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:

  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)
    • 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)
  2. Calculate soil bearing stresses
    • Ps = (D + L)/Lf = (10 + 5) / 1 = 15 kip/ft
    • Pu = max[1.4D, 1.25D + 1.5L] / Lf = max[14, 20] / 1 = 20 kip/ft
    • qs = Ps / B = 15 / (18/12) = 10 ksf --> Service bearing capacity exceeded! Revise design
    • qu = Pu / B = 20 / (18/12) = 13.33 ksf --> Ultimate bearing capacity exceeded! Revise design
  3. Calculate shear capacity at critical section, located 'dv' away from the face of the wall
    • d = h - cover = 30 - 3 = 27 in
    • dv = max[0.9d, 0.72h] = 0.9 (27) = 24.3 in
    • β = 0.21 if h < 14in or (B-tw < 2dv)). Else β = 230/(1000 + dv*25.4). --> β = 0.21
    • tw + dv > B --> Vu = 0 kip
    • Vc = ϕc λ β √fc Lf dv = (0.65) (1) (0.21) (0.852) (1) (24.3) = 33.9 kip
    • Vu / Vr = 0 / 33.9 = 0 --> okay
  4. Calculate the flexural capacity at the critical section, located at the face of the wall
    • Mu = qu Lf (B-tw)2 / 2 = (13.33) (1) (18 - 12)2 / 2 = 240 kip-in = 20 kip-ft
    • Mr = ϕs As fy (d - a/2), where a = (ϕs As fy) / (ϕc α1 fc Lf)
    • Mr = (0.85) (5) (60) (27 - 5.449/2)
    • Mr = (255) (24.276)
    • Mr = 6190 kip-in = 515.85 kip-ft
    • Mu / Mr = 20 / 515.85 = 0.04 --> okay

Notes: