Basic Overturning Question

[jStadts]

I have a basic foundation overturning question. A machine is anchored to a foundation with a 42kip-ft counterclockwise moment on the plate at the left side. The machine weight and concrete weight produce a 21.85 kip force down at the center. Will this foundation overturn? Do I take the moment at the left edge of the foundation or do I take it at the same point as the 42 kip-foot location. The foundation is 99” wide and the plate is 51”.

 

[driftLimiter]

Sum the moments about the most extreme point where tipping can occur. Include foundation weight, use 0.6D Load combinations if the overturning is due to wind or seismic.

The concentrated moment location is not needed when summing the moments simply add the concentrated moment using a rational sign convention.

 

[rlmengineer]

Take moments about the left edge of the foundation because that's where the foundation would rotate if it were to overturn.

Check the bearings capacity of the soil under tge slab and check the required reinforcement in the slab.

 

[jStadts]

Sum the moments about the most extreme point where tipping can occur. Include foundation weight, use 0.6D Load combinations if the overturning is due to wind or seismic.

The concentrated moment location is not needed when summing the moments simply add the concentrated moment using a rational sign convention.

That's what my first thought was but was overthinking it i guess.

So (14.85kips + 7kips) * (4.125ft) = 90.13kip-ft
(42 kip-ft * 2(SF)) = 84 kip-ft

90.13kip-ft > 84kip-ft so it will not overturn.

Thank you!

 

[jStadts]

Take moments about the left edge of the foundation because that's where the foundation would rotate if it were to overturn.

Check the bearings capacity of the soil under tge slab and check the required reinforcement in the slab.

That was my first thought but must've been overthinking it!

Thank you!

 

[BAretired]

 

[BAretired]

I have a basic foundation overturning question. A machine is anchored to a foundation with a 42kip-ft counterclockwise moment on the plate at the left side. The machine weight and concrete weight produce a 21.85 kip force down at the center. Will this foundation overturn? Do I take the moment at the left edge of the foundation or do I take it at the same point as the 42 kip-foot location. The foundation is 99” wide and the plate is 51”.

View attachment 4585

In my last post, I misinterpreted your diagram.

I think you are considering a footing 99" long, loaded with 21,850# at its midpoint and a 42'k moment applied at the left end of the 51" plate. You have a question mark at the left end of the footing, pointing to a curved arrow which presumably represents a moment. Unless there is an applied moment or a support, there can't be a moment at that location. In that case, the footing must resist the vertical load P=21.85k and the moment M=42'k.

The location of the 42'k moment affects the bending moment and deflection of the footing, but if deflections are neglected, it does not affect the distribution of soil pressure.

So (14.85kips + 7kips) * (4.125ft) = 90.13kip-ft
(42 kip-ft * 2(SF)) = 84 kip-ft

90.13kip-ft > 84kip-ft so it will not overturn.

I agree with your conclusion, but your reasoning is wrong. 42*2(SF) gives units of kip-ft^3, which is nonsense.

The usual procedure is to determine eccentricity = M/P = 42'k/21.85k = 1.92' = 23", so move the load P left by 23" and remove moment M = 42'k.
Since the eccentricity is within the 99" length of footing, overturning is not an issue, but the eccentricity exceeds 99/6 = 16.5" (half of the kern), which means that the effective length of footing is reduced from 99" to 3(99/2 - 23) = 79.5". If that is not clear, let me know.

Soil pressure will have an average value of 21,850/79.5 = 275 pounds/lineal foot, but because the load is acting at the kern of the footing, soil pressure varies from 0 to 2*275 = 550 pounds per lineal foot. If the footing has a width of w, the maximum soil pressure is 550/w psf.

 

[EngDM]

In my last post, I misinterpreted your diagram.

I think you are considering a footing 99" long, loaded with 21,850# at its midpoint and a 42'k moment applied at the left end of the 51" plate. You have a question mark at the left end of the footing, pointing to a curved arrow which presumably represents a moment. Unless there is an applied moment or a support, there can't be a moment at that location. In that case, the footing must resist the vertical load P=21.85k and the moment M=42'k.

The location of the 42'k moment affects the bending moment and deflection of the footing, but if deflections are neglected, it does not affect the distribution of soil pressure.


I agree with your conclusion, but your reasoning is wrong. 42*2(SF) gives units of kip-ft^3, which is nonsense.

The usual procedure is to determine eccentricity = M/P = 42'k/21.85k = 1.92' = 23", so move the load P left by 23" and remove moment M = 42'k.
Since the eccentricity is within the 99" length of footing, overturning is not an issue, but the eccentricity exceeds 99/6 = 16.5" (half of the kern), which means that the effective length of footing is reduced from 99" to 3(99/2 - 23) = 79.5". If that is not clear, let me know.

Soil pressure will have an average value of 21,850/79.5 = 275 pounds/lineal foot, but because the load is acting at the kern of the footing, soil pressure varies from 0 to 2*275 = 550 pounds per lineal foot. If the footing has a width of w, the maximum soil pressure is 550/w psf.

I think 2(SF) is what OP is naming safety factor, not square foot to make it ft^3.

Depending on how OP factored his loads, factoring the 42 may be inappropriate (if you factored the 42 but not the others).

 

[BAretired]

I think 2(SF) is what OP is naming safety factor, not square foot to make it ft^3.

Depending on how OP factored his loads, factoring the 42 may be inappropriate (if you factored the 42 but not the others).

Thanks EngDM! I should have thought of that interpretation. I believe you are correct.