809/2004 DE LA COMISIÓN DE 209 DE ABRIL DE 2004)
MOROSIDAD Y COBERTURA INDIVIDUAL
3.7 Administrador, agente de cálculo o equivalente
It is easy to show that the influence coefficient a12 in Eq. (5.8) is equal to the influence coefficient a21. In general, aij = aji. To show this, consider a force F1
applied at point 1, and let d1 be the corresponding displacement. The energy
stored is
U1 = 12F1d1 = 12a11F12
since d1 = a11F1
Next, apply force F2 at point 2. The corresponding deflection at point 2 is a22F2
and that at point 1 is a12F2. During this displacement, force F1 is fully acting and
hence, the additional energy stored is U2= 12F2(a22F2) + F1(a12F2) The total elastic energy stored is therefore
U = U1 + U2 = 12a11F12 + 12a22F22 + a12 F1F2
Now, if F2 is applied before F1, the elastic energy stored is
U¢ = 12a22F22 + 12a11F12 + a21F1F2
Since the elastic energy stored is independent of the order of application of F1
and F2, U and U¢ must be equal. Consequently,
a12= a21 (5.11a)
or in general
148 Advanced Mechanics of Solids
The result expressed in Eq. (5.11b) has great importance in the mechanics of solids, as shown in the next section.
One can obtain an expression for the elastic strain energy in terms of the applied forces, using the above reciprocal relationship. From Eq. (5.10)
U = 12(F1d1 + F2d2 + . . . + Fndn) = 12F1(a11F1 + a12F2 + . . . + a1nFn) + . . . 12 Fn(an1F1 + an2F2 + . . . + annFn) U = 21(a11F12 + a22F22 + . . . + annFn2) + 12(a12F1F2 + a13F1F3 + . . . + a1nF1Fn + . . .) = 12S(a11F12) + S(a12F1F2) (5.12) 5.6 MAXWELL–BETTI–RAYLEIGH RECIPROCAL THEOREM
Consider two systems of forces F1, F2, . . ., and F ′1, F ′2, . . ., both systems having
the same points of application and the same directions. Let d1, d2, . . ., be the
corresponding displacements caused by F1, F2, . . ., and δ1′, δ2′, . . . , the corre-
sponding displacements caused by F ′1, F ′2, . . ., Then, making use of the recipro-
cal relation given by Eq. (5.11) we have
1 F ′d1 + F ′2d2 + . . . + F ′ndn = F ′1(a11F1 + a12F2 + . . . + a1nFn) +F ′2(a21F1 + a22F2 + . . . + a2nFn) + . . . + Fn′(an1F1 + an2F2 + . . . + annFn) = a11F1 F ′1 + a22F2 F ′2 + annFn Fn′ + a12(F ′1F2 + F ′2F1) + a13(F ′1F3 + F ′3F1) + . . . + a1n(F ′1Fn + F ′nF1) (5.13)
The symmetry of the expressions between the primed and unprimed quantities in the above expression shows that it is equal to
F1δ1′ + F2δ2′ + . . . + Fnδn′
i.e. F1 1δ′+F2 2δ′+. . .=F1 1δ′ +F2 2′δ + . . . (5.14) In words:
‘The forces of the first system (F1, F2, . . ., etc.) acting through the corresponding
Energy Methods 149 amount of work as that done by the second system of forces acting through the corresponding displacements produced by the first system of forces’.
This is the reciprocal theorem of Maxwell, Betti and Rayleigh. 5.7 GENERALISED FORCES AND DISPLACEMENTS
In the above discussions, F1 F2, . . . , etc. represented concentrated forces and
d1, d2, . . . , etc. the corresponding linear displacements. It is possible to extend
the term 'force' to include not only a concentrated force but also a bending moment or a torque. Similarly, the term 'displacement' may mean linear or angular displacement. Consider, for example, the elastic body shown in Fig. 5.3, sub- jected to a concentrated force F1 at point 1 and a couple F2 = M at point 2. d1
will now stand for the corresponding linear displacement of point 1 and d2 for
the corresponding angular rotation of point 2. If F1 is a unit force acting alone,
then a11, the influence coefficient, gives the linear displacement of point 1 cor-
responding to the direction of F1. Similarly, a12 stands for the corresponding
linear displacement of point 1 caused by a unit couple F2 applied at point 2. a21
gives the corresponding angular rotation of point 2 caused by a unit concen- trated force F1 at point 1.
