Long-term deflection=three times dead Plus one time live 2

[sundy]

I have come across a papers (“Effects of creep and shrinkage on the deflection of RCC two way flat plates”) which states that, in ACI 318-2000, the total deflection of two-way slab for the creep and shrinkage effect is calculated as three times the deflection due to dead and deflection due to live. Also, I have seen people checking two-way slab long-term deflection with SAFE using three times the deflection due to dead and deflection due to live.

I don’t have ACI 318-2000, it will be appreciated if anybody can point out which clause in ACI 318-2000 dealing with the above deflection calculation.

 

[sdz]

The Australian code has similar provisions so I'm sure that they are somewhere in ACI 318.

 

[asixth]

To say that the total long-term deflection is three times the deflection due to dead load and the deflection due to live load is an approximation only, but I have found it to be a reasonable approximation.

When calculating the deflections in rc structures your approach should follow:

1. Calculate the immediate deflections based on Ieffective, Dead Load and Short-term service live load.

2. Calcalate the long-term deflections based on increasing curvatures due to creep and dead load and long-term service live load.

3. Calculate the incremental deflection as being the difference between the short-term service and long-term load cases.

I think the factor of three comes from the use of a kcs multiplier (creep and shrinkage multiplier). Again, the use of creep-shrinkage multiplier is an approximation magnifier based on testing of heavily reinforced highway girders and should not be used for lightly reinforced (relativley speaking) flat slabs.

The co-efficients comes from the following application where short and long-term deformations are calculated on the following factors:

1. Short-term displacements...

delta_short=G+0.7*Q

2. Long-term displacments...

delta_long=(1+kcs)*G+(1+kcs)*0.4*Q

If the creep-shrinkage multiplier is taken as 2 which it is when no compression steel helps to minimise the effects of shrinkage, then the long-term deformations will approximatley come out to 3*G+1*Q.

Check out 9.5.2.5 in ACI318-08 for the use of long-term deformation multipliers but again, this is only an approximate assumption. I only use it to verify the answers of the rc design package I use.

 

[sundy]

asixth:

Could you provide any references for the caculations:

1. Short-term displacements...

delta_short=G+0.7*Q

2. Long-term displacments...

delta_long=(1+kcs)*G+(1+kcs)*0.4*Q

Especially, for "delta_long=(1+kcs)*G+(1+kcs)*0.4*Q"?

Thanks a lot.

 

[asixth]

The simplified approach for calculating deformations is the same for ACI318. Instead of wtriting kcs, I should have wrote lamda, which can be found in ACI-318 Cl.9.5.2.5.

The long-term deflections above should read:

delta_long1+lamba)*G+lamba*LF_long*Q+LF_short*Q

where:
LF_long is the long-term live load combination factor
LF_short is the short-term live load combination factor

basically the equation above can breakdown to the following:

delta_short=G+LF_short*Q
incremental=lamba*G+lamba*LF_long*Q
delta_long=delta_short+incremental

I will post the relevant section from ACI318 when I go back to work tomorrow.

 

[sundy]

Thanks,asixth

Please include the factors "LF_short","lamba" and "LF_long".

 

[sundy]

I post the paper here.

See page 273 equation(12)

 

[rapt]

Sundy,

Make sure cracking has been allowed for in the initial calculation of deflection before applying this multiplier. Many FEM programs are giving you uncracked deflections.

If you are dealing with uncracked short term deflections as your starting point, the multiplier is in the order of 6 times the short term uncracked deflection to allow for cracking, shrinkage and creep.

But remember this is very approximate as Asixth mentioned.

 

[asixth]

I haven't read through the paper you posted in detail but it looks like it took the equation from the 2000 version of ACI-318. It gives a description of how shrinkage and creep affect the deflections of the slab but doesn't really provide an example. I have posted the relevent section of ACI318-05. If you look at section 9.5.2.5 it says:

"Unless values are obtained from a more comprehensive analysis, additional long-term deflection resulting from creep and shrinkage of flexural members shall be determined from multiplying the immediate deflection cause by the sustained load considered by a factor (lamba)."

So you then need to calculate the immediate deflection caused by load, this is you G+Service Q load case.

The additional deformations that are caused by creep and shrinkage are ESTIMATED by multiplying the sustanined load by a factor lamba.

This will give you a ESTIMATE of the long-term deformations, that is:

Long-term=Instantaneuos + Additional

You will need to find the loading code for your building board to get an idea of what factor of the design live load is included as instantaneous and sustained.

Take your stiffness of the section you are considering as Icracked, this will give you greater flexural deformations and rotations than what you would get if you used Igross.

You should be performing a more comprehensive analysis to calculate creep and shrinkage effects in rc structures. Most design packages these days are very useable and a quick frame run can be set up and modified easily so slab depths can be calculated for your given span and loading. This is for both preliminary and detailed designs.

 

[UcfSE]

sundy, there is no 318-00.

The 318-05 shows what you want in 9.5.2.5. Essentially, you estimate additional long-term deflection as a multiplier of the short-term. The mulitplier is calculated. 2.0 is the worst-case result of that calculation. So to get total deflection, you add long-term plus short-term and get 3 times the short-term.

 

[sundy]

sundy, there is no 318-00.

Please read the papers(pdf format)I post above, it references ACI318-2000. Why?

 

[UcfSE]

Could be a mistake.

 

[rapt]

The normal logic is to use a LT deflection multiplier of 2 and divide the LL into short term and long term portions so

short term deflection = DL + STLL + LTLL

Total long term deflection = 3 * DL + 3 * LTLL + 1 * (STLL - LTLL)
It STLL = 1.0LL and LTLL = .25LL (asixths figures were from the Australian code) as it does for ACI then
Total long term deflection = 3 * DL + 3 * .25 LL + 1 * .75LL

= 3 * DL + 1.5 * LL

The reason for this is that only the LT LL is applied for the creep and shrinkage calculations.

If you want to do it better, you should use a computer program that will do it properly for you, like Sophystic (spelling terrible) or RAPT, because the fudbge factor method above is very unreliable and variable depending on the design! For example, in a bad design I have seen Llambda factors as high as 6 and as low as 2.