Physics > Elasticity > 2.0 Stress
Elasticity
1.0 Basic Definitions
2.0 Stress
3.0 Strain
4.0 Hooke’s Law and Modulus of Elasticity
5.0 Types of Modulus of Elasticity
6.0 Elongation of Rod Under its Self Weight
7.0 Bulk Modulus of Elasticity
8.0 Modulus of Rigidity
9.0 The Stress – Strain Curve
10.0 Analogy of Rod as a Spring
11.0 Potential Energy in a Stretched Wire
12.0 Thermal Stress & Strains
13.0 Poisson’s Ratio
14.0 Relation Between $Y$, $B$, $\eta$ and $\sigma$
2.1 Types of Stress
1. Longitudinal or Normal Stress:
If the restoring force is perpendicular to the area of cross-section, it produces longitudinal or normal stress.
It is of two types:
- Compressive Stress
- Tensile Stress
Note: When the force is applied perpendicular to the cross section area, then it produces normal stress. But when force is applied at some angle, then resolve the force into two components( normal & tangential). This normal component of force will produce longitudinal stress as shown in Fig. E. 3.
Let $\mathop F\limits^ \to $ be applied force to the face which has area $A$. Resolving $\mathop F\limits^ \to $ into two components:
$$\begin{equation} \begin{aligned} {F_n} = F\sin \theta ,{\text{ called normal force}} \\ {F_t} = F\cos \theta ,{\text{ called tangential force}} \\\end{aligned} \end{equation} $$
So, for longitudinal or normal stress we will consider only normal force, $$Normal\ Stress = \frac{{{F_n}}}{A} = \frac{{Fsin\theta }}{A}$$
2. Tangential or Shear Stress:
If the restoring force is tangential to the area of cross section, it produces tangential or shear stress.
Note: When the force is applied tangential to the cross section area, then it produces tangential stress. But when force is applied at some angle, then resolve the force into two components (normal & tangential). This tangential component of force will produce tangential or shear stress as shown in Fig. E. 3.
$$Tangential\ Stress = \frac{{{F_t}}}{A} = \frac{{Fcos\theta }}{A}$$
Example 1. Find out normal stress and shear stress on a fixed block as shown in the Fig. E 4.
Solution: Surface on which force act is $ABCD$. So, cross section area of $ABCD$ is,
$$\begin{equation} \begin{aligned} A = l \times b \\ A = 4 \times 2{m^2} \\ A = 8{m^2} \\\end{aligned} \end{equation} $$
Now resolve force into two components, i.e. normal (${F_t}$ ) and tangential(${F_n}$)
$$\begin{equation} \begin{aligned} {F_t} = Fcos\theta \\ {F_n} = Fsin\theta \\ \therefore NormalStress = \frac{{Fcos\theta }}{A} = \frac{{100cos30^\circ }}{8} = \frac{{25\sqrt 3 }}{4}\frac{N}{{{m^2}}} \\ and \\ TangentialStress = \frac{{Fsin\theta }}{A} = \frac{{100sin30^\circ }}{8} = \frac{{25}}{4}\frac{N}{{{m^2}}} \\\end{aligned} \end{equation} $$
