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# ASCE 7-98, Sect 2.4.1, Comb. 5 Question

## ASCE 7-98, Sect 2.4.1, Comb. 5 Question

(OP)

0.6D + 0.7E + H

Assuming no pressure loading (H), my question is what weight are we to use to calculate E?

I understand that the 0.7 is from NEHRP changing the code to produce factored loads instead of service level, so when using ASD, you must lower it, and for LRFD it is 1.0.

I also understand that the 0.6 is there as a conservative measure because the dead load in this overturning case would be helping you.

What I don't get is why you'd use the full weight of the structure to calculate E and then try and restrain it with 0.6 of the weight.  Can we use 0.6 of the weight to calculate E?  Or is this not permitted?

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

I consider the use of full weight to compute E is an extra precaution to guard against structure with full dead load initially at an earthquake event, but losing some dead load effect on the later stage due to relief of energies from the supports of some dead load sources (structures components no longer synchronize with the motion of the main structure). If this is the case, then the use of reduced weight to compute E would be unwise.

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

I disagree with kslee1000.

This service load load combination is primarily used to maximize any net tension induced due to seismic overturning where "E" is decoupled into the respective horizontal and vertical components E = (rho*Qe)+/-(0.2SDS).

In consideration of foundation elements, the 0.6 boogie factor applied to the dead load basically mandates a safety factor against overturning of 1/0.6 = 1.67.  This is more conservative than the old 1.5 overturning S.F. prescribed in the UBC.

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

Should have read kslee1000's posting better. My apologies!  He/she is totally correct in that one would be using the full seismic weight when determining the horizontal and vertical seismic components.  Then apply the load combo.

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

sundale,
I'm not sure where you disagree with kslee1000?
You seem to say the same thing.

kslee1000 says that the full dead weight should be used to calculate E  (i.e. don't use 0.6D to determine W to determine Qe  - rather use 1.0D to get W to get Qe).

You say that the 0.6 factor on D should be used to get a safety factor against overturning.  Where do you see that you differ?

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

I don't disagree; misread the post and hit submit and not preview...

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

(OP)
I'm having a tough time seeing how the 0.6 is just for OTM safety factor.  Wouldn't a factor for overturning saftey have to be applied to the entire load combination?  By applying it only to the dead load, you leave out the earthquake load.  Let's look at the extremes:

Case 1:  D = 50 'k (righting moment due to dead load)
E = 1000 'k (OTM due to EQ)

Then the equation comes to:  -0.6*50 + 0.7*1000 = 670'k
This is what you're righting moment (RM) needs to be.

Now do it the old way: 1.0D + 1.0E with 1.5 OTM SF

-1.0*50 + 1.0*1000 = 950 'k, but your righting moment needs to be 1.5 times this, or 1425'k!  Doesn't make sense.

Case 2:  D = 500 'k
E = 520 'k
-0.6*500 + 0.7*520 = 64 'k = righting moment

The old way:  -1.0*500 + -1.0*520 = 20'k x 1.5 = 30'k righting moment.  Again, nowhere close.

It seems to me that to be "apples to apples" they had to reduce the dead load to keep up with reducing the earthquake load.  They allude to this in the commentary (see C9.2).  Let's say they kept it even at a 0.6 multiplier and the load combo was:
0.6D + 0.6E
Then whatever percent failure you got at 1.0 would be the same percent failure you get at 0.6.  The earthquake portion has been made 0.7, however, which from what I gather in the commentary is wishy-washy at best.  All this means is that the current code is a little more stringent, i.e., 0.7E vs. 0.6E applied to 0.6D.

Thinking about it this way, I've answered my own question - it would not be logical to apply yet another 0.6 to the W used in the E term, effectively making it 0.6*0.7 = .42.

Now, let's muddy the waters further.  Next question:  if my structure is stable, but my soil pressures are wilting under the load, can I then apply a soil pressure reduction of .67 because of the transient nature of the load?

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

(OP)
Correction:  it's not C9.2 in the commentary.  Turn to section C9.0 under the heading "Use of Allowable Stress Design Standards".

### RE: ASCE 7-98, Sect 2.4.1, Comb. 5 Question

swearingen,

Where do you get your "old way"?  (-1D+1E)

The UBC 97 has -0.9D + E/1.4 - no 1.5 safety factor required by the code explicitly (that I can find).

and the UBC 94 has -.85D + E/1.4 (section 1631) and no 1.5 SF is stated.  (the 1.5 SF was for wind only)

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