Total distance of horizontal section of crane = 68m
Distance front left hand side to intersection point of crane = 18m
Therefore distance from intersection point to centre point of horizontal section of crane = 16m (34m at centre?)
I am assuming that the self weight of the horizontal section of the crane is acting at the centre point. 16m is the distance from the centre of the cranes horizontal section to the intersection point of the crane (point I am calculating moment from)
Thanks for following up.
Ok, So RAy should be worked out as the following:
RAy = (14,000kg x 9.81) + (250kg/m x 68m x 9.81) + (7000kg x 9.81) + (270kg/m x 60m x 9.81)
i.e. all the downward forces including the self weight of the vertical section of the crane.
Then continuing your list:
-14 m ×...
Understood sorry for not including enough details.
First I assumed that RAx = 0 as there are no forces acting in the x directions.
To work out RAy I calculated all the downward forces in Newtons, including the counterweight of 14 tonne, the concrete panel of 700kg and the self weight of the...
I am attempting the work out the forces acting on the base of a crane.
I am yet to come across forces at the base of a crane and am confused as to how to work out the reactionary/moment forces.
Any help would be greatly appreciated.