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Case Study: Structurally sound?

posted Jun 20, 2015, 9:18 AM by jeffery jim   [ updated Jan 14, 2018, 3:42 PM ]

These photos were hot in a few open discussions at IEM and also my Class of 1995 Whatapps group (engineers with 15 years of experience) when earthquake with the magnitude of 6.0 hit Ranau, Sabah. So did the photo projected out about injustice or substandard built? As professional, we cannot simply summarized things that is not proven and therefore let us explore this issue one item at a time. Let us learn how to use visual judgement method. No scientific calculation needed.

1. Size and Dimension
The Column size is estimated to be 250mmx250mm column which is almost the equivalent size of a commonly used 200mmx200mm reinforced concrete pile. Now, by layman reasoning, a single grade G45 250mmx250mm pile can hold up to 75 Tonne. So by parametric estimate, a grade G30 Structure should be able to carry up to 30 tonne (damn underload) of vertical load.

2. Form and Functions
The area where the columns situated is at the middle of the building with intruded section. Besides that, the function of that area is not a heavy duty area, say with loading of slab at 2.5kN/m². From photo, we can give a rough estimate of 16M² of slab is supported by each column. The edge of the column is carrying 10kN or 1 tonne of loading for each floor. As per self weight and safety factor, say 2.5 tonnes.

3. Design reduction factor.
Now for column design, a reduction of 10% is allowed for each floor with maximum of 40% reduction. Three floors mean 30%. So now,
Overall 3 floors of loading to a column is 2.5 tonne x 3 floors
which is 7.5 tonne and the multiple with 70% (30% reduction)
the column will take 5.25 tonne.

So if the design is able to take an estimated 30 tonne and say add safety factor of 2, the design loading is 10.20 tonne against the column capacity of 30 tonne; then what is the problem with this design? Basically, there is no problem with this design, right. So this is a safe design.



   

Whatapps and Facebook posts became viral



    

      

  

Bricked-up or boxed-up columns and the aftermath


The initial parametric estimate by visual (mentioned and elaborated above) was not accepted by layman, therefore a structural analysis was conducted. The bricked-up reinforced concrete column at Ranau was designed appropriately. Somehow, people are still talking about it. Now let me bury this issue to rest. Below are all the calculation from ESTEEM Structural Software based on BS8110.

For your information, instead of using grade G30 as prescribed by JKR, i use G25 instead. Guess what buggers, the column is still okay. Read the design analysis below and if you can understand.


