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

enter image description here

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.

enter image description here

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)

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    $\begingroup$ One side is stronger than the other - just look. $\endgroup$
    – Solar Mike
    Feb 28 at 15:53

3 Answers 3

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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.

enter image description here enter image description here

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  • $\begingroup$ For a doubly symmetric shape, i.e. cross sec is symmetric about two axes (mostly perpendicular to each other), and load acting on the centroid of it, but the direction of the load is somehwere in between the direction of these two axes. Will there be any twisting now or not? Plus, what if the cross sec is symmetric about more than 2 axes, would you still say that only doubly symmetric shapes have shear center on the centroid of the cross sec? $\endgroup$ Mar 1 at 9:54
  • $\begingroup$ @RameezUlHaq 1) "Will there be any twisting now or not?" - No twist, if the force is acting/passing through the shear center, which coincides with the centroid. 2) Yes, no matter how many axes of symmetry a cross-section has, there is only "one pair" of axes that are mutually perpendicular to each other, which defines the cross-section on the plane bounded by these two axes. $\endgroup$
    – r13
    Mar 1 at 19:50
  • $\begingroup$ Not necessarily, square shaped cross sec has 4 axes of symmetries, and has 'two pairs' which are mutually perpendicular to each other. $\endgroup$ Mar 2 at 6:27
  • $\begingroup$ @RameezUlHaq I don't think it is meaningful in arguing, or trying to reinvent, a frequently used engineering jargon. I wish you the best in your engineering journey. $\endgroup$
    – r13
    Mar 2 at 20:40
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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) :

D enter image description here
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.

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  • $\begingroup$ But the beam is actually symmetric about Z axis (or neutral axis). So symmetry about which exact axis are you talking about here? Both Z and Y axis, or a cross sec has to axis-symmetric instead? $\endgroup$ Mar 1 at 9:38
  • $\begingroup$ I am sorry, I can't understand what you're asking me. I've used a fully symmetric cross-section to demonstrate that the twisting is not only a matter of the cross-section but also the positioning of the load. $\endgroup$
    – NMech
    Mar 1 at 10:13
  • $\begingroup$ I mean I was talking about the beam illustrated in the question, which is symmetric about the Z axis. Well your example was a cicular cross sec which is axisymmetric. You stated this, the shear center exist for symmetric beams also. It just then coincides with the symmetry center., so I was just asking symmetric beam about what and how many axes. Because the figure shown in question is still symmetric about Z axis though. I guess r13 user has mentioned in his answer that the cross sec has to symmetric about two axes, and not just one. $\endgroup$ Mar 1 at 10:39
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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

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