Reinforced Concrete Beam Design
This calculation provides the moment and shear capacity of a beam per CSA A23.3
Output:
- Definitions
- ϕc = Resistance factor for concrete
- ϕs = Resistance factor for reinforcing steel
- fc = Specified concrete strength (psi)
- fy = Specified reinforcing steel yield strength (ksi)
- As = Area of reinforcing steel (in2)
- α1 = Stress modification factor for equivalent stress block
- β = Cracked concrete shear factor
- d = Depth to primary reinforcing (in)
- b = Effective width of cross section (in)
- Av = Area of transverse (shear) reinforcing (in2)
- s = Spacing of rebar (in)
- λ = Concrete density modification factor; 1.0 for normal density concrete
- dv = Shear depth, normally taken as 0.9d (in)
- θ = Assumed shear crack inclination (deg)
- Determine the equivalent compression block depth, a
- a = (ϕs As fy) / (ϕc α1 fc b)
- where α1 = 0.85 - 0.0015*fc = 0.8 (fc in MPa)
- a = ((0.85) (5) (60) (1000psi/ksi)) / ((0.65) (0.8) (5000) (12))
- a = (255000) / (31200)
- a = 8.173 in
- Determine the flexural capacity
- Mr = ϕs As fy (d - a/2)
- Mr = (0.85) (5) (60) (24 - 8.173/2)
- Mr = (255) (19.914)
- Mr = 5077.94 kip-in = 423.16 kip-ft
- Determine the shear capacity
- Vr = Vc + Vs ≤ Vrmax
- where:
- dv = 0.9d = 21.6 in
- Vc = ϕc λ β √fc b dv = (0.65) (1) (0.18) (0.852) (12) (21.6) = 25.83 kip
- Vs = ϕs Av fy dv cot(θ) / s = (0.85) (0.155) (60) (21.6) cot(35) / (12) = 20.32 kip
- Vrmax = 0.25 ϕc fc b dv = 0.25 (0.65) (5000) (12) (21.6) / 1000 = 210.6 kip
- Vr = (25.83) + (20.32) ≤ (210.6)
- Vr = 46.15 kip
Notes: