simple stress and strain

Simple Stress and Strain

Mechanics of Deformable Bodies, also known as Strength of Materials, deals with the internal effects of forces acting on a body. The fundamental concepts are stress and strain.

Relevance to Philippine Standards: The National Structural Code of the Philippines (NSCP) 2015, Volume 1, specifies allowable stresses and load factors for various materials (concrete, steel, timber) based on these fundamental principles. For instance, Section 504 details the design of steel members for tension, which relies directly on normal stress calculations.

Stress ($\sigma$)

Stress is the internal resistance of a material to an external force, defined as force per unit area.

Normal Stress (Axial Stress)

Normal stress acts perpendicular to the cross-sectional area. $\sigma = \frac{P}{A}$ where:

• $P$ is the applied axial load (N or lb).
• $A$ is the cross-sectional area (m$^2$ or in$^2$).
• $\sigma$ is the normal stress (Pa or psi).
  • Tensile Stress (+): Pulls the material apart.
  • Compressive Stress (-): Pushes the material together.

Shear Stress ($\tau$)

Shear stress acts parallel to the cross-sectional area. $\tau = \frac{V}{A}$ where:

• $V$ is the shear force.
• $A$ is the area resisting the shear.

Strain ($\epsilon$)

Strain is the measure of deformation representing the displacement between particles in the body relative to a reference length.

Normal Strain

$\epsilon = \frac{\delta}{L}$ where:

• $\delta$ is the total deformation (elongation or contraction).
• $L$ is the original length.

Hooke's Law

For many materials, within the elastic limit, stress is proportional to strain. This relationship is known as Hooke's Law. $\sigma = E \epsilon$ where $E$ is the Modulus of Elasticity (Young's Modulus).

Substituting the definitions of stress and strain: $\frac{P}{A} = E \frac{\delta}{L} \Rightarrow \delta = \frac{PL}{AE}$ This equation calculates the axial deformation of a member under load.

Poisson's Ratio ($\nu$)

When a material is stretched in one direction, it contracts in the lateral directions. Poisson's ratio is the ratio of lateral strain to longitudinal strain. $\nu = -\frac{\epsilon_{lateral}}{\epsilon_{longitudinal}}$

Examples

Example 1: Axial Deformation

Problem: A steel rod with a cross-sectional area of 200 mm$^2$ and a length of 5 m is subjected to an axial tensile load of 50 kN. Determine the elongation of the rod. Assume $E = 200$ GPa.

Solution:

1. Given:

   • $P = 50 \text{ kN} = 50,000$ N
   • $L = 5 \text{ m} = 5000$ mm
   • $A = 200 \text{ mm}^2$
   • $E = 200 \text{ GPa} = 200,000$ MPa ($= 200,000$ N/mm$^2$)

2. Formula: $\delta = \frac{PL}{AE}$

3. Calculation: $\delta = \frac{(50,000 \text{ N})(5000 \text{ mm})}{(200 \text{ mm}^2)(200,000 \text{ N/mm}^2)}$ $\delta = \frac{250,000,000}{40,000,000} = 6.25 \text{ mm}$

Result: The rod elongates by 6.25 mm.

Example 2: Shear Stress in a Bolt

Problem: Two plates are connected by a 20 mm diameter bolt. If a tensile force of 100 kN is applied to the plates, determine the average shear stress in the bolt.

Solution:

1. Identify Shear Type:

   • If the plates are lapped (single shear), the area is the cross-section of the bolt once.
   • If there is a double cover plate (double shear), the area is twice the cross-section.
   • Assumption: Single Shear for this example.

2. Calculate Area: $A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (20 \text{ mm})^2 = 314.16 \text{ mm}^2$

3. Calculate Shear Stress: $\tau = \frac{V}{A}$ $V = P = 100 \text{ kN} = 100,000 \text{ N}$ $\tau = \frac{100,000 \text{ N}}{314.16 \text{ mm}^2} = 318.3 \text{ MPa}$

Result: The average shear stress is 318.3 MPa.