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(Other Names: Unsymmetric Bending, Biaxial Bending)
5.7 Oblique Bending
(5.9.c)
(5.9.b)
(5.9.a)
tvids:
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C
O
We are at this point in the beginning
Radius of Mohr Circle:
The Mohr circle, which is the geometric expression of moments of inertia with respect to all axes passing through a point, is drawn in the following steps:
5.7.2 Reminder-1: (Principal Axes of Inertia from Statics 6.3 topic)
Principal Moment of Inertia:
Angle of Principal Axes of Inertia
2θ
(5.7 4)
(5.7.5)
(5.7.6)
3- M-N line is drawn and point C is determined.
4- A circle centered at C and passing through M and N is drawn.
Note: In Statics 6.3, the area was examined in the x-y plane. However, at bending topics in Strength, the cross-section is examined in y-z plane. For this reason, the formulas in statics 6.3 have been adapted again according to y-z axes.
(5.7.1)
(5.7.2)
(5.7.3)
5.7.1 Our aim in this section: To formulate the normal stress distribution in the beam section when there is biaxial bending in a beam. To understand what biaxial bending is, we will first make two reminders about Statics and Strength:
v
u
y
z
θ
zp
yp
O
Figure 5.7.2
Figure 5.7.1
Oblique (Unsymmetric) Bending
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x
y, yp
B
z, zp
What is Oblique Bending? When does it occur?
As we learned in section 5.1, Simple Bending occurs in the cross-section of a beam in the y-z plane when the following two conditions are present:
If at least one of the above two conditions is not met, oblique bending
(or oblique bending) occurs in the section. Below are 3 examples of oblique bending.
5.7.3 Reminder 2 (Simple Bending )
1st condition: The section must be symmetrical with respect to at least one of the y or z axes.
2nd condition :The direction: of the resultant moment vector in the section must coincide with one of the y or z axes.
x
y, yp
z, zp
B
Normal Stress Formula
simple bending in y direction
simple bending in z direction
Our aim is to obtain the normal stress distribution in the oblique Bending Case.
In symmetrical sections, the principal inertia axes (yp, zp) coincide with the y and z axes.
x
y, yp
z, zp
y
z
G
(2nd condition is not met)
(1st condition is not met)
y
z
G
(1st and 2nd conditions are not met)
They are all oblique Bending
yp
zp
yp
zp
The axis set must always be placed at the center of gravity G of the section.
Why is there no «-» at the beginning of this formula? For the answer, see 5.1.7
Figure 5.7.3
Figure 5.7.4
Figure 5.7.5
Figure 5.7.6
Figure 5.7.7
Oblique (Unsymmetric) Bending
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In symmetrical or non-symmetrical cross-sections, if the direction of the Resultant Bending Moment Vector coincides with one of the principal inertia axes, simple bending occurs according to the principal axis set. Otherwise, oblique bending occurs.
Examples for symmetrical cross-sections
Tip 5.7.1: In symmetrical cross-sections, Cartesian axes are also the principal axes of inertia.
zp ,
,yp
(Resultant moment coincides with one of the principal axes.)
a-) Simple Bending
zp ,
,yp
(Resultant moment does not coincide with any of the principal axes.)
b-) Oblique Bending
Examples for non-symmetrical cross-sections
The principal axes of inertia are different from the y and z axes.
a-) Simple Bending
(Resultant moment coincides with one of the principal axes.)
b-) Oblique Bending
c-) Oblique Bending
(Resultant moment does not coincide with any of the principal axes.)
(The resultant moment does not coincide with any of the principal axes. it coincides with y, but y is not the principal axis of inertia)
Oblique (Unsymmetric) Bending
Figure 5.7.8
Figure 5.7.9
5.7.4 More General Rule in Detecting Oblique Bending:
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5.7.5.1 - Method 1: Reducing Oblique Bending to 2 Simple Bendings (Superposition Method)
Oblique (Unsymmetric) Bending
Simple Bending in z direction
Simple Bending in y direction
Oblique Bending
+
=
+
zp ,
,yp
,yp
zp ,
,yp
zp ,
+
Since the resultant moment does not coincide with the principal axes of inertia
N.A
(Neutral Axis)
N.A
(5.7.7)
Figure 5.7.10.a
Figure 5.7.10.b
Figure 5.7.10.c
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2nd example: Calculation of the normal stress at any point b of the cross-section due to the resultant moment M for a non-symmetrical section:
Oblique Bending
Simple Bending
Simple Bending
(5.7.8)
Figure 5.7.11
(a)
(b)
(c)
Oblique (Unsymmetric) Bending
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In this method, the normal stress equation will be expressed in terms of variables with y and z subscript (Iz, Iy, Izy, Mz, My, z, y). There will be no need to use principal moments of inertia or principal axes of inertia. This method is more practical for non-symmetrical sections.
