CHS Splice

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BCSA

CALCULATION

SHEET

Calcs by

Client

Checked by

Date

SCI/BCSA Connections Group

RS

AW

_{May 2002}

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

25 25 480 diameter 273 x 6.3 CHS Grade S275 273 x 6.3 CHS Grade S275 50 e ₁_e₂ 53.5

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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

CONNECTION DESIGN USING CAPACITY TABLES FROM YELLOW PAGES

Yellow pages

273

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 thickest

section in the range.

See Table H.34

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CHECK 1: Recommended detailing practice

End Plates t_p = 25mm Bolts d = 24mm dia Holes D_h = 26mm Spacing (≥ 2.5d = 60) p = 380 x

π

/8 (< 10d =240) = 149mm Edge distance ( ≥ 1.4D_h = 36) e₂ = 50mm

CHECK 2: Complete end plate yielding

Basic requirement:

t_p2 _p y plate

π

f3 F_t

≤

––––––––––––– 2 1 f₃ = ––

(

k₃ + (k₃2_{– 4 k} 1) 0.5

)

2k₁ k₁ =

l

_n

(

r₂ /r₃

)

[l

_n= natural log_e

]

D 273 r₂ = –– + e₁ = –––– + 53.5 = 190 2 2 D – t_w 273 – 6.3 r₃ = ––––– = ––––––––– = 133.4 2 2 190 k₁ =

l

_n ––––– =

l

_n1.42 = 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 xπx 6.19

∴

Plate capacity = ––––––––––––––––– = 1610kN 2 x 103 F_t= 750kN

≤

1610 kN

∴

O.K. 750kN 750kN t = 6.3 w t = 25 p e = 53.5 1 D = 273

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CHECK 3: Bolt failure with end plate yielding

Basic requirement:

NP_t F_t

≤

––––––––––––––––––––––––––––  1 1       1 – –––– + –––––––––  [l_n= natural log_e

]

 f₃ f₃

l

_n(r₁/r₂)

P_t = Exact bolt capacity = 198kN

f₃ = 6.19 CHECK 2

e_eff = minimum of e₂ and 1.25e₁ = 50 mm

D 273 r₁ = –– + e₁+ e_eff= ––––– + 53.5 + 50 = 240mm 2 2 D 273 r₂ = ––– + e₁ = ––––– + 53.5 = 190mm 2 2 240

l

_n(r₁/r₂) =

l

_n––––– =

l

_n 1.263 = 0.234 190 8 x 198

∴

Bolt group capacity = ––––––––––––––––––––––––––––– = 1036kN

 1 1       1 – –––– + –––––––––––   6.19 6.19 x 0.234  F_t= 750kN < 1036 kN

∴

O.K.

CHECK 4: Bolt failure

Basic requirement for bolt tensile capacity:

F_t

≤

N P_t N P_t = 8 x 198 = 1584 kN F_t = 750kN

≤

1584kN

∴

O.K

750kN

e = 53.5 ₁ e = 50 ₂ t = 6.3 w p = 139

750kN

P_t from Yellow pages Table H.49 P_t from Yellow pages Table H.49

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CHECK 5: Welds

Basic requirement for weld tensile capacity:

(i)

t

_w

≤

a

(continuous full strenghth weld )

or

(ii)

F_t

≤

π D a p_w

p_w = design strength of weld = 220N/mm2

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

(i)

t

_w = 6.3mm

≤

8.4mm πx

273

x

8.4

x

220

π D a p_w

= –––––––––––––––––

=

1585kN

10

3

(ii)

F_t = 750kN <

1585kN

CHECK 6: Member Capacity

Basic requirement:

F_t

≤

A p_y A = 52.8 cm2 p_y = 275 N/mm2

52.8

x

10

2 x

275

A p_y

= ––––––––––––––––

=

1452kN

10

3 F_t = 750kN <

1452kN

e = 53.5 1 t = 6.3 w D = 273mm continuous weld 750kN 750kN p_w from BS 5950 -1 Table 37