The reciprocal relation a12 = a21 can also be
interpreted appropriately. For example, making reference to Fig. 5.3, the above relation reveals that the linear displacement at point 1 in the direction of F1 caused by a unit couple acting
alone at point 2, is equal to the angular rotation at point 2 in the direction of the moment F2
caused by a unit load acting alone at point 1. This fact will be demonstrated in the next few examples.
With the above generalised definitions for forces and displacements, the work done when the forces are gradually increased from zero to their full magnitudes is given by
W = 12
(
F1 1δ +F2 2δ +. . .+Fn nδ)
The reciprocal theorem of Maxwell, Betti and Rayleigh can also be given wider meaning with these extended definitions.
1
2 F1
Fig. 5.3 Generalised forces and displacements
Example 5.1 Consider a cantilever loaded by unit concentrated forces, as shown
a21 2 2 a12 1 F2 = 1 2/3 L F1 = 1 ( b ) (a) Fig. 5.4 Example 5.1
150 Advanced Mechanics of Solids
Solution In Fig. 5.4(a), the unit load F1 acts at point 1. As a result, the deflec-
tion of point 2 is a21. In Fig. 5.4(b) the unit load F2 acts at point 2 and as a result,
the deflection of point 1 is a12. The reciprocal relation conveys that these two
deflections are equal. If L is the length of the cantilever and if point 1 is at a distance of 2
3L from the fixed end, we have from elementary strength of materials d2 due to F1= deflection at 1 due to F1 + deflection due to slope
8 3 4 3
81LEI 54LEI
= +
d1 due to F2 = deflection at 1 due to a unit load at 1 + deflection at 1
due to a moment (L/3) at 1 3 3 8 4 81 54 L L EI EI = +
Example 5.2 Consider a cantilever beam subjected to a concentrated force F at point 1 (Fig 5.5). Let us determine the curve of deflection for the beam.
Fig. 5.5 Example 5.2
F
4 3 2
1
P = F2 at point 2. Point 2 is 2/3 L from the fixed end. Verify the reciprocal theorem.
d2 = Ma21 2 M = F1 1 P 2 1 q1 = Pa12 ( b ) (a) Fig. 5.6 Example 5.3 2/3 L
Solution One obvious method would be to use a travelling microscope and
take readings at points 2, 3, 4, etc. These readings would be very small and consequently, errors would creep in. On the other hand, the reciprocal relation can be used to obtain this curve of deflection more accurately. The deflection at 2 due to F at 1 is the same as the deflection at 1 due to F at 2, i.e. a21 = a12.
Similarly, the deflection at 3 due to F at 1 is the same as the deflection at 1 due to F at 3, i.e. a31 = a13. Hence, one observes the deflections at 1 as F is moved along
the beam to get the required information.
Example 5.3 The cantilever beam shown in Fig. 5.6(a) is subjected to a bending moment M = F1 at point 1, and in Fig. 5.6(b), it is subjected to a concentrated load
Energy Methods 151
Solution From elementary strength of materials the deflection at point 2 due to
the moment M at point 1 is d2=
( )
2 2 2 1 2 3 2 9ML M L EI = EIThe slope (angular displacement) at point 1 due to the concentrated force P at point 2 is q1=
( )
2 2 2 1 2 3 2 9PL P L EI = EIHence, the work done by M through the displacement (angular displacement) produced by P is equal to Mq1= 2 2 9 MPL EI
This is equal to the work done by P acting through the displacement produced by the moment M.
Example 5.4 Determine the change in volume of an elastic body subjected to two equal and opposite forces, as shown. The distance between the points of application is h and the elastic constants for the material are E and n, (Fig. 5.7).
Solution This is a very general
problem, the solution of which is ap- parently difficult. However, we can get a solution very easily by apply- ing the reciprocal theorem. Let the elastic body be subjected to a hydro- static pressure of value s. Every vol- ume element will be in a state of hydrostatic (isotropic) stress. Conse- quently, the unit contraction in any direction from Fig. 5.7(b) is
Fig. 5.7 Example 5.4 ( b ) B P s s s A (a) P e = 2 (1 2 ) E E E σ − νσ = − ν σ
The two points of application A and B, therefore, move towards each other by a distance. Dh = h(1 2 ) E σ ν −
Now we have two systems of forces:
System 1 Force P
Volume change DV
152 Advanced Mechanics of Solids
System 2 Force s
Distance change Dh From the reciprocal theorem
P Dh = s DV or DV = σP h∆ (1 2 ) Ph E ν = −
If n is equal to 0.5, the change in volume is zero.