Frames - The Behaviour and Design of Steel Structures to EC3 - page 381

Solution . For the upper-storey columns, using the value of α cr , su = 22.1
determined in Section 8.5.3(b) and using equation 8.23,
δ su = 1 /( 1 − 1 / 22.1 ) = 1.047, and so
M Ed ,8 = 1.047 × 62.5 = 65.5 kNm,
which is close to the second-order moment of 64.4 kNm determined using the
computer program of [18].
For the lower-storey columns, using the value of α cr , sl = 6.31 determined in
Section 8.5.3(b) and using equation 8.23,
δ sl = 1 /( 1 − 1 / 6.31 ) = 1.188, and so
M Ed ,76 = 1.188 × 119.3 = 141.8 kNm,
which is 9% higher than the value of 130.5 kNm determined using the computer
program of [18].
Forthelowerbeam,thesecond-orderendmoment M Ed ,74 maybeapproximated
by adding the second-order end moments M Ed ,76 and M Ed ,78 , so that
M Ed ,74 = 141.8 + 53.5 /( 1 − 1 / 22.1 ) = 197.8 kNm,
which is 7% higher than the value of 184.7 kNm determined using the computer
program of [18].
Slightlydifferentvaluesofthesecond-ordermomentsareobtainedifthevalues
of α cr , slu and α cr , sl determinedbythestoreydeflectionmethodinSection8.5.3(a)
are used.
8.5.9 Example 9 - plastic analysis of a braced frame
Problem .Determinetheplasticcollapseloadfactorforthebracedframeshownin
Figure 8.3, if the members are all of S275 steel.
Solution . For the horizontal member plastic collapse mechanism shown in
Figure 8.3e,
δ W = 80 α ph × ( δ θ h × 6.0 ) = 480 α ph δ θ h for a virtual rotation δ θ h of 34,
and
δ U = ( 85.0 × δ θ h ) + ( 171.3 × 2 δ θ h ) + ( 171.3 × δ θ h ) = 598.9 δ θ h
so that
α ph = 598.9 / 480 = 1.248.