Types of beams , calculation of support reacions, types of loads

2. Beam:
Beam is a structural member subjected
to lateral loads, that is, forces or moments
having their vectors perpendicular to the axis
of the bar.

3. Types of Beams:
Beams are usually classified by the manner in which they are supported
Simply supported beam
Cantilever beam
Over hanged beam
Continuous beam
Propped cantilever beam
Fixed beam

4. Simply supported beam
A beam with a pin support at one end and a roller support at the other is called a
simply supported beam or a simple beam.

5. Cantilever beam
A beam which is fixed at one end and free at the other, is called a cantilever
beam.

6. Over hanged beam
This beam is simply supported at points A
and B but it also projects beyond the
support at B.

7. Statically Indeterminate Beams
Continuous beam Propped cantilever beam
Fixed beam

8. Support Reactions:
The essential feature of a pin support is that
it prevents translation at the end of a beam
but does not prevent rotation. Thus, end A of
the beam of Fig cannot move horizontally or
vertically but the axis of the beam can rotate
in the plane of the figure.
A pin support is capable of developing a force
reaction with both horizontal and vertical
components (HA and RA), but it cannot
develop a moment reaction.
Simply supported beam

9. Support Reactions:
At end B of the beam the roller support
prevents translation in the vertical direction
but not in the horizontal direction; hence this
support can resist a vertical force (RB) but not
a horizontal force. Of course, the axis of the
beam is free to rotate at B just as it is at A.
Simply supported beam

10. Support Reactions at pin end and roller end:
Simply supported beam

11. Support Reactions:
Cantilever beam
At the fixed support (or clamped support) the
beam can neither translate nor rotate,
whereas at the free end it may do both.
Consequently, both force and moment
reactions may exist at the fixed support.

12. Support Reactions at fixed end:
Cantilever beam

13. Types of loads:
 When a load is applied over a very small area it may be idealized as a concentrated load, which is a single
force.
 When a load is spread along the axis of a beam, it is represented as a distributed load.
 A uniformly distributed load, or uniform load, has constant intensity q per unit distance.
 A varying load has an intensity that changes with distance along the axis; for instance, the linearly
varying load has an intensity that varies linearly from q1 to q2.
 Another kind of load is a couple

14. Calculation of Support Reactions:
Q1. Calculate the support reactions for the beam as shown in the figure

15. Calculation of Support Reactions:
Free body diagram of the
beam

16. Calculation of Support Reactions:
Free body diagram of the beam
𝜮𝑭 𝒙 = 𝟎
⇒ 𝑯 𝑨= 𝟎
𝜮𝑭 𝒚 = 𝟎
𝐑 𝑨 + 𝑹 𝑩 − 𝟑 ⋅ 𝟓 − 𝟕 = 𝟎
𝐑 𝑨 + 𝑹 𝑩 =10.5
𝛴𝑀𝐴 = 0
-3.5 *(0.8)- 7 * (1.2+.8) +RB *3=0
 RB=( 3.5 *0.8 + 7 *2)/3
 RB = 5.6 kN
→ (1)
→ (2)
Substitute (2) in (1)
RA = 10.5- RB
RA = 10.5- 5.6 = 4.9 kN.

17. Q3. Calculate the support reactions for the beam as shown in the figure

18. Free body diagram of the beam
𝜮𝑭 𝒙 = 𝟎
⇒ 𝑯 𝑨= 𝟎
𝜮𝑭 𝒚 = 𝟎
𝐑 𝑨 + 𝑹 𝑩 − 𝟔 − 𝟐 ∗ 𝟏. 𝟓 = 𝟎
𝐑 𝑨 + 𝑹 𝑩 = 𝟗
𝛴𝑀𝐴 = 0
-6 *(1)- 2 *1.5 * (2.5+.75) + RB *4=0
 RB=( 6+ 2*1.5 * 3.25)/4
 RB = 3.94 kN
→ (1)
→ (2)
Substitute (2) in (1)
RA = 9- RB
RA = 9- 3.94= 5.06 kN.

19. Q4. Calculate the support reactions for the beam as shown in the figure

20. Free body diagram of the beam
𝜮𝑭 𝒙 = 𝟎
⇒ 𝑯 𝑨= 𝟎
𝜮𝑭 𝒚 = 𝟎
𝐑 𝑨 − 𝟒 − 𝟏. 𝟓 ∗ 𝟐 = 𝟎
𝐑 𝑨 = 𝟕 𝐤𝐍
𝛴𝑀 𝐹𝐸 = 0
- 𝑀𝐴 − 4*(1)- 1.5 *2 * (2 +1 ) =0
 𝑀𝐴 =- ( 4+ 1.5 *2 * 3)
 𝑀𝐴 = - 13 kN-m

21.  Finding the reactions is usually the first step in the analysis of a beam.
 Once the reactions are known, the shear forces and bending moments can be found. If a beam is
supported in a statically determinate manner, all reactions can be found from free-body diagrams and
equations of equilibrium.
 Internal stress resultants: Shear forces and bending moments are the resultants of stresses distributed
over the cross section. Therefore, these quantities are known collectively as stress resultants. The stress
resultants in statically determinate beams can be calculated from equations of equilibrium.
 When designing a beam, we usually need to know how the shear forces and bending moments vary
throughout the length of the beam. Of special importance are the maximum and minimum values of
these quantities. Information of this kind is usually provided by graphs in which the shear force and
bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as
the abscissa. Such graphs are called shear-force and bending-moment diagrams.

22. Q5. Calculate the support reactions for the beam as shown in the figure