The bending of beams with thin tension flanges Page: 4 of 17
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. A..C.A.. Technical Memorandum Np.. 769
,Subst itut ing-. these terms in the two equilibrium e.quaa-
cox 7 37o
x 4 xz - O zz + z = 0
- + SA Z x . S=
Subtracting the first of these mu.ltiplie:d by sin a, then
the second, multiplied by cos Ca, gives;
cos a + h - sin . 0
and denoting with d an element of the line parallel
with the direction of a, defined, that is, from.
d~ dX a dz
we obtain the equation = 0, which proves our state-
ment. Now assume edge AB of the sheet to be rigidly
fixed. If d and' f are free edges parallel to axes z
and x, we have at d:
S X = Tz = 0 C cos2 C = a sin a cos a = 0
and at f:
Txz = = 0 sin a cos a = 0 sin2 a = 0
If we should exclude the possibility of cos a = 0
along d, and of sin a = 0 at f, the result is a = 0
over the contour. Seeing that the path of the tensions
must be straight, it follows that the field outside of AD
and BC must be unstressed. This condition exists even
if f, instead of being the free contour line, delimits
the symmetrical field.
With the proviso that the field ABCD may enter in
tension, we introduce the assumption that the formation of
the stress trajectories inside this field is so constitut-
ed by the tension lines which emanate from the beam as to
form with it an angle which is constant over its whole
length. This inclination will be discussed by means of
the equation of least work.
The introduced hypothesis is equivalent to assuming
a series of infinite parallel cuts in the sheet which re-
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Cicala, Placido. The bending of beams with thin tension flanges, report, March 1, 1935; (digital.library.unt.edu/ark:/67531/metadc63454/m1/4/: accessed June 18, 2018), University of North Texas Libraries, Digital Library, digital.library.unt.edu; crediting UNT Libraries Government Documents Department.