Load distribution from solid slabs to beams 1

[Gus14]

In school I learned to distribute loads from solid slabs to beams using tributary method i.e,(triangles and trapezoids). However we were always reminded that this method is only applicable while slabs are designed as simply supported but in reality we put slab top reinforcement above the beam in the middle which makes the slab behave as a continuous slab. should we remove the top slab reinforcement because this might increase the beam reaction? in the picture i'm talking about red steel.

 

[Blackstar123]

However we were always reminded that this method is only applicable while slabs are designed as simply supported

Whoever taught you this is wrong or your interpretation of what he taught you is wrong.

The tributary areas and the angles of those lines are based on the yield line theory of slabs. These yield line are formed at points of maximum bending moment and at the axes of rotation. Positive bending moments have yeild lines at the bottom face while negative bending moments have yield lines at top face. When the steel reinforcement is fully yielded, we say slab will get separated into segments according to these the yeild line pattern, for example, triangles and trapezoid for a typical two way slabs. We designed the beams for loads transferred from these formed surfaces.

should we remove the top slab reinforcement

No, unless you want heavy cracks on top face of your slab.

because this might increase the beam reaction?

Loads from slab on beams, will be the same whether you provide top reinforcement or not.

However, in case of edge beams or unequal slab spans, you may get some torsion in beams.

 

[r13]

You need to review your lecture notes, it seems to me you are mixing oranges with apples. The method you mentioned is to gather the loads to the beams, and analyze the beams either with (T beam), or without (rectangle beam) consider the slab's contribution to the stiffness of the beam under consideration. The beam can be assumed fixed, or pinned (rarely applicable to reinforced concrete structures), and you shall provide reinforcement accordingly.

However, in the sketch you provided, it looks like you are design a two span continuous slab with beams as support. You may consider the outer edge beams as pinned, thus no top steel is required (for analysis' sake only), but you can't eliminate the top steel over the middle support, because the resulting negative moment required to make the system "continuous". Otherwise, you need to design the system as two simply support spans (non-continuous but share the middle support), but expecting huge cracks over the middle support region, and large deflection in the mid-span of each simply support system.

 

[r13]

Can you get ride of the top steel in this case?

 

[Blackstar123]

I didn't see the picture earlier.
If the picture is correct, it's not even a two way slab. You should have beams on all 4 sides if you want the slab to have a two way action.

Span ratio<2 but with beam only on two sides is not a two way slab.

 

[Gus14]

I forgot to draw beams on the other sides. But I did mean the slabs are two way slabs.

 

[Gus14]

If I design each slab span as simply supported I can get rid of the steel and suffer from large cracks. However in the picture drawn by retired13, then NO I CANT get rid of the top steel.

 

[Gus14]

Thank you guys for feedback. It's just that I didn't know where the tributary area derivation came from.

 

[BAretired]

Structural engineering is not an exact science. The diagram shows a two span, one way slab. The exact distribution of load to the beams is a judgment call. If the middle beam does not deflect, the tributary area is increased by a factor of 1.25. Since it does deflect, the factor will be more than 1.0 but less than 1.25.

I forgot to draw beams on the other sides. But I did mean the slabs are two way slabs.

I'm not sure why; one way seems to make more sense, but if there are beams on the other sides, what is the condition of continuity in that direction?

Use good engineering judgment in distributing the load. If in doubt, be conservative, and remember that uniform live load does not exist on floors either.

BA

 

[r13]

I think this sketch answers both questions.