### Bending Equation

• State and apply bending equation.

### Solved Example:

#### 42-1-01

In a simple bending of beams, the stress in the beam varies:

### Solved Example:

#### 42-1-02

The beams, one having square cross section and another circular cross section, are subjected to the same amount of bending moment. If the cross-sectional area as well as the material of both beams are the same, then:

### Solved Example:

#### 42-1-03

A simply supported laterally loaded beam was found to deflect more than a specified value. Which of the following measures will reduce deflection?

### Solved Example:

#### 42-1-04

A beam will be in pure bending under a:

### Solved Example:

#### 42-1-05

A simply supported beam of span length 6 m and 75 mm diameter carries a uniformly distributed load of 1.5 kN/m. What is the value of maximum bending stress?

Solution:
$\mathrm{Max.\ Bending\ Moment} = 6.75\ \mathrm{KN.m}$ $\sigma_b = \dfrac{6.75 \times 10^6 \times 64 \times 37.5}{\pi \times 75^4} = 162.97\ \mathrm{MPa}$

### Solved Example:

#### 42-1-06

A wire of circular cross-section of diameter 1 mm is bent into a circular arc of radius 1 m by application of pure bending moments at its ends. The Young’s modulus of the material of the wire is 100 GPa. The maximum tensile stress developed in the wire, in MPA, is:

Solution:
$\dfrac{M}{I} = \dfrac{\sigma}{y} = \dfrac{E}{R}$ \begin{align*} \sigma &= \dfrac{E \cdot y}{R}\\ &= \dfrac{100 \times 10^9 \times 0.5 \times 10^{-3}}{1}\\ &= 50 \times 10^6\ \mathrm{Pa}\\ &= 50\ \mathrm{MPa} \end{align*}

### Solved Example:

#### 42-1-07

In a simple bending theory, one of the assumption is that the material of the beam is isotropic. This assumption means that the:

Solution:

Isotropic materials that have material properties identical in all directions.

Anisotropic materials are materials whose properties are directionally dependent. Properties such as Young's Modulus, change with direction along the object. Common examples of anisotropic materials are wood and composites.