How to Calculate Vertical Reactions and UDLs for a Beam?

[Ry122]

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]

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!

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 )

 

[Ry122]

Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)

 

[tiny-tim]

Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)

Hi Ry122!

Yes … you've taken moments about A …

and everything's correct (except you need a minus in front of the 15/2*80 )

(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 )

 

[badeany]

for the UDL wouldn't 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]

Welcome to PF!

for the UDL wouldn't 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!

ooh … you're right!

I didn't notice the small print ("kN/m") on the diagram!

Yes, 80 is the force-per-length, so it does have to be multiplied by the length.

Thanks, and very well spotted!