Static Equilibrium and Structure problems

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  • #1
dch
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Homework Statement




[PLAIN]http://broadcastsol.com/123.JPG [Broken]

Prob. 1. The structure of massless members is shown in the figure. Determine the reaction forces at the supports C, and forces in the rod EF and DG. Givens: P, a, b, c.

Prob. 2.
The structure of massless members is shown in the figure. Determine the force in the member BE (BE perpendicular to DC) and reaction forces the at the fixed support A, and the hinge D. Givens: P, q, a, alpha, DE = 2EC.


Please anyone give me solutions, many thanks!
 
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Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi dch! welcome to pf! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
dch
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Prob 1, i just found the force EF = P*(a+c)/a and stuck at C & DG
Prob 2, I could not understand what "q" means.
Thanks for your reply :)
 
  • #4
tiny-tim
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hi dch! :smile:
Prob 1, i just found the force EF = P*(a+c)/a and stuck at C & DG
Prob 2, I could not understand what "q" means.
Thanks for your reply :)

For DG, why can't you use the same method you used for EF?

q is the mass (or weight?) of the load … the arrows show that it is evenly distributed along that part of the beam :wink:
 
  • #5
dch
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sorry How can I do this with the angle 45 at DG & DC thanks
 
  • #7
dch
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F(FE)xa=F(DG)xcos(45)x(a+b) right?

Thanks
 
  • #8
tiny-tim
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F(FE)xa=F(DG)xcos(45)x(a+b) right?

(it's clearer if you use * instead of x, or just leave them out completely :wink:)

Right! :biggrin:

and now the force at C? :smile:
 
  • #9
dch
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Can you give me some hints? :D I just thought that :

-P+F(DG)cos(45)+Bx+Cx=0
-F(DG)sin(45)+By+Cy=0

Thanks
 
  • #10
tiny-tim
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Can you give me some hints? :D I just thought that :

-P+F(DG)cos(45)+Bx+Cx=0
-F(DG)sin(45)+By+Cy=0

Thanks

yeees … those are right, but not very helpful, as you've noticed! :smile:

the trick here is that there are two unknown forces (B and C), and you can't really handle that …

so separate them!

just do the forces on CD :wink:
 
  • #11
dch
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so that CD is equilibrium thus,

F(DG)cos(45)-F(EF)=Cx
F(DG)sin(45)=Cy

Is this right :D

In problem 2, the load q is just like the weight of beam AB or not.

Thanks!
 
  • #12
tiny-tim
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so that CD is equilibrium thus,

F(DG)cos(45)-F(EF)=Cx
F(DG)sin(45)=Cy

Is this right :D

Yup! :biggrin:

(make sure you get the signs right :wink:)
In problem 2, the load q is just like the weight of beam AB or not.

Yes. :smile:
 
  • #13
dch
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[PLAIN]http://broadcastsol.com/123.JPG [Broken]

Here's my solution for prob2:

M(D) = -P*DC*sinα-F(BE)cosα*DE*sinα-F(BE)sinα*DE*cosα = 0
-> F(BE)

D:

Dx-F(BE)*cosα=0
Dy-F(BE)*sinα-P=0

-> Dx,Dy

Q= Integral (q*da) at midpoint of AB

A:

Ax-F(BE)cosα=0
Ay-F(BE)sinα=0

->Ax,Ay


Is this right?

Thanks.
 
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  • #14
tiny-tim
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hi dch! :smile:

(try using the X2 icon just above the Reply box :wink:)
M(D) = -P*DC*sinα-F(BE)cosα*DE*sinα-F(BE)sinα*DE*cosα = 0
-> F(BE)

You don't need to split FBE into components …

just use the whole thing … FBE*DE :smile:

(btw, you got it wrong anyway, it should have been F(BE)cosα*DE*cosα + F(BE)sinα*DE*sinα :wink:)
Dx-F(BE)*cosα=0
Dy-F(BE)*sinα-P=0

Fine. :smile:
Q= Integral (q*da) at midpoint of AB

A:

Ax-F(BE)cosα=0
Ay-F(BE)sinα=0

no, you've left q out of Ay.

(btw, I think q is the total load, so you just treat it as a single mass q at the midpoint :wink:)

(and you may need to take https://www.physicsforums.com/library.php?do=view_item&itemid=64" about A also)
 
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  • #15
dch
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M(D) = -P*DC*sinα-FBEcosα*DE*cosα-FBEsinα*DE*sinα = 0
-> FBE=P*DC*sinα/DE=3Psinα/2

D:

Dx-FBE*cosα=0
Dy-FBE*sinα-P=0

-> Dx,Dy

A:

Ax-F(BE)cosα=0
Ay-F(BE)sinα - q =0

M(A) = 0 = -q*AB/2 + F(BE)sinα * AB =0
-> q/2 = FBE sinα

-> Ay = 3q/2


Is this right :D
Thanks.
 
  • #16
tiny-tim
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Yes, that looks fine. :smile:

(though in the first line, FBE*DE is perfectly legitimate, and a lot easier to read and use :wink:)
 
  • #17
dch
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Thank you so much!

Now I have to do 2 problems and I dont want to make a new topic (very sorry about that because I thought I found my kindhearted teacher ^^ ). Please help me do these :


[PLAIN]http:///broadcastsol.com/321.JPG [Broken]

Prob1
The mechanical system of massless members is shown in the figure. Determine the relationship between two forces P1 and P2 to keep the system in equilibrium at position as showing in the figure.

Prob2.
Consider the shaded planar area with the y-axis as the axis of symmetry si shown in the figure. Determine the location (x, y) of the centroid and moment of inertia for the area about the x-axis.
====================================

My solution :

1)

M(A) = -P1*L+FBD*a = 0 -> FBD = P1*L/a

DEF is equilibrium :

Ex+P2=0
Ey+FBD=0

Ey/Ex = tan(30) = FBD/P2 = P1*L / a*P2

Is this right!

Thanks.
 
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  • #18
tiny-tim
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uh-uh, please always start a new thread for a new problem

there really are plenty of other members who can help you (though a lot of them are asleep :zzz: at the moment)!

i'll keep an eye on it, and join in if nobody else does :smile:
 

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