Adequacy Check of Slender Column (USD Method, ACI Code)
Top of the floor
X-X (grid)= 12 ft (left panel) 14
beam geometry: 12 in (width) 12
12 in (depth) 14
Y-Y (grid)= 16 ft (upper panel) 20
beam geometry: 12 in (width) 12
16 in (depth) 20
Lower of the floor
X-X (grid)= 12 ft (left panel) 14
beam geometry: 12 in (width) 12
12 in (depth) 14
Y-Y (grid)= 16 ft (upper panel) 20
beam geometry: 12 in (width) 12
16 in (depth) 20
Dimension (Column of interest) 12 in (X- axis) 15
Immediate upper column 12 in (X- axis) 15
Immediate lower column 12 in (X- axis) 15
Unsupported Length of Column = 132 in
11 ft
From ETABS analysis Dead load Live load
Max.Compression in Column 200 kip 120
Sway Moment on Top of the Column 200 kip-in 100
Sway Moment on Bottom of the Column 130 kip-in 80
Material specification
steel yield strength, fy = 60 ksi
concrete crushing strength, f'c = 4 ksi
Sway/Nonsway Check
If Stability Index, Q=ΣPuΔo/VuLc ≤ 0.05 then Nonsway Frame
Now, Σpu = total factored vertical load of that storey= 2000.0 kip (all columns)
Δo =Relative def. between top & bot.of column= 0.5 in
Vu =Shear force in column = 30 kip
Lc = C/C distance of joints of column in the frame 12 ft
So, Stability Index, Q=ΣPuΔo/VuLc = 0.23 Sway frame (check)
Slenderness Check (Sway Frame)
ACI 10.12.2 (318/318R-128) Criteria klu/r <22
when Ψm = (ΨA + ΨB)/2 < 2 , k = (20 - Ψm)/20 √(1+Ψm)
when Ψm = (ΨA + ΨB)/2 ≥ 2 , k = 0.9√(1+Ψm)
ΨA = At Top of member Column= Σ(EI/L)column / Σ(EI/L)beam
ΨB = At Bot of member Column = Σ(EI/L)column / Σ(EI/L)beam
moment of inertia of beam 0.35 Ig in^4
moment of inertia of column 0.70 Ig in^4
Now
Σ(EI/l)column at Top = 30.00 E in^3
Σ(EI/l)beam at Top = 28.33 E in^3
So, ΨA = 5.58
Again
Σ(EI/l)column at Bot 27.86 E in^3
Σ(EI/l)beam at Bot 28.33 E in^3
So, ΨB = 1.01
Now
Ψm = (ΨA + ΨB)/2 3.29 0
Now, k = 1.864760038182 Don't delete these two cells
Unsupported length ,lu = 132.00 in
Min Radius of Gyration , r = 0.3a 3.60 in
Now, klu/r = 68.37453473333 effect of slenderness must be c
ACI Moment Magnifier Method (For Sway Frame)
The Magnified Sway Moment, δsMs = Ms/(1 -Q)
when δs = 1/(1 -Q) ≤ 1.5
Now, δs = 1/(1 -Q) = 1.30
Another condition, δsMs = Ms/(1 -Pu/(0.75Pc))
Where, Pc = π^2EI/(klu)^2
EI = 0.4EcIg /(1+βd)
Ec = 57000√fc' = 3604996.53 psi
Es = 8Ec = 28839972.26 psi
Now EI = 2002318074.10 in^2
Pc= 300.24 kip
Again δs = 1/(1 -Pu/(0.75Pc))= -1.08879039224
So Magnification facttor of Moment= 1.30
Magnified factored Moment = 520.48192771 kip-in
Column Strength By ACI Code 10.3.5 (on an another note)
As (provided)= 1.24 sq. in
Φ 0.7
Pu (capacity)= 382.02 kip
od, ACI Code)
ft (right panel)
in (width)
in (depth)
ft (lower panel)
in (width)
in (depth)
ft (right panel)
in (width)
in (depth)
ft (lower panel)
in (width)
in (depth)
in(Y-. axis) 12 ft long
in(Y-. axis) 12 ft long
in(Y-. axis) 14 ft long
kip
kip-in
kip-in
choose Sway direction
(insert "1" for preferred one
and "0" for the rest)
X -grid 1
Y -grid 0
`
1
't delete these two cells
of slenderness must be considered
