Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 54

1.9.4 Surface Forces and Boundary Conditions

On the circumferential surfaces of the bar, Eq. (1.57) must be satisfied. There are no pre-
scribed surface tractions on this surface, so that p x = p y = p z = 0, and a x = 0. Since

σ 0, the second and third Eqs. (1.57a) are identically satisfied and the

first equation reduces to

= σ = σ = τ =

x y z yz

a z τ + a y τ = 0 (1.145)

xz xy

where (Fig. 1.17d) a y = dz / ds and a z =− dy / ds , so that

dy
ds τ

dz
ds τ

− + = 0

xz xy

Next consider the conditions on the ends of the bar. The normals to the cross-sections on
the ends are parallel to the x axis. Hence a x =± 1, a y = a z = 0 and the boundary conditions

as represented by Eqs. (1.57) are

p y =± τ xy , p z =± τ xz (1.146)

sign applies to the end of the bar with an external normal in the direction of the
positive x axis. Equation (1.146) indicates that the (shear) forces have the same distribution
as the shear stresses on the ends. It can be shown that the resultant of these stresses at the
end of a bar is a torque and the resultant forces vanish. With the notation of Fig. 1.18, the
resultant forces on an end of area A are
A τ xy dy dz,

Here the +

A τ xz dy dz

(1.147)

in the y and z directions, respectively. Since these quantities can be shown to be zero
(Problem 1.61), the resultant force acting on the end of the bar is zero. However, the resultant
moment is not zero; from moment equilibrium requirements on a cross-section (Fig. 1.18)

A (τ xz y

M t = − τ xy z ) dy dz (1.148)

are related by
Eq. (8) of Example 1.4, which was shown to apply if the cross-section of the bar is circular.
That is

Suppose the twisting moment M t and the angle of twist per unit length φ

φ M t = GJ (1.149)

It follows from Eqs. (1.148) and (1.149) that J can be defined as

A (τ xz y

M t
G

1
G

J =

φ =

− τ xy z ) dy dz (1.150)

φ

The quantity J is known as the torsional constant and GJ is the torsional stiffness of the bar.
For bars of circular cross section, J of Eq. (1.150) reduces to the polar moment of inertia.

1.9.5 Displacement Form of the Governing Differential Equations

As outlined in Section 1.6.1, the displacement form of the governing equations is obtained
by first forming stress-displacement relations by combining Eqs. (1.142) and (1.143), then
placing the result

φ z

,

φ + ∂ω

∂

+ ∂ω
∂

τ xy =− G τ xz =− G − y

y z