Beam Deflection by Superposition

Example: (continued) Note: Beam remains straight from B to C for beam (1) with moment, M.

At C: Strategy: The total deflection is δ_C = δ₃ – δ₁ – δ₂ where δ₂ = (L/2) θ₁ = (L/2) θ_B

From deflection table: δ₃ = PL^{3 /} 3EI and δ₁ = − M(L/2)^{2 /} 2EI = − ML^{2 /} 8EI

Also from deflection table for M: θ_C = − ML/EI so θ₁ = − M(L/2)/EI = − ML/2EI

By superposition δ_C = PL^{3 /} 3EI − ML^{2 /} 8EI − ML²/4EI = PL^{3 /} 3EI − 3ML^{2 /} 8EI (result)

At C the slope is θ_C = θ₃ – θ₁

From deflection table: θ₃ = PL^{2 /} 2EI and from above θ₁ = − ML/2EI

So the slope of the beam by at C by superposition is θ_C = PL^{2 /} 2EI − ML/2EI (result)

At B: Strategy: The total deflection is δ_B = δ₄ – δ₁

From deflection table for P: y₂(x) = Px²/6EI (3L – x)

So at midspan for load P: δ₄ = P(L/2)²/6EI (3L – L/2) = 5PL^{3 /} 48EI

From above δ₁ = − ML^{2 /} 8EI So by superposition δ_B = 5PL^{3 /} 48EI − ML^{2 /} 8EI (result)

At B the slope is θ_B = θ₄ – θ₁ where from above θ₁ = − ML/2EI

From the deflection table: y₂(x) = (Px²/6EI) (3L – x)

So taking the derivative to obtain slope: dy₂(x)/dx = θ₂(x) = (PLx/EI – Px²/2EI)

And the slope in beam 2 at B due to P at x = L/2 is θ₄ = (3/8) PL²/EI

So the slope of the beam at B by superposition is θ_B = (3/8) PL²/EI − ML/2EI (result)

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