I have to calculate the vertical reactions at point A and C.
I know that the sum of all vertical forces is equal to zero but i can't solve the equation since there's two variables.
http://users.on.net/~rohanlal/Q3.jpg
Bv=50
Cv=55
Dv=45
w=80
tiny-tim
Sep5-08, 05:14 AM
I have to calculate the vertical reactions at point A and C.
I know that the sum of all vertical forces is equal to zero but i can't solve the equation since there's two variables.
Hi Ry122! :smile:
general rule … when you have a unknown force you can't get rid of, take moments about a point (strictly, an axis) through the line of that force.
In this case, take moments about A and about C.
(Alternatively, find the point about which the moment of all the given forces is zero :wink:)
Ry122
Sep5-08, 07:37 PM
Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)
tiny-tim
Sep6-08, 07:04 AM
Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)
Hi Ry122! :smile:
Yes … you've taken moments about A …
and everything's correct (except you need a minus in front of the 15/2*80 :wink:) :smile:
(I suggest you take moments about C also, just for practice … but of course, as I expect you've noticed, you don't need to, because you already know the total reaction for A and C together :wink:)
badeany
Sep8-08, 08:43 AM
for the UDL wouldnt the moment created about point A be the load (80) multiplied by the distance it covers (15m) then multipied by the half way point (7.5m) giving you a moment of 9000 KN/M about point a?
rather than what you have with 7.5 x 80 giving you a moment of 600 KN/M about point a?
tiny-tim
Sep8-08, 09:55 AM
for the UDL wouldnt the moment created about point A be the load (80) multiplied by the distance it covers (15m) then multipied by the half way point (7.5m) giving you a moment of 9000 KN/M about point a?
rather than what you have with 7.5 x 80 giving you a moment of 600 KN/M about point a?
Hi badeany! Welcome to PF! :smile:
ooh … you're right!
I didn't notice the small print ("kN/m") on the diagram! :redface:
Yes, 80 is the force-per-length, so it does have to be multiplied by the length. :wink: