In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 223

Checking for shear . Under the collapse loads Q = 954 kN, the end reactions are

R E ={ ( 954 × 4.8 ) − 1979 } / 9.6 = 270.6 kN

and the central reaction is

R C = 2 × 954 − 2 × 270.6 = 1366 kN

so that the maximum shear at collapse is

V = 1366 / 2 = 683 kN.

From Section 5.12.15, the design shear resistance is V c , Rd = 1171 kN > 683 kN
which is satisfactory.
Checking for bending and shear .At the central support, V = 683 kN and

0.5 V c , Rd = 0.5 × 1171 = 586 kN < 683 kN

and so there is a reduction in the full plastic moment M pp . This reduction is from
1979kNmto1925kNm(afterusingClause6.2.8(3)ofEC3).Theplasticcollapse
load is reduced to 942kN > 937.5kN = Q Ed , and so the resistance is adequate.
Checking for bearing . Web stiffeners are required for beams analysed plastically
within h / 2 of all plastic hinge points where the design shear exceeds 10% of the
web capacity, or 0.1 × 1171 = 117.1 kN (Clause 5.6(2b) of EC3). Thus stiffeners
are required at the hinge locations at the points of cross-section change and at the
interior support. Load-bearing stiffeners should also be provided at the points of
concentratedload,butarenotrequiredattheoutersupports(seeSection5.12.15).
An example of the design of load-bearing stiffeners is given in Section 4.9.9.

5.12.18 Example 18 - serviceability of a simply
supported beam

Problem . Check the imposed load deflection of the 610 × 220 UB 125 of
Figure 5.44a for a serviceability limit of L / 360.
Solution . The central deflection w c of a simply supported beam with uniformly
distributed load q can be calculated using

w c = 5 qL 4

5 × 70 × 6000 4
384 × 210000 × 98610 × 10 4 = 5.7 mm. 384 EI y =

(5.73)

(The same result can be obtained using Figure 5.3.)
L / 360 = 6000 / 360 = 16.7 mm > 5.7 mm = w c and so the beam is satisfactory.