ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Topic 6 – Bending
Brief Notes (for class room use)
Beams
Beams are elements that are relatively slender and long with predominantly transverse
loading (perpendicular to the length of the member). There are many types of beams
based on the loading, support conditions, crosssectional properties, materials used, etc.
Simply supported beam with central concentrated load
Cantilever beam with uniformly distributed load
Fixed –Fixed beam with u.d.l.
Two span continuous beam with concentrated loads
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Overhang beam with nonuniform loading
There are a great many real life situations involving bending.
Building frames, door and window lintels,
Furniture, shelves, picture frames, horizontal window blinds,
Handles for frying pans, overhead lighting tracks,
Towers, bridge decks,
Flagpoles, tall grass, pine trees,
Human bones, eyeglass frames,
Automobile axles, steering wheel shafts, aircraft wings, ship decks & frames, etc……
RCH
World’s tallest structure
820m, 160 floors
Dubai
Typical Building frame with beams, columns, floors….
RCH
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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W
Egyptian Lintel
Indian Lintel
Tall
“Toi Toi”
grass
blowing in New Zealand breeze
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Tree bent by the wind
Overhead lighting
Confederation bridge from PEI
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Because of the leaning, extra bending is added to the tower
Car frame bearing the weight of the car
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Uplift on aircraft wings
causes bend
ing
Gymnast on
uneven bars
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Many of these applications require
bending theory to be understood
properly.
The majority of things that we encounter in life have some component of bending in them
and can be analyzed for stresses (at least approximately) if we understand the bending
theory properly.
Gantry 920
To
Taiwan
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Loading
External loadings on the beam can be of various kinds. They are primarily: distributed
loads, concentrated loads, and rarely, directly applied moments. Concentrated loads are
easy to see. Their units are the same as those of force (kN, kip, etc.). Distributed loads
are described using their load intensity and distribution along the beam. The intensity of
load, w(x) is measured as load per unit length at location x along the beam. The units for
w(x) are, kN/m, kip/ft., etc. This intensity may vary as we move from location to
location. For example, a uniformly distributed load (u.d.l.) has the same value of w at all
locations, whereas, a uniformly varying load (u.v.l.) has changing intensity along the
beam. This rate of change is uniform for a u.v.l.
In the figure above, the cross section properties are changing along the length of the
beam. However, the load intensity is not changing. This is a u.d.l. w(x)=15k/ft.
In the figure above, the cross section properties are not changing along the length of the
beam. However, the load intensity is changing. This is a nonuniform load.
w(x)=0 0 < x < 3m
w(x)=200N/m at x = 3m
w(x)=400N/m at x = 6m
In between 3m and 6m, the load intensity is varying uniformly. Through simple
interpolation,
RCH
RCH
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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( )
( )
( )
( )
mxmmN
x
mx
mm
xw 63
3
200
3
36
200400
200 ≤≤=−
−
−
+=
Similarly, for the following beam assuming the origin at point A we get,
( )
( )
( )
( )
mxmN
x
xxw 5.40
5.4
1500
05.4
500
50 ≤≤
−=−
−
−
+=
Similarly,
( )
mxmN
x
xw 95.41
5.4
50 ≤≤
−=
Note that not all loads are distributed. There can be concentrated loads. For a simply
supported beam subject to central concentrated load P, the load intensity w(x) is zero
everywhere except at the middle. At the middle, the value of w(x) is NOT equal to P,
rather, the impulse like integral of the load intensity at the middle is equal to P. The
units for P (kN, kip, etc.) are not the same as those for w(x).
RCH
P
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Bending Moment and Shear Force
Beams are subject to flexure (Bending and Shear). The moments in the beam try to
“bend” or “flex” the element. The shear tries to “skew” the cross section. At any
crosssection of a beam, there is an internal shear force denoted by V and an internal
bending moment denoted by M. Note that internal shear or moment is always a pair of
forces or moments.
In the above, from equilibrium of the element BC, we note that V
B
=P, M
B
=P(Lx). The
same shear force and moment are applied at B in segment AB but in opposite directions.
If we switch the origin from point A to point C in the cantilever beam,
In the new system, from equilibrium of the element BC, we note that V
B
=P, M
B
=Px.
The expressions for shear force and bending moment are different based on the choice of
coordinate axes (and the origin). However, the absolute numerical values for the moment
and shear will be the same irrespective of the choice.
L

