Using 2D Elements in GSA

RC Slab Design

Ian Feltham – R+D, London

• 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

px px

py

py

pv

pv

pv

pv

Reinforcing for in-plane forces

fx

fy

s

s/tan

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

px, py and pv are applied in-plane stressesfx and fy are stresses taken by reinforcement

Reinforcing for in-plane forces

s/tan

py

py

px px

pv

pv

pv

pv

fx

fy fcs/

sin

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

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

fc is the stress in the concrete

s

0

5

10

15

20

15 30 45 60 75theta(degrees)

stress (MPa)

fx

fy

fc

fs = abs(fx) + abs(fy)

px = -4 MPapy = 1 MPapv = 5 MPa

Reinforcing for in-plane forces - effect of varying

check that fc can be taken by concrete

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

x (compression)

y

(compression)

fcu

fcu

fct

fct

compressive strength of concrete with transverse

tension

tensile stress in concrete

Bi-axial strength of concrete

stress taken by Y reinforcement

stress taken by X reinforcement

shear stress

compressive stress

compressive strength of concrete

principal tensile stress

applied stress

X (px,pv)

Y (py,-pv)

stress in concrete

Y (pv,-pv)

X (pv,pv)

Reinforcing for in-plane forces

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)

0.45fcu uncracked 0.30fcu cracked

Reinforcing for in-plane forces - general approach

7

Compression reinforcement in struts

cent

re li

ne o

f stru

t

horizontal steel

vert

ical st

eel

compressive strain

shear strain/2

Principal tensile strain is

approximately 3.6 x design strain of reinforcement

principal compressive strain

0.0035

20

com

pres

sion

ste

el

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

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

Mx MyMxy Mxy

Nx Ny

V V