Bài giảng Strength of materials - Chapter 7: Bending - Trần Minh Tú

STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
1/10/2013 1
7
CHAPTER
1/10/2013
BENDING
Contents
7.1. Introduction
7.2. Bending stress
7.3. Shearing stress in bending
7.4. Strength condition
7.5. Sample Problems
7.6. Deflections of beam
7.7. Statically indeterminate beams
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7.1. Introduction
In previous charters, we considered the stresses in the bars caused
by axial loading and torsion. Here we introduce the third fundamental
loading: bending. When deriving the relationship between the bending
moment and the stresses causes, we find it again necessary to make
certain simplifying assumptions.
We use the same steps in the analysis of bending that we used for
torsion in chapter 6.
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7.1. Introduction
Classification of Beam Supports
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7.1. Introduction
 Limitation
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7.1. Introduction
 Segment BC: Mx≠0, Qy=0
=> Pure Bending
 Segments AB,CD: Mx≠0, Qy≠0
=> Nonuniform Bending
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7.1. Introduction
Pure Bending: Prismatic members subjected to equal and opposite couples
acting in the same longitudinal plane
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7.2. Bending stress
 Simplifying assumptions
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7.2. Bending stress
The positive bending moment causes the
material within the bottom portion of the beam
to stretch and the material within the top portion
to compress. Consequently, between these two
regions there must be a surface, called the
neutral surface, in which longitudinal fibers of
the material will not undergo a change in
length.
Neutral axis
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7.2. Bending stress
Neutral fiber
c d
a b
c d
d 
dz
1 2
1 2
1 2
1 2
y
y
a b
Due to bending moment Mx caused
by the applied loading, the cross-
section rotate relatively to each other
by the amount of d .
 ' '
z
y d ddz c d cd y
dz cd d

The Normal strain of the longitudinal
fiber cd that lies distance y below the
neutral surface.
 – radius of curvature of the neutral fiber.
z
y
 
 Compatibility
Consider a segment of the beam
bounded by two cross-sections that
are separated by the infinitesimal
distance dz.
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12
7.2. Bending stress
 Equilibrium
z
y
E 
Following Hooke’s law, we have.
1
???? 
y
z
x
dA 
x
y
z
K
MxBecause of the loads applied in the
plane yOz, thus: Nz=My=0 and Mx≠0.
0z z
A A
E
N dA ydA  
0x
A
ydA S 
0y z
A A
E
M x dA xydA  
0xy
A
xydA I 
x – neutral axis (the neutral axis
passes through the centroid C of the
cross-section).
y - axis – the axis of symmetry of
the cross-section
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7.2. Bending stress
Mx>0: stretch top portion
Mx<0: compress top portion
=> For convenient:
y
z
x
dA 
x
y
z
K
Mx
2
x z x
A A
E E
M y dA y dA I  
1 x
x
M
EI
EIx – stiffness of beam
Mx – internal bending moment
 – radius of neutral longitudinal fiber
x
z
x
M
y
I
 
y – coordinate of point
x
z
x
M
y
I
 
Belong to tensile zone
Belong to compressive zone
 Flexure formula – section modulus
7.2. Bending stress
• Stress distribution
- Stresses vary linearly with
the distance y from neutral axis
• Maximum stresses at a cross-section
max max
x t
x
M
y
I
 
min max
x c
x
M
y
I
 
ytmax – the distance from N.A to a point farthest away from N.A in the tensile portion
ycmax – the distance from N.A to a point farthest away from N.A in the compressive portion
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7.2. Bending stress
x
y
min
max
h/2
h/2
z
Mx
max
2
x x
x x
M Mh
I W
 
