Clause 7.4.3(2) Clause 7.4.3(2) specifies elastic global analysis to Section 5, allowing for the effects of cracking. The preceding comments on global analysis for deformations apply also to this analysis for bending moments in regions with concrete in tension.
Clause 7.4.3(3)
Paragraph (4) on loading should come next, but it is placed last in clause 7.4.3 because of
the drafting rule that ‘general’ paragraphs precede those ‘for buildings’. It specifies the quasi-permanent combination. Except for storage areas, the values of factors ψ2for floor
loads in buildings are typically 0.3 or 0.6. The bending moments will then be much less than for the ultimate limit state, especially for cross-sections in Class 1 or 2 in beams built unpropped. There is no need to reduce the extent of the cracked regions below that assumed for global analysis, so the new bending moments for the composite members can be found by scaling values found for ultimate loadings. At each cross-section, the area of reinforcement will be already known: that required for ultimate loading or the specified minimum, if greater; so the stresses σs, 0(clause 7.4.3(3)) can be found.
Tension stiffening
A correction for tension stiffening is now required. At one time, these effects were not well understood. It was thought that, for a given tensile strain at the level of the reinforcement, the total extension must be the extension of the concrete plus the width of the cracks, so that allowing for the former reduced the latter. The true behaviour is more complex.
The upper part of Fig. 7.2 shows a single crack in a concrete member with a central reinforcing bar. At the crack, the external tensile force N causes strain εs2= N/AsEain the
bar, and the strain in the concrete is the free shrinkage strain εcs, which is negative, as shown.
There is a transmission length Le each side of the crack, within which there is transfer of
shear between the bar and the concrete. Outside this length, the strain in both the steel and the concrete is εs1, and the stress in the concrete is fractionally below its tensile strength.
Le Le N N es2 esm ec(x) es(x) x es1 ecm ecs Tensile strain 0
Within the length 2Le, the curves εs(x) and εc(x) give the strains in the two materials, with
mean strains εsmin the bar and εcmin the concrete.
It is now assumed that the graph represents the typical behaviour of a reinforcing bar in a cracked concrete flange of a composite beam, in a region of constant bending moment such that the crack spacing is 2Le. The curvature of the steel beam is determined by the mean
stiffness of the slab, not the fully cracked stiffness, and is compatible with the mean longitudinal strain in the reinforcement, εsm.
Midway between the cracks, the strain is the cracking strain of the concrete, corresponding to a stress less than 30 N/mm2in the bar. Its peak strain, at the crack, is much
greater than εsm, but less than the yield strain, if crack widths are not to exceed 0.4 mm. The
crack width corresponds to this higher strain, not to the strain εsmthat is compatible with the
curvature, so a correction to the strain is needed. It is presented in clause 7.4.3(3) as a correction to the stress σs, 0because that is easily calculated, and Tables 7.1 and 7.2 are based
on stress. The strain correction cannot be shown in Fig. 7.1 because the stress σs, 0 is
calculated using the ‘fully cracked’ stiffness, and so relates to a curvature greater than the true curvature. The derivation of the correction101takes account of crack spacings less than
2Le, the bond properties of reinforcement, and other factors omitted from this simplified
outline.
The section properties needed for the calculation of the correction ∆σs will usually be
known. For the composite section, A is needed to find I, which is used in calculating σs, 0, and
Aa and Ia are standard properties of the steel section. The result is independent of the
modular ratio. For simplicity, αstmay conservatively be taken as 1.0, because AI > AaIa.
When the stress σsat a crack has been found, the maximum bar diameter or the maximum
spacing are found from Tables 7.1 and 7.2. Only one of these is needed, as the known area of reinforcement then gives the other. The correction of clause 7.4.2(2) does not apply.
Influence of profiled sheeting on the control of cracking
The only references to profiled steel sheeting in clause 7.4 are in clause 7.4.1(4), ‘… no
account should be taken of any profiled steel sheeting’, and in the definition of hcin clause
7.4.2(1), ‘… thickness … excluding any haunch or ribs’.
The effects of the use of profiled sheeting for a slab that forms the top flange of a continuous composite beam are as follows:
• there is no need for control of crack widths at the lower surface of the slab
• where the sheeting spans in the transverse direction, there is at present no evidence that it contributes to the control of transverse cracks at the top surface of the slab
• where the sheeting spans parallel to the beam, it probably contributes to crack control, but no research on this subject is known to the authors.
For design, the definition of ‘effective tension area’ in clause 7.3.4(2) of EN 1992-1-1 should be noted. A layer of reinforcement at depth c + φ/2 below the top surface of the slab, where c is the cover, may be assumed to influence cracking over a depth 2.5(c + φ/2) of the slab. If the depth of the concrete above the top of the sheeting, hc, is greater than this, it
would be reasonable to use the lower value, when calculating Asfrom equation (7.1). This
recognizes the ability of the sheeting to control cracking in the lower half of the slab, and has the effect of reducing the minimum amount of reinforcement required, for the thicker composite slabs
General comments on clause 7.4
In regions where tension in concrete may arise from shrinkage or temperature effects, but not from other actions, the minimum reinforcement required may exceed that provided in previous practice.
