Frames - The Behaviour and Design of Steel Structures to EC3 - page 382

Checking for reductions in M p using equation 7.17,
M pr 13 = 1.18 × 85.0 × 1 − 89.5 /( 275 × 47.1 × 10 2 / 10 3 )
= 93.4 kNm > 85.0 kNm = M p 13
and so there is no reduction in M p for member 13.
M pr 35 = 1.18 × 171.3 × 1 − 32.1 /( 275 × 51.3 × 10 2 / 10 3 )
= 197.5 kNm > 171.3 kNm = M p 35
and so there is no reduction in M p for member 35.
Therefore α ph = 1.248.
Foraplasticcollapsemechanismintheverticalmemberwithafrictionlesshinge
at 1 and plastic hinges at 2 and 3,
α pv = 85.0 × 2 × δ θ v + 85.0 × δ θ v
20 × 5.0 × δ θ v
= 2.55 > 1.248 = α ph
and so
α p = 1.248.
8.5.10 Example 10 - plastic analysis of an unbraced frame
Problem .Determinetheplasticcollapseloadfactorfortheunbracedframeshown
in Figure 8.4, if the members are all of S275 steel.
Solution . For the beam plastic collapse mechanism shown in Figure 8.4e,
δ W = 40 α p × 6 δ θ = 240 α p δ θ
for virtual rotations δ θ of the half beams 35 and 58, and
δ U = ( 84.2 × δ θ) + ( 84.2 × 2 δ θ) + ( 84.2 × δ θ) = 336.8 δ θ
so that
α p = 36.8 / 240 = 1.403.
Theframeisonlypartiallydeterminatewhenthismechanismforms,andsothe
member axial forces cannot be determined by statics alone. The axial force N 358
determined by a computer first-order plastic analysis (18) is N 358 = 33.8 kN and
the axial force ratio is ( N / N y ) 358 = 33.8 /( 275 × 32.0 × 10 2 / 10 3 ) = 0.0384,
which is less than 0.15, the approximate value at which N / N y begins to reduce
M p , and so α p is unchanged at α p = 1.403.