Department of Civil Engineering, N-W.F.P UET, Peshawar
Design of Retaining Walls
d = 14.5" 3.50 k/ft
M u= 578.8 in-k/ft
5'-3" = 5.25' Figure 11: Factored loads on Heel slab.
Step No 4: Design. i.
Arm: Design for flexure: For h = 13.5′ Mu = 45489.7 ft-lb/ft = 545.87 in-k/ft Asmin = 3{√(4500)}/60000)bd = 0.0034 x 12 x (15 – 2 – 0.5) = 0.51 in2/ft a = Asminfy/ (0.85fc′b) = 0.51 x 60/ (0.85 x 4.5 x 12) = 0.66″ Φ Mn = ΦAsminfy (d-a/2) = 0.9 x 0.51 x 60 x (12.5 - 0.66/2) = 469.7 in-k/ft (39142 ft-lb/ft) Φ Mn calculated from Asmin is less than maximum moment in the arm. Therefore using trial method. Trial i: As = Mu/ 0.9fy(d – a/2) Let a = 0.2d = 0.2 x 12.5 = 2.5″ As = 545.87/ {0.9 x 60 x (12.5 – 2.5/2} = 0.90 in2 Trial ii: a = Asfy/ (0.85fc′b) = 0.90 x 60/ (0.85 x 4.5 x 12) = 1.17 in As = 545.87/ {0.9 x 60 x (12.5 – 1.17/2)} = 0.848 in2
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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