CH8. Deflections
Prepared by: Teng Chhay
1
Introduction
•
•
•
•
The elastic deflection of a beam using the method of double integration and two important
geometrical methods, namely, the moment-area theorems and the conjugate-beam method.
Double integration is used to obtain equations which define the slope and the elastic curve.
The geometric methods provide a way to obtain the slope and deflection at specific points on the
beam.
Each of these methods has particular advantages or disadvantages.
2
Deflection Diagrams and the Elastic Curve
• Deflections of structures can occur from various sources, such as loads,
temperature, fabrication errors, or settlement. In design, deflections must be
limited in order to provide integrity and stability of roofs, and prevent cracking of
attached brittle materials such as concrete, plaster or glass.
• The deflection of a structure is caused by its internal loadings such as normal
force, shear force, or bending moment.
• For beams and frames, the greatest deflections are most often caused by internal
bending.
• For trusses, the internal axial forces cause the deflections.
• Before the slope or displacement of a point on a beam or frame is determined, it
is helpful to sketch the deflected shape of the structure when it is loaded in order
to partially check the results. This deflection diagram represents the elastic curve
which defines the displaced position of the centroid of the cross section along the
members.
• It is necessary to know the restrictions as to slope or displacement that often
occur at a support or a connection.
• With reference to Table 8-1, supports that resist a force, such as a pin, restrict
displacement; and those that resist moment, such as a fixed wall, restrict rotation.
Note also that deflection of frame members that are fixed connected (4) causes
the joint to rotate the connected members by the same amount 𝜃𝜃. On the other
hand, if a pin connection is used at the joint, the members will each have a
different slope or rotation at the pin, since the pin cannot support a moment (5).
3
Deflection Diagrams and the Elastic
Curve (cont.)
• If the elastic curve seems difficult to establish, the moment diagram for the beam or frame should be
drawn first by using sign convention for moments.
• If the shape of the moment diagram is known, it will be easy to construct the elastic curve and vice
versa.
4
Deflection Diagrams and the Elastic Curve
(cont.)
Being able to draw the deflection curve also helps engineers in locating the steel needed to reinforce a
concrete beam, column or wall. Concrete is rather weak in tension, so regions of a concrete structural
member where tensile stresses are developed are reinforced with steel bars, called reinforcing rods. These
rods prevent or control any cracking that may occur within these regions.
5
Example 1
Draw the deflected shape of each of the beam.
The roller at 𝐴𝐴 allows free
rotation with no deflection
while the fixed wall at 𝐵𝐵
prevents both rotation and
deflection.
The pin (internal hinge) at 𝐵𝐵
allows free rotation, and so
the slope of the deflection
curve will suddenly change at
this point while the beam is
constrained by it supports)
The couple moment will rotate
end 𝐴𝐴. This will cause
deflections at both ends of
the beam since no deflections
is possible at 𝐵𝐵 and 𝐶𝐶. Notice
that segment 𝐶𝐶𝐶𝐶 remains
undeformed (a straight line)
since no internal load acts
within it.
No rotation or deflection can
occur at 𝐴𝐴 and 𝐵𝐵.
The compound beam deflects
as shown. The slope abruptly
changes on each side of the
pin at 𝐵𝐵.
Span 𝐵𝐵𝐵𝐵 will deflect concave
upwards due to the load.
Since the beam is continuous,
the end spans will deflect
concave downward.
6
Example 2
Draw the deflected shapes of each of the frames shown.
The vertical loading on this
symmetric frame will bend beam
𝐶𝐶𝐶𝐶 concave upwards, causing
clockwise rotation of joint 𝐶𝐶 and
counterclockwise rotation of joint
When the load 𝑃𝑃 pushes
𝐷𝐷. Since the 90𝑜𝑜 angle at the joints
𝑃𝑃 displaces joints 𝐵𝐵, 𝐶𝐶, and 𝐷𝐷
joints 𝐵𝐵 and 𝐶𝐶 to the right, it to the right, causing each
must be maintained, the columns
will cause clockwise rotation column to bend as shown. The bend as shown. This causes spans
of each column as shown.
fixed joints must maintain their 𝐵𝐵𝐵𝐵 and 𝐷𝐷𝐷𝐷 to be concave
As a result, joints 𝐵𝐵 and 𝐶𝐶
90𝑜𝑜 angles, and so 𝐵𝐵𝐵𝐵 and 𝐶𝐶𝐶𝐶 downwards, resulting in
must rotate clockwise. Since must have a reversed curvature counterclockwise rotation at 𝐵𝐵 and
the 90𝑜𝑜 angle between the
clockwise rotation at 𝐸𝐸. The
with an inflection point near
columns therefore bend as shown.
their midpoint.
connected members must
be maintained at these
The loads push joints 𝐵𝐵 and 𝐶𝐶 to the right,
joints, the beam 𝐵𝐵𝐵𝐵 will
which bends the columns as shown. The
deform so that its curvature
fixed joint B maintains its 90𝑜𝑜 angle;
is reversed from concave up
however, no restriction on the relative
on the left to concave down
rotation between the members at 𝐶𝐶 is
on the right. Note that this
possible since the joint is a pin.
produces a point of
Consequently, only beam 𝐶𝐶𝐶𝐶 does not
7
inflection within the beam.
have a reverse curvature.
Elastic-Beam Theory
When the internal moment 𝑀𝑀
deforms the element of the beam,
each cross section remains plane
and the angle between them
becomes 𝑑𝑑𝑑𝑑. The arc 𝑑𝑑𝑑𝑑 that
represents a portion of the elastic
curve intersects the neutral axis for
each cross section. The radius of
curvature for this arc is defined as
the 𝜌𝜌, which is measured from the
center of curvature 𝑂𝑂𝑂 to 𝑑𝑑𝑑𝑑. Any
arc on the element other than 𝑑𝑑𝑑𝑑 is
subjected to a normal strain.
