Serviceability Limit States - PDF Free Download

Serviceability Limit States www.eurocode2.info 1

Outline Crack control and limitations Crack width calculations Crack width calculation example Crack width calculation problem Restraint cracking Deflection calculations Deflection calculation example 3

Crack control and limitations 4

(Flexural) Crack Width Limits (Table 7.1N) EC 2: Table 7.1(N) Concise: 10.2 Exposure class RC or unbonded PSC members Quasi-permanent load Prestressed members with bonded tendons Frequent load X0,XC1 0.3 0.2 XC2,XC3,XC4 0.3 XD1,XD2,XS1,XS2, XS3 Decompression 5

(Flexural) Crack Width Control EC 2: Cl. 7.3.3 Concise: 10.2 Crack control may be achieved: Limiting the maximum bar diameter using Table 7.2 Limiting the maximum bar spacing using Table 7.3 (this table is not applicable for restraint loading) Calculating cracks to ensure they are within limits 6

Crack width calculations 8

Basis Section Neutral axis Figure 7.2 Slab soffit (h - x) Crack width vs spacing Crack width w Actual crack width Crack width predicted by Expressions (7.8) & (7.14) Crack width predicted by Expressions (7.8) & (7.11) 5(c + /2) 9

Crack width calculation EC2 Exp (7.8) The crack width may be calculated from: w k = s r,max (ε sm - ε cm ) where s r,max ε sm ε cm = maximum crack spacing = mean strain in reinforcement = mean strain in concrete between cracks 10

Maximum crack spacing s r,max c k 1 k 2 Φ = 3.4c + 0.425k 1 k 2 Φ /ρ p,eff = (nominal) cover to the longitudinal reinforcement = factor to take account on bond properties = 0.8 for high bond bars = 1.6 for plain bars = factor to take account of strain distribution = 0.5 for flexure = 1.0 for pure tension = bar diameter When spacing > 5(c + Φ/2); s r,max = 1.3(h x) 11

Crack (ε sm - ε cm ) Difference in concrete and reinforcement strain Strain diagram w k = s.(ε sm - ε cm ) (s r,max includes fos of 1.7) ε cs ε cult S (s r,max includes fos of 1.7) 12

(ε sm - ε cm ) Difference in concrete and reinforcement strain For flexure,(ε sm - ε cm ) may be calculated from: fct, eff s kt ( 1 e p, eff ) p, eff sm cm 0. 6 E E s where: σ s = stress in the tension steel calculated using the cracked concrete section k t = factor that accounts for the duration of loading = 0.6 for short-term load = 0.4 for long-term load α e = E s /E c = modular ratio s s 13

where cont. ρ p,eff, is the effective reinforcement ratio. ρ p,eff where A s A c,eff h c,ef = A s /A c,eff = area of tension reinforcement = effective area of concrete in tension around d the reinforcement Effective = Min{2.5(h - d); (h - x)/3; h/2} h tension area h c,eff Effective tension area for this face d Beam h c,eff Effective tension area Member h c,eff Slab h = lesser of 2.5(h-d), (h-x)/3 or h/2 c,eff Figure 6.12: Typical examples of effective concrete tension 14

(Flexural) Crack width calculation example 15

d = 930 h = 1000 (Flexural) Crack width calculation Calculate the design flexural crack for the beam shown. M QP = 650 knm Concrete class C25/30 A s = 3770 mm 2 (,t0) = 2.63 d = 400 x F s F c (d x/3) 3 No H40 bars 16

Crack width example Step 1 Calculate effective modulus From Table 3.1, E cm = 31 kn/mm 2 E c,eff = E cm /(1 + (,t0) ) = 31 / (1 + 2.63) = 8.54 kn/mm 2 Step 2 Calculate the stress in the tension steel : find x Taking moments about neutral axis: b x 2 /2 = α e A s (d x) 400 x 2 /2 = 200/8.54 x 3770 (930 x) This has the solution, x = 457 mm 17

Crack width example Step 3 Calculate stress in the tension steel Taking moments about the level of force in the concrete: σ s = M QP /(d x/3)a s = 650 x 10 6 /((930 457/3) x 3770) = 222 MPa 18

