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Demonstration of relations between tensions and charges
N. A. Clin*
Abstract
This is a strict demonstration, able to constitute a model, not
only in
Strength of Materials or Mechanics of continuous medium, but
also in all fields
of maths application, anywhere it was used infinitely minutes,
infinitesimal
elements and unspecified approximations in final results.
Lets consider for simplicity a right, rigid bar, being in
equilibrium under
the charges of both concentrated forces ,...,,,21 n
FFF and unit loading linear
),(xp perpendiculars on the axis bar, .xO
Calculation hypotesis: unit loading linear ( distributed load )
function, ),(xp
is continuous on ],[ ba interval, on which there are not
concentrated forces.
Lets cut the bar by two planes which are perpendicular to its
axis, in points
0x and
1x belonging to ),( ba interval, and detach the portion of the
bar
comprised by this planes.
1) Lets consider the point1
x is at the right of .0
x
According to the mechanics of continuous medium axioms, of the
impulse,
namely the angular momentum, for equilibrium,
*Ilfov, Stefanestii de Jos, Ghioceilor 21, 077175, Romania.
Email:[email protected]@clicknet.ro
1
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I) + )()(01
xTxT ,0)(1
0
=x
x
dxxp
II) )()(01
xMxM ,0)()()()(1010
1
0
= xTxxdxxpxxx
x
whwre )(xT is cross ( shear ) force function and )(xM is moment
of deflection
function.
( in (II) momentums relating to fixed point0
x ).
1) To divide (I) by ,01xx
0101
01
1
0
)()()(
xx
dxxp
xx
xTxT
x
x
+
.0=
As )(xp is continuous on ],,[10xx according to theorem of mean,
there is
from the ),,(10xx so that
=1
0
).()()(01
x
x
xxpdxxp
Coming back to the result of the previous equation, we have
01
01 )()(
xx
xTxT
).(p=
Limit at right of this equality in point0
x is
01
limxx
01
01 )()(xx
xTxT ).(lim
01
pxx
=
The left member is forward derivate of )(xT function in point
).(,00xTx
r
2
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When ,)(,0101
xxxx and ;)(01xx because )(xp is continuous in
,0
x we have ),()(lim 0101
xpxpxx
=
therefore of according to the theorem limit of
compound functions, ).()(lim0
01
xppxx
=
Result
).()( 00 xpxTr =
If we are considering point1
x to the left of ,0
x analogously we get
).()(00
xpxTl
=
As ),()()(000
xpxTxTlr
== cross force function, ),(xT is derivable in0
x
and for each0
x from ),( ba
).()(00xpxT =
2) Because )(xp is continuous on ],[10xx and
01xx is positive, according
to theorem of mean, there is from ),,(10
xx so that
==1
0
1
0
)()()()()(00
x
x
x
x
pdxxxpdxxpxx .2
)( 201 xx
Dividing and (II) with01xx and effecting limit at right in point
,
0x we get
)(lim)( 010 pxM xxr = 01limxx 201xx
).(lim 101 xTxx +
Analogous, ).()(lim0
01
xppxx
=
As )(xT is derivable on ),,( ba she it is continuous on this
interval, therefore
).()(lim01
01
xTxTxx
=
Result
).()(00xTxM
r=
If we are considering point 1x to the left of ,0x we get
analogously,
),()(00
xTxMl
=
therefore moment of deflection function, ),(xM is derivable
in0
x and for each
0x from ),( ba
).()(00xTxM =
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Unlike the above demonstration, in the popular
demonstrations:
* is considered an infinitesimal element of bar of length dx
;
* in their drawing, in left point of infinitesimal element of
bar (correspond at
us with0
x ), they consider T and M and in right point of
infinitesimal
element of bar (correspond at us with1
x ), instead of )( 1xT they consider
dxdx
dTT+ or dTT+ , and instead of )(
1xM they consider dx
dx
dMM+ or
dMM + ;* )(xp is supposed constant on dx ;
* equilibrium equations are written;
* infinitely minutes of higher order are neglected;
* known results are obtained.
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