Frames - The Behaviour and Design of Steel Structures to EC3 - page 380

an approximate method of allowing for second-order effects by estimating the
amplified design moments is illustrated below.)

Solution . For the vertical member 123, the first-order moments are M 1 = 0,
M 2 = 23.5 kNm, M 3 =− 53.0 kNm, and for these, the method of [28] leads to
β m = 0.585, and so using equation 8.21,

c m = 0.6 − 0.4 × 0.585 = 0.366

Using N cr ,13 = 836.5 kN from Section 8.5.2, and using equations 8.19 and 8.20,

0.366 × 53.0
1 − 71.6 / 836.5 = 21.2 kNm < 53.0 kNm = M 3

M Ed ,3 =

andso M Ed ,3 = 53.0kNm.Thisisclosetothesecond-ordermomentof53.6kNm
determined using the computer program of [18].
Forthehorizontalmember345,thefirst-ordermomentsare M 3 =− 53.0kNm,
M 4 = 136.6 kNm, M 5 =− 153.8 kNm, and for these, the method of [28] leads to
β m = 0.292, and so

c m = 0.6 − 0.4 × 0.292 = 0.483

Using N cr 35 = 2809.6kNfromSection8.5.2,andusingequations8.19,and8.20,

0.483 × 153.8
1 − 25.3 / 2809.6 = 75.0 kNm < 153.8 kNm = M 5

M Ed ,5 =

andso M Ed ,5 = 153.8kNm.Thisisclosetothesecond-ordermomentof154.6kNm
determined using the computer program of [18].

8.5.7 Example 7 - moment amplification in a portal frame

Problem . The maximum first-order elastic moment in the portal frame of
Figure 8.14 whose buckling load factor was determined in Section 8.5.5 is
151.8 kNm [26]. Determine the amplified design moment.

Solution .UsingtheelasticframebucklingloadfactordeterminedinSection8.5.5
of α cr , s = 13.0 and using equations 8.19 and 8.22,

151.8
1 − 1 / 13.0 = 164.5 kNm

M Ed =

whichisabout6%higherthanthevalueof155.4kNmobtainedusingthecomputer
program of [18].

8.5.8 Example 8 - moment amplification
in an unbraced frame
Problem .Determinetheamplifieddesignmomentsfortheunbracedframeshown
in Figure 8.4.