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For a circular x-section, the polar moment of inertia may be computed in the following manner

**Effect of shape of beam section on stress induced**

**CIRCULAR SECTION :**

For a circular x-section, the polar moment of inertia may be computed in the following manner

Consider any circular strip of thickness dr located at a radius 'r'.

Than the area of the circular strip would be dA = 2pr. dr

Thus

**Parallel Axis Theorem:**

The moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centroid plus the area times the square of the distance between the axes.

If „ZZ' is any axis-sectionintheandplane„XX' ofis centroid G, of the cross-section, then

**Rectangular Section:**

For a rectangular x-section of the beam, the second moment of area may be computed as below :

Consider the rectangular beam cross-section as shown above and an element of area **dA** , thickness **dy** , breadth **B** located at a distance **y** from the neutral axis, which by symmetry passes through the centre of section. The second moment of area **I** as defined earlier would be

Thus, for the rectangular section the second moment of area about the neutral axis i.e., an axis through the centre is given by

Similarly, the second moment of area of the rectangular section about an axis through the lower edge of the section would be found using the same procedure but with integral limits of **0** to **D** .

Therefore

These standards formulas prove very convenient in the determination of INA for build up sections which can be conveniently divided into rectangles. For instance if we just want to find out the Moment of Inertia of an I - section, then we can use the above relation.

**Let us consider few examples to determaine the sheer stress distribution in a given X-sections**

**Rectangular x-section:**

Consider a rectangular x-section of dimension b and d

A is the area of the x-section cut off by a line parallel to the neutral axis. is the distance of the centroid of A from the neutral axis

This shows that there is a parabolic distribution of shear stress with y.

The maximum value of shear stress would obviously beat the location y = 0.

Therefore the shear stress distribution is shown as below.

It may be noted that the shear stress is distributed parabolically over a rectangular cross-section, it is maximum at y = 0 and is zero at the extreme ends.

**I - section :**

Consider an I - section of the dimension shown below.

The shear stress distribution for any arbitrary shape is given as

Let us evaluate the quantity, Ay and Ax the quantity for this case comprise the contribution due to flange area and web area

**Flange area**

**Web Area**

To get the maximum and minimum values of t substitute in the above relation.

y = 0 at N. A. And y = d/2 at the tip.

The maximum shear stress is at the neutral axis. i.e. for the condition y = 0 at N. A.

Hence, ..........(2)

The minimum stress occur at the top of the web, the term bd 2 goes off and shear stress is given by the following expression

............(3)

The distribution of shear stress may be drawn as below, which clearly indicates a parabolic distribution

Note: from the above distribution we can see that the shear stress at the flanges is not zero, but it has some value, this can be analyzed from equation (1). At the flange tip or flange or web interface y = d/2.Obviously than this will have some constant value and than onwards this will have parabolic distribution.

In practice it is usually found that most of shearing stress usually about 95% is carried by the web, and hence the shear stress in the flange is neglible however if we have the concrete analysis i.e. if we analyze the shearing stress in the flange i.e. writing down the expression for shear stress for flange and web separately, we will have this type of variation.

This distribution is known as the "top –hat" distribution. Clearly the web bears the most of the shear stress and bending theory we can say that the flange will bear most of the bending stress.

**Shear stress distribution in beams of circular cross-section:**

Let us find the shear stress distribution in beams of circular cross-section. In a beam of circular cross-section, the value of Z width depends on y.

Using the expression for the determination of shear stresses for any arbitrary shape or a

arbitrary section.

Where òy dA is the area moment of the shaded portion or the first moment of area.

Here in this case „dA' is to be found

The distribution of shear stresses is shown below, which indicates a parabolic distribution

**Principal Stresses in Beams**

It becomes clear that the bending stress in beam sx is not a principal stress, since at any distance y from the neutral axis; there is a shear stress t ( or txy we are assuming a plane stress situation)

In general the state of stress at a distance y from the neutral axis will be as follows.

At some point „P' in the beam, the va

After substituting the appropriate values in the above expression we may get the inclination of the principal planes.

**Illustrative examples: **Let us study some illustrative examples, pertaining to determination** **of principal stresses in a beam

1. Find the principal stress at a point A in a uniform rectangular beam 200 mm deep and 100 mm wide, simply supported at each end over a span of 3 m and carrying a uniformly distributed load of 15,000 N/m.

**Solution: **The reaction can be determined by symmetry

R1 = R2 = 22,500 N

consider any cross-section X-X located at a distance x from the left end.

Hence,

S. F at XX =22,500 –15,000 x

B.M at XX = 22,500 x –15,000 x (x/2) = 22,500 x –15,000 . x2 / 2

Therefore,

S. F at X = 1 m = 7,500 N

B. M at X = 1 m = 15,000 N

Now substituting these values in the principal stress equation,

We get s1 = 11.27 MN/m2

s2 = - 0.025 MN/m2

**Bending Of Composite or Flitched Beams**

A composite beam is defined as the one which is constructed from a combination of materials. If such a beam is formed by rigidly bolting together two timber joists and a reinforcing steel plate, then it is termed as a flitched beam.

The bending theory is valid when a constant value of Young's modulus applies across a section it cannot be used directly to solve the composite-beam problems where two different materials, and therefore different values of E, exists. The method of solution in such a case is to replace one of the materials by an equivalent section of the other.

Consider, a beam as shown in figure in which a steel plate is held centrally in an appropriate recess/pocket between two blocks of wood .Here it is convenient to replace the steel by an equivalent area of wood, retaining the same bending strength. i.e. the moment at any section must be the same in the equivalent section as in the original section so that the force at any given dy in the equivalent beam must be equal to that at the strip it replaces.

Hence to replace a steel strip by an equivalent wooden strip the thickness must be multiplied by the modular ratio E/E'.

The equivalent section is then one of the same materials throughout and the simple bending theory applies. The stress in the wooden part of the original beam is found directly and that in the steel found from the value at the same point in the equivalent material as follows by utilizing the given relations.

**Stress in steel = modular ratio x stress in equivalent wood**

The above procedure of course is not limited to the two materials treated above but applies well for any material combination. The wood and steel flitched beam was nearly chosen as a just for the sake of convenience.

**Assumption**

In order to analyze the behavior of composite beams, we first make the assumption that the materials are bonded rigidly together so that there can be no relative axial movement between them. This means that all the assumptions, which were valid for homogenous beams are valid except the one assumption that is no longer valid is that the Young's Modulus is the same throughout the beam.

The composite beams need not be made up of horizontal layers of materials as in the earlier example. For instance, a beam might have stiffening plates as shown in the figure below.

Again, the equivalent beam of the main beam material can be formed by scaling the breadth of the plate material in proportion to modular ratio. Bearing in mind that the strain at any level is same in both materials, the bending stresses in them are in proportion to the Young's modulus.

**Shear stresses in beams**

When a beam is subjected to non uniform bending, both bending moments, M, and shear forces, V, act on the cross section. The normal stresses, sx, associated with the bending moments are obtained from the flexure formula. We will now consider the distribution of shear stresses, t, associated with the shear force, V. Let us begin by examining a beam of rectangular cross section. We can reasonably assume that the shear stresses t act parallel to the shear force V. Let us also assume that the distribution of shear stresses is uniform across the width of the beam.

**Shear flow**

One thing we might ask ourselves now is: Where does maximum shear stress occur? Well, it can be

shown that this always occurs in the center of gravity of the cross-section. So if you want to calculate the maximum shear stress, make a cut through the center of gravity of the cross-section.

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