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

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

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

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

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

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

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

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

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’

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)

Reinforcing for in-plane forces - general approach 7

Compression reinforcement in struts compressive strain shear strain/2 strain in vertical steel  -0.0022 strain in horizontal steel  -0.0022 40 strain at 20 to strut  0.0022 Principal tensile strain is approximately 3.6 x design strain of reinforcement principal compressive strain  0.0035 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