Answer:
(2PL + yAL²) / (2EA)
Explanation:
For any position x along the bar measured from the top, the stress is:
σ = F(x) / A
where F(x) is the force at position x,
and A is the cross sectional area.
The stress also equals:
σ = E ε(x)
σ = E dδ / dx
where E is the modulus of elasticity,
dx is a short length of bar,
and dδ is the deflection of that short length.
Therefore:
E dδ / dx = F(x) / A
E dδ / dx = (P + yAx) / A
dδ = 1/(EA) (P + yAx) dx
The total deflection is:
δ = ∫ dδ
δ = ∫₀ᴸ 1/(EA) (P + yAx) dx
δ = 1/(EA) ∫₀ᴸ (P + yAx) dx
δ = 1/(EA) (Px + ½yAx²) |₀ᴸ
δ = 1/(EA) (PL + ½yAL²)
δ = (2PL + yAL²) / (2EA)
