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CHAPTER – 5 Bending stresses in Beams.
5.1 Introduction
In Previous chapters we considered the stresses in prismatic
bars subjected to axial loads and Twisting moment. In this chapter we
consider third fundamental loading, bending. A beam is a structural
member that is subjected to loads acting transversely to the
longitudinal axis, as explained in the preceding chapter. Internal loads
develop in beams in the form of shear forces and bending moments to
resist the external loads. Shear stresses and Bending moments
develop with in the cross section due to internally develop shear forces
and bending moments respectively. In this chapter we will restrict
ourselves to study the bending stresses only due to bending moment.
The shear stresses due to shear forces will discuss in the next
chapter. Before stating the discussion on bending strains and stresses
we need to know the concept of Pure bending. Pure bending means a
beam or a portion of the beam under a constant bending moment,
which means that the shear force is zero.
Fig 5.1 Beam/Portion of beam in pure bending
To illustrate the concept pure bending, consider two simply supported
beams as shown in the Fig 5.1 (a) and (b). In Fig 5.1 (a), the region of
beam between the two point loads is constant and the value is ‘Pa’
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and the shear force in this region is zero. Hence the central region is
in pure bending. In Fig 5.1 (b), the beam is loaded only by couples
that produce constant bending moment and zero shear force
throughout the beam. In this chapter we will calculate the normal
strains and stresses in pure bending.
5.2 Bending stress
The stresses caused by the bending moment are known as
bending stresses or flexure stresses. The relationship between bending
moment and bending stresses is called flexure stresses. The following
assumptions are made in deriving the flexure or bending stress
equation.
1. The beam is initially straight or has a very large radius of
curvature compared to its cross section dimensions.
2. The transverse cross section of the beam is symmetrical about
an axis passing through its centroid (in our case it is Y–axis)
and parallel to the plane of the bending.
3. The transverse section of beam, which is plane before bending,
will remain a plane after bending.
4. The cross section dimensions are small compared to its length.
5. The beam is in pure bending. i.e. the beam is subjected to only
bending moment.
6. The material of the beam is homogenous and isentropic.
7. The stress is in elastic and limit and obeys Hook’s Law.
Consider a portion of a beam ab which is in pure bending produced by
couples M as shown in the Fig. 5.2.
Fig. 5.2 A portion of beam in pure bending.
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From the directions of the couples we can say that bending moments
are positive (ref. section 4.5 and Fig. 4.6) and the cross section is
symmetry about y-axis. Consider two plane transverse section mn, pq
separated by a distance of ‘dx’ apart. Under the action of the moments
the beam gets deflected in to a circular curve as shown in the Fig. 5.3.
Fig. 5.3 Deformation of a Beam in pure bending.
The length of the bottom fiber is elongated and top fibers are
contracted. Thus, bottom fibers are in tension and top fibers are in
compression. Somewhere between top and bottom of the beam there is
a surface which does not change in length. This surface indicated in
dashed line in Fig. 5.2 and 5.3, is called neutral surface. The
intersection of the neutral surface with any cross section plane is
called neutral axis of the cross section.
The planes mn, pq get deflected and occupies the positions
m1n1and p1q1 as shown in the Fig. 5.3, being inclined at angle dθ and
intersecting at ‘o’, the center of the curve. Let, R is the radius of
Curvature. The distance between n1, q1 is more than dx and between
m1, p1 less than dx, but along the neutral surface, the initial distance
dx is remains same. Hence,
R.dθ = dx …………………………………………5.1
To evaluate the strains, consider a fiber ‘ef’ at a distance of ‘y’
from the neutral axis as shown in the Fig. 5.2. Initially the length of
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the fiber ef is Li=dx and after deforming the length of the fiber e1f1 is Lf
= (R + y) dθ. So, the strain in the fiber ef is
𝐿𝑓−𝐿𝑖
𝐿𝑓=
(R + y) dθ−𝑑𝑥
𝑑𝑥=
R dθ+y.dθ−𝑑𝑥
𝑑𝑥 (From eq.5.1)
=y.dθ
𝑑𝑥=
𝑦
𝑅 (From eq.5.1)
Thus, Normal strain in beam = 𝑦 𝑅
……………………………… 5.2
This equation shows that longitudinal strain in the beam is inversely
proportional to the radius of curvature and varies linearly with the
distance from the neutral axis. Thus, bending stress is maximum at
the outer surfaces which are at the greatest distance from the neutral
axis. If the distance is measured above the neutral axis, y is negative
and strain is also negative.
according to Hooke’s Law, Normal stress in beam
𝐸𝑦
𝑅 ……………………………………. 5.3
Thus, normal stresses acting on the beam vary linearly with the
distance from neutral axis. This type of stress distribution is shown in
the Fig. 5.4(a). Along the neutral axis bending stress is zero. Top fibers
are in compression (above the neutral axis) and bottom fibers are in
tension (below the neutral axis) for positive bending moment. For
negative bending moment top fibers are in tension and bottom fibers
are in compression.
