This document discusses bending stresses in beams, detailing how to determine stress caused by bending, the distribution of shear and moment, and the application of the pure bending theory. It covers various concepts such as the effects of bending, combined stresses, moment of resistance, and specific formulas for calculating bending stress, highlighting the role of the neutral axis. Examples and problem-solving related to beam design under specified loads are also included.

1. By
Prof. Ramesh R. Rajguru

2. Learning Outcomes L20
Bending Stresses

3.  In this module we will determine the stress in a
beam caused by bending.
 How to find the variation of the shear and
moment in these members.
 Then once the internal moment is determined,
the maximum bending stress can be calculated.

4. 3.1 Bending stresses:
Theory of pure Bending, Assumptions, Flexural formula for straight beams, moment of
resistance, bending stress distribution, Section modulus, beams of uniform strength.
3.2 Direct & Bending Stresses:
Combined stresses, Eccentricity, Stress distribution, Core /kernel of Section.
3.3 Shear Stresses:
Distribution of shear stresses for the section of beam.
Module 3

5. Here is an example of where combined
axial and bending stress can occur.
Beams are important structural members used in
building construction. Their design is often based
upon their ability to resist bending stress.

6. Note the distortion of the lines due to bending of this rubber bar. The top line
stretches, the bottom line compresses, and the center line remains the same
length. Furthermore the vertical lines rotate and yet remain straight.

7. This wood specimen failed in bending due to its fibers being
crushed at its top and torn apart at its bottom.

12. BENDING STRESSES IN BEAMS
Beams are subjected to bending moment and shearing forces
which vary from section to section. To resist the bending
moment and shearing force, the beam section develops
stresses.
Bending is usually associated with shear. However, for
simplicity we neglect effect of shear and consider moment
alone ( this is true when the maximum bending moment is
considered---- shear is ZERO) to find the stresses due to
bending. Such a theory wherein stresses due to bending alone
is considered is known as PURE BENDING or SIMPLE
BENDING theory.

13. Example of pure bending
W W
SFD
-
+
a a
A B
VA= W VB= W
C D
BMD
Wa Wa
+
Pure bending
between C & D

14. BENDING ACTION:
Sagging
M
NEUTRALAXIS
NEUT
RALLAY
ER
σc
Neutral Axis
σt

15. Hogging
σc
Neutral Axis
σt
Neutral layer

16. BENDING ACTION
•Sagging-> Fibres below the neutral axis (NA) get stretched -> Fibres
are under tension
•Fibres above the NA get compressed -> Fibres are in compression
•Hogging -> Vice-versa
•In between there is a fibre or layer which neither undergoes tension
nor compression. This layer is called Neutral Layer (stresses are
zero).
•The trace of this layer on the c/s is called the Neutral Axis.

17. Assumptions made in Pure bending theory
1) The beam is initially straight and every layer is free to
expand or contract.
2) The material is homogenous and isotropic.
3) Young’s modulus (E) is same in both tension and
compression.
4) Stresses are within the elastic limit.
5) The radius of curvature of the beam is very large in
comparison to the depth of the beam.

18. 6) A transverse section of the beam which is plane before bending
will remain plane even after bending.
7) Stress is purely longitudinal.

19. DERIVATION OF PURE BENDING EQUATION
Relationship between bending stress and radius of curvature.
PART I:

20. Consider the beam section of length “dx” subjected to pure
bending. After bending the fibre AB is shortened in length,
whereas the fibre CD is increased in length.
In b/w there is a fibre (EF) which is neither shortened in length
nor increased in length (Neutral Layer).
Let the radius of the fibre E'F′ be R . Let us select one more fibre
GH at a distance of ‘y’ from the fibre EF as shown in the fig.
EF= E'F′ = dx = R dθ
The initial length of fibre GH equals R dθ
After bending the new length of GH equals
G'H′= (R+y) dθ
= R dθ + y dθ

21. Change in length of fibre GH = (R dθ + y dθ) - R dθ = y dθ
Therefore the strain in fibre GH
Є= change in length / original length= y dθ/ R dθ
Є = y/R
If σ ь is the bending stress and E is the Young’s modulus of the material,
then strain
Є = σ ь/E
σ ь /E = y/R => σ ь = (E/R) y---------(1)
σ ь = (E/R) y => i.e. bending stress in any fibre is proportional to the
distance of the fibre (y) from the neutral axis and hence maximum
bending stress occurs at the farthest fibre from the neutral axis.

