Question

Problem review please

Answer 1

When inputting my answers into the boxes it won't accept my answer. I got the same values as you did for d1 and d2.

Answer 2

Are there restrictions on units, precision, etc? like kN for forces, m for displacements, and no units for strain. Scientific notation?

Answer 3

For some reason, the second and third images are just grey. Perhaps it's my Internet connection.

Answer 4

No restriction on unit precision. Your internet connect could be poor. I can see the last two images just fine on my side.

Answer 5

The tolerance for the answers is + or - 2

I can't input the answer as scientific notation. I uploaded a few more images. Hope these are more clear.

Answer 6

d1 = 1.1106mm
d2 = 0.3215mm

F1 = 41.6kN
F2 = 8.87kN

sigma_1 = 54.8MPa
sigma_2 = 18.4MPa

e_1 = 740x10^-6
e_2 = 160x10^-6

Va = downward deflection = 1.66x10^mm

Answer 7

F2 = 8.69kN and sigma_2 = 18.1MPa

Answer 8

Wait, you were too fast!
Have we taken into account of the clearance of 3.7 mm yet?
I think the pinned connection D looks like this:
Oh well, may be it's too early, the drawing tool is not awake yet.
Point D would have a slot 3.7 mm longer than the diameter of the pin, and at rest, the pin would be at the bottom, ready to have an upward movement of 3.7 mm. Any attempted horizontal movement would be nil because horizontally ABCD is free to move.
So for the first 3.7 mm, the bar ABCD is in a statically determinate situation, with the load split between the vertical bars 1 and 2 only. If point D moves up (or down) less than 3.7 mm under the 56 kN load, the whole problem is statically determinate.
On the other hand, if the displacement of D > 3.7 mm, then the 56 kN load must be pro-rated between the statically determinate and indeterminate situation, with just enough load attributed to the statically determinate portion such that the displacement of D is exactly 3.7 mm, and the remaining load used to calculate the displacements of A and C.
Therefore the displacements of A and C will each have a stat. det. and indet. components.

Proceed that way and tell me what you get. You can use the N/mm set of units, or your habitual one, the result should be the same. We can compare notes later.

Answer 9

I got it right when including the 3.70mm.

Answer 10

I have another question. I will post it. I need to find principal stress for x, y, and z. I found part 1, part 2. I need help on part 3 and part 4.

Answer 11

I see the answer sheet, and your work, but I don't think I see the question of the principal stress problem. Could you post it?

Answer 12

^ above my work is the question.

Answer 13

I did not see the question at the top of the answer. I was expecting a diagram or something, lol. My apologies.

I have not done much with 3-D principal stresses, and I am not home these past two weeks. So the next best I could do is to verify the matrix calculations.
I used 66.17 deg. as is (instead of 66+1/6 degrees, which I suspect is the real value), but the difference is minimal.
I get [13.991, 8.714, 2.040] for the first product, and 15.899 for the second.

If it's not resolved by next week, I should be able to work on it further when I get home.

Answer 14

One of my classmates told me I need to use this formula to find the principal stress in x, y, and z.

I am attempting to solve right now.

Answer 15

Here is the page with all the equations.

https://edugen.wileyplus.com/edugen/player/references/index.uni?mode=help&xlinkobject=philpot9781118083475c12-sec-0051&itemid=EAT_1351540446544_0_13094065697046242

Answer 16

I solved for I1, I2, and I3

I_1 = 38ksi
I_2 = 434ksi
I_3 = 1460ksi

How do I utilize the other equations to solve for principal?

Answer 17

The link has explanation as well.

Answer 18

It could be my Internet connection again, the link says
"the content associated with this link is not yet available".
I am in a small village with satellite connection.