Fig. 5.4 Distribution of Normal stresses on a cross section.
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5.3 Flexure Formula
Let us consider the resultant force acting over the cross section.
To calculate the resultant force consider a small elemental area dA in
the cross section at a distance of ‘y’ from the neutral axis as shown in
the Fig. 5.4(b).The force acting on this small elemental area normal to
moment can be calculate from the equation 5.3. Since no external
axial force is acting on the beam, (assumption 5) the equilibrium of
the forces in the x direction leads to equation
∫ σ. dA = 0
∫Ey
R. dA = 0 (from equation 5.3)
Since modulus of elasticity E and radius of curvature R is constant we
can conclude that
∫ y. dA = 0 ………………………………………………..
5.4
This is the equation for first moment of area of cross section with
respect to the neutral axis. That is the neutral axis must pass through
the centroid of the cross section. This property can be used to locate
the position of the neutral axis for a beam of any cross sectional shape
which has symmetry about y- axis.
Let us consider next the moment of the resultant force acting
the over the cross section (Fig. 5.4(b)). The moment acting on this
y. The external moment acting on the
beam is M, the equilibrium of the moment leads to equation
∫ σ. dA.y = M
∫Ey2
R. dA = M (from equation 5.3) ………………………………..
5.5
But we know that ∫ y2. dA is the equation for second moment of
area of cross section with respect to the neutral axis, which is moment
of inertia of cross sectional area with respect to the neutral axis
I = ∫ y2. dA
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Now the equation 5.5 is reduces to 𝐸𝐼
𝑅 = M
𝑀
𝐼=
𝐸
𝑅 ………………………………………………. 5.6
Combining the equation 5.3 and 5.6, we get the flexure formula
𝜎
𝑦=
𝑀
𝐼=
𝐸
𝑅 ……………………………………………. 5.7
From the above, we can write the equation for bending stress
𝑀𝑦
𝐼 ………………………………………………….. 5.8
Thus, bending stress is directly proportional to the bending moment,
inversely proportional to the area moment of inertia about neutral axis
and varies linearly with the distance from neutral axis as shown in the
Fig. 5.4 (a). The maximum stress due to bending moment occurs at
the fiber for which y is maximum, i.e., the extreme fiber from the
neutral axis. max = 𝑀.𝑦𝑚𝑎𝑥
𝐼. The
quantity 𝐼
𝑦𝑚𝑎𝑥 is known as section modulus, which is the property of
the cross section denote with a letter ‘z’.
The section modulus for simple cross sections are shown in the
Fig.5.5
For Rectangular cross section:
I = 𝑏ℎ3
12 and ymax =
ℎ
2
h z = 𝑏ℎ2
6
b
For Circular Cross section
I = 𝜋𝑑4
64 and ymax =
𝑑
2
z = 𝜋𝑑3
32
d
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Fig. 5.5 Section modulus for simple cross sections
5.4 Composite Beams
Beams that are built of more than one material are called
composite beams. Examples are bimetallic beams, sandwich beams,
reinforced concrete beams as shown in the Fig. 5.6. Composite beams
can be analyzed by the same way as that of ordinary beam. The main
advantage with these beams is it can withstand with more bending
load within less space/cross sectional area compared to beam with
single material.
Fig. 5.6 Composite beam cross sections (a) bi metallic beam
(b) sand witch beam (c) reinforced concrete beam
Let us consider a beam made of two materials. Let 1 and 2 are
the suffixes used for material 1 and 2 respectively. At the common
surface, strain in both the material is same.
1 2
But, 1 = 𝜎1
𝐸1 2 =
𝜎2
𝐸2
𝜎1
𝐸1=
𝜎2
𝐸2
𝜎1
𝜎2=
𝐸1
𝐸2
Thus, the ratio of stresses between material 1 and 2 is directly in the
ratio of their modulus of elasticity.
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Let y1 and y2 are the distances of the farthermost fibers of
material 1 and material 2 respectively form the neutral axis.
[𝜀1
𝜀2]
𝑚𝑎𝑥=
y1
y2 (From equation 5.2)
σ1max
E1X
E2
σ2max =
y1
y2
σ1max
σ2max =
y1
y2X
E1
E2 ……………………………….. 5.9
But we know that 𝜎
𝑦=
𝑀
𝐼 (From equation 5.7)
M1 = σ1max. I1
y1 and M2 =
σ2max. I2
y2
The total moment of resistance
M = M1 + M2
= σ1max. I1
y1+
σ2max. I2
y2
= σ2max. I1
y2.
E1
E2+
σ2max. I2
y2 (From eq. 5.9)
M = σ2max
y2[
E1
E2I1 + I2]
Similarly M = σ1max
y1[
E2
E1I2 + I1]
So, for a composite beam the total resisting moment is
M = σ1max
y1[
E2
E1I2 + I1] =
σ2max
y2[
E1
E2I1 + I2] …………… 5.10
[E2
E1I2 + I1] is equivalent moment inertia for material 1 & [
E1E2
I1 + I2]
is equivalent moment inertia for material 2.
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EXAMPLES EXAMPLES EXAMPLES EXAMPLES
5.1. A thin steel rule (E = 206.7GPa) having a thickness 0.734mm
and length 254mm is bent by couples as shown in Fig.P.5.
what is the maximum stress in steel rule.
Fig. P. 5.1
Sol: Given data: t = 0.734mm, E = 206.7x103MPa, l = 254mm,
θ=600=1.047
We know that flexure formula 𝜎
𝑦=
𝑀
𝐼=
𝐸
𝑅.
From the given data we can write the equation from flexure formula
𝜎 =𝐸𝑦
𝑅
Where y = maximum distance of a surface from neutral axis
= 0.734/2 = 0.367
From Fig. P. 5.12 we can write that l = Rθ
Radius of curvature R = 254/1.047=242.6mm
Bending stress 𝜎 =206.7𝑥103𝑥0.367
242.6
= 312.63MPa
5.2. A steel wire of diameter 4mm bent over a drum of radius
0.5m as shown in the Fig. P.5.2. Calculate the maximum
bending stress if E = 200GPa.
Fig. P. 5.2
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Sol: Given data: d = 4mm, E = 200x103MPa, r = 0.5x103m=500mm
We know that flexure formula𝜎
𝑦=
𝑀
𝐼=
𝐸
𝑅.
From the given data we can write the equation from flexure formula
𝜎 =𝐸𝑦
𝑅
Where y = maximum distance of a surface from neutral axis
= 4/2 = 2mm
Radius of curvature R = r+(d/2) = 500+2 = 502mm
Bending stress σ =200x103x2
502
= 796.8 MPa.
5.3. A simple beam AB of length 6.7m supports a uniform load of
intensity 22KN/m and a concentrated load of 53.4KN as
shown in the Fig. P.5.3. The beam cross section is rectangle
with width is 222mm and depth is 686mm. Determine the
maximum compressive and tensile stresses in the beam due
to bending.
Sol: Given data: b = 222mm and d = 686mm.
The maximum bending stress occurs at the cross section due to
maximum bending moment. To find the maximum bending moment
let us construct the shear force and bending moment diagram as
shown in the Fig.P.5.3. From the Shear force and bending moment
diagram shown in the Fig.P.5.3, the maximum bending moment is
204.9KNm.
From the flexure formula, the bending stress is
𝜎 =𝑀𝑦
𝐼
M = Maximum bending moment = 204.9x106 N mm
y = Maximum distance from neutral axis. we know that neutral
axis passes through the centroid. Since the given cross section is
rectangle cross section, the centroid is coincides with the
geometric center.
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y = 686/2 = 343mm.
Fig.P.5.3
I = Moment of Inertia of rectangle cross section
= 𝑏ℎ3
12=
(222)(686)3
12
= 6x109 mm4
The maximum bending stress
𝜎 =204.6𝑥106𝑥343
6𝑥109
= 11.7 MPa.
Since the maximum bending moment is positive, the top fibers are in
compression and bottom fibers are in tension. The bending stress
distribution is as show in the Fig.P.5.3a
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Fig.P.5.3a
5.4. Determine the maximum allowable length of simple beam of
rectangular cross section (Fig.P.5.4) subjected to uniformly
distributed load of 6.5KN/m, if the allowable stress is
8.2MPa.