22. Note: Neutral axis coincides with the horizontal centroidal axis of
the cross section
N A
σc
σt

23. on one side of the neutral axis there are compressive stresses and on
the other there are tensile stresses. These stresses form a couple,
whose moment must be equal to the external moment M. The
moment of this couple, which resists the external bending moment,
is known as moment of resistance.
Moment of resistance
σc
Neutral Axis
σt

24. Moment of resistance
Consider an elemental area ‘da’ at a distance ‘y’ from the neutral axis.
The force on this elemental area = σ ь × da
= (E/R) y × da {from (1)}
The moment of this resisting force about neutral axis =
(E/R) y da × y = (E/R) y² da
da
y
N A

25. Total moment of resistance offered by the beam section,
M'=  (E/R) y² da
= E/R  y² da
 y² da =second moment of the area =moment of inertia about the
neutral axis.
M'= (E/R) INA
For equilibrium moment of resistance (M') should be equal to
applied moment M
i.e. M' = M
Hence. We get M = (E/R) INA

26. (E/R) = (M/INA)--------(2)
From equation 1 & 2, (M/INA)= (E/R) = (σ ь /y) ----
BENDING EQUATION.
(Bernoulli-Euler bending equation)
Where E= Young’s modulus, R= Radius of curvature,
M= Bending moment at the section,
INA= Moment of inertia about neutral axis,
σ ь= Bending stress
y = distance of the fibre from the neutral axis

27.  σ max = the maximum normal stress in the
member, which occurs at a point on the cross-
sectional area farthest away from the neutral axis.
 M = the resultant internal moment, determined
from the method of sections and the equations of
equilibrium, and calculated about the neutral axis
of the cross section
 Y = perpendicular distance from the neutral axis to
a point farthest away from the neutral axis. This is
where σ max acts.
 I = moment of inertia of the cross-sectional area
about the neutral axis

28. (M/I)=(σ ь /y)
or σ ь = (M/I) y
Its shows maximum bending stress occurs at the greatest distance
from the neutral axis.
Let ymax = distance of the extreme fibre from the N.A.
σ ь(max) = maximum bending stress at distance ymax
σ ь(max) = (M/I) y max
where M is the maximum moment carrying capacity of the section,
SECTION MODULUS:
M = σ ь(max) (I /y max)
M = σ ь(max) (I/ymax) = σ ь(max) Z
Where Z= I/ymax= section modulus (property of the section)
Unit ----- mm3 , m3

29. NA
b
b
NA
I
y
M
y
I
M




Moment of resistance or moment
carrying capacity of the beam = M'
External Bending
moment
NA
b
I
y
M
max
max
max


External maximum
Bending moment Maximum Moment of resistance or
maximum moment carrying capacity
of the beam = M'
σbmax
Ymax
σb will be maximum when y = ymax and M = Mmax

30. (1) Rectangular cross section
Z= INA/ ymax
=( bd3/12) / d/2
=bd2/6
section modulus
b
N A
Y max=d/2
d

31. (2) Hollow rectangular section
Z= INA / ymax
=1/12(BD3-bd3) / (D/2)
=(BD3-bd3) / 6D
(3) Circular section
Z= INA / ymax
=(d4/64) / (d/2)
= d3/ 32
B
b
D/2
Ymax=D/2
d/2 D
N A
d
N A
Y max=d/2

32. (4) Triangular section
b
h
N A
Y max = 2h/3
Z = INA / Y max
=(bh3 /36) / (2h/3)
=bh2/24
h/3

33.  The simply supported beam in Fig. a has the
cross-sectional area shown in Fig. b. Determine
the 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?

34. The simply supported beam in Fig. a has the cross-sectional
area shown in Fig. b. Determine the 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?

38. A T beam of span 5m has a flange 125 mm x 12.5 mm and web
187.5mm x8mm. If the maximum permissible stress is 150 Mpa,
find the max. UDL the beam can carry.

39.  An I section girder 250 mm deep has 20mm thick
web. The top flange is 120 mm x 20 mm and the
bottom flange is 160 mm x 20 mm. The UDL on the
girder is 6KN/m and the Max. stress due to
bending is limited to 70 Mpa. Determine the max.
simply supported span on the girder can be
supported. Also determine the % of BM regsisted
by the flanges.

43.  A beam of I section is simply supported over a
span of 4m with an overhang of 1m on either
side of the supports. It carries load W each at
its ends and load 2W at the center. I section
having top flange 100 mm x 20 mm web is 20
mm x 100 mm and bottom flange 180 mm x 20
mm.
 Calculate max. value of W, if allowable
bending stresses are 35 Mpa in tension and 45
Mpa in compression.

46.  A beam with rectangular C/S of dimension 150
mm x 175mm is loaded as shown in Fig.
Determine the maximum tensile and
compressive stresses acting normal to the
section through the beam.