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7/27/2019 Bending stresses Presentation
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3a. Bending Stresses
1Prepared by: Prof. Nabil El-Tayeb
OUTLINE
3a. Bending Stresses
1. Bending Deformation of aStraight Member
2. The Flexure Formula3. Unsymmetrical Bending
4. Curved Beams5. Stress Concentrations
3a. Bending Stresses
2Prepared by: Prof. Nabil El-Tayeb
OBJECTIVES
To establish shear and moment diagrams for a
beam or shaft,
To determine largest shear and moment in a
member, and specify where they occur
To determine stress in the x-sec of memberscaused by bending,
To study special cases of unsymmetrical bending
To study the effect of stress concentrations.
Done
in
Ch3
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3a. Bending Stresses
3Prepared by: Prof. Nabil El-Tayeb
+SFD,N
-
+BMD,
N.m
-
Pure Bending: Prismatic memberssubjected to equal and opposite
couples acting in the same
longitudinal plane. CD
subjected to ZERO SF
3a. Bending Stresses
4Prepared by: Prof. Nabil El-Tayeb
Pri nciple of Superposit ion:The normal
stress due to pure bending may be
combined with the normal stress due to
axial loading and shear stress due to
shear loading to find the complete state
of stress.
Eccentri c L oading: Axial loading
which does not pass through section
centroid produces internal forces
equivalent to an axial force and a
couple.
Transverse Loading:Concentrated or
distributed transverse load produces
internal forces equivalent to a shear
force and a couple.
Other Loading Types
M
P
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3a. Bending Stresses
5Prepared by: Prof. Nabil El-Tayeb
Bending Deformations
length of top layer
decreases and length of
bottom one increases
a neutral sur facemust
exist that is parallel to
the upper and lower
surfaces and for which
the length does not
change
Neutral layer
3a. Bending Stresses
6Prepared by: Prof. Nabil El-Tayeb
BENDING DEFORMATION OF A STRAIGHT MEMBER
A neutral surface is where longitudinal fibers of thematerial will not undergo a change in length.
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3a. Bending Stresses
7Prepared by: Prof. Nabil El-Tayeb
THE FLEXURE FORMULA
By mathematical expression,equilibrium equations of
moment and forces, we get
Ay dA = 0
max
cM =A y
2
dA
The integral represents the moment of inertia of x-sectional area, computed about the neutral axis.
We symbolize its value asI.
3a. Bending Stresses
8Prepared by: Prof. Nabil El-Tayeb
THE FLEXURE FORMULA
Assume that material behaves in a
linear-elastic manner so that
Hookes law applies.
A linear variation of normal strainmustthen be the consequence of
a linear variation in normal stress
Applying Hookes law to Eqn 6-8,
= (y/c)max
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3a. Bending Stresses
9Prepared by: Prof. Nabil El-Tayeb
THE FLEXURE FORMULA
The following Equations are often referred toas the f lexure formula.
Mc
Imax =
max = maximum normal stress in member,at a pt on x-sectional area farthest away fromneutral axis at the edge of the x-sec (c)
M= resultant internal moment, computed aboutneutral axis at distance y in the x-section
I= moment of inertia ofx-sectional area computedabout neutral axis
c = perpendicular distance from neutral axis to a ptfarthest away from neutral axis, where max acts
My
I=
3a. Bending Stresses
10Prepared by: Prof. Nabil El-Tayeb
THE FLEXURE FORMULA
Normal stress at intermediate distancey can bedetermined from:
MyI
a =
is -ve as it acts in the -ve direction (compression)
Equations 6-12 and 6-13 are often referred to asthe flexure formula.
A
B
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3a. Bending Stresses
11Prepared by: Prof. Nabil El-Tayeb
THE FLEXURE FORMULA
IMPORTANT
X-section of straight beam remains plane whenbeam deforms due to bending.
The neutral axis is subjected to zero stress
Due to deformation, longitudinal strain varies
linearlyfrom zero at neutral axis to maximum at
outer fibers of beam
Provided material is homogeneous and Hookeslaw applies, stress also varies linearlyover the x-
section
3a. Bending Stresses
12Prepared by: Prof. Nabil El-Tayeb
FLEXURE FORMULA
Stress Due to Bending (THE FLEXURE FORMULA)
y
xx
y
y=+ve is ve & y=-ve is +ve
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3a. Bending Stresses
13Prepared by: Prof. Nabil El-Tayeb
Example 3.8
A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E= 165 GPa and neglecting
the effects of fillets, (a) Draw the bending stress
distribution in the x-sec., (b) Determine the
maximum tensile and compressive stresses, and
(c) Determine the radius of curvature.
Based on the cross-section
geometry, calculate the location of
the centroid of the section and
moment of inertia.
