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Footing overturning calculations 2
- Thread starter jeffhed
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I understand the 0.6D + W equation and how it results in approximately the old 1.5 factor of safety, but the 0.6D + 0.7E as mentioned in the original post is still unclear to me.
I read this as 0.6D, no live load, 0.7E where E is calculated only for dead load. If the dead load is small compared to a fully loaded structure then I think this condition is generally not significant.
If my assumption is true then what do you do when the maximum uplift is from: D + 0.75L + 0.75*0.7E
It seems to me maximum loads up and down in this case would not include a factor of safety for overturning.
Regards,
-Mike
I read this as 0.6D, no live load, 0.7E where E is calculated only for dead load. If the dead load is small compared to a fully loaded structure then I think this condition is generally not significant.
If my assumption is true then what do you do when the maximum uplift is from: D + 0.75L + 0.75*0.7E
It seems to me maximum loads up and down in this case would not include a factor of safety for overturning.
Regards,
-Mike
mrMikee - I'm not sure what exactly you are asking.
Where do you get the load combination: D + .75L + .75*.7E??
Is this considering that L and E are "transient" loads per IBC? I don't know for sure if L is considered transient, but even if it is, the overturning "check" is found in the .6D + .7E combination only.
Where do you get the load combination: D + .75L + .75*.7E??
Is this considering that L and E are "transient" loads per IBC? I don't know for sure if L is considered transient, but even if it is, the overturning "check" is found in the .6D + .7E combination only.
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- #6
mrMikee,
IBC 2006 ADS load combination EQN16-13 is D+H+F+0.75*(W or 0.7E)+0.75L+0.75*(Lr or S or R). I think in this condition, you will find that the overturning is mcuh less because of all of the additional gravity loads you have in this combination. Your overturning FS would be pretty high, much higher than 0.6D+W or 0.6D+0.7E which typically has the lowest overturning FS.
IBC 2006 ADS load combination EQN16-13 is D+H+F+0.75*(W or 0.7E)+0.75L+0.75*(Lr or S or R). I think in this condition, you will find that the overturning is mcuh less because of all of the additional gravity loads you have in this combination. Your overturning FS would be pretty high, much higher than 0.6D+W or 0.6D+0.7E which typically has the lowest overturning FS.
The 0.75 factor doesn't apply to the 0.7 E earthquake load, even when there are two or more transient loads (IBC 2003 1605.3.1.1).
The 0.6D + W equation does not have a factor of safety included against overturning. The 0.6 D + W equation is part of the allowable stress design set of equations (IBC 2003 1605.3.1). The loads you get from the basic load combinations are service loads. You then size your member for an allowable stress based on those loads. The allowable stress will be approximately 0.6 * stress at failure.
The 0.6 factor in the "0.6D + W" load combination comes from the fact that the Dead Load is often overestimated by engineers (understandably). In the case of this load combination, the more conservative we are with our dead load, the less conservative we are with the uplift from wind load.
Once you have your SERVICE uplift from your "0.6D + W" combination, you then have to come up with an allowable factor of safety against overturning. A typical F.S. against overturning where I work is 2.0, but can be adjusted based on engineering judgement.
The 0.6D + W equation does not have a factor of safety included against overturning. The 0.6 D + W equation is part of the allowable stress design set of equations (IBC 2003 1605.3.1). The loads you get from the basic load combinations are service loads. You then size your member for an allowable stress based on those loads. The allowable stress will be approximately 0.6 * stress at failure.
The 0.6 factor in the "0.6D + W" load combination comes from the fact that the Dead Load is often overestimated by engineers (understandably). In the case of this load combination, the more conservative we are with our dead load, the less conservative we are with the uplift from wind load.
Once you have your SERVICE uplift from your "0.6D + W" combination, you then have to come up with an allowable factor of safety against overturning. A typical F.S. against overturning where I work is 2.0, but can be adjusted based on engineering judgement.
ITHW, respectfully I disagree with you.
IBC (2000) does not specify any ADDITIONAL safety factor for overturning checks. You simply follow the load combinations.
For seismic - see 1617.4.5. This simply refers to overturning moments at each story based upon Fi at each level. You still use the various combinations, plugging in Fi as QE in section 1617.1.1 along with the vertical seismic component of .2SDSD and you have E to use in the combinations.
For wind, see 1609.1.3 for overturning. Again, no mention of any required 1.5 safety factor. Simply use a slightly underestimated dead load and use within the combinations - specifically 0.6D + W. For the alternative load combinations of 1605.3.2, they require you to use 0.67D instead of D within the load combinations.
If you have any specific code reference mandating an additional 1.5 safety factor, I'd love to be shown, but I'm not aware of any in the current codes.
