1.  Topic
:
Solving Statically Indeterminate
Structure : Moment-Area Theorem .
Ikramul Ahasan Bappy
ID:10.01.03.131

2. 
Definition:
A member of any type is classified statically
indeterminate if the number of unknown reactions
exceeds the available number of equilibrium equations,
e.g. a continuous beam having 4 supports

3.  Assumptions:
Contributions for the
theorem .

Beam is initially straight,

Is elastically deformed by
the loads, such that the
slope and deflection of the
elastic curve are very
small, and

Deformations are caused
by bending.
Otto Mohr.(1868)
Charles E. Greene.(1873)

4. 
Moment-curvature relationship:
 Sign convention:

5. B
BA
A
M
dx
EI

6. *12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Theorem 2
The vertical deviation of the tangent at a pt (A) on the elastic curve
w.r.t. the tangent extended from another pt (B) equals the moment
of the area
under the ME/I diagram between these two pts .

This moment is computed about pt (A) where the vertical deviation 6
(tA/B) is to be determined.


7. M/EI Diagram

Determine the
support reactions
and draw the
beam’s M/EI
diagram.

If the beam is
loaded with
concentrated
forces, the M/EI
diagram will consist
of a series of
straight line
segments.

8.  If
the loading
consists
of a series of
distributed loads,
the M/EI diagram
will consist
of parabolic or
perhaps
higher-order curves.

9. 


An exaggerated view of the beam’s elastic
curve.
Pts of zero slope and zero displacement
always occur at a fixed support, and zero
displacement occurs at all pin and roller
supports.
When the beam is subjected to a +ve
moment, the beam bends concave
up, whereas
-ve moment gives the reverse .

10. 


An inflection pt or change in curvature
occurs when the moment if the beam (or
M/EI) is zero.
Since moment-area theorems apply only
between two tangents, attention should be
given as to which tangents should be
constructed so that the angles or
deviations between them will lead to the
solution of the problem.
The tangents at the supports should be
considered, since the beam usually has
zero displacement and/or zero slope at the
supports.

11. # Since axial load neglected, a
there is a vertical force and moment
at A and B. Since only two eqns of
equilibrium are available, problem is
indeterminate to the second
degree.
# Let By and MB are
redundant, #By principle of
superposition, beam is
represented as a
cantilever.
# Loaded separately by
distributed load and
reactions By and MB, as
shown.

12. 0
B
'B
' 'B
1
0
B
'B
' 'B
2
wL3
48EI
B
7 wL4
384EI
B
42
EI
9 kN/m 4 m
48EI
3
7 9 kN/m 4 m
384EI
12
EI
4

13. 2
3
'B
'B
' 'B
PL
2 EI
By 4 m
By 4 m
8By
2 EI
PL
3EI
ML
EI
2
' 'B
2
ML
2 EI
EI
3
21.33B y
3EI
MB 4 m
EI
MB 4 m
2 EI
EI
4M B
EI
2
8M B
EI

14. Substituting these values into Eqns (1) and
(2) and canceling out the common factor
EI, we have
0 12 8B y
0
4M B
42 21.33B y
8M B
Solving simultaneously, we get
By
3.375 kN
MB
3.75 kN m

16. Thank you
for your
attention