Stress Comparison: Sudden versus Gradual Loading

When a load is applied to a body, it causes deformation and internal stress. The rate at which the load is applied significantly affects the resulting stress. This question asks us to compare the stress induced by a sudden load application versus a gradual load application for the same magnitude of load.

Understanding Stress Induction

Stress ($\sigma$) is defined as the internal force per unit area within a material. It's calculated as $\sigma = \frac{F}{A}$, where $F$ is the force and $A$ is the cross-sectional area.

• Gradual Loading: The load is applied slowly, allowing the material to deform progressively without significant acceleration or kinetic energy effects.
• Sudden Loading: The load is applied instantaneously. This imparts kinetic energy to the body, which must be dissipated, typically through internal strain energy, leading to higher peak stresses and deformations compared to gradual loading.

Derivation of Stress Relationship

Let's analyze the energy involved:

Consider a load $P$ applied to a body of length $L$, cross-sectional area $A$, and modulus of elasticity $E$. Let the elongation under gradual load be $\delta_{grad}$ and under sudden load be $\delta_{sud}$.

Stress under Gradual Loading:

When the load $P$ is applied gradually, the maximum stress ($\sigma_{grad}$) is simply:

$$ \sigma_{grad} = \frac{P}{A} $$

The elongation $\delta_{grad}$ is related to stress by Hooke's Law ($\sigma = E \epsilon = E \frac{\delta}{L}$):

$$ \delta_{grad} = \frac{\sigma_{grad} L}{E} = \frac{PL}{AE} $$

The work done by the load is $W_{grad} = \frac{1}{2} P \delta_{grad}$. This work is stored as strain energy ($U_{grad}$).

Stress under Sudden Loading:

When the load $P$ is applied suddenly, the work done by the load up to the maximum deformation ($\delta_{sud}$) is $W_{sud} = P \times \delta_{sud}$.

This work is converted entirely into strain energy ($U_{sud}$) within the material at the point of maximum deformation (where velocity is momentarily zero).

The strain energy stored is given by $U_{sud} = \frac{1}{2} \sigma_{sud} \epsilon_{sud} \times \text{Volume}$. Let the volume be $V = AL$. The stress is $\sigma_{sud}$ and strain is $\epsilon_{sud}$.

$$ U_{sud} = \frac{1}{2} \sigma_{sud} \epsilon_{sud} V = \frac{1}{2} \sigma_{sud} \left( \frac{\sigma_{sud}}{E} \right) AL = \frac{\sigma_{sud}^2 AL}{2E} $$

We also know that the deformation is related to stress: $\delta_{sud} = \epsilon_{sud} L = \frac{\sigma_{sud} L}{E}$.

Equating the work done by the load to the stored strain energy:

$$ W_{sud} = U_{sud} $$

$$ P \times \delta_{sud} = \frac{\sigma_{sud}^2 AL}{2E} $$

Substitute the expression for $\delta_{sud}$:

$$ P \times \left( \frac{\sigma_{sud} L}{E} \right) = \frac{\sigma_{sud}^2 AL}{2E} $$

Simplify the equation:

$$ P L = \frac{\sigma_{sud} AL}{2} $$

Solving for $\sigma_{sud}$:

$$ \sigma_{sud} = \frac{2 P L}{A L} = \frac{2P}{A} $$

Conclusion on Loading Effects

Comparing the stresses:

$$ \sigma_{sud} = \frac{2P}{A} $$

$$ \sigma_{grad} = \frac{P}{A} $$

Therefore:

$$ \sigma_{sud} = 2 \times \sigma_{grad} $$

The stress induced when a load is applied suddenly is twice the stress induced when the same load is applied gradually.