Crack width example Step 4 Calculate difference in concrete and reinforcement strains fct, eff s kt ( 1 e p, eff ) p, eff sm cm 0. 6 E E s s s k t = 0.4 (long-term loading) f ct,eff = f ctm = 2.6 MPa (Table 3.1) α e = E s /E cm = 200 / 31 = 6.45 h c,ef = Min{2.5(h - d); (h - x)/3; h/2} = Min{2.5(1000-930); (1000-457)/3; 1000/2} = Min{175; 181; 500} = 175 mm 19

Crack width example A c,eff = 175 x 400 = 70 000 mm 2 ρ p,eff = A s /A c,eff = 3700 / 70 000 = 0.0539 sm cm 222 0.4 2.6 0.0539 (1 3 20010 222 19.97 0.00067 3 20010 0.001 6.45 0.0539) 222 0.6 20010 3 20

Crack width example Step 5 Calculate the maximum crack spacing s r,max c k 1 k 2 Φ = 3.4c + 0.425k 1 k 2 Φ /ρ p,eff = 1000 930-40/2 = 50 mm = 0.8 (ribbed bars) = 0.5 (flexure) = 40 mm s r,max = 3.4 x 50 + 0.425 x 0.8 x 0.5 x 40 /0.0539 = 296 mm < 5(c + Φ/2) = 350 mm Step 6 Calculate crack width w k = 0.0010 x 296 = 0.30 mm 21

d = 192 h = 250 Workshop problem Calculate the design flexural crack for the slab shown. M QP = 85 knm Concrete class C35/45 A s = 2010 mm 2 /m Assume the depth to neutral axis, x = 63.5 mm x F c (d x/3) H16 bars @ 100 mm CTRS F s HINT: You can start the calculation from step 3 23

Workshop problem Step 3 Calculate stress in the tension steel Taking moments about the level of force in the concrete: σ s = M QP /(d x/3)a s = 85 x 10 6 /((192 63.5/3) x 2010) = 247.5 MPa 24

Workshop problem Step 4 Calculate difference in concrete and reinforcement strains fct, eff s kt ( 1 e p, eff ) p, eff sm cm 0. 6 E E k t = 0.4 (long-term loading) f ct,eff = f ctm = 3.2 MPa (Table 3.1) α e = E s /E cm = 200 / 34 = 5.88 h c,ef = Min{2.5(h - d); (h - x)/3; h/2} = Min{2.5(250-192); (250 63.5)/3; 250/2} = Min{145; 62.2; 125} = 62.2mm A c,eff = 62.2 x 1000 = 62 200 mm 2 s s s 25

Workshop problem A c,eff = 62.2 x 1000 = 62 200 mm 2 ρ p,eff = A s /A c,eff = 2010 / 62 200 = 0.0323 sm cm 247.6 0.4 247.6 47.2 3 20010 0.0010 3.2 0.0323 (1 20010 3 0.00074 5.88 0.0323) 247.6 0.6 20010 3 26

Workshop problem Step 5 Calculate the maximum crack spacing s r,max c k 1 k 2 Φ = 3.4c + 0.425k 1 k 2 Φ /ρ p,eff = 50 mm = 0.8 (ribbed bars) = 0.5 (flexure) = 16 mm s r,max = 3.4 x 50 + 0.425 x 0.8 x 0.5 x 16 /0.0323 = 254.2 mm < 5(c + Φ/2) = 290 mm Step 6 Calculate crack width w k = 0.0010 x 254 = 0.25 mm 27

Restraint cracking 28

Restraint cracking Movement occurs not only due to loading but also due to: Early thermal effects Shrinkage drying autogenous, seasonal/long term temperature drop That becomes a problem when there is Restraint: Edge: End: adjacent slab pours, wall on base, adjacent wall pours infill bays, large area ground slabs (friction, foundations), piled slabs Internal (not covered.. members > say 750 mm th.) 29

Restraint cracking End restraint: restrained strain & stresses Original poured length Free contraction: R = 0 Partial restraint R = 0.0 to 1.0 Restrained strain: Stresses induced Fully restrained: R =1.0 Restrained strain: Large stresses induced 30

