Composite joints in frames for buildings
8.4. Resistance of components
Clause 8.4.2.1(1) This clause supplements clause 6.2 of EN 1993-1-8. The effective width of concrete flange in
tension is the same at a joint as for the adjacent beam (clause 8.4.2.1(1)). Longitudinal bars above the beam should pass either side of the column.
Clause 8.4.2.1(4) Clause 8.4.2.1(4) applies at an external column with a partial- or full-strength joint. The
tensile force in the bars must be transferred to the column; for example, by being looped round it. This applies also at internal columns where there is a change in the tension in the bars (Fig. 8.2, clause 8.4.2.1(3) and Example 8.2).
Clause 8.4.2.2(1) Clause 8.4.3(1) Clauses 8.4.2.2(1) and 8.4.3(1) permit the same 45° spread of force in a contact plate
as used in EN 1993-1-8 for an end plate. The force is assumed in EN 1993-1-8 to spread at tan–12.5 (68°) through the flange and root radius of the column. Where the compressive force relied on in design exceeds the resistance of the steel bottom flange, the length of the contact plate should allow for this (Fig. 8.4).
In EN 1994-1-1, the word ‘connection’ appears only in clause 8.4.3(1), clause A.2.3.2 and Table A.1. It means the set of components that connect a member to another member; for example, an end plate, its bolts and a column flange. Thus, a ‘connection’ is part of a ‘joint’.
Clause 8.4.4.1(2) The model used in clause 8.4.4.1(2) is illustrated in Fig. 8.5. This figure shows an elevation
of the concrete encasement of width h – 2tf (column depth less flange thicknesses) and depth z, the lever arm between the resultant horizontal forces from the beam. A shear force V is transferred through the encasement, which is of thickness bc– tw (column width less web thickness). The concrete strut ABDEFG has width 0.8(h – 2tf)cos θ, where tan θ = (h – 2tf)/z, so its area is
Ac= 0.8(h – 2tf)cos θ (bc– tw) (8.2)
45˚
68˚
Contact plate Column
Beam
Fig. 8.4. Detail of a contact plate between a beam bottom flange and a column
B
G
F E
D q
q h – 2tf
0.4(h – 2tf)
0.8(h – 2tf)cos q
V
V
A
z
NEd C
Fig. 8.5. Strut model for the shear resistance of the concrete encasement to a column web
teff, c
tf, c tp
tf, b
0.212 0
1.3 2.0 kwc, c
scom, c, Ed /fcd
(b) 45˚
68˚ End plate
Beam Column
r
(a)
Fig. 8.6. Model for resistance to compression of the concrete encasement to a column web
Its compressive strength is 0.85νfcd, giving the force C in Fig. 8.5. For horizontal equilibrium at B and F, C sin θ = V. These are equations (8.1) to (8.3) in clause 8.4.4.1.
Clause 8.4.4.1(3) The shear strength of concrete is increased by compression. This is allowed for by the
factor ν in clause 8.4.4.1(3), which ranges from 0.55 for zero axial compression to 1.1 for NEd≥ 0.55Npl, Rd.
Clause 8.4.4.2 The contribution of concrete encasement to the resistance of a column web to horizontal
compression is given in clause 8.4.4.2. For an end-plate joint to a column flange, the depth of encasement assumed to resist compression, teff, c, is shown in Fig. 8.6(a), with the 2.5:1 dispersion, referred to above, extending through the root radius r.
The horizontal compressive strength of the concrete is 0.85 kwc, cfcd, where kwc, cdepends on the vertical compressive stress in the column, σcom, c, Ed, as shown in Fig. 8.6(b).
Example 8.1: end-plate joints in a two-span beam in a braced frame
In development work that followed the publication of ENV 1994-1-1, a set of application rules for composite joints was prepared, more detailed than those now given in Section 8 and Annex A of EN 1994-1-1. These are published by ECCS39as a model annex J. They provide useful guidance in this example, and are referred to, for example, as ‘clause J.1.1 of ECCS TC11’.
