Design of Laterally Unrestrained Beams Design of Steel ...site.iugaza.edu.ps › hbaker › files › 2018 › 02 › Design-of...Design of Laterally Unrestrained Beams Design of Steel

Design of Laterally

Unrestrained

Beams

Design of Laterally Unrestrained Beams Design of Steel Structures to EC3

Page(2) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

4.1 Beam Design Limit States:

1. Bending strength at ULS, including

a) Local buckling of flange or web.

b) LTB if the beam is not fully restrained.

c) Plastic moment capacity.

2. Shear strength at ULS.

3. Resistance to transverse forces at ULS (web bearing and web buckling).

4. Deflection SLS.

4.2 Laterally Unrestrained Beams

Elastic critical moment:

𝑀𝑐𝑟 = 𝐶1𝜋2𝐸𝐼𝑧(𝐾𝑧𝐿)

2 {(√(𝐾𝑧𝐾𝑤

)2

.𝐼𝑤𝐼𝑧

+(𝐾𝑧𝐿)

2𝐺𝐼𝑇𝜋2𝐸𝐼𝑧

+ (𝐶2𝑧𝑔 − 𝐶3𝑧𝑗)2

) − (𝐶2𝑧𝑔 − 𝐶3𝑧𝑗)}



Lateral Torsional Buckling Resistance:

𝑀𝐸𝑑

𝑀𝑏,𝑅𝑑 ≤ 1.0

𝑀𝑏,𝑅𝑑 = 𝜒𝐿𝑇 𝑊𝑦 𝑓𝑦 𝛾𝑀1⁄

Where:

Calculations of 𝜒𝐿𝑇 1- General Method: that can be applied to any type of cross section.



2- Alternative method: rolled or equivalent welded sections.

The relevant buckling curves:

According to this second method, the shape of the bending moment

diagram, between braced sections, can be taken into account by

considering a modified reduction factor 𝜒𝐿𝑇,𝑚𝑜𝑑



The parameter f can be obtained from the following equation, or from an

alternative process provided in the National Annexes:

Where kc is a correction factor, defined according to Table 3.10

Important Note:

The verification of lateral-torsional buckling for a member in bending may

be ignored if at least one of the following conditions is verified (clause

6.3.2.2(4)):



4.3 Solved Problems

Problem (1)

A UKB in grade S275 steel is required to carry loads over a span of 9.0 m. The beam

is laterally restrained at A, B, C and D but unrestrained between these points. The depth

of the beam must not exceed 425 mm, and the ceiling under the beam will be plastered.

Choose a suitable UKB section.

Solution:

𝐹𝐸𝑑 = 𝛾𝐺 𝐺𝐾 + 𝛾𝑄 𝑄𝐾

𝑃𝐵 = 1.35 × 30 + 1.5 × 47 = 111.0 𝐾𝑁

𝑃𝐶 = 1.35 × 23 + 1.5 × 28 = 73.05 𝐾𝑁

𝐹𝑈𝐷𝐿 = 1.35 × 1.3 = 1.755 𝐾𝑁

So design load at ULS:

Shear Force and Bending Moment Diagrams:



Choose a suitable section size:

Assume class 1 x-section, calculate the minimum plastic modulus 𝑊𝑃𝐿,𝑦 about

y-y axis.

𝑊𝑃𝑙,𝑦 = 𝑀𝐸𝑑𝛾𝑀0

𝑓𝑦=

311 × 106 × 1.0

275× 10−3 = 1130.90𝑐𝑚3

From blue book try UKB 406 x178 x 67 where 𝑊𝑃𝐿,𝑦 = 1350𝑐𝑚3.

Geometrical properties:

ℎ = 409.4𝑚𝑚; 𝑏𝑓 = 178.8𝑚𝑚; 𝑡𝑓 = 14.3𝑚𝑚; 𝑡𝑤 = 8.8𝑚𝑚; 𝐴 = 85.5𝑚𝑚2

𝑟 = 10.2𝑚𝑚; 𝑊𝑃𝐿,𝑦 = 1350𝑐𝑚3; 𝐼𝑦 = 24300𝑐𝑚

4; 𝐼𝑧 = 1360𝑐𝑚4;

𝐼𝑇 = 46.1𝑐𝑚4; 𝐼𝑊 = 0.533 𝑑𝑚

6; 𝐸 = 210𝐾𝑁

𝑚𝑚2; 𝐺 = 81

𝐾𝑁

𝑚𝑚2

X-Section Classification:

For flange:

