Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 40

Displacements in the positive coordinate directions are taken to be positive. Rotations
(slopes) are positive if their corresponding vectors, according to the right-hand rule, lie in
a positive coordinate direction. The shear strain γ

xz takes the form

γ xz = ∂

u

z + ∂w x = θ + ∂w x = γ

(1.99)

∂ ∂ ∂

In order for the cross-sections to remain planar, it is necessary that this shear strain be zero.
This corresponds to neglecting the shear deformation effects. Then

d 2

dw
dx

w
dx 2

θ =− or κ =− (1.100)

This is the strain-displacement relation for bending. The component is the displacement

of the beam axis, i.e., of the centerline of the beam. This displacement is referred to as the
deflection of the beam.
The axial displacement of the centerline of the beam is u 0 . The corresponding strain is
0 x =

w

dx . This strain-displacement relation can be combined with that for bending in
the matrix form

du 0 /

0 x
κ

d x u 0

w

0

= (1.101)

d x

0 −

= D u u

where the subscript u has been added so that it is clear that the differential operator D
belongs to the strain-displacement relations.
If shear deformation effects over the cross-section are retained, then γ = 0 and the strain-

displacement relations can be written as

=

d x 00
0 d x 1
00 d x

u 0
w
θ

0 x
γ
κ

(1.102)

= D u u

Engineering beam theory equations will be derived here from the theory of elasticity
without imposing the above deformation assumptions. For strength of material theories
of beams, plates, and shells, the bending due to transverse applied loads is usually consid-
ered to be governed primarily by the extension and contraction of longitudinal “fibers.”
Strains not associated with this extension and contraction may generally be neglected.
Thus, transverse normal and shear strains are set equal to zero. This is equivalent to
assuming that the beam material reacts rigidly when subjected to transverse normal and
shear strains. For our beam, then, the strains z and xz are considered to vanish. Thus,

z = ∂w
∂

γ xz = ∂

u

z + ∂w

z =

0 ,

x =

0 (1.103)

∂ ∂

As mentioned previously, γ xz = 0 corresponds to ignoring shear deformation effects.

Integration of Eq. (1.103) gives

z d

dx

w = w 0 ( x ) , u = u 0 ( x ) − (1.104)

where u 0 and w 0 are the horizontal and vertical displacement components along the x axis

of the beam. As noted in the previous derivation, these components apply for a beam
with a cross-section that remains planar during bending and simply rotates about the
y axis.