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.
Example of pure bending
W
W
C
A
D
a
VA= W
B
a
VB= W
SFD
+
-
Wa
Wa
+
BMD
Pure bending
between C & D
BENDING ACTION:
Sagging
M
NEUTRAL AXIS
NEUTRAL LAYER
σc
Neutral Axis
σt
Hogging
Neutral layer
σt
Neutral Axis
σc
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.
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.
6) A transverse section of the beam which is plane before bending
will remain plane even after bending.
7) Stress is purely longitudinal.
BS 7
Note:
homogeneous: of the same kind throughout
Isotropic: of equal elastic properties in all directions.
w1
Longitudinal axis
A transverse section of the beam = the cross section of the beam
DERIVATION OF PURE BENDING EQUATION
PART I:
Relationship between bending stress and radius of curvature.
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
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.
Note: Neutral axis coincides with the horizontal centroidal
axis of the cross section
σc
N
A
σt
Moment of resistance
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.
σc
Neutral Axis
σt
Moment of resistance
y
da
A
N
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
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
(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,
M= Bending moment at the section,
R= Radius of curvature,
INA= Moment of inertia about neutral axis,
σ ь= Bending stress
y = distance of the fibre from the neutral axis
SECTION MODULUS:
(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,
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
section modulus
(1) Rectangular cross section
b
Z= INA/ ymax
=( bd3/12) / d/2
N
Y max=d/2
A
d
=bd2/6
(2) Hollow rectangular section
B
D/2
Z= INA / ymax
b
d/2
N
Ymax=D/2
A
D
=1/12(BD3-bd3) / (D/2)
=(BD3-bd3) / 6D
(3) Circular section
Z= INA / ymax
d
=(d4/64) / (d/2)
= d3/ 32
Y max=d/2
N
A
(4) Triangular section
Z = INA / Y max
Y max = 2h/3
=(bh3 /36) / (2h/3)
h
N
=bh2/24
A
h/3
b
NUMERICAL PROBLEMS
(1) Calculate the maximum UDL the beam shown in Fig. can carry
if the bending stress at failure is 50 MPa & factor of safety to be
given is 5.
200 mm
w / unit run
5m
Maximum stress = 50 N/mm²
Allowable (permissible) stress = 50/5 =10 N/mm2
300 mm
M


I NA
M 
y

b
y
External Bending
moment
M
max
External maximum
Bending moment
b
I NA
Moment of resistance or moment
carrying capacity of the beam = M'

 b max
y max
σbmax
I NA
Maximum Moment of resistance or
maximum moment carrying capacity
of the beam = M'
σb will be maximum when y = ymax and M = Mmax
Ymax
We have to consider section of the beam where the BM is max, and
stress should be calculated at the farthest fibre from the neutral axis.
E/R=M/INA= σ b/y
M/INA= σ b/y =>
INA = bd³/12= (200× 300³)/12= 45 × 107 mm4
Ymax= d/2=300/2= 150 mm
BMmax =wl²/8= w ×(5000) ²/8
(w × 5000²/8) / 45 × 107 = 10/150
w= 9.6 N/mm = 9.6 kN/m
(2) For the beam shown in Fig. design a rectangular section making the
depth twice the width. Max permissible bending stress = 8 N/mm² .Also
calculate the stress values at a depth of 50mm from the top & bottom at
the section of maximum BM.
9 KN
12 KN/m
A
b
B
2.5 m
3.5 m
d=2b
 MA=0
(12×6 × 3) + (9 × 2.5) -VB × 6 = 0
VB= 238.5/6 =39.75 kN
ΣFy = 0
VA + VB=(12 ×6)+ 9
 VA= 41.25 kN
9 KN
12 KN/m
A
B
C
2.5 m
3.5 m
11.25 kN
41.25 kN
+
2.25 kN
-
39.75 kN
Max. bending moment will occur at the section where the shear force
is zero. The SFD shows that the section having zero shear force is
available in the portion BC. Let that section be X-X, considered at a
distance x from support B as shown below. The shear force at that
section can be calculated as
X
12 KN/m
A
B
2.5 m
3.5 m
X
x
-VB+12 x =0
i.e. -39.75+12x=0
x = 39.75/12 =3.312 m.
BM is max @ 3.312m from B.
BM@xx = 39.75×3.31 - 12×3.31×(3.31)/2
= 65.84 kN-m = 65.84× 10 6 N mm
Now M/I NA= σb/y
65.84×106/(b×(2b)3/12) = 8/b
b³= 1.5×8.23×106
 b= 231.11 mm , d= 2b= 462.22 mm
8 N/mm2
231.11mm
50 mm
231.11
N
σc
462.22mm
A
231.11
σt
8 N/mm2
From similar triangles,
8/ 231.11 = σc/(231.11-50) = σt / (231.11-50)
σc = 6.27 N/ mm2( compressive) & σt = 6.27 N/ mm2(tensile)
50 mm
(3)A Rolled Steel Joist (RSJ) of 200mm × 450 mm of 4m span is
simply supported on its ends. The flanges are strengthened by two
300mm× 20mm plates one riveted to each flange. The second
moment of the area of the RSJ equals 35060×104 mm4. Calculate the
load the beam can carry for the following cases, if the bending stress
in the plates is not to exceed 120 MPa, (a) greatest central
concentrated load (b) maximum UDL throughout the span
300
20
450
RSJ
20
200
4m
INA=I NA(RSJ)+MI due to plates about NA
= (35060 × 104 )+2 [(300 ×(20)³/12+300 × 20 ×(235)²]
=1.01 × 109 mm4
300
20
N
245 mm
A
450
245 mm
20
200
(a)
M/INA= σ b/Y
[Mmax=PL/4]
Ymax = (450+2 ×20) /2= 245mm
σ =120N/mm2
(P ×4000) / 4 (1.01 ×109)=120/245
P = 4.95×105 N
(b)
M/INA= σ b/Y
[Mmax= wl2/8]
Ymax = (450+2 ×20) /2= 245mm
w = 247.35 N/ mm = 247.35 KN/m
(4) An I section beam has 200 mm wide flanges and an overall depth of
500 mm. Each flange is 25 mm thick and the web is 20 mm thick. At a
certain section the BM is ‘M.’ Find what percentage of M is resisted by
flanges and the web.
200
25
500
20
The moment of resistance (moment carrying capacity )of the entire
section is given by
M/INA= σmax /ymax
INA=2[(200 × 25³/12)+200×25 × (237.5)²] +(20× (450)³/12)
=7.14 × 108 mm4
M=( σmax × INA ) /ymax = (σmax × 7.14 × 108) /250
=2.86 × 106 σmax
σmax
250
Consider an element of thickness dy at a distance of y from neutral axis
Let σmax be the extreme fibre stress( maximum bending stress)
σmax
σ
Y
y
250
From similar triangle principle
σmax / σ =250/y
σ =( σmax × y) /250
Area of the element =200× dy
Force on the element = stress × area
P= (σmax ×y/250) ×( 200 dy)
The moment of resistance of this about the NA equals
= (σmax ×y/250) ×( 200 dy) y
=(4/5) y² σmax dy
Therefore moment of resistance of top flange =
250

225
4
5
y  max dy
2
Total moment of resistance of both the flanges
4
250
4
250
y  max dy  2    max  y dy
2255
225
5
M F  2 
2
=2.26x106 σmax
% moment resisted by flanges =(MF/M) × 100
=(2.16 × 106 σmax )/(2.86 × 106 × σmax) × 100 =79.02%
Therefore % moment resisted by the web= 20.98%
2
OR
Total moment of resistance (moment carrying capacity ) of both
the flanges

M 
I
y
where 


max
NA
y
250
INA=2[(200 × 25³/12) +200×25 × (237.5)²] =5.64 × 108 mm4

max
M  250
y
y
I NA
M = 2.26×106 σmax
(5)Locate & calculate the position and magnitude of
maximum bending stress for the beam shown.
X
X
10mm
500 N
x
5mm
80 mm
Let us consider a section X-X at a distance of ‘x’ from the free end.
Bending stress is not maximum at left end (10 mm dia end)
because at that end bending moment may be maximum but Ixx
is also maximum.
 
