Bending stresses and shear stresses

Sanjivani Rural Education Society's
Sanjivani College of Engineering, Kopargaon 423603.
-Department of Civil Engineering-
Course Title: (Solid Mechanics-SY B.TECH Civil)Course Title: (Solid Mechanics-SY B.TECH Civil)
Unit III Bending and Shear StressesUnit III Bending and Shear Stresses
(Introduction)(Introduction)
By
Mr. Sumit S. Kolapkar (Assistant Professor)
Mail Id- kolapkarsumitcivil@sanjivani.org.in
1

 Introduction-
•
•
•
Beam is a structural member which has negligible cross-
section compared to its length.
It carries load perpendicular to the axis in the plane of the
beam. Due to the loading on the beam, the beam deforms and
is called as deﬂection in the direction of loading.
This deﬂection is due to bending moment and shear force
generated as resistance to the bending.
Note- The resistance is in the form of tensile and compressive
stresses in the cross section, known as bending stresses OR
longitudinal stresses.
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2

 Introduction-
•
•
Bending Moment- Is the internal resistance moment
to counteract the external moment due to the loads.
Mathematically it is equal to algebraic sum of
moments of the loads acting on one side of the
section. It can also be deﬁned as the unbalanced
moment on the beam at that section.
Shear force- Is the internal resistance developed to
counteract the shearing action due to external loads.
Mathematically it is equal to algebraic sum of vertical
loads on one side of the section and this act tangential
to cross section.
3

 Moment of force-
•
•
If forces act on a ﬁnite size body they may produce a turning
effect. This is measured by the moment of the force.
Mathematically, the moment of a force F (applied at a point P )
about a point A = ± F.d
where:
• F is the magnitude of the force.
•d is the perpendicular distance from
 Radius of gyration-
The distance from an axis at which the mass or entire area of
a body may be assumed to be concentrated and at which the
moment of inertia will be equal to the moment of inertia of the
actual mass about the axis.
Mathematically,
rxx= √Ixx /A and ryy= √Iyy /A
Note-The unit of radius of gyration is mm, cm etc 4

 Second Moment of Area (Moment of inertia)-
•
•
•
•
Mathematically, we have first moment of a force M = F.d
Second moment of a force M = (F.d) x d
Therefore in term of area we have,
First moment of area = A.x
Second moment of area OR Moment of Inertia = (A.x) . x = Ax2
Where, x = Centre of Gravity
Note- Unit is mm4, m4 ,cm4 etc.
Statement- “The second moment of area is defined as the
summation of areas times the distance squared from a fixed
axis OR ∫y2.dA”
5

 Perpendicular axis theorem-The sum of the moment
of inertia about any two axes in the plane is equal to the
moment of inertia about an axis perpendicular to the plane, the
three axes being concurrent, i.e, the three axes exist together.
Mathematically, J = Ixx+Iyy
Where,
J = Polar Moment of Inertia
 Parallel axis theorem- The moment of inertia about any
axis is equal to the moment of inertia about a parallel axis
through the centroid plus the area times the square of the
distance between the axes. Mathematically, Izz = Ixx+ Ah2
Where,
A = Area of the ﬁgure
6

(MI for various cross sections, theory of simple bending, neutral
layer and neutral axis)
(MI for various cross sections, theory of simple bending, neutral
layer and neutral axis)
By
7

 Moment of Inertia for various solid cross sections-
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8

 Moment of Inertia for various hollow cross sections-
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9

 Theory of simple bending OR pure bending-
•
•
Every cross section of a certain portion of beam is
subjected to only bending moment and no other forces
(either shear or axial).
Value of BM is same at every cross section of that
portion.
10

 Assumptions in Simple Bending-
1.
2.
3.
4.
5.
6.
7.
8.
The beam is initially straight and unstressed.
The material of the beam is perfectly homogeneous and
isotropic, i.e. of the same density and elastic properties
throughout.
The elastic limit is nowhere exceeded.
Young's modulus for the material is the same in tension and
compression.
Plane cross-sections remain plane before and after bending.
Every cross-section of the beam is symmetrical about the
plane of bending, i.e. about an axis perpendicular to the N.A.
There is no resultant force perpendicular to any cross-section.
The radius of curvature is large compared to depth of beam.
11

 Neutral Layer and Neutral Axis- A layer which does
not change its original length even after bending is called as
neutral layer.
The line of intersection of the neutral layer with any normal
section of the beam.
Note-Bending stresses are always zero at neutral layer
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12

