Page 324 - Aircraft Stuctures for Engineering Student

9.4 Shear of closed section beams 305

Fig. 9.25 Shear centre of a closed section beam.
at this stage, it is impossible to equate internal and external moments to produce an
equation similar to Eq. (9.37) as the position of S,, is unknown. We therefore use the
condition that a shear load acting through the shear centre of a section produces zero
twist. It follows that dO/dz in Eq. (9.42) is zero so that

or

which gives

(9.47)

If Gt = constant then Eq. (9.47) simplifies to

(9.48)

The coordinate qs is found in a similar manner by applying S, through S.

Example 9.6
A thin-walled closed section beam has the singly symmetrical cross-section shown in
Fig. 9.26. Each wall of the section is flat and has the same thickness t and shear
modulus G. Calculate the distance of the shear centre from point 4.

The shear centre clearly lies on the horizontal axis of symmetry so that it is only
necessary to apply a shear load Sy through S and to determine &. If we take the x
reference axis to coincide with the axis of symmetry then lT,. = 0, and since S, = 0