Where unpropped construction is used for a continuous beam, the design loading for checking cracking is usually much less than that for the ultimate limit state, so that the quantity of reinforcement provided for resistance to load should be sufficient to control
cracking. The main use of clause 7.4.3 is then to check that the spacing of the bars is not too great.
Where propped construction is used, the disparity between the design loadings for the two limit states is smaller. If cracks are to be controlled to 0.3 mm, a check to clause 7.4.3 is more likely to influence the reinforcement required.
For beams in frames, the preceding comments apply where semi-rigid or rigid connections are used. Where floors have brittle finishes or an adverse environment, simple beam- to-column joints should not be used, because effective control of crack width may not be possible.
Example 7.1: two-span beam (continued) – SLS
Details of this beam are shown in Fig. 6.23. All of the design data and calculations for the ultimate limit state are given in Examples 6.7 to 6.12. For data and results required here, reference should be made to:
• Table 6.2, for characteristic loads per unit length
• Table 6.3, for elastic properties of the cross-sections at the internal support (B in Fig. 6.23(c)) and at mid-span
• Table 6.4, for bending moments at support B for uniform loading on both spans • Fig. 6.28, for bending-moment diagrams for design ultimate loadings, excluding the
effects of shrinkage.
The secondary effects of shrinkage are significant in this beam, and cause a hogging bending moment at support B of 120 kN m (Example 6.7). Clause 7.3.1(8) does not permit shrinkage to be ignored for serviceability checks on this beam, because it does not refer to lightweight-aggregate concrete, which is used here.
Stresses
From clause 7.2.2(1), there are no limitations on stress; but stresses in the steel beam need to be calculated, because if yielding occurs under service loads, account should be taken of the resulting increase in deflections, from clause 7.3.1(7).
Yielding is irreversible, so, from a note to clause 6.5.3(2) of EN 1990, it should be checked for the characteristic load combination. However, the loading for checking deflections depends on the serviceability requirement.96
The maximum stress in the steel beam occurs in the bottom flange at support B. Results for the characteristic combination with variable load on both spans and 15% of each span cracked are given in Table 7.2. The permanent load, other than floor finishes, is assumed to act on the steel beam alone. Following clause 5.4.2.2(11), the modular ratio is taken as 20.2 for all of the loading except shrinkage.
Table 7.2. Hogging bending moments at support B and stresses in the steel bottom flange, for the
characteristic load combination
Loading w(kN/m) Modularratio 10
–6Iy, B (mm4) MEk, B(kN m) 10 –6Wa, bot (mm3) σa, bot(N/mm2) (1) Permanent (on steel beam) 5.78 – 337 104 1.50 69 (2) Permanent (on composite beam) 1.2 20.2 467 18 1.75 10 (3) Variable 17.5 20.2 467 263 1.75 150 (4) Shrinkage – 28.7 467 120 1.75 69
The total bottom-flange compressive stress is
σEk, bot, a= 298 N/mm2(= 0.84fy)
so no allowance is needed for yielding.
Deflections
The maximum deflection of span AB of the beam will occur at about 4.8 m from A (40% of the span), when variable load acts on span AB only. The additional deflection caused by slip of the shear connection is ignored, as clause 7.3.1(4)(a) is satisfied.
Calculated deflections at this point, with 15% of each span assumed to be cracked, are given in Table 7.3. The frequent combination is used, for which ψ1= 0.7, so the variable
loading is
0.7 × 17.5 = 12.3 kN/m
The following method was used for the shrinkage deflections. From Example 6.7 and Fig. 6.27, the primary effect is uniform sagging curvature at radius R = 1149 m, with deflection δ = 45.3 mm at support B. The secondary reaction at B is 20 kN. From the geometry of the circle, the primary deflection at point E in Fig. 7.3(a) is
δ1, E= 45.3 – 5.42× 1000 × (2 × 1149) = 33 mm
The upwards displacement at E caused by the 20 kN reaction at B that moves point B¢ back to B was found to be 26 mm by elastic analysis of the model shown in Fig. 7.3(b), with 15% of each span cracked. The total shrinkage deflection is only 7 mm, despite the high free shrinkage strain, but would not be negligible in a simply-supported span.
