Lesson 05,Bending and Shearing Stresses

1. Bending- and Shearing
Stresses
Lecturer;
Dr. Dawood S. Atrushi
December 2014

2. Chapter 5 Stresses in Beam (Basic Topics)
Deflection Curve
¢ Lateral loads
acting on a beam
cause the beam to
bend, thereby
deforming the axis
of the beam into
curve line, this is
known as the
deflection curve
of the beam.
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Introduction
loads acting transversely to the longitudinal axis
loads create shear forces and bending
moments, stresses and strains due to V
are discussed in this chapter
lateral loads acting on a beam cause the
bend, thereby deforming the axis of
beam into curve line, this is known as
deflection curve of the beam
beams are assumed to be symmetric about x-y plane, i.e. y-axis
axis of symmetric of the cross section, all loads are assumed to act in

3. Plane of bending
The beams are assumed to be
symmetric about x-y plane, i.e. y-axis is
an axis of symmetric of the cross
section, all loads are assumed to act in
the x-y plane, then the bending
deflection occurs in the same plane, it is
known as the plane of bending.
Deflection of beam
It is the displacement of that point from
its original position, measured in y
direction.
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4. known as the plane of bending
the deflection Pure of the Bending beam is the displacement and of Non-uniform
that point from its
original position, measured in y direction
Bending
5.2 Pure Bending and Nonuniform Bending
Pure bending:
M = constant V = dM / dx = 0
pure bending :
M = constant V = dM / dx = 0
pure bending in simple beam and cantilever beam are shown
1
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5. Non-uniform
bending:
M ≠ constant
V = dM / dx ≠ 0
nonuniform bending :
M J constant
Simple beam with
V = dM central / dx J region 0
in
pure bending and
simple beam with central region in pure
end regions in non-uniform
bending and end regions bending.
in nonuniform
bending is shown
5.3 Curvature of a Beam
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consider a cantilever beam subjected to a

6. simple beam with central region in pure
bending and end regions in nonuniform
bending is shown
5.3 Curvature of a Beam
Curvature of a Beam
Consider a cantilever
beam subjected to a load
P. Choose 2 points m1 and
m2 on the deflection curve,
consider a cantilever beam subjected to a
load P
choose 2 points m1 and m2 on the
5.3 Curvature of a Beam
consider a cantilever beam subjected to a
load P
choose 2 points m1 and m2 on the
deflection curve, their normals intersect at
point O', is called the center of curvature,
the distance m1O' is called radius of
curvature !, and the curvature is
defined as
deflection curve, their normals intersect at
point O', is called the center of curvature,
the distance m1O' is called radius of
curvature !, and the curvature is
defined as
l Center of curvature;
intersection of their
normals at point O’.
l radius of curvature ρ;
the distance m1O’
l Curvature κ is defined
as
= 1 /!
= 1 /!
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and we have ! d = ds
and we have d= ds

7. deflection curve, their normals intersect at
point O', is called the center of curvature,
the distance m1O' is called radius of
curvature !, and the curvature is
defined as
Curvature of a Beam
consider a cantilever beam subjected to a
We have; ρdθ = ds
= 1 /!
If the deflection is
small ds ≈ dx,
then
and we have ! d = ds
if the deflection is small ds M dx, then
1 d d
= C = C = C
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P
choose 2 points m1 and m2 on the
deflection curve, their normals intersect at
O', is called the center of curvature,
distance m1O' is called radius of
curvature !, and the curvature is
as
= 1 /!
we have ! d = ds
deflection is small ds M dx, then
!ds dx
sign convention for curvature

8. deflection curve, their normals intersect at
point O', is called the center of curvature,
the distance m1O' is called radius of
curvature !, and the curvature is
We have; ρdθ = ds
defined as
= 1 /!
If the deflection is
small ds ≈ dx,
then
and we have ! d = ds
if the deflection is small ds M dx, then
1 d d
= C = C = C
!ds dx
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sign convention for curvature
1 /!
! d = ds
deflection is small ds M dx, then
1 d d
C = C = C
!ds dx
convention for curvature
bent concave upward (convex downward)
bent concave downward (convex upward)

