DESIGN OF CORBELS(BS CODE).xls

PREPARED BY : DATE :

MLU 10/4/2013

CHECKED BY :

AVT

APPROVED BY :

DCA

REVISION :

A

PROJECT : EXAMPLE STRUCTURE : TITLE : DESIGN OF COREBELS (SEE BS 5.27) I.

II.

Material Properties 40 Mpa fcu = 460 Mpa fy = Es = 200000 Mpa 12000 Mpa Ec = 16.667 n =

design strength of steel material (S 275) reinforced steel design strength modulus of elasticity of steel modulus of concrete (long term loading)

Section Properties 350mm x 400mm Reinforced Concrete Corbels h = 750 mm Vy 350 mm h1 = 400 h2 = mm Vx 700 mm L = 400 mm av = 400 mm b = d' = 704 mm 1.5 Үm = 32 mm ∅m = 10 mm ∅l = 30 mm cc = 2 pcs n-∅ = r = 128 mm 700 mm L = 0.5 β = lb ab

= =

642.37 mm 104 mm

hagg

=

20 mm

length width effective depth partial safety factors for safety of materials diameter of main reinforcement diameter of links vertical distance of centroid of the section to nearest edge number of main bars internal radius of bend (BS 3.12.8.22) r = 4∅ effective length BS 3.8.1.6 { Le = β lo } coefficient dependent on the bar type (BS 3.12.8.4, Type 2 defformed bars) tension bar length at the start of the bend smaller bet. center to center distance between bars or cover plus diameter maximum size of coarse aggregate

III. Shear & Moments 80 KN Vx = 800 KN Vy =

shear along x shear along y

III. Allowable Loads a. Actual force Trial value x = d' - 0.45x z β ∑V Fc Fc x

Ft

= = = = =

tan-1(z/ av) Vy - Fc sin β = 0 Vy / sin β (0.67fcu/Үm) b 0.9x cos β Fc (0.67fcu/Үm) b 0.9 cos β

= Vx + Ft cos β

= =

266.77 mm

=

55.589 º

=

969.69 KN

=

266.77 mm

depth to the neutral axis

584 mm

lever arm

compressive resistance

Vy Vx

Ft =

627.99 KN

s

Fc

b. Check for strain εc = εs x h-x

εs

=

Fc cos

εc( h - x)/ x

b. Steel Area Asreq = Ft/ (0.87*fy) 2 As prov = n pi() d / 4 As min = 0.16% bh As max = 4% bh

= 0.0574 > 0.002 Ok! • steel at yield stress level

1569.2 1608.5 480 12000

c. Check for Shear (BS 3.4.5.2, Table 3.7) p = 100 Asprov bd

=

0.57

= 0.79(p)0.5(400/d)0.25/ 1.25

=

0.61

= V/ bd

=

2 2.84 N/mm

= av bv (v-2d vc/ av) 0.95 fyv

=

v As

c = 0.035

FORCE mm2 mm2 mm2 < As prov Ok! mm2 > As prov Ok!

= = = =

vc

Ft

n =

2 pcs

d. Check for Bearing Stress inside bend

2 140.08 mm

STRAIN

STRESS

reinforcement required reinforcement provided minimum area of reinforcement (BS 3.12.5.4, Table 3.25) maximum area of reinforcement (BS 3.12.6.1)

Fbt

= (Ft/no. bars) (As req/As Prov)

=

fbu

= β √ fcu

=

3.16 Mpa

= Fbt/ (π∅fbu)

=

963.56 mm

length of anchorage

= Fbt. (lx/ la)

=

204.21 mm

tension in bar at start of bend

=

=

49.86 Mpa

la

Fbtb

fb1

Fbtb

306.32 KN

design ultimate anchorage bond stress

bearing stress (BS 3.12.8.25.2)

Fbt r

Fbtb