Derive the bulk modulus (volume modulus of elasticity) for a bar subjected under pressure p.

Answer 1

The bulk modulus (K) is defined as the ratio of the pressure change (Δp) applied to a material to the resulting relative volume change (ΔV/V) that it undergoes.

Mathematically, K is given by the equation:

K = -V(dP/dV)

where V is the initial volume of the material, P is the pressure, and dP/dV represents the derivative of P with respect to V.

To derive this equation, we start with a bar of initial volume V0 and apply a pressure P to it, resulting in a change in volume ΔV. The relationship between pressure and volume change is given by:

(dP/dV) = -K/V

where K/V represents the bulk modulus of the material.

Rearranging the equation, we have:

(dP/dV) * V = -K

Integrating both sides with respect to V, we get:

∫(dP/dV) * V dV = -∫K dV

Simplifying, we have:

∫P dV = -K * ∫dV

Integrating, we obtain:

PV = -KV + C

where C is the constant of integration.

Rearranging the equation, we have:

KV + PV = C

Since P is the external pressure and ΔV is the change in volume, we can write the equation as:

K(V0 + ΔV) + P(V0 + ΔV) = C

Expanding and rearranging the equation, we get:

KV0 + KΔV + PV0 + PΔV = C

Since C is a constant, KV0 + PV0 = C.

Substituting this value back into the equation, we have:

KΔV + PΔV = 0

Factoring out ΔV, we obtain:

(ΔV)(K + P) = 0

Since ΔV cannot be zero, we must have:

K + P = 0

Thus, we find that the bulk modulus (K) is given by:

K = -P

Therefore, the bulk modulus (volume modulus of elasticity) for a bar subjected to pressure P is equal to the negative of the applied pressure.