1. General Use of GSA RC Slab Design
Andrew Fraser
Structural Development & Support
Arup Advanced Technology & Research

2. Interpretation of 2D FE analysis results
FE analysis gives stresses in equilibrium with the applied loads
A linear elastic material will not reflect the cracked nature of concrete
But the structure will have sufficient strength if appropriate reinforcement is provided for the FE stresses
Reference:
‘Guide to the design and construction of reinforced concrete flat slabs,’
- Concrete Society Technical Report 64

3. Reinforcing for in-plane forces
To determine fx and fy:
Resolve horizontally
s/tan.(px+fx) = s.pv
 fx = pv .tan - px
Resolve vertically
s.(py+fy) = s/tan.pv
 fy = pv /tan - py py pv fx fy
s/tan  px s
fx and fy are stresses taken by reinforcement
px, py and pv are applied in-plane stresses

4. Reinforcing for in-plane forces
To determine fx and fy:
Resolve horizontally
s/tan.(px+fx) = s.pv
 fx = pv .tan - px
Resolve vertically
s.(py+fy) = s/tan.pv
 fy = pv /tan - py
s/tan py px pv s 
s/sin fx fy fc
To determine fc:
Resolve horizontally
s.(px+fx) + s/tan.pv = (s/sin.fc).sin
 fc = px+fx + pv/tan
= pv .tan + pv/tan
fc = 2pv/sin2
fc is the stress in the concrete

5. Reinforcing for in-plane forces - effect of varying 
check that fc can be taken by concrete

6. We've looked at the applied forces . . .
. . . now consider the resistances
A bit.

7. Consider tensile stresses in concrete between cracks
tensile strength of concrete will vary along bar
when the tensile stress reaches the local strength, a new crack will form
At some point, a stabilised crack pattern will be reached and no more cracks will form
stress in reinforcement
tensile strength of concrete
stress in concrete

8. Bi-axial strength of concrete
x (compression)
y (compression)
fcu
fct
compressive strength of concrete with transverse tension
tensile stress in concrete

9. Reinforcing for in- and out-of -plane forces in GSA
Applied forces and moments resolved into in-plane forces in sandwich layers
The layers are not generally of equal thickness
Reinforcement requirements for each layer calculated and apportioned to the reinforcement positions
The arrangement of layers and in-plane forces adjusted to determine the arrangement that gives the best reinforcement arrangement Mx My
Mxy Nx Ny V
Reference:
Structures Note 2007NST7 - ‘RCSlab’

11. Reinforcing for in-plane forces
shear stress
compressive stress
stress taken by X reinforcement
principal tensile stress
applied stress
X (px,pv)
Y (py,-pv)
compressive strength of concrete
stress in concrete
Y (pv,-pv)
X (pv,pv)
0.45fcu uncracked 0.30fcu cracked
stress taken by Y reinforcement
Note that the stress taken by the X reinforcement is equal to (pv- px) and that taken by the Y reinforcement is equal to (pv- py)

12. Reinforcing for in-plane forces - general approach7

13. Compression reinforcement in struts
compressive strain
shear strain/2
strain in vertical steel 
strain in horizontal steel 
40
strain at 20 to strut 
Principal tensile strain is approximately 3.6 x design strain of reinforcement
principal compressive strain 
centre line of strut
horizontal steel
vertical steel
20
compression steel
Although compression reinforcement should be avoided, any provided should be within 15 of centre line of strut to ensure strain compatibility