9. Longitudinal Strains in
Beams
Consider a portion AB of a beam in pure bending
produced by a positive bending moment M, the
cross section may be of any shape provided it is
symmetric about y-axis;
5.4 Longitudinal Strains in Beams
consider a portion ab of a beam in pure bending produced by a
positive bending moment M, the cross section may be of any shape
provided it is symmetric about y-axis
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under the moment M, its axis is bent into a circular curve, cross

10. cross section may be of any shape
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11. Under the moment M, its axis is bent into a
circular curve,
pure bending produced by a
section may be of any shape
into a circular curve, cross
normal to longitudinal lines
¢ cross section mn and pq remain
plane and normal to longitudinal lines
¢ the curve is circular
¢ longitudinal lines nq are elongated,
and mp are shortened.
¢ the surface ss do not change in
length and is called the neutral
surface.
¢ its intersection with the cross-sectional
plane is called neutral axis,
for instance, the z axis is the neutral
axis of the cross section.
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12. In the deformed element, denote ρ the
distance from 0’ to N.A., thus
ρdθ = dx
consider consider ¢ Consider the the longitudinal longitudinal the longitudinal line line ef, ef, line the the ef, length length the
L1 L1 after after length L1 after bending is
y
L1 = (! - y) d = dx - C dx
L1 = (! - y) y
then ef = L1 - dx = - C dx
!
and the strain of line ef is
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y
!
then ef =
and the strain of line ef is

13. then ef = L1 - dx = - C dx
and the strain of line ef is
The strain of line ef is:
ef y
x = CC = - C = - y
dx !
x vary linear with y (the distance from N.S.)
εx linear with y.
!
y 0 (above N. S.) = -
y 0 (below N. S.) = +
The longitudinal strains in a beam are
accompanied by transverse strains in the y
and z directions because of the effects of
Poisson's ratio!
the longitudinal strains in a beam are accompanied by in the y and z directions because of the effects of Poisson's 13 Bending Stresses - DAT December, 2014

14. in the y and z directions because of the effects of Poisson's ratio
Example 1
A simply supported
beam AB, L = 4.9 m,
h = 300 mm bent by
M0 into a circular arc.
εbottom = εx = 0.00125
Determine ρ, κ, and δ
(midpoint deflection)
4
Example 5-1
a simply supported beam AB,
L = 4.9 m h = 300 mm
bent by M0 into a circular arc
bottom = x = 0.00125
determine !, , and (midpoint
deflection)
y - 150
! = - C = - CCCC
x 0.00125
= 120 m
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15. then sin = CC = CCCC = 0.020
! 2 x 2,400
Normal Stress in Beams
(Linear Elastic Materials)
= 0.02 rad = 1.146o
then = 120 x 103 (1 - cos 1.146o) = 24 mm
Consider a beam section subjected to a
positive bending moment M.
5.4 Normal Stress in Beams (Linear Elastic Materials)
П x occurs due to bending, Р the longitudinal line of the beam is
subjected only to tension or compression, if the material is linear elastic
¢ The longitudinal
line of the beam
is subjected
only to tension
or compression,
if the material is
linear elastic.
then x = E x = - E y
vary linear with distance y
from the neutral surface
consider a positive bending
moment M applied, stresses are
positive below N.S. and negative
above N.S.
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16. = 0.02 rad = 1.146o
= 120 x 103 (1 - cos 1.146o) = 24 mm
x occurs due to bending, the longitudinal line of the subjected only to tension or compression, if the material is linear elastic
x occurs due to bending, Р the longitudinal line of the beam is
Then;
subjected only to tension or compression, if the material is linear elastic
then x = E x = - E y
vary linear with distance y
from the neutral surface
Stress in Beams (Linear Elastic Materials)
occurs due to bending, Р the longitudinal line of the beam is
only to tension or compression, if the material is linear elastic
then x = E x = - E y
• σx vary linear with distance y from the
vary linear with distance y
neutral surface.
the neutral surface
consider a positive bending
consider a positive bending
x = E x = - E y
linear with distance y
• σx Stresses
moment M applied, stresses are
positive below N.S. and negative
above N.S.
are positive
below the
N.A., and
negative
above the
N.A.
moment M applied, stresses are
positive below N.S. and negative
neutral surface
N.S.
no axial force acts on the cross section, the only resultant is M,
two equations must satisfy for static equilibrium condition
П no axial force acts on the cross section, the only resultant thus two equations must satisfy for static equilibrium condition
a positive bending
M applied, stresses are
below N.S. and negative
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i.e. Fx = Н dA = - НE y dA = 0