The strain in arc 𝑑𝑑𝑑𝑑, located at a
position 𝑦𝑦 from the neutral axis
𝑑𝑑𝑠𝑠 ′ − 𝑑𝑑𝑑𝑑
𝜖𝜖 =
𝑑𝑑𝑑𝑑
However, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = 𝜌𝜌𝜌𝜌𝜌𝜌
And 𝑑𝑑𝑠𝑠 ′ = 𝜌𝜌 − 𝑦𝑦 𝑑𝑑𝑑𝑑
So 𝜖𝜖 =
−
𝜖𝜖
𝑦𝑦
𝜌𝜌−𝑦𝑦 𝑑𝑑𝑑𝑑−𝜌𝜌𝜌𝜌𝜌𝜌
𝜌𝜌𝜌𝜌𝜌𝜌
or
1
=
𝜌𝜌
If the material is homogeneous
and behaves in a linear elastic
manner, then Hooke’s law applies,
𝜎𝜎
𝜖𝜖 =
𝐸𝐸
Since the flexural formula applies,
1 𝑀𝑀
=
𝜌𝜌 𝐸𝐸𝐸𝐸
Since 𝑑𝑑𝑑𝑑 = 𝜌𝜌𝜌𝜌𝜌𝜌
Then 𝑑𝑑𝑑𝑑 =
𝑀𝑀
𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸
If we choose the 𝑣𝑣 axis positive
upward, and if we can express the
1
curvature
in terms of 𝑥𝑥 and 𝑣𝑣
𝜌𝜌
we can then determine the elastic
curve for the beam.
The curvature relationship
1
𝑑𝑑2 𝑣𝑣/𝑑𝑑𝑥𝑥 2
=
2/3
𝜌𝜌
𝑑𝑑𝑑𝑑 2
1+
𝑑𝑑𝑑𝑑
𝑀𝑀
Therefore =
𝐸𝐸𝐸𝐸
𝑑𝑑 2 𝑣𝑣/𝑑𝑑𝑥𝑥 2
1+
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑
2/3
Since the slope of the
elastic curve for most
structures is very small,
we will use small
deflection theory and
assume 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 ≈ 0. So its
square is negligible.
𝑑𝑑2 𝑣𝑣 𝑀𝑀
=
𝑑𝑑𝑥𝑥 2 𝐸𝐸𝐸𝐸
8
The double Integration Method
Sign Convention
The positive deflection 𝑣𝑣 is upward and
Once 𝑀𝑀 is expressed as a
the positive slope angle 𝜃𝜃 will be
function of position 𝑥𝑥, then
measured counterclockwise from the 𝑥𝑥
successive integrations of
2
𝑑𝑑 𝑣𝑣
𝑀𝑀
axis.
= will yield
2
𝑑𝑑𝑥𝑥
𝐸𝐸𝐸𝐸
Positive increases 𝑑𝑑𝑑𝑑 and 𝑑𝑑𝑑𝑑 in 𝑥𝑥 and 𝑣𝑣
the beam’s slope
create an increase 𝑑𝑑𝑑𝑑 that is
𝑑𝑑𝑑𝑑
counterclockwise.
𝜃𝜃 ≈ tan 𝜃𝜃 =
𝑑𝑑𝑑𝑑
Since the slope angle 𝜃𝜃 will be very
𝑀𝑀
small, its value in radians can be
𝑑𝑑𝑑𝑑
=�
𝐸𝐸𝐸𝐸
determined directly from
𝜃𝜃 ≈ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑.
The equation of the elastic
Boundary and Continuity Condition
curve
𝑀𝑀
If the beam is supported by a roller or
𝑣𝑣 = 𝑓𝑓 𝑥𝑥 = �
𝑑𝑑𝑑𝑑
pin, then it is required that the
𝐸𝐸𝐸𝐸
displacement be zero at these points.
For each integration it is
Also, at a fixed support the slope and
necessary to introduce a
displacement are both zero.
“constant of integration” and If a single 𝑥𝑥 coordinate cannot be used
then solve for the constants to express the equation for the beam’s
to obtain a unique solution
slope or the elastic curve, then
for a particular problem.
continuity conditions must be used to
evaluate some of the integration
constants.
9
Procedure for Analysis
The method of double integration is suitable for
Load or Moment Function
elastic deflections for which the beam’s slope is very • For each region in which there is an 𝑥𝑥
small. The method considers only deflections due to
coordinate, express the internal moment 𝑀𝑀 as
bending. Additional deflection due to shear
a function of 𝑥𝑥.
generally represents a few percent of the bending
• Always assume that 𝑀𝑀 acts in the positive
deflection and so it is usually neglected in
direction when applying the equation of
engineering practice.
moment equilibrium to determine 𝑀𝑀 = 𝑓𝑓 𝑥𝑥 .
Elastic Curve
Slope and Elastic Curve
• Draw an exaggerated view of the beam’s elastic
• Provided 𝐸𝐸𝐸𝐸 is constant, apply the moment
curve. Recall that points of zero slope and zero
equation 𝐸𝐸𝐸𝐸𝑑𝑑2 𝑣𝑣/𝑑𝑑𝑥𝑥 2 = 𝑀𝑀(𝑥𝑥), which requires
displacement occur at a fixed support, and zero
two integrations. The constants are
displacement occurs at pin and roller supports.
determined using the boundary conditions for
the supports and the continuity conditions
• Establish the 𝑥𝑥 and 𝑣𝑣 coordinate axes. The 𝑥𝑥 axis
that apply to slope and displacement at
must be parallel to the undeflected beam and its
points where two functions meet.
origin at the left side of the beam, with a positive
• Once the integration constants are
direction to the right.
determined and substituted back into the
• If several discontinuous loads are present,
slope and deflection equations, the slope and
establish 𝑥𝑥 coordinates that are valid for each
displacement at specific points on the elastic
region of the beam between the discontinuities.
curve can be determined. The numerical
values obtained can be checked graphically
• In all cases, the associated positive 𝑣𝑣 axis should
by comparing them with the sketch of the
be directed upward.
elastic curve.
• Positive values for slope are counterclockwise
and positive displacement is upward.
10
Example 3
Each simply supported floor joist shown in the photo is subjected
to an uniform design loading of 4𝑘𝑘𝑘𝑘/𝑚𝑚. Determine the maximum
deflection of the joist. 𝐸𝐸𝐸𝐸 is constant.
𝑑𝑑𝑑𝑑
Solution
𝐸𝐸𝐸𝐸
= 10𝑥𝑥 2 − 0.667𝑥𝑥 3 + 𝐶𝐶1
𝑑𝑑𝑑𝑑
Elastic Curve.
𝐸𝐸𝐸𝐸𝐸𝐸 =
Due to symmetry, the joist’s 3.33𝑥𝑥 3 − 0.1667𝑥𝑥 4 + 𝐶𝐶1 𝑥𝑥 + 𝐶𝐶2
maximum deflection will
Here 𝑣𝑣 = 0 at 𝑥𝑥 = 0
occur at its center. Only a
single 𝑥𝑥 coordinate is needed so that 𝐶𝐶2 = 0,
and 𝑣𝑣 = 0 at 𝑥𝑥 = 10,
to determine the internal
so that 𝐶𝐶1 = −166.7.
moment.
The equation of the elastic
Moment Function
From the free-body diagram, curve is therefore
𝐸𝐸𝐸𝐸𝐸𝐸
we have
𝑥𝑥
= 3.33𝑥𝑥 3 − 0.1667𝑥𝑥 4 − 166.7𝑥𝑥
𝑀𝑀 = 20𝑥𝑥 − 4𝑥𝑥
2
At 𝑥𝑥 = 5𝑚𝑚, note that 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 =
2
= 20𝑥𝑥 − 2𝑥𝑥
0. The maximum deflection is
therefore
Slope and Elastic Curve
521
Applying and integrating
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 = −
𝐸𝐸𝐸𝐸
twice gives
𝑑𝑑2 𝑣𝑣
𝐸𝐸𝐸𝐸 2 = 20𝑥𝑥 − 2𝑥𝑥 2
𝑑𝑑𝑥𝑥
11
Example 4
The cantilevered beam shown is subjected to a couple moment 𝑀𝑀𝑜𝑜 at its
end. Determine the equation of the elastic curve. 𝐸𝐸𝐸𝐸 is constant.
Solution
Using the boundary conditions
𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 = 0 at 𝑥𝑥 = 0 and 𝑣𝑣 = 0 at
Elastic Curve
𝑥𝑥 = 0 then 𝐶𝐶1 = 𝐶𝐶2 = 0.
The load tends to deflect the
Substituting these results into Eqs
beam as shown. By inspection,
(2) and (3) with 𝜃𝜃 = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑, we
the internal moment can be
represented throughout the beam get
𝑀𝑀𝑜𝑜 𝑥𝑥
using a single 𝑥𝑥 coordinate.
𝜃𝜃 =
𝐸𝐸𝐸𝐸
Moment Function
𝑀𝑀𝑜𝑜 𝑥𝑥 2
From the FBD, with 𝑀𝑀 acting in
𝑣𝑣 =
2𝐸𝐸𝐸𝐸
the positive direction, we have
Maximum slope and displacement
𝑀𝑀 = 𝑀𝑀𝑜𝑜
occur at 𝐴𝐴 𝑥𝑥 = 𝐿𝐿 , for which
Slope and Elastic Curve
𝑀𝑀𝑜𝑜 𝐿𝐿
𝜃𝜃
=
𝐴𝐴
Applying and integrating twice
𝐸𝐸𝐸𝐸
yields
𝑀𝑀𝑜𝑜 𝐿𝐿2
𝑣𝑣𝐴𝐴 =
𝑑𝑑 2 𝑣𝑣
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸 2 = 𝑀𝑀𝑜𝑜
(1)
𝑑𝑑𝑥𝑥
The positive result for 𝜃𝜃𝐴𝐴 indicates
𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸 = 𝑀𝑀𝑜𝑜 𝑥𝑥 + 𝐶𝐶1
(2) counterclockwise rotation and the
𝑑𝑑𝑑𝑑
positive result for 𝑣𝑣𝐴𝐴 indicates that
𝑀𝑀𝑜𝑜 𝑥𝑥 2
𝐸𝐸𝐸𝐸𝐸𝐸 =
+ 𝐶𝐶1 𝑥𝑥 + 𝐶𝐶2
(3) 𝑣𝑣𝐴𝐴 is upward.
2
12
Example 5
The beam is subjected to a load 𝑃𝑃 at its end. Determine the
displacement at 𝐶𝐶. 𝐸𝐸𝐸𝐸 is constant.
𝑃𝑃 3
𝑥𝑥 + 𝐶𝐶1 𝑥𝑥1 + 𝐶𝐶2
12 1
(2)
For 𝑥𝑥2 ,
𝑑𝑑2 𝑣𝑣2
𝐸𝐸𝐸𝐸
= 𝑃𝑃𝑥𝑥2 − 3𝑃𝑃𝑃𝑃
𝑑𝑑𝑥𝑥22
𝑑𝑑𝑣𝑣2
𝑃𝑃
= 𝑥𝑥22 − 3𝑃𝑃𝑃𝑃𝑥𝑥2 + 𝐶𝐶3
𝑑𝑑𝑥𝑥2
2
(3)
Solution
Elastic Curve
The beam deflects into the
shape shown. Due to the
loading, two 𝑥𝑥 coordinates must
be considered.
Moment Functions
Using the FBD shown, we have
𝐸𝐸𝐸𝐸𝑣𝑣1 = −
𝐸𝐸𝐸𝐸
𝑃𝑃
3
𝐸𝐸𝐸𝐸𝑣𝑣2 = 𝑥𝑥23 − 𝑃𝑃𝑃𝑃𝑥𝑥22 + 𝐶𝐶3 𝑥𝑥2 + 𝐶𝐶4 (4) Solving, we obtain
6
2
Using boundary condition
0 ≤ 𝑥𝑥1 ≤ 2𝑎𝑎
𝑣𝑣1 = 0 at 𝑥𝑥1 = 0;
𝑃𝑃
3𝑃𝑃
0 = 0 + 0 + 𝐶𝐶2
𝑀𝑀2 = − 𝑥𝑥2 +
𝑥𝑥2 − 2𝑎𝑎
2
2
𝑣𝑣1 = 0 at 𝑥𝑥1 = 2𝑎𝑎;
= 𝑃𝑃𝑥𝑥2 − 3𝑃𝑃𝑃𝑃 2𝑎𝑎 ≤ 𝑥𝑥2 ≤ 3𝑎𝑎
𝑃𝑃
0
=
−
2𝑎𝑎 3 + 𝐶𝐶1 2𝑎𝑎 + 𝐶𝐶2
Slope and Elastic Curve
12
𝑣𝑣2 = 0 at 𝑥𝑥2 = 2𝑎𝑎;
For 𝑥𝑥1 ,
𝑃𝑃
3
2
𝑑𝑑 𝑣𝑣1
𝑃𝑃
0=
2𝑎𝑎 3 − 𝑃𝑃𝑃𝑃 2𝑎𝑎 2 + 𝐶𝐶3 2𝑎𝑎 + 𝐶𝐶4
2
2
𝐸𝐸𝐸𝐸
= − 𝑥𝑥1
2
𝑑𝑑𝑥𝑥12
𝑑𝑑𝑣𝑣1 2𝑎𝑎
𝑑𝑑𝑣𝑣 2𝑎𝑎
= 2
;
𝑑𝑑𝑣𝑣1
𝑃𝑃 2
𝑑𝑑𝑥𝑥1
𝑑𝑑𝑥𝑥2
𝐸𝐸𝐸𝐸
= − 𝑥𝑥1 + 𝐶𝐶1
(1) 𝑃𝑃
𝑃𝑃
𝑃𝑃
𝑀𝑀1 = − 𝑥𝑥1
2
𝑑𝑑𝑥𝑥1
4
−
4
2𝑎𝑎 2 + 𝐶𝐶1 =
2
2𝑎𝑎 2 − 3𝑃𝑃𝑃𝑃 2𝑎𝑎 + 𝐶𝐶3
𝐶𝐶1 =
𝑃𝑃𝑎𝑎2
3
𝐶𝐶2 = 0
10 2
𝑃𝑃𝑎𝑎
𝐶𝐶4 = −2𝑃𝑃𝑎𝑎3
3
Substituting 𝐶𝐶3 and 𝐶𝐶4 into Eqs
(4) gives
𝐶𝐶3 =
𝑃𝑃
3𝑃𝑃𝑃𝑃 2
10𝑃𝑃𝑎𝑎2
3
𝑣𝑣2 =
𝑥𝑥 −
𝑥𝑥 +
𝑥𝑥2 −
3𝐸𝐸𝐸𝐸
6𝐸𝐸𝐸𝐸 2
2𝐸𝐸𝐸𝐸 2
3
2𝑃𝑃𝑎𝑎
𝐸𝐸𝐸𝐸
The displacement at 𝐶𝐶 is
determined be setting 𝑥𝑥2 = 3𝑎𝑎.
