Fatigue assessment based on nominal stress ranges

Clause 6.8.7.1 Comment on the methods referred to from clause 6.8.7.1 will be found in other guides in this

series.

Clause 6.8.7.2(1) For shear connectors, clause 6.8.7.2(1) introduces the partial factors. The recommended

value of γ_{Mf, s}is 1.0 (clause 6.8.2(1)). For γ_Ff, EN 1990 refers to the other Eurocodes. The recommended value in EN 1992-1-1, clause 6.8.4(1), is 1.0. No value has been found in EN 1993-1-1 or EN 1993-1-9. Clause 9.3(1) of EN 1993-2 recommends 1.0 for bridges.

Clause 6.8.7.2(2) Clause 6.8.7.2(2) covers interaction between the fatigue failures of a stud and of the steel

flange to which it is welded, when the flange is in tension. The first of expressions (6.57) is the verification for the flange, from clause 8(1) of EN 1993-1-9, and the second is for the stud, copied from equation (6.55). The linear interaction condition is given in expression (6.56).

It is necessary to calculate the longitudinal stress range in the steel flange that coexists with the stress range for the connectors. The load cycle that gives the maximum value of

∆σE, 2in the flange will not, in general, be that which gives the maximum value of ∆τE, 2in a

Fig. 6.41. Stress ranges in reinforcement in cracked regions

shear connector, because the first is caused by flexure and the second by shear. Also, both

∆σ_{E, 2}and ∆τ_{E, 2}may be influenced by whether the concrete is cracked, or not.

It thus appears that expression (6.56) may have to be checked four times. In practice, it is best to check first the conditions in expression (6.57). It should be obvious, for these, whether the ‘cracked’ or the ‘uncracked’ model is the more adverse. Usually, one or both of the left-hand sides is so far below 1.0 that no check to expression (6.56) is needed.

Example 6.12: fatigue in reinforcement and shear connection

It is assumed that the imposed floor load of 7.0 kN/m²for the two-span beam in Examples 6.7 and 7.1 is partly replaced by a cyclic load. The resistance to fatigue of the reinforcement at the internal support, point B in Figs 6.23 and 6.28, and of the shear connection near the cyclic load are checked. All other data are as before.

Loading and global analysis

The cyclic load is a four-wheeled vehicle with two characteristic axle loads of 35 kN each.

It travels at right angles to beam ABC, on a fixed path that is 2.0 m wide and free from other variable loads. The axle spacing exceeds the beam spacing of 2.5 m, so each passage can be represented by two cycles of point load, 0–35–0 kN, applied at point D in Fig. 6.42(a). For a 25 year design life, 20 passages per hour for 5000 h/year gives NEd= 2.5 × 10⁶cycles of each point load. The partial factor γFfis taken as 1.0.

In comparison with Example 6.7, the reduction in static characteristic imposed load is 7 × 2.5 × 2 = 35 kN, the same as the additional axle load, so previous global analyses for the characteristic combination can be used. These led to the bending moments MEkat support B given in the four rows of Table 7.2. Those in rows 2 and 4 are unchanged, and MEk, B= 263 kN m for loading qk. Analysis for the load Qfatalone, with 15% of each span cracked, gave the results in Fig. 6.42(b), with 31 kN m at support B.

The frequent combination of non-cyclic imposed load is specified, for which ψ1= 0.7.

From Table 6.2, qk= 17.5 kN/m. Therefore, ψ1qk= 0.7 × 17.5 = 12.25 kN/m, acting on span AB and on 10 m only of span BC, giving

M_{Ek, B}= 0.7 × (263 – 31) = 162 kN m

Table 6.5 gives

MEd, min, f= 18 + 162 + 120 = 300 kN m

12 4 1 1 6

Q_fat = 35 kN

B A

0.7q_k

D C

31

2.6 2.6 + 23 12

84 35

(b)

(a)

(c)

10.0 10.0

120

20.0

Fig. 6.42. Fatigue checks for a two-span beam. (a) Variable static and cyclic loads. (b) Design action effects, cyclic load. (c) Design action effects, shrinkage

