What is the Force at Pin E in a Frame and Machine Analysis?

[julz127]

Homework Statement

Find the force acting at pin E

[PLAIN]http://img841.imageshack.us/img841/4983/piczt.png

a = 53.0 mm, b = 212 mm, c = 318 mm, d = 106 mm, and e = 371 mm.

A mass of 515kg acts through H.

Homework Equations

The Attempt at a Solution

By finding the moment about point D, we can find the reaction at G, and since GI is a two force member the same force (opposite sign) acts at the other end, on pin I.

$$F_G = \frac {(F_H \times 2b) }{d_1}$$

where $$d_1 = (2c+d) sin (180 - Atan (\frac{c}{d}))$$

Solving this part I get 5205N, I know this is correct.

To find the forces at E isolate the IKE member. Sum the forces about E, to find the reaction JK then use equilibrium equations to find the x and y forces at E.

$$F_{GIx} = 5205 cos (56.31)$$
$$F_{GIy} = 5205 sin (56.31)$$
$$M_E = (F_{GIy} \times 2a) + (F_{GIx} \times \frac {2a}{cos(30)})$$

This next part I'm having trouble with, can I simply break up the components of $$F_{IG}$$ and then have them act at I? With opposite sign of course.

I've bee working with the assumption that JK acts perpendicular to EI, is this correct?

 

[nvn]

julz127: The answer to both of your last two questions is yes. Dividing 2*a by cos(30 deg) in your last equation currently seems wrong, unless I am misinterpreting. Check that. Also, your last equation seems to erroneously omit F_JK.

 

[julz127]

Yeah, last equation should have been:

$$F_{JK} \times \frac {a} {\cos(30)} = (F_{GIy} \times 2a) + (F_{GIx} \times 2a\tan(30)$$

 

[nvn]

Hint 1: Regarding your equation in post 3, are you sure about 30 deg? Check that.

 

[julz127]

Oh dammit. :P

Do you know how long I spent on this question, before giving up and clicking "show answer"?

Thanks anyway. At least I can do it now.