Shivendra Nandan Galgotias University

1. 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.

2. 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

3. BENDING ACTION:
Sagging
M
NEUTRALAXIS
NEUTRALLAYER
σc
Neutral Axis
σt

4. Hogging
σc
Neutral Axis
σt
Neutral layer

5. 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.

6. 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.

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

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

9. 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

10. 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.

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

12. 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

13. 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

14. 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

15. (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

16. (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

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

18. (2) Hollow rectangular section
Z= INA / ymax
=1/12(BD3
-bd3
) / (D/2)
=(BD3
-bd3
) / 6D
(3) Circular section
Z= INA / ymax
=(πd4
/64) / (d/2)
= πd3
/ 32
B
b
D/2
Ymax=D/2
d/2 D
N A
d
N A
Y max=d/2

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

20. (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.
NUMERICAL PROBLEMS
w / unit run
5 m
200 mm
300 mm
Maximum stress = 50 N/mm²
Allowable (permissible) stress = 50/5 =10 N/mm2

21. NA
b
b
NA
I
y
M
yI
M
σ
σ
=
=
Moment of resistance or moment
carrying capacity of the beam = M'
External Bending
moment
NA
b
I
y
M
max
max
max
σ
=
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

22. 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

23. (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.
b
d=2b
2.5 m 3.5 m
9 KN
12 KN/m
A B

24. Σ 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

25. 9 KN
39.75 kN
41.25 kN
11.25 kN
2.25 kN+
-
2.5 m 3.5 m
12 KN/m
A B
C
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

26. 2.5 m 3.5 m
12 KN/m
A B
X
xX
-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

27. 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

28. 231.11mm
231.11
231.11
462.22mm
8 N/mm2
8 N/mm2
50 mm
50 mm
σc
σt
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)
N A

29. (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
450
20
20
200
4mRSJ

30. 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
450
20
20
200
N A
245 mm
245 mm

31. (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

32. (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
500
25
20

33. section is given by
M/INA= σmax /ymax
M=( σmax × INA ) /ymax = (σmax × 7.14 × 108
) /250
=2.86 × 106
σmax
INA=2[(200 × 25³/12)+200×25 × (237.5)²] +(20× (450)³/12)
=7.14 × 108
mm4
σmax
250

34. Y
σ
σmax
y 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)

35. 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

36. Therefore moment of resistance of top flange =
Total moment of resistance of both the flanges
=2.26x106
σmax
dyydyyMF ∫∫ ××=×=
250
225
2
max
250
225
max
2
5
4
2
5
4
2 σσ
∫
250
225
max
2
5
4
dyy σ
% 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%

37. Total moment of resistance (moment carrying capacity ) of both
the flanges
OR
INA=2[(200 × 25³/12) +200×25 × (237.5)²] =5.64 × 108
mm4
y
I
y
M NA
250
maxσσ
σ
=
=
M = 2.26×106
σmax
where
NAI
y
y
M 250
maxσ
=

38. (5)Locate & calculate the position and magnitude of
maximum bending stress for the beam shown.
10mm
5mm
500 N
80 mm
x
X
X
Let us consider a section X-X at a distance of ‘x’ from the free end.

39. Diameter at X-X , Dx =5 + x/16
Dx=5 + 0.0625 x
Therefore Ixx= πDx
4
/64 = π (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)

40. σb(x-x) = (500 × x × Dx )/ (2 × πDx
4
/64 )
= (5092.96 × x) / Dx
3
=(5092.96 x) / (5+0.0625x) 3
= (5092.96 x ) (5+ 0.0625x)-3
096.5092)0625.05(
)}0625.0)0625.05(3()96.5092{(
3
4
=×+
+×+×−=
−
−
x
xx
96.5092)0625.05(
)0625.0)0625.05(3()96.5092(
3
4
×+=
×+×
−
−
x
xx
Now, to have maximum bending stress, dσb(xx) /dx = 0
5+ 0.0625x =0.1875x
x = 40 mm ∴ Max bending stress = 483.13 Mpa
34
)0625.05(
1
)0625.05(
1875.0
xx
x
+
=
+

41. (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?
250 mm
200mm
25mm
25mm

42. =Σay / Σa
= 200×25 × (250+12.5) + 250 ×25 ×125
(200 ×25) +(250 ×25)
= 186.11mm
INA= (200 × 253
)/12 + 200 × 25 ×(88.89-12.5)2
+(25x2503
)/12 + 25 × 250 ×(186.11-125)2
=85.32x106
mm4
Y
Y
Y
200mm
25mm
25mm
186.11
mm
88.89
mm

43. σt
Let us allow the permissible value of stress in tension σt=100 N/mm2
From similar triangles
σc / σt = 88.89/186.11
σc / 100= 88.89/186.11
∴ σc =47.762N/mm2
< 50 Hence safe.
The actual extreme fibre stress values are σc = 47.762N/mm2
&
σt = 100 N/mm2
σc
88.89 mm
186.11 mm

44. 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

45. 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)
25mm
N
25mm
A
100mm
X
WEB
BS 1 PRACTICE PROBLEMS

46. 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)
40
20
150
120
2020
300mm

47. 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
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)
120mm
80mm
30MM
BS 3

48. 4)A groove 40mm×40mm is cut symmetrically throughout4)A groove 40mm×40mm is cut symmetrically throughout
the length of the circular brass section as shown in fig. Ifthe length of the circular brass section as shown in fig. If
the tensile stress shall not exceed 25 N/mm², find the safethe tensile stress shall not exceed 25 N/mm², find the safe
uniformly distributed load which the brass can carry on auniformly distributed load which the brass can carry on a
simply supported span of 4 meters.simply supported span of 4 meters.
( Ans: 5150 N/m)( Ans: 5150 N/m)
100mm
40
40
BS 4

49. 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)

50. 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)

51. 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.
A B
60KN 30 KN
3m 3m 2m
( Ans: 274mm274mm)