Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Therefore, the total strain energy in the beam is
775
CASTIgLIANO’S SECONd
THEOREM
1 1
1
U + dU = P1D 1 + P2D 2 + Pd 1 D 1 + Pd 2 D 2 + dPd 1 D 1
(b)
2 2
2
If the order of loading is reversed so that the incremental force dP 1 is applied first,
followed by P 1 and P 2 , the resulting strain energy will be
1 1
1
U + dU = P1D 1 + P2D 2 + dP1D 1 + dPd 1 D 1
(c)
2 2
2
Note that, since the beam is linearly elastic, the loads P 1 and P 2 cause the same deflections
D 1 and D 2 regardless of whether or not any other load is acting on the beam. Because
dP 1 remains constant at its point of application during the additional deflection D 1 , the term
dP 1 D 1 does not contain the factor ½.
Since elastic deformation is reversible and energy losses are neglected, the resulting
strain energy must be independent of the order of loading. Hence, by equating Equations
(b) and (c), we obtain
dP1D 1 = Pd 1 D 1 + Pd 2 D 2
(d)
Equations (a) and (d) can then be combined to give
1
dU = dP1D 1 + dP1d
D 1
(e)
2
The term
1
dP dD
2
1 1
is a second-order differential that may be neglected. Furthermore, the strain energy U is a
function of both P 1 and P 2 ; therefore, the change in strain energy, dU, due to the incremental
load dP 1 is expressed by the partial derivative of U with respect to P 1 as
Equation (e) can then be written as
which can be simplified to
U
dU = ∂ P dP 1
∂ 1
∂U
∂P dP
1
= dP D
1 1 1
∂U
=D (f)
∂
P 1
1