(Flexural formula for pure bending)(Flexural formula for pure bending)
By
13

 Flexural Formula for Pure Bending-
From ﬁg. let us consider the one small portion of the
beam, which is subjected to a simple bending.
Now as shown in ﬁgure,
AB and CD: Two vertical sections in a portion of the considered
beam
N.A: Neutral axis of the section of beam
EF: Neutral layer at neutral axis
dx = Length of the beam between sections AB and CD
Let us consider one layer GH at a distance y below the neutral layer EF, it will be equal to
'dx'.
Therefore,Original length of the neutral layer EF = Original length of the layer GH = dx
14

From ﬁg. after bending we have
Section AB and CD will be now section A'B' and C'D'
Similarly, layer GH is increased to G'H' i.e. G'H' > GH
Neutral layer EF will be now E'F', but as per assumptions
length of neutral layer EF = E'F' = dx
A'B' and C'D' are meeting with each other at center O
Radius of neutral layer E'F' is R, known as radius of curvature
Angle made by A'B' and C'D' at center O is 'θ'
Distance of the layer G'H' from neutral layer E'F' is 'y'
Therefore, length of the neutral layer E'F' = R x θ
Original length of the layer GH = Length of the neutral layer EF = Length of the neutral
layer E'F' = R x θ
Length of the layer G'H' = (R + y) x θ
15

Change in length of the layer GH = Length of the layer G'H' -
original length of the layer GH
= [(R + y) x θ] - (R x θ)
= y x θ
Strain in the length of the layer GH = Change in length of the
layer GH/ Originσal length of the layer GH
= y x θ/ R x θ
ε = y/R
Now, we have Hooke's law
Stress (σ) = E. Strain (ε)
Therefore we have,
σ = E x y/R i.e. stress is directly praportional to distance 'y'
R
E
y
σ
 ....................I.
16

(Concept of moment of resistance, section modulus)(Concept of moment of resistance, section modulus)
By
17

 Moment of Resistance-
•
•
•
When a beam is subjected to a pure bending, the layers above the
neutral axis will be subjected to compressive stresses and layers
below the neutral axis will be subjected to tensile stresses.
Therefore, there will be force acting on the layers of the beams
due to these stresses and hence there will be moment of these
forces about the neutral axis too.
Total moment of these forces about the neutral axis for a section
will be termed as moment of resistance of that section.
From ﬁg. consider the rectangular cross-section of the beam.
Let us assume one strip of thickness 'dy' and area 'dA' at a
distance of 'y' from the neutral axis.
Therefore, the force acting on the layer due to bending stress will be
dF = σ x dA
18

Therefore the moment of this layer about the neutral axis is 'dM'
dM = dF x y
dM = σ x dA x y
dM = (E/R) x y x dA x y
dM = (E/R) x y2 dA
Total moment of the forces on the section of the beam around the
neutral axis, also termed as moment of resistance, could be secured
by integrating the above equation and we will have
∫dM = ∫(E/R) x y2 dA
M = (E/R) ∫y2 dA
19

M = (E/R) ∫y2 dA
but, ∫y2 dA = I = Moment of inertia
....................II
Therefore, from equation I and II, we have
R
E
y
σ
I
M
 which is known as bending stress formula OR ﬂexural formula
20

 Bending stress distribution diagram-
R
E
y
σ
I
M

We have ﬂexural formula for bending as follows,
y
R
E
σ
haveeequation wthegrearranginAfter

This equation shows that, for a constant value of 'E' and 'R', stress is directly
praportional to distance 'y'. It means that stress varies linearly with distance 'y' i.e.
when y = 0 bending stress is zero and at the extreme ﬁber where y is maximum
bending stress is also maximum.
21

 Moment of Resistance- Maximum bending moment
resisted by the section. Mathematically,
Where,
z = section modulus
zσ
σ
y
I
M.R


Section modulus OR Modulus of section- Ratio of
M.I of cross section about N-A to the distance of extreme ﬁber
from N-A.
It is denoted by 'z' and its unit is mm3, cm3. Mathematically,
For symmetrical section, σc = σt
(M.R)c = (M.R)t
For unsymmetrical section, σc ≠ σt
max
y
I
z 
22