Parallel-axis theorem applied
2dAII
A
AyY x
(a). Apply the elastic flexural formula to
draw and find the maximum tensile
and compressive stresses.
I
Mcm
(b). Calculate the curvature EI
M
1
SOLUTION:
Y
3a. Bending Stresses
14Prepared by: Prof. Nabil El-Tayeb
Example 3.8 (Cont.)
Based on the cross-section geometry, calculate
the location of the section centroid and
moment of inertia.
mm383000
10114 3
A
AyY
3
3
3
32
101143000
104220120030402
109050180090201
mm,mm,mmArea,
AyA
Ayy
49-43
23
12
123
12
1
23
12
12
m10868mm10868
18120040301218002090
I
dAbhdAIIx
I
Mcm
I=868x103mm4
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3a. Bending Stresses
15Prepared by: Prof. Nabil El-Tayeb
Example 3.8 (Cont.)
Apply the elastic flexural formulato find the maximum tensile and
compressive stresses.
49
49
m10868
m038.0mkN3
m10868
m022.0mkN3
I
cM
I
cM
I
Mc
BB
AA
m
MPa0.76A
MPa3.131B
Calculate the curvature
m7.47
m1095.201 1-3
- +76MPa
131.3MP
49- m10868GPa165mkN31
EI
M
Normal Stress
3a. Bending Stresses
16Prepared by: Prof. Nabil El-Tayeb
Example 3.9
Beam shown has x-sec. area in the shape of a channel.
(a) Determine the normal stresses (maximumstresses) that occurs in the beam at section a-a.
(b) Draw the normal stress distribution over the x-sec area
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3a. Bending Stresses
17Prepared by: Prof. Nabil El-Tayeb
Example 3.9 (Cont.)
Internal loads: N, V, MCalculate the internal loads at sec a-a by using the sectionmethod. Note that the resultant internal forces N passesthrough centroid of x-section.
The resultant internal moment must be computed about thebeams neutral axis a section a-a.
IMcm
3a. Bending Stresses
18Prepared by: Prof. Nabil El-Tayeb
Example 3.9 (Cont.)
Internal moment
Apply moment equation of equilibrium about neutral axis,
+ MNA = 0; 2.4 kN(2 m) + 1.0 kN(0.05909 m) =M
M= 4.859 kNm
I
Mcm
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3a. Bending Stresses
19Prepared by: Prof. Nabil El-Tayeb
Example 3.9 (Cont.)
To find location of neutral axis, x-sectional area divided into 3 compositeparts as shown. Then by using Eqns:
y = = ... = 59.09 mmy A
A
2dAIIA
AyY
x
I
Mcm
I = 42.26x10-6m4Given
3a. Bending Stresses
20Prepared by: Prof. Nabil El-Tayeb
Example 3.9 (Cont.)
Section property
Moment of inertia about neutral axis is determinedusing parallel-axis theorem applied to each of thethree composite parts of the x-sectional area.
I= [1/12(0.250 m)(0.020 m)3
+ (0.250 m)(0.020 m)(0.05909 m 0.010 m)2]
+ 2[1/12(0.015 m)(0.200 m)3
+ (0.015 m)(0.200 m)(0.100 m 0.05909 m)2]
I= 42.26(10-6) m4
I
Mcm
I = 42.26x10-6m4Given
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3a. Bending Stresses
21Prepared by: Prof. Nabil El-Tayeb
Example 3.9 (Cont.)
Maximum bending stress occurs at points farthest awayfrom neutral axis. At bottom of beam,
c= 0.200 m 0.05909 m = 0.1409 m. Thus,
At top of beam, = 6.79MPa. In addition, normal
force of N = 1 kN andshear force V = 2.4 kN willalso contribute additional
stress on x-section.
Mc
Imax = = = -16.2 MPa
4.859 kNm(0.1409 m)
42.26(10-6) m4
3a. Bending Stresses
22Prepared by: Prof. Nabil El-Tayeb
Q? Determine and Plot the Stress distribution acting over entire x-sectional
Example 3.9 (Cont.)
A
P
I
yMx
- + - +
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3a. Bending Stresses
23Prepared by: Prof. Nabil El-Tayeb
Stress due to eccentric loading found bysuperposing the uniform stress due to a centric
load and linear stress distribution due to a pure
bending moment
Eccentric Axial Loading in a Plane of Symmetry
Eccentric loading
PdM
PF
I
yM
A
P
I
yM
A
P
t
C
xxx
)(
)(
bendingcentric
xx
y
M
F
Supplementary Examples
5.10-5.13
3a. Bending Stresses
24Prepared by: Prof. Nabil El-Tayeb
Example 3.10An open-link chain is obtained by bending
low-carbon steel rods into the shape
shown. For a 700-N load, determine (a)
maximum tensile and compressive
stresses, (b) distancebetween section
centroid and neutral axis
SOLUTION:
Find the equivalent centric load and
bending moment.
