50_MoS_Chapter 4_combination of bothPure bending.pdf
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Sep 13, 2024
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About This Presentation
Ductile specimen breaks along a
plane of maximum shear.
+ Brittle specimen breaks along planes perpendicular to the
direction in which tension is a maximum
Slide Content
CHAPTER 4: PURE BENDING
Lecture 1:
1. Introduction
2. Shear and moment diagrams
3. Relation of load, shear and bending moment
Lecture 2:
4. Deformations of symmetric member in pure bending
5. Stress and strain due to pure bending
6. Design beam for bending
Lecture 3:
7. Composite sections
8. Eccentric bending
9. Unsymmetrical bending
Lecture 1:
1. Introduction
2. Shear and moment diagrams
3. Relation of load, shear and bending moment
Lecture 2:
4. Deformations of symmetric member in pure bending
5. Stress and strain due to pure bending
6. Design beam for bending
Lecture 3:
7. Composite sections
8. Eccentric bending
9. Unsymmetrical bending
1. Introduction
- To date we have studied:
+ Axial loading (bars and cables)
+ Torsion loading (shafts)
- Today’s objective:
Member subjected to bending.
- Pure bending:
+ Member subjected to equal and opposite couples
M and M’ acting in the same longitudinal plane.
+ In this topic: Assume that each member possesses a
symmetric plane.
- To date we have studied:
+ Axial loading (bars and cables)
+ Torsion loading (shafts)
- Today’s objective:
Member subjected to bending.
- Pure bending:
+ Member subjected to equal and opposite couples
M and M’ acting in the same longitudinal plane.
+ In this topic: Assume that each member possesses a
symmetric plane.
Example 1:
An athlete holds the barbell.
Distances from the weights to hands are equal (a).
Pure bending occurs on
center portion of bar.
P
A = P
B
M = M’ = P
A.a = P
B.a
a a b
A
B C D
P
A
P
B
R
C R
D
C D M M’
Center portion in pure
bending will form a
circular arc
An athlete holds the barbell.
Distances from the weights to hands are equal (a).
Pure bending occurs on
center portion of bar.
P
A = P
B
M = M’ = P
A.a = P
B.a
a a b
A
B C D
P
A
P
B
R
C R
D
C D M M’
Center portion in pure
bending will form a
circular arc
Beam and shaft:
Important mechanical elements in engineering.
Shaft:
Beam: slender and support loadings that are applied
perpendicular to their longitudinal axis.
General beam: long, straight, constant cross-section area
Important mechanical elements in engineering.
Shaft:
Beam: slender and support loadings that are applied
perpendicular to their longitudinal axis.
General beam: long, straight, constant cross-section area
Classification of beams:
Statically
Determinate
beams
Simply support
beam
Overhanging
beam
Cantilever
beam
Statically
Indeterminate
beams
Continuous
beam
Beam fixed at
one end and
simply supported
at the other end
Fixed beam
Beam
connected
by
hinges
Statically
Determinate
beams
Simply support
beam
Overhanging
beam
Cantilever
beam
Statically
Indeterminate
beams
Continuous
beam
Beam fixed at
one end and
simply supported
at the other end
Fixed beam
Beam
connected
by
hinges
2. Shear and bending-moment diagrams
Sign convection:
Sign convection:
- Analysis of a simply supported beam:
Example 2: Draw shear and moment-bending diagrams
Example 3: Draw shear and moment-bending diagrams
Example 4: Draw shear and moment-bending diagrams
Example 5: Draw shear and moment-bending diagrams
3. Relation of load, shear and bending moment
A simply supported beam AB with distribution load w per
unit length.
�
�=0 V – (V + V) – w x = 0
V = - w x
�??????
��
=−�
the shear curve is negative
V
D – V
C = - ���
�
�
�
�
= - (area under load curve between C&D)
Free body
diagram of
the portion
of beam CC’
A simply supported beam AB with distribution load w per
unit length.
�
�=0 V – (V + V) – w x = 0
V = - w x
�??????
��
=−�
the shear curve is negative
V
D – V
C = - ���
�
�
�
�
= - (area under load curve between C&D)
Free body
diagram of
the portion
of beam CC’
Relations between shear and bending moment
??????
�′=0 (M + M) – M - Vx + w x
x
2
= 0
M = V x -
1
2
� �
2
Dividing by x and letting x approach 0
dM
dx
=V
M
D – M
C = ??????��
�
�
�
�
= area under shear curve between C & D
??????
