In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 218

9.6

Q Q 4.8

M pp

M M M

pp

pu pu

M pp M pp

3.6 1.2 4.8 4.8 3 .6

1.2

(a) C o n t in u ou s b e a m
(le n g t h s

(c) First me chanism

i n

m )

9.6

3.6

M pp

( -

(+ ) ( + )

Q 9 . 6 / 4 ×

M pu M pu

(b) Bending moment (d) Second mechanism

Figure 5.43 Worked example 14.

For the first mechanism shown in Figure 5.43c, inspection of the bending
moment diagram shows that

M pl , p + M pl , p / 2 = Q 1 × 9.6 / 4

so that

Q 1 = ( 3 × 1979 / 2 ) × 4 / 9.6 = 1237 kN.

For the second mechanism shown in Figure 5.43d, inspection of the bending
moment diagram shows that

M pl , u + 3.6 M pl , p / 9.6 = ( 3.6 / 4.8 ) × Q 2 × 9.6 / 4

so that

Q 2 =[ 974 + ( 3.6 / 9.6 ) × 1979 ]× ( 4.8 / 3.6 ) × 4 / 9.6 = 954 kN,

which is less than Q 1 = 1237 kN.
Thus plastic collapse occurs at Q = 954 kN by the second mechanism.

5.12.15 Example 15 - checking a simply supported beam

Problem . The simply supported 610 × 229 UB 125 of S275 steel shown in
Figure 5.44a has a span of 6.0 m and is laterally braced at 1.5 m intervals. Check
theadequacyofthebeamforanominaluniformlydistributeddeadloadof60kNm
together with a nominal uniformly distributed imposed load of 70 kNm.