We get
𝑃𝑃𝑎𝑎3
𝑣𝑣𝑐𝑐 = −
𝐸𝐸𝐸𝐸
13
Moment-Area Theorems
The initial ideas for the two moment-area theorems
were developed by Otto Mohr and later stated
formally by Charles E. Greene in 1873. These
theorems provide a semigraphical technique for
determining the slope of the elastic curve and its
deflection due to bending. They are particularly
advantageous when used to solve problems involving
beams, especially those subjected to a series of
concentrated loading or having segments with
different moments of inertia.
To develop the theorems, reference is made to the
beam shown. If we draw the moment diagram for the
beam and then divide it by the flexural rigidity, 𝐸𝐸𝐸𝐸,
the "𝑀𝑀/𝐸𝐸𝐸𝐸 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 is shown.
Theorem 1: The change in slope between any two
points on the elastic curve equals the area of the
𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between these two points.
𝐵𝐵
𝑀𝑀
𝜃𝜃𝐵𝐵/𝐴𝐴 = �
𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸
𝐴𝐴
This equation forms the basis for the first momentarea theorem.
The notation 𝜃𝜃𝐵𝐵/𝐴𝐴 is referred to as the angle
of the tangent at 𝐵𝐵 measured with respect to
the tangent at 𝐴𝐴. This angle is measured
counterclockwise from tangent 𝐴𝐴 to tangent 𝐵𝐵
is the area of the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram is positive.
14
Moment-Area Theorems
The second moment-area theorem is based on the relative
deviation of tangents to the elastic curve. This deviation is
measured along a vertical line passing through point 𝐴𝐴.
The vertical deviation of the tangent at 𝐴𝐴 with respect to the
tangent at 𝐵𝐵 can be found by integration
𝐵𝐵
𝑀𝑀
𝑡𝑡𝐴𝐴/𝐵𝐵 = � 𝑥𝑥 𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸
𝐴𝐴
Recall from statics that the centroid of an area is determined
from 𝑥𝑥∫
̅ 𝑑𝑑𝑑𝑑 = ∫ 𝑥𝑥𝑥𝑥𝑥𝑥
So
𝐵𝐵 𝑀𝑀
𝑡𝑡𝐴𝐴/𝐵𝐵 = 𝑥𝑥̅ ∫𝐴𝐴
𝐸𝐸𝐸𝐸
𝑑𝑑𝑑𝑑
Here 𝑥𝑥̅ is the distance from the vertical axis through 𝐴𝐴 to the
centroid of the area between 𝐴𝐴 and 𝐵𝐵.
Theorem 2: The vertical deviation of the tangent at a point
𝐴𝐴 on the elastic curve with respect to the tangent extended
from another point 𝐵𝐵 equals the “moment” of the area
under the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between the two points 𝐴𝐴 and 𝐵𝐵.
This moment is computed about point 𝐴𝐴 (the point on the
elastic curve), where the deviation 𝑡𝑡𝐴𝐴/𝐵𝐵 is to be determined.
15
Procedure for Analysis
The following procedure is used to determine the
displacement and slope at point on the elastic curve
of a beam using the moment-area theorems.
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram
• Determine the support reactions and draw the
beam’s 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram.
• If the beam is loaded with concentrated forces,
the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram will consist of a series of
straight line segments, and the areas and their
moments required for the moment-area theorems
will be relatively easy to compute.
• If the loading consists of a series of concentrated
forces and distributed loads, it may be simpler to
compute the required 𝑀𝑀/𝐸𝐸𝐸𝐸 areas and their
moments by drawing the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram in parts
by using the method of superposition.
Elastic Curve
• Draw the beam’s elastic curve. Recall that
points of zero slope occur at fixed supports
and zero displacement occurs at all fixed, pin,
and roller supports.
• If it becomes difficult to draw the general
shape of the elastic curve, use the moment
(or 𝑀𝑀/𝐸𝐸𝐸𝐸) diagram.
• The displacement and slope to be determined
should be indicated on the curve. Since the
moment-area theorems apply only between
two tangents.
Moment-Area Theorems
• Apply theorem 1 to determine the angle
between two tangents, and theorem 2 to
determine vertical deviations between these
tangents.
• Theorem 2 will not yield the displacement of a
point on the elastic curve. It give only the
vertical deviation of a tangent at point 𝐴𝐴 on
the elastic curve from the tangent at 𝐵𝐵.
• After applying either theorem 1 or theorem 2,
the algebraic sign of the answer can be
verified from the angle or deviation as
indicated on the elastic curve.
16
Example 6
Determine the slope at points 𝐵𝐵 and 𝐶𝐶 of the beam shown. Take
𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺 and 𝐼𝐼 = 360 106 𝑚𝑚𝑚𝑚4 .
50
1 100 50
Solution
=−
5−
−
5
𝐸𝐸𝐸𝐸
2
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram. It is easier
375𝑘𝑘𝑘𝑘 � 𝑚𝑚2
to solve the problem in
=−
𝐸𝐸𝐸𝐸
terms of 𝐸𝐸𝐸𝐸 .
