Steel Tutorial for bs code 2

1

CE3166 Tutorial 2: Members under combined axial load and bending moments

Semester 1, AY2009/2010

CE3166 CE Materials and Structural Steel Systems

Tutorial 2:Members under combined axial load

and bending moments

2



Beam-columns – behaviour

Axial stress Bending stress Total stress

Interaction between axial stress and bending stress.

3



Beam-columns – behaviourMembers under combined axial tension and bending moments

LTB due to transverse load Major axis buckling due to axial compression Minor axis buckling due to compression

Check resistance to LTB only.

4



Beam-columns – behaviourMembers under combined axial compression and bending moments

LTB due to transverse load Major axis buckling due to axial compression Minor axis buckling due to compression

How to design the member if ALL these buckling modes can occur?

5



Beam-columns – design procedureMembers under combined axial compression and bending moments1. Determine design strength and section classification – Table 9, Table 11,

Table 12.2. Calculate moment capacities Mcx and Mcy and limits for bending moment to

prevent permanent deformation.3. Determine critical segments about major and minor axes for buckling due to

compression and lateral torsional buckling.4. Calculate compression resistances Pcx and Pcy – for each segment:

- determine effective lengths LEx and LEy about for buckling about both axes –Table 22.

- calculate slenderness about both axes λx and λy.- determine strut curve to use and calculate compressive strengths about both

axes pcx and pcy – Table 23, Table 24.- calculate compression resistances Pcx and Pcy.

6



Beam-columns – design procedure (cont.)5. Calculate buckling resistance moment Mb – for each segment:

- determine effective length LE – Table 13, Table 14.- calculate slenderness λ.- determine slenderness factor v – Table 19.- calculate equivalent slenderness λLT.- determine bending strength pb – Table 16, Table 17.- calculate buckling resistance moment Mb.

6. Determine equivalent uniform moment factors:- mLT – Table 18.- mx, my and myx – Table 26.

7. Perform section capacity check for each critical point – point of maximum bending moments and axial force.

8. Perform buckling resistance check for each segment – length along which buckling can occur.

7



Question 1A typical column in a multi-storey beam-and-column braced frame with full continuity at moment resisting joints is subject to a compressive force Fc = 680 kNwith major-axis bending moments Mx1 and Mx2 causing double-curvature bending and minor-axis bending moments My1 and My2 causing single-curvature bending as shown in Figure Q1. The column is laterally braced at the ends and at 2 m from the top. Assuming that the ends are simply supported, try out the section 203×203×52 UC in S355 steel for the column member and check its adequacy for strength (i.e. cross-section capacity) and stability (i.e. buckling resistance). Use the simplified method.

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

8



Design strengthSection properties of 203×203×52 UC:D = 206.2 mm rx = 8.91 cm u = 0.848B = 204.3 mm ry = 5.18 cm x = 15.8d = 160.8 mm Sx = 567 cm3 Ag = 66.3 cm2

t = 7.9 mm Sy = 264 cm3

T = 12.5 mm Zx = 510 cm3

b/T = 8.17 Zy = 174 cm3

d/t = 20.4

Since T < 16 mm, design strength py = 355 N/mm2.

88.0355275275

===yp

ε

9



Section classificationFlange: Web:b/T = 8.17 > 9ε = 7.92b/T = 8.17 < 10ε = 8.80⇒ Flange is Class 2. ⇒ Use r1 = 1.0

d/t = 20.4 < 80ε/(1+r1) = 35.2Overall classification is ⇒ Web is Class 1.Class 2 (compact).

51.13559.78.160

10680 3

1 =××

×==

yw

c

dtpF

r

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Moment capacity Mc

Assume section is in low shear.

Moment capacities for Class 2 section:Mcx = pySx = 355×(567×103) ×10-6 = 201 kNm.Mcy = pySy = 355×(264×103) ×10-6 = 94 kNm.

Limit to prevent permanent deformation:1.5pyZx = 1.5×355×(510×103) ×10-6 = 272 kNm.1.5pyZy = 1.5×355×(174×103) ×10-6 = 93 kNm.

⇒ Mcx = 201 kNm.Mcy = 93 kNm.

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Critical segments for bucklingBuckling about major axis:One segment- entire length of member.

Buckling about minor axis:Two segments- top (A) and bottom (B).

Need to calculate valuesfor each buckling length:- Compression resistance Pc.- Buckling resistance moment Mb.- Equivalent uniform moment factors m.

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

12



Compression resistance Pc

Buckling about x-axis:From Table 22, effective length LEx = 5.0 m.Slenderness λx = LEx/rx = 56.1From Table 23, use Table 24(b) for buckling about x-axis.Compressive strength pcx = 278 N/mm2.Compression resistance Pcx = Agpcx = (66.3×102)×278 ×10-3 = 1843 kN.

Buckling about y-axis:

Top segment (A) Bottom segment (B)

LEy (m) – Table 22 2.0 3.0

λy = LEy/ry 38.6 57.9

pcy (N/mm2) – Table 24(c) 301 252

Pcy = Agpcy (kN) 1996 1671

13



Buckling resistance moment Mb

• For Class 2 section, βW = 1.0

Top segment (A) Bottom segment (B)

LE (m) – Table 13 2.0 3.0

λ = LE/ry 38.6 57.9

v – Table 19 0.96 0.89

λLT = uvλ√βW 31.4 43.7

pb (N/mm2) – Table 16 341 309

Mb = pbSx (kNm) 193 175

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Equivalent uniform moment factors mBuckling about x-axis:βx = -50/80 = -0.625From Table 26, mx = 0.48βyx = 15/30 = 0.5From Table 26, myx = 0.80

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

LEx LEy

Mx mx mLT

My myx my

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Equivalent uniform moment factors mBuckling about y-axis:Top segment (A):βy = 15/21 = 0.71From Table 26, my = 0.92βLT = 28/80 = 0.35From Table 18, mLT = 0.76

Bottom segment (B):βy = 21/30 = 0.70From Table 26, my = 0.88βLT = -28/50 = -0.35From Table 18, mLT = 0.44

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

LEx LEy

Mx mx mLT

My myx my

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Section capacity check – simplified methodCheck at critical points:

Agpy = (66.3×102)×355 ×10-3 = 2354 kN.Mcx = 201 kNm.Mcy = 93 kNm.

