{SOM- I} [Unit – VIII] Fixed Beams BHCET
A fixed or a build in beam has both of its ends rigidly fixed so that the slope at the ends remains zero. Such a beam is also called as the encastre beam .
The fixed ends give rise to fixing moments there in addition to the reactions.
If perfect end fixing can be achieved, build in beams carry smaller maximum bending moments and have smaller deflections that the corresponding simply supported beams with the same loads applied. Therefore they are stronger and stiffer. However the need for high accuracy in aligning the supports and fixing the ends during erection increases the cost. Small subsidence of either support or temperature changes can set up large stresses. The end fixings are also normally sensitive to vibrations and fluctuations in bending moments.
Fixed beam with isolated load not at the mid span:
Consider a fixed beam as shown in the figure. Let M
A and M
B be the moments due to fixed ends.
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 1

{SOM- I} [Unit – VIII] Fixed Beams BHCET
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 2

{SOM- I} [Unit – VIII] Fixed Beams BHCET
Fixed beam carrying uniformly distributed load:
Consider a fixed beam carrying a uniformly distributed load of intensity w per unit length over the whole span as shown in the figure. Due to the symmetry in loading,
R
A
= R
B
= wl/2
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 3

{SOM- I} [Unit – VIII] Fixed Beams BHCET
Fixed beam subjected to a couple at any point:
Consider a fixed beam subjected to the couple M o at a distance ‘a’ from the left end support. Taking the section at the distance x from A as shown in figure.
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 4

{SOM- I} [Unit – VIII] Fixed Beams BHCET
For a beam of uniform section, the changes in slope is given by
, where
A is the area of the bending moment diagram. In the case of the fixed beam
θ=0 for the whole span and since the product E I is not zero, it follows that A, the resultant area of the bending moment diagram for the beam must be zero.
Figure shows the bending moment diagram for a fixed and the free beam.
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 5

{SOM- I} [Unit – VIII] Fixed Beams
According to the Mohr ’s Theorem
EI(θ
A
– θ
B
) = = A
1
+ A
2
Where A
1
= area of the free bending moment diagram
A
2
= area of the fixed bending moment diagram
Also, EI (y
B
- y
A
) =
Where are the distances of C.G. of area A
1 and A
2 from A.
For a fixed Beam
BHCET
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 6

{SOM- I} [Unit – VIII] Fixed Beams BHCET
.
Consider a fixed beam carrying a central concentrated load W as shown in
Fig. The fixing moment and free moment diagrams are shown in Fig. (b) and
(c) respectively Whereas Fig. (d) is the resultant B.M. diagram.
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 7

{SOM- I} [Unit – VIII] Fixed Beams BHCET
The shear force diagram is shown in the figure (e).
For the fixed beam carrying uniformly distributed load as shown in Fig., due to symmetrical loading.
M
A
= M
B
A
1
= - M
A x l
A
2
= x
L
=
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 8

{SOM- I} [Unit – VIII] Fixed Beams BHCET
The Shear Force Diagram is shown in Figure (e).
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 9

{SOM- I} [Unit – VIII] Fixed Beams BHCET
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 10

{SOM- I} [Unit – VIII] Fixed Beams BHCET
Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET. Page 11