Page 40 - 4560

     . (3.10)
пр n n
After substituting formulas (3.9), (3.10) in the equation (3.8) we
obtain
M l  2M l  l  M l   6EJ      . (3.11)
n 1 n n n n 1 n 1 n 1 z n n
We will get the motion equation for a continuous beam which is
called the equation of three moments. Such equations should be
made up as many as extra nodes the beam has.
If the moment of inertia of a continuous beam varies from a span
to a span, then the equation (3.11) will take the form

l  l l  l
M n  2M n  n  n 1   M n 1   6E      , (3.12)
n 1 n 1 n n
J J J J
n  n n 1  n 1
where J and J - axial moments of inertia of the cross sections
n n 1
n and n +1 spans.
If between two finish resistances of continuous beam there is
rigid fixing, then it is artificially replaced by the span the length
of which is zero (infinite stiffness span), while retaining the
familiar form of the equation of three moments.
Sometimes the right-hand side of (3.11) is convenient to be
introduced by another form of
6 a 6 b
 6EJ        n n  n 1 n 1 ,
z n n
l l
n n 1
where  and  - diagrams’ area of bending moments
n n 1
appearing from a given external load in simple beams with a span
l and l (fig. 3.4), a and b - the distance from the centers of
n n 1 n n 1
gravity of the diagram to n   1 and n   1 resistances.
After defining the reference points M , M , M each span is
n 1 n n 1
considered as a simple beam on two supports, which are loaded
by external load and found resistant moments.
The angles of rotation supporting section  and  you can find
n n
in any way. Very convenient to take advantage the known general
formula for typical loads that are listed in the references.

40