Crane anchor loadings

[landrover77]

Hi, I’m trying to assess the anchor bolt loads for a jib crane that we’re proposing to relocate. Structural calcs are not my field.

I can easily replicate the bolting currently in place to ensure safety, however for my own satisfaction I wanted to calculate the bolt loadings for the anchor bolts to check the epoxy resin we’re proposing to use is adequate. The concrete floor 600m thick with 25mm reinforcing.

However in doing so, I must be missing the approach taken in determining the division of loads on the bolts.

If we assume a floor mounted 1ton jib crane or boom length of 5m with 360º rotation.

This give a moment around the column of 1000*9.81*5 / 1000 = 49.05KNm (unfactored loads). Also for simplicity of the example ignoring rigging and boom weights.

Thus my assumption has been that if the base has 8 bolts, in the worse case rotation position, then 3 bolts are in tension preventing the tipping motion, 2 being on the centre line of rotation and 3 being on the side where the base plate of the column is in compression into the floor. The bolts are at an average distance (S) 0.4m from the columns centre of rotation.

Therefore assuming equilibrium where Ra is tensile force on bolts
Ra x 0.4 - 49.05KNm = 0. Thus Ra = 49.05/0.4 = 122.6KN.

Divide between 3 bolts this gives 41Kn / bolt.

We’re using M20 high tensile bolts, so I’ve no concerns over these, however the epoxy resin design loads are only 56KN. Which by the time apply a SF of 2.0 are nowhere near adequate.

However I know the crane install is fine and so what additions do I need to make to my simplistic analysis? Is there some derating assuming that the area of the base plate in compression takes some of the loading, and does this depend upon the flatness of the floor surface and grouting around the base?

Many Thanks for your help in advance.

 

[chicopee]

Right off the bat, I see a problem "...The concrete floor 600m thick..."

 

[BAretired]

Why is that a problem?

BA

 

[hokie66]

Need to define the centre of rotation. The column doesn't rotate about its centre, it rotates about a point near the edge of the base plate. Depends on the plate stiffness, stiffeners, etc. You could assume the bolt lever arm is the bolt spacing.

I would be very conservative in the depth of embedment when using adhesive bolts on a jib crane. And make sure the holes are cleaned well.

 

[BAretired]

You have one metric ton carried on a 5m boom. Your bolt circle appears to be 1000mm diameter, so the plate diameter is probably about 1200mm.

Your calculation of bolt tension is conservative. There will be 5 bolts in tension, but their tension is not equal. The strain and hence the stress in the bolts varies linearly with their distance from the centroid of compression.

The compression block is located at the outer edge of the plate on the side where the eccentric load is applied. The size of compression block is a trial and error process, but I would guess the centroid of compression block is about 400mm from the column center.

This means you have 2 bolts @ 400, 2 bolts @ 750 and one bolt @ 900 from centroid of compression. The centroid of the tensile force turns out to be 395 from centroid of compression.

M about compression block = 9.81*(5-0.4) = 45.1kN-m.
T (5 bolts) = 45.1/0.395 = 114.2 kN
T (outer bolt) = 114.2/4.33 = 26.4 kN (less than 41)

Using a FS of 2, T(ult) per bolt = 52.8kN.

An M20 A325 bolt has a factored tensile capacity of 156kN. Why would you anchor it with an epoxy with only a fraction of its capacity? I'll bet that the 56kN design load is an allowable load, not a factored load.

BA

 

[truckdesigner]

BAretired - care to elaborate with a rough sketch?


Regards.

 

[landrover77]

BA retired, many thanks i second truck designer, a sketch would be most useful.

Your correct bolt PCD is 1000m base plate 1200mm

should also pont out the typo, the floor is 600mm not 600m depth, as i'm sure most appreciated.

Thanks again

 

[landrover77]

BA retired

Having read through you first post in greater detail.

I wondered could you explain in more detail.

The compression area, is this taken as a cord 0.4m from the edge accross the whole base plate?

Having found the compression area then how do you then find the centroid for tension. is this from geometry alone using the remaining base plate area. ie base plate area - compression area?

Finally the 4.33 factor in the equation below, where is this from?
T (outer bolt) = 114.2/4.33 = 26.4 kN (less than 41)

In reference to the epoxy, i agree with you, its certainly the wak link but my understand is that its the prefered method rather than a menchanical fixing?

thanks again

 

[BAretired]

I found an error in my previous post, so please disregard those calculations.

