Download pdf - COLUMN DESIGN - civil.utm.mycivil.utm.my/iznisyahrizal/files/2017/04/Lecture-3-Column-Design.pdf · Classification of Columns tion Braced Slender Non-slender Unbraced Slender Non-slender

COLUMN DESIGN

Dr. Izni Syahrizal bin IbrahimFaculty of Civil Engineering

Universiti Teknologi Malaysia

Email: [email protected]

mailto:[email protected]

Introduction

• Column: Subjected to axial compressive forces

• Carries load from beams and slabs down to the foundation.

• May also resist bending moment due to continuity of structureand loading eccentricity

• EC2 Clause 5.3.1(7):a) Compression member where the greater cross sectional

dimension does not exceed 4 times the smaller dimension (h 4b)

b) Height is at least 3 times the section depth

Classification of Columns

Co

lum

n C

lass

ific

atio

n

BracedSlender

Non-slender

UnbracedSlender

Non-slender

Lateral stability to the structure as a whole is provided by walls or bracing – resist all lateral forces

Lateral loads are resisted by the bending action of

the column


• Slender or Non-slender column depending on the sensitivity to second order effect (P- effect)

• Use slenderness ratio, to measure column vulnerability by elastic instability or buckling

• Non-Slender:a) Design action are not significantly affected by

deformation (P- effect is small)b) P- effect can be ignored if does not exceed a

particular valuec) P- effect can be ignored if 10% of the

corresponding first order moments


• Short column – , crushing at ultimate strength

• Slender column – , buckling under low compressive load

Compression failure

Buckling failure


Major axis

(x-x)

Minor axis

(y-y)

Plane of

bending

Clear height, l Actual height

Slenderness Ratio

𝝀 =𝒍𝒐𝒊=

𝒍𝒐

𝑰 𝑨

lo = Effective length of the columni = Radius of gyration and the axis considerI = Section moment of area of the section about the axisA = Cross sectional area of the column

Effective Length of Column

a) lo = l b) lo = 2l c) lo = 0.7l d) lo = 0.5l e) lo = l g) lo 2lf) 0.5 lo l

For constant cross section


Braced Column:

𝒍𝒐 = 𝟎. 𝟓𝒍 𝟏 +𝒌𝟏

𝟎. 𝟒𝟓 + 𝒌𝟏𝟏 +

𝒌𝟐𝟎. 𝟒𝟓 + 𝒌𝟐

Unbraced Column:

𝒍𝒐 = 𝒍.𝒎𝒂𝒙 𝟏 + 𝟏𝟎𝒌𝟏𝒌𝟐𝒌𝟏 + 𝒌𝟐

; 𝟏 +𝒌𝟏

𝟏 + 𝒌𝟏𝟏 +

𝒌𝟐𝟏 + 𝒌𝟐

k1, k2 Relative flexibilities of rotational restraints at ends 1 & 2, respectively = 𝜃

𝑀

𝐸𝐼

𝑙

Rotation of restraining members for bending moment, MEI Bending stiffness of compression memberl Clear height of compression member between end restraints at each end

k1, k2 = 𝐶𝑜𝑙𝑢𝑚𝑛 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠,𝐾𝑐𝑜𝑙

Σ𝐵𝑒𝑎𝑚 𝑆𝑡𝑖𝑓𝑛𝑒𝑠𝑠,𝐾𝑏𝑒𝑎𝑚=

𝐸𝐼 𝑙 𝑐𝑜𝑙 2 𝐸𝐼 𝑙 𝑏𝑒𝑎𝑚

EC2: Clause 5.8.3.2(3)


Table 3.19 & 3.20, BS 8110: Part 1: 1997

End Condition at TopEnd Condition at Bottom

1 2 3

Braced Column

1 0.75 0.80 0.90

2 0.80 0.85 0.95

3 0.90 0.95 1.00

Unbraced Column

1 1.2 1.3 1.6

2 1.3 1.5 1.8

3 1.6 1.8 -

4 2.2 - -

Simplified Method


End Condition at TopEnd Condition at Bottom

1 2 3

Braced Column

1 0.75 0.80 0.90

2 0.80 0.85 0.95

3 0.90 0.95 1.00

Unbraced Column

1 1.2 1.3 1.6

2 1.3 1.5 1.8

3 1.6 1.8 -

4 2.2 - -

Condition 1 Column connected monolithically to beams on either side which are at least as deep as the overall dimension of the column in the plane considered. Where column connected to a foundation this should be designed to carry moment

Condition 2 Column connected monolithically to beams or slabs on either side which are shallower than the overall dimension of the column in the plane considered

Condition 3 Column connected to members that do not provide more than nominal restraint to rotation

Condition 4 End of column is unrestrained against both lateral movement and rotation

Table 3.19 & 3.20, BS 8110: Part 1: 1997

Limiting Slenderness Ratio

𝝀𝒍𝒊𝒎 =𝟐𝟎 ∙ 𝑨 ∙ 𝑩 ∙ 𝑪

𝒏

A = 1

1+0.2𝜑𝑒𝑓𝑓: 𝜑𝑒𝑓𝑓 = Effective creep ratio

B = 1 + 2𝜔 : 𝜔 =𝐴𝑠𝑓𝑦𝑑

𝐴𝑐𝑓𝑐𝑑

C = 1.7 − 𝑟𝑚 : 𝑟𝑚 =𝑀𝑜1

𝑀𝑜2

n = 𝑁𝐸𝑑


NEd = Design ultimate axial column loadMo1, Mo2 = First order moments at the end of the column with 𝑀𝑜2 ≥ 𝑀𝑜1fyd = Design yield strength of reinforcementfcd = Design compressive strength of concrete

If 𝝋𝒆𝒇𝒇, & rm is not known, A = 0.7, B = 1.1 & C = 0.7 may be used

EC2: Clause 5.8.3.1

Limiting Slenderness Ratio

Condition apply for C:

(1) If the end moments, Mo1 & Mo2 give rise to tension on the same side of the column, then rm should be taken +ve (follows C 1.7)

(2) If the column is in a state of double curvature, then rm should be taken –ve (follows C 1.7)

(3) For braced members in which the first order moment arise only from or predominantly due to imperfections or transverse loading, then rm should be taken as 1.0 (C = 0.7)

(4) For unbraced member, in general rm should be taken as 1.0 (C = 0.7)