The commentary for ASCE 7-02, section 2, states: [red]"Load combinations (7) and (8) [these are the .6D+W combos] were new to the 1998 edition of ASCE 7. They address the situation in which the effects of lateral or uplift forces counteract the effect of gravity loads. This eliminates an inconsistency in the treatment of counteracting loads in allowable stress design and strength design, and emphasizes the importance of checking stability."[/red]
There used to be confusion between strength and ASD because of the traditional 1.5 safety factor. i.e. how do you utilized a 1.5 factor for OT with factored loads? This was removed in the recent codes and instead, was included in the load combinations.
IBC (2000) does not specify any ADDITIONAL safety factor for overturning checks. You simply follow the load combinations.
For seismic - see 1617.4.5. This simply refers to overturning moments at each story based upon Fi at each level. You still use the various combinations, plugging in Fi as QE in section 1617.1.1 along with the vertical seismic component of .2SDSD and you have E to use in the combinations.
For wind, see 1609.1.3 for overturning. Again, no mention of any required 1.5 safety factor. Simply use a slightly underestimated dead load and use within the combinations - specifically 0.6D + W. For the alternative load combinations of 1605.3.2, they require you to use 0.67D instead of D within the load combinations.
If you have any specific code reference mandating an additional 1.5 safety factor, I'd love to be shown, but I'm not aware of any in the current codes.
The commentary for ASCE 7-02, section 2, states: [red]"Load combinations (7) and (8) [these are the .6D+W combos] were new to the 1998 edition of ASCE 7. They address the situation in which the effects of lateral or uplift forces counteract the effect of gravity loads. This eliminates an inconsistency in the treatment of counteracting loads in allowable stress design and strength design, and emphasizes the importance of checking stability."[/red]
There used to be confusion between strength and ASD because of the traditional 1.5 safety factor. i.e. how do you utilized a 1.5 factor for OT with factored loads? This was removed in the recent codes and instead, was included in the load combinations.
- Thread starter
- #10
I know I originally posted this thread concerning spot footing overturning, but the factor of safety against overturning also applies to retaining walls. IBC 2006 1806.1 states that "Retaining walls shall be designed to ensure stability against overturning, sliding, excessive foundation pressure and water uplift. Retaining walls shall be designed for a safety factor of 1.5 against lateral sliding and overturning." Doesn't mention if this factor of safety is included in the load combinations or is in addition to them. I agree with JAE, seems pretty ultra conservative to apply the factor of safety of 1.5 after using the load combinations.
My load combination is a simplification of equation 6 under section 2.4.1 in ASCE 7-02 which I assume is the same as Equation 16-10 in the IBC when L and E are considered transient loads and the 0.75 factor is used. The ASCE equation also shows that the 0.75 factor is applied to 0.7E too.
My conclusion is that the overturning "checks" are only in the 0.6D equations only as JAE said. I design elevated bins where the ratio of live load to dead load is probably much higher than with typical building structures, and they are relatively tall when compared to plan view dimensions. In higher seismic areas the maximum uplift at the column bases is when fully loaded.
The reason for my concern is that I need to communicate to foundation engineers the loads at the base of my structures that are applied under the different load conditions, and I want to be sure that all assumptions including overturning are clear. I think I need to change how I am currently doing this, but am not yet sure how.
This thread made me think about these issues again.
Thanks,
-Mike
My conclusion is that the overturning "checks" are only in the 0.6D equations only as JAE said. I design elevated bins where the ratio of live load to dead load is probably much higher than with typical building structures, and they are relatively tall when compared to plan view dimensions. In higher seismic areas the maximum uplift at the column bases is when fully loaded.
The reason for my concern is that I need to communicate to foundation engineers the loads at the base of my structures that are applied under the different load conditions, and I want to be sure that all assumptions including overturning are clear. I think I need to change how I am currently doing this, but am not yet sure how.
This thread made me think about these issues again.
Thanks,
-Mike
- Thread starter
- #12
mrMikee,
Would it be possible to give your loads for uplift, horizontal, etc for each load type (D, L, W and E)? That way the foundation designer can take the loads and put them in the load combinations he is using (strength or ASD). I recently dealt with the same kind of issues on an asphalt plant foundation. The equipment included 75 foot tall silos that were 10' diameter and also some cold feed bins over a conveyer belt that had a dynamic load at the top. The problem I had was the equipment designer was giving me maximum uplift, gravity, and horizontal loads only. But when it came time to design the footings and anchor bolts (strength design for footings and ACI cracked concrete calculations for the bolts), I needed the individual loads so I could put them in the strength design load combinations. This works out great for our firm for projects like this and prefabricated steel buildings when we can get the loads broken out like that. Steel building compnaies always provide load outputs like that in their calculation books, you just have to dig to find them. However, sometimes the equipment designers have limited ability to provide these loads.