Oblique Bending
Oblique Bending
Oblique Bending
Now we will investigate
the answer to this question..>>
?
?
?
Figure 5.7.12.a
Figure 5.7.12.b
Figure 5.7.12.c
Oblique (Unsymmetric) Bending
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3
5.7.6 Reminder:
Neutral axis
y
z
x
G
Neutral Plane
Figure 5.7.14
Figure 5.7.13.a
Figure 5.7.13.b
Oblique (Unsymmetric) Bending
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5.7.7 Oblique Bending Calculations According to Cartesian Coordinates:
3
x is the axis of the beam and the normal of the cross-section. Lines perpendicular to the axis in the x-y' plane remain vertical after bending. Plane cross-sections remain plane. As a result of bending, the shape of the rod takes the form of an arc with center 𝐶1 and radius r
Final length of fiber JK:
Total elongation of fiber JK:
Unit elastic elongation (normal strain) of JK:
The normal stress in the elastic region at a point c on the JK fiber is:
The y' coordinate of any point c:
y’ is the axis perpendicular to N.A (neutral axis). The line of force is perpendicular to the resultant moment M. The assumptions made for simple bending are valid for oblique bending in the x-y' plane.
(5.7.9)
Force line
y
z
y
Cross-section of beam
Neutral axis
β
y’
G
c
z
β
y.sinβ
β
y’
z.cosβ
Figure 5.7.15
Figure 5.7.16.a
Figure 5.7.16.b
Oblique (Unsymmetric) Bending
DE is the fiber on the neutral plane. Its length does not change.
In this case DE = D'E’
JK is the fiber at a distance of 𝑦′ from the neutral plane. Its final length is 𝐽′ 𝐾′.
(We think that point c in the cross section is on the JK fiber.)
Initial lengths of all fibers are equal. JK=DE
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b
If we open and edit the last equation 5.7.9:
Since there is no external normal force, the internal normal force in the section must be zero (due to static balance):
internal diff. force = stress x diff. area
Multiplying dF by the perpendicular distance z from G is equal to the differential internal moment in the y direction:
Total internal moment in y direction:
It is positive according to the right hand rule.
Multiplying dF by the perpendicular distance y from G is equal to the differential internal moment in the z direction:
Total internal moment in z direction:
Let's define coefficients:
(5.7.10)
(5.7.11)
y
G
z
x
y
z
dA
Cross-section
c
Figure 5.7.17
Oblique (Unsymmetric) Bending
It is positive according to the right hand rule.
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From equations 5.7.10 and 5.7.11, the coefficients are obtained as follows:
1. Special case: In symmetrical sections with respect to at least one axis, the product moment of inertia is zero 𝐼𝑧𝑦=0. Note that in this case, equation 5.7.13 turns into equation 5.7.7 in the 1st method.
y
G
z
x
y
z
Cross-section of beam
c
dA
We also defined normal stress as:
Accordingly, the normal stress at any point with y, z coordinates of any cross-section of a beam subjected to oblique bending can be calculated from the following equation:
(5.7.13)
Oblique Bending
2. Special case: In addition to the 1st special case, if My = 0, a simple bending case is obtained and in this case, equation 5.7.13 turns into the normal stress equation in simple bending.
Simple Bending
Equation 5.7.13 is valid for isotropic bodies in the elastic region and is a general equation covering all bending situations. From the two special cases below, we can see that 5.7.13 is a general equation:
(5.7.12a,b)
Simple Bending is actually a special case of oblique bending.
Figure 5.7.18
Figure 5.7.19
Figure 5.7.20
Oblique (Unsymmetric) Bending
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z
y
Kiriş kesiti
Tarafsız Eksen
β
y’
G
The coefficients in the stress formula were previously defined as follows:
Also from equations 5.7.12a,b:
Neutral axis direction angle in case of oblique bending:
(5.7.14)
If 𝑀𝑦=0 is taken in cross-sections that are symmetrical with respect to at least one axis, simple bending occurs.
5.7.7.2 Special Case Review :
Simple Bending
In this case, from equation 5.7.14
This shows that the neutral axis is the z axis, which was said to be the case in simple bending.
(Equation 5.7.14 is valid for elastic loading for isotropic materials.
It is the angle counterclockwise from the +y axis. Covers all bending types.
Figure 5.7.21
Figure 5.7.22
Oblique (Unsymmetric) Bending
From the above equations::
We also know that 𝐼𝑧𝑦=0 in this type of cross-section
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When viewed from the x-y' plane, we can also obtain the radius of curvature of the beam relative to center C. Namely:
If we add the squares of the stress coefficients:
Also from equations 5.7.12a,b :
The general equation of the radius of curvature in case of oblique bending is:
(5.7.15)
5.7.7.4 Special Case Review :
Simple Bending
(The last equation was also derived for Simple Bending, which was explained in Chapter 5.1.)
