Learning Objectives:

• State and apply bending equation.

\[\dfrac{M}{I} = \dfrac{\sigma}{c} = \dfrac{E}{R}\]

where,
M= Bending Moment
I = Moment of Inertia of the cross section about the neutral axis.
\(\sigma\) = stress
c = distance from the neutral axis
For symmetric sections, c = depth/2
E = Modulus of Elasticity
R = Radius of Curvature
This formula is called flexture formula.

Solved Example: 42-1-01

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

A. Linearly

B. Parabolically

C. Hyperbolically

D. Elliptically

Correct Answer: A

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:

A. Maximum bending stress developed in both beams are same.

B. The circular beam will experience more bending stress than the square one.

C. The square beam will experience more bending stress than the circular one.

D. As the material is same, both beams will experience same deformation.

Correct Answer: B

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?

A. Increase the area moment of inertia

B. Increase the span of the beam

C. Select a different material having lesser modulus of elasticity

D. Magnitude of the load to be increased.

Correct Answer: A

Solved Example: 42-1-04

A beam will be in pure bending under a:

A. Constant shear force and a constant bending moment

B. Constant shear force and zero bending moment

C. Constant bending moment and zero shear force

D. None of the above

Correct Answer: C

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?

A. 162.98 MPa

B. 325.95 MPa

C. 625.95 MPa

D. 651.90 MPa

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

Correct Answer: A

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:

A. 25

B. 50

C. 75

D. 100

\[\dfrac{M}{I} = \dfrac{\sigma}{y} = \dfrac{E}{R}\] \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*}

Correct Answer: B

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:

A. Normal stress remains constant in all directions

B. Normal stress varies linearly in the material

C. Elastic constants are same in all the directions

D. Elastic constants varies linearly in the material

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.

Correct Answer: C

Learning Objectives:

• Explain the concept of the neutral axis

• Determine the depth of the neutral axis for a given cross-section.

At some point between these two surfaces, there must be a plane where the normal stresses and strains are zero. We call this plane the Neutral Plane (N.P.) represented in diagram by Neutral Axis (N.A.).
All layers above the neutral axis suffer compression whereas those below the neutral axis suffer tension. The amount by which a layer is compressed or stretched depends upon its position with respect to the neutral axis.

Solved Example: 42-2-01

The neutral axis of the cross-section a beam is that axis at which the bending stress is:

A. 0

B. Min

C. Max

D. $\infty$

The axial stress is zero on the neutral surface and increases linearly as one moves away from the neutral axis.

Correct Answer: A

Learning Objectives:

• Review simple beam theory.

• Generalize simple beam theory to three dimensions and general cross sections.

• Consider combined effects of bending, shear and torsion.

• Study the case of shell beams.

The section modulus of the cross-sectional shape is of significant importance in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.
It includes the idea that most of the work in bending is being done by the extreme fibres of the beam, ie the top and bottom fibres of the section. The distance of the fibres from top to bottom is therefore built into the calculation.
The elastic modulus is denoted by Z. To calculate Z, the distance (y) to the extreme fibres from the centroid (or neutral axis) must be found as that is where the maximum stress could cause failure.

\[s = \dfrac{I}{c}\]

Solved Example: 42-3-01

A beam of uniform strength has:

A. Same cross-section throughout the beam

B. Same bending stress at every section

C. Same bending moment at every section

D. Same shear stress at every section

In general, beams have uniform cross section throughout their length. When they are loaded, there is a variation in bending moment from section to section along the length. The stress in extreme outer fibre (top and bottom) also vary from section to section along their length. The extreme fibres can be loaded to the maximum capacity of permissible stress (say $\sigma_{max}$), but they are loaded to less capacity. Hence, in beams of uniform cross section there is a considerable waste of material. When a beam is suitably designed such that the extreme fibres are loaded to the same maximum permissible bending stress by varying the cross section it will be known as a beam of uniform strength.

Correct Answer: B

Solved Example: 42-3-02

Find the section modulus of a circular section whose section diameter is 28 mm.

A. 2155

B. 6310

C. 1177

D. 650

\[ Z = \dfrac{\pi d^3}{32} = \dfrac{\pi (28)^3}{32} = 2155.1\ mm^3\]

Correct Answer: A