Sol: Given data:
Fig.P.5.4
Since it is a simply supported beam with UDL the maximum bending
moment is WL2
8 = 0.8125L2. (See Fig. 4.14, Chapter 4)
From the flexure formula, the bending stress is
𝜎 =𝑀𝑦
𝐼
M = Maximum bending moment = 0.8125L2 N mm
y = Maximum distance from neutral axis. We know that neutral
axis passes through the centroid. Since the given cross section is
rectangle cross section, the centroid is coincides with the
geometric center.
y = 240/2 = 120mm.
I = Moment of Inertia of rectangle cross section
= 𝑏ℎ3
12=
(140)(240)3
12
= 1.61x108 mm4
8.2 = 0.8125𝐿2𝑥120
1.61𝑥108
Length of the beam L = 3680mm = 3.68m
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5.5. The beam shown in the figure Fig.P.5.5 is subjected to
positive bending by couples M. Determine the ratio of the
maximum tensile and compressive stresses if the cross
section is (a) an equilateral triangle (b) a semi circle
Fig.P.5.5
Sol: (a) Equilateral triangle:
Fig.P.5.5a
For triangle the centroid is at a distance of 2h/3 from the big end and
h/3 from the small end (Fig.P.5.5a). The bending moment is positive,
so top fibers are in compression and bottom fibers are in tension.
Maximum compression stress at the top fibers:
M= M, y = 2h/3 and I = I
c = 𝑀.2ℎ
3𝐼=
2𝑀ℎ
3𝐼
Maximum tension stress at the bottom fibers:
M= M, y = h/3 and I = I
c = 𝑀.ℎ
3𝐼=
𝑀ℎ
3𝐼
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠=
𝑀ℎ
3𝐼.
3𝐼
2𝑀ℎ = 0.5
(b) Semi Circle:
For Semi circle the centroid is at a distance of 4r/3π (0.212d) from the
big end and 0.282d from the small end. The bending moment is
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positive, so top fibers are in compression and bottom fibers are in
tension.
Maximum compression stress at the top fibers:
M= M, y = 0.282d and I = I
c = 𝑀.(0.282𝑑)
𝐼
Maximum tension stress at the bottom fibers:
M= M, y = 0.212d and I = I
c = 𝑀.(0.212𝑑)
𝐼
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠=
𝑀(0.212𝑑)
𝐼.
𝐼
𝑀(0.282𝑑) = 0.752
5.6. A beam ABC is loaded as shown in the Fig.P.5.6. The cross
section is a channel section as shown in the Fig. Calculate
the maximum tensile and compressive stresses in the beam.
Sol: To find the maximum bending moment, let us construct shear
force and bending moment diagram as shown in the Fig.P.5.6
Fig.P.5.6
From Fig.P.5.6, we can understand that the maximum bending
moment is 3.375KN.m and it is negative bending moment which
results top fibers are in tension and bottom fibers are in compression.
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To locate the neutral axis, let us find the centroid, since neutral axis
is passes through centroid. For this divide the given cross section into
three areas as shown in the Fig.P.5.6.
A1 = 300x12 = 3600mm2, y1 =74mm,
A2 = 68x12 = 816mm2, y2=34mm,
A3 = 68x12 = 816mm2, y3=34mm.
Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2+𝐴3𝑦3
𝐴1+𝐴2+𝐴3 = 61.52mm.
To find the Moment of Inertia let us use parallel axis theorem
I = [(300)(12)3
12+ (300)(12)(74 − 61.52)2] +
[(12)(68)3
12+ (68 )(12)(34 − 61.52)2] + [
(12)(68)3
12+ (68 )(12)(34 − 61.52)2]
= 43200+560701.44+314432+617997.9+314432+617997.9
= 2468761mm4.
Fig.P.5.6 (a)
Maximum compressive stress (point B in Fig. P.5.6)
y = 61.52mm, M = 3.375x106 N mm and I = 2468761mm4
𝑀𝑦
𝐼=
3.375𝑥106𝑥61.52
2468761 = 84.10 MPa (Compressive)
Maximum tensile stress (point A in Fig. P.5.6)
y = 18.48mm, M = 3.375x106 N mm and I = 2468761mm4
𝑀𝑦
𝐼=
3.375𝑥106𝑥18.48
2468761 = 25.26 MPa (Tensile)
The stress distribution and neutral axis is shown in the Fig.P.5.6 (a)
5.7. Determine the maximum bending stress caused by a
concentrated load 5.4KN on a simple beam of cross section as
shown in the Fig.P.5.7
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Sol: The given beam is a simply supported beam with a point load.
