Corbel
2
PREPARED BY : MLU DATE : 7/19/2022 CHECKED BY : AVT APPROVED BY : DCA REVISION : A PROJECT : EXAMPLE STRUCTURE : TITLE : DESIGN OF COREBELS (SEE BS 5.27) I. Material Properties fcu = 40 Mpa design strength of steel material (S 275) fy = 460 Mpa reinforced steel design strength Es = 200000 Mpa modulus of elasticity of steel Ec = 12000 Mpa modulus of concrete (long term loading) n = 16.667 II. Section Properties 350mm x 400mm Reinforced Concrete Corbels h = 750 mm h1 = 350 mm h2 = 400 mm L = 700 mm length av = 400 mm b = 400 mm width d' = 704 mm effective depth Үm = 1.5 partial safety factors for safety of materials ∅m = 32 mm diameter of main reinforcement ∅l = 10 mm diameter of links cc = 30 mm vertical distance of centroid of the section to nea = 2 pcs number of main bars r = 128 mm L = 700 mm β = 0.5 coefficient dependent on the bar type (BS 3.12.8.4, Type 2 defformed bars) lb = 642.37 mm tension bar length at the start of the bend ab = 104 mm smaller bet. center to center distance between bars plus diameter hagg = 20 mm maximum size of coarse aggregate III. Shear & Moments Vx = 80 KN shear along x Vy = 800 KN shear along y III. Allowable Loads a. Actual force Trial value x = 266.77 mm depth to the neutral axis z = d' - 0.45x = 584 mm lever arm = = 55.589 º V ∑ = Fc = = 969.69 KN compressive resistance Fc = x = Fc = 266.77 mm Ft = = 627.99 KN b. Check for strain = x h - x = = 0.0574 > 0.002 Ok! • steel at yield stress level b. Steel Area Asreq = Ft/ (0.87*fy) = 1569.2 reinforcement required As prov = n = 2 pcs = 1608.5 reinforcement provided As min = 0.16% bh = 480 < As prov Ok! As max = 4% bh = 12000 > As prov Ok! c. Check for Shear (BS 3.4.5.2, Table 3.7) p = 100 Asprov = 0.57 b d vc = = 0.61 v = V/ bd = 2.84 n-∅ internal radius of bend (BS 3.12.8.22) r = 4∅ effective length BS 3.8.1.6 { Le = β lo } β tan -1 (z/ av) Vy - Fc sin β = 0 Vy / sin β (0.67fcu/Үm) b 0.9x cos β (0.67fcu/Үm) b 0.9 cos β Vx + Ft cos β εc εs εs εc( h - x)/ x mm 2 n pi() d 2 / 4 mm 2 mm 2 minimum area of reinforcement (BS 3.12.5.4, Table 3.25) mm 2 maximum area of reinforcement (BS 3.12.6.1) 0.79(p) 0.5 (400/d) 0.25 / 1.25 N/mm 2 Vy Vx Vy Vx Ft Fc c = 0.035 s Fc cos Ft FORCE STR AIN STRESS
description
BS CODE
Transcript of Corbel
PREPARED BY : MLU
DATE : 4/18/2023
CHECKED BY : AVT
APPROVED BY : DCA
REVISION : A
PROJECT : EXAMPLESTRUCTURE :TITLE : DESIGN OF COREBELS (SEE BS 5.27)
I. Material Propertiesfcu = 40 Mpa design strength of steel material (S 275)
fy = 460 Mpa reinforced steel design strengthEs = 200000 Mpa modulus of elasticity of steelEc = 12000 Mpa modulus of concrete (long term loading)
n = 16.6667
II. Section Properties 350mm x 400mm Reinforced Concrete Corbelsh = 750 mm
h1 = 350 mmh2 = 400 mm
L = 700 mm lengthav = 400 mmb = 400 mm widthd' = 704 mm effective depth
Үm = 1.5 partial safety factors for safety of materialsm∅ = 32 mm diameter of main reinforcement
l∅ = 10 mm diameter of linkscc = 30 mm vertical distance of centroid of the section to nearest edge
= 2 pcs number of main barsr = 128 mmL = 700 mmβ = 0.5 coefficient dependent on the bar type
(BS 3.12.8.4, Type 2 defformed bars)lb = 642.371 mm tension bar length at the start of the bend
ab = 104 mm smaller bet. center to center distance between bars or coverplus diameter
hagg = 20 mm maximum size of coarse aggregate
III. Shear & MomentsVx = 80 KN shear along xVy = 800 KN shear along y
III. Allowable Loadsa. Actual forceTrial value x = 266.77 mm depth to the neutral axis
z = d' - 0.45x = 584 mm lever arm
= = 55.5894 º∑V =Fc = = 969.686 KN compressive resistanceFc =
x = Fc = 266.775 mm
Ft = = 627.989 KN
b. Check for strain=
x h - x
= = 0.05736 > 0.002 Ok!• steel at yield stress level
b. Steel Area Asreq = Ft/ (0.87*fy) = 1569.19 reinforcement required
As prov = n = 2 pcs = 1608.5 reinforcement provided
As min = 0.16% bh = 480 < As prov Ok!
As max = 4% bh = 12000 > As prov Ok!
c. Check for Shear (BS 3.4.5.2, Table 3.7)p = 100 Asprov = 0.57
b d
vc = = 0.61
v = V/ bd = 2.84
n-∅internal radius of bend (BS 3.12.8.22) r = 4∅effective length BS 3.8.1.6 { Le = β lo }
β tan-1(z/ av)Vy - Fc sin β = 0Vy / sin β(0.67fcu/Үm) b 0.9x cos β
(0.67fcu/Үm) b 0.9 cos β
Vx + Ft cos β
εc εs
εs εc( h - x)/ x
mm2
n pi() d2/ 4 mm2
mm2 minimum area of reinforcement (BS 3.12.5.4, Table 3.25)
mm2 maximum area of reinforcement (BS 3.12.6.1)
0.79(p)0.5(400/d)0.25/ 1.25
N/mm2
Vy
Vx
VyVxFt
Fc
c = 0.035
s
Fccos
Ft
FORCE STRAIN STRESS
As = av bv (v-2d vc/ av) 0.95 fyv = 140.08
d. Check for Bearing Stress inside bend Fbt = (Ft/no. bars) (As req/As Prov) = 306.32 KN design ultimate anchorage bond stress
fbu = = 3.16 Mpa
la = = 963.56 mm length of anchorage
Fbtb = Fbt. (lx/ la) = 204.21 mm tension in bar at start of bend
fb1 = Fbtb = 49.86 Mpa
mm2
β √ fcu
Fbt/ (π fbu)∅
bearing stress (BS 3.12.8.25.2)
Fbt
rFbtb