Top end (point 1):

Bottom end (point 2):

Section capacity check passes.

0.1≤++cy

y

cx

x

yg

c

MM

MM

pAF

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

848.09315

20180

2354680

=++

860.09330

20150

2354680

=++

17



Buckling resistance check – simplified methodCheck for each segment:

pyZx = 355×510 ×10-3 = 181 kNm. ⇒ Buckling resistance pyZy = 355×174 ×10-3 = 61.8 kNm. check fails.

Top segment (A):

Bottom segment (B):

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

0.1≤++yy

yy

xy

xx

c

c

ZpMm

ZpMm

PF 0.1LTLT ≤++

yy

yy

bcy

c

ZpMm

MMm

PF

894.08.612192.0

1818048.0

1843680

=×

+×

+

mx = 0.48myx = 0.80

Pcx = 1843 kN

my = 0.92mLT = 0.76

Pcy = 1996 kNMb = 193 kNm

my = 0.88mLT = 0.44


968.08.612192.0

1938076.0

1996680

=×

+×

+

05.18.613088.0

1818048.0

1671680

=×

+×

+

960.08.613088.0

1755044.0

1671680

=×

+×

+

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Buckling resistance check – more exact approachCheck for each segment:Major axis buckling:

Lateral torsional buckling:

Interactive buckling:

0.15.05.01 ≤+⎟⎟⎠

⎞⎜⎜⎝

⎛++

cy

yyx

cx

c

cx

xx

cx

c

MMm

PF

MMm

PF

0.11LTLT ≤⎟⎟⎠

⎞⎜⎜⎝

⎛+++

cy

c

cy

yy

bcy

c

PF

MMm

MMm

PF

0.11

1

1

5.01≤

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

cy

ccy

cy

cyy

cx

ccx

cx

cxx

PFM

PFMm

PFM

PFMm

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

mx = 0.48myx = 0.80

Pcx = 1843 kN

my = 0.92mLT = 0.76


my = 0.88mLT = 0.44


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Buckling resistance check – more exact approachTop segment (A):Major axis buckling: Lateral torsional buckling:


934.019966801

932192.0

1938076.0

1996680

=⎟⎠⎞

⎜⎝⎛ +

×+

×+

781.0

1996680193

199668012192.0

18436801201

18436805.018048.0

=⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +×

+

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ ×+×

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

mx = 0.48myx = 0.80

Pcx = 1843 kN

my = 0.92mLT = 0.76


my = 0.88mLT = 0.44


724.093

3080.05.0

18436805.01

2018048.0

1843680

=×

×+

⎟⎠⎞

⎜⎝⎛ ×+

×+

20



Buckling resistance check – more exact approachBottom segment (B):Major axis buckling: Lateral torsional buckling:


932.016716801

933088.0

1755044.0

1671680

=⎟⎠⎞

⎜⎝⎛ +

×+

×+

03.1

1671680193

167168013088.0

18436801201

18436805.018048.0

=⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +×

+

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ ×+×

Fc = 680 kN

5 m

Fc = 680 kN

Mx2 = 50 kNm

Mx1 = 80 kNm Mx1

Mx2

Fc = 680 kN

My1

Fc = 680 kN

My2 My2 = 30 kNm

My1 = 15 kNm

1

2

2 m

3 m

Lateral restraint

A

B

28 21

mx = 0.48myx = 0.80

Pcx = 1843 kN

my = 0.92mLT = 0.76


my = 0.88mLT = 0.44


724.093

3080.05.0

18436805.01

2018048.0

1843680

=×

×+

⎟⎠⎞

⎜⎝⎛ ×+

×+

21



Design handbooksResistances and capacities for standard members can be obtained from handbooks (e.g. Handbook of Structural Steelwork).

Mcx = 201 kNmMcy = 74.1 kNmAgpy = 2350 kN

(Note: Limit for moment capacity of 1.2pyZ is used).

22



Design handbooks (cont.)Resistances and capacities for standard members can be obtained from handbooks (e.g. Handbook of Structural Steelwork).

Pcx = 1870 kNPcy,top = 2020 kNMb,top = 200 kNmPcy,bottom = 1670 kNMb,bottom = 178 kNm

pyZx = 181 kNmpyZy = 61.8 kNm

23



Questions to ponder1. What if shear in member is high? How would you account for high

shear in the section capacity and buckling resistance checks?

2. Since it is possible for a member to pass the buckling resistance check using the simplified method but FAIL the same check using the more exact method, is it necessary to perform the more exact check even if the simplified check passes?

3. Suppose you have two members of the same section which are subjected to the same axial force, bending moment distribution, support and restraint conditions. However, these members are ofdifferent lengths. If the shorter member passes the section capacity check, will the longer member also pass the same check?

24



Questions to ponder4. Should you use the algebraic or absolute values for moments Mx

and My to determine the equivalent uniform moment factors mx, my,mLT and myx? How about for the section capacity and buckling resistance checks?

25



Table 18

26



Table 26

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Steel Tutorial for bs code 2