�′=0 (M + M) – M - Vx + w x
x
2
= 0
M = V x -
1
2
� �
2
Dividing by x and letting x approach 0
dM
dx
=V
M
D – M
C = ??????��
�
�
�
�
= area under shear curve between C & D
Example 7: Draw shear and moment-bending diagrams
1/ Draw shear and bending moment diagrams
2/ Example 4
2/ Example 4
Symmetric member in pure bending
x: Normal stress
xy,
xz: Shear stress
The system of the elementary internal forces exerted on the
section is equivalent to the couple M.
Tension on the
bottom
Compression on
the top
x: Normal stress
xy,
xz: Shear stress
The system of the elementary internal forces exerted on the
section is equivalent to the couple M.
Tension on the
bottom
Compression on
the top
�
�=
��??????=0
??????
�
= �
��??????=0;
??????
�
= −�
��??????=??????
�=
��??????=0
??????
�
= �
��??????=0;
??????
�
= −�
��??????=??????
4. Deformations in a symmetric member in pure bending
A prismatic member processing a
symmetric plane and subjected to
couples M and M’.
+ M: same in any cross section
bend uniformly
+ Top portion symmetric plane
= AB (a portion of a circle of center C).
+ AB and A’B’ change in length.
A prismatic member processing a
symmetric plane and subjected to
couples M and M’.
+ M: same in any cross section
bend uniformly
+ Top portion symmetric plane
= AB (a portion of a circle of center C).
+ AB and A’B’ change in length.
Consider a rubber (highly deformable material) straight
prismatic member subjected to a bending moment.
prismatic member subjected to a bending moment.
3 assumptions:
-The longitudinal axis x (within the neutral surface) does not
experience any change in length.
-All cross sections of the beam remain plane and
perpendicular to the longitudinal axis during deformation.
-Any deformation of the cross section within its own plane
will be neglected.
-The longitudinal axis x (within the neutral surface) does not
experience any change in length.
-All cross sections of the beam remain plane and
perpendicular to the longitudinal axis during deformation.
-Any deformation of the cross section within its own plane
will be neglected.
5. Stress and Strain due to bending
- Neutral surface:
x = 0;
x = 0
- Neutral axis: Neutral surface
Intersects a transverse section.
Determine longitudinal strain
of JK (above neutral surface)?
Length of original JK (DE):
L =
Length of JK: L’ = (-y)
Elongated deformation of JK:
= L’ – L = -y
x =
�
=
;�
=
;�
(Geometry of deformation) Transverse section
- Neutral surface:
x = 0;
x = 0
- Neutral axis: Neutral surface
Intersects a transverse section.
Determine longitudinal strain
of JK (above neutral surface)?
Length of original JK (DE):
L =
Length of JK: L’ = (-y)
Elongated deformation of JK:
= L’ – L = -y
x =
�
=
;�
=
;�
(Geometry of deformation) Transverse section
- Longitudinal normal strain
x varies linearly with distance y
from the neutral surface.
x = −
�
The largest distance from the neutral surface: c
Maximum absolute value:
�=
�
�=−
�
�
�
Transverse section
x varies linearly with distance y
from the neutral surface.
x = −
�
The largest distance from the neutral surface: c
Maximum absolute value:
�=
�
�=−
�
�
�
Transverse section
Transverse Strain
Poisson’s ratio effect:
�=
�=−
�
Consider transverse strain
�:
Above neutral axis, beam is compressed
� < 0
� > 0
Below neutral axis, beam is elongated
� > 0
� < 0
�=−
�
;
�=
�
The neutral axis of transverse section
will be bent into a circle of radius
′
=
Poisson’s ratio effect:
�=
�=−
�
Consider transverse strain
�:
Above neutral axis, beam is compressed
� < 0
� > 0
Below neutral axis, beam is elongated
� > 0
� < 0
�=−
�
;
�=
�
The neutral axis of transverse section
will be bent into a circle of radius
′
=
Stress distribution:
How stress distribution in a beam relates to the internal
resultant bending moment acting on the beam’s cross
section?
Assumptions:
-Stress under bending remains below yield strength
y
(proportional limit and elastic limit)
-Homogeneous material
�=�
�
with
�=−
�
�
�
� = −
�
�
�
� = −
�
�
�
(Stress distribution over cross-
sectional area)
How stress distribution in a beam relates to the internal
resultant bending moment acting on the beam’s cross
section?
Assumptions:
-Stress under bending remains below yield strength
y
(proportional limit and elastic limit)
-Homogeneous material
�=�
�
with
�=−
�
�
�
� = −
�
�
�
� = −
�
�
�
(Stress distribution over cross-
sectional area)
The location of neutral surface:
�
�=
��??????=0 −
??????
�
��??????=0 ��??????=0
In elastic range, neutral axis pass through the centroid of the
section.
Elastic flexure formula
??????
�= −�
��??????=??????;
−�−
�
??????
�
�??????=
??????
