5 bending stress
TRANSCRIPT
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DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE
SEMESTER 4 / 6
COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.
VTOS NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI
PAGE 22
TOPICBENDING STRESS
SUB TOPIC5.1 Simple Bending Theory5.2 Non-Symmetry Bending5.3 Second Moment of Area5.4 Mohrs Circle5.5 Parallel Axes Theorem5.6 Stress and Deflection
Chapter 5 BENDING STRESS p1
INFORMATION SHEET
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5.1 Simple Bending Theory
In this chapter we continue our study of beams by determining how the stressresultants, the bending moment M(x) and the transverse shear force V(x), are
related to the normal stress and the shear stress at section x. Loads (transverse forces or
couple) applied to a beam cause it to deflect laterally, as illustrated in
Figure 5.1.
FIGURE 5.1 Transverse deflection of a beam
This lateral deflection, or bending, changes the initially straight longitudinal
axis of the beam into a curve that is called the deflection curve, shown dashed in
Figure 5.1. By relating the curvature of the deflection curve to the bending moment
M, we can determine the distribution of the normal stress x. You will discover that thisderivation includes all three of the fundamental types of equations: geometry of
deformation (in the strain-displacement analysis), material behavior (in the stress-
strain relations), and equilibrium (in the definition of stress resultants and in relating
stress resultants to the external loads and reactions). ,
Beam-Deformation Terminology.
To simplify this study of beams, we initially consider only straight beams that have a
longitudinal plane of symmetry (LPS), and for which the loading and support aresymmetric with respect to this plane, as illustrated in Figure 5.2.
FIGURE 5.2 Illustration of some beam-deformation terminology
Under these conditions, this longitudinal plane of symmetry is the plane of bending.
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Pure Bending.
Let us begin our analysis of beams by examining the deformation of a uniform beam
segment subjected to pure bending, that is,
a segment whose material properties are constant along its length, and
for which M(x) is constant.
If equal couples MOare applied to the ends of an otherwise unloaded segment of beam, as
in Figure 5.3, the moment is constant along the segment and the segment is said to be in
pure bending.
(a) The cross
section before
deformation
(b) The undeformed beam (c) Plane section remain plane
FIGURE 5.3 A uniform beam segment undergoing pure bending
We can see that changes in length (strain, ), occurs at D*G* (compression) and A*E*
(tension). However, BF retains it original length; which is called neutral surface (NS).
Lines ABD and EFG in Figure 5.3b represent the edges of typical cross sections in the
undeformed beam; lines A*B*D* and E*F*G* in Figure 5.3c represent these same cross
sections after deformation, as seen from the front face of the beam.
From Figure 5.3c we can determine the following characteristics of a uniform beam
undergoing pure bending:
1. Since M(x) = MO = const, pure-bending deformation of a beam is uniform along the
length of the segment undergoing pure bending: so whatever happens at a typical
cross-sectionABD also happens at section EFG.
2. Pure bending has front-to-back symmetry. The only way that this can be possible is
for all cross sections like ABD and EFG, to remain plane and remain
perpendicular to the deflection curve.
3. In summary, when a beam undergoes pure bending, its deflection curve forms a
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circular arc, and its cross sections remain plane and remain perpendicular to
the deflection curve. Experiments show that this is, indeed, the way that beams
deform whensubjected to pure bending.
Strain-Displacement Analysis
Let us continue our analysis of the deformation of a uniform beam segment subjected to
pure bending, that is, a segment for which M(x) is constant.
Kinematic Assumptions of Bernoulli-Euler Beam Theory. The previous discussion of
pure bending can be summarized in the following four deformation assumptions of
Bernoulli-Euler beam theory:
1. The beam possesses a longitudinal plane of symmetry, and is loaded and supported
symmetrically with respect to this plane. This plane is called the plane of bending.
2. There is a longitudinal plane perpendicular to the plane of bending that remains free
of strain (i.e., x = 0) as the beam deforms. This plane is called the neutral surface
(NS). The intersection of the neutral surface with a cross section is called the neutral
axis (NA) of the cross section. The intersection of the neutral surface with the plane
of bending is called the axis of the beam: it forms the deflection curve of the
deformed beam.
