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Static equilibrium problems

by laura001
Tags: equilibrium, static
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laura001
#19
Oct19-06, 05:45 PM
P: 24
Q4

http://img141.imageshack.us/my.php?image=q4wp7.jpg

m = 17kg
L= 6m
theta = 85 degrees

Calculate the force (in N) required for equilibrium.
radou
#20
Oct19-06, 05:52 PM
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These are all very similar (and typical) statics assignments, so, if you solved the one at the beginning of the thread (the one without virtual work), then you should know how to solve these one. As said before, use the fact that a system is in equilibrium if the sum of all moments of the forces acting on it (including couples, as in Q3) with respect to any point must vanish, as must the sum of all horizontal and vertical forces (i.e. compoments of forces). Pick the 'moment equation point' wisely - it's a way to elliminate unknowns.
laura001
#21
Oct19-06, 06:01 PM
P: 24
ok here is Q5:

http://img138.imageshack.us/my.php?image=q5ayq9.jpg

q = 2kN/m
a= 0.350

Calculate the moment reaction in A (in kNm) counter clock-wise positive.

PS i really didn't mange to work out the 1st one either... please please please! do it for me, the thing is... if i dont get these Q's all right then i can't sit the exam... which is a week away. And i would definetely be revising and doing all these things on my own in the exam, i just need the opportunity to sit the exam...
laura001
#22
Oct19-06, 06:07 PM
P: 24
q6

http://img138.imageshack.us/my.php?image=q5ayq9.jpg

a = 2m
F = 25kN
q = 50kN/m

Calculate the vertical reaction force in support C in kN (upwards positive)
radou
#23
Oct19-06, 06:10 PM
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Q5.

You can separate the two disks and look only at the part from the hinge S to the end. Use the sum of moments in the point of the hinge to obtain the reaction in B (which is vertical, of course). Then you can use the sum of all vertical forces to obtain the vertical force in the hinge C (the horizontal force equals zero, since there is no horizontal load applied to the disk). Further on, you can look at the part A-S now separately; the force acting on it is the vertical force from the hinge C (but in the opposite direction of that one which you got while looking at the right disk). Use the sum of moments at point A to obtain the value of the reaction moment in A.
laura001
#24
Oct19-06, 06:11 PM
P: 24
q7

http://img88.imageshack.us/my.php?image=q7ath5.jpg

F = 20kN
q= 25 kN
a= 7m

Calculate the internal moment at B (in kNm) use the sign convention in the figure (which is U = positive)
laura001
#25
Oct19-06, 06:14 PM
P: 24
q8

http://img172.imageshack.us/my.php?image=q8qp8.jpg

F = 25kn
q = 30kN
a = 5m

calculate the reaction force at A in kn, upwards positive.

PS. if u wont give me the answers then could u possibily make sure i have the right answers if i post what i think is right at the end of this thread? :)
laura001
#26
Oct19-06, 06:17 PM
P: 24
q9

http://img384.imageshack.us/my.php?image=q9iw9.jpg

a = 0.800m
F=20kN
M=14kNm
q=3kN/m

Calculate the normal force in C (in kN). Use correct signs for tension and compression.
laura001
#27
Oct19-06, 06:19 PM
P: 24
q10 (last question!)

http://img217.imageshack.us/my.php?image=q10mg1.jpg

a=2m
F=11kN
M=8kNm
q=4kN/m

Calculate the shear force in C (in kN_ with correct sign of deformation. Adopt the sign convention from the figure .
radou
#28
Oct19-06, 06:23 PM
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P: 3,224
Basically, these are all the same. If you understood Q5, then you should be able to solve any of them. The point is that you can always 'disconnect' the system at the place of the hinge and set up equilibrium equations for each part separately. If there is something specific that you don't understand, you should rather PM me, since this thread has become quite a mess.
laura001
#29
Oct19-06, 06:27 PM
P: 24
ok what im gonna do is quickly try to work them all out, and post my results in one reply... and if u could maybe just check that i have them right that would be so cool. ty
laura001
#30
Oct19-06, 06:36 PM
P: 24
for Q5 i got that that the reaction moment at A is 2.6133333 kNm.
for Q6 i got that Cv = 75kN.
for Q7 i got that the internal bending moment at B is equal to -860kNm.
for Q8 i got that the vertical reaction force at A is 9.357kN.
Q9 and Q10 are different from Q5 so im not sure how to solve them...

for Q3 i got -10.2991kNm
radou
#31
Oct19-06, 06:39 PM
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Quote Quote by laura001
ok 1st one i've solved is Q5, i got that the reaction moment at A is

2.6133333 kNm... is that right?
Yes, seems correct. But do post all the results on one post.
manan1
#32
Oct25-11, 10:37 AM
P: 7
Quote Quote by radou View Post
http://usera.imagecave.com/polkijuhz...ystem1.bmp.jpg

Ok, as said, the system is replaced with a mechanism, where a hinge is put in the place where you have to find the moment, and there is a couple of moments M added to that same place. Now, you have to construct an initial relative rotation between the two diskd connected by the hinge, so do it as is done in the displacement sketch. Now all you have to do is apply the principle of virtual work to get the moment M: [tex]F\cdot d_{F}+M(A_{1}+A_{2}) = 0[/tex], where A1 and A2 are the angles of rotation of the two disks.
I believe I am in the same course and am having difficulties with this problem. The imagecave server is down so i cant have a look at your diagram. If you would reupload the image if you have it I would be very grateful.

I want to ask certain questions that arose in my head from this incomplete understanding of your solution. I was wondering if you introduced a virtual displacement at x (the point where the hinge is added) and a virtual rotation is created between A and B, wont it also lead a displacement of the GIVEN hinge "s"? If this displacement were to exist wont we also have to add the virtual work carried out by vertical force acting on the hinge "s"?
manan1
#33
Oct25-11, 10:41 AM
P: 7
Quote Quote by radou View Post
M should equal -0.54 kNm, I checked on it. You obviously didn't do the trigonometry right. [tex]\frac{0.6}{4}=\frac{d_{F}}{4-2.2}[/tex], [tex]A_{1} = \tan A_{1} = \frac{3.4}{4}[/tex], and [tex]A_{2} = \tan A_{2} = \frac{0.6}{4}[/tex]. I forgot to point this out - A1 and A2 are differential rotations, so in the theory of small displacements you can use the identity [tex]\alpha \approx \tan(\alpha)[/tex].
Also I do not understand how one could possibly arrive at the above trigonometric relations. Maybe if I saw the diagram it would be clear, but to the extent of my understanding of your words, it does not make sense . I guess without the diagram I wont understand anything properly.
manan1
#34
Oct25-11, 06:30 PM
P: 7
Ok, forget my first post and the question therein. I got how the problem works. However I still would like an answer to the question I posed in my second post, it would clarify my understanding of these things. I used some other trigonometric ratios. Thanks


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