The maximum bending moment is 𝑊𝑎𝑏
𝐿=
5.4𝑥103𝑥1.2𝑥1.8
3.0 = 3888Nm.
Position of neutral axis
A1 = 25x75=1875mm2, y1 = 12.5mm, A2 = 1875mm2, y2 = 62.5mm
Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2
𝐴1+𝐴2 = 37.5mm.
Fig.P.5.7
Moment of Inertia I = [(75)(25)3
12+ (75)(25)(37.5 − 12.5)2] +
[(25)(75)3
12+ (75 )(25)(37.5 − 62.5)2]
= 97656.25+1171875+878906.25+1171875
= 3320312.5 mm4 =33203125x10 -12 m4
Since the maximum bending moment is positive top fibers are in
compression and bottom fibers are in tension.
Maximum tensile stress (point B in Fig. P.5.7)
y = 37.5x10-3 m, M = 3888 Nm and I =332.312.5x10 -12 m4
𝑀𝑦
𝐼=
3888𝑥37.5𝑥10−3
3320312.5x10−12 = 43.91 MPa (Tensile)
Maximum compressive stress (point A in Fig. P.5.7)
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y = 62.5x10-3 m, M = 3888 Nm and I =332.312.5x10 -12 m4
𝑀𝑦
𝐼=
3888𝑥62.5𝑥10−3
3320312.5x10−12 = 73.2 MPa (Compressive)
5.8. An overhang beam ABC of T cross section supports a
concentrated load as shown in the Fig.P.5.8. Calculate the
maximum permissible value of load P based upon allowable
stress in material 40MPa in tension and 70MPa in
compression.
Sol: To find the maximum bending moment let us construct the shear
force and bending moment diagram as shown in the Fig.P.5.8. The
maximum bending moment is – P.
Position of neutral axis
A1 = 20x80=1600mm2, y1 = 40mm,
A2 = 100x20=2000mm2, y2 = 90mm
Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2
𝐴1+𝐴2 = 67.8mm.
Fig.P.5.8
Moment of Inertia I = [(20)(80)3
12+ (20)(80)(67.8 − 40)2] +
[(100)(20)3
12+ (100 )(20)(67.8 − 90)2]
= 853333.33+1236544+66666.67+985680
= 3142224 mm4 = 3142224x10-12m4
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Since the maximum bending moment is negative, top fibers are in
tension and bottom fibers are in compression.
Maximum tensile stress (Point B in Fig.5.8)
y = 33.3x10-3 m, M = P Nm and I =3142224x10-12m4
𝑀𝑦
𝐼=
𝑃𝑥33.3𝑥10−3
3142224x10−12 = 0.01054P MPa (Tensile)
Given that maximum tensile stress = 40MPa
0.01054 = 40
P = 3.774KN.
Maximum compressive stress (Point A in Fig.5.8)
y = 67.7x10-3 m, M = P Nm and I =3142224x10-12m4
𝑀𝑦
𝐼=
𝑃𝑥67.7𝑥10−3
3142224x10−12 = 0.0215P MPa (Compressive)
Given that maximum compressive stress = 70MPa
0.0215P = 70
P = 3.26 KN.
So, The maximum possible load is 3.26KN
5.9. A cantilever beam AB, loaded as shown in the Fig. P.5.9 is
constructed of a section as shown. Find the maximum
compressive and tensile stresses in the cross section.
Sol: To find the maximum bending moment let us construct the shear
force and bending moment diagram.
Fig.P.5.9
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From the Fig.P.5.9, maximum bending moment is 1.8x106Nmm and it
is negative. That is top fibers are in tension and bottom fibers are in
compression.
Position of neutral axis:
A1 = 400x120=48000mm2, y1 = 200mm,
A2 = (π/4)x602=2827.4mm2, y1 = 300mm
Centroid �̅� = 𝐴1𝑦1−𝐴2𝑦2
𝐴1−𝐴2 = 193.74mm.
