Strip Footing Capacity

This tool calculates the bearing, shear, and flexural capacity of a strip footing in accordance with CSA A23.3

Input:

Materials:
psi
ksi
Geometry:
in
in
in
in
in2
ft
Loads:
kip
kip
ksf
ksf
concrete beam

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)
    • λ = Concrete density modification factor; 1.0 for normal density concrete
    • 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 = Width of footing (in)
    • Lf = Footing length (ft)
    • tw = Wall thickness (in)
    • Cover = concrete cover to centroid of reinforcing (in)
    • qa = Allowable bearing resistance for service conditions (ksf)
    • qult = Ultimate bearing resistance (ksf)
    • qs = Service bearing demand (ksf)
    • qu = Factored ultimate bearing demand (ksf)
    • dv = Shear depth, normally taken as 0.9d (in)
    Calculate soil bearing stresses
    DCR = 91%
    • Ps = (D + L)/Lf = (4 + 3) / 1 = 7 kip/ft
    • Pu = max[1.4D, 1.25D + 1.5L] / Lf = max[5.6, 9.5] / 1 = 9.5 kip/ft
    • qs = Ps / B = 7 / (30/12) = 2.8 ksf
    • qu = Pu / B = 9.5 / (30/12) = 3.8 ksf
    Calculate shear capacity at critical section, located 'dv' away from the face of the wall
    DCR = 25%
    • d = h - cover = 12 - 3 = 8.6875 in
    • dv = max[0.9d, 0.72h] = 0.72 (12) = 8.64 in
    • β = 0.21 if h < 14in or (B-tw < 2dv)). Else β = 230/(1000 + dv*25.4). --> β = 0.21
    • Vu,crit = qu Lf (B - tw - dv) = (3.8) (1) (30 - 12 - 8.64)/12 = 2.96 kip
    • Vc = ϕc λ β √fc Lf dv = (0.65) (1) (0.21) (0.852) (1) (8.64) = 12.05 kip
    • Vu / Vr = 2.96 / 12.05 = 0.25 --> okay
    Calculate the flexural capacity at the critical section, located at the face of the wall
    DCR = 3%
    • Mu = qu Lf (B-tw)2 / 2 = (3.8) (1) ((30 - 12)/12)2 / 2 = 51.36 kip-in = 4.28 kip-ft
    • Mr = ϕs As fy (d - a/2), where a = (ϕs As fy) / (ϕc α1 fc Lf)
    • Mr = (0.85) (5) (60) (8.6875 - 3.269/2)
    • Mr = (255) (7.053)
    • Mr = 1799 kip-in = 149.88 kip-ft
    • Mu / Mr = 4.28 / 149.88 = 0.03 --> okay
    User Notes: