Bending Stress

BENDING STRESSES IN BEAMS

Bending is usually combined with the shearing action. However, to simplify, the effect of shear may be neglected ( this is true when the maximum bending moment is considered---- shear is ZERO) in calculating the stresses due to bending only. Such a theory wherein stresses due to bending alone is considered is known as PURE BENDING or SIMPLE BENDING theory.

BS 1

Example of pure bendingBS 2

SFD-

+

a aA B

VA= W

W W

VB= W

C D

BMD

Wa Wa

+Pure bending between C & D

BENDING ACTION:

BS 3

Sagging

M

NEUTRAL AXIS

NEUTRAL LAYER

Neutral Axis

σc

σt

Hogging

Neutral Axis

BS 4

σc

σt

Neutral layer

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.

BS 5

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.

BS 6

Assumptions made in Pure bending theory

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

DERIVATION OF PURE BENDING EQUATION

Relationship between bending stress and radius of curvature

BS 8

PART I:

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.

BS 9

After bending the new length of GH equals

The initial length of fibre GH equals R dθ

EF= E’F’ = dx = R dθ

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.

In b/w there is a fibre (EF) which is neither shortened in length nor increased in length (Neutral Layer).

GlHl = (R+y) dθ = R dθ + y dθ

BS 10

σ ь

EЄ =

If σь is the bending stress and E is the Young’s modulus of the material, then strain

=> Є =y dθ

R dθ

y

R=Є =

change in length

original length

Therefore the strain in fibre GH

Change in length of fibre GH = (R dθ + y dθ) - R dθ = y dθ

BS 11

i.e. bending stress is proportional to distance y, and hence maximum bending stress occurs at the farthest fibre from the neutral axis.

σь

E=

y

Rσь = y

E

R=> --------(1)

Part II : Moment of resistance

day

N A

Consider an elemental area ‘da’ at a distance ‘y’ from the neutral axis.

BS 12

The force on this elemental area = σь × da

day

N A

{from (1)}= × y × daE

R

the moment of resistance offered =

moment of this force about neutral axis.

∴

Therefore, dM = y da × y E

R= y² da

E

R

y² da = second moment of the area = moment of inertia about the neutral axis.

BS 13

Total moment of resistance offered by the beam section,

E

R= y² da

E

RM = y² da

M = INA E

R

= --------(2)E

R

M

INA

(Bernoulli-Euler bending equation)

Where E= Young’s modulus R= Radius of curvature, M= Bending moment at the section

BS 14

Note: (1) For equilibrium the moment of resistance equals bending moment at that section

(2) Neutral axis coincides with the horizontal centroidal axis of the cross section

From equation 1 & 2,

σь

y==

E

R

M

INA

----BENDING EQUATION.

Unit ----- mm3 , m3

BS 15

INA= Moment of inertia about neutral axis

σ ь= Bending stress

y = distance of the fibre from the neutral axis

SECTION MODULUS:

M

INA

=σь

y=> M = σь ×

INA

y=> M = σь(max) ×

INA

ymax

=> M = σь(max) × Z

Where, Z =INA

ymax

= section modulus (property of the section)

Where σb(max) = maximum permissible bending stress

M = maximum bending moment on the section

(1) Rectangular cross section

b

d/2

N A

Ymax=d/2

BS 16

Z =INA

ymax

=bd3

12

d

2

bd2

6=

(2) Hollow rectangular section

(3) Circular section

B

bD/2

Ymax=D/2

d/2

DN A

d

N A

Ymax=d/2

BS 17

Z =INA

ymax

=D

2

BD3 - bd3

12

BD3 - bd3

6D=

Z =INA

ymax

D

2=

πd4

64

πd3

32=

(4) Triangular section

b

h

N A

Y max = 2h/3

BS 18

Z =INA

ymax

=bh3

36

2h

3

bh2

24=

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

A100mm

X

WEB

BS 1 PRACTICE PROBLEMS

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

1202020

300mm

BS 2

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

4)A groove 40mm×40mm is cut symmetrically throughout 4)A groove 40mm×40mm is cut symmetrically throughout the length of the circular brass section as shown in fig. If the length of the circular brass section as shown in fig. If the tensile stress shall not exceed 25 N/mm², find the safe the tensile stress shall not exceed 25 N/mm², find the safe uniformly distributed load which the brass can carry on a uniformly 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

BS 5

5) A simply supported beam of rectangular cross section 100mm200mm 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 200mm200mm 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.

A B

60KN 30 KN

3m 3m 2m

( Ans: 274mm274mm)