Analysis of Columns in ULS strength check

1.       For cross section subjected the pure longitudinal compression the compressive strain in concrete is limited to ε_c2.

2.       For cross section subjected to axial compressive force and bending moment where the cross section is not fully in compression and neutral axis lies within the section the strain at the most compressed face shall be taken as ε_cu2 with parabolic rectangular – stress strain diagram.

3.       For cross section subjected to axial compression and bending moment and the neutral axis lies out side the section the strain at the out most fiber shall be reduced maintaining the stress at a distance of (1- ε_c2/ ε_cu2)h  from the most compressed force as ε_c2.

The particular case arrives when the columns suffers only compressive strain and no tensile strain is developed. The strain at the highly compressed face is .0035-0.75 times the strain at the least compressive face.

In this case column extreme fiber shall ne be allowed to reach a strain of .0035 unless the other face strain is zero.

For a columns.

                 $$N={{f_{av}bx+f'{A'_s}-{f_s}{A_s}-(1)}}$$

Taking moments about center

                   $$M_{rd}={\beta_1f_{cd}bx_u}\left({{h\over 2} - \beta_2 x}\right) + \sum{f'_s A'_s \left({h\over 2} - d' \right)} + \sum{f_s A_s \left({h\over 2} - d \right)}$$

Step 1:          Assume neutral axis 

Step 2:          Determine the strain in steel and stress corresponding to assumed neutral axis

$${f_c}={f_{cd}}{\left(1-{\left(1-{{\epsilon_c}\over {\epsilon_c}}\right)}^2\right)}$$

Step 3:          Substitute in equation 1 and arrive at the value of

Step 4:          Estimate the moment capacity and ensure it is greater than M applied.

Step 5:          If it does not satisfy change the value of and next trial can be started.

or 

Use following VBA function code to calculate iterative Neutral axis automatically which fulfills the criteria of 

C + T = P

Where C = force in Compression above neutral axis

            T = force in Tension above neutral axis

            P = Axial force in column

For Circular Pier

Public Function xulc(D, cov, dia, N, Fcd, fyd, pu)

Dim min As Double

Dim max As Double

Dim mid As Double

min = 1

max = 12 * D

abc:

mid = (min + max) / 2

If (pulc(D, cov, dia, N, Fcd, fyd, mid) > pu) Then

max = mid

Else

min = mid

End If

If ((Application.Round(pulc(D, cov, dia, N, Fcd, fyd, mid), 5)) = Application.Round(pu, 5)) Then

xulc = mid

GoTo enda

Else:

GoTo abc

enda:

End If

End Function

For Rectangular Pier

Public Function xu(b, D, cov, dia, nb, nd, Fcd, fyd, spacd, pu)

Dim min As Double

Dim max As Double

Dim mid As Double

min = 1

max = 12 * D

abc:

mid = (min + max) / 2

If (pua(b, D, cov, dia, nb, nd, Fcd, fyd, spacd, mid) > pu) Then

max = mid

Else

min = mid

End If

If ((Application.Round(pua(b, D, cov, dia, nb, nd, Fcd, fyd, spacd, mid), -1)) = Application.Round(pu, -1)) Then

xu = mid

GoTo enda

Else:

GoTo abc

enda:

End If

End Function

Biaxial Bending

Step 1:          Design the column independently in each direction.  

Step 2:          If the column satisfies both the equation 8.1 and 8.2 of IRC 112: 2011 then no further check is necessary.

Step 3:          If equations are not satisfied then Biaxial bending to be considered and Equ. 8.3 of IRC 112: 2011 to be satisfied. 