max min  
/ 2
x
x
I
W
h
max max
2
t c hy y 
min
2
x x
x x
M Mh
I W
 
with called the section modulus of the beam
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7.2. Bending stress
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7.2. Bending stress
Properties of American Standard Shapes
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7.2. Bending stress
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7.2. Bending stress
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7.2. Bending stress
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7.3. Shearing stress in Bending
In general, a beam will support both shear and moment – non-uniform
bending. The shear force Qy is the results of a transverse shear-stress
distribution that acts over the beam’s cross-section.
As a results of the shearing stress, the shear
strain will be developed and (these will) tend to
distort the cross-section. This non-uniform
shear-strain will cause the cross-section warp
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7.3. Shearing stress in Bending
 The Shear formula
When a beam is subjected to loads that
produce both bending moment and shear force,
then both normal and shear stresses are
developed in the beam.
Consider a beam of rectangular cross section
subjected to a positive shear force. We assume
• The Shear stresses acting on the cross-
section are paralell to the shear force, that is,
parallel to the vertical sides of the cross-section
• The Shear stresses are uniformly distributed
across the width of the beam, although they
may vary along the height
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7.3. Shearing stress in Bending
Let’s consider the horizontal force equilibrium of a portion (of the
element) taken from the beam in fig. (a). A FBD of this element is shown in
fig. (b). This distribution is caused by the bending moment M an M+dM.
We have excluded the effect of V and V+dV and the transverse loading on
the FBD, because they are not involved in a horizontal force summation.
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7.3. Shearing stress in Bending
Now consider the shaded top portion of the
element that has been sectioned at y’ from
the neutral axis. This segment has the width
of bs, and the cross-sectional area of As. The
Longitudinal shear  stress acts over the
bottom face of the segment (constant across
the width bs of the bottom face).
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7.3. Shearing stress in Bending
 ' 0
s s
s s s
z
A A
F dA dA b dz    
 0
s s
s s s
x xA A
M dM M
ydA ydA b dz
I I
 
s
s s
x A
dM
ydA b dz
I
 
Have: 
y
x
dM
Q
I
s
s s
x
A
ydA S 
then: 
 
s
y x
zy s
x
Q S
I b
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 
s
y x
zy s
x
Q S
I b
7.3. Shearing stress in Bending
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7.3. Shearing stress in Bending
x
y
h
b=
y
b
c
max
As
2
21
2 2 2 2 4
s
x
h h b h
S y y b y
2 2
2 2
3 3
12 6
.
. 2 4 4
y y
zy
Q Qb h h
y y
bh b bh

 0
2
zy
h
y  
ax
3
0 
2
y
m
Q
y
bh
 
bs=b; Ix=bh
3/12;
Sx
s=yC.A
s =>
• Shearing stress distributions at
the rectangular cross-section:
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7.3. Shearing stress in Bending
• Shearing stress distributions at
the wide-flanged cross-section:
21 .
2
s
x xS S d y 
21 .
2
y
zy x
x
Q
S d y
I d

max0 => 
y x
x
Q S
y
I d
 
2
1
1
=>
2 2 2
y
x
x
Qh h
y s S d s
I d

y
x
h
s
b
y
d
max
 1
 1
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7.3. Shearing stress in Bending
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7.4. Strength condition
x
y
N
K
C
B
Mx
z
max
min
max
h
/2
h
/2
maxmax

minmin
max
max



B
B
B
B
Consider a rectangular cross-section of the beam. Normal and shearing
stresses distribution on the cross-section are shown in figure.
K, N – uniaxial stress
C- pure shearing stress
B- special plane stress
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7.4. Strenght condition
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7.4. Strenght condition
• Bending stress criterion
Cross-section need to control: cross-section have maximum bending
moment (ductile: absolute magnitude, brittle : maximum positive an
negative bending momet)
- Ductile materials:
- Brittle materials :
   max minmax , allow    
   max min= ; 
tensile compressive
allow allowk n
      
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• Shear stress criterion
Cross-section need to control : cross-section have maximum shear
force
- Ductile materials :
 axmmax   
  0
n

