A while ago, I found an old textbook at a used bookstore, Vector Mechanics for Engineers (Statics). It’s old; there’s an actual 5.25" floppy disk inside the back cover. I studied a fair bit of Physics and Math back in college, but never quite to the level this book is at. I’m at an age where I don’t want my brain to start getting fluffy, and I wanted something to do when I go out to eat by myself (I’ve run out of crossword puzzles), so I bought the book and I’ve been gradually working my way through it. It’s been going reasonably well, and it’s kinda nice to learn something at my own pace, but I also don’t have a professor I can go to when I don’t understand something or get a problem wrong. So this is kind of a homework question, but not for an actual class I’m taking.
I’m stuck on one of the problems. I’ve worked through most of it. I’m trying to find the value of an angle, Φ, for which TF is at a maximum in the following equation:
(3.125 TD sin Φ) - (2.5 TD cos Φ) = 2TF
Is there some trigonometric equation I can use so that both of the terms on the left use the same function, or some other technique? I think I know the answer from the way the problem is written, and the answer is in the back of the book, I’m just curious if there’s a purely mathematical way to work it out.
Do you know calculus? The general technique for maximizing or minimizing a function is to take the derivative, and then set the derivative equal to zero.
That said, a sum of any two sinusoidal functions (sine or cosine) with the same frequency is always some other single sinusoidal function, with some amplitude and phase. It’s been a long time since I’ve done it, but… If I recall correctly, it relies on the fact that sine and cosine are orthogonal? So the amplitude would be the Pythagorean sum of the amplitudes of the sine and cosine parts, and the phase would be the inverse tangent of the ratio of the amplitudes.
First off, you can simplify down to:
1.25 sin Φ - cos Φ = x
Since you’re maximizing TF, dividing through by positive constants doesn’t change the answer.
You can guess that you want an answer between pi/2 and pi. That way, sin(Φ) is positive, and cos(Φ) is negative (so subtracting it makes it positive).
If you graph that as a point at (-1, 1.25) (reflecting the fact that cos is an x-axis thing, while sin is a y-axis thing), you can see that the angle it makes with respect to +X is (pi/2 + arctan(1.25)), or 2.467*. Which is the answer, I believe. Any other angle would give a smaller answer due to mixing in “too much” of either the sin or cos components.
* ETA: Hmm, this isn’t quite right. I probably should have actually drawn it out. I believe Saint_Cad’s answer is correct; it’s about 2.246 radians.
Offhand a max/min problem would involve setting the derivative to 0
dividing by 2 and differentiating
dTF/dphi = 1.5625 Td cos(phi) + 1.25 Td sin(phi)
Set equal to 0 and we find that Td does not affect the answer and that 1.5625 cos = -1.25 sin. So we are in the second and fourth quadrant. Divide by 1.5625 and we get cos = -0.8 sin
Since cos^2 + sin^2 = 1 this is now 0.64 sin^2 + sin^2 = 1 which cleans up to sin^2 = 1 ÷ 1.64 or sin = 0.78086881 which is right at 51.3402 degrees. Since this is in the 2nd and 4th Quadrant it would be 128.6598 and 308.6508 degrees.
As @Chronos said, the ‘easy’ way to do this is take the derivative, rearrange the terms so that you have sin(Φ)/cos(Φ) = -3.125/2.5, and take the arctangent of both sides to get the value of Φ. You could do a sum of amplitudes but frankly if you are going to go to that effort just revert to Euler’s formula and do the algebra and solve directly, but this is what calculus was invented to do.
I did study calculus, but it’s been a while. I’m on chapter 4 of the book, and haven’t needed any calculus yet, so it wasn’t the first thing I thought of for attacking this problem.
I think I can simplify that a bit. From
cos Φ = -0.8 sin Φ
it becomes
-1.25 = tan Φ
Φ = -51.3 degrees, which is the answer in the book. (I probably screwed up a + or - in there somewhere; I’ve been doing that a lot.)
The question appears to have been answered by other means, but it is possible to combine two sine and cosine terms into a single term. Here’s how.
Suppose we have the expression \mathcal{Q} = a \sin \phi + b \cos \phi. Draw a right triangle with angle \alpha so that a is the adjacent leg to \alpha and b is the opposite leg to \alpha. (This implies that \tan \alpha = b/a.) Let c = \sqrt{a^2 + b^2} be the hypotenuse of the triangle; we then have \cos \alpha = a/c and \sin \alpha = b/c.
Then we have \mathcal{Q} = c ( \frac{a}{c} \sin \phi + \frac{b}{c} \cos \phi ) = c (\cos \alpha \sin \phi + \sin \alpha \cos \phi). If you remember your trig identities, this becomes \mathcal{Q} = c \sin (\phi + \alpha).
The point of calculus to make hard questions easy. For example, if you wanted to find the largest rectangle with a given perimeter, you might cleverly deduce that A=\frac{P^2}{16}-\frac{(x-y)^2}{4} from which you conclude that it is maximized when x=y. But calculus does it with no cleverness involved.
In the present instance you are trying to maximize x=2T_f=a\sin\phi-b\cos\phi. Then dx/d\phi=a\cos\phi+b\sin\phi. Set it to 0 and find \tan\phi=-a/b=-1.25, so that \phi=\tan^{-1}(-1.25)=-53.34^o.