FLOOR PATHNAME : T:\Design esteem\RANAU COLUMN\Ranau\rf\rf.ccd
COLUMN DETAILED DESIGN CALCULATION:
CODE OF PRACTICE USED IS: BS8110:1985
Floor D.L. L.L. fcu fy cover Load incre.
rf 1.40 1.60 25 460 35 10
Rebar maximum spacing = 250 mm, Minimum spacing = 40 mm
Rebar maximum size = 25 mm, Minimum bar size = 12 mm
Link maximum size = 20 mm, Minimum link size = 6 mm
Minimum rebar percentage used for Column Design = 1.00
Design for Braced Column in X-direction
Design for Braced Column in Y-direction
Yield strain = Yield strength/Young Modulus =0.87*460/200000 =0.0020
Location of Column: 1-A
Floor No. = 5; Live load reduction = 0
Column Fixity: Top X = Fixed; Top Y = Fixed; Bottom X = Fixed; Bottom Y = Fixed
Column Effective Height Coefficient: X = 0.75; Y = 0.75
Column X-Dimension,A = 250 mm; X-Effective Height = 2250 mm
Column Y-Dimension,B = 250 mm; Y-Effective Height = 2250 mm
Factored Upper Moment,Mx = 17.7 kNm; My = 17.7 kNm
Factored Lower Moment,Mx = 15.1 kNm; My = 15.1 kNm
DL=1.40 & LL=1.60 Factored Moment,Mx = 17.7 kNm; My = 17.7 kNm
Dead Load,DL = 94.6 kN; Live Load,LL = 54.4 kN
Load Combination of DL = 94.6 kN & LL = 54.4 kN
Total Ultimate Load,UL = (1+Allowable increase)*(DLFac*DL+LLFac*LL) kN
= (1+0.10)*(1.40* 94.6+1.60* 54.4) = 241.4 kN
So, design for Ultimate Load, N = 241 kN= 241401 N
Ultimate Mx = 17.7 kNm; My = 17.7 kNm
Design For X-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in X-Dimension = Hef/A =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Design For Y-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in Y-Dimension = Hef/B =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Location : 1-A
X-Slenderness ratio = 0.0
X-Slenderness ratio = 0.0 <= 15.0 --> Slender moment = 0.0
Y-Slenderness ratio = 0.0 < 15.0 --> Slender moment = 0.0
Notation: A = Rebar area in mm^2 ; TransY = Transformed Rebar Location in mm
: fs = Rebar stress in N/mm^2 ; fsA = fs x A in kN ; Fs = Sum of fsA
: fd = Rebar lever arm in mm ; fdA = fsA x fd in kNm ; Fsd = Sum of fdA
Calculate Moment Capacity in X direction
DESIGN AXIAL LOAD = 241.4 kN ; NEUTRAL AXIS DEPTH, x = 97.5 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 391.3 78.7 78.7 82.0 6.45 6.45|
| 82.0 -82.0 201 207.0 -400.2 -80.5 -1.8 -82.0 6.60 13.05|
| 82.0 82.0 201 207.0 -400.2 -80.5 -82.3 -82.0 6.60 19.65|
| -82.0 82.0 201 43.0 391.3 78.7 -3.6 82.0 6.45 26.10|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.6 kN ; Fsd = 26.10 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.5 N = 248.6 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.6*(125.0-88.9/2) kNmm = 20.03 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.6+248.6)*1.00 = 245.0 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.10+20.03)*1.00 = 46.13 kN
-------------------------------------------------------------------------------------------
Calculate Moment Capacity in Y direction
DESIGN AXIAL LOAD = 241.4 kN ; NEUTRAL AXIS DEPTH, x = 97.5 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 391.3 78.7 78.7 82.0 6.45 6.45|
| 82.0 -82.0 201 43.0 391.3 78.7 157.3 82.0 6.45 12.90|
| 82.0 82.0 201 207.0 -400.2 -80.5 76.9 -82.0 6.60 19.50|
| -82.0 82.0 201 207.0 -400.2 -80.5 -3.6 -82.0 6.60 26.10|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.6 kN ; Fsd = 26.10 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.5 N = 248.6 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.6*(125.0-88.9/2) kNmm = 20.03 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.6+248.6)*1.00 = 245.0 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.10+20.03)*1.00 = 46.13 kN