�
�
2
�??????=??????
I
z = �
2
�??????
�
: Area moment of inertia
�=
�.�
??????
�
�=−
��
??????
�
�
�=
��??????=0 −
??????
�
��??????=0 ��??????=0
In elastic range, neutral axis pass through the centroid of the
section.
Elastic flexure formula
??????
�= −�
��??????=??????;
−�−
�
??????
�
�??????=
??????
�
�
2
�??????=??????
I
z = �
2
�??????
�
: Area moment of inertia
�=
�.�
??????
�
�=−
��
??????
�
�=
??????.�
??????
�
where
�: 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 equations of equilibrium.
c: The perpendicular distance from the neutral axis to
a point farthest away from the neutral axis.
I
z: The moment of inertia of the cross-sectional area
about the neutral axis.
�=
??????.�
??????
�
where
�: 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 equations of equilibrium.
c: The perpendicular distance from the neutral axis to
a point farthest away from the neutral axis.
I
z: The moment of inertia of the cross-sectional area
about the neutral axis.
Area moment of inertia of a rectangular area:
I
z = �
2
�??????
�
= �
2
(���)
�/2
;�/2
=
��
3
12
Maximum absolute value:
�=
�
Definition: Curvature of the neutral surface
1
=
�
�
=
�/�
�
=
??????�
??????
��
�
=
??????
??????
��
I
z = �
2
�??????
�
= �
2
(���)
�/2
;�/2
=
��
3
12
Maximum absolute value:
�=
�
Definition: Curvature of the neutral surface
1
=
�
�
=
�/�
�
=
??????�
??????
��
�
=
??????
??????
��
Example 1:
If the beam is subjected to a bending moment of M = 50
kN.m, determine the maximum bending stress in the beam?
Solution:
If the beam is subjected to a bending moment of M = 50
kN.m, determine the maximum bending stress in the beam?
Solution:
The centroid moment of inertia:
Example 2:
Bending in xz plane
determine I
z in yz plane.
A
1 = A
3 = 2400 mm
2
A
2 = 3060 mm
2
Q
1 = Q
3 = 2400 x 50 = 120000 mm
3
Q
2 = 3060 x 15 = 45900 mm
3
� =
2�
1:�
2
2�
1:�
2
= 36.37 mm
I
z,1 = I
z,3 =
24 ∗ 100
3
12
+ 2400*13.63
2
I
z,2 =
102∗30
3
12
+ 3060*21.37
2
I
z = 2I
z,1 + I
z,2 = 6.519 x 10
6
[mm
4
]
Example 2:
Bending in xz plane
determine I
z in yz plane.
A
1 = A
3 = 2400 mm
2
A
2 = 3060 mm
2
Q
1 = Q
3 = 2400 x 50 = 120000 mm
3
Q
2 = 3060 x 15 = 45900 mm
3
� =
2�
1:�
2
2�
1:�
2
= 36.37 mm
I
z,1 = I
z,3 =
24 ∗ 100
3
12
+ 2400*13.63
2
I
z,2 =
102∗30
3
12
+ 3060*21.37
2
I
z = 2I
z,1 + I
z,2 = 6.519 x 10
6
[mm
4
]
Example 3: Determine centroid moment of inertia?
A
1 = 1800 mm
2
; A
2 = 1200 mm
2
Q
1 = 50 x 1800 = 90000 mm
3
Q
2 = 20 x 1200 = 24000 mm
3
?????? =
90000:24000
1800:1200
= 38 mm
I
x,1 = I
1 + A
1d
1
2
I
x,2 = I
2 + A
2d
2
2
I = I
x,1 + I
x,2
d
1
d
2
A
1 = 1800 mm
2
; A
2 = 1200 mm
2
Q
1 = 50 x 1800 = 90000 mm
3
Q
2 = 20 x 1200 = 24000 mm
3
?????? =
90000:24000
1800:1200
= 38 mm
I
x,1 = I
1 + A
1d
1
2
I
x,2 = I
2 + A
2d
2
2
I = I
x,1 + I
x,2
d
1
d
2
6. Design of prismatic beam for bending
The design of beam is controlled by the maximum absolute
value ??????
��� of bending moment.
�=
�
????????????�.�
??????
�=
�
????????????�
??????
For a safe design:
�≤
���<
?????? (elastic range)
where
Y is yield strength,
���=
??????
������ �� ������
,
U is ultimate stress
The design of beam is controlled by the maximum absolute
value ??????
��� of bending moment.
�=
�
????????????�.�
??????
�=
�
????????????�
??????
For a safe design:
�≤
���<
?????? (elastic range)
where
Y is yield strength,
���=
??????