In this case final version of equation 5.7.15:..>>
(Equation 5.7.15 is valid for isotropic materials in elastic loading and includes simple bending..)
Figure 5.7.16.a
Figure 5.7.16.b
Figure 5.7.22
Oblique (Unsymmetric) Bending
If 𝑀𝑦=0 is taken in cross-sections that are symmetrical with respect to at least one axis, simple bending occurs. We also know that 𝐼𝑧𝑦=0 in this type of cross-section.
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5.7.7.5 Normal Stress Distribution in Oblique Bending:
z
y
Beam cross-section
N.A
β
G
A
B
c
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
y’
y’
x
G
B
A
Stress distribution viewed from the x-y' plane:
Figure 5.7.23
Figure 5.7.24
Oblique (Unsymmetric) Bending
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M
My
Mz
N.A
The resultant bending moment occurring in any section of an L-profile bar is M = 20kNm and the angle of this moment with the +z axis is calculated as 300. If the yield strength of the material is 200MPa and the safety coefficient n=2, determine whether this section will exceed the safety limits.
Solution with Method 2: using y-z Cartesian coordinates
Moments of Inertia:
Coordinates of the center of gravity relative to the lower left corner:
M
Since the section is not symmetrical, oblique bending occurs due to the M moment. From equation 5.7.13 :
Moment Components:
Stresses are zero at points on the Neutral Axis (N.A):
N.A equation :
The maximum and minimum stresses are at the points A (115; 35) and C (-65; -85), which are farthest from the neutral axis.
safe
When the center of gravity and moment of inertia values are calculated, the following values are found.
(See the example at the end of the example 8.5 at statics lecture notes for calculations.)
Example5.7.1
Figure 5.7.25
Figure 5.7.26
Oblique (Unsymmetric) Bending
Allowable stress:
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In the example 8.5 at the static lecture notes, the calculated values for the same geometry are:
Solution with Method 1: Using Principal Axes yp - zp
angle of principal axes of inertia
principal moments of inertia :
(Angle between +z axis and max Principal axis) :
zp
yp
Oblique Bending (M)
=
+
safe.
(We found the same results as Method 1)
In Neutral Axis:
(N.A Equation)
zp
yp
a(z,y)
N.A
M
zp
yp
Clockwise
Equ. 2.14.
Oblique (Unsymmetric) Bending
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As shown in Figure 5.7.28, a force of F=10kN is applied from its free end to a rectangular cross-section beam fixed to a wall at its left end. Find the intensities and locations of the maximum and minimum normal stresses occurring in the beam and draw the normal stress distribution in the critical section.
z
x
y
60 mm
80 mm
L=1 m
F
A
B
C
O
2nd way: We take the moment with the vector operation relative to point O and find the components with their signs in this way.
point where
moment is taken
A point on the line of force
z
60 mm
80 mm
L=1 m
F
A
B
C
O
E
y
First of all, we can move the force F to point E on its own line. The maximum moment components that the F force will create in the beam will occur in the build-in section (the section in contact with the wall) that is farthest from the force.
Solution:
Calculation of Moment components at the built-in end:
1st way: From perpendicular distances (Scalar)
The signs are found using the right hand rule.
x
Components of force F:
E
a
b
O
E
1m
x
Example 5.7.2
Figure 5.7.28
Figure 5.7.29
Oblique (Unsymmetric) Bending
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(N.A equation)
A(-40,30)
B(40,-30)
Normal stress calculation for built-in section:
We can use the 1st method, the Superposition method.
z, zp
y, yp
O
z, zp
O
y, yp
Oblique Bending
Simple Bending
Simple Bending
z, zp
y, yp
O
N.A
A
B
The points farthest from N.A are A and B, and max and min stresses occur at these points.
Stresses are zero on N.A.
According to this equation, N.A is drawn as in Figure 5.7.31.a.
0
0
O
z, zp
y, yp
60 mm
80 mm
Figure 5.7.31.a
Figure 5.7.30
Figure 5.7.31.b
Figure 5.7.31.c
Oblique (Unsymmetric) Bending
Since the section is symmetrical:
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A
B
N.A
z
y
O
y’
x
beam
Cross section
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
It is the cross-section plane
y’
x
O
A
B
N.A
A
B
y'
y
z
O
Normal stress distribution from perspective view
View from x-y' plane:
O: center of gravity of the cross-section
Figure 5.7.32
Figure 5.7.33
Figure 5.7.34
Oblique (Unsymmetric) Bending