 Design Criteria- The condition of strength OR of design
is. Mathematically,
Where,
σσandσσ ec.allowablc.maxet.allowablt.max

max
max
c.maxc.maxt.maxt.max
y
I
.z....................
z
M
σ
general,In
y
I
M
σandy
I
M
σ


Note- Greater the value of 'z', greater will be the strength of the
section
23

(Numericals based on ﬂexural formula)(Numericals based on ﬂexural formula)
By
24

 Problem 1- A rectangular lamina is 40mm wide and 120mm
deep. Calculate its moment of inertia with respect to
a) horizontal and vertical centroidal axis
b) 40mm side
c) 120mm side
d) horizontal axis 60mm below the bottom
e) vertical axis 20mm away from 120mm side
25

For rectangular lamina, b=40mm, d=120mm
‫؞‬A=40×120=4800mm2
46
33
yy
46
33
xx
mm100.64
12
40120
12
db
I
and
mm1076.5
12
12040
12
bd
I






I)
II)
46
33
AB mm1004.23
3
12040
3
bd
Ii.e
side40mmaboutinertiaofMoment



46
33
BC mm1056.2
3
40120
3
db
Ii.e
side120mmaboutinertiaofMoment



III)
26

For rectangular lamina, b=40mm, d=120mm
‫؞‬A=40×120=4800mm2
IV)
46
HH
26
HH
2
xxHH
mm1074.88I
120)(4800)10(5.76I
120mm6060hhere,.........wAhIIi.e
theoremaxisparallelbyTherefore
axis.horizontalrequiredtheisH-Hﬁg.inshownAs



46
VV
26
VV
2
YYVV
mm108.32I
40)(4800)10(0.64I
40mm2020hhere,.........wAhIIi.e
theoremaxisparallelbyTherefore
axis.verticalrequiredtheisV-Vﬁg.inshownAs



V)
27

 Problem 2- A beam of span 4m carries u.d.l. of 15 KN/m. The
c/s of the beam is as shown in ﬁgure. Find maximum stress
induced. Draw bending stress diagram.
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st=1594111758391000&source=images&cd=vfe&v
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ABAK
28

N.mm103030kN.m
8
415
8
wl
M
is,momentbendingmaximumFor UDL
6
22



Let, Ӯ be the distance between CG of the section and
bottom face i.e. position of N.A
29

 Problem 3- A T-section is formed by cutting the bottom ﬂange
of an I-section. The ﬂange is 100 mm x 20 mm and the web is
150 mm x 20 mm. Draw the bending stress distribution
diagrams if bending moment at a section of the beam is 10 kN.
m
Soln
31

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stresses-in-
beams&psig=AOvVaw1L486yDfFgLqJGokjTtB4F&ust=1594279864373000&source=images&cd=vfe&ved=
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32

 Problem 4- A simply supported beam of 5m span carries a
UDL of 2kN/m over entire span. The cross section of the beam
is shown in ﬁg. Find out maximum bending stress induced in
the beam
N.mm106.25kN.m6.25
8
52
8
wl
M
is,momentbendingmaximumFor UDL
6
22



Let, Ӯ be the distance between CG of the section and
bottom face i.e. position of N.A




























2
3
2
3
xx
xx2xx1xx
21
2211
2
2
2
1
2
1
)25.101145(1000
12
10100
)7025.101(1400
12
14010
I
III
:inertiaofmomentofnCalculatio
mm25.101
10001400
)1451000()70(1400
AA
YAYA
Y
mm145
2
10
140Y,mm100010100A
mm70
2
140
Y,mm140010140A
is,axisneutralofPosition
33

 Contd...Problem 4- A simply supported beam of 5m span
carries a UDL of 2kN/m over entire span. The cross section of
the beam is shown in ﬁg. Find out maximum bending stress
induced in the beam
4
46
66
mm5576250
mm10576.5
10922.110654.3



2
6
6
t
6
6
c
cc
c
c
tc
N/mm113.483isbeamin theinducedstressbendingmaximum
N/mm483.11325.101
105.576
106.25
σ
and
N/mm640.5475.48
105.576
106.25
σ
Y
I
M
σ
Y
σ
I
M
haveweformulastressbendingfrom
mm48.75Y
101.25150Y
Y150Y
ﬁber,extremeupperthefromCGofdistanceTherefore















34

 Problem 5- Maximum bending moment of simply supported
beam is 10kN.m. The cross section of the beam is as shown in
ﬁg. Find the maximum bending stress in tension and
compression.
35