 - Experimental determine 0
 
 
2

 - Failure hypothesis 3
 
 
3

 - Failure hypothesis 4
7.4. Strenght condition
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Problem 7.5.1: 
For the timber beam and loading shown 
• Draw the shear and bend-moment diagrams
• Determine the maximum normal stress due to bending
• Check strength condition of the beam, known []=1,5
kN/cm2
7.5. Sample Problem
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SOLUTION:
• Support reactions:
 0 : 40kN 14kNy B B DF M R R  
• Using method of section
• Section 1 - 1
1 1
1 1 1
0 20 kN 0 20kN
0 20kN 0m 0 0
yF Q Q
M M M


• Section 2 - 2
2 2
2 2 2
0 20 kN 0 20kN
0 20kN 2.5m 0 50kN m
yF Q Q
M M M
 


• Section 3 – 3
• Section 4 – 4
• Section 5 – 5
• Section 6 – 6
3 3
4 4
5 5
6 6
26kN 50kN m
26kN 28kN m
14kN 28kN m
14kN 0
Q M
Q M
Q M
Q M
 
 
 
7.5. Sample Problem
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• From shear force and bending moment
diagram:
max max26kN 50kN mBQ M M 
Q
kN
M
kNm
• Maximum bending moment
221 1
6 6
6 3
3
max
max 6 3
0.080m 0.250m
833.33 10 m
50 10 N m
833.33 10 m
x
x
W b h
M
W

 
6 2
max 60.0 10 Pa=60MPa=6kN/cm 
• smax > [s] => Strength condition isn’t
satistify
7.5. Sample Problem
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7.5. Sample Problem
A simply supported steel beam is
to carry the distributed and
concentrated loads shown.
Knowing that the allowable normal
stress for the grade of steel to be
used is 160 MPa, select the wide-
flange shape that should be used.
SOLUTION:
• Considering the entire beam as a
free-body, determine the reactions
at A and D.
• Develop the shear diagram for the
beam and load distribution. From
the diagram, determine the
maximum bending moment.
• Determine the minimum
acceptable beam section modulus.
Choose the best standard section
which meets this criteria.
Problem 7.5.2: 
38
 5m 60kN 1.5m 50kN 4m 0
58.0kN
58.0kN 60kN 50kN=0
52.0kN
A D
D
y A
A
M V
V
F V
V


52.0kN
60kN
8kN
A A
B A q
B
Q V
Q Q S
Q
• Maximum bending moment: 
Q = 0 => z = 2,6 m.
max
( ) 67.6kNmQM S AE 
Q
kN
58
66Mmax
M
kNm
7.5. Sample Problem
Support reactions:
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• Bending stress criterion:
 
max
min 2
6 3 3
67.6kN m
16kN/cm
422.5 10 m 422.5 m
M
W
c


 
 
