PROJECT :TITLE:

BS8110 : CORBELS DESIGN 1) ULTIMATE LOADS ON THE CORBEL

Ult Vertical Vu = 2926 kN

Ult Horizontal Tu = 293 kN

Fcu = 40 N/mm2

Fy = 460 N/mm2

Ec = 28 kN/mm2

Es = 200 kN/mm2

2) DETERMINATION OF CORBEL GEOMETRY

Check: Max Bearing Stress

= 0.8 Fcu = 32.0 N/mm2

Width of Bearing Plate, w = 1400 mm

Length of Bearing Plate, l = 700 mm

Vu/(l.w) = 3.0 N/mm2 , < 0.8 Fcu

av = 350 mm

Cover @ Top/End, C = 30 mm

Cover @ Side, Cside = 72 mm

Diameter of Rebar, D = 25 mm

Diameter of Link, Dl = 10 mm

Le = D+5D+Dl+C >= 190 mm Width of Corbel, b = 1800 mm

Length of Corbel, req >= 890 mm Length of Corbel, prov = 700 mm

h, prov @ col face = 1600 mm

= min(0.8Fcu^0.5, 5N/mm2) = 5.00 N/mm2 d @ column face = 1557.5 mm

Check: av < d --> CORBEL

Vu/bd = 1.04 N/mm2

hy @ outer face >= 0.5h = 800 mm

hy, prov = 800 mm

Check: Max Shear Stress vmax

< Vmax

3) EVALUATION OF INTERNAL FORCES

Assume x (1st trial = 0.4d) = 421 mm es = 0.0035(d-x)/x = 0.0094 , > 0.002 ?

z = d - 0.45x = 1368 mm fs1(es>=0.002) = 0.87fy = 400 N/mm2

sin A = 0.9688 fs2(es< 0.002) = Es.es = 1890 N/mm2

cos A = 0.2479 fs = min(fs1, fs2) = 400 N/mm2

Fc = Vu/sin A = 3020241 N Ft = Vu.av/z+Tu = 1041184 N

x = Fc/(0.402Fcu.b.cosA) = 421.0 mm

As = Ft/fs = 2602 mm2

>= (0.5Vu+Tu)/0.87fy = 4387 mm2

As, req = 4387 mm2

As, prov: T 25 491 mm2

No. = 24

Total As = 11775 mm2

11520 mm2

115200 mm2

4) CHECK SHEAR

p = 100As,prov/bd = 0.420 Horizontal Links:

Spacing Sh = 200 mm

vc = 0.554 N/mm2 Ash = b.Sh(v-vc')/0.87fy = -3494 mm2

vc' = (2d/av).vc = 4.927 N/mm2 2193 mm2

v = Vu/bd = 1.04 N/mm2 Ash, req = 2193 mm2

Ash, prov: T 10 78.5 mm2

No. of legs = 6

No. of sets = 7

Total Ash = 3297 mm2

Ash for Upper 2/3.d = 1038 mm

5) CHECK BEARING STRESS INSIDE BEND

Tensile Force in As, Fbt Tension in Bar at Start of Bend

= (Ft/No. As)(As,req/As,prov) = 16162 N/Bar Fbt' = (L1/Lb).Fbt = -107957 N

Ultimate Anchoage Bond Stress

= fbu = 0.5Fcu^0.5 = 3.16 N/mm2

Check: Asmin = 0.004bh =

Check: Asmax = 0.040bh =

Ashmin = 0.5 Asreq =

Int. Radius, r = 4D = 100 mm

Anchorage Bond Length Req.

= Lb = Fbt/(3.14.D.fbu) = 65 mm 70.91304348 mm

97 mm

Straight Length of Bar before Start of Bend 70.91304348 mm

= L1 = 500 mm

Check:

Additional Length after Standard Bend Bearing Stress

= La = Lb - L1-12D + 4D = -635 mm = Fbt/(rD) = -43.18 N/mm2

46.92 N/mm2, OK

6) CHECK SPACING OF BARS

Min. Aggregate Size, AS = 20 mm

Min. Horz Space = AS + 5 = 25 mm Check: Actual Clear Spacings

>= D = 25 mm Side = 72 mm < 80mm, OK

Internal = 46 mm , OK

Max. Horz Space = 47000/fs = 781 mm

<= 300 mm

7) CRACK WIDTH CALCULATION

Service Vertical V = 2070 kN 3.009E-04

Service Horizontal T = 207 kN 3.114E-04

Moment @ Col. Face

= M = T.av = 724.6590909 kN-m = -8.091E-04

Assume As' = 0. 82.1 mm

42.8 mm

m = Es/Ec = 7.142857143 82.1 mm

x = mAs/b{[1+ 2bd/(As.m)]^0.5 -1}

= 338 mm

z = d - x/3 = 1445 mm Check : Crack Width, Wcr

fsb (M) = M/(As.z) = 42.6 N/mm2 = -0.184 mm

fsh (T) = T/As = 17.6 N/mm2 < 0.200 mm, OK

fs = fsb + fsh = 60.2 N/mm2

ab1 (Ctr-Ctr Bar) =

ab2 (Cside + D) =

ab = min(ab1 , ab2) =

<= 2Fcu/(1+2.D/ab) =

es = fs/Es =

eh = (h-x)/(d-x).es =

em = eh - b(h-x)^2/{3EsAs(d-x)}

ac1 =

ac2 =

acr = max(ac1, ac2) =

= 3acrem/{1+2[(acr-Cmin)/(h-x)]}