The bar has a length l and cross-sectional area a. determine its elongation
due to the force p and its own weight. the material has a specific weight y
(weight/volume) and a modulus of elasticity e.

(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)

answer: 500 wph.

explanation: