2. Using 2D Elements in GSA RC Slab Design Ian Feltham – R+D, London

3. FE analysis gives stresses in equilibrium with the applied loads A linear elastic material will not reflect the cracked nature of concrete But structure will have sufficient strength if appropriate reinforcement is provided for the FE stresses Interpretation of 2D FE analysis results

4. pxpx pxpx pypy pypy pvpv pvpv pvpv pvpv Reinforcing for in-plane forces fxfx fyfy  s s/tan  To determine f x and f y : Resolve horizontally s/tan .(p x +f x ) = s.p v  f x = p v.tan  - p x Resolve vertically s.(p y +f y ) = s/tan .p v  f y = p v /tan  - p y p x, p y and p v are applied in-plane stressesf x and f y are stresses taken by reinforcement

5. Reinforcing for in-plane forces s/tan  pypy pypy pxpx pxpx pvpv pvpv pvpv pvpv fxfx fyfy fcfc s/sin  To determine f c : Resolve horizontally s.(p x +f x ) + s/tan .p v = (s/sin .f c ).sin   f c = p x +f x + p v /tan  = p v.tan  + p v /tan  f c = 2p v /sin2  To determine f x and f y : Resolve horizontally s/tan .(p x +f x ) = s.p v  f x = p v.tan  - p x Resolve vertically s.(p y +f y ) = s/tan .p v  f y = p v /tan  - p y f c is the stress in the concrete s 

6. Reinforcing for in-plane forces - effect of varying  check that f c can be taken by concrete

7. stress in concrete stress in reinforcement tensile strength of concrete 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

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

9. stress taken by Y reinforcement stress taken by X reinforcement shear stress compressive stress compressive strength of concrete principal tensile stress applied stress X (p x,p v ) Y (p y,-p v ) stress in concrete Y (p v,-p v ) X (p v,p v ) Reinforcing for in-plane forces Note that the stress taken by the X reinforcement is equal to (p v - p x ) and that taken by the Y reinforcement is equal to (p v - p y ) 0.45f cu uncracked 0.30f cu cracked

10. Reinforcing for in-plane forces - general approach 7

11. Compression reinforcement in struts centre line of strut horizontal steel vertical steel compressive strain shear strain/2 Principal tensile strain is approximately 3.6 x design strain of reinforcement principal compressive strain  0.0035 20  compression steel 40  strain at 20  to strut  0.0022 strain in vertical steel  -0.0022 strain in horizontal steel  -0.0022 Although compression reinforcement should be avoided, any provided should be within 15  of centre line of strut to ensure strain compatibility

12. 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 Reinforcing for in- and out-of -plane forces MxMx MyMy M xy NxNx NyNy V V