-------------------------------------------------------------------------------------------
Eccentricity 20 mm = 0.02*UL = 0.02*241.4 = 4.8 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*241.401 = 3.0 kNm
X-Moment Eccentricity, Mxe = 3.0 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*241.401 = 3.0 kNm
Y-Moment Eccentricity, Mye = 3.0 kNm
Total steel area provided, As = 804 mm^2; i.e. 1.3%
Pure Axial load capacity, Nuz = fac*fcu*Ac+fyy*fy*As
= 0.45*25*62500 + 0.87*460*804 = 1024985 N = 1025.0kN
Alpha, x(>=1 && <=2) = 5*N/(3*Nuz) + 2/3 = 5*241.4/(3*1025.0) + 2/3 = 1.06
(Mx/Mux)^x + (My/Muy)^x = (17.7/46.1)^1.06 + (17.7/46.1)^1.06 
= (0.384)^1.06 + (0.384)^1.06 = 0.73
Biaxial Iteration No. = 4 ; pfy_fcu = 0.24; Mux_bh2fcu = 0.12; Muy_bh2fcu = 0.12;
nXYIt XIt YIt MLayer SLayer Nos As % Nuz Alpha Mx Mux RX My Muy RY Ratio
3 2 2 1x 2T16 1x 2T16 4 804 1.3 1025 1.06 18 46 0.38 18 46 0.38 0.73
Vu, Vu at 1-A = 0.0 kN
Vu Stress, v = V/bd = 0.0*1000/(250*207) = 0.000 N/mm^2
Refer to BS8110:Part 1:1985 Table 3.9
Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/207) = 1.932
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.78) = 0.78
vc = ( 0.79*(0.78)^1/3*(1.932)^1/4*(1.000)^1/3 )/1.25 = 0.685 N/mm^2
Because of Tension force due to Transfer Wall --> Shear Capacity = 0.000 N/mm^2
Vu Stress - Vu Capacity = v - vc = vd = 0.000 - 0.000 = 0.000 N/mm^2
vd < 0.40 N/mm^2 --> Design for vd = 0.40 N/mm^2
Steel area provided by Link size 6, Asv = 2*pie*dia*dia/4 = 2*3.1416*6*6/4 = 56.5 mm^2
Link spacing required, Sv = 0.87*460*56.5/(250*0.400)= 226
Vu Capacity provided by Link = 0.87*460*56.5/(226*250) = 0.400 N/mm^2
Asv/Sv = 56.5/226= 0.250
---------------------------------------------------------------------------------------
Location of Column: 2-A
Floor No. = 5; Live load reduction = 0
Column Fixity: Top X = Fixed; Top Y = Fixed; Bottom X = Fixed; Bottom Y = Fixed
Column Effective Height Coefficient: X = 0.75; Y = 0.75
Column X-Dimension,A = 250 mm; X-Effective Height = 2250 mm
Column Y-Dimension,B = 250 mm; Y-Effective Height = 2250 mm
Factored Upper Moment,Mx = 17.7 kNm; My = 17.7 kNm
Factored Lower Moment,Mx = 15.1 kNm; My = 15.1 kNm
DL=1.40 & LL=1.60 Factored Moment,Mx = 17.7 kNm; My = 17.7 kNm
Dead Load,DL = 94.0 kN; Live Load,LL = 54.2 kN
Load Combination of DL = 94.0 kN & LL = 54.2 kN
Total Ultimate Load,UL = (1+Allowable increase)*(DLFac*DL+LLFac*LL) kN
= (1+0.10)*(1.40* 94.0+1.60* 54.2) = 240.2 kN
So, design for Ultimate Load, N = 240 kN= 240157 N
Ultimate Mx = 17.7 kNm; My = 17.7 kNm
Design For X-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in X-Dimension = Hef/A =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Design For Y-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in Y-Dimension = Hef/B =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Location : 2-A
X-Slenderness ratio = 0.0
X-Slenderness ratio = 0.0 <= 15.0 --> Slender moment = 0.0
Y-Slenderness ratio = 0.0 < 15.0 --> Slender moment = 0.0
Notation: A = Rebar area in mm^2 ; TransY = Transformed Rebar Location in mm
: fs = Rebar stress in N/mm^2 ; fsA = fs x A in kN ; Fs = Sum of fsA
: fd = Rebar lever arm in mm ; fdA = fsA x fd in kNm ; Fsd = Sum of fdA
Calculate Moment Capacity in X direction
DESIGN AXIAL LOAD = 240.2 kN ; NEUTRAL AXIS DEPTH, x = 97.3 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 390.5 78.5 78.5 82.0 6.44 6.44|
