Finding Tension and Shear Loads of Seismic Snubbers

[vidividi12]

Hi,

I am asking this question as thinking that maybe there is someone who knows the answer.

When I want to calculate tension and shear loads of cables at picture below, I am using equations below;

Cable Load (Fc = Horizontal seismic force / cos60°)
Tension Load (T = Fc x sin60x1,3)
Shear Load (V = Fc * cos60x1,3)

However, I want to use seismic snubbers to mount my panelboard on wall surface like picture below. How can I calculate tension and shear loads?

 

[Carultch]

I will approach this problem, only considering the dead weight of the fixture. In reality, there is a seismic calculation with an equivalent lateral force, that also governs the reaction loads on the anchors.

It is reasonable to assume shear is equally distributed among the anchors. Thus, assuming symmetric snubbers on the left, the shear in each anchor = weight/8. One reason why shear wouldn't be equally distributed, is if you had a load that applied a rotating moment to the fixture CW or CCW, as viewed from the front. This would happen if the contents of the fixture were either left-heavy or right-heavy.

Number the left/right fastener pairs 1 thru 4, from bottom to top. Call the tension in each fastener, T1 thru T4. Two fasteners have tension T1, and so forth, so you'll see a factor of 2 in front of T1 thru T4, when we use them. Call their vertical positions from the bottom, y1 thru y4.

The fixture will compress against its bottom edge. Call the total compression force C. We ultimately don't care what C is, but it may be an important placeholder for the algebra.

Force balance in the front-back direction:
C = 2*(T1 + T2 + T3 + T4)

Take moments about the bottom edge, so that C doesn't appear in the following equation. The center of mass of your fixture, is offset from the wall a distance d. The total moment about the bottom edge due to the fixture weight (W), is W*d.
W*d = 2*(T1*y1 + T2*y2+ T3*y3 + T4*y4)

The approximation with the tension calculation, is that the fixture is infinitely rigid, compared to the stiffness of the fasteners, and that the fasteners are linear-elastic in their response to tension. We're also neglecting rotation between the fixture and the snubbers. If we did consider the fixture fastener rotating within the snubbers, we'd have the same reasoning, but in two layers of calculations. This all means that stretch distance in each fastener is proportional to its y-value, for geometric consistency. With the linear elastic assumption, it also means the y-value is proportional to the the tension in the fastener. Define a constant of proportionality k, to relate tension to y-position. The value of k is not necessarily the spring constant of the fastener, don't let the name deceive you.

This generates the following equations:
T1 =k*y1
T2 =k*y2
T3 =k*y3
T4 =k*y4

Replace T1 thru T4 in the moment equation:
W*d = 2*k*(y1^2 + y2^2+ y3^2 + y4^2)

Solve for k:
k = (W*d)/(2*(y1^2 + y2^2+ y3^2 + y4^2))

Plug in your given information for W, d, and the y's. Get your value of k. Then multiply by the y-positions to get each tension. This gives the tension in each fastener.

 

[PaulMmn]

I have a question/comment about the 4 guy wires on the upright cabinet. It appears that the guy wires are all at a 60 degree angle with respect to vertical. Are they also at a 45 degree angle to the corners of the cabinet?

I'm imagining that during a 'seismic event' they are designed to keep the cabinet from moving out of the vertical position. This assumes the ceiling they're anchored to doesn't move with respect to the floor.

Wouldn't having a tension spring in each cable run provide some flexibility in case the building doesn't move as a unit? Something like this:

 

[SceneryDriver]

I will approach this problem, only considering the dead weight of the fixture. In reality, there is a seismic calculation with an equivalent lateral force, that also governs the reaction loads on the anchors.

It is reasonable to assume shear is equally distributed among the anchors. Thus, assuming symmetric snubbers on the left, the shear in each anchor = weight/8. One reason why shear wouldn't be equally distributed, is if you had a load that applied a rotating moment to the fixture CW or CCW, as viewed from the front. This would happen if the contents of the fixture were either left-heavy or right-heavy.

Number the left/right fastener pairs 1 thru 4, from bottom to top. Call the tension in each fastener, T1 thru T4. Two fasteners have tension T1, and so forth, so you'll see a factor of 2 in front of T1 thru T4, when we use them. Call their vertical positions from the bottom, y1 thru y4.

The fixture will compress against its bottom edge. Call the total compression force C. We ultimately don't care what C is, but it may be an important placeholder for the algebra.

Force balance in the front-back direction:
C = 2*(T1 + T2 + T3 + T4)

Take moments about the bottom edge, so that C doesn't appear in the following equation. The center of mass of your fixture, is offset from the wall a distance d. The total moment about the bottom edge due to the fixture weight (W), is W*d.
W*d = 2*(T1*y1 + T2*y2+ T3*y3 + T4*y4)

The approximation with the tension calculation, is that the fixture is infinitely rigid, compared to the stiffness of the fasteners, and that the fasteners are linear-elastic in their response to tension. We're also neglecting rotation between the fixture and the snubbers. If we did consider the fixture fastener rotating within the snubbers, we'd have the same reasoning, but in two layers of calculations. This all means that stretch distance in each fastener is proportional to its y-value, for geometric consistency. With the linear elastic assumption, it also means the y-value is proportional to the the tension in the fastener. Define a constant of proportionality k, to relate tension to y-position. The value of k is not necessarily the spring constant of the fastener, don't let the name deceive you.

This generates the following equations:
T1 =k*y1
T2 =k*y2
T3 =k*y3
T4 =k*y4

Replace T1 thru T4 in the moment equation:
W*d = 2*k*(y1^2 + y2^2+ y3^2 + y4^2)

Solve for k:
k = (W*d)/(2*(y1^2 + y2^2+ y3^2 + y4^2))

Plug in your given information for W, d, and the y's. Get your value of k. Then multiply by the y-positions to get each tension. This gives the tension in each fastener.

Your math looks correct in theory. In practice however, the four lines will never share the load equally. Minuscule differences in line length mean that in reality, three lines will carry the load and the fourth will be functionally slack. If this is to ensure the cabinet doesn't tip over in a seismic event, it really shouldn't matter. Why not just brace the cabinet with some Unistrut? Seems easier and probabaly cheaper than all that rigging hardware.


SceneryDriver

 

[Carultch]

Your math looks correct in theory. In practice however, the four lines will never share the load equally. Minuscule differences in line length mean that in reality, three lines will carry the load and the fourth will be functionally slack. If this is to ensure the cabinet doesn't tip over in a seismic event, it really shouldn't matter. Why not just brace the cabinet with some Unistrut? Seems easier and probabaly cheaper than all that rigging hardware.

SceneryDriver

The tension lines are not part of the definition of the problem in question. It was an example of calculating forces for which the OP knew the formula. The problem was defined in second image showing side brackets, instead of overhead cables, as the method of securing the equipment.