where c is the distance from the neutral axis to the outermost fiber. For a W-flange beam, assume c = 10 inches:
Iₓ = 14.81 in³ * 10 in
Iₓ = 148.1 in⁴
None of the provided choices match with the calculated value. It is crucial to verify the assumptions made about the factor of safety and beam dimensions for specific applications.