Bolt tension on a cantilever beam?

[pete_gibson]

New here - so hi to all.

I have a cantilever beam that's 650mm long that has a mass of 100kg suspended from it. At the fixed end of the beam there is a plate that is bolted in place by 4 bolts. The top 2 bolts are in tension and I'm trying to work out the approximate force/ tension these bolts would see. The bolt centres are above the neutral axis by 75mm.

It's been a long time since doing calcs and I can't remember the equation(s) for this.

Any help appreciated.
Pete.

[Kelly_Bramble]

Here's the axial bolt force exerted by torque calculator.

https://www.engineersedge.com/calcul...orque_calc.htm

Shear force/stress on pin or bolt.

https://www.engineersedge.com/materi...hear_calcs.htm


Look here for your loading configuration.. Note that bolt torque created axial force should exceed any external loading induced in service.

https://www.engineersedge.com/beam_calc_menu.shtml

If the beam loading configuration you need to calculate is not there model it with the 2D Statics Modeler and Calculator.

https://www.engineersedge.com/calcul...d-modeling.htm

[Hudson]

A ball park method:

You have a moment on the beam that is transferred to the plate. Presumably you know the distance from the bolt holes to the bottom edge of the plate.

I would calculate the tension force on the bolts by assuming the plate wants to pivot about the bottom edge. That assumption in mind, the strain on the top bolts is going to be proportionally more than the bottom bolts as the distance from the bottom edge varies to the bottom bolts. (Similar triangles and all.) So, if the top bolts are twice as far from the edge as the bottom bolts the strain is twice as much – put in your own proportions.


The strain and load are proportional in the bolts. You know the total moment on the beam is equal to the moment on the mounting plate. You know the distance of the bottom edge to the bolt holes and the proportion of load carried by each set of bolts.

You can set up the equations, right?

I am always as concerned with the wall anchors as the bolts. Apply safety factors as the situation requires. We're also assuming the plate is stiff enough not to bend significantly. I make no guarantees regarding this method.

[pete_gibson]

Thanks for the replies. The ball park method suits as I'm trying to make sure I have a reasonable safety factor for the bolts. The plate is sufficiently stiff and is bolted into a steel upright via a nut. I know the plate geometry (see picture) but am not sure of the equation I would need here.
Capture.PNG

[jboggs]

Its a simple lever. Sum of moments about the lower bolt centerline = zero.

[Kelly_Bramble]

Use this calculator:

https://www.engineersedge.com/beam_b...d-stress-4.htm

Set Fx = 0 determine the worst case stress at a and b use bolt with FOS at least 2X.

[Hudson]

You need help with the equations?

Moment from arm = 100 kg x 650 mm = 65000 kg mm

(22.5 mm x Force on bottom bolts) + (152.5 mm x Force on top bolts) = 65000 kg mm in the opposite direction as the sum of the moments equals zero if there is not rotation.

Comparing the bolt hole distances to the bottom edge, 22.5 mm and 152.5 mm

152.5/22.5 = 6.778 by similar triangles, strain on top bolts is 6.778 times the force on the bottom bolt. Hence the stress is 6.778 times as much. Substituting:

(22.5 x Force on bottom bolts) + (152.5 mm x 6.778 x Force on Bottom bolt) = 65000 kg mm

Combining, (22.5 + 1033.65) x Force on bottom bolts = 65000 kg mm

Force on bottom bolts = 65000 kg mm/(22.5 + 1033.65) mm = 61. 6 kg

There are two bottom bolts each has 30.8 kg of estimated tension load from the moment. There is also a significant tension load in the bolt from tightening the bolt to the wall that must be estimated from the tightening torque and a friction factor or some other method.

The top bolts have an estimated tension load from the moment 6.778 times the bottom bolts or (6.778 x 30.8 kg) or 209 kg each. This too is added to the installation tension.

The shear load at the wall is in the vertical direction and is 100 kg shared by all four bolts or is carried by the clamping friction (preferred).

If you tighten an SAE grade 5.8, M10x1.5 bolt to the 380MPa proof load the clamp load will be over 21000 N. or 2140 kg. The load from the moment is less than 10% additional. With larger bolts, better grades and finer threads bolt stress drops.

Now you owe me a beer and whatever your local specialty is. Plus, you have to turn your attention to the holding power of the wall anchor.

[pete_gibson]

Now you owe me a beer and whatever your local specialty is. Plus, you have to turn your attention to the holding power of the wall anchor.

You're right, I do. Thanks for your help. Much appreciated.