simply supported beams subjected to concentrated and udls

1. ANKIT SOOD
CHRISTINA MAYINI
DHEER VEER VIKRAM SINGH
HITESH SOOD
SIDDHARTH MAHAJAN
VISHAL MISHRA

2. The purpose of this calculation is to obtain information about shear, bending
moment, and deflection distribution over the length of a beam, which is under
various transverse loads: couples, concentrated and linearly distributed loads.
The result of calculation is represented by shear force, bending moment and
deflection diagrams.
The material of the beam is linear-elastic and
isotropic with elasticity modulus E.
All loads are lateral (forces or moments have
their vectors perpendicular to the beam axis) and
acting at the same plane. All deflections occur in
this plane of bending.
Deflections are small compared to the length of
the beam. In this case we apply equilibrium
equations to the unreformed beam axis (or its
parts) and assume that the curvature of the
deformed beam axis is equal to the second
derivative of the deflection function.

3. RELATION BETWEEN LOADING, S.F, B.M, SLOPE AND DEFLECTION

4. Consider a simply supported uniform section
beam with a single load F at the centre.
The beam will be deflect
symmetrically about the
centre line with zero slope
(dy/dx) at the centre
line. It is convenient to
select the origin at the
centre line.

5. Consider a simply supported uniform section beam with a Concentrated Load and UDL
The B.M Equation is:
w
W1
b c
3
b c a
w b
6
c
l
b c
+
Note that Macaulay terms are integrated with respect to, for example, (x -a) and
they must be ignored when negative. Substitution of end conditions will then yield the
values of the constants A and B in the normal way and hence the required
values of slope or deflection.

6. Simply supported beams when subjected to multiple loadings
yields graphs of these nature