Restraint cracking 1) Cracking will occur if r ctu CIRIA C660: Cl 3.2 r = R ax free = K 1 { c T 1 + ca R 1 + c T 2 R 2 + cd R 3 } where: K 1 = allowance for creep = 1.0 to BS EN 1992-3 or = 0.65 to CIRIA C660 c = coefficient of thermal expansion (typical design value 12 m) T 1 = Peak to ambient temperature o C (See CIRIA C660). (e.g. 500 mm thick wall formed using 18 mm ply, using C30/37 concrete with 40%ggbs = 29 o C) ca = Autogenous shrinkage strain (typical design values using C30/37 concrete 15 m @ 3days, 50 m long term) R 1, R 2, R 3 = appropriate restraint factor for the short-term, medium term and long term see figure L1 of BS EN 1992-3 (includes for creep) or calculated for base wall restraint in accordance with CIRIA C660 (excludes for creep) 31

Restraint cracking (cont d) 1) Cracking will occur if r ctu CIRIA C660: Cl 3.2 r = R ax free = K 1 { c T 1 + ca R 1 + c T 2 R 2 + cd R 3 } where: cd = drying shrinkage strain, a function of time, thickness, RH, cement Class (BS EN 1992-1-1 or CIRIA C660) (e.g. 500 mm thick wall, using C30/37 concrete with 40% ggbs Class N = 340 m) T 2 = long-term drop in temperature after concreting. Recommended values: of 20 o C for concrete cast in the summer and 10 o C for concrete cast in winter. (See CIRIA C660), ctu = tensile strain capacity of the concrete. A function of concrete strength and type of aggregate used. (Typical design values of 76 m @ 3 days and 108 m for 28 days may be used for initial calculations. See CIRIA C660. 32

Restraint factors Table 1 Values of restraint factor R for a particular pour configuration BS EN 1992-3 Annex L Pour configuration Thin wall cast on to massive concrete base usually 0.5 Massive pour cast onto blinding Massive pour cast onto existing concrete Suspended slabs Infill bays, i.e. rigid restraint R 0,6 to 0,8 at base 0,1 to 0,2 at top 0,1 to 0,2 0,3 to 0,4 at base 0,1 to 0,2 at top 0,2 to 0,4 0,8 to 1,0 Beware: effects of creep included 33

Restraint cracking Short term load strength Long term load strength Stress due to early thermal & shrinkage & Stress due to early seasonal thermal & drying shrinkage Stress due to early thermal allowing for creep CS TR 67 34

Restraint cracking 1) Cracking will occur if r ctu r = R ax free = K 1 { c T 1 + ca R 1 + c T 2 R 2 + cd R 3 } CIRIA C660: Cl 3.2 Short term ( 3 days) Medium term ( 28 days) long term ( > 10000 days) 35

Restraint cracking 1) Cracking will occur if r ctu CIRIA C660: Cl 3.2 2) Minimum reinforcement, (with respect to restraint to movement): A s,min = k c ka ct (f ct,eff / f yk ) Where: k c = coeff. for stress distribution = 1.0 for full tension k Crack = coeff. width for thickness 1.0 for h < w300 k = smm and r,max 0.75 for h > 800 mm cr (interpolation allowed) A ct where = area of concrete in the tension zone just prior to onset of cracking. Most Maximum often crack based spacing on full thickness s r,max = 3.4c of the + 0.425 section. (k 1 / ( p,eff ) f ct,eff f ctm = mean tensile strength when cracking may be first expected to Crack inducing strain cr occur (Typical design values for a C30/37 concrete, 1.73 MPa @ 3 days and 2.9 MPa @ 28 days See BS EN 1992-1-1 = 500 MPa 3) Controlled cracking: f yk BS EN 1992-1-1 Exp (7.1) BS EN 1992-1-1 Exp (7.8) 36

Restraint cracking Watchpoints: Ensures rebar does not yield Typically 0.58% for C30/37 in a 300 mm wall 0.8 factor on f ct,eff for sustained loading? or 0.67 to TR 59 Early age only? B Hughes Revised CIRIA C660? 37

Restraint cracking As before where: c = nominal cover, c nom in mm k 1 1) = Cracking 0.8 for high will bond occur bars (CIRIA if C660 r suggests a value 1.14 to account ctu for poor bond conditions) CIRIA C660: Cl 3.2 k 2 = 1.0 for tension (e.g. from restraint), 0.5 for bending, ( 2) Minimum reinforcement, 1 + 2 )/ 2 1 for combinations BS EN 1992-1-1 Exp (7.1) = bar (with diameter, respect mm to restraint to movement): p,eff = A s /A c,eff, where A c,eff is calculated for each face = min0.5h or 2.5(c + 0.5) x b A s,min = k c ka ct (f ct,eff /w f yk ) 3) Controlled cracking: Crack width w k = s r,max cr where Maximum crack spacing s r,max = 3.4c + 0.425 (k 1 / p,eff ) Crack inducing strain cr ( sm - cm )...... BS EN 1992-1-1 Exp (7.8) BS EN 1992-1-1 Exp (7.11) 38

