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?

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Dorothea · Answer

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.

Calculating Required Moment of Inertia for Simply Supported Beam

Calculation for Required Moment of Inertia

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