Elliptical concrete roof

[jdw]

I am in the process of analysing a curved (elliptical)roof with the following dimensions etc.
span 13'4"
internal dim. -principal axis of ellipse 6'-2" (to centre)
secondary axis of ellipse 3'-11" (from top of roof to centre)
thickness 6",
reinforcement 3/8"h.t. bars longt. @10" c.c longt., top and bot. bars aligned radially /3/8" h.t. bars (curved) @10" c.c ( all top and bottom)
Please verify if these dimensions and reinf. are adequate or advise on how I should approach the design.
regards,
jdw

 

[engcomp]

I can't visualize the shape from your description.
Span 13'4" - are you talking about a dome shape with an oval base?

 

[Snatch]

Engcomp,
Thanks for responding.
Not exactly, the shape is similar to a cylindrical shell roof. However, instead of being an arc of a circle as in the cylindrical shell roof, the curve forming the roof is half of an ellipse. The building is 43' long and 13'4" wide.

jdw

 

[engcomp]

And the "shell" is supported along the 43' length?

 

[engcomp]

Hi, jdw

With continuous support along the length of the "shell", what we have is actually an elliptic arch. The thrust line is a shape governed by the selfweight and superimposed loads, for example a parabola if the loads were uniformly distributed. You will note that the shape corresponds to the bending moment line of a freely supported member carrying the given loads. Asymetrical snow and live loads usually have worse effect, even though total load is less.

This is how I would go about it:

1. Overlay the shape (parabola) on the ellipse so that its maximum offsets above and below are equal.

2. From the geometry of the overlaid shape calculate the thrust at the points of maximum offset.

3. Multiply the thrust with the offset and check the capacity of your 6" thick slab to take that moment.

Hope this helps? Helmut
engcomp@pbq.com.au

 

[engcomp]

Hi, Janine

Thanks for your email. The answer should go here because it might benefit others.

The thrust line is any line you can draw that links all given loads (I presume you know how to draw force diagrams). You will note that it requires a horizontal component to draw the line. That component must be supplied by the arch's supports. If the supports cannot supply the horizontal component, then the optimum thrust line cannot be achieved.

For example, if one support is freely sliding, the moments in the arch are the same as in a straight, freely supported member. The only difference is that the self weight is different from a straight member.

Check your 6" section for that first. If it's OK, don't worry about arch action.

Regards, Helmut