M
I
Y
Diameter at X-X ,
Dx =5 + x/16
Dx=5 + 0.0625 x
Therefore Ixx= Dx4/64 =  (5 + 0.0625x) 4 /64
M/I= σb/y
Mxx= 500x,
y= ymax @ section x-x = Dx/2
σb(x-x) = (Mxx× ymax) / Ixx
Dx/2
σb(x-x) = (500 × x × Dx) / (2 × Ixx)
σb(x-x) = (500 × x × Dx )/ (2 × Dx4/64 )
= (5092.96 × x) / Dx3
=(5092.96 x) / (5+0.0625x) 3
= (5092.96 x ) (5+ 0.0625x)-3
Now, to have maximum bending stress, dσb(xx) /dx = 0
 {( 5092 . 96 x ) (  3  ( 5  0 . 0625 x )
(5  0 . 0625 x )
3
4
 0 . 0625 )} 
 5092 . 96  0
( 5092 . 96 x ) ( 3  ( 5  0 . 0625 x )
 0 . 0625 )
4
 (5  0 . 0625 x )  5092 . 96
3
0 . 1875 x
( 5  0 . 0625 x )
4

1
( 5  0 . 0625 x )
3
5+ 0.0625x =0.1875x
x = 40 mm
 Max bending stress = 483.13 Mpa
(6) The beam section shown in fig. has a simple span of 5 m. If the
extreme fibre stresses are restricted to 100 MPa & 50 MPa under
tension & compression respectively, calculate the safe UDL
(throughout the span) the beam can carry inclusive of self weight. What
are the actual extreme fibre stresses?
200mm
25mm
250 mm
25mm
200mm
Y =ay / a
Y = 200×25 × (250+12.5) + 250 ×25 ×125
25mm
88.89
mm
(200 ×25) +(250 ×25)
Y = 186.11mm
INA= (200 × 253)/12 + 200 × 25 ×(88.89-12.5)2
+(25x2503 )/12 + 25 × 250 ×(186.11-125)2
=85.32x106mm4
186.11
mm
25mm
Let us allow the permissible value of stress in tension σt=100 N/mm2
From similar triangles
σc
σc / σt = 88.89/186.11
σc / 100= 88.89/186.11
88.89 mm
186.11 mm
σt
 σc =47.762N/mm2 < 50 Hence safe.
The actual extreme fibre stress values are σc = 47.762N/mm2 &
σt = 100 N/mm2
Mmax=wl2/8 = w × 50002/8
y=186.11 for σt=100
y=88.89 for σc =47.762
M/INA= σb/y
(wl2)/(8 × 85.32 × 106) = 100/186.11= 47.72/ 88.89
w =14.67 N/mm=14.67 KN/m
BS 1
PRACTICE PROBLEMS
1) Find the width “x” of the flange of a cast iron beam having
the section shown in fig. such that the maximum compressive
stress is three times the maximum tensile stress, the member
being in pure bending subjected to sagging moment.
( Ans: x= 225 mm)
25m m
100m m
A
N
WEB
25m m
X
2)A cast iron beam has a section as shown in fig. Find
the position of the neutral axis and the moment of inertia
about the neutral axis. When subjected to bending
moment the tensile stress at the bottom fibre is 25
N/mm². Find, a) the value of the bending moment b) the
stress at the top fibre.
( Ans: M= 25070 Nm, σc =33.39 N/mm²)
120
20
20
20
150
40
300m m
BS 3
3)A cast iron beam has a section as
shown in fig .The beam is a simply
supported on a span of 1.25 meters
and is used to carry a downward point
load at midspan. Find the magnitude 120m m
of the load if the maximum tensile
stress on the beam section is 30
N/mm². Determine also the maximum
compressive stress.
(Ans. W= 174.22 N, σc =40.73 N/mm²)
30MM
80m m
BS 4
4)A groove 40mm×40mm is cut symmetrically throughout
the length of the circular brass section as shown in fig. If
the tensile stress shall not exceed 25 N/mm², find the safe
uniformly distributed load which the brass can carry on a
simply supported span of 4 meters.
( Ans: 5150 N/m)
40
100mm
40
BS 5
5) A simply supported beam of rectangular cross section
100mm200mm has a span of 5m. Find the maximum safe
UDL, the beam can carry over the entire span, if the
maximum bending stress and maximum shear stress are
not to exceed 10 MPa & 0.60 MPa respectively.
( Ans: w = 2.13 KN/m)
BS 6
6) A cantilever beam of square cross section 200mm200mm
which is 2m long, just fails at a load of 12KN placed at its
free end. A beam of the same material and having
rectangular cross section 150mm 300mm is simply
supported over a span of 3m.Calculate the central point
load required just to break this beam.
(Ans: P = 27KN)
BS 7
7) In an overhanging beam of wood shown in Fig., the
allowable stresses in bending and shear are 8MPa &
0.80MPA respectively. Determine the minimum size of a
square section required for the beam.
30 KN
60KN
A
B
3m
( Ans: 274mm274mm)
3m
2m