������ �� ������
,
U is ultimate stress
1/ Determine the centroidal area moment of inertia (I
,z)
?
,z)
?
2/ Knowing that the allowable normal stress for the steel
used is 24 ksi. Determine the minimum flange width b that
can be used.
used is 24 ksi. Determine the minimum flange width b that
can be used.
7. Composite sections
- Member’s material is made of homogeneous, with
modulus E:
- The member subjected to pure bending is made of 2 or
more materials with different moduli of elasticity:
x = −
�
bonded
E
1
E
2
- Member’s material is made of homogeneous, with
modulus E:
- The member subjected to pure bending is made of 2 or
more materials with different moduli of elasticity:
x = −
�
bonded
E
1
E
2
8. Eccentric bending
Centric loading: Line of action of loads P and P’ passes
through the centroid of the cross section
Eccentric loading: Line of action of loads P and P’ does not
pass through the centroid of the cross section.
P P
Vertical forces exerted on the press cause an eccentric loading
Centric loading: Line of action of loads P and P’ passes
through the centroid of the cross section
Eccentric loading: Line of action of loads P and P’ does not
pass through the centroid of the cross section.
P P
Vertical forces exerted on the press cause an eccentric loading
Member with eccentric loading: F = P’ and M = P’d
Internal forces in member of eccentric loading
Stress distribution in eccentric loading:
��������+
��������=
�
Internal forces in member of eccentric loading
Stress distribution in eccentric loading:
��������+
��������=
�
�=
��������+
��������
�=
??????
??????
−
??????.�
??????
�
- The distribution of stresses across the section is linear but
not uniform.
Another example of stress distribution in eccentric loading:
- The neutral axis does not coincide with the centroid axis
of the section.
At centroid axis: y = 0 but
� 0
Neutral axis corresponds to points
�= 0
Neutral axis
�=
��������+
��������
�=
??????
??????
−
??????.�
??????
�
- The distribution of stresses across the section is linear but
not uniform.
Another example of stress distribution in eccentric loading:
- The neutral axis does not coincide with the centroid axis
of the section.
At centroid axis: y = 0 but
� 0
Neutral axis corresponds to points
�= 0
Neutral axis
Example 3: Determine max. value of P?
all = 30 Mpa (in tension),
all = 120 Mpa (in compressive)
P uniform stress distribution
0=
�
�
M= Pd linear stress distribution:
1=
�.�
�
??????
;
2=
�.�
�
??????
all = 30 Mpa (in tension),
all = 120 Mpa (in compressive)
P uniform stress distribution
0=
�
�
M= Pd linear stress distribution:
1=
�.�
�
??????
;
2=
�.�
�
??????
Superposition:
�= -
�
�
+
�.�
�
??????
= 377P < 30
�= -
�
�
-
�.�
�
??????
= - 1559P < -120
�= -
�
�
+
�.�
�
??????
= 377P < 30
�= -
�
�
-
�.�
�
??????
= - 1559P < -120
Example 4: P = 8 kN
Determine the stress at point A.
Solution:
Determine the stress at point A.
Solution:
9. Unsymmetrical bending
Symmetrical bending:
+ Member possesses at least one plane of symmetry
+ Member is subjected to couples acting in that plane.
+ The members remained symmetric and bend in the plane
of symmetry
+ The neutral axis of the cross section coincided with the
axis of the couple.
Symmetrical bending:
+ Member possesses at least one plane of symmetry
+ Member is subjected to couples acting in that plane.
+ The members remained symmetric and bend in the plane
of symmetry
+ The neutral axis of the cross section coincided with the
axis of the couple.
Un-symmetric bending:
+ Bending couples do not act in the plane of symmetry of
the member.
Bending couples act in
different plane
Member does
not possess any
plane of
symmetry
+ Bending couples do not act in the plane of symmetry of
the member.
Bending couples act in
different plane
Member does
not possess any
plane of
symmetry
An unsymmetrical case: - Resolve M into M
z and M
y
M
z = Mcos; M
y = Msin
�=−
�
�.�
??????
�
�=+
�
�.�
??????
�
�=−
�
�.�
??????
�
+
�
�.�
??????
�
z and M
y
M
z = Mcos; M
y = Msin
�=−
�
�.�
??????
�
�=+
�
�.�
??????
�
�=−
�
�.�
??????
�
+
�
�.�
??????
�
Example 5:
Couple M is applied to a beam of the cross section shown in
a plane forming an angle with the vertical.
Determine stress at:
(a)Point A
(b)Point B
(c)Point D
Couple M is applied to a beam of the cross section shown in
a plane forming an angle with the vertical.
Determine stress at:
(a)Point A
(b)Point B
(c)Point D
1/ Determine the stress at points A, B, D and E.
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