Substituting numerical data for
Elastic Curve. The 10𝑘𝑘𝑘𝑘
𝐸𝐸 and 𝐼𝐼
load causes the beam to
375
deflect as shown. (The beam −
200 106 � 360 106 10−12
is deflected concave down,
= −0.00521𝑟𝑟𝑟𝑟𝑟𝑟
since 𝑀𝑀/𝐸𝐸𝐸𝐸 is negative). By
The negative sign indicates that
the construction, the angle
the angle is measured clockwise
between tan 𝐴𝐴 and tan 𝐵𝐵
from 𝐴𝐴
𝜃𝜃𝐵𝐵 = 𝜃𝜃𝐵𝐵/𝐴𝐴
Also
𝜃𝜃𝐶𝐶 = 𝜃𝜃𝐶𝐶/𝐴𝐴
Moment-Area Theorem.
Applying theorem 1, 𝜃𝜃𝐵𝐵/𝐴𝐴 is
equal to the area under the
𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between
points 𝐴𝐴 and 𝐵𝐵; that is
𝜃𝜃𝐵𝐵 = 𝜃𝜃𝐵𝐵/𝐴𝐴
In the similar manner, the area
under 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between
points 𝐴𝐴 and 𝐶𝐶 equals 𝜃𝜃𝐶𝐶/𝐴𝐴 . We
have
𝜃𝜃𝐶𝐶 = 𝜃𝜃𝐶𝐶/𝐴𝐴
1
100
500𝑘𝑘𝑘𝑘 � 𝑚𝑚2
=
−
10 = −
2
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
Substituting numerical values for
𝐸𝐸𝐸𝐸, we have
500
−
200 106 � 360 106 10−12
= −0.00694𝑟𝑟𝑟𝑟𝑟𝑟
17
Example 7
Determine the deflection at points 𝐵𝐵 and 𝐶𝐶 of the beam shown.
Values for the moment of inertia of each segment are indicated in
the figure. Take 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺.
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 Diagram. By
inspection, the moment
diagram for the beam is
rectangle. Here we construct
the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram relative to
𝐼𝐼𝐵𝐵𝐵𝐵 , realizing that 𝐼𝐼𝐴𝐴𝐴𝐴 = 2𝐼𝐼𝐵𝐵𝐵𝐵 .
Elastic Curve. The couple
moment 𝐶𝐶 causes the beam
to deflect as shown. The
tangents at 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 are
indicated.
Δ𝐵𝐵 = 𝑡𝑡𝐵𝐵/𝐴𝐴
Also
Δ𝐶𝐶 = 𝑡𝑡𝐶𝐶/𝐴𝐴
Moment-Area Theorem.
Applying theorem 2, 𝑡𝑡𝐵𝐵/𝐴𝐴 is
equal to the moment of the
area under the 𝑀𝑀/𝐸𝐸𝐸𝐸_𝐵𝐵𝐵𝐵
diagram between 𝐴𝐴 and 𝐵𝐵
computed about point 𝐵𝐵,
since this is the point where
the tangential deviation is to
be determined.
Δ𝐵𝐵 = 𝑡𝑡𝐵𝐵/𝐴𝐴 =
250
𝐸𝐸𝐼𝐼𝐵𝐵𝐵𝐵
4 2 =
2000
𝐸𝐸𝐼𝐼𝐵𝐵𝐵𝐵
Substituting the numerical
data
2000 × 10003
Δ𝐵𝐵 =
200 103 × 4 106
= 2.5𝑚𝑚𝑚𝑚
Likewise, for 𝑡𝑡𝐶𝐶/𝐴𝐴 we must
compute the moment of the
entire 𝑀𝑀/𝐸𝐸𝐼𝐼𝐵𝐵𝐵𝐵 diagram from
𝐴𝐴 to 𝐶𝐶 about point 𝐶𝐶. We
have
Δ𝐶𝐶 = 𝑡𝑡𝐶𝐶/𝐴𝐴
250
500
=
4 5 +
(3)(1.5)
𝐸𝐸𝐼𝐼𝐵𝐵𝐵𝐵
𝐸𝐸𝐼𝐼𝐵𝐵𝐵𝐵
7250 × 10003
=
= 9.06𝑚𝑚𝑚𝑚
200 103 × 4(106 )
Since both answers are positive,
they indicate that points 𝐵𝐵 and 𝐶𝐶
lie above the tangent at 𝐴𝐴.
18
Example 8
Determine the slope at point 𝐶𝐶 of the beam shown. 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 =
6 106 𝑚𝑚𝑚𝑚4
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram. See the figure.
Elastic Curve. Since the loading is applied symmetrically to the
beam, the elastic curve is symmetric as shown. We are required
to find 𝜃𝜃𝐶𝐶 . This can easily be done, realizing that the tangent at 𝐷𝐷
is horizontal, and therefore, by the construction, the angle
𝜃𝜃𝐷𝐷/𝐶𝐶 between tan 𝐶𝐶 and tan 𝐷𝐷 is equal to 𝜃𝜃𝐶𝐶 ; that is
𝜃𝜃𝐶𝐶 = 𝜃𝜃𝐷𝐷/𝐶𝐶
Moment-Area Theorem. Using theorem 1, 𝜃𝜃𝐷𝐷/𝐶𝐶 is equal to the
shaded area under the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between points 𝐶𝐶 and 𝐷𝐷.
We have
30
1
60 30
𝜃𝜃𝐶𝐶 = 𝜃𝜃𝐷𝐷/𝐶𝐶 = 3
+ 3
−
𝐸𝐸𝐸𝐸
2
𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸
135𝑘𝑘𝑘𝑘 � 𝑚𝑚2
=
𝐸𝐸𝐸𝐸
Thus,
135000 × 10002
𝜃𝜃𝐶𝐶 =
= 0.112𝑟𝑟𝑟𝑟𝑟𝑟
200 103 × 6 106
19
Example 9
Determine the slope at point 𝐶𝐶 of the beam shown. 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺,
𝐼𝐼 = 360 106 𝑚𝑚𝑚𝑚4 .
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram.
see the figure.
Elastic Curve.
We are required to find 𝜃𝜃𝐶𝐶 .
Establish tangents at 𝐴𝐴, 𝐵𝐵
(the supports), and 𝐶𝐶. The
angle 𝜙𝜙 can be found using
𝑡𝑡𝐵𝐵/𝐴𝐴
𝜙𝜙 =
𝐿𝐿𝐴𝐴𝐴𝐴
This equation is valid since
𝑡𝑡𝐵𝐵/𝐴𝐴 is actually very small, so
that 𝑡𝑡𝐵𝐵/𝐴𝐴 can be
approximated by the length
of a circular arc defined by a
radius of 𝐿𝐿𝐴𝐴𝐴𝐴 = 8𝑚𝑚 and
sweep of 𝜙𝜙.