From Fig. 6.42(b),

MEd, max, f= 300 + 31 = 331 kN m

Verification for reinforcement at cross-section B

From equation (D7.5), the allowance for tension stiffening is ∆σ_s= 52 N/mm². From σ_{s, 0} given in Table 6.5,

σ_{s, max, f}= 201 + 52 = 253 N/mm² From Fig. 6.41,

σs, min, f= 253 × 300/331 = 229 N/mm² so

∆σ_{s, f}= 253 – 229 = 24 N/mm²

From clause 6.8.4 of EN 1992-1-1, for reinforcement,

m = 9 for NE> 10⁶ N^*= 10⁶ ∆σRsk(N*) = 162.5 N/mm²

By analogy with equation (D6.41), the damage equivalent stress range for N_Ed= 5 × 10⁶ cycles of stress range 24 N/mm²is given by

24⁹× 5 × 10⁶= (∆σ_{E, equ})⁹× 10⁶ whence

∆σE, equ= 29 N/mm² Using equation (D6.40),

29 £ 162.5/1.15

so the reinforcement is verified.

If the axle loads had been unequal, say 35 and 30 kN, the stress range for the lighter axle would be a little higher than 24 × 30/35 = 20.6 N/mm², because its line OJ in Fig. 6.41 would be steeper. Assuming this stress range to be 21 N/mm², the cumulative damage check, for γMf= 1.15, would be

2.5 × 10⁶× (24⁹+ 21⁹) £ (162.5/1.15)⁹× 10⁶ which is

8.6 × 10¹⁸£ 2.25 × 10²⁵

Verification for shear connection near point D

Vertical shear is higher on the left of point D in Fig. 6.42, than on the right. From Fig.

6.42(b), ∆V_{Ed, f}= 23 kN for each axle load.

Table 6.5. Stresses in longitudinal reinforcement at support B

Action n Load

(kN/m) MEk

(kN m) 10^–6Ws, cr

(mm³) σs, 0

(N/mm²)

Permanent, composite 20.2 1.2 18 1.65 11

Variable, static (ψ1= 0.7) 20.2 12.25 162 1.65 98

Shrinkage 28.7 – 120 1.65 73

Cyclic load 20.2 – 31 1.65 19

Totals 331 201

The maximum vertical shear at this point, including 10.0 kN from the secondary effect of shrinkage, Fig. 6.42(c), is 59.5 kN, Table 6.6. The shear forces VEdare found from the loads and values MEkin Table 6.5. The resulting maximum longitudinal shear flow, for the uncracked unreinforced composite section, is 110 kN/m, of which the cyclic part is 43.2 kN/m.

Clause 6.8.1(3) limits the shear per connector under the characteristic combination to 0.75PRd. For this combination, the shear flow from the non-cyclic variable action increases by 19 from 44.7 to 44.7/0.7 = 63.9 kN/m, so the new total is 110 + 19 = 129 kN/m. The shear connection (see Fig. 6.30), is 5 studs/m, with PRd= 51.2 kN/stud. Hence,

PEk/PRd= 129/(5 × 51.2) = 0.50

which is below the limit 0.75 in clause 6.8.1(3).

The range of shear stress is

∆τ_E= 43 200/(5π × 9.5²) = 30.5 N/mm²

The concrete is in density class 1.8. From clause 6.8.3(4),

∆τc= 90 × (1.8/2.2)²= 60 N/mm² From equation (D6.41),

∆τ_{E, 2}= 30.5[5 × 10⁶/(2 × 10⁶)]^1/8= 34.2 N/mm² As γMf, s= 1.0,

∆τc, d= 60 N/mm²

so the shear connection is verified.

Table 6.6. Fatigue of shear connectors near cross-section D

Action 10^–6Iy

(mm⁴) 10^–3Ac/n

(mm²) z (mm) VEd(kN) VEdAcz/nIy

(kN/m)

Permanent, composite 828 9.90 157 2.7 5.0

Variable, static (ψ1= 0.7) 828 9.90 157 23.8 44.7

Shrinkage 741 6.97 185 10.0 17.4

Cyclic load 828 9.90 157 23.0 43.2

Totals 59.5 110

CHAPTER 7