If lim: Short (Non-slender) columnIf lim: Slender column. Second order effects must be considered

in design

EC2: Clause 5.8.3.1

Example 1

SLENDERNESS

Example 1: Slenderness

4 m

4 m

4 m

250 500

250 500

250 500

250 500

25

0

40

0

25

0

40

0

25

0

40

0

275 350

A B C

1

2

3

4

• Braced column• Axial load = 1050 kN• Bending moment (major axis) = 40 kNm (top) & 12 kNm (bottom)• Bending moment (minor axis) = 15 kNm (top) & 10 kNm (bottom)• Concrete grade = C25• Steel = 500 N/mm2

l1 = 6 m l2 = 8 m


l1 = 6 m l2 = 8 m

4 m

5 m

1.5 m250 500

250 500

250 500

275 350

A B C

Roof

1st Floor

Ground

Example 1: SlendernessEC2: Clause 5.8.3.2(3) Method

z

z

y y

Secondary beam250 400

L = 4 m 5 m

Main beam250 500

L = 6 m Main beam250 500

L = 8 m

Secondary beam250 400

L = 4 m

-12 kNm

40 kNm

NEd = 1050 kN

Mz

15 kNm

-10 kNmMy

27

5 m

m

Mz

350 mm

My

Moment & Axial Force


Dimension & Size

Column:b h = 275 300 mmActual length: lz = 5000 – 500 = 4500 mm

ly = 5000 – 400 = 4600 mm

Beam:Main beam, b h = 250 500 mmActual length: l1 = 6000 mm

l2 = 8000 mm

Secondary beam, b h = 250 400 mmActual length: l1 = l2 = 4000 mm


Moment of Inertia, I = bh3/12

Column:

𝐼𝑧𝑧 =275×3003

12= 0.98 × 109 mm4

𝐼𝑦𝑦 =300×2753

12= 0.61 × 109 mm4

Beam:

Main beam, 𝐼𝑚𝑏 =250×5003

12= 2.60 × 109 mm4

Secondary beam, 𝐼𝑠𝑏 =250×4003

12= 1.33 × 109 mm4


Stiffness, K = EI/l

Column:

𝐾𝑧𝑧 =0.98×109

4500= 2.18 × 105 mm3

𝐾𝑦𝑦 =0.61×109

4600= 1.32 × 105 mm3

Beam:

Main beam 𝐾𝑚𝑏1 =2.60×109

6000= 4.34 × 105 mm3

𝐾𝑚𝑏2 =2.60×109

8000= 3.26 × 105 mm3

Secondary beam 𝐾𝑠𝑏1 = 𝐾𝑠𝑏2 =1.33×109

4000= 3.33 × 105 mm3


Relative Column Stiffness, k = Kcol/2(Kbeam)

z-axis:

Top end: 𝑘2 =2.18×105

2 4.34×105+3.26×105= 0.14 > 0.1 k2 = 0.14

Bottom end: 𝑘1 =2.18×105

2 4.34×105+3.26×105= 0.14 > 0.1 k1 = 0.14

y-axis:

Top end: 𝑘2 =1.32×105

2 3.33×105+3.33×105= 0.10 < 0.1 k2 = 0.10

Bottom end: 𝑘1 =1.32×105

2 3.33×105+3.33×105= 0.10 < 0.1 k1 = 0.10



𝑙𝑜 = 0.5𝑙 1 +𝑘1

0.45 + 𝑘11 +

𝑘20.45 + 𝑘2

𝑙𝑜,𝑧 = 0.5𝑙𝑧 1 +0.14

0.45+0.141 +

0.14

0.45+0.14= 2795mm

𝑙𝑜,𝑦 = 0.5𝑙𝑦 1 +0.10

0.45+0.101 +

0.10

0.45+0.10= 2718mm


Radius of Gyration, 𝒊 =𝑰

𝑨

𝑖𝑧𝑧 =𝐼𝑧𝑧𝐴=

0.98 × 109

275 × 350= 101

𝑖𝑦𝑦 =𝐼𝑦𝑦

𝐴=

0.61 × 109

275 × 350= 79.4

Slenderness Ratio, = lo/i

𝜆𝑧𝑧 =2795

101= 𝟐𝟕. 𝟕

𝜆𝑦𝑦 =2718

79.4= 𝟑𝟒. 𝟐


Slenderness Limit, 𝝀𝒍𝒊𝒎 =𝟐𝟎∙𝑨∙𝑩∙𝑪

𝒏

A = 0.7 (eff NOT known)B = 1.1 ( NOT known)C = 1.7 – rm (where rm = (Mo1/Mo2)

z-axis: 𝑟𝑚,𝑧 =−12

40= −0.30

Cz = 1.7 – (0.30) = 2.00

y-axis: 𝑟𝑚,𝑦 =−10

15= −0.67

Cy = 1.7 – (0.67) = 2.37

𝑛 =𝑁𝐸𝑑𝐴𝑐𝑓𝑐𝑑

𝑓𝑐𝑑 =0.85𝑓𝑐𝑘

𝛾𝑐=0.85×25

1.5= 14.17 N/mm2

𝑛 =1050×103

275×350 ×14.17= 0.77

𝜆𝑙𝑖𝑚,𝑧𝑧 =20×0.7×1.1×2.00

0.77= 35.1 > 𝜆𝑧𝑧 = 27.7 Non-slender about z-axis

𝜆𝑙𝑖𝑚,𝑦𝑦 =20×0.7×1.1×2.37

0.77= 41.5 > 𝜆𝑦𝑦 = 34.2 Non-slender about y-axis


Effective Length, lo = Factor Clear Height

z-axis: End condition: Top & Bottom = Condition 1Factor = 0.75 𝑙𝑜,𝑧 = 0.75 × 4500 = 3375mm

y-axis: End condition: Top & Bottom = Condition 1Factor = 0.75 𝑙𝑜,𝑦 = 0.75 × 4600 = 3450mm

Slenderness Ratio, = lo/i

𝜆𝑧𝑧 =3375

101= 33.4 < 𝜆𝑙𝑖𝑚,𝑧𝑧 = 35.1 Non-slender about z-axis

𝜆𝑦𝑦 =3450

79.4= 43.5 > 𝜆𝑙𝑖𝑚,𝑦𝑦 = 41.5 Slender about y-axis

Simplified Method

Axial Load & Moment in Column

For analysis without full frame analysis:

a) Axial loads may generally be obtained by increasing the loads obtained by 10% by assumption that beams & slabs are simply supported. Higher percentage may be required when adjacent spans and/or loadings on them are grossly dissimilar.

b) Bending moments may be calculated using the simplified one free-joint sub-frame. The arrangement of the design ultimate variable action should be such as that to cause maximum moment in the column.