𝜃𝜃𝐶𝐶 = 𝜙𝜙 − 𝜃𝜃𝐶𝐶/𝐴𝐴 =
𝑡𝑡𝐵𝐵/𝐴𝐴
8
− 𝜃𝜃𝐶𝐶/𝐴𝐴
Moment-Area Theorem
Using theorem 1
1
20𝑘𝑘𝑘𝑘 � 𝑚𝑚
𝜃𝜃𝐶𝐶/𝐴𝐴 = (2𝑚𝑚)(
)
2
𝐸𝐸𝐸𝐸
20𝑘𝑘𝑘𝑘 � 𝑚𝑚2
=
𝐸𝐸𝐸𝐸
Applying theorem 2
1
1 60
𝑡𝑡𝐵𝐵/𝐴𝐴 = 2 + 6
6
3
2 𝐸𝐸𝐸𝐸
2
1
60
+ 2 [ 2
]
3
2
𝐸𝐸𝐸𝐸
So
So that
𝜃𝜃𝐶𝐶 =
𝜃𝜃𝐶𝐶 =
800
20
80
− =
8𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
80
200 106 × 360(106 )(10−12 )
𝜃𝜃𝐶𝐶 = 0.0011𝑟𝑟𝑟𝑟𝑟𝑟
20
Example 10
Determine the deflection at 𝐶𝐶 of the beam shown. Take 𝐸𝐸 =
200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 = 4 106 𝑚𝑚𝑚𝑚4 .
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram.
See the figure
Elastic diagram. We are
required to find Δ𝐶𝐶 . This is
not necessarily the maximum
deflection of the beam, since
the loading and hence the
elastic curve are not
symmetric. By constructing
the tangents at 𝐴𝐴, 𝐵𝐵 (the
supports), and 𝐶𝐶, 𝑡𝑡𝐴𝐴/𝐵𝐵 is
determined, then Δ′ can be
found from proportional
triangles
𝑡𝑡𝐴𝐴/𝐵𝐵
Δ′
=
3
6
𝑡𝑡𝐴𝐴/𝐵𝐵
Or
Δ′ =
2
𝑡𝑡𝐴𝐴/𝐵𝐵
Δ𝐶𝐶 =
− 𝑡𝑡𝐶𝐶/𝐵𝐵
2
Moment-Area Theorem
Applying theorem 2 to
determine 𝑡𝑡𝐴𝐴/𝐵𝐵 and 𝑡𝑡𝐶𝐶/𝐵𝐵 .
1
1
5
𝑡𝑡𝐴𝐴/𝐵𝐵 = 6
6
3
2
𝐸𝐸𝐸𝐸
30𝑘𝑘𝑘𝑘 � 𝑚𝑚3
=
𝐸𝐸𝐸𝐸
1
1
2.5
𝑡𝑡𝐶𝐶/𝐵𝐵 = 3
3
3
2
𝐸𝐸𝐸𝐸
3
3.75𝑘𝑘𝑘𝑘 � 𝑚𝑚
=
𝐸𝐸𝐸𝐸
1 30
3.75 11.25
Δ𝐶𝐶 =
−
=
2 𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
Δ𝐶𝐶 =
11.25
200 106 × 4 106 10−12
Δ𝐶𝐶 = 0.0141𝑚𝑚 = 14.1𝑚𝑚𝑚𝑚
21
Example 11
Determine the deflection at 𝐶𝐶 of the beam shown. Take 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺,
𝐼𝐼 = 250 106 𝑚𝑚𝑚𝑚4 .
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram. Consists of a
triangular and a parabolic
segment.
Elastic curve. The loading
causes the beam of deform as
shown. We are required to find
Δ𝐶𝐶 . By constructing tangents at
𝐴𝐴, 𝐵𝐵 (the supports), and 𝐶𝐶, it is
seen that
Δ𝐶𝐶 = 𝑡𝑡𝐶𝐶/𝐴𝐴 − Δ′
However
Or
Hence
𝑡𝑡𝐵𝐵/𝐴𝐴
Δ′
=
16
8
Δ′ = 2𝑡𝑡𝐵𝐵/𝐴𝐴
Δ𝐶𝐶 = 𝑡𝑡𝐶𝐶/𝐴𝐴 − 2𝑡𝑡𝐵𝐵/𝐴𝐴
Moment-area theorem
Applying theorem 2 to
determine 𝑡𝑡𝐶𝐶/𝐴𝐴 and 𝑡𝑡𝐵𝐵/𝐴𝐴 .
𝑡𝑡𝐶𝐶/𝐴𝐴 =
3
1
192
8
8 −
4
3
𝐸𝐸𝐸𝐸
1
1
192
8 +8
8 −
+
3
2
𝐸𝐸𝐸𝐸
11264𝑘𝑘𝑘𝑘 � 𝑚𝑚3
=−
𝐸𝐸𝐸𝐸
1
1
192
𝑡𝑡𝐵𝐵/𝐴𝐴 = 8
8 −
3
2
𝐸𝐸𝐸𝐸
2048𝑘𝑘𝑘𝑘 � 𝑚𝑚3
=−
𝐸𝐸𝐸𝐸
11264
2048
Δ𝐶𝐶 = −
−2 −
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
3
7168𝑘𝑘𝑘𝑘 � 𝑚𝑚
=−
𝐸𝐸𝐸𝐸
Thus,
−7168
Δ𝐶𝐶 =
200 106 × 250(106 )(10−12 )
Δ𝐶𝐶 = −0.143𝑚𝑚
22
Example 12
Determine the slope at the roller 𝐵𝐵 of the double overhang beam shown. Take
𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 = 18 106 𝑚𝑚𝑚𝑚4 .
Solution
𝑴𝑴/𝑬𝑬𝑬𝑬 diagram. The 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram can be simplified by drawing it in
parts and considering the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagrams for the three loadings each
acting on a cantilever beam fixed at 𝐷𝐷.
Elastic Curve. If tangents are drawn at 𝐵𝐵 and 𝐶𝐶 the slope 𝐵𝐵 can be
determined by finding 𝑡𝑡𝐶𝐶/𝐵𝐵 , and for small angles,
𝑡𝑡𝐶𝐶/𝐵𝐵
𝜃𝜃𝐵𝐵 =
2
Moment Area Theorem. To determine 𝑡𝑡𝐶𝐶/𝐵𝐵 we apply the moment
area theorem by finding the moment of the 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram between 𝐵𝐵𝐵𝐵
about point 𝐶𝐶. This only involves the shaded area under two of the
diagram. Thus
30
2𝑚𝑚 1
10
53.33𝑘𝑘𝑘𝑘 � 𝑚𝑚3
𝑡𝑡𝐶𝐶/𝐵𝐵 = 1𝑚𝑚 2𝑚𝑚 −
2𝑚𝑚
+
=
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
3 2
𝐸𝐸𝐸𝐸
Thus,
𝜃𝜃𝐵𝐵 =
53.33
= 0.00741𝑟𝑟𝑟𝑟𝑟𝑟
2 200 106 [8 106 10−12
23
Conjugate-Beam Method
The conjugate-beam method was
developed by H. Muller-Breslau in
1865. It requires the same amount of
computation as the moment-area
theorems to determine a beam’s
slope or deflection. This method
relies only on the principles of
statics, and hence its application will
be more familiar.
Here the shear 𝑉𝑉 compares with the
slope 𝜃𝜃, the moment 𝑀𝑀 compares
with the displacement 𝑣𝑣, and the
external load 𝑤𝑤 compares with the
𝑀𝑀/𝐸𝐸𝐸𝐸 diagram. To make use of this
comparison we will now consider a
beam having the same length as the
real beam, but referred to here as
the “conjugate beam”. The conjugate
beam is “loaded” with the 𝑀𝑀/𝐸𝐸𝐸𝐸
diagram derived from the load 𝑤𝑤 on
the real beam.
From the comparisons, we can
state two theorems related to the
conjugate beam:
Theorem 1: The slope at a point
in the real beam is numerically
equal to the shear at the
corresponding point in the
conjugate beam.
Theorem 2: The displacement of
a point in the real beam is
numerically equal to the moment
at the corresponding point in the
conjugate beam.
24
Conjugate-Beam supports
When drawing the conjugate beam, it is important
that the shear and moment developed at the
supports of the conjugate beam account for the
corresponding slope and displacement of the real
beam at its supports, a consequence of theorems 1
and 2.
25
Procedure for Analysis
The following procedure provides a method that
may be used to determine the displacement and
slope at a point on the elastic curve of a beam
using the conjugate-beam method.
Conjugate Beam
• Draw the conjugate beam for the real beam. This
beam has the same length as the real beam and
has corresponding supports as listed in Table 8-2.
• If the real support allows a slope, the conjugated
support must develop a shear; and if the real
support allows a displacement, the conjugate
support must develop a moment.
• The conjugate beam is loaded with the real
beam’s 𝑀𝑀/𝐸𝐸𝐸𝐸 diagram. This loading is assumed
to be distributed over the conjugate beam and is
directed upward when 𝑀𝑀/𝐸𝐸𝐸𝐸 is positive and
downward when 𝑀𝑀/𝐸𝐸𝐸𝐸 is negative. In other
words, the loading always act away from the
beam.
Equilibrium
• Using the equations of equilibrium,
determine the reactions at the conjugate
beam’s supports.
• Section the conjugate beam at the point
where the slope 𝜃𝜃 and displacement Δ of the
real beam are to be determined. At the
section show the unknown shear 𝑉𝑉𝑉 and
moment 𝑀𝑀𝑀 acting in their positive sense.
• Determine the shear and moment using the
equations of equilibrium. 𝑉𝑉𝑉 and 𝑀𝑀𝑀 equal 𝜃𝜃
and Δ, respectively, for the real beam. In
Particular, if these values are positive, the
slope is counterclockwise and the
displacement is upward.
26
Example 13
Determine the slope and deflection at point 𝐵𝐵 of the steel beam
shown. The reactions have been computed. 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 =
475 106 𝑚𝑚𝑚𝑚4 .
Solution
Conjugate Beam. The
conjugate beam is shown.
The supports are 𝐴𝐴𝐴 and 𝐵𝐵𝐵
correspond to supports 𝐴𝐴 and
𝐵𝐵 on the real beam. The
𝑀𝑀/𝐸𝐸𝐸𝐸 diagram is negative, so
the distributed load acts
downward away from the
beam.
Equilibrium. Since 𝜃𝜃𝐵𝐵 and Δ𝐵𝐵
are to be determined, we
must compute 𝑉𝑉𝐵𝐵𝐵 and 𝑀𝑀𝐵𝐵𝐵 in
the conjugate beam.
250
Σ𝐹𝐹𝑦𝑦 = 0; −
− 𝑉𝑉𝐵𝐵𝐵 = 0
𝐸𝐸𝐸𝐸
250
𝜃𝜃𝐵𝐵 = 𝑉𝑉𝐵𝐵𝐵 = −
𝐸𝐸𝐸𝐸
−250
=
200 106 × 475 106 10−12
= −0.00263𝑟𝑟𝑟𝑟𝑟𝑟
Σ𝑀𝑀𝐵𝐵𝐵 = 0;
250
8.33 + 𝑀𝑀𝐵𝐵𝐵 = 0
𝐸𝐸𝐸𝐸
2083
Δ𝐵𝐵 = 𝑀𝑀𝐵𝐵𝐵 = −
𝐸𝐸𝐸𝐸
−2083
=
200 106 × 475 106 10−12
= −0.0219𝑚𝑚 = −21.9𝑚𝑚𝑚𝑚
The negative signs indicate
the slope of the beam is
measured clockwise and the
displacement is downward.
27
Example 14
Determine the maximum deflection of the steel beam shown. The
reactions have been computed. 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 = 60 106 𝑚𝑚𝑚𝑚4 .
We require 𝑉𝑉 ′ = 0 so that
Solution
Conjugate Beam. The conjugate Σ𝐹𝐹 = 0; − 45 + 1 2𝑥𝑥 𝑥𝑥 = 0
𝑦𝑦
𝐸𝐸𝐸𝐸 2 𝐸𝐸𝐸𝐸
beam loaded with the 𝑀𝑀/𝐸𝐸𝐸𝐸
diagram is shown. Since the 𝑀𝑀/𝐸𝐸𝐸𝐸 𝑥𝑥 = 6.71𝑚𝑚 (0 ≤ 𝑥𝑥 ≤ 9𝑚𝑚) OK
diagram is positive, the distribution
Using this value for 𝑥𝑥, the
acts upward (away from the
maximum deflection in the
beam).
real beam corresponds to the
Equilibrium. The external
moment 𝑀𝑀𝑀. Hence
reactions on the conjugate beam
Σ𝑀𝑀 = 0;
are determined first and are
45
6.71
indicated on the FBD. Maximum
𝐸𝐸𝐸𝐸
deflection of the real beam occurs
1 2 6.71
1
−
6.71
6.71
at the point where the slope of the
2
𝐸𝐸𝐸𝐸
3
beam is zero. This corresponds to
+ M′ = 0
the same point acts within the
−2012
Δ𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑀𝑀′ =
region 0 ≤ 𝑥𝑥 ≤ 9𝑚𝑚 from 𝐴𝐴𝐴, we can
𝐸𝐸𝐸𝐸
isolate the section shown.