max max
max x
x
M M
W
W
 

7.5. Sample Problem
Choose the best standard section which
meets this criteria.
4481.46W200
5358.44W250
5497.38W310
4749.32W360
63738.8W410
mm10 33
 SShape
9.32360 W
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7.6. Deflection of Beam
Because the design of beams is frequently governed by rigidity rather
than strength, the computation of deflection is an integral component of
beam analysis. For example, building codes specify limits on deflections
as well as stresses.
7.4.1. Differential equation of elastic curve
• The deformed axis of the beam is called
elastic curve.
• Consider a cantilever beam with concentrated
load acting downward at the free end. Chosen
coordinates are shown in fig.
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7.6. Deflection of Beam
Assumption.
- Displacements and slopes are very small.
- The stresses are below the elastic limit
• The deflection y is the displacement in the y
direction of any point on the axis of the beam.
Deflection is positive when downward.
m
m
m1
m1
y
• The angle of slope (angle of rotation) is
angle between cross-section before and after
deformation.
• From geometry can obtain
 (z) = tg = y’(z) 
42
7.6. Deflection of Beam
• From equilibrium of bending beam – the curvature of neutral
longitudinal fiber (relationship between bending moment and
curvature for pure bending)
( )1 x
x
M z
EI 
3
2 2
1 "( )
"( )
(1 ' )
y z
y z
y 
• From elementary 
calculus :
'' ( )x
x
M z
y
EI
(small deformation)
But + or - ???
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• Prismatic beams
BENDING MOMENT 
EQUATION
SHEAR FORCE 
EQUATION
LOAD EQUATION
7.6. Deflection of Beam
z
M
M>0
''( ) 0y z 
z
M ''( ) 0y z 
M<0
( )
"( ) x
x
M z
y z
EI
 - Differential equation of elastic curve
2
2x x
d y
EI M
dz
3
3x y
d y
EI Q
dz
4
4
( )x
d y
EI q z
dz
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7.4.2. Double - integration method
7.6. Deflection of Beam
x
x
Mdy
z dz C
dz EI
x
x
M
y(z) dz C .dz D
EI
• Bending moment is function of coordinate z. Then first integration
give the angle of slope function
• Second integration give the deflection’s function
• C an D are constants of integration to be determined from the
prescribed constrains (boundary conditions)
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7.6. Deflection of Beam
• Boundary conditions
• Continuity conditions
C C
y y 
C C
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7.6. Deflection of Beam
Problem 7.4.1: The cantilever beam shown
in figure is subjected to a vertical load P at
its end. Determine the deflection and angle
of slope at free end. EI = const
SOLUTION
 M F L z 
B
F
L-z
L
EIx
z
 '' x
x x
F L zM (z)
y (z)
EI EI
2
x x
F L z) F z
z dz C Lz C
EI EI 2
2 3
x
F z z
y z L Cz D
EI 2 6
0 0 0z C 
0 0 0z y D 
2
B
x
FL
z L
2EI
Boundary condition
3
B
x
FL
y y z L
3EI
Bending moment function:
Substitution to diff. equation of elastic curve:
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• For a beam subjected to a distributed load,
2
2
dM d M dQ
Q z q z
dz dz dz
• Equation for beam displacement becomes
2 4
2 4
d M d y
EI q z
dz dz
3 21 1
1 2 3 46 2
EI y z dz dz dz q z dz
C z C z C z C
• Integrating four times yields
• Constants are determined from boundary 
conditions.
 Direct Determination of the Elastic Curve From the Load Distribution
7.6. Deflection of Beam
• Consider a beam subjected non-uniform bending consist n
segments, is numbered 1,2,,i, i+1,..,n in order from left to right.
Bending rigidities of each segments are: E1I1, E2I2,, EnIn.. Consider
two adjacent segments (i) and (i+1). Between them there is a
special connection that deflection and slope have “jump”. In the
cross-section between two segments there are concentrated loading
and moment, and distributed loading also have “jump”
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7.7. Initial Parameters Method
 0
0F
0M
y0
y
Fa
aM
q0
iq
qi+1
z=a
i
y
i+1
y
(a)