Would it be possible to give your loads for uplift, horizontal, etc for each load type (D, L, W and E)? That way the foundation designer can take the loads and put them in the load combinations he is using (strength or ASD). I recently dealt with the same kind of issues on an asphalt plant foundation. The equipment included 75 foot tall silos that were 10' diameter and also some cold feed bins over a conveyer belt that had a dynamic load at the top. The problem I had was the equipment designer was giving me maximum uplift, gravity, and horizontal loads only. But when it came time to design the footings and anchor bolts (strength design for footings and ACI cracked concrete calculations for the bolts), I needed the individual loads so I could put them in the strength design load combinations. This works out great for our firm for projects like this and prefabricated steel buildings when we can get the loads broken out like that. Steel building compnaies always provide load outputs like that in their calculation books, you just have to dig to find them. However, sometimes the equipment designers have limited ability to provide these loads.
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- #13
The bottom line is that the ASCE 7 wind loads are service loads.
If you were designing per LRFD, it is the classic:
Resistance Factor * Actual Capacity > Load Factor * Service Load
which is the same as:
Actual Capacity > (Load factor / resistance factor) * Service Load
Because the (Load factor / resistance factor) = F.S., you have:
Actual Capacity > F.S. * Service Load
If you are designing per ASD, it is the classic:
Actual Capacity / F.S. > Service Load
Made up example
Solved by LRFD:
Service overturning Moment: 10 k-ft
F.S. = 1.6 (This factor of safety is inluded in the load combinations)
Ultimate overturning moment = 1.6 * 10 k-ft = 16 k-ft
Required actual resisting moment capacity > Ultimate overturning moment, so:
Required actual resisting moment capacity = 16 k-ft
Solved by ASD:
Service overturning resisting moment = 10 k-ft
F.S. = 1.6 (The engineer has to come up with this factor of safety)
Required actual resisting moment capacity / F.S. > Service overturning moment, so:
Required actual resisting moment capacity = Service overturning moment * F.S, so:
Required actual resisting moment capacity = 10 k-ft * 1.6 = 16 k-ft
Summary, you need to use an allowable resistance to overturning, which requires you providing the appropriate factor of safety.
If you were designing per LRFD, it is the classic:
Resistance Factor * Actual Capacity > Load Factor * Service Load
which is the same as:
Actual Capacity > (Load factor / resistance factor) * Service Load
Because the (Load factor / resistance factor) = F.S., you have:
Actual Capacity > F.S. * Service Load
If you are designing per ASD, it is the classic:
Actual Capacity / F.S. > Service Load
Made up example
Solved by LRFD:
Service overturning Moment: 10 k-ft
F.S. = 1.6 (This factor of safety is inluded in the load combinations)
Ultimate overturning moment = 1.6 * 10 k-ft = 16 k-ft
Required actual resisting moment capacity > Ultimate overturning moment, so:
Required actual resisting moment capacity = 16 k-ft
Solved by ASD:
Service overturning resisting moment = 10 k-ft
F.S. = 1.6 (The engineer has to come up with this factor of safety)
Required actual resisting moment capacity / F.S. > Service overturning moment, so:
Required actual resisting moment capacity = Service overturning moment * F.S, so:
Required actual resisting moment capacity = 10 k-ft * 1.6 = 16 k-ft
Summary, you need to use an allowable resistance to overturning, which requires you providing the appropriate factor of safety.
I agree with ITHW. The intent of the .6W in the load combination is to account for the fact that we may overestimate the deadload. For example, a roof system in which we account for future roofing (2nd layer of shingles), would be overestimated if there is only one layer of shingles on the roof. And, the more likely scenario is that the shingles will be ripped off anyway in the case of the wind loading we would be designing for, further reducing the dead load.
I understand there are cirtain circumstances where the dead load is very precisely known and also unlikely to change. In those circumstances, one may decide to not use any factor of safety in conjunction with the load case. I think it depends on the level of confidence that the actual dead load will be there at the time of loading. I think it would have been foolish of the code writers to try to incorporate a "cover all" factor of safety into one load combination.
I understand there are cirtain circumstances where the dead load is very precisely known and also unlikely to change. In those circumstances, one may decide to not use any factor of safety in conjunction with the load case. I think it depends on the level of confidence that the actual dead load will be there at the time of loading. I think it would have been foolish of the code writers to try to incorporate a "cover all" factor of safety into one load combination.
jeffhed,
What I have done in the past was to list the individual loads as you suggest, but also the maximum load down and uplift. I'm starting to get a little concerned about the uplift loads because of the confusion that is possible regarding overturning factors of safety and stability of foundations. I work for a concrete plant manufacturer and don't get involved with the foundation design myself but want to make sure there is a clear understanding of what the information is that I provide on drawings.
I've suggested that we just put on the basic loads and let the foundation engineer do the load combinations himself based on local codes etc., but there is resistance to this because we will frequently get asked for the maximum loads if we don't. I suspect this is because someone unqualified is designing the foundation. Perhaps some guy in the maintenance department.