3. Cross sections, which are plane and are perpendicular to the axis of the
undeformed beam, remain plane and remain perpendicular to the deflection
curve of the deformed beam.
4. Deformation in the plane of a cross section (i.e., transverse strains y and z) may
be neglected in deriving an expressionforthe longitudinal strain x.
The third of the preceding assumptions is crucial to the development of the Bernoulli-Euler
beam theory; it leads to a practical theory of bending ofbeams that is comparable to the
theories of axial deformation and torsion covered previously.
Strain-Displacement Analysis; Longitudinal Strain.Because of Assumptions 1 and 4, the fibers in any plane parallel to the xy plane behave
identically to the corresponding fibers that lie in the xy plane (i.e., the plane of bending).
Therefore, bending deformation is independent of the coordinate z, so the drawing in
Figure 5.4 represents the deformation of any plane in the beam parallel to the xy plane.
Using Figure 5.4 and the preceding four deformation assumptions, we can develop an
expression for the extensional strain x, in a longitudinal fiberatcoordinates (x, y, z) in the
beam.
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(a) The undeformed beam segment (b) The deformed beam segmentFIGURE 5.4 The geometry of deformation of a beam segment, showing the plane of
bending.
In Figure 5.4a, points A and P lie in the cross-sectional plane at coordinate x in the
undeformed beam: similarly, points B and Q lie in lie in the cross-sectional plane at (x + x)
in the undeformed beam. Line segment PQ is parallel to the x axis and lies at distance +y
above the NS (xz plane). Therefore, in the undeformed beam the infinitesimal fibers AB and
PQ are both of Iength x. From Assumption 3, points A* and P* lie in a plane that is
perpendicular to the neutral surface of the deformed beam, and points B* and Q* lie in a
plane that also is perpendicular to the deformed neutral surface. According to Assumption 2,
however, the length ofA*B*, a fiber lying in the neutral surface, is unchanged; that is, the
length of A*B* is still x, as indicated in Figure 5.4b. Finally, by virtue of Assumption 4, A*P*
= AP = y, and B*Q* = BQ = y. Figure 5.4 therefore, embodies all four deformation
assumptions of Bernoulli-Euler beam theory, so we can use it in deriving an expression for
the extensional strain of a longitudinal fiber.
From the general definition of extensional strain, we can express the extensional strain in the
longitudinal fiber PQ as
x x (x, y, z) =lim
(P*Q* -
PQ)=
lim x* - x
Q P PQ x 0 x
Considering A*B* to be the arc of a circle of radius (x) subtending an angle *, we get
A*B* =
x= *
Similarly,
P*Q* = x* = ( y) *
Combining these two equations, we get
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I = A y2 dA :
Figure 5.6
Iz = A y2 dA
So,
=
h/2y2 bdy
-h/2
= by3 h/2
3 -h/2
Iz =bh3
12
From Figure 5.6;
dA = bdy
Area moment of inertia
Recall
M = - Ay
dA
and = - Ey /
M = - A y (- Ey / ) dA
I= E / - A y2 dA
M =EI
= EIk Where EI = flexural rigidity
To get rid of radius of curvature which requires longer calculation to find, we have tocombine these two equations:
M =EI
So,
E =M
I
x =- E y
= - M yI
x =- M y
Flexural FormulaI
S is elastic section modulus where S = I / c,
c = h / 2
then
S= bh2 / 6
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Then max at ymax = c
max =
- M
c =
- M
Flexural Formula
I S
Where
max =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 method of
sections and the equations of equilibrium, and computed about the
neutral axis of the cross section
I =The moment of inertia of the cross sectional area computed about
the neutral axis
c =The perpendicular distance from the neutral axis to a point farthest
away from the neutral axis, where max acts
EXAMPLE 5.1
A simple cast iron (E = 175 GPa) beam of rectangularcross section carries a load of 5 kN/m. Determine:
(a) The maximum tensile and compressive stresses
at the midspan,
(b) The normal stress and strain at a point A, and
(c) The radius of curvature of the beam at B.
