Portlandjib.com Righting Moment
 Home new on-line STEM curriculum NEW STEM CURRICULUM Arbor School Crane Class about Roosevelt HS STEM PROGRAM BRIDGES CRANES BRIDGE BEAMS ENGINEERING middle school projects crane projects Engineer's kit Portland Transformer montessori bridge beams Montessori Children's House scissors lift new crane new bridges contact FAQ

What's keeping the crane from overturning? The RIGHTING MOMENT.

Think of it as the turning force on one side of a big see-saw- the load and the boom being

on the other side.

The Righting Moment has 2 parts here. The moment of the crane body and the moment of the

counter weight-ct.wt.

First the crane body- it weighs 30# and the CG is 1.6ft from the pivot

M = F x D

M = 30# x 1.6ft

M = 48ft#

,

Now for the ctwt. The 4 bricks weigh 20# and are 42" (3.5ft) from the pivot.

So M = F x D

M = 20# x 3.5 ft = 70 ft#

TOTAL MOMENT   48 + 70 = 118ft#

That's a lot more than the 59.39ft# of OTM.

The ratio of RM/OTM = 1.96 In the crane industry a ratio of 1.5 is the minimum,

so with a ratio of almost 2, the crane will be stable with a ctwt. of 20#.

Now let's say we have to land (or place) the load farther away-right now

3.46ft is the farthest reach we have, so let's lengthen the boom by adding the jib.

First we release the catch that holds the jib to the underside of the boom,

and the jib swings down.

Then we lower the boom a bit and the jib rolls forward

We've removed the 4 load lines and replaced them with a single line that goes from

the load windlass, over the single sheave mounted on the small mast, and then down

to the tip of the jib. Now we can raise the jib.

Next we'll bolt it on to the boom and secure it with lines running from the tip of the jib,

over the small mast, to an anchor point in the middle of the boom.

Now we've got a boom that's 6.5 ft long, but will that change the Moment?

The force is the same -the 10# bucket, but the distance has increased.

cos 30 = Dist load /6.5ft

cos 30 = .866

M = F x D

M = 10# x 5.63 ft

M = 56.3 ft#

What about the boom? It still weighs 11#, but the CG, now that the jib has been extended, is 3.5ft

cos 30 = D boom / 3.5ft

cos 30 = .866

.866 = D boom/ 3.5ft

D boom = 3.03 ft

M = F x D

M = 11# x 3.03ft

M = 33.33ft#

TOTAL MOMENT  33.33 + 56.3 = 89.6ft#.     That's a lot more than the 60ft# we had before.

The RM is still 118ft#, so our new ratio of  RM/OTM is 1.37.

That's fine for a static situation- if the load is just hanging.

But if the load is moving up or down, or if the crane is slewing (swiveling) or if the wind

catches the load and makes it swing or ?????.

Then we've got a DYNAMIC load, and the 118ft# of RM might not be enough.

So we'll add 2 more bricks to make the ctwt. 30#

The moment of the crane body hasn't changed-it's still 48ft#, but the

moment of the ctwt. is now

M = F x D

M = 30# x 3.5ft

M = 105Ft#

TOTAL MOMENT  = 48ft# + 105ft# = 153 ft#

and the ratio of    RM/OTM 153/86 = 1.77  well above the minimum 1.5

Did you notice that the center of mass pointer in the photo was at 26".

That wasn't right.  I'll have a new page "how to find the CG" soon.