Example 2

AXIAL LOAD & MOMENT IN COLUMN

Example 2: Axial Load & Moment in Column

Level BeamCharacteristics Action (kN/m)

Permanent Variable

Roof 2/A-C 20.0 6.0

B/1-4 8.0 3.0

First Floor 2/A-C 27.0 12.0

B/1-4 15.0 7.0

Ground Floor 2/A-C 7.8 -

B/1-4 7.8 -

Determine axial force and bending moment in column B/2


Column: 275 350 mm

Main Beam: 250 500 mm𝐿1𝑦 = 𝐿2𝑦 = 4 𝑚

Secondary Beam: 250 400 mm𝐿1𝑧 = 8 𝑚𝐿2𝑧 = 6 𝑚

z

z

y y

x

x

4 m

2/A-C

B/1-4

5 m

1.5 m

w1Z w2Z

w2y

w1y

Roof

1st Level

Ground


Roof w1z w2z w1y w2y

Gk 20 20 8 8

Qk 6 6 3 3

Max 36 36 15.3 15.3

Min 27 27 10.8 10.8

1st Floor w1z w2z w1y w2y

Gk 27 27 15 15

Qk 12 12 7 7

Max 54.5 54.5 30.8 30.8

Min 36.45 36.45 20.25 20.25

Ground w1z w2z w1y W2y

Gk 7.8 7.8 7.8 7.8

Qk 0 0 0 0

Max 10.5 10.5 10.5 10.5

Min 10.5 10.5 10.5 10.5


Roof to 1st Level:Main Beam: (36 8 0.5) + (36 6 0.5) 252 kNSec. Beam: (15.3 4 0.5) + (15.3 4 0.5) 61 kNSelfweight: 1.35 (0.275 0.35 4 25) 13 kNN1st-Roof: 326 kN

1st Level to Ground:Load from above: 326 kNMain Beam: (54.5 8 0.5) + (54.5 6 0.5) 381 kNSec. Beam: (30.8 4 0.5) + (30.8 4 0.5) 123 kNSelfweight: 1.35 (0.275 0.35 5 25) 16 kNN1st-Gnd 847 kN

Ground to Footing:Load from above: 847 kNMain Beam: (10.5 8 0.5) + (10.5 6 0.5) 74 kNSec. Beam: (10.5 4 0.5) + (10.5 4 0.5) 42 kNSelfweight: 1.35 (0.275 0.35 1.5 25) 5 kNNGnd-Foot: 967 kN

Axial Loads


Bending Moments at z-z Axis

Stiffness, K = I/L

Beam: 6 𝑚: 𝐾𝑏1 =250×5003

12×6000= 4.34 × 105 𝑚𝑚3

8 𝑚: 𝐾𝑏2 =250×5003

12×8000= 3.26 × 105 𝑚𝑚3

Column:4 𝑚:275×3503

12×4000= 2.46 × 105 𝑚𝑚3

5 𝑚:275×3503

12×5000= 1.97 × 105 𝑚𝑚3

1.5 𝑚:275×3503

12×1500= 6.55 × 105 𝑚𝑚3


Fixed End Moment, FEM

𝑀1 =𝑤𝐿2

12=27 × 62

12= 81.0 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=36 × 82

12= 192.0 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟏𝟏𝟏 𝒌𝑵𝒎6 m 8 m

4 m

wmin = 27.0 kN/mwmax = 36.0 kN/m

Kc

0.5Kb1 0.5Kb2

Roof to 1st Level

Moment in Column

𝑀 = ∆𝐹𝐸𝑀 ×𝐾𝑐

𝐾𝑐+0.5𝐾𝑏1+0.5𝐾𝑏2= 111 ×

2.46

2.46+0.5×4.34+0.5×3.26= 𝟒𝟑. 𝟔 kNm

M1 M2



𝑀1 =𝑤𝐿2

12=36.5 × 62

12= 109.4 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=54.5 × 82

12= 290.4 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟏𝟖𝟏 𝒌𝑵𝒎

1st Level to Ground

Moment in upper column, Mupper

𝑀𝑢𝑝𝑝𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 181 ×

2.46

2.46+1.97+0.5×4.34+0.5×3.26= 𝟓𝟒. 𝟏 kNm

Moment in lower column, Mlower

𝑀𝑙𝑜𝑤𝑒𝑟 = ∆𝐹𝐸𝑀 ×𝐾𝑐,𝑢

𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 181 ×

1.97

2.46+1.97+0.5×4.34+0.5×3.26= 𝟒𝟑. 𝟑 kNm

M1 M2

6 m 8 m

4 m

5 m


Kc,l

Kc,u

0.5Kb1 0.5Kb2



𝑀1 =𝑤𝐿2

12=10.5 × 62

12= 31.6 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=10.5 × 82

12= 56.2 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟐𝟒. 𝟔 𝒌𝑵𝒎

Ground to Footing



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 24.6 ×

1.97

1.97+6.55+0.5×4.34+0.5×3.26= 𝟑. 𝟗 kNm



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 24.6 ×

6.55

1.97+6.55+0.5×4.34+0.5×3.26= 𝟏𝟑. 𝟏 kNm

M1 M2

6 m 8 m

5 m

1.5 m

wmax = 10.5 kN/m wmax = 10.5 kN/m

Kc,l

Kc,u

0.5Kb1 0.5Kb2


Bending Moments at z-z Axis, Mz-z (kNm)

43.6

54.1 43.3

3.9 13.1

6.5


Bending Moments at y-y Axis

Stiffness, K = I/L

Beam: 4 𝑚: 𝐾𝑏1 =250×4003

12×4000= 3.33 × 105 𝑚𝑚3

4 𝑚: 𝐾𝑏2 =250×4003

12×4000= 3.33 × 105 𝑚𝑚3

Column:4 𝑚:350×2753

12×4000= 1.52 × 105 𝑚𝑚3

5 𝑚:350×2753

12×5000= 1.21 × 105 𝑚𝑚3

1.5 𝑚:350×2753

12×1500= 4.04 × 105 𝑚𝑚3



𝑀1 =𝑤𝐿2

12=10.8 × 42

12= 14.4 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=15.3 × 42

12= 20.4 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟔 𝒌𝑵𝒎4 m 4 m

4 m

wmax = 10.8 kN/mwmax = 15.3 kN/m

Kc

0.5Kb1 0.5Kb2

Roof to 1st Level

Moment in Column

𝑀 = ∆𝐹𝐸𝑀 ×𝐾𝑐

𝐾𝑐+0.5𝐾𝑏1+0.5𝐾𝑏2= 111 ×

1.52

1.52+0.5×3.33+0.5×3.33= 𝟏. 𝟗 kNm

M1 M2



𝑀1 =𝑤𝐿2

12=20.3 × 42

12= 27 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=30.8 × 42

12= 41 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟏𝟒 𝒌𝑵𝒎

1st Level to Ground



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 14 ×

1.52

1.52+1.21+0.5×3.33+0.5×3.33= 𝟑. 𝟓 kNm



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 14 ×

1.21

1.52+1.21+0.5×3.33+0.5×3.33= 𝟐. 𝟖 kNm

M1 M2

4 m 4 m

4 m

5 m


Kc,l

Kc,u

0.5Kb1 0.5Kb2



𝑀1 =𝑤𝐿2

12=10.5 × 42

12= 14 𝑘𝑁𝑚

𝑀2 =𝑤𝐿2

12=10.5 × 42

12= 14 𝑘𝑁𝑚

∆𝐹𝐸𝑀 = 𝑀2 −𝑀1 = 𝟎 𝒌𝑵𝒎

Ground to Footing



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 𝟎. 𝟎 kNm



𝐾𝑐,𝑢+𝐾𝑐,𝑙+0.5𝐾𝑏1+0.5𝐾𝑏2= 𝟎. 𝟎 kNm

M1 M2

4 m 4 m

5 m

1.5 m

wmax = 10.5 kN/m wmax = 10.5 kN/m

Kc,l

Kc,u

0.5Kb1 0.5Kb2


Bending Moments at y-y Axis, My-y (kNm)