Solution
The neutral axis z passes through the centroid C and
I =bh3
=(0.08m)(0.12m)3
= 11.52 x 10-6 m412 12
The section modulus of the cross sectional area
S =bh2
=(0.08m)(0.12m)2
= 192 x 10-6 m3
6 6
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a) At the midspan, the bending moment is M = 5 (4)2 / 8 = 10 kN.m. Because M is
positive, the maximum tensile and compressive stresses occur at the bottom and top
of fibers, respectively:
max =My
=(10 kN.m)(0.06m) = 52.1 MPa
I 11.52 x 10-6 m4
Or
These stresses act on infinitesimal elements at D and E
b) At a section through point A, the bending moment in M = 10(1) 5(1)2 / 2 = 7.5 kN.m,
and we have
A =- MyA =
- (7.5 kN.m) (-
0.02m) = 13 MPa
I 11.52 x 10-6 m4
The normal strain at point A is thus
A =A =
(13 x 106 N/m2)= 74.3
E 175 x 109 N/m2
c) The radius of curvature,
= - yA =
-(-0.02
m) = 269 mA 74.3
Chapter 5 BENDING STRESS p9
max =M
=(10 kN.m)
= 52.1 MPaS 192 x 10-6 m4
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5.2 Non-Symmetry Bending
In previous section, we have been considering flexural stress and strain in beams whose
cross-sectional shape and whose loading and support conditions produce bending that is
confined to a longitudinal plane of symmetry (LPS) of the beam. Where we assume:
a) the deflection of [be beam can be characterized by a deflection curve in the LPS
b) there is no tendency of the beam to twist
However, we also need to be able to analyze the behavior of beams that are not loaded and
supported in this simple manner.
(a) Components of load in two planes
of symmetry
(b) Bending moments due to an inclined load
(positive My and positive Mz shown)FIGURE 5.7 A doubly symmetric beam with inclined loading
If a beam is subjected to a non symmetry loadings such as in Figure 5.7, then the stress at
the defined center C should be contributed by the total of stress in the other two axes on the
plane which the load acting. Where, stress at x, x
x =Myz
-Mzy Flexural Formula
(non-symmetry bending)Iy Iz
This is illustrated in Figure 5.8 below.
FIGURE 5.8 Flexural stress due to inclined loading of a doubly symmetric beam.
It is noted that in Figure 5.8c that the stress at center C is zero. So;x = Myz* - Mzy* = 0
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EXAMPLE 5.2
A 1600 lb.in couple is applied to a wooden beam, of
rectangular cross section 1.5 x 3.5 in., in a plane forming
an angle of 30 with the vertical. Determine:
(a) The maximum stress in the beam
(b) The angle that the neutral surface forms with the
horizontal plane
Solution
a) The maximum stress in the beam, max
max = 1 + 2
=Mzy +
MyzIz Iy
We need to find My and Mz
Mz = M cos = 1.6 kNm (cos 30)= 1.386 kNm
My = M sin = 1.6 kNm (sin 30)= 0.8 kNm
Maximum stress
1 = Mz y =
(1.386 kNm)
(0.175m) = 452.6 kPa
Iz 535.9 x 10-6 m4
2 =My z =
(0.8 kNm)(0.075m)= 609.5 kPa
Iy 98.44 x 10-6 m4
max = 1 + 2= 452.6 kPa + 609.5 kPa= 1.062 Mpa
The distribution of the stresses
across the section.
Chapter 5 BENDING STRESS p12
Iz =bh3
=(0.15m)(0.35m)3
= 535.9 x 10-6 m412 12
Iy =bh3
=(0.35m)(0.15m)3
= 98.44 x 10-6 m412 12
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c) Angle of Neutral Surface with the horizontal plane.
tan
= (
Iz) tan
Iy
= 535.9 x 10-6 m4 tan
3098.44 x 10-6 m4
= 3.143
= tan -1 3.143= 72.4
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5.3 Second Moment of Area
The moment-area method provides a semigraphical technique for finding the slope and
displacement at specific points on the elastic curve of a beam or shaft.
