Elasticity_ Barber

96 7 Body forces
The astute reader will notice that the case of gravitational loading can be
recovered as a special case of these results by writing a 0y = g and setting all
the other terms to zero. In fact, a reasonable interpretation of the gravitational
force is as a D’Alembert force consequent on resisting the gravitational
acceleration. Notice that if a body is in free fall — i.e. if it is accelerating
freely in a gravitational field and not rotating — there is no body force and
hence no internal stress unless the boundaries are loaded.
Substitution of (7.15, 7.16) into (7.6) shows that the inertia forces due
to rigid-body accelerations are conservative if and only if ˙Ω = 0 — i.e. if
the angular velocity is constant. We shall determine the body force potential
for this special case. Methods of treating the problem with non-zero angular
acceleration are discussed in §7.4 below.
From equations (7.4, 7.15, 7.16) with ˙Ω =0 we have
∂V
∂x = ρ(a 0x − Ω 2 x) (7.17)
∂V
∂y = ρ(a 0y − Ω 2 y) , (7.18)
and hence, on partial integration of (7.17)
V = ρ
(a 0x x − 1 )
2 Ω2 x 2 + h(y) , (7.19)
where h(y) is an arbitrary function of y only. Substituting this result into
(7.18) we obtain the ordinary differential equation
dh
dy = ρ(a 0y − Ω 2 y) (7.20)
for h(y), which has the general solution
h(y) = ρ
(a 0y y − Ω2 y 2 )
+ C , (7.21)
2
where C is an arbitrary constant which can be taken to be zero without loss
of generality, since we are only seeking a particular potential function V .
The final expression for V is therefore
(
V = ρ a 0x x + a 0y y − 1 )
2 Ω2 (x 2 + y 2 ) . (7.22)
The reader might like to try this procedure on a set of body forces which do
not satisfy the condition (7.6). It will be found that the right-hand side of the
ordinary differential equation like (7.20) then contains terms which depend
on x and hence this equation cannot be solved for h(y).