CIV206
Mechanics of Materials
Pure Bending
Prof. Osama Ahmed Mohamed, Ph.D., P.E.,M.ASCE
Contents
Pure Bending
Other Loading Types
Symmetric Member in Pure Bending
Bending Deformations
Strain Due to Bending
Beam Section Properties
Properties of American Standard Shapes
Deformations in a Transverse Cross Section
Reading:
Chapter 4, Class textbook
4- 2
Internal Forces in Structural Members
• Beams subjected to transverse
loading develop internal shear
forces and bending moments.
Free body
diagram
• Frames subjected to external
loading develop internal
shear forces, bending
moments, and axial forces.
Pure Bending
Pure bending occurs
in section where
there is not shear
force.
Pure Bending: Prismatic
members subjected to
equal and opposite
couples acting in the
same longitudinal plane
Pure bending
between C & D.
Free body diagram
between C & D shows no
4- 4
shear force.
Pure Bending: No shear force (C to D)
• Because the beam is
symmetric, the reactions
share the transverse load
equally,
• Shear force at a section just
to the left of point C is
80 Ibs – V = 0, V = 80 Ibs
• Shear force at a section just
to the right of C
80 Ib – 80 Ib – V =0, V = 0
Therefore, there is no shear
force from C to D.
Span CD is in Pure Bending
• Bending moment at C (+cw):
M = - 80 Ib x 12 in = -960 Ib.in
• Bending moment at D (+cw):
M = -80 Ib x (12+26) in + 80 Ib x 26
in = - 960 Ib.in
So moment didn’t change from CD
and remained equal to 960 Ib.in.
Transverse and Eccentric Loading
Transverse Loading:
Concentrated or
distributed transverse
load produces internal
forces equivalent to a
shear force and a
couple
Eccentric Loading: Axial
loading which does not pass
through section centroid
produces internal forces
equivalent to an axial force
and a couple
4- 6
Load on a Beam
Transverse loading with respect
to the axis of the beam
Cantilever beam: free at one end
(A) and fixed at the other end (B),
or continuous.
Free body diagram produced by cutting a
beam produces the internal forces,
which are shear force and bending
moment.
Shear force
Convention: this moment is
assumed to compress the
bottom of the beam
Symmetric Member in Pure Bending
This is the plane
of symmetry: it
divides the
section into two
identical
segments
This is called the
axis of bending
(perpendicular to
the plane
containing the
internal bending
moments)
The internal bending
moments are
contained in in the
plane of symmetry
The moment is the same about
any axis perpendicular to the
plane of the couple and zero
about any axis contained in the
plane.
4- 8
Symmetric Member in Pure Bending
Plane of the
couple
Axis of bending
(z-axis)
Compression normal stress occurs
above axis of bending and tension
normal below axis of bending
M z    y x dA  M
Fx    x dA  0
M y   z x dA  0
Forces parallel to x-axis must be
in equilibrium
No bending exist about any axis
except about z-axis
4- 9
Bending Moment and Elastic Stresses
https://youtu.be/asBW0Ojc0bY
M z    y x dA  M
There is no
moment applied
about the y-axis
Fx    x dA  0
M y   z x dA  0
This moment is applied
about the z-axis
Bending Deformations and neutral axis
Beam with a plane of symmetry in pure
bending:
member remains symmetric and bend uniformly to
form circular arc
cross-sectional planes before deformation,
passes through arc center and remains planar
after deformation
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
4- 11
Strain Due to Bending
Consider a beam segment of original length L.
After deformation, the length of the neutral surface
remains L. At other sections,
Such as surface A-B
L     y
  L  L     y     y

y
y
x     
(strain varieslinearly)
L


c
c
m 
or ρ 

y
 x   m
c
Strain at distance y from shape centroid
m
Maximum strain
m 
c

y
x   m
c
4- 12
Where is the Location of Neutral Axis?
For a linearly elastic material,
y
 x  E x   E m
c
y
   m (stress varies linearly)
c
For static equilibrium,
y
Fx  0    x dA     m dA
c
0
m
y dA

c
First moment with respect to
neutral plane is zero.
Therefore, the neutral
surface must pass through
the section centroid.
Stress Due to Bending
For static equilibrium,
 y 
M   y x dA   y   m  dA
 c 
m 2
 mI
M
y dA 
c
c
For a linearly elastic
Mc M
material,
m 

y
I
S
 x  E x   E m
c
y
Substituti
ng  x    m
y
   m (stress varies linearly)
c
c
My
x  
My
I



x  
I
4- 14
Beam Section Properties
The maximum normal stress due to bending,
Mc M

I
S
I  section moment of inertia
I
S   section modulus
c
A beam section with a larger section modulus will
have a lower maximum stress
m 
Consider a rectangular beam cross section,
3
1
I 12 bh
S 
 16 bh3  16 Ah
c
h2
Between two beams with the same cross sectional
area, the beam with the greater depth will be
more effective in resisting bending.
Structural steel beams are designed to have a
large section modulus.
4- 15
Properties of American Standard Shapes
flange
web
4- 16
Example
Knowing that the couple shown acts in
a vertical plane, determine the stress
at (a) point A, (b) point B.
Here is the flexural stress
formula
My
x  
I
Example - Solution
For the grey area
Example - Solution
Neutral axis
Principle of Superposition
•The normal stress due to pure bending
may be combined with the normal stress
due to axial loading and shear stress due
to shear loading to find the complete
state of stress.
Axial
compression
stress
Summary
•Neutral axis coincides with centroid of the
shape for elastic materials. Strain at the
neutral axis zero.
•In elastic materials, the stress is inversely
proportional to the moment of inertia and the
elastic section modulus.