Technical
28 NSC
July/Aug 19
elastic modulus, Z = 111 cm3
plastic modulus, S = 199 cm3
With some careful spreadsheet work:
v = 1.05
w = 0.00449 (includes the warping constant)
βw = 111 ⁄ 199 = 0.558
λ = 4000/32.3 = 123.8
Then LT = 0.648 × 1.05 × 123.8 × 0.558 = 62.9
The bending strength can then be calculated from B.2.1, with the
result that
pb = 105 N/mm2
The buckling resistance moment Mb = 105 × 111 × 10-3 = 11.7 kNm
Method 2 – BS 5950 effective section method
Given that the section is slender, an effective section may be
calculated. Clause 3.6.2.2 prescribes that the effective width of a
class 4 slender outstand should be taken as equal to the class 3
limiting value (18ε, as above).
The overall depth of the effective section is therefore
18 × 0.88 × 8.1 = 128.3 mm. The dimensions of the effective section
are shown in Figure 2.
Calculations are required to determine the position of the neutral
axis (accounting for the root radii if doing a ‘proper’ job!), and
calculating the effective elastic modulus of the section. The
effective elastic modulus is calculated as 36.3 cm3.
w = 3 6 . 3 = 0.18
199
Then LT = 0.648 × 1.05 × 123.8 × 0.18 = 35.7
Following the same process from B.2.1, the bending strength,
pb = 339 N/mm2
The buckling resistance moment Mb = 339 × 36.3 × 10-3 = 12.3 kNm
Method 3 – BS EN reduced stress method
The ratio for local buckling is defined differently in the Eurocode,
which species c/t as the dimensions of the outstand, not overall
depth.
(227.2 – 13.3 – 10.2)
c/t = = 25.2
8.1
The limiting value depends on the stress ratio between the stress
at the tip of the web, and at the root radius (refer to Table 5.2 in
BS EN 1993-1-1). To evaluate the limit, BS EN 1993-1-5 must be
consulted to calculate the buckling factor, kσ .
If the neutral axis is at 58.4 mm from the face of the flange (from
section property tables), the stress ratio may be calculated from
the dimensions shown in Figure 3.
= - 3 4 . 9 = -0.207
168.8
From Table 4.2 of BS EN 1993-1-5, then
kσ = 0.57 – 0.21ψ + 0.07ψ2
kσ = 0.57 – 0.21 × (-0.207) + 0.07 × (-0.207)2 = 0.616
Back in BS EN 1993-1-1 Table 5.2,
the limit is 21 k = 21 × 0.81 × 0.616 = 13.3
25.2 > 13.3, so the section is class 4 (not surprisingly, given the BS
5950 classification)
To ensure the section remains class 3, the reduced design strength
is given by 25.2
235
( )2 = 100.5 N/mm2 21 × 0.616
26
Figure 2: BS 5950 effective section
Figure 3: Elastic stresses in the web of the gross section