Crack (ε sm - ε cm ) Difference in concrete and reinforcement strain Strain diagram w k = s.(ε sm - ε cm ) (s r,max includes fos of 1.7) ε cs ε cult S (s r,max includes fos of 1.7) 39

Crack inducing strain, cr a) Edge restraint:- early thermal effects cr = k c T 1 + ca R 0.5 ctu b) Edge restraint:- long term restraint effects cr = k { c T 1 + ca R 1 + c T 2 R 2 + cd R 3 } 0.5 ctu c) End restraint Can be critical! CIRIA C660: Cl 3.2 CIRIA C660: Cl 3.2 cr = 0.5 e k c kf ct,eff 1 + (1/ e ) /E s BS EN 1992-3 Exp (M.1) where: k c = coeff. for stress distribution = 1.0 for full tension k = coeff. for thickness 1.0 for h < 300 mm and 0.75 for h > 800 mm f ct,eff = f ctm for long-term effects, 28 day value considered to be reasonable e.g. 2.9 Mpa for C30/37. NB Possible 0.8 factor for sustained load in CIRIA C660 e = modular ratio, E s /E c. Typical values are 6 @ 3 days, 7 @ 28 days and 12 longterm. When cracking occurs, no creep has taken place so a modular ratio of 7 should be used. = ratio of total area of reinforcement to the gross section in tension. Note that this different from p,eff.

Deflection calculations 41

Deflection limits Deflections are limited for the following reasons: 1. Excessive deflections are unsightly and alarming. EC2 restricts total deflections to span/250. 2. To avoid damage to cladding, partitions and finishes due to increments in deflection following their construction. EC2 limits deflections after construction of finishes to span/500. 3. Both construction tolerances and deflections need to be considered in the design of fixings for cladding systems and partitions. In practice it can be difficult to separate construction tolerances from deflections. 42

Deflection limits The EC2 deflection limits are guidelines. It is the designers responsibility to agree suitable deflection limits with his client taking into account the nature of the structure and finishes. 43

Introduction Four factors need to be considered in the calculation of deflections 1. criteria defining the limiting deflections 2. appropriate design loads 3. appropriate design material properties 4. means of predicting behaviour Deflections cannot be predicted exactly before construction since neither the loading or material properties are accurately known at the design stage. 44

Load Deflection calculation Calculated assuming no cracking Actual behaviour Calculated assuming concrete has no tensile strength Deflection 45

Basic behaviour = ( II ) + (1 - )( I ) where: = deformation parameter considered (e.g. strain, curvature) I is the calculated uncracked parameter II is the calculated cracked parameter = distribution coefficient allowing for tension stiffening at a section. e.g. total curvature = S(cracked curvature + uncracked curvature) for each effect considered 46

Basic behaviour S Crack Crack Crack s2 Idealised steel stress Steel stress s1 0 S (1 - )S S Concrete stress Where = distribution coefficient allowing for 47 tension stiffening at a section.

Basic behaviour where: = distribution coefficient = 1 - ( sr / s ) 2 where: = coefficient taking account of the influence of the duration of the loading or of repeated loading on the average strain = 1.0 for first loading = 0.5 for long-term loading But always use 0.5 s = stress in tension steel based on cracked section sr = stress in tension steel based on cracked section at first cracking NB sr / s M cr /M for flexure 48

Basic behaviour = 1 - (first crack result/cracked analysis result) 2 Load no cracking M Ed M cr Actual cracked = 1.0 for fully cracked sections (in theory) = 0.0 for un-cracked sections Deflection 49

Concrete material properties for deflection calculation It is only possible to estimate concrete material properties at the design stage. Actual material properties may differ significantly from those assumed in design. Therefore, it is prudent to assume a range of material properties in deflection calculations. EC2 relates all the concrete properties required for deflection prediction to the concrete grade and cement type. In practice, properties are influenced by the aggregate type, curing etc. Mean values should be used for the tensile strength and elastic modulus of concrete to obtain a best estimate of the actual deflection. 50

