The steel plate girder shown in Fig. 6.2-60 is on the upper limit for a Class 3 cross-section. It has the following properties in the absence of shear:
. Plastic section modulus of girder, Wpl;y¼ 4:436 107mm3
. Elastic section modulus of girder, Wel;min ¼ 3:750 107mm3(based on the mid-plane
of the flange as allowed by 3-1-1/clause 6.2.1(9).
2060 30 30 20 400 400
Fig. 6.2-60. Plate girder for Worked Example 6.2-13
All plates are Grade S355 to EN 10025 and the girder is restrained against lateral torsional buckling and stable against shear buckling due to the presence of closely spaced transverse stiffeners. The thickness-dependent yield stresses are taken from 3-1- 1/Table 3.1 which gives a constant yield stress of 355 MPa throughout. (The UK National Annex to EN 1993-2 requires the values in EN 10025 to be used.) The maximum bending moment that the section can withstand in conjunction with a shear force of 7871 kN is calculated.
Web area Aw¼ hwtw¼ ð2060 60Þ 20 ¼ 40 000 mm2
Plastic shear resistance: Vpl;Rd¼ Awð fy= ffiffiffi 3 p Þ M0 ¼ 1:2 40 000ð355=pffiffiffi3Þ 1:00 ¼ 9838 kN
with taken as 1.2 as recommended in 3-1-5/clause 5.1(2). VEdis greater than 0.5Vpl;Rd
so shear will reduce moment resistance to My;V;Rd. From expression 3-1-1/(6.29): ¼ 2VEd Vpl;Rd 1 2 ¼ 2 7871 9838 1 2 ¼ 0:360
As the beam is symmetric and the yield strength is the same everywhere, expression 3-1- 1/(6.30) can be used.
6.2.9.2. Sections susceptible to shear buckling
If the section’s shear resistance is limited by shear buckling as discussed in section 6.2.6 of this guide, 3-1-1/clause 6.2.8(2) effectively requires section 7 of EN 1993-1-5 to be used to perform the interaction between shear and bending.
6.2.9.2.1. Class 1 and 2 cross-sections
The approach is similar to that for no shear buckling. 3-1-5/clause 7.1(1) allows the designer to neglect the interaction between shear and bending moment when the design shear force is less than 50% of the shear buckling resistance based on the web contribution alone. Where the design shear force exceeds this value, the following interaction has to be satisfied:
1þ 1 Mf;Rd Mpl;Rd ð23 1Þ2 1:0 3-1-5/(7.1)
where 3is the ratio VEd=Vbw;Rdand 1is the usage factor for bending, MEd=Mpl;Rd, based on
the plastic moment resistance of the section. Mf;Rd is the design plastic bending resistance
based on a section comprising the flanges only. For unequal flanges, this may, for simplicity, be taken as the smaller plastic resistance of the two flanges multiplied by the distance between the centroids of the flanges according to 3-1-5/clause 7.1(3). The interaction produced is illustrated in Fig. 6.2-61. The full web shear resistance contribution Vbw;Rdis obtained at a
moment of Mf;Rd. For smaller moments, the coexisting shear can increase further due to
the additional flange shear contribution, Vbf;Rd, from 3-1-5/clause 5.4.
The expressionð23 1Þ2can be rewritten as:
2VEd Vbw;Rd 1 2 3-1-1/clause 6.2.8(2) 3-1-5/clause 7.1(1) 3-1-5/clause 7.1(3) My;V;Rd¼ Wpl;y A2w 4tw fy M0 ¼ 4:436 1070:36 ð2000 20Þ 2 4 20 355 1:00 ¼ 13 192 kNm but not greater than:
Mc;Rd¼
Wel;minfy
M0 ¼
3:750 107 355
1:0 ¼ 13 317 kNm > My;V;Rd
just. Therefore, the moment resistance of the section reduces from 13 317 kNm to 13 192 kNm with a coexistent shear force of 7871 kN.