Superpose the uniform stress due to
the centric load and the linear stress
due to the bending moment.
Evaluate the maximum tensile and
compressive stresses at the inner
and outer edges, respectively, of thesuperposed stress distribution.
Find the neutral axis by
determining the location where thenormal stress is zero.
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3a. Bending Stresses
25Prepared by: Prof. Nabil El-Tayeb
Example 3.10 (Cont.)
Equivalent centric load
and bending moment
Normal stress due to acentric load
Normal stress due to
bending moment
3a. Bending Stresses
26Prepared by: Prof. Nabil El-Tayeb
Example 3.10 (Cont.)
Maximum
tensile and
compressive
stresses
Neutral axis
location
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3a. Bending Stresses
27Prepared by: Prof. Nabil El-Tayeb
Example 3.10 (Cont.)
The largest allowable stresses for the castiron link are 30 MPa in tension and 120
MPa in compression. Determine the largest
force P which can be applied to the link.
SOLUTION:
Determine equivalent centric load and
bending moment.
Evaluate the critical loads for the
allowable tensile and compressive stresses.
The largest allowable load is the
smallest of the two critical loads.
From Sample Problem 4.2,
49
23
m10868
m038.0
m103
I
Y
A
Superpose the stress due to a centric
load and the stress due to bending.
3a. Bending Stresses
28Prepared by: Prof. Nabil El-Tayeb
Example 3.10 (Cont.)
Determine equivalent centric and bending loads.
momentbending028.0
loadcentric
m028.0010.0038.0
PPdM
P
d
Evaluate critical loads for allowable stresses.
kN0.77MPa1201559
kN6.79MPa30377
PP
PP
B
A
The largest allowable load
Superpose stresses due to centric and bending loads
P
PP
I
Mc
A
P
PPP
I
Mc
A
P
A
B
A
A
155910868
022.0028.0
103
37710868
022.0028.0
103
93
93
kN0.77P
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3a. Bending Stresses
29Prepared by: Prof. Nabil El-Tayeb
UNSYMMETRIC BENDING
Moment arbitrarily applied If a member is loaded such that
resultant internal moment does not act
about one of the principal axes of x-
section, resolve the moment into
components directed along the
principal axes
Use flexure formula todetermine normal stress
caused by each momentcomponent
Use principle ofsuperposition to
determine resultantnormal stress at the pt
Mz
My
3a. Bending Stresses
30Prepared by: Prof. Nabil El-Tayeb
UNSYMMETRIC BENDINGMoment arbitrarily appliedResultant general normal stress at any pt on x-section is:
Mzy
Iz=- +
Myz
Iy
= normal stress at the pt
y, z= coordinates of ptmeasured fromx, y, zaxes
having origin at centroid of x-
sectional area and forming a
right-handed coordinatesystem
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3a. Bending Stresses
31Prepared by: Prof. Nabil El-Tayeb
UNSYMMETRIC BENDING
Orientation of neutral axis
My, Mz = resultant internal moment components alongprincipal yand zaxes. Positive if directed along +yand +z
axes. Can also be stated asMy = M sin and Mz = M cos ,
where is measured positive from +zaxis toward +y axis
Iy, Iz = principal moments of inertia computed about the y and
zaxes, respectively
0
y
y
z
z
I
zM
I
yM
From which Iz
Iytan = tan
3a. Bending Stresses
32Prepared by: Prof. Nabil El-Tayeb
UNSYMMETRIC BENDINGOrientation of neutral axisAngle of neutral axis can be
determined by applying Eqn 6-17
with = 0, since no normal stress
acts on neutral axis.
Finally, we get
For unsymmetrical bending, angle defining direction of
moment M is not equal to angle , angle defining
inclination of neutral axis unless
Iz= Iy. Thus, 90o
Equation 6-19Iz
Iytan = tan 0
y
y
z
z
I
zM
I
yM
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3a. Bending Stresses
33Prepared by: Prof. Nabil El-Tayeb
Unsymmetrical Bending
Superposition is applied to determine stresses inthe most general case of unsymmetric bending.
Resolve the couple vector into components along
the principle centroidal axes.
sincos MMMM yz
Superpose the component stress distributions
y
y
z
zx
I
yM
I
yM
Along the neutral axis,
tantan
sincos
0
y
z
yzy
y
z
zx
I
I
z
y
I
yM
I
yM
I
yM
I
yM
3a. Bending Stresses
34Prepared by: Prof. Nabil El-Tayeb
General Case of Eccentric Axial Loading Consider a straight member subject to equal
and opposite eccentric forces.