Data
The subject of Examples 6.7 and 7.1 is a two-span beam ABC continuous over its central support (see Figs 6.23–6.28). It is now assumed that this beam is one of several similar beams in a multistorey braced frame (Fig. 8.7). Its joints with the external columns are nominal pins. The spans of the composite-slab floors are 2.5 m, as before. For simplicity, in the work on beams AB and BC, column EBF will be treated as fixed at nodes E and F. These beams are attached to the column at B by the end-plate connections shown in Fig. 8.8, which also gives dimensions of the column section. Its other properties are as follows: HEB 240 cross-section, Aa= 10 600 mm2, fy= 355 N/mm2, 10–6Iy= 112.6 mm4, 10–6Iz= 39.23 mm4.
The end plates are of mild steel, fy= 275 N/mm2, and relatively thin, 12 mm, to provide the plastic behaviour required. They are attached to the beam by 10 mm fillet welds to the flanges, and 8 mm welds to the web. They are each attached to the column by four Grade 8.8 M20 bolts with properties: fub= 800 N/mm2, fyb= 640 N/mm2, net area at root of thread As, b= 245 mm2per bolt.
The only other change from the data used in Examples 6.7 and 7.1 on geometry, materials and loadings concerns the reinforcement in the slab.
Longitudinal reinforcement at support B
These partial-strength joints need rotation capacity. Its value cannot be found at this stage, but it is known to increase with both the diameter of the reinforcing bars in the slab
C B
A
12 3
3
12
D
F E
Fig. 8.7. Model for a two-span beam ABC, with an internal column EBF
and the area of reinforcement provided. However, the amount of top reinforcement should be limited, so that the whole of the compressive force across the joint can be resisted by the beam bottom flange and the unstiffened column web.
Detailed guidance is given in Couchman and Way.104For a steel beam of depth 450 mm in S355 steel, the recommended minimum areas are 3000 mm2 for bars with 5%
elongation and 860 mm2 for bars with 10% elongation. The recommended maximum amount depends on the size of the column and the details of the bolts in tension, and is about 1200 mm2for this example. The recommended bar diameters are 16 and 20 mm.
For these reasons, the previous reinforcement (13 No. 12 mm bars, As= 1470 mm2) is replaced by six No. 16 mm hot-formed bars (minimum elongation 10%): As= 1206 mm2, fsk= 500 N/mm2.
Classification of the joints
It is assumed initially that flexural failure of a joint will occur in a ductile manner, by yielding of the reinforcement in tension and the end-plate or column flange in bending;
160
383
m2 = 35.4 + 2 240
10 14.6 12
450
25 200
60
25
60
90
= = A
9.4 A
270 230 130
800 240
17 10
21
100 mm 16 dia.
30
100
Fig. 8.8. Details of the beam-to-column end-plate connections
(a) Elevation (b) Section A–A
(c) Column section (d) Slab reinforcement
and that at bottom-flange level the compressive resistance of the column web will be sufficient. As the spans are equal, it is unlikely that shear of the column web will be critical.
The joint is expected to be ‘partial-strength’. This can be checked by comparing the tension resistance of the top two bolts, from Table 3.4 of EN 1993-1-8, with the force to yield the beam top flange:
FT, Rd, bolts= 2(k2fubAs/γM2) = 2 × 0.9 × 0.8 × 245/1.25 = 282 kN (D8.3)
FRd, flange= bftffyd= 190 × 14.6 × 0.355 = 985 kN
Thus, the resistance moment Mj, Rdfor the joint will be much less than Mpl, Rdfor the beam, and lateral buckling will be less critical than before.
There is no need to find the stiffness of the joint at this stage, because it is clearly either
‘rigid’ or ‘semi-rigid’. Either type may be treated as ‘semi-rigid’.