(𝑐

𝑡) = 5.23

9𝜖 = 9 × 0.92 = 8.28

→ 5.23 < 8.28 →→ 𝑐𝑙𝑎𝑠𝑠 1

For web:

(𝑐

𝑡) = 41.0

72𝜖 = 72 × 0.92 = 66.24

→ 41.0 < 66.24 →→ 𝑐𝑙𝑎𝑠𝑠 1

→ 𝑡ℎ𝑒 𝑥 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑙𝑎𝑠𝑠 1(𝑃𝑙𝑎𝑠𝑡𝑖𝑐)

Flexural Strength Verification:

𝑀𝐸𝑑 = 259. 2 𝐾𝑁. 𝑚

𝑀𝑐,𝑅𝑑 = 𝑀𝑃𝑙,𝑅𝑑 = 𝑊𝑃𝑙 𝑓𝑦

𝛾𝑀0=

1350 × 103 × 275

1.0× 10−6 = 371.25 𝐾𝑁. 𝑚

> 311𝐾𝑁. 𝑚 → 𝑂. 𝐾

Bulking Resistance:

- The effective length of the segment is 3.0m.

- The ratio of the moment segment:

𝜓 =272.9

311= 0.878

From Table (3.6):

𝐼𝑓 𝜓 = 1.00 → 𝐶1 = 1.00

𝐼𝑓 𝜓 = 3/4 → 𝐶1 = 1.14



𝑆𝑜 𝑏𝑦 𝑙𝑖𝑛𝑒𝑎𝑟 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑤ℎ𝑒𝑛 𝜓 = 0.878 → 𝐶1 = 1.068

𝑀𝑐𝑟 = 𝐶1𝜋2𝐸𝐼𝑧(𝐾𝑧𝐿)

2 {(√(𝐾𝑧𝐾𝑤

)2

.𝐼𝑤𝐼𝑧

+(𝐾𝑧𝐿)

2𝐺𝐼𝑇𝜋2𝐸𝐼𝑧

+ (𝐶2𝑧𝑔 − 𝐶3𝑧𝑗)2

) − (𝐶2𝑧𝑔 − 𝐶3𝑧𝑗)}

𝑀𝑐𝑟 = 1.068 ×𝜋2 × 210 × 103 × 1360 × 104

(1 × 3000)2{(√(

1

1)

2

.0.533 × 1012

1360 × 104+

(1 × 3000)2 × 81 × 103 × 46.1 × 104

𝜋2 × 210 × 103 × 1360 × 104+ (0)2 ) − 0.0} × 10−6

𝑀𝑐𝑟 = 756.233 𝐾𝑁. 𝑚

𝜆𝐿𝑇̅̅ ̅̅ = √𝑊𝑃𝑙 𝑓𝑦

𝑀𝑐𝑟= √

1350 × 103 × 275

756.233 × 106= 0.70

Check:

𝜆𝐿𝑇̅̅ ̅̅ = 0.7 > 𝜆𝐿𝑇,0̅̅ ̅̅ ̅̅ = 0.4

OR

𝑀𝐸𝑑

𝑀𝑐𝑟=

311

756.233= 0.411 > 𝜆𝐿𝑇,0̅̅ ̅̅ ̅̅

2= 0.16

So we need to consider LTB.

- In order to calculate 𝜒𝐿𝑇 we will use the alternative method.

ℎ

𝑏=

409.4

178.8= 2.29 > 2 → 𝐵𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝐶𝑢𝑟𝑣𝑒 𝑐 → 𝛼𝐿𝑇 = 0.49

𝛽 = 0.75; 𝜆𝐿𝑇,0̅̅ ̅̅ ̅̅ = 0.4

∅𝐿𝑇 = 0.5[1 + 𝛼𝐿𝑇(𝜆𝐿𝑇̅̅ ̅̅ − 𝜆𝐿𝑇,0̅̅ ̅̅ ̅̅ ) + 𝛽𝜆𝐿𝑇̅̅ ̅̅2

∅𝐿𝑇 = 0.5[1 + 0.49(0.7 − 0.4) + 0.75 × 0.72 = 0.757

𝜒𝐿𝑇 =1

∅𝐿𝑇 + √∅𝐿𝑇2 − 𝛽𝜆𝐿𝑇̅̅ ̅̅

2

=1

0.757 + √0.7572 − 0.75 × 0.72= 0.826

Check

𝜒𝐿𝑇 = 0.826 ≤ 1.0 → 𝑂. 𝐾

𝜒𝐿𝑇 = 0.826 ≤1

𝜆𝐿𝑇̅̅ ̅̅ ̅2 =

1

0.72= 2.04 → 𝑂. 𝐾

- Calculate 𝜒𝐿𝑇,𝑚𝑜𝑑

From Table (3.10) we calculate Kc



𝜓 = 0.878 → −1 ≤ 𝜓 ≤ 1 → 𝐾𝐶 =1

1.33 − 0.33𝜓

𝐾𝐶 =1

1.33 − 0.33 × 0.878= 0.961

𝑓 = 1 − 0.5(1 − 𝐾𝐶)[1 − 2.0(𝜆𝐿𝑇̅̅ ̅̅ − 0.8)2

]

