Design Example-Beam Compact Section

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Steelwork design guide to BS 5950-1:2000 Volume 2: Worked examples (2003 Edition)

Job No.

CDS 153

Sheet

1

of

10

Rev


Job No.

CDS 153

Job Title

Example no. 16

Subject

Beam subject to combined biaxial bending and compression (compact section)

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 623345 Fax: (01344) 622944 Client

©

16

, 8 0 0 2 / 2 0 / 8 0


Jun 2003

Checked by

ASM

Date

Oct 2003

300

A

B

C

600

K U

R E N R E A V K

Date

The beam shown in Figure 16.1 is subject to axial compression applied at its ends, a vertical point load at mid-span and a horizontal uniformly distributed load out of plane. The beam is restrained against lateral movement and torsion at its mid-span by a secondary beam, but is otherwise unrestrained. It is assumed that the beam is pinned at its ends in the major and minor axis directions. Design the beam in S355 steel for the loading shown below. Two solutions are presented; the full calculation and the Blue Book approach. 300

Figure 16.1

16.1.1 Loading (factored) Factored compression force Major axis point load Major axis distributed load Minor axis distributed load

F c =

300 kN W = 100 kN w = 2 kN/m (assumed self-weight) w = 3 kN/m

16.1.2 Bending moment and shear force Major axis

The shear force and bending moment diagrams for major axis bending are as shown in Figure 16.2. Maximum moment occurs at the centre: M

wL

2

=

WL =

+

8

2.0 × 6.0

2 +

100 × 6.0

8

4

= 159 kNm

4

Shear force at the ends: F ve

wL =

2

W +

2

=

2.0 × 6.0 2

+

100

= 56 kN

2

Shear force at the centre: F vc

=

F ve

wL −

2

= 50 kN

89

Rev

MDH

16.1 Introduction

, D T L

S S E C O R P

10

of

Beam subject to combined biaxial bending and compression (compact section)

, y p o C d e l l o r t n o c n U

1

Made by

SCI

CALCULATION SHEET I C S

Sheet

5 0 0 2 / 8 / 5 1 n o e t u t i t s n I n o i t c u r t s n o C l e e t S e h T m o r f e c n e c i l / r g e r d o . n z u i b d l e t e L t s s . e p x o e h d n s I / / : l p a t c t i h n o h t c e o T g r S o H 5 I 7 r o 7 f 2 d 7 e 8 c 4 u 4 d 3 o 1 r 0 p e l l R a . c t d n e e v r m e u s c e o r s d t h s i g i h t r f l l o a n o t i h s g r i e r v y y p p o o c s c i d r l a a i h r e a t a y u m b i s o h T T


Example 16 Biaxial bending and compression (compact section) 3000

Sheet

2

of

3000 100 kN

300 kN

2 kN/m

A

B

300 kN D Factored loads

6000

I C S

56

50

© , y p o C

Shear forces kN

50

d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0 , D T L

Bending moments kNm

159

Figure 16.2

The shear force and bending moment diagrams for minor axis bending are as shown in Figure 16.3. 3000

3000

300 kN

B

3 kN/m

A

300 kN D Factored loads

S S E C O R P

6000

5.63

3.38

R E N R E A V K

d e s n e c i L

Major axis bending

Minor axis

K U

, s s e c o r p r e n r e a v k : y p o c

56

Shear forces kN 3.38

5.63 3.38

1.89

1.89

Figure 16.3

Bending moments kNm

Minor axis bending

Maximum moment occurs at the central support: M

wL =

2

=

8

3.0 × 3.0

2

= 3.38 kNm

8

From the shear force diagram, the maximum shear force also occurs at the central support. F v