Approximate global analysis
Tables in Appendix B of Couchman and Way104enable a rough check to be made on this initial design, without much calculation. They give resistances Mj, Rd in terms of the cross-section of the steel beam, its yield strength, the thickness and grade of the end plate, the number and size of bolts in tension, and the area of reinforcement. Even though the beam used here is an IPE section, it can be deduced that Mj, Rdis about 400 kN m.
For both spans fully loaded, it was found in Example 6.7 that MEdat B was 536 kN m from loading (see Fig. 6.28) plus 120 kN m from shrinkage. The flexural stiffness of the joint is not yet known, but it will be between zero and ‘fully rigid’. If fully rigid, the joint will obviously be ‘plastic’ under ultimate loading, and there will then be no secondary shrinkage moment. At mid-span, for the total load of 35.7 kN/m (see Table 6.2), the sagging bending moment is then
357 ×122/8 – 400/2 = 443 kN m
If the joint acts as a pin, the mid-span moment is 443 + 200 = 643 kN m
It is recommended in Couchman and Way104that mid-span resistances should be taken as 0.85Mpl, Rd, to limit the rotation required at the joints. From Example 6.7, Mpl, Rdwith full shear connection is 1043 kN m, so the bending resistance of the beam is obviously sufficient.
Vertical shear
For MEd= 400 kN m at B, the vertical shear at B is Fv, Ed, B= 35.7 × 6 + 400/12 = 247 kN
The shear resistance of the four M20 bolts is now found, using Table 3.4 of EN 1993-1-8.
Two of the bolts may be at yield in tension. The shear applied to these bolts must satisfy
Fv, Ed/Fv, Rd+ Ft, Ed/(1.4Ft, Rd) £ 1.0 (D8.4)
The net shear area of each bolt is As, b= 245 mm2, so from Table 3.4 of EN 1993-1-8, Fv, Rd= 0.6fubAs, b/γM2= 0.6 × 800 × 0.245/1.25 = 94.1 kN
From equation (D8.3) with Ft, Ed= Ft, Rd, Fv, Ed£ (1 – 1/1.4)Fv, Rd= 27 kN For four bolts,
Fv, Rd= 2 × (94.1 + 27) = 242 kN (D8.5)
This shows that it may be necessary to add a second pair of bolts in the compression region of the joint.
Bending resistance of the joint, excluding reinforcement
Unpropped construction was used in Example 6.7. From Table 6.2, the design ultimate load for the steel beam is 7.8 kN/m. For the construction phase, this is increased to 9.15 kN/m, to allow for the higher density of fresh concrete and the construction imposed loading. For rigid joints at the internal support between two 12 m spans,
MEd, B= wL2/8 = 9.15 × 122/8 = 165 kN m
The plastic resistance of the joint during construction is required. An upper limit is easily obtained. The lever arm from the top bolts to the centre of the bottom flange is
zbolts= 450 – 60 – 7.3 = 383 mm (D8.6)
The resistance cannot exceed
FT, Rd, boltszbolts= 282 × 0.383 = 108 kN m (D8.7)
so the previous hogging bending moment of 165 kN m cannot be reached. It is assumed that the stiffness of the joint is sufficient for its plastic resistance, found later to be 83 kN m, to be reached under the factored construction loading.
Resistance of T-stubs and bolts in tension
The calculation of the bending resistance consists of finding the ‘weakest links’ in both tension and compression. In tension, the column flange and the end plate are each modelled as T-stubs, and prying action may occur. Some of the dimensions required are shown in Fig. 8.9. From Fig. 6.8 of EN 1993-1-8, the dimensions m overlap with 20% of the corner fillet or weld. Thus, in Fig. 8.9(a), for the end plate:
m = 45 – 4.7 – 0.8 × 8 = 33.9 mm (D.8.8)
It is evident from the geometry shown in Fig. 8.9 that the end plate is weaker than the column flange, so its resistance is now found.
tp = 12 tf = 17
0.8rc = 16.8 m = 23.2
e = 75
rc = 21
m = 33.9
55 8
m2 = 37.4
55 33.9
Fig. 8.9. Dimensions of T-stubs, and the yield line pattern
(a) Plan details of T-stubs
(b) Yield line pattern in end plate
Clause 6.2.4.1 of EN 1993-1-8 gives three possible failure modes:
(1) yielding of the plate
(2) a combination of (1) and (3) (3) failure of the bolts in tension.