17. moment M applied, stresses are
positive below N.S. and negative
above N.S.
moment M applied, stresses are
positive below N.S. and negative
above N.S.
¢ No axial force acts on the cross section,
the only resultant is M, thus two
equations must satisfy for static
equilibrium condition;
П no axial force acts on the cross section, the thus two equations must satisfy for static equilibrium condition
П no axial force acts on the cross section, the only resultant thus two equations must satisfy for static equilibrium condition
i.e. Fx = Н dA = - НE y dA = 0
i.e. Fx = Н dA = - НE y dA = 0
П E and are constants at the cross section, 5
П E and are constants at the cross section, thus we have
E and κ are constants at the cross section,
thus we have
l Нy dA = 0
5
we conclude that the neutral axis passes through section, also for the symmetrical condition in 17 Bending Stresses - DAT December, 2014

18. we conclude that the neutral axis passes through the controid section, also for the symmetrical condition in y axis, the pass through the centroid, hence, the origin of coordinates the centroid of the cross section
we conclude that the neutral axis passes through section, also for the symmetrical condition in y pass through the centroid, hence, the origin of coordinates the centroid of the cross section
section, also for the symmetrical condition in y axis, the pass through the centroid, hence, the origin of coordinates O the centroid of the cross section
we conclude that the neutral axis passes through the controid section, also for the symmetrical condition in y axis, the pass through the centroid, hence, the origin of coordinates the centroid of the cross section
conclude that the neutral axis passes through the controid of the cross
We conclude that the neutral axis passes
through the centroid of the cross section.
for the symmetrical condition in y axis, the y axis must
the centroid, hence, the origin of coordinates O is located at
the moment resultant of stress is
the cross the section
moment The moment resultant resultant of stress of stress x the moment resultant of is
σx is;
x the moment resultant of stress is
resultant of stress dM = - is
x y stress dA
x is
x dM = x dM - x y = dA
- x y dA
dM = - x y dA
then M dM = = - - Нx x y y dA
dA = НE y2 dA = E НM = E I
where I = Нy2 dA is the moment of inertia of the area w. r. t. z axis
then M = - Нx y dA = НE y2 dA = E НM = E I
where I = Нy2 dA is the moment of inertia of the area w. r. t. z axis
then M = - Нx y dA = НE y2 dA M = E I
where I = Нy2 dA is the moment of inertia area w. r. t. z axis
then M = - Нx y dA = НE y2 dA = E НM = E I
where I = Нy2 dA is the moment of inertia of the area w. r. t. z axis
= - Нx y dA = НE y2 dA = E Нy2 dA
= E I
= Нy2 dA is the moment of inertia of the cross-sectional
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- DAT December, 2014
axis
thus = C = CC
1 M
I is the moment
of inertia of the
cross-sectional
Then;

19. M = E I
where I = Нy2 dA is the moment of inertia of the cross-area w. r. t. z axis
where I = Нy2 dA is the moment of inertia of the cross-sectional
area w. r. t. z axis
1 M
thus = C = CC
1 M
! E I
thus = C = CC
! E I
this is the moment-curvature equation,
and EI is called flexural rigidity
this is the moment-curvature equation,
and EI is called flexural rigidity
+ M = + curvature
- M = - curvature
the normal stress is
M M y
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x = - E y = - E y (CC) = - CC
E I I
Thus;
This is the Moment-
Curvature Equation.
EI is called Flexural
Rigidity.
+ M = + curvature
- M = - curvature
the normal stress is
M M y
x = - E y = - E y (CC) = - CC

20. - M = - curvature
the normal stress is
The normal stress is;
M M y
x = - E y = - E y (CC) = - CC
E I I
this is called the flexure Flexture formula, Formula
the stress x stresses or flexural stresses
6
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6
curvature
curvature
M M y
y = - E y (CC) = - CC
E I I
flexure formula, the stress x is called bending
stresses
σx is called bending stresses or flextural
stresses
σx is varying linearly
6
+ M = + curvature
- M = - curvature
the normal stress is
M M y
x = - E y = - E y (CC) = - CC
E I I
this is called the flexure formula, the stress x stresses or flexural stresses