Concrete tensile strength Deflections in slabs depend significantly on the effective concrete flexural strength which governs the cracking moment. The flexural strength of concrete is calculated from the peak failure load of unreinforced concrete beams with engineers bending theory. EC2 defines the flexural strength of concrete as follows: f fl =(1.6-h/1000)f ctm >f ctm where fctm = 0.3fck 2/3 is the mean tensile strength of concrete which can be estimated indirectly in the splitting test. fck is the characteristic cylinder compressive strength of concrete. The flexural strength is greater than the tensile strength since the tensile stress distribution is not linear at failure as assumed in its derivation. 51

Flexural strength of concrete The flexural strength is greater due the assumptions implicit in its derivation as illustrated below. f ct Strain f fl >f ct Stress at peak load Stress assumed in calculation f fl 52

Concrete tensile strength In reinforced concrete structures the effective flexural strength of concrete is reduced by tensile stresses induced by restraint of shrinkage by reinforcement and restraining elements such as stiff columns and shear walls. It is conservative to use the tensile strength f ctm in deflection calculations. The How to Leaflet suggests that the design value concrete tensile strength for a low restraint layout is taken as the mean of the tensile and flexural strengths. 53

Long-term deflections Three additional factors must be considered in the long term calculation of deflections. 1.Loading 2.Creep 3.Shrinkage 54

Design loads: Permanent loads In concrete structures, deflections increase with time under sustained load. The greater part of the deflection normally occurs under sustained loads. Therefore, long-term deflections are calculated under a best estimate of the sustained load during the lifetime of the structure. The design load for calculating long-term deflections is the permanent load: Permanent load = G k + 2 Q k where G k = dead load Q k = imposed load Recommended values for 2 are: 0.3 for residential and offices, 0.6 for parking 0.8 for storage 55

Design loads: Frequent load Cracking is irreversible. Therefore, it is prudent to calculate longterm deflections using a modified flexural strength which corresponds to the worst cracking during the lifetime of the structure. The How to Leaflet suggests that the frequent load combination is used to calculate the deflection affecting cladding. The frequent load is given by: Frequent load = G k + 1 Q k Recommended values for 1 are: 0.5 for residential and offices, 0.7 for parking 0.9 for storage 56

Time dependent deformation Creep is the continuous deformation of a member under sustained load. Shrinkage consists of autogenous (due to hydration) and drying shrinkage. 57

Creep EC2 uses the effective modulus method to model creep in which creep is modelled as a delayed elastic strain. The creep strain at time t is given by: cc (t) = (t0) * where (t0) = the strain at the time of first loading t 0 and = /Ec(t0) where Ec(t0) is the elastic modulus of the concrete at time t 0. * = the true creep coefficient. = EC2 [E c (t 0 )/E c28 ] 58

E c EC2 defines the creep coefficient in terms of the 28 day tangent modulus of concrete, E c E c = 1.05 E cm where E cm = secant modulus = 22[f cm /10] 0.3 See table 3.1 MPA 59

EC2 Annex B or... Figure 3.1 60

Creep So total strain: (t) = (t0) (1+ *) = [/Ec(t0)] (1+ *) = /E ceff where E ceff = effective elastic modulus = E c (t0)/(1+ *) For practical purposes E ceff = E c28 /(1+ EC2 ) In practice, there are usually several loads placed at different times. In that case long term modulus, E LT for n serviceability loads, W i :- E LT = SW/{ (W 1 /E ceff,1 ) + (W 2 /E ceff,2 ) + (W 3 /E ceff,3 ) +.... +(W n /E ceff,n )} 61

Shrinkage induced curvature Shrinkage induces curvatures in asymmetrically reinforced sections that can increase deflections by as much as 25%. The reinforcement restrains the shortening of the member due to shrinkage which induces tension in the concrete. Consequently, the cracking moment is reduced. Tensile stress 62

Shrinkage induces a curvature that is given by: 1/rcs = cs e S/I where c = free shrinkange strain e = E s /E ceff S = Ase = the first moment of area of the reinforcement about the centroid of the transformed section I = second moment of area of the section EC2 extends the distribution coefficient approach to cover cracked sections by applying to S cr /I cr and (1- ) to S uncr /I uncr. 63

