Exams WiSe 15/16

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Lilly
Beiträge: 1
Registriert: 12.10.2015, 15:25

Exams WiSe 15/16

Beitrag von Lilly »

These were the exam questions, as far as I could remember them...
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tola99
Beiträge: 102
Registriert: 28.03.2012, 16:24

Re: Exams WiSe 15/16

Beitrag von tola99 »

Did anybody already solve the example 6 of the first test? would be nice to compare the solution because i'm not sure with my solution!
thanks in advance!

Nedaa
Beiträge: 2
Registriert: 17.11.2015, 12:09

Re: Exams WiSe 15/16

Beitrag von Nedaa »

I got Fa to be = 3Fg/(4sin theta)

the curve is going downward from 0 to 90, max value at 0 and lowest at 90,, but some how not logical because if you stand at theta 90, you should have the max tension in the Tendon.. !! what did you get?

tola99
Beiträge: 102
Registriert: 28.03.2012, 16:24

Re: Exams WiSe 15/16

Beitrag von tola99 »

my result is F_{AT}=3 *cos*F_{G/2}. So i will calculate it again today, but we both have the same course of the curve.....

Nedaa
Beiträge: 2
Registriert: 17.11.2015, 12:09

Re: Exams WiSe 15/16

Beitrag von Nedaa »

you are correct, I was mixing the Fa cos with Fa sin..... I considered that Fg as a whole divided by 2... .

Good Luck

reox
Beiträge: 14
Registriert: 30.11.2015, 20:46

Re: Exams WiSe 15/16

Beitrag von reox »

F_{AT} = 3 \cdot cos(\theta) \cdot \frac{F_G}{2} is also my result.

Does anyone has a solution to 7.3? My considerations are, that the parallel muscle will simply shorten by the amount of each fiber, e.g. \Delta l_{muscle,parallel} = \Delta l_{fiber} \cdot L while the pennate muscle will only shorten \Delta l_{muscle,pennate} = \Delta l_{fiber} \cdot sin(\alpha) \cdot L where L is the number of Elements in Series (or l in the slides e.g. the height of the section), e.g the ratio will be \frac{\Delta l_{muscle,parallel} }{\Delta l_{muscle,pennate}} = \frac{1}{sin(\alpha)}

clstn_d
Beiträge: 2
Registriert: 20.11.2016, 22:26

Re: Exams WiSe 15/16

Beitrag von clstn_d »

Hi, did anyone found a solution for questions 6.1 and 6.2 of exam II ?
Thanks !

reox
Beiträge: 14
Registriert: 30.11.2015, 20:46

Re: Exams WiSe 15/16

Beitrag von reox »

my guess would be to use the engineers bending theory: sigma = M*x / I
I is pi/4 * (R_o^4 - R_i^4) for hollow rods, and I assume we should calculate the bending moment as sin theta * F * L?
I would not know why the L is given otherwise...
So sin theta * F * L gives the bending moment as the normal component does not have any effect on the moment (i assume).
thus inserting this in the formular, you get a formular depding on R_o, R_i and x for the stress distribution in the rod.
then you can insert the values and i get to 3.7544MPa for the 32/27mm rod which gives 3.75% R_F, and 4.183MPa for the 33/29 rod which gives 5.2% of R_F. This is an increase in Risk of about 40%

But I'm not 100% sure if this is correct...
edit: mhh is sigma_11 in the direction of L and A is the neutral surface? Then the above calculation is wrong...
I took it from the slides 09 slide 52, where sigma_11 would be the calculations above. But what is A then?
edit2: I think there is a minus missing for the moment, but you get the idea...

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