# Static equilibrium problems

1. Oct 19, 2006

### laura001

hey every1... i'm stuck on this problem and unless i get it and a few others right im gonna fail my engineering statics course :( im desperate! I study aerospace engineering but because i've been ill i missed a few classes and i really dont 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

2. Oct 19, 2006

### 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

3. Oct 19, 2006

Working on it.

4. Oct 19, 2006

### laura001

5. Oct 19, 2006

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.

Last edited: Oct 19, 2006
6. Oct 19, 2006

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.

7. Oct 19, 2006

### laura001

thanks for the effort, but i dont 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?

Last edited: Oct 19, 2006
8. Oct 19, 2006

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 $$\alpha \approx \tan(\alpha)$$.

9. Oct 19, 2006

### laura001

double post

Last edited: Oct 19, 2006
10. Oct 19, 2006

### laura001

yea i made a mistake with the trigonometry, am gonna try the 2nd Q now but i dont have much confidence that i'll get the right answer. pleaseee stick around to correct me im certain i'll make a mistake, thx 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 im 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?

Last edited: Oct 19, 2006
11. Oct 19, 2006

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.

12. Oct 19, 2006

### laura001

is there anybody, out there? (pink floyd)

13. Oct 19, 2006

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.

14. Oct 19, 2006

### laura001

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

15. Oct 19, 2006

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

16. Oct 19, 2006

### laura001

thankyou! btw, i know this will sound rude as some1 can sound... but i have 8 more Q's and if i dont get them all right i really am gonna 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...

17. Oct 19, 2006

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.

18. Oct 19, 2006

### 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.

Last edited: Oct 19, 2006
19. Oct 19, 2006

### laura001

20. Oct 19, 2006