FEM-Design performs design calculations for reinforced concrete-, steel- and timber structures according to Eurocode. The following design considers EC2 (standard) and the National Annex (NA) for Denmark, Finland, Germany, Hungary, Romania, Norway, Sweden, Poland and United Kingdom.

Design forces

The design forces are the forces that the reinforcements should be designed for in the reinforcement directions. The term design forces have meaning only in surface structures like plate, wall or 3D plate. In beam structures the design forces are equivalent to the internal forces. The necessary reinforcement calculations are based on the design forces.

The way of calculating the design forces is common in all modules and in all standards.

In FEM-Design the design forces calculation is based on the mechanism of optimal reinforcement calculation for skew reinforcements made by M.P. Niel- sen, Wood-Armer and Dr. Ferenc Németh, see [2]. The following description will show the way of calculation for moments but the way of the calculation is the same for normal forces too. Just substitute the m signs with n and you will have the calculation for normal forces.

For the calculation of the design forces we have given:

  • ξ, η reinforcement directions,
  • α, β angle of global x direction and the ξ, η reinforcement directions,
  • mx, my, mxy internal forces.

The results will be the design moments:

  • 1543503447758-943.png

In the first step we are taking a ξ-ϑ coordinate system and transform the internal forces into this system:

1543503517958-895.png

Now the design forces will be chosen from four basic cases called a), b), ξ) and η). The possible design moment pairs of the cases:

a) case:

1543503652914-266.png

b) case:

1543503750827-594.png

ξ) case:

1543503838340-661.png

η) case:

1543503816725-369.png

From the four cases the one is invalid where:

  • the signs are different: mξ*mη < 0
  • the crack tensor invariant is less than the internal forces invariant:

1543503891591-532.png

The valid positive pair will be the design moment for bottom reinforcement; the valid negative pair will be the design moments for the top reinforcement (positive means positive and zero values; negative means negative and zero values).

So the result will be four values in a certain point: two moment values for each reinforcement directions. It can sound strange that the reinforcements are used for both positive and negative moment in one direction at the same time, but if we are looking at a plate where the mx is positive and the my is negative and the reinforcements have an angle of 45 degree to the x direction we could imagine that the bottom reinforcement bars make equilibrium to the mx and the top reinforcement bars make equilibrium to the my. So a certain reinforcement direction takes positive and negative loads at the same time.


Shrinkage as load action

In the Plate and 3D Structure modules the shrinkage behaviour of reinforced concrete slabs can be taken into consideration as load action. The program add this movement effect (specific rotation) calculated from the formulas written be- low to the structure as invisible load (one load case must be defined as Shrinkage type, see User’s Guide [1]).

Note: The shrinkage effect has to be used together with applied reinforcement.

The effect of the shrinkage for the surface reinforcement bars in one direction (here X) (it is also valid in other bar directions):

1543504152702-119.png

The specific normal force causing the given shrinkage value (εcs [‰] at concrete materials) in the concrete zone of the section is (here in X direction):

NX = EAc εcs [k&Nu;/m]

The position change of centre of gravity considering reinforcement bars (here X-direction; see dashed line):

1543504296967-109.png

where:   

n = Es / Eand Ss is the statical moment of (here) X-directional bars around the Y axis of the calculation plane.

The moment around the Y axis of the calculation plane from NX because of the position change of centre of gravity:

MY = NX zs

The specific rotation (curvature) from MY for 1 meter wide section:

1543504426046-612.png


Design calculations for surface structures

Ultimate limit state

The design of the slab is performed with respect to the design moments described in Design forces.

In order to minimize cracking in the slab a good way is to reinforce according to the elastic moments which normally also leads to good reinforcement economy. The required bending reinforcement is designed according to EC2 3.1.7, where a rectangular stress distribution as shown below has been assumed.

1543932761308-583.png

λ = 0,8                                              for  fck ≤ 50 MPa

λ = 0,8 - (fck - 50)/400                     for  50 < fck ≤ 90 MPa

and:

η = 1,0                                              for  fck ≤ 50 MPa

η = 1,0 - (fck - 50)/200                     for  50 < fck ≤ 90 MPa

 

If the current moment is larger than the moment representing balanced design, compression reinforcement will be provided if allowed by the user otherwise an error message will be displayed. If the spacing regulations for the reinforcement are exceeded before adequate moment capacity can be reached a warning mes- sage will be displayed.

Note, that the required bending reinforcement is at design level primary not affected of the presence of user defined reinforcement. However, when user defined applied reinforcement is selected the stiffness will be effected, which in most cases will influence the moment distribution and thus secondary the required bending reinforcement.

Shear capacity

The shear capacity is calculated according to EC2 6.2.2 and 6.2.3 considering applied bending reinforcement when the option Checking has been selected. Otherwise, the required bending reinforcement according to this chapter. The design criteria for the shear capacity is:

VSd < VRd1
 

where:   

VSd is the design shear force;

VSd = Q, which is calculated as  1543933319172-750.png

VRd is the shear capacity.

If the section in which the shear force is acting has an angle with respect to the reinforcement directions the shear capacity is calculated as: 1543933343993-360.png

Punching

The punching capacity is calculated according to EC2 6.4.3 - 6.4.5.
 

Checking

Punching without shear reinforcement

A concrete compression check on u0 is made according to 6.4.5 (6.53). A concrete shear check on u1  is made for a capacity calculated according to 6.4.4 (6.47).

Punching with shear reinforcement

A concrete compression check on u0 is made according to 6.4.5 (6.53). Reinforcement is calculated with regard to critical perimeters u1, u2, ... unReinf  according to 6.4.5 (6.52 ).

(ui are control perimeters above the reinforced region, distance between them is ”Perimeter distance”, defined in the calculation parameter).