Regards,
-Mike
What I have done in the past was to list the individual loads as you suggest, but also the maximum load down and uplift. I'm starting to get a little concerned about the uplift loads because of the confusion that is possible regarding overturning factors of safety and stability of foundations. I work for a concrete plant manufacturer and don't get involved with the foundation design myself but want to make sure there is a clear understanding of what the information is that I provide on drawings.
I've suggested that we just put on the basic loads and let the foundation engineer do the load combinations himself based on local codes etc., but there is resistance to this because we will frequently get asked for the maximum loads if we don't. I suspect this is because someone unqualified is designing the foundation. Perhaps some guy in the maintenance department.
Regards,
-Mike
mrMikee,
We always break out the loads for each condition. For a bin and seismic we do the following. We give the seismic load due to dead load and the seismic load due to live load material. The seismic will include the vertical gravity, vertical effects (+/-), vertical due to lateral (+/-), and the shear. I know it seems like a lot of information but it's the only way IMHO to give the information out to others. We break this out this way on all designs because it just makes it easier to understand and I can apply the load combinations as requested. When I run into trouble is getting this information from suppliers. Generally they are giving me some standard information that was generated with the original design in the 70's and if I ask for something else I get the dear in the headlights look because the engineer who originally did the design retired 10 years ago and everyone left on staff is just a tech person. Man I love working in material handling!
We always break out the loads for each condition. For a bin and seismic we do the following. We give the seismic load due to dead load and the seismic load due to live load material. The seismic will include the vertical gravity, vertical effects (+/-), vertical due to lateral (+/-), and the shear. I know it seems like a lot of information but it's the only way IMHO to give the information out to others. We break this out this way on all designs because it just makes it easier to understand and I can apply the load combinations as requested. When I run into trouble is getting this information from suppliers. Generally they are giving me some standard information that was generated with the original design in the 70's and if I ask for something else I get the dear in the headlights look because the engineer who originally did the design retired 10 years ago and everyone left on staff is just a tech person. Man I love working in material handling!
- Thread starter
- #18
mrMikee,
Our firm would comletely ignore the maximum uplift and vertical gravity loads because we don't know what load combinations they are using. In the asphalt plant foundation project I mentioned, I designed the footings and when talking to the engineer that supplied me with the loads I was told that the loads he gave me for dead+wind actually had live loads in them and the actual uplift could/would be higher. It is reasons like this that we ignore any type of maximum loads and ask specifically for broken out loads like you provide. This is the only way I feel confident that I am designing the foundation as the equipment/steel building designer requires.
As far as the overturning factor of safety issue goes, it appears that we are split. Some of us feel that the load combination includes the 1.5 factor of safety. Others feel that this factor of safety would be satisfied if the dead loads counted on were well defined (weight of concrete, etc.). ITHW makes and intersting point with the example he gave, I have never seen it laid out like that.
Our firm would comletely ignore the maximum uplift and vertical gravity loads because we don't know what load combinations they are using. In the asphalt plant foundation project I mentioned, I designed the footings and when talking to the engineer that supplied me with the loads I was told that the loads he gave me for dead+wind actually had live loads in them and the actual uplift could/would be higher. It is reasons like this that we ignore any type of maximum loads and ask specifically for broken out loads like you provide. This is the only way I feel confident that I am designing the foundation as the equipment/steel building designer requires.
As far as the overturning factor of safety issue goes, it appears that we are split. Some of us feel that the load combination includes the 1.5 factor of safety. Others feel that this factor of safety would be satisfied if the dead loads counted on were well defined (weight of concrete, etc.). ITHW makes and intersting point with the example he gave, I have never seen it laid out like that.
ITHW-
In your example you state,
"Required actual resisting moment capacity = 10 k-ft * 1.6 = 16 k-ft"
I agree with that. However, if you are looking at it from that perspective, then you needn't use the 0.6*D factor - you would use 1.0*D. Since you are using the "Required actual resisting moment capacity", no reduction factor is required. You are simply taking the 0.6 from the left side and making it 1/0.6 (or whatever FS you decide on) to the left hand side.
In your example you state,
"Required actual resisting moment capacity = 10 k-ft * 1.6 = 16 k-ft"
I agree with that. However, if you are looking at it from that perspective, then you needn't use the 0.6*D factor - you would use 1.0*D. Since you are using the "Required actual resisting moment capacity", no reduction factor is required. You are simply taking the 0.6 from the left side and making it 1/0.6 (or whatever FS you decide on) to the left hand side.
The 0.6D was specifically to include the overturning factor and as jt12 pointed out still also accounts for an overestimate of the dead loads.
See:
For a good discussion on load combos etc...near end of first page it discusses this case.
See:
For a good discussion on load combos etc...near end of first page it discusses this case.
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