# Reinforced Concrete Beam Design

This calculation provides the moment and shear capacity of a beam per CSA A23.3

## Input:

in
in
psi
ksi
in2
in2
in
deg ## Output:

1. 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)
2. 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
3. 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
4. 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