Interaction to 3-1-5/clause 7.1 Shear resistance Vbw,Rd to 3-1-5/clause 5 VEd Vbw,Rd Mpl,Rd Mf,Rd MEd Vbw,Rd 2
which is the same form as the web strength reduction factor: ¼ 2VEd Vpl;Rd 1 2
which is used when there is no shear buckling. For no shear buckling and symmetrical sections, expression 3-1-5/(7.1) would therefore give the same result as the method in section 6.2.9.1.1 above. For sections with no shear buckling and unequal flanges, expression 3-1-5/(7.1) would give a slightly more conservative result than the method in section 6.2.9.1.1 above. It is worth noting that expression 3-1-5/(7.1) is not used for Class 1 and 2 cross-sections in EN 1994 when there is shear buckling. Instead, the web strength is reduced by the factorð1 Þ where:
¼ 2VEd Vb;Rd 1 2
and the plastic moment resistance is recalculated. Unfortunately, the two Eurocode parts have not been reconciled but interchanging methods will generally have little practical consequence. 3-1-5/clause 7.1(4)requires that where axial force is present such that the whole web is in compression, Mf;Rd should be taken as zero in accordance with 3-1-5/clause 7.1(5). It is unclear what to do if there is no external axial force but the whole web is still in compression, as could occur with an asymmetric beam. A safe interpretation, given the relatively small amount of testing on asymmetric sections, would be to take Mf;Rd as zero in this case also.
This is likely to be conservative at high shear, given the weak interaction between bending and shear found in the tests on composite beams discussed in section 6.2.9.1.2 above.
3-1-5/clause 7.1(2)does not require the interaction in 3-1-5/clause 7.1(1) to be verified at sections nearer than hw=2 to a support, where it is assumed that there is a bearing stiffener
present. This is because the effect of buckling is small adjacent to a stiffener. However, the cross-section resistance should still be verified at the support. It is therefore recommended here that 3-1-5/clause 7.1(1) should be applied at the support, but using the plastic shear resistance in place of the shear buckling resistance.
6.2.9.2.2. Class 3 cross-sections
The approach is identical to that above for Class 1 and 2 sections, except that the resulting bending resistance must additionally not exceed the elastic bending resistance. This effectively truncates the interaction diagram in Fig. 6.2-61 in the same way as in Fig. 6.2- 58. The plastic bending resistance is again used in the interaction because of the weakness of interaction between bending and shear found in the studies identified in section 6.2.9.1.2 above. This ensures that shears well in excess of 50% of the web contribution to shear resistance can be accommodated before any reduction is made to the elastic bending resistance. In earlier drafts of EN 1993-1-5, 1 in expression 3-1-5/(7.1) was taken as MEd=Mel;Rd, based on the elastic bending resistance. This had the disadvantage that the bending resistance predicted was less than that of the flanges alone when the shear force was equal to Vbw;Rd.
For composite beams where the cross-section is built up in stages, the same interaction can be applied and guidance on the relevant value of MEdto use is given in the Designers’ Guide
to EN 1994-2.7 A separate check must be made of the accumulated elastic stresses, via 1
from 3-1-5/clause 4.6. In general, it will always be conservative to base 1 on the ratio of
accumulated stress to the allowable stress i.e. 1.
The comments made above for Class 1 and 2 sections regarding the use of 3-1-5/clause 7.1(4) for asymmetric sections and on 3-1-5/clause 7.1(2) for sections close to supports also apply to Class 3 sections.
6.2.9.2.3. Class 4 cross-sections, including beams with longitudinal stiffeners
Two methods are possible for Class 4 cross-sections. If the required geometric constraints are met as discussed in section 6.2.2.5.1 of this guide, it will usually be most economic to use the
3-1-5/clause 7.1(4)
3-1-5/clause 7.1(2)
same interaction method as above for Class 1, 2 and 3 sections. Expression 3-1-5/(7.1) again applies but the calculation of Mf;Rd and Mpl;Rd must consider effective widths for flanges,
allowing for plate buckling. Mpl;Rd is, however, calculated using the gross web, regardless
of any reduction that might be required for local buckling under direct stress. The reason for allowing plastic properties to be used in the interaction is again due to the weakness of interaction found in the tests on beams with Class 4 webs identified in section 6.2.9.1.2 above. It is still necessary to verify the girder under direct stresses alone to 3-1-5/clause 4.6, using elastic design and appropriate effective sections for flanges and webs. This again truncates the interaction.
While the interaction of expression 3-1-5/(7.1) applies to beams with longitudinally stiffened webs, the authors are not aware of similar test justification to support the use of plastic properties in the interaction. Such webs have less post-buckling strength when overall web buckling is critical, but the approach once again leads to an interaction with shear only at very high percentages of the web shear resistance. A safer option is to replace 1 by 1 in the interaction until such time as there have been further studies to
confirm this to be unnecessary. In these cases, if the section is built up in stages, 1 is the
usage factor based on accumulated stress.
The comments made above for Class 1 and 2 cross-sections regarding the use of 3-1-5/ clause 7.1(4) for asymmetric sections and on 3-1-5/clause 7.1(2) for sections close to supports also apply to Class 4 cross-sections. In the latter case, some interpretation is required for longitudinally stiffened webs. It is suggested here that the distance hw=2 be
replaced by bmax=2 (where bmax is the height of the largest sub-panel) when checking buckling of sub-panels.
Expression 3-1-5/(7.1) should also be used to verify flanges in box girders. However, in this case, Mf;Rdis taken equal to zero according to 3-1-5/clause 7.1(5), 1is replaced by 1and 3 is determined as the greater value obtained for overall flange shear buckling (based on the average shear stress in the flange but not less than half the maximum flange shear stress) and for sub-panel buckling (based on the average shear stress in the most critical sub- panel, determined from the elastic shear flow distribution).
For a single-cell box girder with vertical shear only, the flange shear stress varies linearly from a maximum positive value shearat one web to a negative valueshearat the other web.
The average shear stress is therefore zero. The relevant shear stress to use for overall flange buckling is then governed by the requirement to be not less than half the maximum value, which occurs at a web junction, i.e. 0.5shear. It is not entirely clear if this sign change is to
be considered. If the sign is not considered, only the magnitude, the average shear stress is equal to half the maximum value (i.e. 0.5shear) and the two requirements are the same.
When torsional shear stress tor, which is uniform throughout the flange, is included, consideration of sign of the shear stress does make a difference. If it is considered, the average stress is tor and half the maximum is 0:5shearþ 0:5tor. This is probably the
intended interpretation. If it is not considered, the average stress is 0:5shearþ tor and half
the maximum is 0:5shearþ 0:5tor. This is more conservative, whereupon the shear stress is
50% of the shear stress at the web–flange junction due to the beam vertical shear force plus 100% of the torsional shear stress, which was the requirement in BS 5400: Part 3.4 This latter interpretation has been conservatively used in Worked Example 6.2-15 but it was probably not the drafters’ intended interpretation. If shear stress from distortional warping or transverse loading on the box is present, this must also be included.
The interaction for a box girder flange becomes:
1þ ð23 1Þ2 1:0 (D6.2-52)
This means that there is no interaction between direct stress and shear in the flange when 3 0:5 but that no direct stress can be carried when 3¼ 1:0 as shown in Fig. 6.2-62.
Worked Example 6.2-15 illustrates the check of a box flange. It is noted that closed stiffeners are not explicitly covered in 3-1-5/Annex A.3 when determining shear buckling resistance. If closed stiffeners are provided on the flange, it is suggested here that the effective stiffener second moment of area is derived for a section which comprises:
. the stiffener itself, with reduced area derived in accordance with 3-1-5/clause 4.4 if
necessary;
. an attached width of flange plate at each connection to the stiffener of 15"t each side of
the connecting stiffener leg (or half the distance to an adjacent stiffener leg if smaller) plus the thickness of the stiffener leg as provided in 3-1-5/Fig. 5.3.
The formulae in 3-1-5/Annex A.3 are very conservative for closed stiffeners as they do not allow for their significant torsional stiffness.
Where the geometric constraints discussed in section 6.2.2.5.1 are not met, the method of 3-1-5/clause 10 as discussed in section 6.2.2.6 of this guide may be used. This will, however, be much more conservative as there is no allowance made for plastic redistribution, and shear stresses reduce the allowable resistance moment, whatever their magnitude.
1.0
1.0
η3
0.5
η1
Fig. 6.2-62. Interaction to 3-1-5/clause 7.1 for flanges