| 82.0 -82.0 201 207.0 -400.2 -80.5 -2.0 -82.0 6.60 13.04|
| 82.0 82.0 201 207.0 -400.2 -80.5 -82.4 -82.0 6.60 19.63|
| -82.0 82.0 201 43.0 390.5 78.5 -3.9 82.0 6.44 26.07|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.9 kN ; Fsd = 26.07 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.3 N = 248.0 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.0*(125.0-88.6/2) kNmm = 20.01 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.9+248.0)*1.00 = 244.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.07+20.01)*1.00 = 46.08 kN
-------------------------------------------------------------------------------------------
Calculate Moment Capacity in Y direction
DESIGN AXIAL LOAD = 240.2 kN ; NEUTRAL AXIS DEPTH, x = 97.3 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 390.5 78.5 78.5 82.0 6.44 6.44|
| 82.0 -82.0 201 43.0 390.5 78.5 157.0 82.0 6.44 12.88|
| 82.0 82.0 201 207.0 -400.2 -80.5 76.6 -82.0 6.60 19.47|
| -82.0 82.0 201 207.0 -400.2 -80.5 -3.9 -82.0 6.60 26.07|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.9 kN ; Fsd = 26.07 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.3 N = 248.0 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.0*(125.0-88.6/2) kNmm = 20.01 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.9+248.0)*1.00 = 244.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.07+20.01)*1.00 = 46.08 kN
-------------------------------------------------------------------------------------------
Eccentricity 20 mm = 0.02*UL = 0.02*240.2 = 4.8 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*240.158 = 3.0 kNm
X-Moment Eccentricity, Mxe = 3.0 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*240.158 = 3.0 kNm
Y-Moment Eccentricity, Mye = 3.0 kNm
Total steel area provided, As = 804 mm^2; i.e. 1.3%
Pure Axial load capacity, Nuz = fac*fcu*Ac+fyy*fy*As
= 0.45*25*62500 + 0.87*460*804 = 1024985 N = 1025.0kN
Alpha, x(>=1 && <=2) = 5*N/(3*Nuz) + 2/3 = 5*240.2/(3*1025.0) + 2/3 = 1.06
(Mx/Mux)^x + (My/Muy)^x = (17.7/46.1)^1.06 + (17.7/46.1)^1.06 
= (0.385)^1.06 + (0.385)^1.06 = 0.73
Biaxial Iteration No. = 4 ; pfy_fcu = 0.24; Mux_bh2fcu = 0.12; Muy_bh2fcu = 0.12;
nXYIt XIt YIt MLayer SLayer Nos As % Nuz Alpha Mx Mux RX My Muy RY Ratio
3 2 2 1x 2T16 1x 2T16 4 804 1.3 1025 1.06 18 46 0.38 18 46 0.38 0.73
Vu, Vu at 2-A = 0.0 kN
Vu Stress, v = V/bd = 0.0*1000/(250*207) = 0.000 N/mm^2
Refer to BS8110:Part 1:1985 Table 3.9
Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/207) = 1.932
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.78) = 0.78
vc = ( 0.79*(0.78)^1/3*(1.932)^1/4*(1.000)^1/3 )/1.25 = 0.685 N/mm^2
Because of Tension force due to Transfer Wall --> Shear Capacity = 0.000 N/mm^2
Vu Stress - Vu Capacity = v - vc = vd = 0.000 - 0.000 = 0.000 N/mm^2
vd < 0.40 N/mm^2 --> Design for vd = 0.40 N/mm^2
Steel area provided by Link size 6, Asv = 2*pie*dia*dia/4 = 2*3.1416*6*6/4 = 56.5 mm^2
Link spacing required, Sv = 0.87*460*56.5/(250*0.400)= 226
Vu Capacity provided by Link = 0.87*460*56.5/(226*250) = 0.400 N/mm^2
Asv/Sv = 56.5/226= 0.250
---------------------------------------------------------------------------------------
Location of Column: 1-B
Floor No. = 5; Live load reduction = 0
Column Fixity: Top X = Fixed; Top Y = Fixed; Bottom X = Fixed; Bottom Y = Fixed
Column Effective Height Coefficient: X = 0.75; Y = 0.75
Column X-Dimension,A = 250 mm; X-Effective Height = 2250 mm
Column Y-Dimension,B = 250 mm; Y-Effective Height = 2250 mm
Factored Upper Moment,Mx = 17.7 kNm; My = 17.7 kNm