(Concept of shear stress)(Concept of shear stress)
By
38

 Concept of Shear stress-
Is a stress that acts tangential to the surface of the material.
When a beam is in pure bending, the only stress resultants
are the bending moments and the only stresses are the
normal stresses acting on the cross sections.
However, most beams are subjected to loads that produce
both bending moments and shear forces (nonuniform
bending). In these cases,both normal and shear stresses are
developed in the beam.
From, ﬁgure 'ꞇ' is shear stress distributed uniformly over the
right-hand face. For the equilibrium in the 'y' direction, the
total shear force i.e (shear stress x cross sectional area)
acting on the right-hand face must be balanced by an equal
but oppositely directed shear force i.e (shear stress x cross
sectional area) on the left-hand face.
Since the areas of these two faces are equal, then the shear
stresses on the two faces must be equal.
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39

The forces acting on the left- and right-hand side faces form
a couple having a moment about the 'z' axis, acting
clockwise in the ﬁgure.
For the equilibrium of the element this moment must be
balanced by an equal and opposite moment resulting from
shear stresses acting on the top and bottom faces of the
element. Therefore corresponding horizontal shear forces
equal to (shear stress x cross sectional area) form an
anticlockwise couple.
We have following observations regarding shear stresses
acting on a rectangular element:
1. Shear stresses on opposite (and parallel) faces of an
element are equal in magnitude and opposite in direction.
2. Shear stresses on adjacent (and perpendicular) faces of
an element are equal in magnitude.
40

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FOPTI_222_W10.
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QjRxqFwoTCJj8pcKruOoCFQAAAAAdAAAAABAJ
Now suppose that the two beams are glued along
the contact surface, so that they become a single
solid beam.
When this beam is loaded, horizontal shear
stresses must develop along the glued surface in
order to prevent the sliding shown in Fig.b.
Because of the presence of these shear stresses,
the single solid beam is much stiffer and stronger
than the two separate beams.
41

 Concept of complimentary shear stress-
•
•
Let us consider a rectangular block ABCD as shown in ﬁgure. Assume
that a set of shear stresses (ꞇ) of opposite direction, as displayed in
ﬁgure, is applied over the surfaces of rectangular block i.e. AB and CD.
Now if we consider the effect of this set of shear stresses (ꞇ) of opposite
direction, we can easily say that there will be zero net force acting over
the rectangular block but there will be one couple acting over the
rectangular block in anticlockwise direction.
• In order to maintain the equillibrium of rectangular block, there must be one more couple of
similar intensity acting over the rectangular block in opposite direction i.e. in clockwise direction.
Therefore, there will be one more set of shear stresses (ꞇ’) of same intensity acting over the rest
two opposite surfaces of rectangular block i.e side AD and BC and this set of shear stresses will
be termed as complementary shear stress.
• Therefore we can say that according to principle of complementary shear stresses,
“A set of shear stresses acting across a plane will always be accompanied by a set of balancing
shear stresses of similar intensity across the plane and acting normal to it” 42

Concept of complimentary shear stress-
Let us consider that thickness of the rectangular block ABCD, normal to the plane of paper, is one
or unity.
We have applied set of shear stresses 'ꞇ' and set of complementary shear stresses ꞇ’.
Let us determine the force acting on surface AB and CD of rectangular block
Force acting on surface AB = Shear stress x area
Force acting on surface AB = ꞇ x AB x 1 = ꞇ x AB
Direction of force (ꞇ x AB) acting on surface AB will be towards right.
Similarly, force acting on surface CD = ꞇ x CD x 1 = ꞇ x CD
Direction of force (ꞇ x CD) acting on surface CD will be towards left.
As these two forces i.e. (ꞇ x AB) and (ꞇ x CD) are equal in magnitude and acting in opposite
direction and therefore these two forces will develop one couple which will be acting in
anticlockwise direction.
Couple developed by set of shear stresses (ꞇ) = ꞇ x AB x BC
Similarly, Couple developed by set of complementary shear stresses (ꞇ’) = ꞇ’ x BC x AB
43

 Concept of complimentary shear stress-
In order to balance the rectangular block, couple developed by applied set of
shear stresses (ꞇ) and complementary set of shear stresses (ꞇ’) must be equal.
ꞇ x AB x BC = ꞇ’ x BC x AB
ꞇ = ꞇ’
Therefore we can say from above equation that a set of shear stresses acting
across a plane will always be accompanied by a set of balancing shear
stresses of similar intensity across the plane and acting normal to it.
44

