Beams - Mechanics of Structures: Variational and Computational Methods - page 735

FIGURE 12.7
Coordinate systems.

This point is the center of twist or the shear center. The shear center is the point on the cross-
section through which the plane of the resultant of the applied loads passes such that no
twisting mome nt s a re developed. From I y ∗ ω ∗ = 0 and I z ∗ ω ∗ = 0 , the distance of the shear

center from the y, z principal axes is

I z ω I y y − I y ω I y z

y S = in the y direction

−

I y z

I y y I z z

(12.20b)

z S = −

I y ω I z z + I z ω I y z

in the z direction

I y y I z z − I y z

If the y ∗ , z ∗ (Fig. 12.7), which corresponds to the

principal axes for the centroid, the distance of the shear center from the

axes are rotated through an angle α

y, z axes is

I z ω
I zz

I y ω
I y y .

y S = z S =− (12.21)

Shear centers will be treated later in this chapter.
If the

y ∗ , z ∗ with the origin at the shear ω coordinate, corresponding to the principal axes

ω ∗ = ω +

z S

−

y S center, is taken as y z, then

A ω ∗ 2 dA

2 dA

I ω ∗ ω ∗ = =

A (ω +

z S y − y S z )

Substitution of Eqs. (12.20a) and (12.21) into this expression leads to

I
y

I
z

ω
I y y −

ω
I zz

I ω ∗ ω ∗ = I ω ω − (12.22a)

Since I y y > 0 and I zz > 0 , it can be seen from Eq. (12.22a) that I ω ∗ ω ∗ is a minimum for all

choices of the ω coordinate. Rewrite Eq. (12.22a) as

I ω ∗ ω ∗ = I ω ω + z S I y ω −

y S I z ω (12.22b)