Factored Lower Moment,Mx = 15.1 kNm; My = 15.1 kNm
DL=1.40 & LL=1.60 Factored Moment,Mx = 17.7 kNm; My = 17.7 kNm
Dead Load,DL = 94.0 kN; Live Load,LL = 54.2 kN
Load Combination of DL = 94.0 kN & LL = 54.2 kN
Total Ultimate Load,UL = (1+Allowable increase)*(DLFac*DL+LLFac*LL) kN
= (1+0.10)*(1.40* 94.0+1.60* 54.2) = 240.2 kN
So, design for Ultimate Load, N = 240 kN= 240157 N
Ultimate Mx = 17.7 kNm; My = 17.7 kNm
Design For X-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in X-Dimension = Hef/A =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Design For Y-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in Y-Dimension = Hef/B =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Location : 1-B
X-Slenderness ratio = 0.0
X-Slenderness ratio = 0.0 <= 15.0 --> Slender moment = 0.0
Y-Slenderness ratio = 0.0 < 15.0 --> Slender moment = 0.0
Notation: A = Rebar area in mm^2 ; TransY = Transformed Rebar Location in mm
: fs = Rebar stress in N/mm^2 ; fsA = fs x A in kN ; Fs = Sum of fsA
: fd = Rebar lever arm in mm ; fdA = fsA x fd in kNm ; Fsd = Sum of fdA
Calculate Moment Capacity in X direction
DESIGN AXIAL LOAD = 240.2 kN ; NEUTRAL AXIS DEPTH, x = 97.3 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 390.5 78.5 78.5 82.0 6.44 6.44|
| 82.0 -82.0 201 43.0 390.5 78.5 157.0 82.0 6.44 12.88|
| 82.0 82.0 201 207.0 -400.2 -80.5 76.6 -82.0 6.60 19.47|
| -82.0 82.0 201 207.0 -400.2 -80.5 -3.9 -82.0 6.60 26.07|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.9 kN ; Fsd = 26.07 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.3 N = 248.0 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.0*(125.0-88.6/2) kNmm = 20.01 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.9+248.0)*1.00 = 244.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.07+20.01)*1.00 = 46.08 kN
-------------------------------------------------------------------------------------------
Calculate Moment Capacity in Y direction
DESIGN AXIAL LOAD = 240.2 kN ; NEUTRAL AXIS DEPTH, x = 97.3 mm
-------------------------------------------------------------------------------------------
| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
-------------------------------------------------------------------------------------------
| -82.0 -82.0 201 43.0 390.5 78.5 78.5 82.0 6.44 6.44|
| 82.0 -82.0 201 207.0 -400.2 -80.5 -2.0 -82.0 6.60 13.04|
| 82.0 82.0 201 207.0 -400.2 -80.5 -82.4 -82.0 6.60 19.63|
| -82.0 82.0 201 43.0 390.5 78.5 -3.9 82.0 6.44 26.07|
-------------------------------------------------------------------------------------------
| Axial Load, fsA & Bending, Fsd: Fs = -3.9 kN ; Fsd = 26.07 kNm |
-------------------------------------------------------------------------------------------
Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.3 N = 248.0 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 248.0*(125.0-88.6/2) kNmm = 20.01 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-3.9+248.0)*1.00 = 244.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.07+20.01)*1.00 = 46.08 kN
-------------------------------------------------------------------------------------------
Eccentricity 20 mm = 0.02*UL = 0.02*240.2 = 4.8 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*240.158 = 3.0 kNm
X-Moment Eccentricity, Mxe = 3.0 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*240.158 = 3.0 kNm
Y-Moment Eccentricity, Mye = 3.0 kNm
Total steel area provided, As = 804 mm^2; i.e. 1.3%
Pure Axial load capacity, Nuz = fac*fcu*Ac+fyy*fy*As
