Moments

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Can anyone help me with these moments I take about A and get different results to taking them about B :s-smilie:

Reply 1

bcrazy

Are you considering the reaction forces at A and B? If so, can you show what working you have done?

Reply 2

craglyboy

Original post by bcrazy

Are you considering the reaction forces at A and B? If so, can you show what working you have done?

Yeah trying to find RA and RB I call em Ay and By
sum of moments about B=0

(-3x1)+(5x2)-(Ayx3)+(4x4)

Ay=7.6KN

but taking moments about A I get By=1KN and then Ay=5KN

(edited 12 years ago)

Reply 3

bcrazy

Original post by craglyboy

Yeah trying to find RA and RB I call em Ay and By
sum of moments about B=0

(-3x1)+(5x2)-(Ayx3)+(4x4)

Ay=7.6KN

but taking moments about A I get By=1KN and then Ay=5KN

You need to check your working when taking moments about A...show it if you can't find your mistake. You should find that the force at B is actually acting downwards.

Reply 4

craglyboy

Original post by bcrazy

You need to check your working when taking moments about A...show it if you can't find your mistake. You should find that the force at B is actually acting downwards.

So my working for taking moments about B is correct?

For moments about A I got:

(4x1)+(5x1)-(3x2)-(3xBy)=0

to get By=1KN

(edited 12 years ago)

Reply 5

bcrazy

Original post by craglyboy

So my working for taking moments about B is correct?

Yes

Reply 6

craglyboy

Original post by bcrazy

Yes

For moments about A I got:

(4x1)+(5x1)-(3x2)-(3xBy)=0

to get By=1KN

but you say this is wrong so is By from equating the forces using Ay=7.6Kn

Sum of Fx=4-7.6+5-3-By=0

By=-1.6Kn

(edited 12 years ago)

Reply 7

bcrazy

Original post by craglyboy

For moments about A I got:

(4x1)+(5x1)-(3x2)-(3xBy)=0

to get By=1KN

If you are taking clockwise as positive then then 4kN force should be negative (make sure you always think carefully in terms of clockwise and anticlockwise rather than up and down)

Reply 8

craglyboy

Original post by bcrazy

If you are taking clockwise as positive then then 4kN force should be negative (make sure you always think carefully in terms of clockwise and anticlockwise rather than up and down)

Is the bottom part of my last post wrong then as well

For moments about A I got:

(4x1)+(5x1)-(3x2)-(3xBy)=0

to get By=1KN

but you say this is wrong so is By from equating the forces using Ay=7.6Kn

Sum of Fx=4-7.6+5-3-By=0

By=-1.6Kn

Reply 9

bcrazy

Original post by craglyboy

Sorry, have got confused with your edits which bit you are asking about at the start but your working giving By=-1.6Kn is correct (ignoring the slightly dodgy rounding you did earlier on...should be Ay=23/3 and By=-5/3)

Reply 10

craglyboy

Taking anti-clockwise as positive in all this I hope :s-smilie:

Reply 11

craglyboy

Original post by bcrazy

I was on about resolving the Fx, thanks for the help I think my working is all alright now apart from shoddy rounding as you say.

Can you show me how you would take moments about A?

Reply 12

bcrazy

Original post by craglyboy

Taking anti-clockwise as positive in all this I hope :s-smilie:

Sorry, yes! I would have taken moments about A the same as you did when you corrected it...but actually I would have only looked to take moments about B and then resolved vertically- the only reason I would be tempted to take moments about A would be to double check.

Reply 13

craglyboy

Original post by bcrazy

Yeah that was the only reason I did to wasn't very confident and seemingly rightly so thanks anyways now I can start to answer the question, got to find shear force and bending moment now haha thanks man!