We have to assume:
1) The beam is initially straight,
2) It is elastically deformed by the loads,
3) Slope and deflection of the elastic curve is very small,
4) Deformation are caused by bending.
Moment area method is based on 2 theorems
- Theorem 1 : First Moment Area
- Theorem 2 : Second moment of Area
The vertical deviation of the tangent at a point (A) on the elastic curve with respect to the
tangent extended from another point (B) equals the moment of area under the M/EI diagram
between these two points (A and B). This moment is computed about point (A) where the
vertical deviation (tA/B) is to be determined.
Consider the following beam:
(a)
(c) dt is the vertical deviation of the tangent
on each side of the differential element dx.(b)
FIGURE 5.9 A beam experiencing distributed load
From Figure 5.9, we are ought to find tA/B as shown in (c). Mathematically we know that, s
= r;
So, ds = x d = dt
From the 3rd
assumption, we assume the slope and deflection of the elastic curve is verysmall, which mean ds = dt = x d
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From Theorem 1, which is not covered in this lecture:
d =M
dxEI
Hence
dt = d =M
dxEI
t = xM
dxEI
So,
tA/B = B
xM
dxA EI
Since the centroid of an area is found from x dA = x dA, and M/EI dx represent the areaunder the M/EI diagram, (Figure 5.9 b), we can also write:
tA/B = x B M
dxA EI
Where:
X is the distance from A to the centroid of the area under the M/EI diagram between A
and B.
Note that:
tA/B = tB/A
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EXAMPLE 5.3
Determine the displacement of points B and C of
the beam shown. EI is constant.
Solution:Construct the M/EI diagram
B = tB/A
C = tC/A
Moment Area Theorem
Consider area from A to B:
B = tB/A = (L
) [( -MO )(
L)] = -
MOL2
4 EI 2 8EIConsider area from A to C:
C = tC/A = (L
) [( -MO )( L )] = -
MOL2
2 EI 2EI
Both tB/A and tC/A is negative, showing that point B and C is below tangent at A.
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5.4 Mohrs Circle
It is proven that:
R = (Ix - Iy )2 + Ixy22
Procedure to construct Mohrs Circle
1) Compute Ix, Iy and Ixy
2) Construct the circle,
- abscissa represents the moment of inertia I
- ordinate - represents the product of inertia Ixy
3) Determine the coordinate of center of the circle, C from origin where
C =(Ix + Iy)
2
4) Plot reference point A having coordinate A (Ix, Ixy)
- Ix is always positive
- Ixy will be either positive or negative
5) Connect reference point A to the center C, where AC = radius of circle, R
6) Determine AC by trigonometry
7) Determine Imin and Imax : points that intersect abscissa (product of inertia Ixy = 0)
8) Determine 2p1
FIGURE 5.10 Mohrs Circle showing the important point.
EXAMPLE 5.4
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Use Mohrs circle to determine the principal moments
of inertia for the beams cross-sectional area shown
below, with respect to axes passing through the
centroid.
Solution
Compute Ix, Iy and Ixy
Using parallel axes theorem (Section 5.5), we get
Ix = 2.90 x 109 mm4
Iy = 5.60 x 109 mm4
Ixy= -3.00 x 109 mm4
C =(Ix + Iy) =
(2.90 + 5.60)= 4.25
2 2
A (Ix, Ixy) ; A (2.90, -3.00)
(a)
(b)
Imax = C + R = 4.25 + 3.29 = 7.54 x 109 mm4
Imin = C - R = 4.25 - 3.29 = 0.960 x 109 mm4
From figure (b):
2p1 = 180 - tan
-1
(|BA|/|BC|)= 180 - tan-1 (|3.00|/|1.35|)
= 114.2
p1 = 57.1
The major principal axis (for Imax = 7.54 x 109 mm4) is
therefore oriented at an angle p1 = 57.1, measured
counterclockwise, from the positive x axis. The minor
axis is perpendicular to this axis. The results are shown
in Figure (a).