1.9

3.5 2.8

0.0

Design Moments

• Elastic moment from frame action – should NOTconsider any redistribution

• Slender column:a) Non-linear analysis to determine second order

momentb) Nominal stiffness method (EC2: Clause 5.8.7), orc) Nominal curvature method (EC2: Clause 5.8.8)

Design Moments

Slender ColumnNominal Curvature Method (EC2: Clause 5.8.8)

𝑴𝑬𝒅 = 𝑴𝟎𝑬𝒅 +𝑴𝟐

M0Ed = First order moment including effect imperfectionMEd = Nominal second order moment

Design Moments

Braced Slender Column

𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝑬 +𝑴𝟐,𝑴𝟎𝟏 + 𝟎. 𝟓𝑴𝟐, 𝑵𝑬𝒅 ∙ 𝒆𝒐

M02

M NEd ei

M01

M0E

First Order Moment

0.5M2

0.5M2

M2 = Ned e2

Additional Second Order Moment

+ =

M02

M0E + M2

M01 + 0.5M2

Total Moment Diagram

Design Moments

Unbraced Slender Column

𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟏 +𝑴𝟐,𝑴𝟎𝟐 +𝑴𝟐, 𝑵𝑬𝒅 ∙ 𝒆𝒐

M02

M NEd ei

M01

First Order Moment

M2

M2 = NEd e2

Additional Second Order Moment

+ =

M02 + M2

M01 + M2

Total Moment Diagram

Design Moments

𝑀01 = 𝑚𝑖𝑛 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 + 𝑁𝐸𝑑𝑒𝑖𝑀02 = 𝑚𝑎𝑥 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 + 𝑁𝐸𝑑𝑒𝑖NEd = Ultimate axial load

𝑒𝑜 = 𝑚𝑎𝑥ℎ

30, 20 𝑚𝑚

𝑒𝑖 =𝑙𝑜400

Mtop, Mbot = Moment at top and bottom of column𝑀0𝐸 = 0.6𝑀02 + 0.4𝑀01 ≥ 0.4𝑀02 (M01 & M02 should have the same sign if they produce tension on the same side, otherwise opposite sign)M2 = Nominal second order moment = 𝑁𝐸𝑑𝑒2

e2 = Deflection = 𝑙 𝑟 𝑙𝑜

2

𝑐

lo = Effective lengthc = Factor depending on curvature distribution, normally 𝜋2 ≈ 10

Design Moments

𝑙

𝑟= Curvature = 𝐾𝑟𝐾𝜑

𝑙

𝑟𝑜

Kr = Axial load correction factor = 𝑛𝑢−𝑛

𝑛𝑢−𝑛𝑏𝑎𝑙< 1

𝑛 =𝑁𝐸𝑑

𝐴𝑐𝑓𝑐𝑑𝑛𝑢 = 1 + 𝜔 𝑛𝑏𝑎𝑙 = 0.4 𝜔 =

𝐴𝑠𝑓𝑦𝑑


K = Creep correction factor = 1 + 𝛽𝜑𝑒𝑓 ≥ 1

ef = Effective creep factor = 𝜑𝑀𝑜𝐸𝑞𝑝

𝑀𝑜𝐸𝑑= 0 if 𝜑 < 2,

𝑀

𝑁> ℎ, 𝑙 < 75)

𝛽 = 0.35 +𝑓𝑐𝑘200

−𝜆

150

𝑙

𝑟𝑜=𝜀𝑦𝑑

0.45𝑑=

𝑓𝑦𝑑

𝐸𝑠0.45𝑑

EC2: Clause 5.8.8.3: Curvature

Design Moments

EN 1992-1-2: Figure 3.1: Creep Correction Factor

Design Moments

EN 1992-1-2: Figure 3.1: Creep Correction Factor

Design Moments

Non-Slender Column

• Ignored second order moment effects• Ultimate design moment, MEd = M02

𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝟏, 𝑵𝑬𝒅 ∙ 𝒆𝒐

M02

M NEd e1

M01

M0E

First Order Moment

Design of Longitudinal Reinforcement

Design ChartSolution of Basic

EquationApproximate

Method

• For columns having rectangular or circular cross section

• Symmetrical arrangement of reinforcement

• Unsymmetrical arrangement of reinforcement

• Cross section is non-rectangular


For practical purpose as with BS8110, the rectangular stress block used for design of beam may also be used

for the design of columns

FccFsc

Fst

st

sc

cc

As

As’

fcc = ccfck/c

x

z

d2

h d

x

Section Strain Stress

Neutral Axis


• However, unlike with BS8110 the maximum compressive strain when designing to EC2 will be less than 0.0035 if the whole section is in compression.

• This compressive strain will further fall to 0.00175 (fck < 50 N/mm2) if the section is subject to pure compression.

• This will affect the steel strains and hence forces which the reinforcement can carry


EC2 strain relationship at ULS for fck 50 N/mm2

Stress

0.0035 max

h d

x

General Relationship

When x hPure

Compression

Hinge point

h/2

0.00175 min

x

𝟎. 𝟎𝟎𝟏𝟕𝟓𝒙

𝒙 −𝒉𝟐 0.00175

0.00175

FccFsc

Fst

d2x


Equilibrium Equations

Equilibrium of Load:

𝑁 = 𝐹𝑐𝑐 + 𝐹𝑠𝑐 − 𝐹𝑠𝑡

Equilibrium of Moment:

𝑀 = 𝐹𝑐𝑐 0.5ℎ − 0.5𝜆𝑥 + 𝐹𝑠𝑐 0.5ℎ − 𝑑2 − 𝐹𝑠𝑡 𝑑 − 0.5ℎ

where

𝐹𝑐𝑐 =𝜂𝑓𝑐𝑘

𝛾𝑐𝜆𝑥 𝑏; 𝐹𝑠𝑐 =

𝐴𝑠𝑐𝑓𝑦𝑘

𝛾𝑠; 𝐹𝑠𝑡 =

𝐴𝑠𝑡𝑓𝑦𝑘

𝛾𝑠


Rearranged to:

𝑵

𝒃𝒉= 𝒇 𝝀; 𝜼; 𝒇𝒄𝒌; 𝜸𝒄;

