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Bending Stresses in Beams

Bending Stresses in Beams

Theory of Pure Bending

Pure bending refers to the state of bending in a beam in which bending moment is present but shear force is negligible over the length considered. The principal observations and conclusions from the theory of pure bending are given below.

  • Plane sections originally parallel to each other remain plane and parallel after bending (they do not warp). This is the plane sections remain plane assumption.
  • Fibres on the convex side elongate and fibres on the concave side shorten. There exists an intermediate layer where there is no change in length; this layer is the neutral axis (N.A.). The neutral axis passes through the centroid of the cross-sectional area for homogeneous sections.
  • The longitudinal strain in a fibre is proportional to its distance from the neutral axis. Therefore strain varies linearly with distance from N.A., and hence stress (within elastic limit) also varies linearly with distance from N.A.
  • Under bending the beam axis is assumed to bend into an arc of a circle so that each fibre remains straight but changes its length according to its radius from the neutral axis.

Let s be the longitudinal stress produced by bending at a distance y from the neutral axis. Then (see Theory of Pure Bending below for expression):

Theory of Pure Bending

From equilibrium and Hooke's law the standard flexure formula (result of simple bending) is obtained:

σ = (M y) / I

where σ is the bending stress at distance y from the neutral axis, M is the bending moment at the cross-section, and I is the second moment of area (moment of inertia) of the section about the neutral axis. The maximum bending stress occurs at the extreme fibre where |y| is maximum (denoted c), so σ_max = M c / I. The section modulus Z = I / c, therefore M = σ_max Z.

Assumptions in the theory of simple bending

  • The beam is initially straight and the material is continuous.
  • The beam has constant cross-sectional area and an axis of symmetry (where applicable).
  • The material is homogeneous and isotropic.
  • Hooke's law is valid (material behaviour is linear elastic) so stress is proportional to strain.
  • Plane transverse sections remain plane and perpendicular to the bent axis after bending.
  • Adjacent layers are free to expand or contract longitudinally and laterally; effects such as Poisson's ratio interactions between layers are neglected in simple bending theory.
  • The value of Young's modulus (E) is assumed the same in tension and compression.

Some Standard Results And Examples

Orientation of a square section: strength comparison

Consider a square section placed in two different orientations: (I) with sides horizontal and vertical, and (II) with a diagonal horizontal (so the diagonal is vertical). For a given extreme fibre stress, the section modulus depends on orientation. The following figures illustrate the two positions:

Orientation of a square section: strength comparison
Orientation of a square section: strength comparison

Calculated values show that arrangement (I) is 41.4% stronger than arrangement (II) for bending in the vertical plane shown. That is, the section modulus in (I) is larger by approximately 41.4% compared to (II).

Strongest rectangular beam cut from a circular log

Given a circular log of diameter D, find the rectangle of breadth b and depth d (d measured in the bending direction) that gives the maximum section modulus (hence maximum bending strength for given material).

The section modulus for a rectangle of breadth b and depth d about its neutral axis is Z = (b d^2) / 6 (for bending about the horizontal centroidal axis). For a given circular boundary the rectangle must fit inside the circle, so relationship between b and d follows from geometry (see).Strongest rectangular beam cut from a circular logFor maximum Z subject to b and d constrained by the circle, differentiation leads to the optimal ratio:

Strongest rectangular beam cut from a circular log

Hence the strongest rectangle that can be cut from a circle satisfies

b = d / √2

Geometrical and algebraic steps that lead to this result are shown in the following diagrams and expressions:

Strongest rectangular beam cut from a circular log
Strongest rectangular beam cut from a circular log
Strongest rectangular beam cut from a circular log

Therefore b = d / √2 gives the largest section modulus for a rectangular section inscribed in a circle of given diameter.

Comparing weights for sections of the same bending strength

Problem statement (text preserved in content): Three beams have the same length, same allowable bending stress and are subjected to the same maximum bending moment. The cross section of the beams are a circle, a square and rectangle with depth twice the width. Find the ratio of the weights of the circular and rectangular beams with respect to the square beams.

Let the circular section be of diameter d.

Let the square section be of side x.

Let the rectangular section be of width b and depth 2b.

For equal bending moment under the same allowable stress, the section modulus must be equal for all three sections (M = σ Z and M, σ same for all). Therefore Z_circle = Z_square = Z_rectangle.

Using standard expressions for section modulus:

Comparing weights for sections of the same bending strength
Comparing weights for sections of the same bending strength

From these relations one obtains

Comparing weights for sections of the same bending strength
Comparing weights for sections of the same bending strength

Numerical relationships derived give d = 1.193 x and b = 0.6299 x.

  • Cross-sectional areas (which are proportional to weight for equal lengths and same material) then compare as: W_rectangular < W_square < W_circular.

Hence for the same bending strength under the same allowable stress, the rectangular section (depth twice width) is the lightest and the circular section is the heaviest.

Improving section modulus by trimming a square section

A square-section beam carrying a vertical bending moment can have its section modulus increased slightly by trimming off triangular corners (shaded portions in the figure) oriented parallel to the neutral axis. The geometry of the parts to be cut and the resulting increase in section modulus are shown in the diagrams:

Improving section modulus by trimming a square section

Key points:

  • For maximum section modulus, the portion EB to be cut off satisfies the relation shown in Improving section modulus by trimming a square section.
  • The corresponding increase in section modulus is given in Improving section modulus by trimming a square section.
  • For a circular section a small increase in section modulus (about 0.7%) is possible by cutting segmental portions as shown, with the depth of the segment d approximately 0.011 times the diameter of the section.
Improving section modulus by trimming a square section
Improving section modulus by trimming a square section
Improving section modulus by trimming a square section

Flitched Beams

A flitched beam is a composite beam formed by placing steel plates (flitches) on a timber beam and fastening them together so that they act compositely. The composite action increases bending strength and stiffness. The steel and wood share bending stresses according to their stiffnesses.

