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ɸVc = 4ɸ √𝑓′𝑐 bod = 4 x 0.75 x √4000 x 4 x (20 + d) x d
1000
ɸVc ≥ Vu, Vu = ɸVc
3.059 x [(9.5 x 9.5) - (
(20 + d)2
144
)] = 4 x 0.75 x √4000 x 4 x (20 + d) x d
1000
d = 10.9”
Take d= 11”
Check one way shear
Vu = qu x b [(b − c
2
) - d]
Vu = 3.059 x 9.5 [(9.5 − 1.75
2
) – 0.917] = 86 kip
ɸVc = 2ɸ √𝑓′𝑐 bod = 2 x 0.75 x √4000 x 4 x (20 + 11) x 11
1000
= 129.4 kip
ɸVc > Vu
Flexural Moment
Mu = wL2
/8 = 3.059 x 9.5 x (9.5 – 2.667) /8 = 24.82 kip-ft
Assume a = 0.3”
As = Mu / ɸfy (d - a/2)
= 24.82 x 12 / 0.9 x 50 (11 – 0.3/2) = 0.61 in2
9’-6”
20” + d”
9’-6”
Check a
a = Asfy/0.85f’cb = 0.06 x 50000 / 0.85 x 4000 x 9.5 = 0.94” (not ok)
Assume a = 0.95”
As = Mu / ɸfy (d - a/2)
= 24.82 x 12 / 0.9 x 50 (11 – 0.95/2) = 0.628 in2
Check a
a = Asfy/0.85f’cb = 0.628 x 50000 / 0.85 x 4000 x 9.5 = 0.97” (Satisfied)
Use 4 # 4 (As, provided = 0.8 in2 > As, actual = 0.628 in2
)
Dowel Steel Area, As = 0.005 (31” x 31”) = 4.8 in2
Use 4 # 10 (As, provided = 1.27 x 4 = 5.08 in2 > As, actual = 4.8 in2
)
Determination of tension development length Ld
Required development length, Ld = 41db = 41 x 0.5 = 20.5”
Available development length = b/2 - 12/2 - 3 (d/2)
= ( 9.5 x 12/2 ) - 20/2 - 2.5
= 44.5” > 20.5”
Determination of dowel steel requirement
Use #10 bar (i.e. same size of column bar) for dowel steel.
Requirement development length in compression, Ldc = 20 in (Table A.11)
Min. lapping splice length = 0.0005 x 50000 x (1.27) = 31.75”
Splice length < required development length (Satisfied)
Dowel bar length = Ldc + min lap splice length = 20 + 31.75” = 51.75”
Min. area of dowels = 0.005 x gross area of the supported column = 0.005 x 400
= 2 in2
Overall depth of footing, h = d + 1.5 db + 3 = 11 + (1.5 x 0.5) + 2.5 = 14.25”
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