FIXED BEAMS AND ARCH SYSTEM.pdf

2.FIXED BEAM AND ANALYSIS OF
THREE-HINGED ARCHES
THEORY OF STRUCTURES
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Prepared by Goodluck Mwakalobo

FIXED BEAM a beam whose both ends are fixed. A
fixed beam also is called built in or encaster beam.
2.1 GENERAL REMARKS
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Fig. 2.1

In case of simply supported, The deflection is zero at the
ends. But slope is not zero at the ends as shown in fig.
2.1 (a). In case of fixed beam deflection and slope are
zero at the fixed ends as shown in fig. 2.1(b). slope will
be zero at the ends if the deflection curve is horizontal at
the ends.
-To bring slope back to zero(i.e to make deflection curve
horizontal at the fixed ends). The ends moments MA and
MB will be acting in which MA will be acting anti-clockwise
and MB will be acting clockwise as shown in Fig. 2.1 (b).
2.1 GENERAL REMARKS
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Continous beam a beam which is suported on more
than two supports.
2.1 GENERAL REMARKS
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Fig. 2.2

Let us consider a fixed beam carrying a point
load at the centre.
.4 FIXING MOMENT OF BEAMS
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Fig. 2.

In order to find the fixed moment for that beam we
need to know how to draw bending moment diagram
and the moment area method.
Since it is very difficult to find the unknown[reaction
force, moments] for indeterminant structure so the
following are
procedure how to calculate the bending
moment of fixed beams.
a] Make a simply supported beam subjected to a
given vertical loads.
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i] Then find reaction R’A and R’B by using EQE
ii] Draw the shear force [SFD] and bending moment
diagram [BMD].
b].A simply supported beam is subjected to end moments
only [without given loading].let the reactions to be R due
to moments. Since the load is symetrical then MA = MB
- Draw the bending moment diagram.
c] Add the bending moment diagrams drawn from
procedure [a] and procedure [b] then you will get
resultant bending moment diagram
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So the resultant RA = R’A - R and RB = R’B + R
To find the value of R, you need to find the value of
MA and MB .
To find the value of MA and MB .
The resultant moment at any section at a distance X
from A. = Mx - M’x but also we know moment at any
section equal to
Thus
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Integrate the above the above equation for entire
length, we get.
But represents the slope. And the slope at the
fixed ends i.e at A and B are zero. The above
equation can be written as
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Now represents the area of B.M.
diagram.
due to vertical loads and represents
the area of B.M. due to end moments.
Substituting the value in the equetion of slope.
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0

The above equation shows that the area of BMD due
to vertical load is equal to BMD due to ends moment.
Again consider the equation below
Multiplying the above equetion by x, we get
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1

Integrating for the whole length of the beam i.e from
o to L. we get
In the above equation , represent the area of
BMD due to vertical loads at a distance X from the
end A. The term represents the moment of the
area of BMD about the end A.
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Hence represents the moment of the total area of
BMD. Due to vertical loads about A, and it is equal to
total area BMD due to vertical vertical loads
multiplied by the distance of C.G of area from A.
Where distance of C.G. of BMD due to vertical load
Similarly
Where distance of C.G. of BMD due to vertical load
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Substituting the above values in second order
equation which has multiplied by x
After integration then, we get
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Enter boundary conditions
Since slope and deflection at A and B are zero, Hence
and are zero thus
but we know that hence
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Hence the distance C.G. of BMD due to vertical load
from A is equal to the distance C.G. of BMD due to
end moments from A.
Therefore MA and MB calculated by
i. Equating the area of BMD due to vertical loads to
the area of BMD due to end moments.
ii. Equating the distance of C.G. of BMD due to
vertical loads to the distance of C.G. of BMD due
to end moments
NB: The distance of C.G. must be taken from the
same end in both case.
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A fixed beam AB, 6m long, is carrying a point load of
5o kN at its centre the moment of inertia of the beam
is and value of E for beam material is
Determine the fixed end moment at A nd B.
solution:
EXAMPLE
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solution:
a]simply supported beam
EXAMPLE
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i] Then find reaction RA and RB by using EQE
RA = RB = 5/2 = 25 kN. (Since load is symetrical).
ii] TO draw shear force [SFD] and bending moment
diagram [BMD].
NB: These are the crucial points to note when you draw
shear force diagram by using integration method.
we
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First of all you need to know the following relation.
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Second the following points.
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So starting with shear force diagram
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2

now dM = VdX from dV = dM/dX
-From support moment is zero since simply supported
HENCE
- from 3m length distance
dM = 25dX = 75KNM
after the force dM = VdX. dM = -25dX
dM= -25dX =
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3

SO it will be
M - 75= -25(6) - -25(3)
M-75 = -150 + 75
M = -75 + 75 = 0 kNM
b].Then A simply supported beam is subjected to end
moments only [without given loading]. After that
draw the bending moment diagram.
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-After that draw the bending moment diagram.
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c] lastly add the bending moment diagrams drawn
from procedure [a] and procedure [b] then you will
get resultant bending moment diagram
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Now Equating area of two BMD
(1/2 X 75 X 3) X 2 = MA X 6
MA = 37.5 kNM
Since load is symetrical hence MA = MB
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THANK YOU

FIXED BEAMS AND ARCH SYSTEM.pdf

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