How Do You Calculate Bending Moment Using Virtual Work Principles?

[laura001]

hey every1... I'm stuck on this problem and unless i get it and a few others right I am going to fail my engineering statics course :( I am desperate! I study aerospace engineering but because I've been ill i missed a few classes and i really don't have time to learn the theory to answer these... thank you for your time.

Here's the diagram for the 1st Q:

http://img383.imageshack.us/my.php?image=q1cm6.jpg

The Q that goes with it is:

Determine the bending moment at a distance x from the roller support on the left hand side, using the principles of virtual work.

The values are: F = 2kn, d=2.200m and x=0.600m

 

[laura001]

2nd Question:

http://img407.imageshack.us/my.php?image=q2un0.jpg

Each of the two uniform hinged bars have a mass m and length L. Both bars are connected by hinge S and supported as shown. The structure is loaded by a mass of 4m, applied at the hinge S. Gravitational acceleration is g = 9.81m/s2.

m= 15kg
L=5.500m
angle theta= 130 degrees

Calculate the torque Ma (in Nm) required for equilibrium

 

[radou]

Working on it.

 

[laura001]

your saving my academic life! thankyou!

 

[radou]

http://usera.imagecave.com/polkijuhzu322/system1.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: F\cdot d_{F}+M(A_{1}+A_{2}) = 0, where A1 and A2 are the angles of rotation of the two disks.

 

[radou]

Regarding the second assignment, replace the system with a system where the force G = mg acts in the center of each bar, and where G' = 4mg acts on the hinge. Try this: \sum M_{(A)}=0 \Rightarrow R_{B} = \cdots ; \sum F_{yi}=0 \Rightarrow R_{A} = \cdots ; \sum M_{(B)}=0 \Rightarrow M_{A} = \cdots I hope this works.

 

[laura001]

thanks for the effort, but i don't fully 'get it' :(

so what i have so far is:

F . df + M(A1 + A2)=0 which i think is

2 X df + M(40.36 + 8.53) = 0

ok and using trigonometry i got that df is 0.606711 m, so

i get that M = 0.024819 kNm... is that right?

 

[radou]

thanks for the effort, but i don't fully 'get it' :(

so what i have so far is:

F . df + M(A1 + A2)=0 which i think is

2 X df + M(40.36 + 8.53) = 0

ok and using trigonometry i got that df is 0.606711 m, so

i get that M = 0.024819 kNm... is that right?

M should equal -0.54 kNm, I checked on it. You obviously didn't do the trigonometry right. \frac{0.6}{4}=\frac{d_{F}}{4-2.2}, A_{1} = \tan A_{1} = \frac{3.4}{4}, and A_{2} = \tan A_{2} = \frac{0.6}{4}. I forgot to point this out - A1 and A2 are differential rotations, so in the theory of small displacements you can use the identity \alpha \approx \tan(\alpha).

 

[laura001]

double post

 

[laura001]

yea i made a mistake with the trigonometry, am going to try the 2nd Q now but i don't have much confidence that i'll get the right answer. pleaseee stick around to correct me I am certain i'll make a mistake, thanks again.

ok so i drew a free body diagram of the system, with mg acting downwards on the centre of each member, and 4mg acting downwards on hinge A. Then i calculated the vertical reaction force at B from moment equilibrium about point A, and got that Bv = 73.582kN

then summing forces in the y direction gives that Av = 809.318kN and summing forces in the x direction gives that Ax = 0. But now I am not sure what to do! lol... i know all the forces but am not sure how to calculate the torque at Ma... i mean, i think i can do it without using virtual work... by summing all the moments about point A, and then the reaction moment would be negative of that right... but have no idea how to do it with virtual work, the idea of displacements confuses me.

do u have maybe msn or yahoo radou?

 

[radou]

yea i made a mistake with the trigonometry, am going to try the 2nd Q now but i don't have much confidence that i'll get the right answer. pleaseee stick around to correct me I am certain i'll make a mistake, thanks again.

P.S. If you're certain you'll make a mistake, then you will make a mistake. Conslusion: don't be certain you'll make a mistake.

 

[laura001]

is there anybody, out there? (pink floyd)

 

[radou]

is there anybody, out there? (pink floyd)

Unfortunately, no msn neither yahoo.

Btw, it doesn't say anywhere that you have to use virtual work in assignment 2. Your way of thinking about 2 seems correct.

 

[laura001]

hey are u still out there radou? could u possibily help me on another couple questions? :!)

 

[radou]

hey are u still out there radou? could u possibily help me on another couple questions? :!)

As long as I'm online, I'll help.

 

[laura001]

thankyou! btw, i know this will sound rude as some1 can sound... but i have 8 more Q's and if i don't get them all right i really am going to fail this course :( could i maybe just have the answers? believe me when i say that, i will learn the theory behind all of this stuff but i reallly am in need of nothing less than a miracle just now...

 

[radou]

... could i maybe just have the answers? ...

Depends on how big the questions are. Btw, answers won't help you if you don't understand anything. But nevermind, let's give it a try.

 

[laura001]

ok this is actually Q11

http://img120.imageshack.us/my.php?image=qmc9.jpg

a= 0.200m
b=2m
theta= 35 degrees

Determind the torque M (in kNm) on the activating lever of the dump truck necessary to balance the load at the given dump angle theta, g = 9.81m/s2.

 

[laura001]

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]

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]

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 don't get these Q's all right then i can't sit the exam... which is a week away. And i would definitely be revising and doing all these things on my own in the exam, i just need the opportunity to sit the exam...

 

[laura001]

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]

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]

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]

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 won't 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]

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]

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]

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]

ok what I am going to 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]

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 I am not sure how to solve them...

for Q3 i got -10.2991kNm

 

[radou]

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]

http://usera.imagecave.com/polkijuhzu322/system1.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: F\cdot d_{F}+M(A_{1}+A_{2}) = 0, 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 can't 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, won't it also lead a displacement of the GIVEN hinge "s"? If this displacement were to exist won't we also have to add the virtual work carried out by vertical force acting on the hinge "s"?

 

[manan1]

M should equal -0.54 kNm, I checked on it. You obviously didn't do the trigonometry right. \frac{0.6}{4}=\frac{d_{F}}{4-2.2}, A_{1} = \tan A_{1} = \frac{3.4}{4}, and A_{2} = \tan A_{2} = \frac{0.6}{4}. I forgot to point this out - A1 and A2 are differential rotations, so in the theory of small displacements you can use the identity \alpha \approx \tan(\alpha).

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 won't understand anything properly.

 

[manan1]

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