Worked Example Reinforced concrete pier Biaxial bending

RCC pier of size = 1800 x 1500 mm 

Reinforcement = 34 Nos of 32 mm dia

P =17300 KN axial load

M about major axis = 8000  KNm 

M about minor axis = 6000 KNm  

Grade of concrete = M60 and 

Grade of steel  = Fe 500

 

Section	Bars
B = 1800 mm	18
D = 1500 mm	16

ω=Asfy d/(Acfc d)	Pt/Fc k
0.161	0.017

Ac (mm2)	RH
2,672,656	70.00%

LOAD COMB=	ULS
Axial Load Pu (kN) =	17300.0
MoEd Current axis (kN.m) =	6000.0
MoEd traffic axis (kN.m) =	8000.0
φefy=φ(∞,t0)(MoEqpy/MoEdy)=	0.00
φefz=φ(∞,t0)(MoEqpz/MoEdz)=	0.11
A Y=1/(1+0.2φefy)=	1.00
A Z=1/(1+0.2φefz)=	0.98
n = N ED / (A c f cd)=	0.235
λlim Y =20 A B C /sqrt(n)=	33.25
Slander Ratio λ y =	4.4
λ lim z =20 A B C /sqrt(n)=	32.53
Slender Ratio λ z =	5.3
n u = 1 + ω=	1.161
K r=(n u - n ) / ( n u - n bal )=	1.217
(1/r0)z=εyd/(0.45 dz)=	3.7E-06
βz=0.35+fck/200-λ/150=	0.615
Kφz=1+βz φef=	1.068
(1/r)z=K r Kφz (1/r0)z=	4.8E-06
e2z=(1/r) (le2)/c=	2.57
M2z=Ned e2z=	0.00
(1/r0)y=εyd/(0.45 dy)=	3.1E-06
βy=0.35+fck/200-λ/150=	0.620
Kφy=1+βy φef=	1.000
(1/r)y=K r Kφy (1/r0)y=	3.8E-06
e2y=(1/r) (le2)/c=	2.02
M2y=Ned e2y=	0.00
Xuz =	528
XuY =	641
MEdz (kN.m) =	6000
MEdy (kN.m) =	8000
MuZ1 (kN.m) =	16696
MuY1 (kN.m) =	19939
Pu / Puz =	1.0001
α =	1.11
(MY/MY1)α+ (MZ/MZ1)α =	0.68
	OK

Important VBA codes

Axial Load Capacity for given Neutral axis position

For Circular Pier

Public Function pulc(D, cov, dia, N, Fcd, fyd, i)

Dim j As Double

Dim force As Double

Dim pua As Double

Dim ef As Double

Dim stress As Double

Dim strain As Double

Dim y As Double

Dim mo As Double

Dim moa As Double

Dim mua As Double

strain = 0

stress = 0

force = 0

For j = 1 To N

y = D / 2 - ((D / 2 - cov) * Cos(360 / N * (j - 1) * 3.14159 / 180))

strain = 0.0035 * (1 - y / i)

If strain > 0 Then

stress = Application.min((strain * 200000), (fyd))

Else

stress = Application.max((strain * 200000), (-fyd))

End If

force = stress * 3.1415 / 4 * dia ^ 2

mo = force * (D / 2 - y)

ef = force + ef

moa = mo + moa

Next j

pua = (betac1(i / D, D, Fcd) * Fcd * area(i, D) + ef) / 1000

pulc = pua

End Function

For Rectangular Pier

Public Function pua(b, D, cov, dia, nb, nd, Fcd, fyd, spacd, i)

Dim j As Double

Dim force As Double

Dim ef As Double

Dim stress As Double

Dim strain As Double

Dim y As Double

strain = 0

stress = 0

force = 0

For j = 1 To (nd + 2)

y = cov + (j - 1) * spacd

strain = 0.0035 * (1 - y / i)

If strain > 0 Then

stress = Application.min((strain * 200000), (fyd))

Else

stress = Application.max((strain * 200000), (-fyd))

End If

If j = 1 Then

force = 3.14159 / 4 * dia ^ 2 * (nb + 2) * stress

Else

If j = (nd + 2) Then

force = 3.14159 / 4 * dia ^ 2 * (nb + 2) * stress

Else

force = 3.14159 / 4 * dia ^ 2 * 2 * stress

End If

End If

ef = force + ef

Next j

pua = (beta1(D, Fcd, i / D) * Fcd * b * Application.min(i, D) + ef) / 1000

End Function