chs splice

Example 5 - Splices - CHS tension splice SheetTitle

CHS Tension Splices - Worked Example 5

259

BCSA

CALCULATIONSHEET

Calcs by

Title

Client

Checked by Date

Job No SheetJoints in Steel Construction - Simple Connections

SCI/BCSA Connections Group

RS AW May 2002

1 of 5

DESIGN EXAMPLE 5 TENSION SPLICE

Check the following connection for the design forces shown:

Design Information:

End Plates: 25mm thick

Bolts: M24 8.8

Material: S275 steel

Weld: 12mm fillet

Example 5 - Splices - CHS tension splice

750kN

750kN

25 25

480 diameter

273 x 6.3 CHS Grade S275

273 x 6.3 CHS Grade S275

50 e 1 e 2

53.5

Total number of bolts N = 8

Joints in Steel Construction: Simple Connections - Example 5 - column splices - CHS tension splice

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260

SUMMARY 0F FULL DESIGN CHECKS FOR EXAMPLE 5

Sheet CHECK

Capacity Applied Comments

Nos kN kN

3 CHECK 1 Recommended detailing Not Not All recommendations practice Applicable Applicable adopted

3 CHECK 2 Complete end plate yielding 1610 750

4 CHECK 3 Bolt failure with end plate yielding 1036 750 CRITICAL CHECK

4 CHECK 4 Bolt failure 1584 750

5 CHECK 5 Weld

- Tension Capacity 1585 750 Full strength welds

5 CHECK 6 Member - Tension capacity 1452 750

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CONNECTION DESIGN USING CAPACITY TABLES FROM YELLOW PAGES

Yellow pages273 X 6.3 CHS Grade S275 Standard connection Table H.36

Total number of bolts N = 8

End plate thickness = 20mm

Tying capacity = 946 kN

Applied force = 750 kN

750kN < 946kN O.K.

Note: The capacity from Table H.36 is conservatively based on the thickestsection in the range.

SeeTable H.34

note (2)


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261

3 of 5

CHECK 1: Recommended detailing practice

End Plates tp = 25mm

Bolts d = 24mm dia

Holes Dh = 26mm

Spacing ( 2.5d = 60) p = 380 x pi /8 (< 10d =240) = 149mm

Edge distance ( 1.4Dh = 36) e2 = 50mm

CHECK 2: Complete end plate yielding

Basic requirement:

tp2 py plate pi f3Ft 2

1 f3 = ( k3 + (k32 4 k1)

0.5)2k1

k1 = ln (r2 /r3) [ln = natural loge]

D 273r2 = + e1 = + 53.5 = 190 2 2

D tw 273 6.3r3 = = = 133.4 2 2

190k1 = ln = ln1.42 = 0.354 133.4

k3 = k1 + 2 = 0.354 + 2 = 2.354

1 f3 = ( 2.354 + (2.3542 4 x 0.354)

0.5)2 x 0.354

= 6.19

252 x 265 x pi x 6.19 Plate capacity = = 1610kN

2 x 103

Ft = 750kN 1610 kN O.K.

750kN 750kN

t = 6.3 w

t = 25 p

e = 53.5 1

D = 273


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262

CHECK 3: Bolt failure with end plate yielding

Basic requirement:

NPt Ft 1 1 1 + [ln = natural loge] f3 f3 ln(r1/r2)

Pt = Exact bolt capacity = 198kN

f3 = 6.19 CHECK 2

eeff = minimum of e2 and 1.25e1 = 50 mm

D 273r1 = + e1+ eeff = + 53.5 + 50 = 240mm 2 2

D 273r2 = + e1 = + 53.5 = 190mm 2 2

240ln(r1/r2) = ln = ln 1.263 = 0.234 190

8 x 198 Bolt group capacity = = 1036kN 1 1 1 + 6.19 6.19 x 0.234

Ft = 750kN < 1036 kN O.K.

CHECK 4: Bolt failure

Basic requirement for bolt tensile capacity:

Ft N Pt

N Pt = 8 x 198

= 1584 kN

Ft = 750kN 1584kN O.K

4 of 5

750kN

e = 53.5 1e = 50 2

t = 6.3 w

p = 139

750kN

Pt fromYellow pagesTable H.49

Pt fromYellow pagesTable H.49


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263

5 of 5

CHECK 5: Welds

Basic requirement for weld tensile capacity:

(i) tw a (continuous full strenghth weld )

or

(ii) Ft pi D a pw

pw = design strength of weld = 220N/mm2

weld throat a = 0.7 s = 0.7 x 12mm = 8.4mm

(i) tw = 6.3mm 8.4mm

pi x 273 x 8.4 x 220 pi D a pw = = 1585kN 103

(ii) Ft = 750kN < 1585kN

CHECK 6: Member Capacity

Basic requirement:

Ft A py

A = 52.8 cm2

py = 275 N/mm2

52.8 x 102 x 275A py = = 1452kN 103

Ft = 750kN < 1452kN

e = 53.5 1 t = 6.3 w

D = 273mm

continuous weld

750kN750kN

pw fromBS 5950 -1 Table 37

O.K

O.K

py fromBS 5950 -1

Table 9

O.K

Yellow pagesTable H.70


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to the

term

s and

cond

itions

of th

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chs splice

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