4

Pure Bending

Objectives

✓ Determine the internal moment at a section of a beam

✓ Determine the stress in a beam member caused by
bending
✓ Determine stresses in composite beams
✓ Determine stresses in curved beams
The girders of this bridge have been
designed on the basis of their ability to
resist bending stress.
EXAMPLE

The bar carries equal weights at equal

distances from the hands of the weight
lifter.
The center portion of the rear axle of the
sport buggy is in pure bending.

Copyright ©2018 Pearson Education Ltd. All Rights Reserved.

Pure Bending: Prismatic members subjected to equal and
opposite couples acting in the same longitudinal plane
 Free-body diagram of a clamp

Photo 4.2 Clamp used Centric load and bending moment

to glue lumber pieces in section CC’ is seen.
together.
 SHEAR AND MOMENT DIAGRAM

• Shear is obtained by summing forces perpendicular to the

beam’s axis up to the end of the segment.

• Moment is obtained by summing moments about the end of

the segment.

• Note the sign conventions are

opposite when the summing
processes are carried out with
opposite direction.
EXAMPLE 1

Draw the shear and moment

diagrams for the beam shown
EXAMPLE 2

Draw the shear and moment

diagrams for the beam shown
SYMMETRIC MEMBERS IN PURE BENDING

Assumptions:
1. Plane section remains plane
2. Length of longitudinal axis remains unchanged
3. Plane section remains perpendicular to the
longitudinal axis
4. In-plane distortion of section is negligible
Internal Moment and Stress Relations

The minus sign in equation is due to the fact that a tensile

stress leads to a negative moment (clockwise) of the normal
force x dA about the z axis.
 Deformation

Since all the faces represented in the

E and E’ will lie in the plane defined by CD two projections are at 900 to each
and, therefore, that the section remains other, that
plane
Deformation
Beam with a plane of symmetry in pure
bending:
• member remains symmetric
• bends uniformly to form a circular arc
• cross-sectional plane passes through
arc center and remains planar
• length of top decreases and length of
bottom increases
• a neutral surface must exist that is
parallel to the upper and lower
surfaces and for which the length does
not change
• stresses and strains are negative
(compressive) above the neutral plane
and positive (tension) below it
Deformation
 Deformation

or

the value of the strain is valid anywhere, and the longitudinal

normal strain x varies linearly with the distance y from the neutral
surface. The maximum absolute value of the strain at c
STRESSES AND DEFORMATION IN THE
ELASTIC RANGE

Hooke’s law for uniaxial stress

This result shows that, in the elastic range,

the normal stress varies linearly with the
distance from the neutral surface
STRESSES AND DEFORMATION IN THE
ELASTIC RANGE

This equation shows that the first moment of the cross section
about its neutral axis must be zero. Thus, for a member
subjected to pure bending and as long as the stresses remain in
the elastic range, the neutral axis passes through the centroid of
the section.
 STRESSES AND DEFORMATION IN THE
ELASTIC RANGE

I is the moment of inertia or second

moment of area of the cross
section

The maximum absolute value of the stress

The normal stress at any distance y from

the neutral axis

called the elastic flexure formulas, and the normal stress x caused by the
bending or “flexing” of the member is often referred to as the flexural stress.
 STRESSES AND DEFORMATION IN THE
ELASTIC RANGE

The ratio I/c depends only on the geometry of the cross section. This
ratio is defined as the elastic section modulus S, where

Since the maximum stress m is inversely proportional

to the elastic section modulus S, beams should be
designed with as large a value of S as is practical.
STRESSES AND DEFORMATION IN THE
ELASTIC RANGE

The deformation of the member caused by the bending moment M is

measured by the curvature of the neutral surface. The curvature is
defined as the reciprocal of the radius of curvature ρ
DEFORMATIONS IN A TRANSVERSE CROSS
SECTION
 EXAMPLE 3

The simply supported beam in Fig. a has the cross-sectional area shown in
Fig. b. Determine the absolute maximum bending stress in the beam and
draw the stress distribution over the cross section at this location. Also,
what is the stress at point B?
Solution
EXAMPLE 4

The beam shown in Fig. a

has a cross-sectional area in
the shape of a channel,
Fig.b.

