Page 47 - 4749

N l  R l
l   x 1 1  A 1 ; (2.40)
1
E A E A
c c c c
N l ( R  ) P l
l  x 2 2  A 2 . (2.41)
2
E A E A
s s s s
6. Make synthesis. Substitute absolute elongations (2.40) and
(2.41) into the equation of compatibility of deformations (2.39)
and consider the equilibrium equation (2.38), we obtain the system
of equations:

 R l 1  R A  P l 2
A
   0;
E F c E F s (2.42)

s
c

 R  P R A  0.
B

Its solution will:
P P
R  l E A ; R  l E A .
A
B
1 1 s s 1 2 c c
l E A l E A
2 c c 1 s s

Static uncertainties are disclosed. We see that the force
distribution between the elements of static undetectable system
depends on the ratio between the cross-sectional area, and modulus
of elasticity of the lengths of these elements. The harder the
element is, the greater the load it takes over.
In the calculations we should bear in mind that in static
determined systems internal force factors can arise only from the
active and reactive external forces, and in static undetectable
systems - for other reasons, for example, temperature changes or
inaccuracies of fabrication and installation of the system elements.

Questions for self-assessment

1. What case of beam deformation is defined as a central tension or
compression?
47