(c)
Chapter 5 BENDING STRESS p18
R = CA = (Ix - Iy )2 + Ixy22
= (1.35)2 + (-3.00)2= 3.29
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5.5 Parallel Axes Theorem
From Figure 5.10 below, it is known that the moment of inertia is
Ix = A y2 dA
Iy = A x2 dA
FIGURE 5.10
However, if the moment of inertia for an area is known about a centroidal axis we can
determine the moment of inertia of the area about a corresponding parallel axis using the
parallel axis theorem. Consider the following Figure 5.11.
FIGURE 5.11
In this case, a differential element dA is located at an arbitrary distance y from the centroidal
x axis, whereas the fixed distance between the parallel x and x axes is defined as d y. Since
the moment of inertia of dA about the x axis is dIx = (y + dy)2 dA, then for the entire area,
Ix = A (y + dy)2 dA = A y2 dA + 2dy A ydA + dy2 A dA= Ix = 0
Finally we get
Ix = Ix + A dy2
Iy = Iy + A dx2
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FIGURE 5.12
The product of inertia
Ixy = A xy dA
FIGURE 5.13
Consider the shaded area shown in Figure 5.13, where x and y represents a set of
centroidal axes, and x and y represent a corresponding set of parallel axes. Since the
product of inertia of dA with respect to the x and y axes is dIxy = (x + dx)(y + dy) dA, then for
the entire area,
Ixy = A (x + dx)(y + dy)dA = A xy dA + dx A ydA + dxdy A dA
The first term on the right represents the product of inertia of the area with respect to the
centroidal axis, Ixy. The second and third terms are zero since the moments of area are
taken about the centroidal axis. Realizing that the fourth integral represent the total area A,
we therefore have the final result
Ixy = Ixy + A dxdy
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EXAMPLE 5.5
A T beam with dimension shown is given moment at the cross section, M = 4 kMn.
Determine
a) The location of the neutral axis of the cross section
b) The moment of inertia with respect to the neutral axis, and
c) The maximum tensile stress and the maximum compressive stress on the
cross section
Solution
a) Locate the neutral axis
We can use first moment of area method where
y A = y1 A1 + y2 A2 = (0.55m)(0.5m)(0.1m) + (0.25m)(0.1m)(0.5m)= 0.04m3
Where
A = A1 + A2 = (0.5m)(0.1m) + (0.1m)(0.5m) = 0.1m2
Then
y =yA
=0.04m3
= 0.4mA 0.1m2
b) The moment of inertia with respect to the neutral axisThe moment of inertia of a rectangle about an axis through its own centroid is
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Ix =bh3
12
And, from parallel axes theorem, the moment of inertia about an axis through C
parallel to the axis through the centroid C is
I = Ix1 + A dy12 + Ix2 + A dy22
=(0.1m)(0.5m)3
+ (0.5m)(0.1m)(0.15m)2 +(0.5m)(0.1m)3
+ (0.1m)(0.5m)(0.15m)212 12
= 3.33 x 10-3 m4
c) The maximum tensile stress and the maximum compressive stress on the cross
section
x =- My
I
(max)C =My
=-(4kNm)(0.2m)
= -240 kPaI 3.33 x 10-3 m4
(max)T =My
=-(4kNm)(-0.4m)
= 480 ksiI 3.33 x 10-3 m4
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5.6 Stress and Deflection
The fundamental assumptions of the technical theory for slender beams are based
upon the geometry of deformation. We can state them as follows:
1. The deflection of the beam axis is small compared with the span of the beam. The
slope of the deflection curve is therefore very small and the square of the slope is
negligible in comparison with unity. If the beam is slightly curved initially, the
curvature is in the plane of the bending, and the radius of curvature is large in
relation to its depth (p > 10/z).
2. Plane sections initially normal to the beam axis remain plane and normal to that
axis after bending (for example, a-a). This means that the shearing strains y is
negligible. The deflection of the beam is thus associatedprincipallywith the axial or
bending strains ex. The transverse normal strains ey and the remaining strains
(z, xz,yz) may also be ignored.
3. The effect of the shearing stresses xy on the distribution of the axial or bending
stress x is neglected. The stresses normal to the neutral surface, y, are small
compared with xand may also be omitted. This supposition becomes unreliable in
the vicinity of highly concentrated transverse loads.