−2012
=
Note that the peak of the
200 106 × 60(106 )(10−12 )
distributed loading was determined
= −0.0168𝑚𝑚 = −16.8𝑚𝑚𝑚𝑚
from proportional triangles 𝑤𝑤/𝑥𝑥 =
The negative sign indicates the deflection is downward.
(18/𝐸𝐸𝐸𝐸)/9.
28
Example 15
The girder is made from a continuous beam and reinforced at its center with
cover plates where its moment of inertia is larger. The 4𝑚𝑚 end segments have
a moment of inertia of 𝐼𝐼 = 270 106 𝑚𝑚𝑚𝑚4 and the center portion has a
moment of inertia of 𝐼𝐼′ = 540 106 𝑚𝑚𝑚𝑚4 . Determine the deflection at the
center 𝐶𝐶. Take 𝐸𝐸 = 200𝐺𝐺𝐺𝐺𝐺𝐺. The reactions have been calculated.
segment 𝐴𝐴′ 𝐶𝐶𝐶 is isolated and the
Solution
resultants of the distributed loads
Conjugate Beam. The moment
and their locations are determined.
diagram for the beam is
Σ𝑀𝑀𝐶𝐶𝐶 = 0;
′
620
400
200
determined first. Since 𝐼𝐼 = 2𝐼𝐼, we
6 −
3.333 −
1
can express the load on the
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
20
conjugate beam in terms of the
0.667 + 𝑀𝑀𝐶𝐶𝐶 = 0
−
𝐸𝐸𝐸𝐸
constant 𝐸𝐸𝐸𝐸.
2173
Equilibrium. The reactions on
𝑀𝑀𝐶𝐶𝐶 = −
𝐸𝐸𝐸𝐸
the conjugate beam can be
calculated by the symmetry of the Substituting the numerical data for
𝐸𝐸𝐸𝐸 and converting units, we have
loading or using the equations of
Δ𝐶𝐶 = 𝑀𝑀𝐶𝐶𝐶
equilibrium. The results are
−2173
shown.
=
200 106 × 270 106 10−12
Since the deflection at 𝐶𝐶 is to be
= −0.0402𝑚𝑚 = −40.2𝑚𝑚𝑚𝑚
determined, we must compute
The negative sign indicates that
the internal moment at 𝐶𝐶𝐶. Using
the deflection is downward.
the method of sections,
29
Example 16
Determine the displacement of the pin at 𝐵𝐵 and the slope of each beam
segment connected to the pin for the compound beam shown. 𝐸𝐸 =
200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 = 18 106 𝑚𝑚𝑚𝑚4 .
Distributed load and positive
Solution
regions have a distributed load
Conjugate Beam. The elastic
that acts upward.
curve for the beam is shown in
Equilibrium. The external
order to identify the unknown
reactions at 𝐵𝐵𝐵 and 𝐶𝐶𝐶 are
displacement Δ𝐵𝐵 and the slope
calculated first and the results
𝜃𝜃𝐵𝐵 𝐿𝐿 and 𝜃𝜃𝐵𝐵 𝑅𝑅 to the left and
are indicated. In order to
right of the pin. The 𝑀𝑀/𝐸𝐸𝐸𝐸
determine 𝜃𝜃𝐵𝐵 𝑅𝑅 , the conjugate
diagram has been drawn in
beam is sectioned just to the
parts using the principle of
right of 𝐵𝐵𝐵 and the shear force
superposition. In this regard,
𝑉𝑉𝐵𝐵′ 𝑅𝑅 is computed. Thus
the real beam is thought of as
Σ𝐹𝐹𝑦𝑦 = 0;
cantilevered from the left
125 250 2
support, 𝐴𝐴. The moment
𝑉𝑉𝐵𝐵′ 𝑅𝑅 +
−
−
=0
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
diagrams for the 40𝑘𝑘𝑘𝑘 load, the
reactive force 𝐶𝐶𝑦𝑦 = 10𝑘𝑘𝑘𝑘, and
the 50𝑘𝑘𝑘𝑘 � 𝑚𝑚 loading are given.
Notice that negative regions of
this diagram develop a
downward distributed load and
positive regions have a
30
Example 16
Determine the displacement of the pin at 𝐵𝐵 and the slope of each beam
segment connected to the pin for the compound beam shown. 𝐸𝐸 =
200𝐺𝐺𝐺𝐺𝐺𝐺, 𝐼𝐼 = 18 106 𝑚𝑚𝑚𝑚4 .
The slope 𝜃𝜃𝐵𝐵 𝐿𝐿 can be found from
Solution (cont.)
a section of beam just to the left
127𝑘𝑘𝑘𝑘 � 𝑚𝑚2
of 𝐵𝐵𝐵. Thus
𝜃𝜃𝐵𝐵 𝑅𝑅 = 𝑉𝑉𝐵𝐵′ 𝑅𝑅 =
𝐸𝐸𝐸𝐸
Σ𝐹𝐹𝑦𝑦 = 0;
127
=
127 125 250 2
200 106 × 18 106 10−12
𝑉𝑉𝐵𝐵′ 𝐿𝐿 +
+
−
−
=0
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
= 0.0353𝑟𝑟𝑟𝑟𝑟𝑟
𝜃𝜃𝐵𝐵 𝐿𝐿 = 𝑉𝑉𝐵𝐵′ 𝐿𝐿 = 0
The internal moment at 𝐵𝐵𝐵
Obviously, Δ𝐵𝐵 = 𝑀𝑀𝐵𝐵𝐵 for this
yield s the displacement of
segment is the same as previously
the pin. Thus,
calculated, since the moment arms
Σ𝑀𝑀𝐵𝐵𝐵 ;
are only slightly different .
125
250
−𝑀𝑀𝐵𝐵′ +
1.67 −
2.5
𝐸𝐸𝐸𝐸
𝐸𝐸𝐸𝐸
2
5 =0
−
𝐸𝐸𝐸𝐸
427
Δ𝐵𝐵 = 𝑀𝑀𝐵𝐵𝐵 = −
𝐸𝐸𝐸𝐸
−427
=
200 106 × 18 106 10−12
= −0.119𝑚𝑚 = −119𝑚𝑚𝑚𝑚
31
Q&A
32