Concrete Calculation AS3600:2018
Concrete Data
The concrete grade is 40 MPa
The steel yield strength is 500 MPa
Section data
The section depth is 400 mm
The section width is 1000 mm
The concrete cover is 50 mm
The longitudinal reinforcement diameter is 16 mm
The number of layers of longitudinal reinforcement is 1
The transverse reinforcement diameter is 16 mm
The area of longitudinal reinforcement is N16-200 = 1005 mm²/m
The area of transverse reinforcement is N16-200 = 1005 mm²/m
Design Actions
The ULS bending moment is M* = 185 kNm
The ULS shear V* = 255 kN
Bending - Calculated As
The effective depth is d =400-50-16-(0.5+(1-1) x 16,0) = 326 mm
The effective depth of the outermost layer of the tension bars is do =400-50-16-0.5 x 16 = 326 mm
The parameter γ = MAX(0.97-0.0025 x 40,2) = 0.87
The parameter α₂ = MAX(0.85-0.0015 x 40,2) = 0.79
The depth of the compression block is a = γ ku d = 326-(326^2-(2 x 185 x 1000000/(0.85 x 0.79 x 40 x 1000))^0.5,1) = 21.9 mm
The neutral axis depth of the section is dn = 21.9/0.87 = 25.2 mm
The strain at the tension reinforcement is εs = 0.003 x (326-25.2/18.3,4) = 0.0493
The required area of steel reinforcement is As = 185 x 1000000/(0.85 x 500 x (326-21.9/2),0) = 1382 mm²
Bending - Moment section capacity
The effective depth is d =400-50-16-(0.5+(1-1) x 16,0) = 326 mm
The effective depth of the outermost layer of the tension bars is do =400-50-16-0.5 x 16 = 326 mm
The parameter γ = MAX(0.97-0.0025 x 40,2) = 0.87
The parameter α₂ = MAX(0.85-0.0015 x 40,2) = 0.79
The neutral axis depth of the section is dn = 1005 x 500/(0.87 x 0.79 x 40 x 1000,1) = 18.3 mm
The strain at the tension reinforcement is εs = 0.003 x (326-18.3/18.3,4) = 0.0504
The parameter ku = 18.3/326 = 0.056
The parameter kuo = 0.056 x 326/326 = 0.056
The ultimate strength in bending is Muo = 1005 x 500 x 326 x (1-0.87 x 0.056/2/1000000,1) = 159.8 kNm
The strength reduction factor in bending is φ = MIN(0.85,MAX(0.65,1.24-13 x 0.056/12)) = 0.85
The design capacity is φMuo = 0.85 x 159.8 = 135.8 kNm
The utilisation ratio is UR = 185/135.8 = 1.36
Beam Shear Action
The shear strength reduction factor is φ = 0.75
The kv value is considered as 0.15
The compression struct angle is ϑv = 36 ˚
The effective shear depth is dv = MAX(0.72 x 400, 0.9 x 326) = 293.4 mm
The concrete shear strength is fcv = MIN(40^0.5, 8 MPa) = 6.32 MPa
The concrete shear design capacity is φVuc = 0.75 x 0.15 x 6.32 x 1000 x 293.4/1000 = 208.6 kN
Excess shear carried by reinforcement φVus = MAX(255.0-208.6,1) = 46.4 kN
The required reinforcement in shear is (Asv / s) req'd = 1000 x (1000 x 46.4 x TAN(36 x PI(/180))/(0.75 x 500 x 293.4),0)=306 mm²/m
The reinforcement shear capacity is (1005 x 500 x 293.4/TAN(36 x PI(/180))/1000000,1)=202.9 kN
The concrete shear design capacity is φVu = 208.6+202.9=411.5 kN
The utilisation ratio is UR = 255.0/411.5=0.62
The footing thickness in beam shear action is OK
Punching Shear Action (to be reviewed)