Moment of Inertia I = [(120)(400)3
12+ (400)(120)(200 − 193.74)2] −
[(𝜋)(60)4
64+ (𝜋 4⁄ )(602)(300 − 193.74)2]
I = 6.4x108 + 1.88x106 – 6.3x105 – 3.2x107
= 6.1x108 mm4
Maximum compressive stress (Point B in Fig.P.5.9)
y = 193.7 mm, M = 1.8x106Nmm and I =6.1x108 mm4
𝑀𝑦
𝐼=
1.8𝑥106𝑥193.7
6.1x108 = 0.57 MPa (Compressive)
Maximum tension stress (Point A in Fig.P.5.9)
y = 206.3 mm, M = 1.8x106Nmm and I =6.1x108 mm4
𝑀𝑦
𝐼=
1.8𝑥106𝑥206.3
6.1x108 = 0.61 MPa (Tensile)
5.10. A beam of I section shown in Fig.P.5.10 is subjected to a
bending moment of 10KNm at its neutral axis. Find the
maximum stress induced in the beam.
Sol: Bending stress = 𝑀𝑦
𝐼
Where M = bending moment = 10x106 N.mm.
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Fig.P.5.10
Position of neutral axis:
A1 = 100x20=2000mm2, y1 = 10mm,
A2 = 20x100=2000mm2, y1 = 70mm
A3 = 60x20=1200mm2, y3 = 130mm,
Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2+𝐴3𝑦3
𝐴1+𝐴2+𝐴3 = 60.77mm
Moment of Inertia I = [(100)(20)3
12+ (100)(20)(10 − 60.77)2] +
[(20)(100)3
12+ (100)(20)(70 − 60.77)2] +
[(60)(20)3
12+ (60)(20)(130 − 60.77)2]
= 66666.67+5155185.5+1666666.7+170385.8+40000+5751351.5
= 12850256 mm4
Since the bending moment is positive, top fibers are in compression
and bottom fibers are in compression.
Maximum tensile stress (at Point B in Fig.P.5.10)
y = 60.77mm, M = 10x106 N.mm, I = 12.58x106 mm4
𝑀𝑦
𝐼=
10𝑥106𝑥60.77
12.58x106 = 48.31 MPa (Tensile)
Maximum compressive stress (at Point T in Fig.P.5.10)
y = 79.23mm, M = 10x106 N.mm, I = 12.58x106 mm4
𝑀𝑦
𝐼=
10𝑥106𝑥79.23
12.58x106 = 62.98 MPa (Compressive)
Department of Mechanical Engineering, K L University 21
5.11. An I section beam has following dimensions. Top flange 6cm
wide, 1cm thick. Bottom flange 12cm wide, 1cm thick. Web
1cm thick. Total depth of the section is 12cm. The beam is
5m long simply supported over a span 3m, overhanging both
supports by the same amount and it carries a point load of
2KN each end. Find the maximum stress in the material due
to bending.
Sol: The beam with loading and the cross section dimensions are
shown in the Fig.P.5.11. The maximum bending moment is 2KNm and
is positive. The top fibers are in compression and bottom fibers are in
tension.
Position of neutral axis
A1 = 120x10=1200mm2, y1 = 5mm,
A2 = 10x100=1000mm2, y1 = 60mm
A3 = 60x10=600mm2, y3 = 110mm,
Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2+𝐴3𝑦3
𝐴1+𝐴2+𝐴3 = 47.14mm
Moment of Inertia I = [(120)(10)3
12+ (120)(10)(47.14 − 5)2] +
[(10)(100)3
12+ (100)(10)(47.14 − 60)2] +
[(60)(10)3
12+ (60)(10)(47.14 − 110)2]
= 10000+2130935.5+833333.33+165379.6+5000+2370827.8
= 5.5x106 mm4
Department of Mechanical Engineering, K L University 22
Fig.P.5.11
Maximum tensile stress (at Point B in Fig.P.5.11)
y = 47.14 mm, M = 2x106 N.mm, I = 5.5x106 mm4
𝑀𝑦
𝐼=
2𝑥106𝑥47.14
5.5x106 = 17.14 MPa (Tensile)
Maximum compressive stress (at Point T in Fig.P.5.11)
y = 72.86mm, M = 2x106 N.mm, I = 5.5x106 mm4
𝑀𝑦
𝐼=
2𝑥106𝑥72.86
5.5x106 = 26.49 MPa (Compressive)
5.12. The bending moment acting on the triangular cross section
of a beam is 3.6KNm. Determine the maximum tensile and
compressive stresses on the cross section.
Sol: The bending moment is a positive bending moment. So, the top
fibers are in compression and bottom fibers are in tension.