The eccentric force is equivalent to the system
of a centric force and two couples.
PbMPaMP
zy forcecentric
By the principle of superposition, the
combined stress distribution is
y
y
z
zx
I
zM
I
yM
A
P
If the neutral axis lies on the section, it may
be found from
A
PzI
M
yI
M
y
y
zz
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3a. Bending Stresses
35Prepared by: Prof. Nabil El-Tayeb
Example 3.11
A 180 N . m couple is applied to a
rectangular wooden beam in a plane
forming an angle of 30 deg. with the
vertical. Determine (a) the maximum
stress in the beam, (b) the angle that the
neutral axis forms with the horizontal
plane.
SOLUTION:
Resolve the couple vector into
components along the principle
centroidal axes and calculate the
corresponding maximum stresses.
sincos MMMM yz
Combine the stresses from the
component stress distributions.
y
y
z
zx
I
zM
I
yM
Determine the angle of the neutral
axis.
tantany
z
I
I
z
y
3a. Bending Stresses
36Prepared by: Prof. Nabil El-Tayeb
Example 3.11 (Cont.) Resolve the couple vector into components and calculate
the corresponding maximum stresses.
The largest tensile stress due to the combined loading
occurs atA.
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3a. Bending Stresses
37Prepared by: Prof. Nabil El-Tayeb
Example 3.11 (Cont.)
Determine the angle of the neutral axis.
3a. Bending Stresses
38Prepared by: Prof. Nabil El-Tayeb
Stress Concentrations
Stress concentrations may occur:
in the vicinity of points where the
loads are applied I
McKm
in the vicinity of abrupt changes
in cross section
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3a. Bending Stresses
39Prepared by: Prof. Nabil El-Tayeb
Example 3.12
NmmNmmNmm
Nmm
(c):I=13.92x10-6m
4
z=0.089mM=5.11Nm
c=15 kN/m2
t=32.7kN/m2
IMc
a a
bbc c
d d
e e
cbottom
ctop
M=5.1Nm, I=13.92x10-6m4,
cbottom=.089m, ctop=.041m,
bQ
IV
IbVQ
bQ
V=5.15N, I=13.92x10-6m4,
to be calculated at aa, bb,
cc, dd, and ee.b
Q
c, I
Mc
3a. Bending Stresses
40Prepared by: Prof. Nabil El-Tayeb
(c):I=13.92x10-m
z=0.089mM=5.11Nm
c=25.7kN/m2
t=32.7kN/m2
I
Mc
Ib
VQ
a a
bbc c
d d
e e
a
b cd
e
, N/m2 x10
6
b
Q
I
V
Ib
VQ
b
Q7x103 6
0,0b
Q
b
Q
eeaa
eeaa
2666 /10017.016.0
)2/03.0)(03.016.0(1037
b
Q1037 mNx
xxx
bb
bb
2666/1007.0
04.0
)2/03.0)(03.016.0(1037
b
Q1037 mNx
xxx
cc
cc
2666 /10015.004.0
)2/089.0)(04.0089.0(1037bQ1037 mNxxxx
bb
dd
V= 515N, I=13.92x10-
6m4
Example 3.12 Cont.
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3a. Bending Stresses
41Prepared by: Prof. Nabil El-Tayeb
EXAMPLE 3.13 Combined stresses
A force of 15,000 N isapplied to the edge of themember shown. Neglect theweight of the member anddetermine the state of
stress at pts Band C.
3a. Bending Stresses
42Prepared by: Prof. Nabil El-Tayeb
EXAMPLE 3.13 (SOLN)
Internal loadingsSectioning the member through B and C. For equilibrium,axial force of 15 kN acting through the centroid and abending moment of 750. kNmm about the centroidal or
principal axis.
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3a. Bending Stresses
43Prepared by: Prof. Nabil El-Tayeb
EXAMPLE 3.13 (SOLN)
Stress components
1. Normal force
Uniform normal-stress distributiondue to normal force is shown.
=P/A = = 3.75 MPa
2. Bending moment
Normal stress distribution due tobending moment is shown.
max =Mc/I= = 11.25 MPa
3a. Bending Stresses
44Prepared by: Prof. Nabil El-Tayeb
EXAMPLE 3.13 (SOLN)
Superposition
If above normal-stress distributions
are added algebraically, resultantstress distribution is shown. Although
not needed here, the location of the
line of zero stress can be determined
by proportional triangles, i.e.,
7.5 MPa
x=
15 MPa
(100 mmx)
x = 33.3 mm
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3a. Bending Stresses
45Prepared by: Prof. Nabil El-Tayeb
EXAMPLE 3.13 (SOLN)
Superposition
Elements of material atB and Care subjected only
to normal or uniaxial stress as shown. Hence
B = 7.5 MPa C= 15 MPa