I received a tricky bonus question today in my engineering class:

Bonus Question

I've begun by equating all moments and forces to zero, about an arbitrary point, which I labeled at $C$. Finally, I attempted to solve for $C_x$ and $C_y$ without examining $M_c$. However, my answers for $C_x$ and $C_y$ were incorrect, and I'm unsure as to the process for solving $M_c$.

How do I solve for the reactions at the base of a T-frame that is subject to various point loads, including a fixed tether at one end of the top bar?

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    $\begingroup$ I've edited your question with the details you provided in your comment, but there is one thing about the image that I think still needs a little bit of explicit clarification: Is this a fixed frame, or are the vertical and horizontal members attached with a pin connection? Please edit your question to make this clear. $\endgroup$
    – Air
    Mar 25 '15 at 19:52
  • $\begingroup$ @Air - The vertical and horizontal members must be connected to each other with a fixed (i.e. moment) connection, or equilibrium is not satisfied for moments about the connection (based on P and Tbd). Also, I'd never assume a pinned connection unless it was specifically stated. I am therefore satisfied that the question is clear; voting to reopen. $\endgroup$
    – AndyT
    Mar 26 '15 at 13:00
  • $\begingroup$ @AndyT Never mind - I was looking at the FBD and not the known values above it. $\endgroup$
    – Air
    Mar 26 '15 at 14:34

The reactions at $C$ can be determined by resolving vertically, horizontally, and for moments. For equilibrium the total vertical force, horizontal force and moment must be zero.

The (slight) complication in this question is the inclined force $T_{BD}$. This needs to be resolved into a vertical and a horizontal force. You can use trigonometry to work out the angle of the force to the vertical (I will call it $\theta$). The vertical component is then $T_{BD} \times \cos (\theta)$ and the horizontal component is $T_{BD} \times \sin (\theta)$.

Once $T_{BD}$ has been resolved into components, summing vertical forces, horizontal forces should be relatively easy.

For moments: the horizontal component of $T_{BD}$ will be acting clockwise about $C$ at a lever arm of 600 mm; the vertical component will be acting clockwise about $C$ with a lever arm of 150 mm.

Alternative method for moments:

I find this way more complicated, but it is valid. Instead of resolving Tbd into horizontal and vertical components, you can treat it was one force. The lever arm is the minimum distance from $C$ to a passing through $B$ and running in the direction of $T_{BD}$, i.e. it is measured as the perpendicular distance from $T_{BD}$ to $C$. Note that this is not necessarily the distance from $C$ to $B$!


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