𝒙

𝒉;𝑨𝒔𝒄𝒃𝒉

;𝑨𝒔𝒕𝒃𝒉

; 𝒇𝒚𝒌; 𝜸𝒔

𝑴

𝒃𝒉𝟑= 𝒇 𝝀; 𝜼; 𝒇𝒄𝒌; 𝜸𝒄;

𝒙

𝒉;𝑨𝒔𝒄𝒃𝒉

;𝑨𝒔𝒕𝒃𝒉

; 𝒇𝒚𝒌; 𝜸𝒔;𝒅

𝒉;𝒅𝟐𝒉

These equations form the basis for the N-M interaction charts used for the design of columns


Area and Number of Reinforcement

𝐴𝑠,𝑚𝑖𝑛 =0.10𝑁𝐸𝑑𝑓𝑦𝑑

𝑜𝑟 0.002𝐴𝑐

𝐴𝑠,𝑚𝑎𝑥 = 0.04𝐴𝑐 (Outside lap location)

𝐴𝑠,𝑚𝑎𝑥 = 0.08𝐴𝑐 (At lap location)

EC2: Clause 9.5.2

Requirements for Links

EC2: Clause 9.5.3

• The diameter of links should not be less than 6 mm or one-quarter of the diameter of the longitudinal bar.

• The maximum spacing, Smax should not be less than:

a) 20 times the minimum diameter of longitudinal barsb) the lesser dimension of the columnc) 400 mm

• At a distance greater than the larger dimension of the column above or below a beam these spacing can increased by a factor of 1.67

Design Procedure

Step Task Standard

1 Determine design life, exposure class, fire resistanceEN 1990 Table 2.1EN 1992-1-1: Table 4.1EN 1992-1-2: Sec. 5.6

2 Determine material strengthBS 8500-1: Table A.3EN 206-1: Table F1

3 Select size of column EN 1992-1-2: Table 5.5

4 Calculate min. cover for durability , fire and bond requirements EN 1992-1-1: Sec. 4.4.1

5 Analyze structure to obtain critical moments and axial forces EN 1992-1-1: Sec. 5

6 Check slenderness EN 1992-1-1: Sec. 5.8.3

7 Determine final design moment EN 1992-1-1: Sec. 5.8.8

8 Determine area of reinforcement required EN 1992-1-1: Sec. 6.1

9 Detailing EN 1992-1-1: Sec. 8 & 9

Example 3

DESIGN OF NON-SLENDER COLUMN BENT ABOUT

MAJOR AXIS

Classification: Braced non-slender columnConcrete, fck: 25 N/mm2

Reinforcement, fyk: 500 N/mm2

Exposure Class: XC1Fire Resistance: 1 hourDesign Life: 50 yearsEffective Length, lo: 4.2 mBar Size: bar = 20 mm, links = 6 mm

Example 3: Design of Non-Slender Column Bent About Major Axis

NEd = 1200 kN

M02 = 35 kNm

25 kNm

Mz

M

z

z

y y 250

300

Cross Section


Cover: Durability, Bond & Fire Resistance

Minimum concrete cover regard to bond, Cmin, bond = 20 mm (EN 1992-1-1: Table 4.2)

Minimum concrete cover regard to durability, Cmin, dur = 15 mm (EN 1992-1-1: Table 4.4N)

Minimum required axis distance for R60 fire resistance, asd = 36 mm assumed fi = 0.50 (EN 1992-1-2: Table 5.2a)Nominal concrete cover regard to fire, Cnom, fire = asd - links - bar/2

= 36 – 6 – 20/2 = 20 mm

Allowance in design for deviation, Cdev = 10 mm

Nominal cover, Cnom = max{Cmin, bond, Cmin, dur} + Cdev = 20 + 10 = 30 mm

Use Cnom = 30 mm


Design Moment

For non-slender column:𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝟏, 𝑵𝑬𝒅 ∙ 𝒆𝒐

𝑀02 = 𝑚𝑎𝑥 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 + 𝑁𝐸𝑑𝑒𝑖 = 35 + 1200 × 0.0105 = 𝟒𝟕. 𝟔 𝒌𝑵𝒎𝑚𝑎𝑥 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 = 35 𝑘𝑁𝑚


=4200

400= 10.5 𝑚𝑚 = 0.0105 𝑚

𝑁𝐸𝑑 ∙ 𝑒𝑜 = 1200 × 0.002 = 𝟐𝟒 𝒌𝑵𝒎


30, 20 𝑚𝑚 = 𝑚𝑎𝑥

300

30, 20 𝑚𝑚 = 20 𝑚𝑚 = 0.020 𝑚

𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝟏, 𝑵𝑬𝒅 ∙ 𝒆𝒐 = 𝟒𝟕. 𝟔 𝒌𝑵𝒎


Reinforcement

𝑑2 = 𝑐 + ∅𝑙𝑖𝑛𝑘 +∅𝑏𝑎𝑟

2= 30 + 6 +

20

2= 46mm

𝑑2ℎ=46

300= 0.15

𝑁

𝑏ℎ𝑓𝑐𝑘=

1200 × 103

250 × 300 × 25= 0.64

𝑀

𝑏ℎ2𝑓𝑐𝑘=

47.6 × 106

250 × 3002 × 25= 0.08

From design chart:𝐴𝑠𝑓𝑦𝑘

𝑏ℎ𝑓𝑐𝑘= 0.35 𝐴𝑠 =

0.35𝑏ℎ𝑓𝑐𝑘

𝑓𝑦𝑘=0.35×250×300×25

500= 𝟏𝟑𝟏𝟑mm2

𝐴𝑠,𝑚𝑖𝑛 =0.10𝑁𝐸𝑑

𝑓𝑦𝑑=0.10×1200×103

0.87×500= 𝟐𝟕𝟔mm2 or 0.002𝐴𝑐 = 150mm2

𝐴𝑠,𝑚𝑎𝑥 = 0.04𝐴𝑐 = 0.04 × 250 × 300 = 𝟑𝟎𝟎𝟎mm2

Provide 4H20 + 2H12 (As = 1483 mm2)

0.35


Links

links, min = the larger of:i. 0.25 20 mm = 5 mm orii. 6 mm Use links = 6 mm

Smax = the lesser of:i. 20 (12 mm) = 240 mm orii. the lesser dimension of the column = 250 mm oriii. 400 mm

Use Smax = 240 mm

Provide H6-240

At section 300 mm below & above beam & at lap joints, Smax = 0.6 240 mm = 144 mm

Provide H6-140

25

0

300

H6

-24

0H

6-1

40

Example 4

DESIGN OF SLENDER COLUMN BENT ABOUT

MINOR AXIS

Classification: Braced slender columnConcrete, fck: 25 N/mm2


Exposure Class: XC1Fire Resistance: 1 hourDesign Life: 50 yearsEffective Length, lo: 4.13 mSlenderness ratio, : 52Bar Size: bar = 20 mm, links = 6 mm