= 0.45*25*62500 + 0.87*460*804 = 1024985 N = 1025.0kN
Alpha, x(>=1 && <=2) = 5*N/(3*Nuz) + 2/3 = 5*240.2/(3*1025.0) + 2/3 = 1.06
(Mx/Mux)^x + (My/Muy)^x = (17.7/46.1)^1.06 + (17.7/46.1)^1.06 
= (0.385)^1.06 + (0.385)^1.06 = 0.73
Biaxial Iteration No. = 4 ; pfy_fcu = 0.24; Mux_bh2fcu = 0.12; Muy_bh2fcu = 0.12;
nXYIt XIt YIt MLayer SLayer Nos As % Nuz Alpha Mx Mux RX My Muy RY Ratio
3 2 2 1x 2T16 1x 2T16 4 804 1.3 1025 1.06 18 46 0.38 18 46 0.38 0.73
Vu, Vu at 1-B = 0.0 kN
Vu Stress, v = V/bd = 0.0*1000/(250*207) = 0.000 N/mm^2
Refer to BS8110:Part 1:1985 Table 3.9
Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/207) = 1.932
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.78) = 0.78
vc = ( 0.79*(0.78)^1/3*(1.932)^1/4*(1.000)^1/3 )/1.25 = 0.685 N/mm^2
Because of Tension force due to Transfer Wall --> Shear Capacity = 0.000 N/mm^2
Vu Stress - Vu Capacity = v - vc = vd = 0.000 - 0.000 = 0.000 N/mm^2
vd < 0.40 N/mm^2 --> Design for vd = 0.40 N/mm^2
Steel area provided by Link size 6, Asv = 2*pie*dia*dia/4 = 2*3.1416*6*6/4 = 56.5 mm^2
Link spacing required, Sv = 0.87*460*56.5/(250*0.400)= 226
Vu Capacity provided by Link = 0.87*460*56.5/(226*250) = 0.400 N/mm^2
Asv/Sv = 56.5/226= 0.250
---------------------------------------------------------------------------------------
Location of Column: 2-B
Floor No. = 5; Live load reduction = 0
Column Fixity: Top X = Fixed; Top Y = Fixed; Bottom X = Fixed; Bottom Y = Fixed
Column Effective Height Coefficient: X = 0.75; Y = 0.75
Column X-Dimension,A = 250 mm; X-Effective Height = 2250 mm
Column Y-Dimension,B = 250 mm; Y-Effective Height = 2250 mm
Factored Upper Moment,Mx = 17.7 kNm; My = 17.7 kNm
Factored Lower Moment,Mx = 15.1 kNm; My = 15.1 kNm
DL=1.40 & LL=1.60 Factored Moment,Mx = 17.7 kNm; My = 17.7 kNm
Dead Load,DL = 93.3 kN; Live Load,LL = 54.1 kN
Load Combination of DL = 93.3 kN & LL = 54.1 kN
Total Ultimate Load,UL = (1+Allowable increase)*(DLFac*DL+LLFac*LL) kN
= (1+0.10)*(1.40* 93.3+1.60* 54.1) = 238.9 kN
So, design for Ultimate Load, N = 239 kN= 238914 N
Ultimate Mx = 17.7 kNm; My = 17.7 kNm
Design For X-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in X-Dimension = Hef/A =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Design For Y-Braced Column
Effective height,Hef = 2250 mm
Slenderness ratio,sr in Y-Dimension = Hef/B =2250/ 250 = 9.0
Slenderness ratio = 9.0 < 15 ---> No additional moment
Location : 2-B
X-Slenderness ratio = 0.0
X-Slenderness ratio = 0.0 <= 15.0 --> Slender moment = 0.0
Y-Slenderness ratio = 0.0 < 15.0 --> Slender moment = 0.0
Notation: A = Rebar area in mm^2 ; TransY = Transformed Rebar Location in mm
: fs = Rebar stress in N/mm^2 ; fsA = fs x A in kN ; Fs = Sum of fsA
: fd = Rebar lever arm in mm ; fdA = fsA x fd in kNm ; Fsd = Sum of fdA
Calculate Moment Capacity in X direction
DESIGN AXIAL LOAD = 238.9 kN ; NEUTRAL AXIS DEPTH, x = 97.0 mm
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| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
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| -82.0 -82.0 201 43.0 389.7 78.4 78.4 82.0 6.42 6.42|
| 82.0 -82.0 201 207.0 -400.2 -80.5 -2.1 -82.0 6.60 13.02|
| 82.0 82.0 201 207.0 -400.2 -80.5 -82.6 -82.0 6.60 19.62|
| -82.0 82.0 201 43.0 389.7 78.4 -4.2 82.0 6.42 26.05|
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| Axial Load, fsA & Bending, Fsd: Fs = -4.2 kN ; Fsd = 26.05 kNm |
