I hope I'm asking this in the right place. I have the following problem:

enter image description here

and am to calculate A,B & C so that the system is in static equilibrium. I calculated A without too many problems.

My problem is with B & C

What I have done so far:

I realised that my placing my moment at the point ABF I can eliminate B as an unknown and solve for C.

I calculated the resultant for the diagonal part ($R_1=q_0\cdot l$) and for the vertical part ($R_2=\frac{q_0\cdot 2l}{2}=q_0\cdot l$).

I calculated the distances of $R_1$ and $R_2$ from my moment $M^{ABF}$ using the Pythagorean theory and set them both negative as they are turning my system in a mathematically negative direction.

The force C I added to my moment as $C\cdot 2l$.

My moment calculation is at this point:

$$\Sigma M^{ABF}=2l\cdot C-(\sqrt{(2l)^2+(2l)^2}+\frac{l}{2})\cdot R_1-\sqrt{(2l)^2+(\frac{2}{3}2l})^2\cdot R_2$$

My problem: I don't know what to do with the moment $M$ which is at a distance of $l$ from $M^{ABF}$. I have tried adding as it is, adding it and multiplying by $l$, but both times the answer comes out wrong (I have a numerical answer).

I'm obviously doing something wrong but I don't know what.

In case it helps to have values:
$F=4q_0l$, $M=2q_0l^2$, $\alpha=45°$
$q_0=3kN/m$, $l=2m$

  • $\begingroup$ Exactly what *distance" should you be using when you write "moment = force x distance"? In your equation for $\sum M^{ABF}$, one of the "distances" is wrong. $\endgroup$ – alephzero Dec 3 '16 at 18:18
  • $\begingroup$ Distance from $M^{ABF}$?Well, I took $R_1$ to be in the middle of the diagonal part, so at $\frac{l}{2}$. Then I added the hypotenuse of the triangle with side length $2l$ and $2l$. Then I took $R_2$ to be at a point 2/3 away from the point if the triangular force so calculated the length of that hypotenuse from the triangle with sides $2l$ and $\frac{2}{3}2l$. I've realised that the position of M doesn't matter but can't see what I've done wrong.. $\endgroup$ – Andy Grey Dec 3 '16 at 18:25
  • $\begingroup$ Does the force need to be perpendicular to the 'line of distance' from the moment? $\endgroup$ – Andy Grey Dec 3 '16 at 19:08
  • $\begingroup$ Ok. So I calculated a new angle $\beta$ with which $R_2$ meets the line of distance $\overline{M^{ABF}R_2}$ from $M^{ABF}$ and then calculated the component of $R_2$ which is perpendicular to $\overline{M^{ABF}R_2}$ . That seems to have worked, but that would have been a problem with the force, not the distance. So have I still missed something, or is there a simpler way to come to the same point? $\endgroup$ – Andy Grey Dec 3 '16 at 19:27

Bending moments due to forces and distances are found by the product of the force and the perpendicular distance of its projection to the point of study.

You have the following structure (ignore the dimensions, they are just placeholders):

enter image description here

Transforming the distributed loads into equivalent concentrated loads, you get:

enter image description here

That figure already shows you the dimensions you need to consider for each force:

  • The linear load along the $2\ell$ bar is horizontal. Therefore, the perpendicular distance to use is the vertical distance.
  • The uniform load along the $\ell$ bar is at a 135 degree angle with the horizontal. Therefore, the distance must be at a 45 degree angle.

The concentrated bending moment $M$ in the middle of the horizontal $2\ell$ bar is simply added to your sum. Such loads are always just added to the equilibrium equation.

So, we can get:

$$\begin{align} M_{ABF} = 0 &= M + 2C\ell - \dfrac{2q_0\ell}{2}\cdot\dfrac{2}{3}2\ell - q_0\ell\left(2\ell\sqrt2 + \dfrac{\ell}{2}\right) \\ &= M + 2C\ell - q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) \\ \therefore C &= \dfrac{q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) - M}{2\ell} \\ \end{align}$$

Knowing this, it is then trivial to find the value of $B$ via the equilibrium of vertical forces $\left(\sum F_y = 0\right)$ equation. This is left as an exercise for the reader.

All diagrams were done with Ftool, a free 2D frame analysis tool.

| improve this answer | |
  • $\begingroup$ Ahhh! Thank you! Of course, I can move the force along the line of force. I had totally missed that $\endgroup$ – Andy Grey Dec 4 '16 at 12:50

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