8.3 Linear Theory for Growing Beams and Plates
413
Expanding these relations and dropping products of the small terms θ and
(dN/dx)dx = dN (but not N ) yield
dV
dx
= −q + N
dθ
dx
dM
dx
= V
(8.12)
in the limit dx → 0.
Boundary Value Problem In the linear theory, a general beam problem can be
separated into stretching and bending parts. For the stretching problem, Eqs. (8.10)
give either 0 or N , depending on whether the axial force N or displacement u is
prescribed at the ends. With N known, the bending problem can then be solved.
The primary dependent variables for bending are the displacement v(x), rotation
θ(x), bending moment M(x), and shear force V (x). Using Eqs. (8.3), (8.4), (8.11) 1 ,
and (8.12) 2 , the last three quantities can be expressed in terms of v 0 = v(x) and the
growth moment M g (x), giving
v = v 0 (x)
θ = −v
M = −EI v
− M g
V = −EI v
− M
g ,
(8.13)
where prime denotes differentiation with respect to x.
Substituting Eqs. (8.13) 2,4 into (8.12) 1 yields
EI
d 4 v
dx 4 − N
d 2 v
dx 2 = q −
d 2 M g
dx 2
(8.14)
to be solved for the vertical displacement v(x). For N = M g = 0, this relation
reduces to the classical fourth-order beam equation.
Equation (8.14) requires four boundary conditions. In most cases, either v or V
and θ or M are prescribed at each end, with (8.13) then used to write the conditions
in terms of v. 6 Common boundary conditions are shown in Fig. 8.8. It is important
to be sure signs are correct when specifying nonzero boundary conditions. For
example, the negative sign in the concentrated force is needed because F points
in the opposite direction to the positive direction defined for V on the left side of
an element (see Fig. 8.5b). The last boundary condition shown in Fig. 8.8 is a mixed
6 These boundary conditions are analogous to specifying either displacement or force for a
traditional spring and either rotation angle or moment for a torsional spring.
413
Expanding these relations and dropping products of the small terms θ and
(dN/dx)dx = dN (but not N ) yield
dV
dx
= −q + N
dθ
dx
dM
dx
= V
(8.12)
in the limit dx → 0.
Boundary Value Problem In the linear theory, a general beam problem can be
separated into stretching and bending parts. For the stretching problem, Eqs. (8.10)
give either 0 or N , depending on whether the axial force N or displacement u is
prescribed at the ends. With N known, the bending problem can then be solved.
The primary dependent variables for bending are the displacement v(x), rotation
θ(x), bending moment M(x), and shear force V (x). Using Eqs. (8.3), (8.4), (8.11) 1 ,
and (8.12) 2 , the last three quantities can be expressed in terms of v 0 = v(x) and the
growth moment M g (x), giving
v = v 0 (x)
θ = −v
M = −EI v
− M g
V = −EI v
− M
g ,
(8.13)
where prime denotes differentiation with respect to x.
Substituting Eqs. (8.13) 2,4 into (8.12) 1 yields
EI
d 4 v
dx 4 − N
d 2 v
dx 2 = q −
d 2 M g
dx 2
(8.14)
to be solved for the vertical displacement v(x). For N = M g = 0, this relation
reduces to the classical fourth-order beam equation.
Equation (8.14) requires four boundary conditions. In most cases, either v or V
and θ or M are prescribed at each end, with (8.13) then used to write the conditions
in terms of v. 6 Common boundary conditions are shown in Fig. 8.8. It is important
to be sure signs are correct when specifying nonzero boundary conditions. For
example, the negative sign in the concentrated force is needed because F points
in the opposite direction to the positive direction defined for V on the left side of
an element (see Fig. 8.5b). The last boundary condition shown in Fig. 8.8 is a mixed
6 These boundary conditions are analogous to specifying either displacement or force for a
traditional spring and either rotation angle or moment for a torsional spring.