Determine the maximum

bending stress that occurs in
the beam at section a–a.
Solution
MEMBERS MADE OF COMPOSITE MATERIALS

Beams constructed of two or more different

materials are referred to as composite beams.
An example is a beam made of wood with
straps of steel at its top and bottom

Consider a composite beam formed from

two materials with E1 and E2.
MEMBERS MADE OF COMPOSITE MATERIALS

• Consider a composite beam formed from

two materials with E1 and E2.
• Normal strain varies linearly.

• Piecewise linear normal stress variation.

Neutral axis does not pass-through

section centroid of composite section.
• Elemental forces on the section are

• Define a transformed section such that

MEMBERS MADE OF COMPOSITE MATERIALS

y is the distance from the neutral

surface, and I is the moment of
inertia of the transformed section with
respect to its centroidal axis.

he curvature of the composite member

EXAMPLE

The composite beam in figure is made of wood and

reinforced with a steel strap located on its bottom side. If
the beam is subjected to a bending moment of M = 2 kN.m,
determine the normal stress at points B and C. Take Ew =
12 GPa and Est = 200 GPa.
Solution

Since steel has a greater stiffness than wood (Est > Ew), the width of the
wood is reduced to an equivalent width for steel.
Solution

The location of the centroid (neutral axis), calculated from the bottom of
the section

The moment of inertia about the neutral axis

Solution

Normal Stress.

The normal stress in the wood at B

EXAMPLE

The reinforced concrete beam has the cross-sectional area shown.

If it is subjected to a bending moment of M = 60 kN.m, determine
the normal stress in each of the steel reinforcing rods and the
maximum normal stress in the concrete. Take Est = 200 GPa and
Econc = 25 GPa.
Solution
Solution
Solution

Normal Stress.
ECCENTRIC AXIAL LOADING IN A PLANE OF
SYMMETRY
The flexure formula applies to a straight member and loading is
centric. If the member is curved, however, the loads does not pass
through the centroid of the cross section, i.e, when the loading is
eccentric.

Walkway light. Bench press.

ECCENTRIC AXIAL LOADING IN A PLANE OF
SYMMETRY
When the loading is eccentric.
ECCENTRIC AXIAL LOADING IN A PLANE OF
SYMMETRY
The stress distribution due to the original eccentric loading can be
obtained by superposing the uniform stress distribution corresponding
to the centric load P and the linear distribution corresponding to the
bending couple M.
EXAMPLE

Knowing that for the cast

iron link shown the
allowable stresses are 30
MPa in tension and 120
MPa in compression,
determine the largest
force P which can be
applied to the link.
Solution

Centroid.
Solution

Centroidal Moment of Inertia.

Solution

Force and Couple at C.

Solution Superposition

Largest Allowable Force

The allowable tensile stress of 30 MPa

The allowable compressive stress of

120 MPa
Maximum
UNSYMMETRIC BENDING ANALYSIS
Our analysis of pure bending has been limited so far to members
possessing at least one plane of symmetry and subjected to couples
acting in that plane.

Consider the bending couples do not act in a plane of symmetry of the

member. However, since the vertical plane is not a plane of symmetry,
we cannot expect the member to bend in that plane or the neutral axis
of the section to coincide with the axis of the couple.
UNSYMMETRIC BENDING ANALYSIS
Consider first a member with a vertical plane of symmetry subjected to
bending couples M and M’

Applied moment is
resolved into y and z components

According to resolved moment

bending stress can be calculated at
any point on the cross section using
superposition method.
UNSYMMETRIC BENDING ANALYSIS

Neutral axis for unsymmetric bending.

GENERAL CASE OF ECCENTRIC AXIAL
LOADING ANALYSIS

The stresses due to the centric load P

 Example
(a) Determine the stress at points A, B, C, and D.
(b) Locate the neutral axis of the cross section.
Solution
Solution

b. Neutral Axis. The stress will be zero at a point G between B and C, and at a
point H between D and A. Since the stress distribution is linear,