How exactly the beam is twisting?

Question

Consider an I section beam shown below, which is unsymmetric about the x-y plane.

I apply a load P, at the centroid of this section. The textbook I follow, says that the member will twist apart from bending. I don't understand exactly how did we conclude it is twisting?

To better understand my point consider that I cut the beam through a section to expose its intermediate c/s as shown in (a). If twisting is to occur there must be forces developed in the cross section such that they form an internal resistive twisting moment about some axis. This internal twisting moment, should be equal and opposite to the twisting moment produced by P, about the same axis.

You know like how we get an internal resistive torque in a shaft and the applied torque is opposite to it (as shown in b)

Answer 1

Under gravity load, a cantilever beam will deflect but not twist only if the load is acting through the shear center of the cross-section. For doubly symmetrical shapes, the shear center coincides with the geometric center (centroid), but for the I beam with unequal flanges, there is an offset distance between the shear center and geometric center, thus, there exists a twisting moment. The direction of rotation depends on the relative position of the shear center and the centroid along the axis.

Answer 2

The beam is twisting because the force P produces a twisting moment.

That would occur in symmetric and unsymmetric beam crosssections.

Take for example the following symmetric cross-sections of a beam (assuming the load is P, the radius is R , and the length of the beam is L) :

Bending moment	$P\cdot L$	$P\cdot L$
Twisting moment	$0 $	$P\cdot R$

Since there is a twisting moment on the beam the beam will twist.

The idea of the shear center is really useful when working with concentrated loads because then you can calculate precisely the twisting moment. Usually, the twisting moment is given, and the shear center is usually used to determine the twist rotation.

Finally, the shear center exist for symmetric beams also. It just then coincides with the symmetry center.

Answer 3

An approximate explanation for why this particular section will both twist and turn is as follows.

Let's call the big flang F1, with thickness t1, and the second flange F2, t2 with corresponding I1 and I2 as the 2nd area moments and the depth of the beam in Z direction D.

This beam is composed of two unsymmetrical flanges and a web.

If we assume the contribution of the web is neutral, which is a reasonable assumption, the center of stiffness of the beam in the Y direction measured from the face of F1 flange is,

$$\bar Dz= \left( \frac{\frac{t1}{2}I_1+(D-\frac{t2}{2})I_2}{I_1+I_2} \right)$$

This is obviously not the same as the centroid of the beam because it's biased toward the big flange due to "I" being related to the square of the height of the flange.

So we have an offset loading which will cause both bending moment and twist