Example 4: Design of Slender Column Bent About Minor Axis

NEd = 1200 kN

35 kNm

25 kNm

Mz

M

z

z

y y 250

300

Cross Section


Cover: Durability, Bond & Fire Resistance

Minimum concrete cover regard to bond, Cmin, bond = 20 mm (EN 1992-1-1: Table 4.2)

Minimum concrete cover regard to durability, Cmin, dur = 15 mm (EN 1992-1-1: Table 4.4N)

Minimum required axis distance for R60 fire resistance, asd = 36 mm assumed fi = 0.50 (EN 1992-1-2: Table 5.2a)Nominal concrete cover regard to fire, Cnom, fire = asd - links - bar/2

= 36 – 6 – 20/2 = 20 mm

Allowance in design for deviation, Cdev = 10 mm

Nominal cover, Cnom = max{Cmin, bond, Cmin, dur} + Cdev = 20 + 10 = 30 mm

Use Cnom = 30 mm


Design Moment

For slender column:𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝑬 +𝑴𝟐,𝑴𝟎𝟏 + 𝟎. 𝟓𝑴𝟐, 𝑵𝑬𝒅 ∙ 𝒆𝒐

𝑀01 = 𝑚𝑖𝑛 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 + 𝑁𝐸𝑑𝑒𝑖 = 25 + 1200 × 0.0124 = 𝟑𝟕. 𝟒 𝒌𝑵𝒎

𝑀02 = 𝑚𝑎𝑥 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 + 𝑁𝐸𝑑𝑒𝑖 = 35 + 1200 × 0.014 = 𝟒𝟕. 𝟒 𝒌𝑵𝒎𝑚𝑎𝑥 𝑀𝑡𝑜𝑝 , 𝑀𝑏𝑜𝑡 = 35 𝑘𝑁𝑚


=4130

400= 12.4 𝑚𝑚 = 0.0124 𝑚

𝑀0𝐸 = 0.6𝑀02 + 0.4𝑀01 ≥ 0.4𝑀02𝑀0𝐸 = 0.6 47.4 + 0.4 37.4 ≥ 0.4 47.4 = 13.5 𝑘𝑁𝑚 ≥ 19.0 𝑘𝑁𝑚 MOE = 19.0 kNm


Second Order Moment, M2

𝑀2 = 𝑁𝐸𝑑𝑒2 = 1200 × 0.0484 = 𝟓𝟖. 𝟏 𝒌𝑵𝒎

𝑒2 = 𝑙 𝑟 𝑙𝑜

2

𝑐=2.84×10−5×41302

10= 48.4mm = 0.0484 m

lo = 4130 mm c = 10𝑙

𝑟= 𝐾𝑟𝐾𝜑

𝑙

𝑟𝑜= 1 × 1.20 × 2.37 × 10−5 = 2.84 × 10−5

Effective depth to the minor axis, d = 250 – 30 – 6 – (20/2) = 204 mm

𝑲𝒓 =𝒏𝒖−𝒏

𝒏𝒖−𝒏𝒃𝒂𝒍= 𝟏 (𝒂𝒔𝒔𝒖𝒎𝒆𝒅) < 𝟏

𝐾𝜑 = 1 + 𝛽𝜑𝑒𝑓 ≥ 1 = 1 + (0.128 × 1.54) = 1.20

𝜑𝑒𝑓 =𝜑𝑀𝑜𝐸𝑞𝑝

𝑀𝑜𝐸𝑑= 0.67 × 2.3 = 1.54

𝑀𝑜𝐸𝑞𝑝

𝑀𝑜𝐸𝑑= 0.67(assumed)

(, to) = 2.3 (Fig. 3.1: EN 1992-1-1)

𝛽 = 0.35 +𝑓𝑐𝑘

200−

𝜆

150= 0.35 +

25

200−

52

150= 0.128

𝑙

𝑟𝑜=

𝑓𝑦𝑑

𝐸𝑠

0.45𝑑=

0.87×500200000

0.45×204= 2.37 × 10−5

Class RRH = 80%Age = 3 days2

3

4

2.3

5

1



𝑁𝐸𝑑 ∙ 𝑒𝑜 = 1200 × 0.020 = 𝟐𝟒 𝒌𝑵𝒎


30, 20 𝑚𝑚 = 𝑚𝑎𝑥

300

30, 20 𝑚𝑚 = 20 𝑚𝑚 = 0.020 𝑚

𝑴𝑬𝒅 = 𝒎𝒂𝒙 𝑴𝟎𝟐,𝑴𝟎𝑬 +𝑴𝟐,𝑴𝟎𝟏 + 𝟎. 𝟓𝑴𝟐, 𝑵𝑬𝒅 ∙ 𝒆𝒐

47.4 kNm

77.0 kNm 66.4 kNm 24 kNm

𝑴𝑬𝒅 = 𝟕𝟕. 𝟎 𝒌𝑵𝒎


Reinforcement


2= 30 + 6 +

20

2= 46mm

𝑑2𝒉=46

𝟐𝟓𝟎= 0.18 ≅ 0.20

𝑁


1200 × 103

300 × 250 × 25= 0.64

𝑀


77.0 × 106

300 × 2502 × 25= 0.16


𝑏ℎ𝑓𝑐𝑘= 0.72 Kr = 0.58

𝑀0𝐸 +𝑀2 = 19 + 0.58 × 58.1 = 52.6 kNm𝑀01 + 0.5𝑀2 = 37.4 + 0.5 0.58 × 58.1 = 54.2 kNm

𝑀


54.2 × 106

300 × 2502 × 25= 0.12

First assumption, Kr = 1.0

54.2 kNm max

0.72Kr = 0.58


Reinforcement


2= 30 + 6 +

20

2= 46mm

𝑑2𝒉=46

𝟐𝟓𝟎= 0.18 ≅ 0.20

𝑁


1200 × 103

300 × 250 × 25= 0.64

𝑀


77.0 × 106

300 × 2502 × 25= 0.16


𝑏ℎ𝑓𝑐𝑘= 0.72 Kr = 0.58

𝑀0𝐸 +𝑀2 = 19 + 𝟎. 𝟓𝟖 × 58.1 = 52.6 kNm𝑀01 + 0.5𝑀2 = 37.4 + 0.5 𝟎. 𝟓𝟖 × 58.1 = 54.2 kNm

𝑀


𝟓𝟒. 𝟐 × 106

300 × 2502 × 25= 0.12

First assumption, Kr = 1.0

54.2 kNm max

0.55

Kr = 0.58



𝑏ℎ𝑓𝑐𝑘= 0.45 Kr = 0.48

𝑀0𝐸 +𝑀2 = 19 + 0.48 × 58.1 = 46.8 kNm𝑀01 + 0.5𝑀2 = 37.4 + 0.5 0.48 × 58.1 = 51.3 kNm

