Question

review my work please.

Answer 1

@mj

Answer 2

@mjdennis

Answer 3

so i attempt this another way and doesn't give me the right answer... I have to input my answer in a small box...

Answer 4

Sorry, not my area of expertise.

Answer 5

i thought you were a mechanical engineer?

Answer 6

Aero and elecronic

Answer 7

well I guess I will have to talk to a classmate and see what they got. I swear my solultion is correct...

Answer 8

@CandyCove

Answer 9

@TheSmartOne

Answer 10

I am posting my other solution for the same problem... I did this problem two different ways, but still not getting the right answer.

Answer 11

@mathmate @DanJS might be able to help you :)

Answer 12

sorry my laptop just shut off for some odd reason...

Answer 13

let me get my other solutions up.

Answer 14

@MrCoolGuy

Answer 15

sorry i'm not good at engineering

Answer 16

I attempted the problem two different ways...

Answer 17

@hartnn

Answer 18

@raffle-snaffle
Your approach is a good start.
Summing moments about the hinge is correct.
The compatibility of deflections equation is also correct.
However, the compatibility of deflections equation is missing a component, namely the bending deflection of the horizontal bar AD. In a situation like this, F1 and F2 are likely to be substantial, hence the bar AD is subject to substantial vertical forces. It seems that the section properties of AD is not given, nor does it say that the AD is infinitely rigid. The safe thing to do, in exams, is to write down "assume AD is rigid" signifying that you did not forget the effect. In real practice, you need to find out and taking that into account.
The next step you need to do is to equate delta-a and delta-c using the equation delta=PL/EA respectively for each, where P=F1 or F2.
My computer battery is running low. I send this and continue later.

Answer 19

@mathmate thanks. I will be on here.

Answer 20

I solved this problem another way using deflection...

Answer 21

I will attach the image...

Answer 22

even solving this way I can't obtain the correct solution.

Answer 23

Is this what the spacing would look like in the beam? At the pin it states there is 3.70mm clearance.

Answer 24

let me post the pic of the problem

Answer 25

Sorry, I'm back now. Battery all charged, and plugged in. I am on vacation, so not on too often.
What I was trying to say was, there are four variables, delta-a, delta-c, F1 and F2.
You have already ide http://openstudy.com/whats-newntified two equations,
\(\Sigma\) moments = 0.......................(1)
delta-a/3.8=delta-c/1.2......................(2)
Then you need two more equations, namely
delta=\(\epsilon\)L=FL/EA, or more precisely,
F1=\(\epsilon_1/E_1A_1\)=(\(\delta_a\)/L1)E1A1...........................(3)
F2=\(\epsilon_2/E_2A_2\)=(\(\delta_c\)/L2)E2A2...........................(4)

Substitute equations (3) and (4) into the moment equation (1) to eliminate F1 and F2 [(1) then becomes (1a)]. After that, solve for delta-a and delta-c using equations (1a) and (2). Backsubstitue to find F1 & F2

Answer 26

delta - a and delta - c ?

Answer 27

delata refers to the bar 1 and 2 right?

Answer 28

OK, so the question does specify that ABCD is rigid, so you can gladly ignore bending on the horizontal bar. Otherwise, things complicate a little, although still manageable. Are you on introduction to indeterminate structures, or is it a statics problem?

Answer 29

Yeah, delta is either a or c. I elaborated the equation to (3) and (4) below.

Answer 30

well we are reviewing in this course. I already took strength of materials. This is the second half of the course. Yes, I am familiar with indeterminate structures.

Answer 31

Yes, this is a good review of indeterminate structures, based on the principle of compatibility of displacements.

Answer 32

why do we take delta_A - a and delta_B - b ? I thought delta => was the change length as which the bar stretched beyond its original length.

Answer 33

I am sorry for the sloppy notation. I wasn't at my usual keyboard, so I just got it out as soon as possible. delta-a actually meant \(\delta_a\) and similarly for delta-c.
I have not referred at any time to the linear dimensions a, b and c. I substituted the numerical values, like 3.8 and 1.1.

Answer 34

Oh okay thank fully that is what you meant.

Answer 35

I a eating something. I will be back in about 5min.

Answer 36

I have to go get supper going. Will hopefully be back later on, and if not, please feel free to post your queries and I will answer them the next time I'm on.

Answer 37

okay no worries. I will finish the problem and post the solutions up. Thanks.

Answer 38

Hmmm I get very large numbers for delta. It doesn't make sense to me.

Answer 39

And if you go back to solve for F1 and F2 those values are going to be huge!

Answer 40

Which of the following cases do you think is the initial position of the pin?

I will look into the deflections soon.

Answer 41

use mm, mm^2, and N/mm^2=MPa N
It is a weird but consistent set of units.

L1:1500;E1:76000;A1:740;
L2:2000;E2:115000;A2:480;
P:56000;
a:0.8;b:1.9;c:1.1;
F1:d1*E1*A1/L1; F2:d2*E2*A2/L2;

[3.8*F1+1.1*F2=3.0*P, d1/3.8=d2/1.1], solve for d1,d2

d1=1.110647696890604 mm
d2=0.32150328067886 mm

Above does not take into account of clearance of pin.

Answer 42

I am starting to get confused with my units for some reason.... I know the difference between SI and English. When I am dealing with m, mm, N, and kN.

I know 1kN =1000N and 1m = 1000mm

Am I allowed to deal with kN and mm or you can only deal with m and kN. Usually I deal with everything in meters. If I have mm I just convert to m in my head by moving the decimal over. Thanks clarification up above.

Answer 43

Wait 76GPa = 76,000? I thought G = giga = 1,000,000,000?

Answer 44

When solving for d1 and d2 I can have both mm and m in same equation? I thought we are suppose to stay consistent with one set of units, either mm or m?

Answer 45

d1 = 1.1106mm and d2 = 0.3215mm

F1 = 41.6kN
F2 = 8.87kN

sigma1 = 54.8MPa
sigma2 = 18.4MPa

e1 = 740x10^-6
e2 = 160x10^-6

part (c)
va = v1 = 1.66x10^3mm

Answer 46

The set of weird units I used is actually what you would use in practice, because MPa=N/mm^2, N, mm are all practical units. Most SI drawings show mm (without indicating the unit), so 76 GPa=76000MPa=76000 N/mm^2

However, if you are not used to this set of units, I suggest you stick to the conventional one, namely, N, m, seconds, hence N/m^2=Pa. Convert all kN into N though, and as you have done, mm into metres, and mm^2 into 10^-6 m^2. This way, you will likely to be stuck with a 10^(-6) or even a 10^(-12) along the way, but that's ok. Also, 76GPa is still 76000 MPa.

I think the best to do at this point is to use the standard SI units, and see if you get the same deflections as I got. Then we'll take it from there to worry about the clearance business.

Answer 47

Okay that makes more sense that you used 76000"MPa" than just 76000. I already solved the problem using mm and kN. I will use my units of measurement later. I kind of prefer to know both ways anyways because that way I know how to solve all type of problems with any variation of units.

How are my solutions for the next set of questions?