Department of Mechanical Engineering, K L University 23
Fig.P.5.12
The Moment of Inertia = (60)(120)3
36 = 2.88x106 mm4
Maximum tensile stress (at point B) (Fig.P.5.12)
M = 3.6x106 N.mm, y = 40mm, I = 2.88x106 mm4
𝑀𝑦
𝐼=
3.6𝑥106𝑥40
2.88x106 = 50 MPa (Tensile)
Maximum compressive stress (at Point T in Fig.P.5.11)
y = 80mm, M = 3.6x106 N.mm, I = 2.88x106 mm4
𝑀𝑦
𝐼=
3.6𝑥106𝑥80
2.88x106 = 100 MPa (Compressive)
5.13. A timber beam of rectangular cross section of length 8m is
simply supported. The beam carries a UDL of 12KN/m over
the entire length and a point load of 10KN at 3m from the
left support. If the depth is two times the width and the
stress in the timber is not to exceed 8MPa, find the suitable
dimensions of the section.
Sol: Let the width of the cross section is b and thus, depth of the
cross section is 2b.
Department of Mechanical Engineering, K L University 24
Fig.P.5.13
From the above, Fig. P.5.13 shear force and bending diagram, M =
111.586 KNm.
Moment of Inertia I = b(2b)3
12 = 0.67b4
Maximum bending stress:
M = 111.586x106 N.mm, y = b mm, I = 0.67b4mm4
Bending stress = 8 = 𝑀𝑦
𝐼=
111.586𝑥106𝑥𝑏
0.67b4
Width b = 275mm
Depth 2b = 550mm
5.14. A steel beam having an I- section as shown in the Fig.P.5.14
is 4m long and is simply supported at the ends. If the safe
stress in tension for the beam is 30MPa, determine the
permissible uniformly distributed load acting on the whole
span of the beam.
Sol: Given data: , l = 4m. Let the value of the UDL is
w/unit length. So, the maximum bending moment is M = wl2
8 = 2w.
Position of neutral axis:
Since the given cross section is symmetry about both the axis, the
centroid is at the center of the web.
Department of Mechanical Engineering, K L University 25
Fig.P.5.14
Moment of Inertia
I = 2. [(200)(20)3
12+ (200)(20)(290 − 150)2] + [
(20)(260)3
12]
= 1.57x108 + 29.29x106
= 1.86x108 mm4
Maximum bending stress:
M =2w, y = 150 mm, I = 1.86x108 mm4
30 = 𝑀𝑦
𝐼=
2𝑤𝑥150
1.86x108
UDL value w = 18.6x106 N/mm
W = 18.6 KN/m
5.15. A beam of 2m length is simply supported at the ends and
carries a UDL of 30KN/m over its entire length. If the cross
section of the beam as shown in the Fig.P.5.15, determine
the maximum tensile and compressive stresses in the beam.
Sol: Given data: l = 2m, w = 30KN/m.
The maximum bending moment M = wl2
8 = 15x106 N.mm
Position of neutral axis:
A1 = 160x120 = 19200mm2, y1 = 80mm
A2 = 80x60 = 4800mm2, y2 = 110mm
Department of Mechanical Engineering, K L University 26
Centroid �̅� = 𝐴1𝑦1−𝐴2𝑦2
𝐴1−𝐴2 = 70mm
Moment of Inertia I = [(120)(160)3
12+ (120)(160)(80 − 70)2] −
[(80)(60)3
12+ (80)(60)(110 − 70)2]
= 40.96x106+1.92x106 – 1.44x106 – 7.68x106 = 33.76x106 mm4
Fig.P.5.15
The maximum bending moment is a positive. So, the top fibers are in
compression and bottom fibers are in tension.
Maximum tensile stress (Point B in Fig.P.5.15)
M =15x106, y = 70 mm, I = 33.76x106 mm4
𝑀𝑦
𝐼=
15𝑥106𝑥70
33.76x106 = 31.10MPa (Tensile)
Maximum compressive stress (Point T in Fig.P.5.15)
M =15x106, y = 90 mm, I = 33.76x106 mm4
𝑀𝑦
𝐼=
15𝑥106𝑥90
33.76x106 = 39.98MPa (Compressive)
5.16. Determine the ratio of weights of three beams having same
lengths, made of same material, subjected to same maximum
bending moment and having same maximum normal stress, if
their cross sections are (i) a rectangular with height equal to
twice the width (ii) a square (iii) a circle.