(Shear stress distribution in beam)(Shear stress distribution in beam)
By
45

 Shear stress distribution in beams-
From ﬁg. consider a small portion ABDC of length 'dX' of beam loaded with UDL. As beam
is subjected to UDL, the shear force and bending moment vary at every point along the
length of beam. Therefore, let
M = Bending moment at AB
M + dM = Bending moment at CD
F = Shear force at AB
F + dF = Shear force at CD
I = Moment of inertia about N.A
46

Now consider a small strip at a distance of 'y'
from N.A. Therefore, let
σ = Bending stress at 'y' from N.A
a = Cross sectional area of strip
Therefore, by using ﬂexure formula
y
I
M
σ
y
σ
I
M


y
I
dM)(M
)σd(σ 


.........on face AB
.........on face CD
Now, we know that
Force = stress x area
‫؞‬ ay
I
M
FAB

Similarly,
47

Similarly,
ay
I
dMM
FCD


‫؞‬
Therefore, net unalanced force on strip
dF = FCD - FAB
ay
I
M
-ay
I
dMM



ay
I
dM

yA
I
dM
a.y
I
dM
ay.dy
I
dM
F
2d
0
2d
0





‫؞‬Total unbalanced force (F) above N.A
48

Where,
Shear stress(ꞇ) = Total Force / c.s area
‫؞‬
bdx
yA
I
dM
τ



‫؞‬ Ib
yA
dx
dM
τ 
areaofmomentFirstyA 
ForceShearF
dx
dM
........
Ib
yF.A
τ 
Where, F = Shearr force at the section under consideration
A = Area above OR below the layer under consideration
ӯ = Distance of the centroid of area under consideration from N.A
b = Width
I = Moment of inertia of the section @ X-X axis
49

(Shear stress distribution in various c/s of beam)(Shear stress distribution in various c/s of beam)
By
50

 Shear stress distribution for various cross
sections-
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51

sections-
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sections-
solids&psig=AOvVaw3TLDTo1TpEe60nAwD9vYMr&ust=159411679691
2000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCICTmYmxuOoCFQ
AAAAAdAAAAABAN
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solids&psig=AOvVaw3TLDTo1TpEe60nAwD9vYMr&ust=1594116796912000&source=im
ages&cd=vfe&ved=0CAIQjRxqFwoTCICTmYmxuOoCFQAAAAAdAAAAABAi
55

(Shear stress distribution in circular c/s of beam)(Shear stress distribution in circular c/s of beam)
By
56

 Shear stress distribution in circular section-
Consider the circular section of a beam as
shown in ﬁgure.
Assume one layer EF at a distance y1 from
the neutral axis of the circular beam section.
Let,
R = Radius of the circular section of the
beam
F = Shear force acting on the circular section
of the beam
N.A = Neutral axis of the beam section
EF = Layer of the beam at a distance 'y1' from the N.A of the beam
section
A= Area of section, where shear stress is to be determined
ȳ = Distance of C.G of the area from neutral axis of the beam
section 57

Shear stress at a section will be given by,
Where,
F = Shear force (N)
τ = Shear stress (N/mm2)
A = Area of section, where shear stress is to be determined (mm2)
ȳ = Distance of C.G of the area, where shear stress is to be determined,
from neutral axis of the beam section (m)
A. ȳ = Moment of the whole shaded area about the neutral axis
I = Moment of inertia of the given section about the neutral axis (mm4)
For circular cross-section, Moment of inertia, I =
b= Width of the given section where shear stress is to be determined (m)
4
D
64
π

58

Let us consider one strip of thickness 'dy' and area 'dA' at a
distance 'y' from the neutral axis of the section of the beam.
We can see here that width 'b' will be dependent over the value
of 'y' and let us ﬁrst determine the value of 'b' and after that we
will determine the value of area of small strip of thickness dy i.
e. dA.
(b/2)2 = R2- y2
Therefore, area of shaded strip 'dA' is
59

Moment of this small strip area 'dA' about the neutral axis is
As we have all values such as value for A. ȳ and width b, therefore we will have following
expression for shear stress for a beam with circular cross-section.
Above equation shows that shear stress 'ꞇ' increases as 'y' decreases i.e
i.) Maximum shear stress will occur at y1 = 0 or maximum shear stress will occur at
neutral axis.
ii.) Shear stress will be zero at the extreme ends because at extreme ends y1 = R and
therefore shear stress will be zero at extreme ends.
60