= 5.63 kN

90

10

Rev



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5 0 0 2 / 8 / 5 1 n o e t u t i t s n I n o i t c u r t s n o C l e e t S e h T m o r f e c n e c i l / r g e r d o . n z u i b d l e t e L t s s . e p x o e h d n s I / / : l p a t c t i h n o h t c e o T g r S o H 5 I 7 r o 7 f 2 d 7 e 8 c 4 u 4 d 3 o 1 r 0 p e l l R a . c t d n e e v r m e u s c e r o s d t s h i g h r t i f l l o a n o t i h s g r i e r v y y p p o o c s c i d r l a a i h r e a t a y u m b i s o h T T


Example 16 Biaxial bending and compression (compact section)

Sheet

3

of

10

Rev

16.2 Member checks 16.2.1 Trial section

I C S © , y p o C d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0 , D T L K U S S E C O R P R E N R E A V K , s s e c o r p r e n r e a v k : y p o c d e s n e c i L

Try 406 x 140 x 46 UB in grade S355 From section property tables: D = 403.2 mm Depth B = 142.2 mm Width t = 6.8 mm Web thickness T = 11.2 mm Flange thickness d = 360.4 mm Depth between fillets Ag = 58.6 cm2 Area of cross-section S x = 888 cm3 Plastic modulus S y = 118 cm3 Plastic modulus Z x = 778 cm3 Elastic modulus Z y = 75.7 cm3 Elastic modulus r x = 16.4 cm Radius of gyration r y = 3.03 cm Radius of gyration u = 0.872 Buckling parameter x = 39.0 Torsional index

Local buckling ratios: Flange Web

Vol 1 Page B-4

Vol 1 Page B-5

Vol 1 Page B-4

b/T = 6.35 d/t = 53.0

16.2.2 Section classification Grade of steel = S355 T < 16 mm Therefore py = 355 N/mm2

ε =

275

=

p y

275 355

3.1.1 Table 9

= 0.88

For the outstand element of a compression flange, the limiting b/T for a class 1 plastic flange is 9ε . Limiting b/T = 9 × 0.88 = 7.92

3.5.2 Table 11

The actual b/T = 6.35 < 7.92 Therefore, the flange is class 1 plastic . For the web of an I- or H-section under axial compression and bending (the “generally” case in Table 11), the limiting d/t for a class 1 plastic web is 80 ε but ≥ 40ε 1 + r 1 r 1 =

F c

=

dtp y

Limiting d/t

300 × 10

3

360.4 × 6.8 × 355

=

80 × 0.88 1 + 0.34

= 34, but –1 < r 1 ≤ 1, therefore r 1 = 0.34

= 52.5

The actual d/t = 53.0 > 52.5 Therefore, the web is not class 1 plastic.

91

3.5.2 Table 11

3.5.5



Example 16 Biaxial bending and compression (compact section) The limiting d/t for a class 2 compact web is

Limiting d/t I C S © , y p o C d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0 , D T L K U S S E C O R P R E N R E A V K , s s e c o r p r e n r e a v k : y p o c d e s n e c i L

=

100 × 0.88 1 + ( 1.5 × 0.34 )

100 ε 1 + 1.5 r 1

Sheet

4

of

but ≥ 40ε

10

Rev

3.5.2 Table 11

= 58.3

The actual d/t = 53.0 < 58.3 Therefore, the web is class 2 compact. Since the flange is class 1 plastic and the web is class 2 compact , the cross section is class 2 compact.

16.2.3 Determine the effective length The member is pinned at both ends about both axes, with an additional restraint against lateral (minor axis) deflection at its mid-span. LEx = 6000 mm LEy = 3000 mm

16.2.4 Slenderness λ x

=

λ y

=

L Ex

=

r x L Ey r y

=

6000 164 3000 30.3

4.7.2 = 36.6

= 99.0

16.2.5 Check the compression resistance Basic requirement F c P c P c = Ag pc (for a compact cross section)

4.7.4

The compressive strength pc is obtained from the relevant strut curve for buckling about the x-x and y-y axes. Buckling about the x-x axis

Use strut curve a For λ x = 36.6 and py = 355 N/mm2, pcx = 336 N/mm2 P cx = 5860 × 336 × 10–3 = 1969 kN

Table 23 Table 24a

Buckling about the y-y axis

Use strut curve b For λ y = 99.0 and py = 355 N/mm2, pcy = 157 N/mm2 P cy = 5860 × 157 × 10–3 = 920 kN

Table 23 Table 24b

Therefore, P c = 920 kN F c = 300 kN 300 kN < 920 kN Therefore, the compression resistance is adequate.