𝑓 = 1 − 0.5(1 − 0.961)[1 − 2.0(0.7 − 0.8)2] = 0.981 ≤ 1.0

𝜒𝐿𝑇,𝑚𝑜𝑑 =𝜒𝐿𝑇𝑓

=0.826

0.981= 0.842

𝑀𝑏,𝑅𝑑 = 𝜒𝐿𝑇,𝑚𝑜𝑑 𝑊𝑦 𝑓𝑦 𝛾𝑀1⁄ =0.842 × 275 × 1350 × 103

1.0× 10−6

= 𝐾𝑁312.59. 𝑚

𝑀𝐸𝑑𝑀𝑏,𝑅𝑑

=311

312.59= 0.99 < 1

→ 𝑇ℎ𝑒 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑟𝑡𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑈𝐾𝐵406 × 178 × 67 𝑖𝑠 𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

Shear Strength Verification:

𝑉𝐸𝑑 = 106.3 𝐾𝑁

Shear area:

𝐴𝑣 = 𝐴 − 2𝑏𝑡𝑓 + (𝑡𝑤 + 2𝑟)𝑡𝑓 ≥ ƞℎ𝑤𝑡𝑤

= 8550 − 2 × 178.8 × 14.3 + (8.8 + 2 × 10.2) × 14.3 = 3853.88𝑚𝑚2

> 1.0 × (409.4 − 2 × 14.3) × 8.8 = 3351.04𝑚𝑚2

𝑉𝑃𝐿,𝑅𝑑 = 𝐴𝑣(𝑓𝑦 √3⁄ )

𝛾𝑀0=

3853.88 × 275

√3 × 1.0× 10−3 = 611.886 𝐾𝑁 > 106.3 𝐾𝑁

𝑉𝐸𝑑 = 106.3𝐾𝑁 < 50% 𝑉𝑃𝐿,𝑅𝑑 = 305.94𝐾𝑁

→ 𝑁𝑜 𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 𝑜𝑛 𝑑𝑒𝑠𝑖𝑔𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 𝑏𝑒𝑛𝑑𝑖𝑛𝑔

Shear buckling:

Shear buckling of the unstiffened web need not be considered provided:

ℎ𝑤𝑡𝑤

≤ 72∈

ƞ

ℎ𝑤𝑡𝑤

=380.8

8.8= 43.27

72∈

ƞ= 72 ×

0.92

1.0= 66.24

66.24 < 66.24 →Therefore, shear buckling check need not be considered

→ 𝑇ℎ𝑒 𝑈𝐾𝐵406 × 178 × 67 𝑖𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑡𝑜𝑟𝑦 𝑖𝑛 𝑠ℎ𝑒𝑎𝑟.



Deflection Verification: SLS unfactored imposed actions.

A simple approach is to find the distributed load that would cause the same

maximum bending moment, then to check whether the deflection caused

by that distributed load is acceptable.

122.1𝐾𝑁. 𝑚 =𝑤𝑙2

8=

𝑤 × 92

8→ 𝑤 = 12.06 𝐾𝑁/𝑚

𝛿𝑚𝑎𝑥,𝑚𝑖𝑑 𝑠𝑝𝑎𝑛 =5

384×

𝑤𝑙4

𝐸𝐼𝑦=

5

384×

12.06 × 90004

210000 × 24300 × 104= 20.19𝑚𝑚

Vertical deflection limit:

𝑠𝑝𝑎𝑛

360=

9000

360= 25𝑚𝑚

→ 20.19𝑚𝑚 < 25𝑚𝑚

→ 𝑇ℎ𝑒 𝑈𝐾𝐵406 × 178 × 67 𝑚𝑒𝑒𝑡𝑠 𝑡ℎ𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑎.

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Design of Laterally Unrestrained Beams Design of Steel ...site.iugaza.edu.ps › hbaker › files › 2018 › 02 › Design-of...Design of Laterally Unrestrained Beams Design of Steel