Question

problem

Answer 1

@mathmate

Answer 2

I have to review a little on the square and rectangular bars. I don't do that often.

Answer 3

I think I know how to solve each of these.

Answer 4

so we would use tau = Tc/J?

Answer 5

I remember they are not super complicated, it's just there's warping, so equations are different.

Answer 6

lets just do the first one. I am sure I can figure out the second one and third one.

Answer 7

What if we work separately, and compare notes.
ok we'll do the first one.

Answer 8

J is going to be different for the 2 and 3 one because they are both squares.

Answer 9

tau_allowed = T*c/J then we solve for c and multiply by 2 to get diameter?

Answer 10

exactly!

Answer 11

okay cool well we can compare notes tomorrow. I will finish it tonight. What time will you be around tomorrow?

Answer 12

You going to be around in the morning?

Answer 13

I will be on and off all day, so far.
The best is post what you have, and I will post what I have whenever I get it done. For discussions (probably not necessary), we'll see then.

Answer 14

okay sounds good.

Answer 15

I have <3" for the first one, but will compare the exact value tomorrow.

Answer 16

sounds good, take care.

Answer 17

tc you too!

Answer 18

@mathmate

my answers are:


B1 = 0.1095in
B2 = 1.165in
B3 = 0.924in

Brb going to eat some breakfast.

Answer 19

Are you there?

Answer 20

I am, but I have to go.
Here are mine:
(a) b=1.10335
(b) b=1.08235 (using alpha=0.208, according to Popov).
(c) b=0.81233 (using alpha=0.231, ditto).
I'll let you check that out in the mean time.

Answer 21

okay thanks.

Answer 22

what did you use for J in first one?

Answer 23

what was c? I used 1/2b

Answer 24

Yep, for (a), I did
T*(b/2)/J=tau
where J=pi b^4/32
solving for b gives
b=(16T/(pi*tau))^(1/3)
=(16*2.4/(pi*9.1))^(1/3)
=1.10335

Answer 25

that is weird I got that answer now.

Answer 26

so for the second one I used J = 1/12*b*b^3 and used c = 1/2*b

Answer 27

For the third one I used c = 1/2*b and J = 1/12*2*b*b^3. What did you use? I might heading out but will return later this evening.

Answer 28

I am eating something I will be checking back regularly.

Answer 29

k i am here

Answer 30

so what is up? what did you use for J?

Answer 31

popov and ditto?

Answer 32

For the second and third one, I use parameters set out by Popov:
\(\tau_{max}=\frac{T}{\alpha bc^2}\) for different values of b and c
where
\(\alpha = 0.208\) for b/c=1
\(\alpha =0.246\) for b/c=2
b/c=2 means b=2c in the above equation.
Note that question defines dimesions as b*2b, and in the equation, we use b*2c, or 2c*c.
So if we solve for "c", it's equivalent to "b" in the given question.

Egor P. Popov, Mechanics of Materials, second SI edition. (c) 1976.

Answer 33

The reason that parameters are required is because with a non-circular section, the section does NOT remain plane (warping), which makes it a very complicated analysis.
Popov gave the empirical values above to enable engineers to move on with the daily routines without having to study a highly complex analysis.

Answer 34

By the way, alpha ranges from 0.208 for a square section to 0.246 for b/c=2.0, 0.299 for b/c=6, and 0.333 for b/c=inf.

Answer 35

"ditto" is short for "as above".

Answer 36

okay, I didn't know that and didn't know anything about popov lol

Answer 37

I don't know what is going on here.

Answer 38

Perhaps your Mech. of Materials book would have a similar table.

Answer 39

let me look.

Answer 40

would something like this be in the appendices?

Answer 41

If you use the Beer and Johnson book, it should be there somewhere.

Answer 42

we don't use that textbook. We uses Philpot 3rd edition. My textbook is online too through the wileyplus site.

Answer 43

I can't find it anywhere. I will email my instructor this problem. I have one more under torsion problem I'd like to discuss. I have a solution.

Answer 44

tau_max = Tc/J = T(1/2D)/pi/32*D^4

A_median = (D)*(2D) = 2D^2

tau_max for the square = T/2At = T/4D^2t

set them equal and solve for t, thickness

Answer 45

t = D*pi/64

Answer 46

ok, that's the next section on Popov, shear on thin-walled sections. lol

Answer 47

Ohhh I remember this graph now.

Answer 48

I forgot how to use it though. lol

Answer 49

So you can now finish (b) and (c) and see if we agree on numbers.
gtg, but will check a little later before I disappear for the day! lol

Answer 50

okay sounds good.

Answer 51

For the second and third one, I use parameters set out by Popov:
τmax=Tαbc2 for different values of b and c
where
α=0.208 for b/c=1
α=0.246 for b/c=2
b/c=2 means b=2c in the above equation.
Note that question defines dimesions as b*2b, and in the equation, we use b*2c, or 2c*c.
So if we solve for "c", it's equivalent to "b" in the given question.

Egor P. Popov, Mechanics of Materials, second SI edition. (c) 1976.

I am confused about this. Let's do the second one together.

Answer 52

so I have a bxb square soooo....

Answer 53

τmax=T/(αbc^2)
so T=2.4, b=c
For (b)
9.1=2.4/(0.208*b*b^2)
give b=1.0823.
gtg
post what you have!

Answer 54

So what I see is b = 2c get that and c = 1/2*b get this, but where do you get the idea of b/c =2 is this saying D/d on the graph. I got to go. I solved the second part but still want to clear this up before I proceed. My buddy just got here so I am going to study thermo. I will work on it a little later.

Answer 55

Above refer to the question
Below refer to Popov's table on the second page.

It may be confusing because we use c for radius in formulas, b for the smaller dimension (in the question). Popov uses b for the longer dimension, and c for the smaller dimension in his table, and more importantly in the equation for tau, namely
\(\Large \tau_{max}=\frac{T}{\alpha b c^2}\).
So I use, for the second one part (b)
\(\Large 9.1=\frac{2.4}{0.208 (b)b^2}\). [ since b=c ]
and
\(\Large 9.1=\frac{2.4}{0.246 (2b) (b)^2}\), since the smaller dimension is squared.