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Corbel Design For Column : Project

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Section : 3B14, 3B15

Type : Corbel Design


Design Code : BS8110
Prepared By : Date : 7/4/2020
Project Checked By : Job ID :

Corbel Design For Column (Based on BS8110:Part1)


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Column Name - C8a
1. DESIGN DATA d2
Applied Ultimate Load, Pu = 460.0 KN
Lower Column Depth, d1 = 200 mm Pu
Lower Column Width, b1 = 500 mm av
Upper Column Width, d2 = 200 mm
Corbel Width, bc = 500 mm
h1
Corbel Projection from Lower Column, Lp = 200 mm
Corbel Height, h1 = 700 mm
h2
Corbel Height, h2 = 0 mm
Effective Depth of Corbel, d = 650 mm [OK]
Lp
Distance b/t Load & Up Col. Edge, av = 75 mm < d
Angle of Inclination of Concrete Strut, β =
Concrete Strength, fcu
d1
= 30 N/mm2
Steel Strength of Stirrups, fyv = 460 N/mm2
Steel Strength of Tension Bars, fys = 460 N/mm2
Elastic Modulus of Steel, Es = 205 KN/mm2
Concrete Ultimate Strain, ϵcc = 0.0035
Concrete Ultimate Strain, ϵys = 0.002
2. DESIGN FOR TENSION REINFORCEMENT
To Check Tensile Steel Yield or Not Yield
k = Pu / 0.9 fcu * bc * av = 0.45
Lever arm, z = d(0.5+√(0.25-(1+k)kav^2/d^2)) / (1+k)
= 443.0 mm
= 0.68 d
Neutral Axis Depth, x=(d-z)/0.45 = 460.05 mm
= 0.71 d > 0.636d [NOT OK]
Steel Strain, ϵs = ϵcc*((d-x)/x) = 0.00145 < 0.002
Therefore, tension steel will not yield.
When Tensile Steel Yields,
Steel Stress, fs = 0.87*fys = 400.20 N/mm2
WhenTensile Steel Does Not Yield,
Steel Stress, fs = Es * ϵs = 296.25 N/mm2
Tensile Force, Ft = Pu* (av/z) = 77.88 KN < 0.5*Pu = 230 KN
For Ft < 0.5*Pu,
Take Ft = 230.00 KN
Required As = Ft/fs = 776.38 mm2
Provide 5T 20 = 1570.8 mm2 [OK]
Section : 3B14, 3B15
Type : Corbel Design
Design Code : BS8110
Prepared By : Date : 7/4/2020
Project Checked By : Job ID :
For Ft > 0.5*Pu,
Take Ft = KN
Required As = Ft/fs = mm2
Min. Reinforecment , As,min = 0.004 bd = 1300 mm2
Max. Reinforecment, As,max = 0.4 bd = 130000 mm2
Provide 5T 20 = mm2
3. CHECK SHEAR DESIGN
Shear Stress, v = Pu / bd = 1.42 N/mm2 < 0.8*√fcu [OK]
100As/bd = 0.48
400/d = 1.00
Concrete Shear Stress, vc = (fcu/25)^(1/3) * (0.79*(100As/bd)^(1/3) * (400/d)^0.25 / 1.25)
= 0.53 N/mm2
Enhanced Shear Stress, vc' = vc*(2d/av)
= 9.14 N/mm2 > v= 1.42 [OK]
vs = v-vc' = -8.65 N/mm2 < 0.4
Take vs = 0.4 N/mm2
Required Asv = (vs*av*b / 0.87 fyv) = 37.48 mm2 < 0.5As
Min. As = 0.5 As = 388.19 mm2
Provide Links 1T 10 @ 100 (In the Upper 2/3 of Effective Depth)
Upper 2/3 of Effective depth of Corbel = 433 mm
Total No. of Links = INT((2/3)*d / sv)+1
= 5.0 Nos
Total Link Area, Asv = 785.4 mm2 [OK]
4. CALCULATE ANCHORAGE LENGTH
Bearing Stress inside the bend, Fbt = (Ft / No. of bars)*(As,req / As,Prov)
= 22.74 KN
Ultimate Anchorage Bond Stress, fbu = 0.5 √fcu (BS8110-Part1, Table-3.28)
= 2.74 N/mm2
Required Anchorage Length, Lb=Fbt/πφfbu= 140 mm
Anchorage Length for Standard Bend, Lb = 12*φ
= 240 mm
Therefore, Provide Anchorage Length = 240 mm

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