16.2.6 Check for shear buckling If the d /t ratio exceeds 70ε for a rolled section, the web should be checked for shear buckling in accordance with cl ause 4.4.5 of BS 5950-1:2000 [1]. 70ε = 70 × 0.88 = 61.6 d/t = 53.0 < 61.6, so there is no need to check for shear buckling . 92

4.2.3




Sheet

5

of

10

Rev

16.2.7 Check the shear capacity Major axis


Basic requirement F v P v P v = 0.6 py Av p y = 355 N/mm2 A v = tD = 6.8 403.2 = 2742 mm2 P v = 0.6 × 355 × 2742 × 10–3 = 584 kN Maximum shear force F v = 56 kN. 56 kN < 584 kN Therefore, the major axis shear capacity is adequate.

4.2.3

Minor axis

Basic requirement F v P v P v = 0.6 py Av py = 355 N/mm2 Av = 0.9 Ao, where Ao is the combined area of the two flanges. Av = 0.9 × 2 × 142.2 × 11.2 = 2867 mm2 P v = 0.6 × 355 × 2867 × 10–3 = 611 kN Maximum shear force F v = 5.63 kN. 5.63 kN < 611 kN Therefore, the minor axis shear capacity is adequate.

4.2.3

16.2.8 Check the moment capacity Major axis

Basic requirement M x M cx Check whether the shear is “high” (i.e. F v > 0.6 P v ) or “low” at the point of maximum moment.

4.2.5.2

At the centre of the member, F vc = 50 kN 0.6 P v = 0.6 × 584 = 350 kN 50 kN < 350 kN Therefore, the shear is low . For low shear, the moment capacity for a class 2 section is given by: M cx = py S x M cx = 355 × 888 × 10 –3 = 315 kNm Check limit to avoid irreversible deformation under serviceability loads. For a simply supported beam M cx ≤ 1.2 py Z x 1.2 py Z x = 1.2 × 355 × 778 × 10 –3 = 331 kNm

4.2.5.2

4.2.5.1

Therefore M cx = 315 kNm From Figure 16.2, M = 59 kNm x 159 kNm < 315 kNm Therefore, the major axis moment capacity is adequate. Minor axis

Basic requirement M y M cy By inspection, the shear is low (i.e. F v < 0.6 P v ). For low shear, the moment capacity for a class 2 section is given by M cy = py S y M cy = 355 × 118 × 10 –3 = 41.9 kNm

93

4.2.5.2




Sheet

6

of

Check limit to avoid irreversible deformation under serviceability loads. For a continuous beam M cy ≤ 1.5 p Z y y –3 1.5 py Z y = 1.5 × 355 × 75.7 × 10 = 40.3 kNm


10

Rev

4.5.2.1

Therefore M cy = 40.3 kNm From Figure 16.3, M = 3.38 kNm y 3.38 kNm < 40.3 kNm Therefore, the minor axis moment capacity is adequate.

16.2.9 Lateral-torsional buckling The beam is restrained against lateral-torsional buckling at its supports and at its mid-span. There are, therefore, two segments to consider. However, due to symmetry, it is sufficient to design one segment only in this example. Consider segment length AB. Basic requirements: M x M b / mLT and M x M cx (already checked in 16.2.8 above)

4.3.6.2

The buckling resistance moment M b for a class 2 compact section is given by: M b

4.3.6.4

= p b S x

where p b is the bending strength and is dependent on the design strength p y and the equivalent slenderness λ LT.