Accuracy of deflection calculations Many factors influence the accuracy of deflection calculations including: 1. actual loading relative to design loading 2. early age striking and loading from constructing slabs above 3. differences between actual and assumed material properties 4. Composite action between floor slabs and floor screeds and partitions 5. Temperature effects 64

Use of finite element analysis to calculate deflections Two approaches are commonly used: 1.Cracked section analysis in which the plate stiffness is reduced to account for cracking 2.Elastic analysis with reduced stiffness to allow for cracking creep and shrinkage. In this case, the effective E value can be taken as: E* ceff =0.5E c /(1+) where the factor of 0.5 accounts for the effects of cracking and shrinkage 65

Deflection The deflection may be calculated: Either by calculating the curvatures (due to load, shrinkage, creep) at a number of sections and then double integrating numerically Or by the simplified formula: δ = kl 2 (1/r) k depends on the shape of the bending moment diagram. Both methods are described in detail in How to design concrete structures using Eurocode 2: Deflection 66

Rigorous method 67

Rigorous method 68

Rigorous method 69

Rigorous method This is the approach used in the Rigourous RC Spreadsheets 70

Rigorous method TCC41R 71

Simpler method (outline) Essentially add curvature due to SLS moments: 1 1 1 ( 1 ) r r r sls. moments To curvature due to shrinkage: cracked, slsmoments 1 1 1 ( 1 ) So total curvature: 1 1 1 r r r uncracked, sls, moments r shrinkage r cracked, shrinkage r uncracked, shrinkage slsmoments Calculate deflection: shrinkage δ = kl 2 (1/r) k from chart depends on shape of BMD 72

Simpler method (in detail) 73

Simpler method 74

Simpler method 75

Simpler method 76

Simpler method 77

Deflection calculation example 78

d = 600 h = 700 Worked example Estimate the long-term deflection for the beam shown. Span = 9.5 m M QP = 200 knm Concrete class C25/30 A s = 2450 mm 2 x c = 329 mm I cr = 7976 x 10 6 mm 4 x u = 350 mm (ignoring reinforcement) I u = 8575 x 10 6 mm 4 (ignoring reinforcement) (,t0) = 2.8 ε cs = 470 x 10-6 d = 300 5 No H25 bars 79

Deflection calculation example Step 1 Calculate cracking moment M cr 0.9f ctmi h x f ctm = 2.6 MPa (Table 3.1) u u M cr 0.9 2.6 857510 700 350 57.3 knm 6 If uncracked section properties are used, M cr = 57.3 knm Section is cracked, therefore: ζ = 1 0.5(57.3/200) 2 = 0.95 80

Deflection calculation example Step 2 Calculate flexural curvature 1 M QP r u E c, eff I u E c,eff = E cm /(1 + (,t0) ) = 31 / (1 + 2.8) = 8.15 kn/mm 2 6 1 20010 6 2.8610 3 6 r u 8.1510 857510 M /mm 6 1 QP 20010 6 3.0810 3 6 r c E c, eff I c 8.1510 797610 /mm 1 r n 1 r c 1 (1 ) r 0.95 3.0810 u (1 0.95) 2.8610 3.0710 6 6 6 /mm 81

Deflection calculation example Step 3 Calculate shrinkage curvature 1 cs S e r I where: S c = A s (d x) = 2450 (600 329) = 664 x 10 3 mm 3 6 3 1 47010 (200/8.15) 66410 6 0.9610 /mm 6 r 797610 s S u = A s (d x) = 2450 (600 350) = 612.5 x 10 3 mm 3 6 3 1 47010 (200/8.15) 612.5 10 6 0.8210 /mm 6 r 857510 su 1 r n 1 r c 1 (1 ) r 0.95 0.9610 u (1 0.95) 0.8210 0.9510 6 6 6 /mm 82

Deflection calculation example Step 4 Calculate deflection Total curvature = 3.07 x 10-6 + 0.95 x 10-6 = 4.02 x 10-6 /mm For a simply supported slab, k = 0.104 δ = kl 2 (1/r) = 0.104 x 9500 2 (4.02 x 10-6 ) = 37.8 mm 83

Serviceability Limit States www.eurocode2.info 84