Yield line theory is used for bending of the plate. The critical mechanism in this case will be either that shown in Fig. 8.9(b) or a circular fan, for which the perimeter is
leff, cp= 2πm
from Table 6.6 in clause 6.2.6.5 of EN 1993-1-8.
For the non-circular pattern, dimension m2in Fig. 8.8(a) is also relevant, and
leff, nc= αm £ 2πm
where α is given by Fig. 6.11 in EN 1993-1-8 or in Fig. 4.9 of Couchman and Way.104In this case, α = 6.8, so the circular pattern governs for mode (1), and
leff, 1= 2πm = 6.28 × 33.9 = 213 mm
The plastic resistance per unit length of plate is
mpl, Rd= 0.25tf2fy/γM0= 0.25 × 122× 0.275/1.0 = 9.90 kN m/m (D8.9)
From equation (D8.3), the tensile resistance of a pair of bolts is 282 kN.
Mode 1. For mode 1, yielding is confined to the plate. From Table 6.2 of EN 1993-1-8, the equation for this mode is
FT, 1, Rd= 4Mpl, 1, Rd/m with
Mpl, 1, Rd= 0.25leff, 1tf2fy/γM0= leff, 1mpl, Rd= 0.213 × 9.9 = 2.11 kN m From equation (D8.8) for m,
FT, 1, Rd= 4 × 2.11/0.0339 = 249 kN
Mode 2. This mode is more complex. The equation for the tension resistance FT, 2, Rdis now explained. The effective length of the perimeter of the mechanism is leff, 2, and the work done for a rotation θ at its perimeter is 2mpl, Rdleff, 2θ, from yield line theory. With each bolt failing in tension, the work equation is
Ft, Rd
FT, 2, Rd Q/2
m n
q
Fig. 8.10. Plan of a T-stub, showing failure mode 2
FT, 2, Rd(m + n)θ = n
S
Ft, Rdθ + 2mpl, Rdleff, 2θ (D8.10)where n is shown in Fig. 8.10.
In Table 6.2 of EN 1993-1-8 there is the further condition that n £ 1.25m = 1.25 × 33.9 = 42.4 mm
The effective length
leff, 2= αm = 6.8 × 33.9 = 231 mm
from Table 6.6 of EN 1993-1-8. For two bolts,
S
Ft, Rd= 282 kN, from equation (D8.3).Substituting in equation (D8.10):
FT, 2, Rd= (2mpl, Rdleff, 2+ n
S
Ft, Rd)/(m + n)= (2 × 9.9 × 231 + 42.4 × 282)/76.3 = 217 kN (D8.11) Mode 3. Failure of the bolts – mode 3 – has
FT, 3, Rd= 282 kN
from equation (D8.3), so mode 2 governs. From Fig. 8.10, the prying force is Q = 282 – 217 = 65 kN
Beam web in tension
The equivalent T-stub in Fig. 8.10 applies a tensile force of 217 kN to the web of the beam.
Its resistance is given in clause 6.2.6.8 of EN 1993-1-8 as Ft, wb, Rd= beff, t, wbtwbfy, wb/γM0
(equation (6.22) of EN 1993-1-8), and beff, t, wbis taken as the effective length of the T-stub, leff, 2= 231 mm. Hence,
Ft, wb, Rd= 231 × 9.4 × 0.355/1.0 = 771 kN (D8.12)
so this does not govern.
Column web in tension
The effective width of the column web in tension, to clause 6.2.6.3 of EN 1993-1-8, is the length of the T-stub representing the column flange. The resistance is
Ft, wc, Rd= ωbeff, t, wctwcfy, wc/γM0
where ω is a reduction factor to allow for shear in the column web. In this case, the shear is zero, and ω = 1. The column web is thicker than the beam web, so from result (D8.12), its resistance does not govern.
