A simply-supported W14x43 beam (A36 steel) spans 24 ft and supports a 15 kip point load at midspan and a 54 k-ft clockwise \”point couple\” (applied moment) at Roller A. Determine the deflection of the beam at midspan.
The Correct Answer and Explanation is:
Correct Answer:
The deflection of the beam at midspan is 0.331 inches downward.
Explanation
To determine the total deflection at the midspan of the beam, we can use the principle of superposition. This engineering principle allows us to analyze a complex loading scenario by breaking it down into simpler, individual loading cases. We can calculate the deflection caused by each load separately and then algebraically add the results to find the total deflection. The two loading cases are:
- A 15 kip point load (P) applied at the midspan of the simply supported beam.
- A 54 k-ft clockwise moment (M₀) applied at the left support (Roller A).
First, we must gather the necessary properties for the calculation.
- Beam Section: W14x43. From a standard steel properties table, the moment of inertia (I) about the strong axis is 428 in⁴.
- Material: A36 steel. The modulus of elasticity (E) for steel is 29,000 ksi (kips per square inch).
- Span Length (L): 24 ft.
For consistency, all units must be converted to kips and inches.
- L = 24 ft × 12 in/ft = 288 in
- M₀ = 54 k-ft × 12 in/ft = 648 k-in
Case 1: Deflection due to the 15 kip point load (P)
The standard formula for the midspan deflection of a simply supported beam with a concentrated load at its center is:
Δ₁ = P * L³ / (48 * E * I)
The 15 kip load acts downward, so this deflection will be downward.
Δ₁ = (15 kips * (288 in)³) / (48 * 29,000 ksi * 428 in⁴)
Δ₁ = 358,318,080 kip-in³ / 594,816,000 kip-in²
Δ₁ = 0.6024 inches (downward)
Case 2: Deflection due to the 54 k-ft moment (M₀)
The standard formula for the midspan deflection of a simply supported beam with a moment applied at one end is:
Δ₂ = M₀ * L² / (16 * E * I)
The applied 54 k-ft moment is clockwise at the left support. This action will cause the center of the beam to bend upward, creating an upward deflection.
Δ₂ = (648 k-in * (288 in)²) / (16 * 29,000 ksi * 428 in⁴)
Δ₂ = 53,747,712 kip-in³ / 198,272,000 kip-in²
Δ₂ = 0.2711 inches (upward)
Total Deflection
Finally, we superimpose the two results. Taking downward deflection as positive and upward deflection as negative:
Total Deflection = Δ₁ – Δ₂
Total Deflection = 0.6024 in – 0.2711 in = 0.3313 inches
Since the result is positive, the net deflection is downward.