21. The maximum tensile and compressive
stresses occur at the points located
farthest from the N.A.
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M compressive
farthest
x M 1 / I
maximum tensile and compressive
at the points located farthest
M c1 M
- CC = - C
I S1
M c2 M
CC = C
I S2
I I
= C , S2 = C are known as the section moduli
c1 c2

22. M c M
x vary linearly with y
x M M x M 1 / I
thus S1 = S2 and 1 = - 2 = - CC = - C
the maximum tensile and compressive
stresses occur at the points located farthest
from the N.A.
the maximum tensile and compressive
stresses for rectangular occur cross at the section
points located farthest
from The the maximal N.A.
stresses;
the maximum tensile and compressive
stresses occur at the points located farthest
from the N.A.
b h3 b h2
M c1 M
I = CC S = CC
M c1 M
1 = - CC = - C
12 6
1 = - CC = - C
I S1
M c2 M
I S1
M c2 M
for circular cross section
M c1 M
2 = CC = C
1 = - CC = - C
2 = CC = C
d 4 d 3
I S2
I S2
I S1
M c2 M
I = CC S = CC
64 32
I I
I I
I S
2 = CC = C
where S1 = C , S2 = C are known as the section moduli
where S1 = C , S2 = C are known as the section moduli
the I preceding S2
analysis of normal stress in beams concerned pure bending,
c1 c1 c2
c2
no shear force
I I
in the case of nonuniform bending (V J 0), shear force produces warping
if the cross section is symmetric w.r.t. z axis (double symmetric section), then c1 = c2 = c
if the cross section is symmetric w.r.t. z axis (double symmetric section), then c1 = c2 = c
where S1 = C , S2 = C are known as the section moduli
if the cross section is symmetric w.r.t. z axis (double symmetric M c cross
M
section), then c1 = c2 = c
M c M
c1 c2
S1 and S2 are known as the
7
section moduli
thus S1 = S2 and 1 = - 2 = - CC = - C
thus S1 = S2 and 1 = - 2 = - CC = - C
I S
I S
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M c M
thus S1 = S2 and = - = - = -

23. when we have nonuniform bending
the flexure formula gives results in the beam where the stress distribution
not disrupted by Example irregularities in the shape, 2
or by discontinuous in loading
otherwise, stress concentration occurs)
A steel wire of diameter d = 4 mm is bent
around a cylindrical drum of radius R0 = 0.5
m, E = 200 Gpa. Determine M and σmax.
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example 5-2
steel wire of diameter d = 4 mm
bent around a cylindrical drum of radius
= 0.5 m
= 200 GPa pl = 1200 MPa
determine M and max
the radius of curvature of the wire is
d
! = R0 + C

24. MPa 1,200 MPa (OK)
Example 3
= 796.8 MPa 1,200 MPa (OK)
= 6.7 m
50 kN
700 mm
tensile and
A simple beam AB of length L = 6.7 m
q = 22kN/m, P = 50kN
b = 220mm, h = 700mm
Determine the maximum tensile and comprssive
stresses due to bending
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25. = 12 mm b = 300 mm h = 80 mm
determine = 0.018 the m3
Example maximum tensile 4
and
compressive stresses in the beam
An overhanged beam ABC subjected
uniform load of intensity q = 3.2 kN/m for the
cross section (channel section) t = 12 mm, b
= 300 mm, h = 80 mm. Determine the
maximum tensile and compressive stresses
in the beam.
139.9 kN-m
CCCCC = 10.8 MPa
0.018 m3
construct the V-dia. and M-dia. first
we can find + Mmax = 2.205 kN-m
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10.8 MPa
subjected
m
- Mmax = - 3.6 kN-m
next, we want to find the N. A. of the section
A(mm2) y(mm) A y (mm3)
A1 3,312 6 19,872
A2 960 40 38,400
A3 960 40 38,400
total 5,232 96,672

26. Thank You
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