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Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.0 N = 247.4 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 247.4*(125.0-88.4/2) kNmm = 19.98 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-4.2+247.4)*1.00 = 243.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.05+19.98)*1.00 = 46.03 kN
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Calculate Moment Capacity in Y direction
DESIGN AXIAL LOAD = 238.9 kN ; NEUTRAL AXIS DEPTH, x = 97.0 mm
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| Rebar Coord/Area,A TransY fs fsA Fs fd fdA Fsd| 
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| -82.0 -82.0 201 43.0 389.7 78.4 78.4 82.0 6.42 6.42|
| 82.0 -82.0 201 43.0 389.7 78.4 156.7 82.0 6.42 12.85|
| 82.0 82.0 201 207.0 -400.2 -80.5 76.2 -82.0 6.60 19.45|
| -82.0 82.0 201 207.0 -400.2 -80.5 -4.2 -82.0 6.60 26.05|
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| Axial Load, fsA & Bending, Fsd: Fs = -4.2 kN ; Fsd = 26.05 kNm |
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Concrete Axial Load Capacity,Fcc = k1*bx*x = 10.20*250*97.0 N = 247.4 kN
Concrete Bending Capacity, Fcd = Fcc*(by2- a/2) = 247.4*(125.0-88.4/2) kNmm = 19.98 kNm
Total Axial Load Capacity = (Fs + Fcc)*Fac = (-4.2+247.4)*1.00 = 243.1 kN
Total Bending Capacity = (Fsd + Fcd)*Fac = (26.05+19.98)*1.00 = 46.03 kN
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Eccentricity 20 mm = 0.02*UL = 0.02*238.9 = 4.8 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*238.914 = 3.0 kNm
X-Moment Eccentricity, Mxe = 3.0 kNm
Eccentricity 5 percent = 0.05*h*UL = 0.05*250*238.914 = 3.0 kNm
Y-Moment Eccentricity, Mye = 3.0 kNm
Total steel area provided, As = 804 mm^2; i.e. 1.3%
Pure Axial load capacity, Nuz = fac*fcu*Ac+fyy*fy*As
= 0.45*25*62500 + 0.87*460*804 = 1024985 N = 1025.0kN
Alpha, x(>=1 && <=2) = 5*N/(3*Nuz) + 2/3 = 5*238.9/(3*1025.0) + 2/3 = 1.06
(Mx/Mux)^x + (My/Muy)^x = (17.7/46.0)^1.06 + (17.7/46.0)^1.06 
= (0.385)^1.06 + (0.385)^1.06 = 0.73
Biaxial Iteration No. = 4 ; pfy_fcu = 0.24; Mux_bh2fcu = 0.12; Muy_bh2fcu = 0.12;
nXYIt XIt YIt MLayer SLayer Nos As % Nuz Alpha Mx Mux RX My Muy RY Ratio
3 2 2 1x 2T16 1x 2T16 4 804 1.3 1025 1.06 18 46 0.39 18 46 0.39 0.73
Vu, Vu at 2-B = 0.0 kN
Vu Stress, v = V/bd = 0.0*1000/(250*207) = 0.000 N/mm^2
Refer to BS8110:Part 1:1985 Table 3.9
Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/207) = 1.932
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.78) = 0.78
vc = ( 0.79*(0.78)^1/3*(1.932)^1/4*(1.000)^1/3 )/1.25 = 0.685 N/mm^2
Because of Tension force due to Transfer Wall --> Shear Capacity = 0.000 N/mm^2
Vu Stress - Vu Capacity = v - vc = vd = 0.000 - 0.000 = 0.000 N/mm^2
vd < 0.40 N/mm^2 --> Design for vd = 0.40 N/mm^2
Steel area provided by Link size 6, Asv = 2*pie*dia*dia/4 = 2*3.1416*6*6/4 = 56.5 mm^2
Link spacing required, Sv = 0.87*460*56.5/(250*0.400)= 226
Vu Capacity provided by Link = 0.87*460*56.5/(226*250) = 0.400 N/mm^2
Asv/Sv = 56.5/226= 0.250
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SUMMARY OF TOTAL LOADING WITHOUT LOAD ALLOWANCE OF 10 PERCENT:
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Total number of Columnn Section for Analysis = 4
Total number of Columnn Section being Analysed = 4
Total Column Selfweight = 18 kN
Total Dead Load in key plan = 358 kN
Total Dead Load in the floor = 376 kN
Total Live Load in the floor = 217 kN
Total Dead Load in the floor & floors above = 376 kN
Total Live Load in the floor & floors above = 217 kN



Tabulation of RC Columns design





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