x
x
P
B
A
C
L

x
A
B
V
B
M
B
V
B
x
P
B
C
M
B
M
B
=
Px
x
L

x
P
B
A
C
x
A
B
V
B
M
B
V
B
L

x
P
B
C
M
B
M
B
=

P
(
L

x
)
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Sign convention:
Previous conventions in “statics” considered positive (or negative) moment based on
whether it is clockwise or counterclockwise. That convention is appropriate when
dealing with just the external moments applied on a structure or component. However,
the theory of bending deals with internal moments. Internal moments always occur as a
conjugate pair of equal and opposite moments. Therefore, at any location in the beam,
the internal moment is a pair of moments (one clockwise and one counterclockwise on
either side of the location). Therefore, the practice followed in the “statics course” is not
valid here. Instead, we shall consider the internal moment that causes “sagging”
(concave up) as the positive moment. Conversely, the internal moment that causes
“hogging” (convex up) as the negative moment.
As in the case of moment, the internal shear force is also a pair of equal and opposite
forces (on either side of the location under consideration). Therefore, the simple “statics”
convention that “up” is positive does not work. We shall consider that shear force that
has “left side force up and right side force down” as positive (left up, right down is +ve).
Similarly, the shear force that is “down on the left” and “up on the right” is negative
shear.
Positive
shear
and
moment
Negative
shear
and
moment
Sagging
Hogging
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Relationship between Shear, Moment and Loading
Both the shear force and bending moment of a beam are directly related to the external
loading (load and its distribution along the beam). Let there be a distributed loading of
intensity w(x). This may vary along the length. There may also be other types of loading
on the beam, an example of which is shown in the figure.
Consider a small elemental length ∆x in the beam (at any location x from the origin).
Note that the beam segment parallel to the length is prismatic and shows up as a rectangle
in elevation. However, the beam crosssection is not necessarily a rectangle. It can be
any shape (preferably a shape that is symmetric about the vertical axis to avoid
longitudinal twisting & warping under the action of loads).
Let the load intensity at the left end of the segment be w while the load intensity at the
right end of the segment be w+∆w. Since the segment is small in length, the total load on
it can be taken as,
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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( )
( )
2
2
xwxwx
www
∆∆+∆=∆
∆++
Since all quantities are infinitensimal, ∆w∆x is much smaller than w∆x and hence can be
ignored.
Let this load act at a location k∆x from the right end. We note that the proportionality
factor k is such that 0 ≤ k ≤ 1.
The segment under consideration has shear and moment acting on either side as shown.
The shear and moment gradually change as we travel along the length and that s reflected
in the infinitensimal changes shown in the quantities at the right compared to those at the
left.
Applying equilibrium equations to the elemental length,
( )
2
0
2
:0
w
w
x
V
VV
xw
xwVF
y
∆
−−=
∆
∆
=∆+−
∆∆
+∆−=↑
∑
In the limit, as ∆x tends to zero, ∆w also tends to zero. Note that although the values of
∆x and ∆w tend to zero, the ratio of ∆V and ∆x does not tend to zero. The ratio becomes
the derivative.
w
dx
dV
x
V
Lim
w
x
−==
∆
∆
→∆
→∆
0
0
The slope of the shear force diagram is equal to the negative of load intensity.
On the same elemental length of beam, we apply moment equilibrium equation about
point O on the right side c/s.
( )
xk
w
wV
x
M
MMxk
xw
xwxVMM
O
∆
∆
+−=
∆
∆
=∆++∆
∆∆
+∆+∆−−=
∑
2
0
2
:0
In the limit, as ∆x tends to zero, ∆w also tends to zero. Note that although the values of
∆x and ∆w tend to zero, the ratio of ∆M and ∆x does not tend to zero. The ratio becomes
the derivative.
V
dx
dM
x
M
Lim
w
x
==
∆
∆
→∆
→∆
0
0
The slope of the moment diagram is equal to the value of shear force at that point.
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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For a general loading, the trends of shear force and bending moment diagrams can be
established by using the relationships found above.
Integral relationships between Shear, Moment & Loading
The equations above define relationships in derivative form. We can easily reverse them
to obtain relationships in integral form. Thus, between any two locations x=a and x=b,
we get,
∫∫∫
−=−∴−=⇒−=⇒−=
b
a
ab
b
a
b
a
wdxVVwdxdVwdxdVw
dx
dV
The change in shear force between two points is equal to the –ve of area under the load
diagram between the two points.
Similarly, we can integrate the moment equation.
∫∫∫
=−∴=⇒=⇒=
b
a
ab
b
a
b
a
VdxMMVdxdMVdxdMV
dx
dM
The change in bending moment between two points is equal to the area under the shear
force diagram between the two points.
We can use these relationships to sketch the shape of bending moment diagrams (BMD)
and shear force diagrams (SFD). These diagrams are of great importance in numerous
applications. They give us an understanding of the behaviour of the beams and help us
greatly in the design of beams for all practical situations.
It is extremely important for students to know how to sketch these diagrams easily.
Concentrated forces on the beam segment
Let there be a concentrated force P on the beam segment under consideration. We apply
equilibrium equations to the elemental length.
( )
0:0 =∆+−−=↑
∑
VVPVF
y
PV −=∆∴
This implies that at the concentrated load, the shear force diagram suddenly changes
value by the amount –P.
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Similarly, we apply moment equilibrium equation for the segment.
( ) ( )
PkV
x
M
MMxkPxVMM
O
−=
∆
∆
=∆++∆+∆−−=
∑
0:0
In the limit, as ∆x tends to zero, ∆x and k∆x merges with ∆x and hence the value of k
tends to 1.
PV
dx
dM
x
M
Lim
k
x
−==
∆
∆
→
→∆
1
0
The slope of the moment diagram suddenly changes on either side of P by a value of P!
Similarly, if there is a concentrated moment applied at any point on the beam, we can
show that there will be no change on the shear diagram but the moment diagram will
suddenly drop by the value of the moment.
All these observations are summarised in the following table.
P
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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For determinate beams, we can easily find the bending moment and shear force at any
location by following simple steps:
a. Obtain any necessary reactions at the supports through statics,
b. Choose the origin (of xaxis) such that computing M and V is made easy.
c. Slice the beam at the required location (say, x),
d. Draw the free body diagram of the left side (or right side) of the slice,
e. Place symbols of shear V
x
and moment M
x
at the slice,
f. Apply simple statics (equations of equilibrium) to find V
x
and M
x
.
g. Plot SFD & BMD
h. Check if they make sense using the shearmomentloading relationships.
RCH
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Bending Moment Diagram (BMD) and Shear Force Diagram (SFD)
There is no distributed load on the
beam. Therefore, the shear force must
be constant.
Looking at the segment BC, we see
that shear is equal to P and it is
downwards on the right side.
Therefore, it is a positive shear.
V=P
At a distance x from the origin, the
moment is
M
B
=Px
1
`
The negative sign is because the
bending causes a hogging type
deformation. Since the shear is
constant, the slope of the BMD line
must be constant and positive.
L