λ LT

= uvλ

x λ/

=

99 39.0

4.3.6.7

β W = 2.5

Table 19

For a section with equal flanges and λ / x = 2.5, v = 0.93 For a class 2 compact section, β W = 1.0 Therefore, λ LT = 0.872 × 0.93 × 99.0 = 80.3 For py = 355 N/mm2 and λ LT = 80.3, Table 16 gives p b = 189 N/mm2 –3 M = 168 kNm b = 189 × 888 × 10 Since there is no major axis loading between the restraints (apart from the self-weight of the beam, which is considered insignificant for the purpose of evaluating m LT), mLT can be obtained directly from Table 18 for a known value of β . β is the ratio of the bending moments at points A and B, i.e. β = 0 From Table 18, mLT = 0.6 M b / mLT

= 168 / 0.6

= 280 kNm

From Figure 16.2, M = 159 kNm x 159 kNm < 280 kNm Therefore, the buckling resistance moment is adequate.

94

4.3.6.9 Table 16

4.3.6.6 Table 18




16.2.10

Sheet

7

of

10

Rev

Interaction between axial load and bending

Cross section capacity

Basic requirement:

F c

+

A g p y I C S

d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0 , D T L K U S S E C O R P R E N R E A V K , s s e c o r p r e n r e a v k : y p o c d e s n e c i L

+

M cx

M y

4.8.3.2 ≤1

M cy

At point B

© , y p o C

M x

300 × 10 58.6 × 10

2

3

+

× 355

159

+

315

3.38 40.3

= 0.144 + 0.505 + 0.084 = 0.73 < 1 Therefore, the cross section capacity is adequate. Member buckling resistance

BS 5950-1:2000 presents two methods for checking the member buckling resistance: the simplified method (Clause 4.8.3.3.1) and the more exact method (Clause 4.8.3.3.2). In the simplified method, the following relationships must be satisfied: F c

+

P c F c P cy

m x M x

+

p y Z x

+

m LT M LT M b

m y M y

≤ 1

4.8.3.3.1

and

p y Z y

+

m y M y

≤ 1

p y Z y

M LT is the maximum major axis moment in segment AB

= 159 kNm.

4.8.3.3.1

mx is determined between restraints on the x axis, i.e. A and C, according to the shape of the bending moment diagram (see Figure 16.2).

As this is one of the specific cases listed in BS 5950-1:2000 [1], the value of mx can be obtained directly from Table 26. mx = 0.9

4.8.3.3.4 Table 26

my is determined between restraints on the y axis, i.e. A and B or B and C, according to the shape of the bending moment diagram between those restraints as shown in Figure 16.4. 3.38 A B

1.89

3000

Figure 16.4

Minor axis moment

95




Sheet

8

of

10

Rev

In this case, my must be obtained from the following general formula. my = 0.2 +

I C S © , y p o C d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0

0.1 M 2 + 0.6 M 3 + 0.1 M 4 M max

but m y ≥

M max

The moments M 2 and M 4 are the values at the quarter points and the moment M 3 is the value at mid-length. M max is the maximum moment in the segment and M 24 is the maximum moment in the central half of the segment. Location

M 2

Moment kNm

m y = 0.2 +

M 3

1.69

M 4

1.69

M 24

0

0.1 × 1.69 + 0.6 × 1.69 + 0 3.38

1.89

M max

3.38

= 0.55 but m y ≥

300

+

920

0.9 × 159 × 10

6

355 × 778 × 10

3

+

= 0.326 + 0.518+ 0.069

300 920

+

0.6 × 159 168

+

0.55 × 3.38 × 10

6

355 × 75.7 × 10

3

0.55 × 3.38 × 10

6

355 × 75.7 × 10

3

K U

Therefore, the member buckling resistance is adequate.

= 0.45

= 0.96 <1

Adopt 406 x 140 x 46 UB in S355

Note: 1. If appropriate, the web should be checked for bearing and buckling at the supports and at the point of load application, as in Example 2. 2. The deflections should be checked at the serviceability limit state in accordance with the recommendations in clause 2.5.2.


