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Salam

Dear Sefpians!!

I have to design a foundation for a 15m high floodlight pole, dia of the pole is 10 inches, pole is welded with the baseplate (provided by manufacturer), how the forces will be transferred to anchors. what forces will be on anchors. how forces will be transferred from anchors to foundation. anchors will be embedded in-situ concrete.

Regards

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Hello Waqar,

Do you have the baseplate size from vendor. If yes, what is the bolt pattern (Bolt holes and distance). Also, did the vendor give you design loads for anchor bolts? If yes, great. If you don't have anchor bolt loads, you need to manually calculate wind on the floodlights and pole and the reaction at the base and design anchor bolts and foundation accordingly.

If you have the bolt pattern, I can show you how to resolve the forces to anchor bolts and other things that you should keep in mind. For foundation, a square foundation with a concrete pedestal would do the job.

Thanks.

 

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Thanks Sir umar

pole has 10 inches dia and base plate is 15in x 15in, with four anchors option on four sides. no loads for anchors are provided by manufacturer, an approx weight of pole is given 350Kgs mounted with 2x lights panels weight 80Kgs . wind speed 100mph for islamabad area. i am not clear if wind will be applied on one side of the pole (half area) or suction may also be applied.

Regards

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So it's no very heavy. Applying 100mph, what is the factored shear and moment that you get at the base of the pole (Use the exposed area of lights and pole to calculate that). I would also like to know the center to center distance of proposed bolts as well.

Thanks.

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i applied a wind pressure of 30psf on one side of the pole and got 1.6M=40k-ft and shear 1.6V=1.76, m not sure about the factor 1.6. anchors are 12" apart at center. 

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You should use the load combination with wind loads. I don't know what the latest is for ACI so make sure your factors reflect wind not live load.

Assuming we go with 4 bolts as proposed at 12" c/c, You need to design your anchor bolts for tension : T = 40 kip ft / ( 1 ft * 2 bolts) = 20 kips per bolt on tension side. I have used 2 bolts as the other two bolts would be in compression.

Shear per bolt would be (assuming 2 bolts take shear) = 1.76/ 2 = Almost 1 kips per bolt.

Now with these forces, you need to size your bolt based on ACI Appendix D. Start with a 1" bolt. Have you used Appendix D before? If yes, great. If no, give it a try for a day and if you can't figure out let me know.

Thanks.

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5 hours ago, UmarMakhzumi said:

Shear per bolt would be (assuming 2 bolts take shear) = 1.76/ 2 = Almost 1 kips per bolt.

Why r u asumimg only two bolts effective in shear? R u assuming bolts in tension would not be effective in shear or what? 

Also why earth quack load is not being used to calculated base forces?

@waqar saleem as umer said your bolts will not only be designed for tension only but also for some failure moods associated with bolts i.e. breakout failure, pryout failure etc. so appendix D of ACI deals with it. 

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3 hours ago, Engr Waqas said:

Why r u asumimg only two bolts effective in shear? R u assuming bolts in tension would not be effective in shear or what?

Most of the time, base plate holes are oversized for easier installation. That makes it hard to determine how many bolts would be effective in shear so that is why 50% of the bolts are ignored. The one in tension are considered to take shear. For the case here, shear is very small and can also be ignored.

3 hours ago, Engr Waqas said:

Also why earth quack load is not being used to calculated base forces?

Earthquake load can be checked and the max of earthquake or wind should be considered for all design situations. However, looking at the weight here, you can tell that earthquake would not govern.

Thanks.

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In calculating design forces, have you considered the application of point load on a panel, which will support lights, at the top of the pole, in addition to the UDL on the pole itself.

Is it possible that only two bolts will be effective in resisting moment for a wind direction in which two bolts will lie on the neutral axis, and the moment arm will be sqrt(2) x 12.

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4 hours ago, BAZ said:

In calculating design forces, have you considered the application of point load on a panel, which will support lights, at the top of the pole, in addition to the UDL on the pole itself.

Is it possible that only two bolts will be effective in resisting moment for a wind direction in which two bolts will lie on the neutral axis, and the moment arm will be sqrt(2) x 12.

This is a very good point and would result in higher tension. This is a very likely limit state.

Thanks.

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Eq will likely not govern, as a rough value 1kpax1.6 is the wind load. You get mu=wl2/2 and vu=1.6kn. T=C on each side will be mu/arm between bolts. For each bolt tension will be T/2 and shear will be Vu/2 (only 2 bolts resist shear). Remember there will be tension shear interaction for bolt design plus use Appendix D of ACI. This is a simple example. For complex base plate patterns use RisaBase software for biaxial design or make a model with assumption that base plate is stiff (use very thick shell) and frame elements as bolts. Or use the formula to calculate tension compression on bolts from applied moment and P which depends on distance of bolt from centroid divided by group x2 or y2. See any foundation book.

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