Solving Statics Truss Problems: Plotting Force in Member BC & Angle ø Range

[LewisNL]

The question is stated below. I found out an answer but it doesn't seems to be correct to me. I made a few sketches to clear things up a bit, sorry for bad quality, too much compression.

The simple truss of Prob. 4/2 is modified as shown in the figure, in that the angle of the support surface at C can be varied from 0 (vertical) to 90◦ (horizontal).

(a) Plot the force in member BC as a function of ø over this range. Note any unusual conditions.
(b) For what value of ø, if any, is the force in member BC zero?
(c) If member BC is designed to fail at a load of 5 kN in either tension or compression, what is the allowable range for the angle ø?

Answers:
(a)
Fa = 500N
Fcx = Fbx + Fcx = 0
Fcx = -1200N
Fcy = tan(ø) = Fcy / Fcx
Fcy = -tan(ø) * -1200
Fbx = (Fa / 1,25) * 3 = 1200N
Fa + Fby + Fcy = 0
Fby = 500 – 1200 * tan(ø)

So if we want to plot this, x would be x = a vector from 0 to 90. And y would be y = –1200 * tan(ø).

If I plot this it shows some kind of weird looking tangens-function. Is this normal or did I make a miscalculation somewhere?

The function has a lot of 0-points so question (b) seems a bit strange too.

Maybe someone can help to clear things for me up a bit?

Thanks in advance :)

 

[Soveraign]

As a sanity check, given your diagram, it is clear that when theta is zero, the force pushing at C should be only in the -x direction and with magnitude you calculated above. Now, at theta approaches 90 degrees, I would expect that the force normal to the support surface C is sitting on to approach infinity.

Specifically in the y direction (the direction relevant to parts b. and c. of your question) the force would be zero when theta is zero and approach infinity as theta approaches 90 degrees.

Your equation for Fcy seems to follow this idea (although I'm not sure what you had declared as + and - in the y direction). Drawing out the vectors for myself, I get the following:
$$F_{cy} = 500N \frac{3m}{1.25m} \tan(\theta)$$

...which I think agrees with what you have. So, to solve the questions parts b. and c. it seems to me you would want to solve for when $F_{cy}$ is zero and 5 kN.

Since you mentioned that your plot of the function had a lot of zero points, I feel I should ask if your plotting program is calculating tangent in degrees or radians?

 

[wais]

Your calculations and approach seem to be correct. However, there is a small error in your calculation for Fcx. It should be -1000N, not -1200N. This is because Fbx is acting in the opposite direction of Fcx, so they cancel each other out.

As for the plot, it is normal to have a weird-looking tangent function. This is because the force in member BC is directly proportional to the angle ø, so as ø increases, the force also increases at a faster rate. This results in a steeply increasing curve.

For question (b), the force in member BC will be zero when Fcy is equal to -500N. This occurs at an angle of ø = 26.57°. This makes sense as at this angle, the vertical component of Fcy (Fcy = -500N) cancels out with the vertical component of Fby (Fby = 500N).

For question (c), we can use the formula for Fby that you have calculated: Fby = 500 – 1200 * tan(ø). If we set this equal to 5kN and solve for ø, we get ø = 15.3° or ø = 74.7°. This means that the allowable range for the angle ø is from 15.3° to 74.7°, as long as the force in member BC does not exceed 5kN in either direction.

I hope this helps clarify things for you. Keep up the good work!