3.38

= 0.91 <1

= 0.326 + 0.568+ 0.069

R E N R E A V K

0.8 × 1.89

Therefore, m y = 0.55

, D T L

S S E C O R P

4.8.3.3.4 Table 26

0.8 M 24

96




Sheet

9

of

10

Rev

16.3 Blue Book approach The section capacities and member resistances calculated in section 16.2 could have been obtained directly from Volume 1 [2]. Try 406 x 140 x 46 UB in grade S355 I C S © , y p o C d e l l o r t n o c n U , 8 0 0 2 / 2 0 / 8 0 , D T L K U S S E C O R P R E N R E A V K , s s e c o r p r e n r e a v k : y p o c d e s n e c i L

16.3.1 Compression resistance For LEx = 6.0 m, P cx = 1970 kN For LEy = 3.0 m, P cy = 925 kN P c Therefore, = 925 kN F c = 300 kN 300 kN < 925 kN Therefore, the compression resistance is adequate.

Vol 1 Page D-115

16.3.2 Shear Capacity (major axis) P v = 584 kN (no value given for minor axis shear) F v = 56 kN 56 kN < 584 kN Therefore, the major axis shear capacity is adequate.

Vol 1 Page D-79

16.3.3 Moment capacity For F /P z = 0.144 M cx = 315 kNm M x = 159 kNm 159 kNm < 315 kNm Therefore, the major axis moment capacity is adequate.

Vol 1 Page D-114

For F /P z = 0.144 M cy = 32.2 kNm M y = 3.38 kNm 3.38 kNm < 32.2 kNm Therefore, the minor axis moment capacity is adequate.

Vol 1 Page D-114

The value of M cy given by Volume1 is considerably lower than that calculated in Section 16.2 above (40.3 kNm) due to the fact that Volume 1 uses the limit of 1.2 p Z is more appropriate in this example, since the y , whereas the higher limit of 1.5 py Z beam is continuous over a central support for minor axis bending.

16.3.4 Buckling resistance moment For LE M b

= 3.0 m and F /P z = 167 kNm

= 0.144

Vol 1 Page D-115

For a quick approximation, mLT may conservatively be taken as 1.0. However, more accurately, mLT should be obtained from Table 18 of BS 5950-1:2000 as in Section 16.2.9 of this example. mLT = 0.6 M = 278 kNm b / mLT = 167 / 0.6 M = 159 kNm x 159 kNm < 278 kNm Therefore, the buckling resistance moment is adequate.

97

Table 18




Sheet

10

of

10

Rev

16.3.5 Interaction between axial load and bending Cross section capacity at point B

Ag py = 2080 kN


F c

+

A g p y

M x M cx

Vol 1 Page D-115

M y

+

300

=

+

2080

M cy

= 0.144 + 0.505 + 0.105

159

+

315

3.38 32.2

= 0.75 < 1

Therefore, the cross section capacity is adequate. 4.8.3.3.4

Member buckling resistance

For a quick approximation, mx and my may conservatively be taken as 1.0. However, more accurately, mx and my should be obtained from Table 26 of BS 5950-1:2000 as in Section 16.2.10 of this example. = = = =

mx my py Z x py Z y F c

m x M x

+

P c

Table 26

0.9 0.55 276 kNm 26.9 kNm +

p y Z x

Vol 1 Page D-115

m y M y

=

p y Z y

300

+

0.9 × 159

925

+

0.55 × 3.38

276

26.9

= 0.324 + 0.518+ 0.069 = 0.91 <1 F c P cy

+

m LT M LT M b

+

m y M y p y Z y

=

300 925

+

0.6 × 159

+

0.55 × 3.38

167

26.9

= 0.324 + 0.571+ 0.069 = 0.96 <1

Therefore, the member buckling resistance is adequate. Adopt 406 x 140 x 46 UB in S355

Note: 1. If appropriate, the web should be checked for bearing and buckling at the supports and at the point of load application, as in Example 2. 2. The deflections should be checked at the serviceability limit state in accordance with the recommendations in clause 2.5.2.

98


Design Example-Beam Compact Section

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