In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 217

5.12.13 Example 13 - fully plastic moment of a tee-section

Problem . The welded tee-section shown in Figure 5.40b is fabricated from two
S275 steel plates. Compare the first yield and fully plastic moments.
First yield moment . The minimum elastic section modulus was determined in
Section 5.12.2 as

W el , y = 16.93 cm 3 .
For t max = 10 mm, f y = 275 N / mm 2 .

EN10025-2

M el , y = f y W el , y = 275 × 16.93 × 10 3 Nmm = 4.656 kNm.

Full plastic moment . For the plastic neutral axis to divide the cross-section into
two equal areas, then using the thin-walled assumption leads to

( 80 × 5 ) + ( z p × 10 ) = ( 82.5 − z p ) × 10

so that z p = ( 82.5 × 10 − 80 × 5 )/( 2 × 10 ) = 21.3 mm.
Using equation 5.67,

M pl , y = 275 ×[ ( 80 × 5 ) × 21.3 + ( 21.3 × 10 ) × 21.3 / 2
+ ( 82.5 − 21.3 ) 2 × 10 / 2 ] Nmm
= 8.117 kNm.

Alternatively, the section properties W el , y and W pl , y can be calculated using
Figure 5.6, and used to calculate M el , y and M pl , y .

5.12.14 Example 14 - plastic collapse of a non-uniform beam

Problem . The two-span continuous beam shown in Figure 5.43a is a 610 × 229
UB 125 of S275 steel, with 2 flange plates 300 mm × 20 mm of S275 steel
extending over a central length of 12 m. Determine the value of the applied loads
Q at plastic collapse.
Solution . The full plastic moments of the beam (unplated and plated) were
determined in Section 5.12.12 as M pl , u = 974 kNm and M pl , p = 1979 kNm.
The bending moment diagram is shown in Figure 5.43b. Two plastic collapse
mechanismsarepossible,bothwithaplastichinge( − M pl , p ) atthecentralsupport.
For the first mechanism (Figure 5.43c), plastic hinges ( M pl , p ) occur in the plated
beam at the load points, but for the second mechanism (Figure 5.43d), plastic
hinges ( M pl , u ) occur in the unplated beam at the changes of cross-section.