x
x
1
P
B
A
C
M
B
=

Px
1
ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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ENG 4312, Mechanics of Solids I  Dr. Seshu Adluri Bending
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Leonhard Euler developed the beam theory in its present form along with Daniel
Bernoulli following the contributions of Daniel’s uncle Jakob Bernoulli (1654–1705).
Daniel also developed the famous Bernoulli principle in fluid mechanics. Daniel was a
mentor for Euler. Euler introduced and popularized the concept of a mathematical
function f(x), notation for trigonometric functions that we use today, the use of
exponential function “e”, notations Σ and π, hyperbolic functions and a great many more
things. The contributions of Euler and Bernoulli to engineering, mathematics,
economics, statistics, physics, operations research, philosophy and many allied subjects
are astoundingly large and remain important even to this day.
The beam theory developed by Euler and Bernoulli has become the corner stone of
Engineering in the 19
th
and 20
th
centuries and led to the wide acceptance of mathematical
rigour in engineering sciences. Once the beam theory had been successfully applied to
large engineering projects (among them the Eiffel Tower, the Ferris wheel, etc.), all other
branches of engineering started developing at a furious pace.
We can safely say that Engineering would not have been what it is today, without the
efforts of Euler and the various members of the Bernoulli family.
Leonhard
Euler
(1707

1783)
Daniel Be
rnoulli (1700
–
17
82)
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