Attached is a sketch showing a plan of the baseplate and the position of the load P. The anchor bolts are numbered from 1 to 8. The bolt circle is 1000mm in diameter and the plate is 1200mm diameter.

The applied moment is 5000P about the center of plate. The compression block is a circular segment shown hatched in the diagram. Under elastic conditions, the stress would vary linearly from the straight line to the curve. Under ultimate conditions, the stress would be uniform throughout the segment.

The centroid of the compressive force is estimated (conservatively) to be at 400mm from the center of plate and the total compression is C. The applied moment about the force C is 4600P.

Resisting the applied moment is the force in Bolts 1, 2, 8, 3 and 7. Bolts 4 and 6 are ignored. The force in each bolt is designated as Tn where n is the number of the bolt. T1 is the maximum bolt force for the orientation of the jib crane shown. Let's just call that T.

T2 = T8 = 7.5T/9
T3 = T7 = 4T/9

The total bolt force is T(1 + 15/9 + 8/9) = 32T/9 = 3.56T (this is where I went wrong before)

The c.g. of bolt force is at T(500 + 15/9 * 350)/3.56T = 304mm from center of plate or 704mm from force C.

3.56T * 704 = 4600P
T = 1.84P
C = P + 3.56T = 7.53P

If P = 10kN, T = T1 = 18.4kN (maximum anchor bolt force)

and C = 75.3kN (spread over shaded area).

The maximum tension is less than previously calculated. I believe the location of C is probably located near Bolt 5, so this calculation is conservative.

BA

 

[BAretired]

creed 12,

I don't disagree with the use of epoxy to anchor your bolts, but I was under the impression that some epoxies can develop the full tensile capacity of an A325 bolt with less than 600mm embedment.

BA

 

[landrover77]

BA retired, much appreciated.

I understand you approach.

Reference the epoxy i've been unable to find any porduct which provides tensile strengths approaching M16-24studs.

thanks again.

 

[BAretired]

creed12,

For a large baseplate like that, I assume you will be using levelling nuts under the plate. Are you planning to use a flowable grout? Do you know if there are grout holes in the baseplate?

I am just thinking it may be a little tricky to attempt to pack grout in from the perimeter of the plate.

BA

 

[ToadJones]

I believe using leveling nuts will change the approach BA used here.
Your anchor rods will take compressive loads on the compression side.

 

[BAretired]

Toad,

I don't think leveling nuts will change the approach very much. The c.g. of the three compression anchor bolts will still be 400mm from the center of the plate.

BA

 

[ToadJones]

If using leveling nuts, he can just use MC/I to determine the bolts loads, right?
M = moment
C = distance from centerline to bolt in question
I = moment of inertia of the bolt group = Sum of D^2 terms

 

[ToadJones]

...I would be hesitant to count on compression on the grout with leveling nuts being used

 

[csd72]

The maximum tension in a ring of bolts on levelling nuts is easy to get a formula for.

Treat the bolts as a continuous ring and then divide this area by the number of bolts.

From memory you get P = 4M/ND where M is the moment, N is the number of bolts and D is the diameter of the centreline of the bolts.

as others have said, i am not sure about using the epoxy anchors for compression.

 

[BAretired]

Toad,

By your method,

M = 5000P

A = 8
I = 2(500^2 + 2(350^2)) = 990,000

P/A = P/8 = 0.125P
My/I = 5000P*500/990,000 = 2.525P

Max. bolt tension = 2.4P
Max. bolt compression = 2.65P
Grout pressure = 0

I don't know why you are hesitant to consider compression on the grout, particularly if you use a non-shrink or slightly expansive, flowable grout.

csd72,

Using your method,

T = 4M/ND (shouldn't there be a pi somewhere?)
so T = 4*5000P/8*1000 = 2.5P nearly the same as Toad's ans.

So for a load of 10kN, the maximum bolt tension would be about 25kN if the grout is not participating in the compression.

BA

 

[ToadJones]

BA-
I am hesitant, because:

1). That's the type of guy I am
2). I have seen grout on large plates improperly installed, so for calc purposes, I ignore it if i can.

Having said that, on my normal columns I specify that leveling nuts and 1/4" leveling plate be installed and grouted to the proper elevation prior to steel erection. If done properly, column installation is a snap.

 

[ToadJones]

and yes, that is precisely how I'd calculate the tension....seems more conservative than your approach (not necessarily more correct)