Column web in transverse compression
The resistance is given in clause 6.2.6.2 of EN 1993-1-8. It depends on the plate slenderness λpand the width of the column web in compression, which is
beff, c, wc= tf, b+ 2÷2ap+ 5(tfc+ s) + sp
(equation (6.11) of EN 1993-1-8), where apis the throat thickness of the bottom-flange welds, so ÷2ap= 10 mm here; spallows for 45° dispersion through the end plate, and is 24 mm here; and s is the root radius of the column section (s = rc= 21 mm). Hence,
beff, c, wc= 14.6 + 20 + 5 × (17 + 21) + 24 = 248 mm For web buckling, the effective compressed length is
dwc= hc– 2(tfc+ rc) = 240 – 2 × (17 + 21) = 164 mm
The plate slenderness is
= 0.932(beff, c, wcdwcfy, wc/Eatwc2)0.5
= 0.932 × [248 × 164 × 0.355/(210 × 100)]0.5= 0.773 The reduction factor for plate buckling is
ρ = ( – 0.2)/ 2= 0.573/0.7732= 0.96 The factor ω for web shear is 1.0, as before.
It is assumed that the maximum longitudinal compressive stress in the column is less than 0.7fy, wc, so from clause 6.2.6.2(2) of EN 1993-1-8, the reduction factor for this, kwc, is 1.0. From equation (6.9) in EN 1993-1-8,
Fc, wc, Rd= ωkwcρbeff, c, wctwcfy, wc/γM1
= 0.96 × 248 ×10 × 0.355/1.0 = 845 kN (D8.13)
Clearly, the tensile force of 217 kN governs the resistance of the steel connection.
Bending resistance of the steel joint, for both beams fully loaded
From equation (D8.5), the lever arm is 383 mm, so the resistance, excluding the reinforcement, is
Mj, Rd, steel= 217 × 0.383 = 83 kN m (D8.14)
governed by bending of the end plate. The critical mode 2 includes failure of the top row of bolts in tension. However, the joint is closely based on a type given in Couchman and Way,104which is confirmed by ECCS TC1139as having ‘ductile’ behaviour.
From Example 6.7, the plastic bending resistance of the steel beam, an IPE 450 section, is
Mpl, a, Rd= 1.702 × 355 = 604 kN m
This exceeds four times Mj, Rd, so clause 5.2.3.2(3) of EN 1993-1-8 permits this joint to be classified as ‘nominally pinned’ for the construction stage.
Resistance of the composite joint
For the composite joint, the reinforcement is at yield in tension. Its resistance is Ft, s, Rd= 1206 × 0.500/1.15 = 524 kN
This increases the total compressive force to
Fc= 217 + 524 = 741 kN (D8.15)
This is less than the compressive resistance of 845 kN, found above. The bars act at a lever arm of 543 mm (Fig. 8.8(a)), so the bending resistance of the composite joint is
Mj, Rd, comp= 83 + 524 × 0.543 = 83 + 284 = 367 kN m (D8.16)
Check on vertical shear
For the maximum design beam load of 35.7 kN/m and a hogging resistance moment at B of 367 kN m, the vertical shear in each beam at B is 244 kN, which just exceeds the shear resistance found earlier, 242 kN. It will probably be found from elastic–plastic global analysis that the vertical shear at B is reduced by the flexibility of the joints. If necessary, two extra M20 bolts can be added in the lower half of each end plate. This has no effect on the preceding results for resistance to bending.
Maximum load on span BC, with minimum load on span AB
This loading causes maximum shear in the column web. There is an abrupt change in the tension in the slab reinforcement at B. The load acting on the steel members is equal for
λp
λp λp
the two spans, and is assumed to cause a hogging bending moment at node B equal to the resistance of the joints, 83 kN m from equation (D8.11). The ultimate loads on the composite member are 1.62 kN/m on AB and 27.9 kN/m on BC (see Table 6.2).
