The effects of geometric imperfections should be considered in combination with the effects of wind loads (i.e. not as an alternative load combination). For global analysis, the imperfections may be represented by an inclination 0i.
0 i = (1/200) x a h X a m where a h = (2/R/), to be taken as not less than 2/3 nor greater than 1.0 a m = [0.5 (1 + 1/rn)]°.5 l is the height of the building in metres m is the number of vertical members contributing to the horizontal force in the bracing system.
The effect of the inclination may be represented by transverse forces at each level and included in the analysis along with other actions (see Figure 5):
Effect on roof diaphragm: Hi = 0 i Na where Na and Nb are longitudinal forces contributing to Hi.
In most cases, an allowance for imperfections is made in the partial factors used in the design of elements. However for columns, the effect of imperfections, which is similar in principle to the above, must be considered (see Chapter 5, originally published as Columns^).
Crack widths should be limited to ensure appearance and durability are satisfactory. In the absence of specific durability requirements (e.g. water tightness) the crack widths may be limited to 0.3 mm in all exposure classes under the quasi-permanent combination. In the absence of requirements for appearance, this limit may be relaxed (to say 0.4 mm) for exposure classes X0 and XC1 (refer to Table 7). The theoretical size of the crack can be calculated using the expressions given in Cl 7.3.4 from Eurocode 2-1-1 or from the 'deemed to satisfy' requirements that can be obtained from Table 11, which is based on tables 7.2N and 7.3N of the Eurocode. The limits apply to either the bar size or the bar spacing, not both.
Figure 5
Examples of the effect of geometric imperfections
Figure 5
Examples of the effect of geometric imperfections
Figure 6
Determination of steel stress for crack width control
Table 11
Maximum bar size or spacing to limit crack width
Table 11
Maximum bar size or spacing to limit crack width
Wmax = 0.4 mm |
Wmax = 0.3 mm | ||||||
stress (s s)MPa |
Maximum bar size (mm) |
Maximum bar spacing (mm) |
Maximum bar size (mm) |
Maximum bar spacing (mm) | |||
16O |
4O |
300 |
BZ |
300 | |||
ZOO |
BZ |
OR |
300 |
¿S |
OR |
250 | |
¿4O |
¿O |
250 |
16 |
200 | |||
¿8O |
16 |
200 |
1Z |
150 | |||
BZO |
1Z |
150 |
1O |
100 | |||
B6O |
1O |
100 |
8 |
The steel stress may be estimated from the expression below (or see Figure 6): ss = fyk mAs,req Note The steel stress may be estimated from the expression below (or see Figure 6): ss = fyk mAs,req where fyk 7ms m n A gms n As,prov d = characteristic reinforcement yield stress = partial factor for reinforcing steel = total load from quasi-permanent combination = total load from ULS combination = area of reinforcement at the ULS s,req As,prov = area of reinforcement provided Figure 6 Determination of steel stress for crack width control To determine stress in the reinforcement (ss), calculate the ratio Gk/Qk, read up the graph to the appropriate curve and read across to determine ssu. ss can be calculated from the expression: ss = ssu s To determine stress in the reinforcement (ss), calculate the ratio Gk/Qk, read up the graph to the appropriate curve and read across to determine ssu. ss can be calculated from the expression: ss = ssu s = ratio of redistributed moment to elastic moment |
Was this article helpful?
Get All The Support And Guidance You Need To Be A Success At Living Green. This Book Is One Of The Most Valuable Resources In The World When It Comes To Great Tips on Buying, Designing and Building an Eco-friendly Home.