I can follow the first part of the solution, so I know the forces at F and H. Then ABF is assumed to be a free body, and the sum of the forces on it in the y direction must be 0. From that they calculate FBG. But there’s another force on the body that they just left out, FCF. Is there some reason that member can just be omitted, or assume to be 0?
I can see that the sum of the vertical components of FBG and FCF must be 3.6 kips upward, but not why FCF must be 0.
I haven’t worked through the problem but by inspection FCF would be in compression, and since these members are only allowed to act in tension it must be zero.
Why would it have to be in compression, though; is there some other constraint or free body for which FCF can’t also be in tension? Is it just that two diagonals across a rectangle can’t both be under tension, or something like that?
Interesting that quoting a post in Discourse doesn’t preserve the subscript tags.
Honestly, I did enough of these problems in school (and also worked on a project developing tutorial software that specifically calculated loads in arbitrary truss structures that the user would interactively define) that I can just look at a structure this simple and just intuit which members are in tension and compression without turning a crank. I can also integrate shear and moment diagrams mentally without writing out the equations and an generally identify the inflection point (where the shear is zero and moment is maximum) just by inspection with pretty good reliability, which is just one of those things you can do with experience, and I’m far from the best at it; I’ve known civil structural engineers who can spot an error in a matrix of dozens of equations like it was in blinking neon lights.
Generally speaking, if you have two crossing members in a truss, one will be in tension and one in compression unless there are net outward forces on both opposing base members because that is just the way structures respond to typical external loads. It helps if you think about the truss being a unitary beams in bending and then think about whether the “outer fiber” at the top or bottom should be in tension and in which direction the load from bending increases. That works for a single ‘story’ truss but once you start stacking trusses together all of that intuition kind of goes out the window and you have to start using finite difference methods to work it out because you then end up with a statically indeterminate structure.
Now let me tell you how you can distinguish between ‘local’ and ‘body’ bending modal frequency in beam models…
Sometimes I love this board. I may have more questions before I finish the book.
I’d still like to find some some specific, definable rule why FCF can’t be under tension. The link I posted is to a different version of the book than the one I’m working through. I found a PDF with answers to my version, and I’ve been using it more often than I’d like to admit… Both answers do the same thing, though; they just assert that one of the diagonals is 0 without explaining why. After that, the problem is easy to solve.
That’s the problem with studying this on my own; but I’m well into chapter 6, and I’ve only been really stuck twice.
They note that they are tension only members. So usefully tell you they can’t be under compression. After that, as you note, the rule is indeed simple - you can’t have both cross members in tension at the same time. At best you can have both members with no tension. The rest falls out from inspection and a lot by simple symmetries, which abound in these sorts of problems.
You could probably prove to yourself the lack of simultaneous tension by analysing the triangles in an arbitrary rectangle and deriving a closed form expression for the tensions. Then the symmetry of the system should make it clear.
Mainly, Its when the top of the truss shifts left or right. Its not locked in place that way, except by the counters. Here, if the length of the counter BG is increased x , then CF must be reduced x …ie, compressed around as much as the other is stretched. its made of cable, or flexible rod, so force required to compress is zero.
There is another contribution to the change in force of the counters, and that is stretching or compressing of the truss members. Could this put tension on, and result in the increase of length of CF being MORE than x ? Yes, but only if the stretch of the truss members was as much as x.
Does the shape of the truss cause a lot of growth of the counters ,more easily than the change in angle at the truss joints ? no the compression of the vertical members offsets the stretch of any horizontal member, so the growth of the counters is less for the same forces…
The counter is going to stretch much more than the truss members… while the forces are of the same order, 5 to 15 … its got to be at most a tenth the cross sectional area ? So even if the counter was taking a low force, like 3 , well its going to stretch like it was a truss member taking 30… So the truss members do not stretch more than the counters, and so the angle change is more significant, and net result, one counter must be shorter than its original unstressed length, ie one must be 0 tension.
Actually, to make it easy, just take a square frame, put two tension-only crossmembers inside of it, and apply a single arbitrary load to one corner and you’ll see that in order to resolve the loads you have to assume one of them is zero (in compression). If you actually had two compression-bearing load members I believe you’d have a statically-indeterminate condition which would then require doing a deflection or virtual work analysis.
They’re just units. The first thing you learn in doing real world structural analysis is to put all loads, factors, and geometric/material properties into a consistent set of units and then ignore them when working the problem. I can’t count the number of times I’ve looked at a finite element analysis and declared, “Your results are off by a factor of 386.4,” to be met with looks of amazement about how I could be so precise about the error. Ditto with dynamic analysis of vibration and shock data and 39.48.
Why can’t the diagonals both be under tension if all four sides are under compression?
I appreciate all the comments so far, and they are helping, but none of them are things that have been covered in the textbook so far. I should be able to find a subset of the truss, or a single joint within it, for which the forces only balance if that diagonal counter is 0. I can’t find that case yet.
What I may do is assume FCF is under some arbitrary tension (10 kips, for example), solve for the remaining forces, and see where I run into a contradiction.
I’m beginning to suspect that it is statically-indeterminate.
I know. But usually the problems are tweaked to make things drop out of easily, but only if you are comfortable with the usual magic ratios in those units. If you are not, all the wired-in pattern recognition fails to fire. It doesn’t matter for getting the gist of the problem, but makes evaluating the answer a pain, when it should be doable in one’s head.