Department of Mechanical Engineering, K L University 27
Sol: We know that bending stress =𝑀𝑦
𝐼. Since, all the beams are
having same lengths, same material, same maximum bending
moment and same normal stresses we can say that all the beams
must have same 𝑦
𝐼 values.
Fig.P.5.16
[y
I]
rectangle=
a
(a)(2a)3
12
=1.5
a3
[y
I]
square=
b2
(b)(b)3
12
=6
b3
[y
I]
circle=
c2
(π)(c)4
64
=10.18
c3
1.5
a3 =6
b3 =10.18
c3
b = 1.587a and c = 1.89a
Weight of rectangular beam: Weight of square beam: Weight of
circular beam is
(ρ).(2a2.l) : (ρ).(b2.l) : (ρ).(𝜋
4c2 .l)
2a2 : 2.52a2 : 2.82a2
1:1.26:1.41
Department of Mechanical Engineering, K L University 28
5.17. Find the dimensions of the strongest rectangular beam that
can be cut out of log of a wood 180mm diameter.
Sol: Let‘d’ is the diameter of wood and b, h are dimensions of
rectangular cross sections. Given that d = 180mm.
Fig.P.5.17
For the strongest beam the induced bending stress should be very
less. That means 𝑦
𝐼 value should be small. For the above rectangular
cross section y = ℎ
2 and I =
𝑏ℎ3
12
𝑦
𝐼 =
6
𝑏ℎ2 .For small values of 𝑦
𝐼 ,bh2 value should be high.
Let Z = bh2, but d = √𝑏2 + ℎ2
h2 = d2 – b2.
Z = (b)( d2 – b2)
Z = bd2 – b3
Let us take 𝑑𝑍
𝑑𝑏= 0
0 = d2 – 3b2
Width b = 𝑑
√3 = 103.92mm
and height h = √d2 – b2 = 146.96mm
Department of Mechanical Engineering, K L University 29
Problems for Practice
5.1. A rectangular beam 20cm deep by 10cm wide is subjected to
maximum bending moment of 500KNm. Determine the maximum
stress in the beam. If the value of E for the material is
200GN/m2, find the radius of curvature for that portion of the
beam where the bending moment is maximum
[750MN/m2, 26.67m]
5.2. The moment of inertia of a symmetrical section of a beam about
its neutral axis is 2640cm4 and its depth is 20cm. Determine the
longest span over which, when simply supported, the beam would
carry a UDL of 6KN/m run without the stress due to bending
exceeding 120MPa. [5.93m]
5.3. A T-section beam having flange 2 cm X 10 cm and web 10 cm X 2
cm is simply supported over a span of 6m. It carries a UDL of 3
KN/m run including its own weight over its entire span; together
with a load of 2.5 KN at its mid span. Find the maximum tensile
and compressive stresses occurring in beam section.
[12.94MPa (comp.), 25.87MPa (tensile)]
5.4. Find the dimensions of the strongest rectangular beam that can
be cut out of log of a wood 450mm diameter.
[breadth = 259.8mm, depth = 367.43mm]
5.5. A CI beam of I- section with top flange 8 cm X 2 cm thick, bottom
flange 16 cm X 4 cm thick and the web is 20 cm deep and 2 cm
thick. The beam is freely supported on a span of 5m. If the tensile
stress is not to exceed 20MPa, find the safe uniformly distributed
load which the beam can carry. Find also the maximum
compressive stress. [6.82KN/m,37.34MPa]
5.6. Compare the bending strength of a solid circular section with that
of hollow with internal diameter equal to 2/3 the external
diameter, if both sections have the same cross sectional areas
[1:1.938]
5.7. A horizontal beam of section shown in Fig.P.5.7 is 3 m long and is
simply supported at the ends. Find the maximum UDL it can
Department of Mechanical Engineering, K L University 30
carry if the compressive and tensile stresses must not exceed
56MPa and 30MPa respectively.
5.8. A beam having a cross section in the form of channel (Fig.P.5.8) is
subjected to bending moment. Calculate the thickness of the
channel in order that the bending stress at the top and bottom of
the beam will be 7:3. [50.8mm]
5.9. Fig.P.5.9 shows the section of a beam. Determine the ratio of
its resistance to bending in y-y plane to that in the x-x plane
if the maximum bending stresses is remains same in both the
cases. (Hint: Ratio between Moment of Inertias about y- and x-
axis) (0.33)
Fig.P.5.9
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