We will also ﬁnd the value of maximum shear stress
and it could be easily calculated by using the value of y1
= 0
 
3:4avg:max.
2
avg....................avg.......max.
max.
2
max.
4
2
max.
4
22
max.
ττ
D
4
π
F
Area
Force
ττ
3
4
τ
A
F
3
4
τ
D
4
π
F
3
4
τ
D
64
π
3
D
4
F
τ
64
3
2
I3
RF
τ















D
DF

61
Shear stress distribution diagram
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shear-stress-distribution-diagram-for.html

(Shear stress distribution in rectangular c/s of beam)(Shear stress distribution in rectangular c/s of beam)
By
62

 Shear stress distribution in rectangular section-
Consider the rectangular section ABCD of
a beam as shown in following ﬁgure.
Assume one layer EF at a distance y from
the neutral axis of the beam section. Let,
b= Width of the rectangular section
d= Depth of the rectangular section
N.A = Neutral axis of the beam section
EF = Layer of the beam at a distance 'y' from the N.A of the beam
section
A= Area of section CDEF, where shear stress is to be determined
ȳ = Distance of C.G of the area CDEF from neutral axis of the beam
section
63

Shear stress at a section will be given by,
Now, consider the value of the area of
section, where shear stress is to be
determined
A= b x (d/2-y)
Distance of C.G of the area CDEF from neutral axis of the beam
section, ȳ could be written as mentioned here
ȳ = y + (d/2-y)/2
ȳ = (d/2+ y)/2
I = bd3/12 64

Let us use the value of above parameters
in equation of shear stress and we will
have
..............Shear stress at layer EF
65

From above equation it is observed that,
1. At y = d/2 (at top or bottom), shear
stress = 0
2. At y = 0 (at N.A), maximum shear stress
is
bd
F
Area
ForceShear
τ
τ1.5τ
bd
F
1.5τ
avg
Where,
avgmax
max



Therefore we can say that for a rectangular section, value of maximum shear
stress will be equal to the 1.5 times of mean (average) shear stress
66

From the equation of shear stress we can say that shear stress
distribution diagram will follow parabolic curve as shown in
following ﬁg.
67
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shear-stress-distribution-in.html

(Numericals on shear stress distribution)(Numericals on shear stress distribution)
By
68

 Problem1- A beam of symmetrical I section is 200mm x
400mm in size. The thickness of ﬂange is 20mm and web is
15mm. The beam is carrying maximum shear force 80 kN.
Draw shear stress distribution diagram over the depth of
section.
69

71

 Problem2- A channel section is shown in ﬁg. It is used as a
beam and carries a shearing force of 40 kN. Obtain the
maximum shear stress and draw shear stress distribution
diagram over the depth of section.
72

73
section)al(symmetricbottomandtop...frommm........75
2
150
y
mm106475.11I
10)4775.16125.28(I
XaxisX@mmetrical........sy..........
12
13090
12
150100
I
12
bd
12
BD
I
Ib
yFA
τ
haveWe
46
6
33
33











74

 Shear connector-
•
•
When two or more parts are required to be
connected by means of bolts, screws or
nails to prevent relative motion between
them, the connectors are called as shear
connector.
Note- The longitudinal shear along the
shearing surface are resisted by shear
connectors.
75

1.
2.
Design steps-
Shear ﬂow- Is the longitudinal force per unit length
transmitted across the section.
‫؞‬Horizontal shear per unit length = ꞇ = FAӮ/I.b
Shear force at each connector (F) = Horizontal shear x
spacing OR pitch

 Shear connector-
76
boltofarea
Ib
yFA

3. Shear strength of connector = shear force at each connector
‫؞‬F = shear stress x area of connector

 Concept of shear flow-
•
1.
2.
It measures the shear force per unit length. Is useful in
applications of
Built up members- Sections that are built up and joint together
using nails/bolts. Shear flow help us to decide appropriate
spacings between nails/bolts to ensure sufficient strength.
Thin walled members- Members where thickness is very small
compared to other dimensions. Shear flow helps to determine
the shear force distribution in each portion of the cross
section and is necessary to work out shear centre.
Therefore mathematically,
Shear flow = FAӮ/I
Note- It is similar to shear stress only width 'b' is eliminated.
77

Bending stresses and shear stresses

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