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.

**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 in
the plane of cross-section and „XX' is a parallel axis
through the 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,
thequantity 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 out using the
Pythagoras theorem

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 value of bending
stresses is given as

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 nonuniform 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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