# Calculating upright loads to base plate

1. Jul 13, 2009

### 3thanol

Hello PF I hope you can give me some guidance. This is something that I am getting more and more on my desk, to be honest I don't understand it enough to give an answer. I have searched the net looking for an answer but can only find details on fulcrum points class 1,2 & 3 which does not seem to be what I receive. Maybe I am overlooking or missing something. I don't claim to be brilliant at maths but this is really bugging me, I am sure it has a relatively simple formula.

Anyway I have attached a picture of a quick drawing I done. I have i.e. square base plate with 4 fixings points and an upright in the centre x high in this example 3mtrs high. The upright has a force of 10kN applied pushing it directly horizontal at the very top.

What is the load transmitted to the base plate and ultimately how can I then workout the load at the relative fixings points, I understand the 2 fixings opposite to the direction of force will be taking most Tensile load but am at a lose to know where to really start.

I would hazard a guess that the load is transmitted to the base plate and would become a value of xxxKNm and would pivot around the fulcrum, lifting the opposite side which would then become a tensile load again, with maybe slight shear given the slight oblique motion possibly?

Any help would be greatly appreciated, Thank you.

#### Attached Files:

• ###### Fulcrum on base plate.jpg
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2. Jul 13, 2009

### PhanthomJay

I'm not sure of what you mean by the fulcrum point. The 10 kN load applied 3 m from the base produces a 30 kN-m couple at the base plate. But what is the base plate attached to? If it were a plate bolted at the 4 locations shown (0.7m spacing) and double nutted (one nut above and one nut below the base plate) to a foundation, them 2 of the bolts would be in tension, and the other 2 in compression. The tensile force in the two bolts in tension would be 30/0.7) = 43 kN or about 21.5 kN/bolt. The compressive force in each of the other 2 bolts would have the same magnitude. And each bolt would see 2.5 kN shear. But your question needs clarification.

Last edited: Jul 13, 2009
3. Jul 14, 2009

### 3thanol

Thanks Phantomjay. The Base plate is steel which is fixed to concrete with bolts, the fulcrum or pivot point, I would have thought is the edge of the base plate where the application would pivot around should it be toppled over.

Would the measurement not be from the back fixing centres to that edge 0.85 and if the case would this take the other two bolts out of compression and useful and a measure of 0.15 be obtainable, which could contribute to clamp the whole thing down.

so you would have 30/0.85/2 = 17.62kN per back fixing 30/0.15/2 = 100kN per front fixings. then would have to work out what percentage each bolt is clamping.

Also how did you come to that shear load if I may ask.

Many thanks.

4. Jul 14, 2009

### PhanthomJay

I assumed double nutted bolts (the lower nuts under the base plate are usually called leveling nuts); with this assumption, the base plate sits on the nuts, not the concrete, and bolt forces are determined as I noted, using the center of the base plate as the neutral axis, to yield the 21.5 kN tension in each of the 2 tension side bolts, and 21.5 kN compression in each of the 2 compression side bolts.
If one were to assume that the base plate is directly resting on the concrete, and bolted to the concrete with just single nuts on the bolts above the plate, the analysis woul be a bit different, as you'd have tensile forces on two of the bolts, and a triangular distribution of compressive stress acting on the other side, from the concrete bearing stresses. Such analysis for this case would be simuilar to reinforced concrete design with tension in the rebar and compression in the concrete, with the compressive force in the bolts ignored. The first method yields a conservative result if this is the case. Resist the temptation to use the far edge of the plate as the pivot for this latter case; all bolts will not be in tension.

The shear load is assumed equally distributed amongst each bolt (10 kN/4 = 2.5 kN).