This is the question

I have trouble solving part A of the question. I thought the resulting torque would just be 10k * L of the beam.


Your suggested solution indicates that you seem to be confusing torque with bending moment. This is understandable since they are often used interchangeably in common parlance.

However, they mean very different things in engineering. To understand the difference, it's important to clarify the standard notation for a beam's local axes:

  • The x-axis is the beam's longitudinal axis. That is, the axis which describes the beam's length.
  • The y- and z-axes are the beam's transversal axes, describing it's width and height.

So, given that, we can then state that:

  • Bending moment is the internal stress state along a beam due to rotations around the two transversal (y and z) axes. When you think of a beam or cantilever deflecting downwards due to a load, that beam is under bending moment.
  • Torque is the internal stress state along a beam due to rotations around its longitudinal (x) axis. When you think of a screw being twisted, a torque has been applied to it, causing torsion.

So, the calculation you expected of $\text{Force}\times\text{Length}$ would give you the bending moment of the beam, not the torque.

However, the general gist of your solution is correct: both bending moment and torque are equal to $\text{Force}\times\text{Distance}$. However, to calculate the torque you need to use another distance.

The question states that the force is "acting on the outer edge of the cross-section in the direction of the symmetry axis." From this we can conclude where on the section the force is applied and in which direction it's pointing:

  • I personally find "outer edge" to be unclear, but it's fair to assume they mean the left- or right-most point of the section (one of the top corners).
  • Likewise, "in the direction of the symmetry axis" is also unclear since triangles have three axes of symmetry, but it's fair to assume they mean the vertical one, which therefore means the force is vertical.

So now we just need to calculate the torque itself. The force is vertical and on the left- or right-most edge of the section. Since the force is vertical, the torque is calculated by multiplying it by a horizontal distance. The relevant distance in this case is to the beam's centroid, which runs along that vertical axis of symmetry. Therefore the distance is equal to half the section's side, so $120/2=60\text{ mm}$. Therefore

$$T = F\times d = 10 \times 60 = \pm600\text{ kNmm}$$

I had to put that $\pm$ sign because the problem as stated doesn't give us enough information to know whether the resultant torque is positive or negative. That depends on where the load is applied (left or right-most corner) and the force's direction (positive or negative, though I suppose one could argue it should be in the positive "direction of the symmetry axis"... which itself is unclear since we can't necessarily assume the positive direction is upwards).


In a thin walled open tube of uniform thickness J is;

$J = \frac{1 } {3 }\times Ut^3 $

t = thickness of the wall of tube.

U is is the length of the median boundary (perimeter of median cross section)

So the rotation is $ \theta = \frac {TL} {GJ} $

$ T = 10kN \times x,\space x \space is\space 1/3 \space of\space triangle\space height. $

$ Height of triangle =3x=60 \times \sqrt 3= 103.92mm \space x = 34.64mm $

$T = 10kN\times 34.64mm $

Torque = force times distance to center of rotation.

So this is the answer to first part (a).

  • $\begingroup$ Why is $x=h/3$? The force is vertical ("acting in the direction of the axis of symmetry"), so the relevant distance to the centroid is in the horizontal direction, which would make $x=h/2$ (horizontal distance from edge to centroid). $\endgroup$
    – Wasabi
    Oct 27 '18 at 17:54
  • $\begingroup$ @Wasabi, the equilateral triangle has three axis of symmetries, perpendiculars from one of the corners to the middle of the base across from it. May be you are thinking of the vertical line through Geometrical Center of triangle but that is not an axis of symmetry. $\endgroup$
    – kamran
    Oct 27 '18 at 18:19
  • $\begingroup$ Sure it is. It's not the only one, sure, but it certainly is an axis of symmetry. $\endgroup$
    – Wasabi
    Oct 27 '18 at 18:20
  • $\begingroup$ no, it is not. when I get back from work I explain. I $\endgroup$
    – kamran
    Oct 27 '18 at 19:28

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