In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 222

5.12.17 Example 17 - checking a plastically analysed beam

Problem. Check the adequacy of the non-uniform beam shown in Figure 5.43
and analysed plastically in Section 5.12.14. The beam and its flange plates are
of S275 steel, and the concentrated loads have nominal dead load components of
250kN(whichincludeallowancesforselfweight),andimposedloadcomponents
of 400 kN.
Classifying the sections . The 610 × 229 UB 125 was found to be Class 1 in
Section 5.12.15.
For the 300 × 20 plate, t p = 20 mm, f y = 265 N / mm 2

EN10025-2

ε = √ ( 235 / 265 ) = 0.942 T5.2

For the 300 × 20 plate as two outstands

c p /( t p ε) = ( 300 / 2 )/( 19.6 × 0.942 ) T5.2

= 8.13 < 9 and the outstands are Class 1. T5.2

For the internal element of the 300 × 20 plate,

c p /( t p ε) = 229 /( 20 × 0.942 ) T5.2

= 12.2 < 33 and the internal element is Class 1. T5.2

Checking for plastic collapse . The design loads are Q Ed = ( 1.35 × 250 ) + ( 1.5 ×
400 ) = 937.5kN.
The plastic collapse load based on the nominal full plastic moments of the
beam was calculated in Section 5.12.14 as

Q = 954 kN > 937.5 kN = Q Ed ,
and so the resistance appears to be adequate.
Checking for lateral bracing . For the unplated segment,

L stable = ( 60 − 40 × 0 ) × 0.924 × 4.97 × 10 = 2808 mm. 6.3.5.3

Thusabraceshouldbeprovidedatthechangeofsection(aplastichingelocation),
and a second brace between this and the support.
For the plated segment,

( 3932 × 10 4 ) + ( 2 × 300 3 × 20 / 12 )
( 159 × 10 2 ) + ( 2 × 300 × 20 )

i z = = 68.1 mm

and so,

L stable ={ 60 − 40 × 974 /( − 1979 ) }× 0.924 × 68.1 = 5109 mm. 6.3.5.3

Thus a single brace should be provided in the plated segment. A satisfactory
location is at the load point.