The total deflection, 31 mm, is span/390. This ratio appears not to be excessive. However, the functioning of the floor may depend on its maximum deflection relative to the supporting columns. It is found in Example 9.1 that the deflection of the composite slab, if cast unpropped, is 15 mm for the frequent combination. This is relative to the supporting beams, so the maximum floor deflection is
(a) R 45.3 5.4 33 E 1.8 10.2 B' B A A 26 mm 20 kN 10 kN 10 kN 45.3 mm B E (b)
Fig. 7.3. Sagging deflection at point E caused by shrinkage
Table 7.3. Deflections at 4.8 m from support A, for the frequent combination
Load Modular ratio Deflection (mm)
Dead, on steel beam – 9
Dead, on composite beam 20.2 1
Imposed, on composite beam 20.2 14
Primary shrinkage 27.9 33
31 + 15 = 46 mm
or span/260. For the characteristic combination, this increases to 37 + 16 = 53 mm
or span/230.
The limiting deflections given in the UK’s draft National Annex to EN 199096depend
on the serviceability requirement. For floors with partitions, they range from span/300 to span/500. For this floor, some combination of using propped construction for the composite slabs and/or the beams, and cambering the beams, will be necessary.
Reducing the modular ratio for imposed loading to 10.1 makes little difference: the value 14 mm in Table 7.3 becomes 12 mm. The extensive calculations for shrinkage lead to a net deflection of only 7 mm, because the secondary effect cancels out most of the primary effect. This benefit would not occur, of course, in a simply-supported span.
Control of crack width
Clause 7.4 applies to reinforced concrete that forms part of a composite member. In
the beam considered here, the relevant cracks are those near support B caused by hogging bending of the beam, and cracks along the beam caused by hogging bending of the composite slab that the beam supports. The latter are treated in Example 9.1 on a composite slab.
Clause 7.4.1(1) refers to exposure classes. From clause 4.2(2) of EN 1992-1-1, Class
XC3 is appropriate for concrete ‘inside buildings with moderate humidity’. For this class, a note to clause 7.3.1(5) of EN 1992-1-1 gives the design crack width as 0.3 mm. The method of clause 7.4.1(3) is followed, as clause 7.4.1(4) does not apply.
Minimum reinforcement
The relevant cross-section of the concrete flange is as shown in Fig. 6.23(a), except that effective widths up to 2.5 m should be considered.
From clause 5.4.1.2, the effective width is assumed to increase from 1.6 m at support B to 2.5 m at sections more than 3 m from B. It may be difficult to show that sections 3 m from B are never ‘subjected to significant tension’ (clause 7.4.2(1)). Calculations are therefore done for both effective widths, assuming uncracked unreinforced concrete. From the definition of z0in that clause, n0= 10.1.
It is found for both of these flange widths that z0is such that kc> 1, so, from equation
(7.2), it is taken as 1.0.
A value is required for the strength of the concrete when cracks first occur. As unpropped construction is used, there is at first little load on the composite member, so from clause 7.4.2(1), conservatively, fct, eff= 3.0 N/mm2. Assuming that 10 mm bars
are used for the minimum reinforcement, Table 7.1 gives σs= 320 N/mm2. Then, from
equation (7.1),
100As/Act= 100 × 0.9 × 1 × 0.8 × 3.0/320 = 0.675%
However, clause 5.5.1(5) also sets a limit, as a condition for the use of plastic resistance moments. For this concrete, flctm= 2.32 N/mm2, and fsk= 500 N/mm2. Hence, from
equation (5.8) with kc= 1.0,
100ρs= 100 × (355/235)(2.32/500) = 0.70% (D7.4)
This limit governs; so for a slab 80 mm thick above the sheeting the minimum reinforcement is
As, min= 7 × 80 = 560 mm2/m
Cracking due to direct loading
Only the most critical cross-section, at support B, will be considered. Clause 7.4.3(4) permits the use of the quasi-permanent combination, for which the variable loading is
ψ2qk, with ψ2= 0.6, from clause A1.2.2(1) of EN 1990.
From Table 7.2, the bending moment at B that stresses the reinforcement is
ME, qp, B= 18 + 263 × 0.6 + 127 = 303 kN m
The neutral axis for the cracked section is 313 mm below the top of the slab (Table 6.12), so the section modulus for reinforcement at depth 30 mm is
10–6W
s= 467/(313 – 30) = 1.65 mm3
Hence, from clause 7.4.3(3),
σs, 0= 303/1.65 = 184 N/mm2
The correction for tension stiffening, equation (7.5), is now calculated, assuming that the reinforcement used in Example 6.7, 12 mm bars at 125 mm spacing, will be satisfactory. This gives ρs= 0.0113.
Using values obtained earlier,
αst= AI/AaIa= 11 350 × 467/(9880 × 337) = 1.59
From equation (7.5),
∆σs= 0.4 × 2.32/(1.59 × 0.0113) = 52 N/mm2 (D7.5)
From equation (7.4),
σs= 184 + 52 = 236 N/mm2
From Table 7.1, φs£ 16 mm. From Table 7.2, the bar spacing £ 200 mm.
The use of 12 mm bars at 125 mm spacing at support B satisfies both conditions. Finding the cross-sections of the beam at which this reinforcement can be reduced to the minimum found above may require consideration of the bending-moment envelopes both for ultimate loads and for the quasi-permanent combination.