(a)
(a)
i
(a)
i+1
z
 y
a
1 2 i i+1 n
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7.7. Initial Parameters Method
• After mathematical manipulation (Fourier expansion deflection’s
function at z=a), using relations among bending moment, shear force
and transverse distributed load, we obtained recurring formula of
deflection’s function (deflection (i+1)-th segment is calculated through
deflection of i-th segment)
1
2 3 4 5
'
( ) ( ) ( )
1 ( ) ( ) ( ) ( )
...
2! 3! 4! 5!a
i i a a
a a a
y z y z y z a
z a z a z a z a
 M Q q q
EI
a aM M 
a aQ Q 
1( ) ( )a i iq q a q a 
' ' '
1( ) ( )a i iq q a q a 
where
Note: only for EI = const
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7.7. Initial Parameters Method
2 3 4 5
'
1 0 0 0 0 0 0
1
( ) ...
2! 3! 4! 5!
z z z z
y z y z M Q q q
EI
• Deflection of first segment:
'
0 0 0 0 0 0, , , , , ,...y M Q q q - initial parameters:
NOTES:
 Positive sign of couples, concentrated load, and distributed
load is shown in figure.
 If connection between (i)-th and (i+1)-th segment is pinned,
then
 If (i)-th và (i+1)-th segments are one-piece then
0ay 
0a ay 
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Problem 7.8:
Use initial parameters method
to determine deflection at point
C and slope at point D of the
beam subjected by loading as
shown in figure
Solution:
1. Support reactions:
2. Table of initial parameters
B
11
V qa
4
D
9
V qa
4
1 2 3
DB
M=qaP=4qa
aaa
2q
A C
2a
3a
z = 0 z = a z = 2a
0 0y 
0 0 
0 0M 
0 0Q 
0q q 
,
0 0q 
0ay 
0a 
0aM 
a BQ V 
aq q 
, 0aq 
0ay 
0a 
0aM 
aQ P 
, 0aq 
0aq 
To find out yC => determine y2(z)
VB VD
7.8. Sample Problems
To find out D => determine y3’z)
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7.8. Sample Problems
1
2 3 4 5
'
( ) ( ) ( )
1 ( ) ( ) ( ) ( )
...
2! 3! 4! 5!a
i i a a
a a a
y z y z y z a
z a z a z a z a
 M Q q q
EI
Recurring formula
 Consider segment 1(AB):
0 ≤ z ≤ a
4
1 o o
x
qz
y (z) y z
24EI
z = 0 z = a z = 2a
0 0y 
0 0 
0 0M 
0 0Q 
0q q 
,
0 0q 
0ay 
0a 
0aM 
a BQ V 
aq q 
, 0aq 
0ay 
0a 
0aM 
aQ P 
, 0aq 
0aq  Consider segment 2 (BC):
a ≤ z ≤ 2a
4 4 3
B
2 o o
x x x
qz q(z a) V (z a)
y (z) y z
24EI 24EI 6EI
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7.8. Sample Problems
z = 0 z = a z = 2a
0 0y 
0 0 
0 0M 
0 0Q 
0 0q 
,
0 0q 
0ay 
0a 
0aM 
a BQ V 
aq q 
, 0aq 
0ay 
0a 
0aM 
aQ P 
, 0aq 
0aq 
1
2 3 4 5
'
( ) ( ) ( )
1 ( ) ( ) ( ) ( )
...
2! 3! 4! 5!a
i i a a
a a a
y z y z y z a
z a z a z a z a
 M Q q q
EI
 Consider segment 3 (CD):
2a ≤ z ≤ 3a
4 4 3 3
B
3 o o
x x x x
qz q(z a) V (z a) P(z 2a)
y (z) y z
24EI 24EI 6EI 6EI
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7.8. Sample Problems
Equations of elastic curve of every segment :
4
1 o o
x
qz
y (z) y z
24EI
4 4 3
B
2 o o
x x x
qz q(z a) V (z a)
y (z) y z
24EI 24EI 6EI
4 4 3 3
B
3 o o
x x x x
qz q(z a) V (z a) P(z 2a)
y (z) y z
24EI 24EI 6EI 6EI
y0, 0 ???
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7.8. Sample Problems
4
o
x
5qa
y
24EI
3
o
x
qa
6EI
4
C 2
x
7qa
y y (z 2a)
24EI
3
D 3
x
qa
y' (z 3a)
6EI
 To determine 2 initial parameters y0 and 0 , let’s consider boundary
conditions:
z = a => y1(z=a) = 0 
z = 3a => y3(z=3a) = 0
 From equations of elastic curve of two segments y1(z) và y3(z), using
boundary condition, we have:
 Thus
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7.7. Statically Indeterminate Beams
• Consider beam with fixed support at A and roller
support at B.
• From free-body diagram, note that there are four
unknown reaction components.
• Conditions for static equilibrium yield
0 0 0z y AF F M   
The beam is statically indeterminate.
 1 2
0 0
z z
EI y dz M z dz C z C 
• Also have the beam deflection equation,
which introduces two unknowns but provides
three additional equations from the boundary
conditions:
At 0, 0 0 At , 0z y z L y 
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THANK YOU FOR 
ATTENTION !
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