𝑀


51.3 × 106

300 × 2502 × 25= 0.11


𝑏ℎ𝑓𝑐𝑘= 0.40 Kr = 0.45

𝐴𝑠 =0.40𝑏ℎ𝑓𝑐𝑘

𝑓𝑦𝑘=0.40×300×250×25

500= 𝟏𝟓𝟎𝟎mm2


𝑓𝑦𝑑=0.10×1200×103

0.87×500= 𝟐𝟕𝟔mm2 or 0.002𝐴𝑐 = 150mm2

𝐴𝑠,𝑚𝑎𝑥 = 0.04𝐴𝑐 = 0.04 × 250 × 300 = 𝟑𝟎𝟎𝟎mm2

Provide 4H20 + 2H16 (As = 1659 mm2)

51.3 kNm max



𝑏ℎ𝑓𝑐𝑘= 0.45 Kr = 0.48

𝑀0𝐸 +𝑀2 = 19 + 𝟎. 𝟒𝟖 × 58.1 = 46.8 kNm𝑀01 + 0.5𝑀2 = 37.4 + 0.5 𝟎. 𝟒𝟖 × 58.1 = 51.3 kNm

𝑀


51.3 × 106

300 × 2502 × 25= 0.11


𝑏ℎ𝑓𝑐𝑘= 0.40 Kr = 0.45

𝐴𝑠 =0.40𝑏ℎ𝑓𝑐𝑘

𝑓𝑦𝑘=0.40×300×250×25

500= 𝟏𝟓𝟎𝟎mm2


𝑓𝑦𝑑=0.10×1200×103

0.87×500= 𝟐𝟕𝟔mm2 or 0.002𝐴𝑐 = 150mm2

𝐴𝑠,𝑚𝑎𝑥 = 0.04𝐴𝑐 = 0.04 × 250 × 300 = 𝟑𝟎𝟎𝟎mm2

Provide 4H20 + 2H16 (As = 1659 mm2)

51.3 kNm max


Links

links, min = the larger of:i. 0.25 16 mm = 4 mm orii. 6 mm Use links = 6 mm


Use Smax = 250 mm

Provide H6-250


Provide H6-150

25

0

300

H6

-25

0H

6-1

50

y y

Biaxial Bending

• EC2: Clause 5.8.9 states that separate design in each principal direction (disregarding biaxial bending) may be the first step. NO further checking is necessary if:

(a)𝜆𝑦

𝜆𝑧≤ 2 and

𝜆𝑧

𝜆𝑦≤ 2

(b)𝑒𝑦

ℎ𝑒𝑞

𝑒𝑧

𝑏𝑒𝑞≤ 0.2 or

𝑒𝑧

𝑏𝑒𝑞

𝑒𝑦

ℎ𝑒𝑞≤ 0.2

b, h = Width and depth of section

𝑏𝑒𝑞 = 𝑖𝑦 12 & ℎ𝑒𝑞 = 𝑖𝑧 12 for an equivalent rectangular section

y, z = Slenderness ratio with respect to y- and z- axis, respectively

𝑒𝑦 =𝑀𝐸𝑑,𝑧

𝑁𝐸𝑑: Eccentricity along y-axis

𝑒𝑧 =𝑀𝐸𝑑,𝑦

𝑁𝐸𝑑: Eccentricity along z-axis

Med,y = Design moment about y-axis including second order momentMed,z = Design moment about z-axis including second order momentNEd = Design axial load in the respective load combination

Biaxial Bending

z

y

h

b

NEd

ey

ez

iy

iy

iz iz

Biaxial Bending

• If the above conditions are NOT fulfilled, biaxially bent columns may be designed to satisfy the following simplified criterion:

𝑴𝑬𝒅,𝒛𝑴𝑹𝒅,𝒛

𝒂

+𝑴𝑬𝒅,𝒚

𝑴𝑹𝒅,𝒚

𝒂

≤ 𝟏. 𝟎

MRd,y = Moment resistance in y-axisMRd,z = Moment resistance in z-axisa = Exponent;

For circular & elliptical cross section, a = 2For rectangular cross section:

NRd = Acfcd + Asfyd

Ac = Gross area of concrete sectionAs = Area of longitudinal reinforcement

NEd/NRd 0.1 0.7 1.0

a 1.0 1.5 2.0

Biaxial Bending

Adaptation from BS 8110

Column may be designed for a single axis bending BUT with an increased moment:

(a) If𝑀𝐸𝑑,𝑧

ℎ′≥𝑀𝐸𝑑,𝑦

𝑏′then𝑀𝐸𝑑𝑧

′ = 𝑀𝐸𝑑,𝑧 + 𝛽ℎ′

𝑏′𝑀𝐸𝑑,𝑦

(b) If𝑀𝐸𝑑,𝑧

ℎ′<𝑀𝐸𝑑,𝑦

𝑏′then𝑀𝐸𝑑𝑦

′ = 𝑀𝐸𝑑,𝑦 + 𝛽𝑏′

ℎ′𝑀𝐸𝑑,𝑧

𝛽 = 1 −𝑁𝐸𝑑

𝑏ℎ𝑓𝑐𝑘(0.3 1.0)

hh’

bb’

z z

y

y

MEdz

MEdy

Example 5

DESIGN OF NON-SLENDER COLUMN BENT ABOUT BOTH

AXIS

Classification: Braced non-slender columnConcrete, fck: 25 N/mm2


Effective Length loz: 3.17 mloy: 3.00 m

Slenderness ratio z: 27.7y: 34.2

Bar Size: bar = 25 mm, links = 6 mmNominal cover: 30 mm

Example 5: Design of Non-Slender Column Bent About Both Axis

Mz

z

z

y y 300

350

Cross Section

My NEd = 1800 kNMz = 55 kNmMy = 32 kNm


DESIGN MOMENT

Imperfection moment, 𝑀𝑖𝑚𝑝 = 𝑁𝐸𝑑 ∙ 𝑒𝑖 = 𝑁𝐸𝑑 ∙𝑙𝑜

400

𝑀𝑖𝑚𝑝,𝑧 = 1800 ×3.70

400= 16.7 kNm

𝑀𝑖𝑚𝑝,𝑦 = 1800 ×3.00

400= 13.5 kNm

Design moment including the effect of imperfection:MEd,z = 55 + 16.7 = 71.7 kNmMEd,y = 32 + 13.5 = 45.5 kNm