Flitched Beams

The modular ratio (ratio of moduli of elasticity) is used to convert steel parts into equivalent wood width when analysing as an equivalent wooden section. The modular ratio is defined as:

Flitched Beams

Moment of resistance of the section

Case - 1: Flitches attached symmetrically at the sides.

Consider a timber beam of breadth b and depth d with two identical steel plates each of thickness t attached symmetrically to the vertical sides. Let M_r be the total moment of resistance of the composite section. Let M_w and M_s be the moments of resistance contributed by the wood and steel respectively under the same curvature.

The steel contribution can be treated by transforming steel into an equivalent width of wood using the modular ratio m. The equivalent wooden section then has width b + m(2t) and depth d. Therefore the composite section's moment of resistance equals that of a wooden member of breadth b + m(2t) and depth d. This rectangular section is called the equivalent wooden section.

Moment of resistance of the section
Moment of resistance of the section
Moment of resistance of the section

Case - 2: Flitches attached symmetrically at the top and bottom.

When flitches are attached at top and bottom, the steel plates are placed furthest from the neutral axis; this increases the section modulus more effectively than side flitches for the same steel area. Therefore top-and-bottom flitching usually gives higher moment carrying capacity than side flitching (for the same quantity of steel).

Beams Of Uniform Strength

For economy of material, a beam may be designed so that the extreme fibre stress is equal to the permissible stress at every section along its length. Such a beam is called a beam of uniform strength. Because bending moment varies along the span, the cross-section must vary accordingly. There are three common design approaches:

  • Vary width while keeping depth constant.
  • Vary depth while keeping width constant.
  • Vary both width and depth.

Case 1 - Constant depth, varying width

Let permissible bending stress be σ. For any section at distance x from a reference (for example a support), bending moment is M(x). For a rectangular section of constant depth d and width b(x), the section modulus is Z = b(x) d^2 / 6. Requiring σ = M(x) / Z gives the width at x:

Case 1 - Constant depth, varying width

Therefore

Case 1 - Constant depth, varying width

and the width at mid-span (or any particular x of interest) can be obtained similarly, see Case 1 - Constant depth, varying width and Case 1 - Constant depth, varying width.

Case 1 - Constant depth, varying width
Case 1 - Constant depth, varying width

Case 2 - Constant width, varying depth

For constant width b, the required depth d(x) at a section with bending moment M(x) is obtained by equating M(x) = σ Z and using Z = b d(x)^2 / 6. Thus d(x) is proportional to √M(x). The detailed expressions and illustrative diagrams are shown below.

Case 2 - Constant width, varying depth
Case 2 - Constant width, varying depth
Case 2 - Constant width, varying depth

Applications, Remarks And Practical Notes

  • The flexure formula σ = M y / I is the foundation for bending design in beams of many materials within the elastic limit.
  • Section modulus Z = I / c is a convenient design parameter: for a given allowable stress and bending moment, choose a section with sufficient Z.
  • Flitched beams are practical where combining timber and steel gives an economical solution: transform steel to equivalent timber using modular ratio for strength calculations.
  • Beams of uniform strength are useful to save material; practical manufacturing and handling constraints may limit how much variation in cross-section is economical.
  • When trimming sections to increase section modulus, the gains are often small and must be weighed against fabrication complexity and reduction in other properties (e.g., shear capacity).

Summary

The theory of pure bending leads to the flexure formula which relates bending stress to bending moment and section geometry. Section modulus is the primary geometric property used in bending design. Practical measures-such as choosing optimal orientation, cutting optimal rectangular sections from logs, adding flitches, or designing beams of uniform strength-allow more efficient use of material. All designs must respect the assumptions of simple bending and remain within elastic limits for the theory to be applicable.

The document Bending Stresses in Beams is a part of the Mechanical Engineering Course Mechanical Engineering SSC JE (Technical).
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FAQs on Bending Stresses in Beams

1. What are bending stresses in beams?
Ans. Bending stresses in beams refer to the internal stresses that develop within a beam when subjected to bending moments. These stresses occur due to the variation in the beam's cross-sectional shape and the resulting distribution of forces along its length.
2. How do bending stresses affect the performance of beams?
Ans. Bending stresses can significantly impact the performance of beams. Excessive bending stresses can lead to deformation or failure of the beam, affecting its load-carrying capacity and structural integrity. It is crucial to analyze and design beams considering the bending stresses to ensure their safe and efficient operation.
3. What factors influence the magnitude of bending stresses in beams?
Ans. Several factors influence the magnitude of bending stresses in beams. These include the magnitude and location of the applied loads, the beam's material properties, its cross-sectional shape, and the beam's length and supports. Understanding these factors is essential for accurately predicting and managing bending stresses in beam designs.
4. How can bending stresses in beams be calculated or analyzed?
Ans. Bending stresses in beams can be calculated using various analytical methods, such as the Euler-Bernoulli beam theory or the Timoshenko beam theory. These methods involve determining the beam's moment of inertia, calculating the bending moment distribution, and then using these values to calculate the maximum bending stress. Finite element analysis (FEA) is also commonly used to analyze bending stresses in complex beam structures.
5. What are some common techniques to reduce bending stresses in beams?
Ans. There are several techniques to reduce bending stresses in beams. Increasing the beam's cross-sectional area, selecting materials with higher strength or stiffness, and modifying the beam's shape can help reduce bending stresses. Additionally, using reinforcement techniques like adding flanges or ribs to the beam, or using composite materials, can also effectively reduce bending stresses and enhance beam performance.
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