J. Construct. Steel Research 7 (1987) 75-77
Design Data Sheet
D. A. Nethercot Department of Civil and Structural Engineering, The University of Shef...
D. A. Nethercot Department of Civil and Structural Engineering, The University of Sheffield, UK
A method of allowing for restraint from adjacent segments when assessing lateral-torsional buckling strength according to BS 5950 Part 1.
NOTATION
EIM(EIR) minor axis flexural rigidity of the critical (adjacent) segment KM(KR) kl, k2
LM(LR) Pm(Px0
stiffness parameter for critical (adjacent) segment restraint parameters unbraced length of critical (adjacent) segment bending strength of critical (adjacent) segment neglecting interaction, i.e. determined on basis of LM(La)but allowing for pattern of moments using m-factor or n-factor as appropriate (see Table 13 of BS 5950 Part 1).
NOTES (1) The usual method of determining bending strength for laterally continuous beams of the type illustrated in Fig. 1 neglects the beneficial effect of interaction, basing design on the resistance of the weakest segment considered in isolation. (2) Where segment lengths vary and/or the pattern of moments within segments is quite different significant restraint may be available to the most critical segment from adjacent (more stable) segments. The 75 J. Construct. Steel Research (7) (1987)---O Elsevier Applied Science Publishers Ltd,
England, 1987. Printed in Great Britain
76
D. A. Nethercot
magnitude of this restraint may be estimated by an adaptation of Fig. 23 of BS 5950 Part 1 in which the restraint coefficients are taken as: KM kl -
(1)
K M + KRB
EIM/LMfor the critical segment, KR = 0"5(EIR/LR) (1 - Pm/PRs) for an adjacent segment. T h e effective length L E thus obtained may then be used directly with cl.4.3 of B S 5950 Part 1 to calculate hLr and hence to obtain an enhanced bending strength p b. This provides an improved estimate of the bending resistance of the whole beam allowing for the fact that instability must involve the interaction of all segments. in which
(3)
KM
K M + KRA ' k2 -
KM =
Example Assuming the use of a 457 x 152 UB52 in Grade 43 steel determine the value of P for the beam illustrated in Fig. 1 that just causes failure. All loads are factored values. 4 0 PkN
6 0 PkN ..o
J
4m
J
6m
~
8m
~
227 PkN. m
Fig. 1.
F r o m tables Sx ry [y u x
= = = = =
1090 c m 3 3.11 cm 645 cm' 0.859 43.9
/
2 ~ I~N. m
Design data sheet
I-2 2-3 3--4
77
L (m)
h
kLr
pb (N/mm 2)
Mb (kN m)
fl
m
~Ib (kN m)
4 6 8
129 193 257
101 139 172
124 75 53
135 82 58
0 0.82 -0.88
0.57 0.91 0-43
237 90 135
/14b v a l u e s a r e M J m to allow for beneficial e f f e c t o f m o m e n t g r a d i e n t . C l e a r l y s e g m e n t 2 - 3 is critical; t h e r e f o r e estimate restraining effects o f s e g m e n t s 1-2 a n d 3-4. KM = E x 645/6 = 441 kN/m
.'. for segment 2-3 hLT -----0"859 X 0"88 X (0"8_3X 193) = 121 hence Pb = 95 N/mm 2, Mb = 104 kN m, Mb = 114 kN m .'. ignoring interaction P = 90/227 = 0.40 including interaction P = 114/227 = 0.50 and gain in buckling resistance is 25%
REFERENCES 1. BS 5950: Structural use of steel work in building, Part 1, Code of Practice for design in simple and continuous construction: hot rolled sections, British Standards Institution, 1985. 2. Nethercot, D. A. and Trahair, N. S., Lateral buckling approximation for elastic beams, The Structural Engineer, 54 (1976) 197-200.