The flexibility of the joints and cracking of concrete both reduce hogging bending moments, so both are neglected in these checks on shear in the column web and anchorage of the reinforcement. The moment on the composite joint at B in span BC is taken as the additional resistance provided by the slab reinforcement, which is 284 kN m (equation (D8.16)). For the other three members meeting at node B, elastic analysis gives the bending moments shown in Fig. 8.11(a). The total bending moments at B, including construction, are shown in Fig. 8.11(b). The shear forces in columns DB and BE are
(75 + 37.5)/3 = 37.5 kN
If the end plate in span AB is plastic under ultimate construction loading, the whole of the difference between the beam moments shown is caused by change of tension in the reinforcement. Thus, the relevant lever arm, z, is 543 mm.
From clause 5.3(3) of EN 1993-1-8, the shear force on the web panel is Vwp, Ed= (Mb1, Ed– Mb2, Ed)/z – (Vc1, Ed– Vc2, Ed)/2
= (367 – 217)/0.543 – [37.5 – (–37.5)]/2 = 276.2 – 37.5 = 239 kN
The sign convention used here is given in Fig. 5.6 of EN 1993-1-8. It is evident from Fig.
8.11(b) and the equation above that the web shear from the beams, 276 kN, is reduced by the shear forces in the column. The change of force in the reinforcement is 276 kN.
Shear resistance of the column web
From clause 6.2.6.1(1) of EN 1993-1-8, the shear resistance of an unstiffened column web panel is
Vwp, Rd= 0.9fy, wcAvc/(÷3γM0)
where Avc is the shear area of the column web. This is given in EN 1993-1-1, and is 3324 mm2here. Hence,
Vwp, Rd= 0.9 × 0.355 × 3324/(÷3× 1.0) = 613 kN (> Vwp, Ed)
This load arrangement therefore does not govern the design of the joint.
z 134 + 83
37.5 75 D
B C A
284
37.5 75
E 12
3 134
12
37.5 75
284 + 83
1.62 kN/m 27.9 kN/m
Fig. 8.11. Analyses for unequal design loadings (ultimate limit state) on spans AB and BC
(a) Bending-moment diagram (kN m) (b) Action affects on joint (kN and kN m)
Anchorage of the force from the reinforcement
The force of 276 kN (above) has to be anchored in the column. The strut-and-tie model shown in Fig. 8.2 requires transverse reinforcement to resist the force Ftqshown in the figure, and the force depends on the directions chosen for the struts.
The mean distance of the three 16 mm bars shown in Fig. 8.8(d) from the centre-line of the column is 420 mm. The two concrete struts AB and AD shown in Fig. 8.12 can, for calculation, be replaced by line AC. Resolution of forces at point A gives the strut force as 184 kN. The depth of concrete available is 80 mm. With fck= 25 N/mm2, the total width of the struts is
bc= (184 × 1.5)/(0.08 × 0.85 × 25) = 163 mm
This width is shown to scale in Fig. 8.12, and is obviously available.
The existing transverse reinforcement (Example 6.7) is 12 mm bars at 200 mm spacing.
Insertion of three more bars (As= 339 mm2) at 200 mm spacing provides an extra resistance
TRd= 339 × 0.5/1.15 = 147 kN
which exceeds the tie force of 122 kN shown in Fig. 8.12. Three extra bars are provided on each side of the column.
The available area of column flange to resist bearing stress is 80 × (120 + 115) = 18 800 mm2
so the mean stress is
120
420
50 A
300 120
B C
184 kN
163 138 kN
D
3 No. 12 bars at 200 mm 122 kN
Fig. 8.12. Strut-and-tie model for anchorage of unbalanced tension in slab reinforcement
Table 8.1. Initial stiffnesses of joints, Sj, ini(kN m/mrad)
No shear Joint BA,
No shear Joint BA,