Tension members - The Behaviour and Design of Steel Structures to EC3 - page 47

2.7.3 Example 3 - checking a bolted universal beam
section member

Problem. A610 × 229UB125tensionmemberofS355steelisconnectedthrough
both flanges by 20 mm bolts (in 22 mm diameter bolt holes) in four lines, two
in each flange as shown in Figure 2.11b. Check the member for a design tension
force of N t , Ed = 4000 kN.
Solution.

t f = 19.6 mm, f y = 345 N / mm 2 , f u = 490 N / mm 2 EN 10025-2

A = 15900 mm 2
N pl , Rd = Af y /γ M 0 = 15900 × 345 / 1.0 = 5486 kN 6.2.3(2)a
A net = 15900 − ( 4 × 22 × 19.6 ) = 14175 mm 2 6.2.2.2(3)
N u , Rd = 0.9 A net f u /γ M 2 = 0.9 × 14175 × 490 / 1.1 = 5683 kN 6.2.3(2)b
N t , Rd = 5486 kN ( the lesser of N pl , Rd and N u , Rd )> 4000 kN = N t , Ed
6.2.3(2)

and so the member is satisfactory.

2.7.4 Example 4 - checking an eccentrically connected
single (unequal) angle
Problem. A tension member consists of a 150 × 75 × 10 single unequal angle
whose ends are connected to gusset plates through the larger leg by a single
row of four 22 mm bolts in 24 mm holes at 60 mm centres. Use the method of
Section2.3tocheckthememberforadesigntensionforceof N t , Ed = 340kN,ifthe
angle is of S355 steel and has a gross area of 21.7 cm 2 .
Solution.

t = 10 mm, f y = 355 N / mm 2 , f u = 490 N / mm 2 EN 10025-2

Gross area of cross-section, A = 2170 mm 2

N pl , Rd = Af y /γ M 0 = 2170 × 355 / 1.0 = 770.4 kN 6.2.3(2)a

A net = 2170 − ( 24 × 10 ) = 1930 mm 2

6.2.2.2(3)
β 3 = 0.5 ( since the pitch p 1 = 60 mm = 2.5 d 0 ) EC3-1-8 3.10.3(2)
N u , Rd = β 3 A net f u /γ M 2 = 0.5 × 1930 × 490 / 1.1 = 429.9 kN 6.2.3(2)b
N t , Rd = 429.9 kN ( the lesser of N pl , Rd and N u , Rd )> 340 kN = N t , Ed

6.2.3(2)

and so the member is satisfactory.