Plate centre stress governing design moment

[alphaxy]

Fellow Engineers,

I am modelling a concrete slab using plate elements. I am looking for the moments in the slab at the center of each element. I noticed that the output gives the bending moments per unit width. What is the per unit width? Would that be the thickness of the plate element?

I have a mesh size f 0.65x0.65, and some of the mesh varies. Do I have to multiply or divide the bending moment in kNm/m to 0.65m?

 

[jhardy1]

If you have a reported slab moment of 20 kNm/m, say, this means a one metre wide strip of slab has a total moment of 20 kNm; a 500 mm wide strip has a total moment of 10 kNm; a two metre wide strip has a total moment of 40 kNm; and so on.

Hope this helps!

 

[alphaxy]

Thanks Julianhardy

Is the strip width you are referring to, depends on the mesh of plates in STAAD? Or just the normal assumption we used to do when we design slab (which is always 1 meter strip)

 

[alphaxy]

Sirs,

I am confused with the design moment for plates in STAAD.

When STAAD generate bending forces, it gives kNm/m.

say my mesh would be 0.65m and a 20kNm/m. Does it mean that I have a design moment of 13kN/m?

"OR"

Use 20kNm, and when you obtain rho from equation Mu = (phi)bd^2f'cq(1-0.59q) u will use 0.65m for width? in this case, divide.

This is really confusing. I hope you could help me out

CHeeRS!

 

[accountclosed]

The reason STAAD reports plate moments in terms of moment per unit width is so you don't have to worry about that kind of accounting. The moment is normalized to a unit width, regardless of the element size. So in JulianHardy's example of 20 kNm/m:

- a one meter wide strip carries a moment of 20 kNm,
- a 10 m wide strip carries a moment of 200 kNm,
- a 0.65 m wide strip carries a moment of 13 kNm.

Notice that there is no mention of mesh size in those sample numbers.

 

[alphaxy]

STAADicEquilibrium,

Thank you very much..You made it clear..=) What I did in my analysis is I used 20kN/m...my mesh size is all less than 1 so I guess I am in conservative side.

The moment is dependent on size of mesh, and if we have a wall with mesh size of 0.65, 0.7 and 0.8..we can make use of 0.8 for a conservative result..

Now, in checking the moment capacity of the wall, we can also make use of 0.8 as the value of width..

I hope you all agree with me..

thanks

 

[accountclosed]

You could do your concrete design calcs for 20 kNm on a 1 m width of slab, or you could apply 16 kNm to a 0.8 m width of slab. It's directly proportional. The result is going to be some amount of required reinforcing steel.

If you applied 20 kNm to a 1 m width of slab, then your resulting required area of steel will be on a "per meter" basis.

If you applied 16 kNm to a 0.8 m width of slab, then your resulting required area of steel is "per 0.8 meters", so you would just need to be sure that you specify a rebar pattern that considers this.

If you applied 20 kNm to a 0.8 m width of slab, then you have been conservative.

Hope this continues to clarify.

Cheers!