Question

Prove that elastic potential energy bar unit volume is 1/2stress×strain​

Answer 1

$\Huge{\sf{Elastic \ Potential \ Energy \colon }}$

When a deformation force is acted on a spring. The work done in contracting or expanding the spring would be stored as elastic potential energy

Mathematically,

$\boxed{\sf{U = \frac{1}{2}K {e}^{2}}}$

$\sf{Here}\begin{cases}\sf{K : Spring \ Constant}\\\sf{e : extension \ in \ the \ spring} \end{cases}$

But

$\sf{K = \dfrac{Y A}{l}}$

$\sf{Here}\begin{cases}\sf{Y : Young's \ Modulus }\\ \sf{l : Length\ of \ Spring }\\ \sf{A : Area \ of \ Cross \ Section} \end{cases}$

Implies,

$\longrightarrow \ \boxed{\boxed{\sf{U = \dfrac{1}{2} \dfrac{YA}{l} {e}^{2}}}}$

$\rule{300}{2}$

• Elastic Potential Energy per unit volume is known as Energy Density

Now,

$\sf{D = \dfrac{U}{V}}$

Volume can be expressed as a product of area and length of the spring » V = A × l

$\sf{\longrightarrow \ D = \dfrac{\dfrac{YA}{2l}{e}^{2}}{A \times l} } \\ \\ \sf{\longrightarrow \ D = \dfrac{1}{2} \times \dfrac{e}{l} \times Y \dfrac{e}{l}} \\ \\ \sf{\longrightarrow \ D = \dfrac{1}{2} \times strain \times (Y \times strain)}$

$\rule{300}{2}$

From Hooke's Law,

$\sf{Modulus \ of \ Elasticity = \dfrac{Stress}{Strain}} \\ \\ \leadsto \sf{Strain \times Y = Stress}$

$\rule{300}{2}$

Hence,

$\large{\longrightarrow \ \boxed{\boxed{\sf{D = \dfrac{1}{2} \times stress \times strain }}}}$

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

solly...I don't know the answer

#StayHomeStaySafe