CHECK BIAXIAL MOMENT

𝑒𝑧 =𝑀𝐸𝑑,𝑦

𝑁𝐸𝑑=71.7×106

1800×103= 40mm

𝑒𝑦 =𝑀𝐸𝑑,𝑧

𝑁𝐸𝑑=45.5×106

1800×103= 25mm

𝑒𝑦

ℎ

𝑒𝑧

𝑏=

25

350

40

300= 0.54 > 0.2

𝑒𝑧

𝑏

𝑒𝑦

ℎ=

40

300

25

350= 1.84 > 0.2

Check Biaxial Bending

𝜆𝑦

𝜆𝑧=34.2

27.7= 1.2 < 2

𝜆𝑧

𝜆𝑦=27.7

34.2= 0.8 < 2

Ignore Biaxial BendingCheck Biaxial Bending


REINFORCEMENT DESIGN

Effective depth: h’ = 350 – 30 – 6 – (0.5 25) = 301.5 mmb’ = 300 – 30 – 6 – (0.5 25) = 251.5 mm

𝑀𝐸𝑑,𝑧

ℎ′=71.7×106

301.5= 238 kN

𝑀𝐸𝑑,𝑦

𝑏′=45.5×106

251.5= 181 kN

Since 𝑀𝐸𝑑,𝑧

ℎ′>𝑀𝐸𝑑,𝑦

𝑏′𝑀𝐸𝑑,𝑧

′ = 𝑀𝐸𝑑,𝑧 + 𝛽ℎ′

𝑏′𝑀𝐸𝑑,𝑦

𝑁𝐸𝑑


1800×103

300×350×25= 0.69

𝛽 = 1 −𝑁𝐸𝑑

𝑏ℎ𝑓𝑐𝑘= 1 − 0.69 = 0.31 0.30

𝑀𝐸𝑑,𝑧′ = 71.7 + 0.31

301.5

251.545.5 = 𝟖𝟖. 𝟖 kNm


REINFORCEMENT DESIGN


2= 30 + 6 +

25

2= 48.5mm

𝑑2ℎ=48.5

350= 0.14 ≈ 0.15

𝑁


1800 × 103

300 × 350 × 25= 0.69

𝑀


88.8 × 106

300 × 3502 × 25= 0.10


𝑏ℎ𝑓𝑐𝑘= 0.48 𝐴𝑠 =

0.48𝑏ℎ𝑓𝑐𝑘

𝑓𝑦𝑘=0.48×300×350×25

500= 𝟐𝟓𝟐𝟎mm2


𝑓𝑦𝑑=0.10×1800×103

0.87×500= 𝟒𝟏𝟒mm2 or 0.002𝐴𝑐 = 210mm2

𝐴𝑠,𝑚𝑎𝑥 = 0.04𝐴𝑐 = 0.04 × 300 × 350 = 𝟒𝟐𝟎𝟎mm2

Provide 4H25 + 2H20 (As = 2592 mm2)

0.48


LINKS

links, min = the larger of:i. 0.25 25 mm = 6.3 mm orii. 6 mm Use links = 8 mm


Use Smax = 300 mm

Provide H8-300


Provide H8-175

30

0

350

H8

-30

0H

8-1

75

z

z


CHECK BIAXIAL BENDING

Steel Area:All: 4H25 + 2H20 As = 2592 mm2

z-z: 4H25 + 2H20 As,z = 2592 mm2

y-y: 4H25 + 0H20 As,y = 1964 mm2

𝑑2,𝑧

ℎ=48.5

350= 0.14 &

𝑑2,𝑦

𝑏=48.5

300= 0.16

𝑁


1800 × 103

300 × 350 × 25= 0.69

𝐴𝑠,𝑧𝑓𝑦𝑘


2592 × 500

300 × 350 × 25= 0.49

𝑀𝑅𝑑,𝑧𝑏ℎ2𝑓𝑐𝑘

= 0.10

MRd,z = 0.10 300 3502 25 10-6 = 91.9 kNm

0.49

0.10


CHECK BIAXIAL BENDING (continued)

𝐴𝑠,𝑦𝑓𝑦𝑘


1964 × 500

350 × 300 × 25= 0.37

𝑀𝑅𝑑,𝑦

𝑏ℎ2𝑓𝑐𝑘= 0.07

MRd,z = 0.07 350 3002 25 10-6 = 55.1 kNm

NRd = Acfcd + Asfyd = 0.567fckAc + 0.87fykAs

= (0.567 25 300 350) + (0.87 500 2592)= 2616 kN

𝑁𝐸𝑑

𝑁𝑅𝑑=1800

2616= 0.69 a = 1.49 (from Table)

Imperfection only be taken in one direction where they have the most unfavourable effects:𝑀𝐸𝑑,𝑧

𝑀𝑅𝑑,𝑧

𝑎

+𝑀𝐸𝑑,𝑦

𝑀𝑅𝑑,𝑦

𝑎

=71.7

91.9

1.49+

𝟑𝟐

55.1

1.49= 𝟏. 𝟒𝟒 > 𝟏. 𝟎 FAIL

0.37

0.07


NEW ARRANGEMENT OF REINFORCEMENT

Steel Area:All: 4H25 + 4H20 As = 3221 mm2

z-z: 4H25 + 2H20 As,z = 2592 mm2

y-y: 4H25 + 2H20 As,y = 2592 mm2

𝑑2,𝑧

ℎ=48.5

350= 0.14 &

𝑑2,𝑦

𝑏=48.5

300= 0.16

𝑁


1800 × 103

300 × 350 × 25= 0.69

𝐴𝑠,𝑧𝑓𝑦𝑘


2592 × 500

300 × 350 × 25= 0.49

𝑀𝑅𝑑,𝑧𝑏ℎ2𝑓𝑐𝑘

= 0.10

MRd,z = 0.10 300 3502 25 10-6 = 91.9 kNm

0.49

0.10


NEW ARRANGEMENT OF REINFORCEMENT

𝐴𝑠,𝑦𝑓𝑦𝑘


2592 × 500

350 × 300 × 25= 0.49

𝑀𝑅𝑑,𝑦

𝑏ℎ2𝑓𝑐𝑘= 0.10

MRd,z = 0.10 350 3002 25 10-6 = 78.8 kNm

NRd = Acfcd + Asfyd = 0.567fckAc + 0.87fykAs

= (0.567 25 300 350) + (0.87 500 3221)= 2889 kN

𝑁𝐸𝑑

𝑁𝑅𝑑=1800

2889= 0.62 a = 1.44 (from Table)

Imperfection only be taken in one direction where they have the most unfavourable effects:𝑀𝐸𝑑,𝑧

𝑀𝑅𝑑,𝑧

𝑎

+𝑀𝐸𝑑,𝑦

𝑀𝑅𝑑,𝑦

𝑎

=71.7

91.9

1.44+

𝟑𝟐

78.8

1.44= 𝟎. 𝟗𝟕 ≤ 𝟏. 𝟎 PASS

0.49

0.10


H8

-30

0H

8-1

75

30

0

350

z

z

y y