A concrete shear check on uout is made for a capacity calculated according to 6.4.4 (6.47)

(uout is either the first perimeter that does not need reinforcement, or if it is not found, the perimeter that is k deff distance from the outer perimeter of the reinforcement).

Warnings

A warning message is shown, if reinforcement does not comply with the detailing rules in 9.4.3.

Design

  1. Check, if reinforcement is needed at all,
  2. If reinforcement is needed, it is designed to satisfy the detailing rules in 9.4.3, if possible,
  3. Design fails and a warning message is displayed if, uout is not found within 6 deff distance from the column perimeter.

Comments, limitations

- openings are not considered when control perimeters are generated.
- the position of the column relative to the plate is considered only in the generation of control perimeter. It means, (user or program defined) reinforcement may be partly out of the plate, but it won't affect the calculation.
- If ”Calculate β automatically” is set in the calculation parameter, β is calculated according to equation 6.4.3 (6.39).

Serviceability limit state

Method of solution

The program performs crack- and deflection control for all load combinations according to EC2 7.3 and 7.4. Two limiting conditions are assumed to exist for the calculations: Stadium I (the uncracked condition) and Stadium II (the fully cracked condition).

Stadium I Uncracked condition

If the user does not activate the option Cracked section analysis, the calculation will be performed with respect to the total stiffness of the slab.

Stadium II Fully cracked condition

If the option Cracked section analysis, is activated the program will consider the decrease in slab stiffness on behalf of cracking. This means an iterative calculation where the slab in the beginning is assumed to be uncracked when the section forces are calculated. Sections which are not loaded above the level which would cause the tensile strength of the concrete to be exceeded will be considered to be uncracked (Stadium I). Sections which are expected to crack will behave in a manner intermediate between the uncracked and fully cracked conditions and an adequate prediction of behavior used in the program is shown below.

The stiffness calculation is performed considering the required or the applied reinforcement depending on what option has been selected. If applied reinforcement has been selected this is used in all load combinations. If applied reinforcement is not present or not selected the required reinforcement is used instead. In the latter case the required reinforcement in every element is calculated as the maximum value from all load combinations, which means that all calculations of serviceability limit values are performed with the same reinforcement.

In the next step a new calculation based on the new stiffness distribution is performed and so on. When the deflection values resulting from two calculations does not differ more than a defined percentage of the first one or the maximal number of allowed calculations has been reached the calculation is stopped.

Crack width

Crack width is according to EC2 7.3.4 calculated as:

wk = Sr,maxsm - εcm

where:

Sr,max is the maximum crack spacing,

εsm is the mean strain in the reinforcement under the relevant com- bination of loads, including the effect of imposed deformations and taking into account the effects of tension stiffening. Only the additional tensile strain beyond the state of zero strain of the concrete at the same level is considered,

εcm is the mean strain in the concrete between cracks.

εsm - εcm may be calculated from the expression:

1557391677004-144.png

where:

σs is the stress in the tension reinforcement assuming a cracked section. For pretensioned members, σs may be replaced by &Delta;σp the stress variation in prestressing tendons from the state of zero strain of the concrete at the same level,

αe= is the ratio Es / Ecm

Ap and Ac,eff are as defined in 7.3.2 (3),

ξ1 according to Expression (7.5),

kt is a factor dependent on the duration of the load. 

kt = 0,6 for short term loading

kt = 0,4 for long term loading. For long term loads (kt = 0,4):

Ap´ = 0,0 (pre or post-tensioned tendons)

Ac,eff:

1557392170606-753.png

1557392181561-817.png

hcef = min (2,5 * (h - d), (h - x) / 3, h / 2 )

sr,max = k3c + k1 k2 k4φ /ρp,eff

where:

φ is the bar diameter. Where a mixture of bar diameters is used in a
section, an equivalent diameter, φeq, should be used. For a section with n1 bars of diameter φ1 and n2 bars of diameter φ2, the following expression should be used,

1557392495138-639.png

is the cover to the longitudinal reinforcement,

k1 is a coefficient which takes account of the bond properties of the bonded reinforcement:

k1 = 0,8 for high bond bars,

k1 = 1,6 for bars with an effectively plain surface (e.g. prestressing tendons),

k2 is a coefficient which takes account of the distribution of strain:

k2 = 0,5 for bending,

k2 = 1,0 for pure tension,

k2 = (ε1 + ε2) / 2ε1

where ε1 is the greater and ε2 is the lesser tensile strain at the boundaries of the section considered, assessed on the basis of a cracked section.
Recommended values of k3 = 3,4 and k4 = 0,425 are used.

Maximum crack spacing:

sr,max = 1,3 (h - x)

Equivalent quantities perpendicular to crack direction:

  • Reinforcement area:

1557393319443-687.png

  • Number of bars:

1557393330997-858.png

  • Diameter:

1557393343977-995.png

Deflections

The calculations is performed according to EC2 7.4.3.

Stadium I Uncracked condition

Load depended curvature is calculated as:

1 / rf = M / Ec,ef I1

where:

M is current moment,

I1 is Moment of Inertia in Stadium I,

Ec,ef is the modulus of elasticity with respect to creep.

The modulus of elasticity is calculated as:

Ec,eff = Ecm / (1 + φ)

where φ is the creep coefficient.

Curvature with respect to shrinkage is considered according to 2.2.2 above.

Stadium II Fully cracked condition

Load depended curvature is calculated as:

1 / rf = M / Ec,ef I2

where:

Ec,ef is the modulus of elasticity as shown above,

I2 is the moment of Inertia in Stadium II,

M is current moment.

Curvature with respect to shrinkage is considered according to 2.2.2 above.

Sections which are expected to crack will behave in a manner intermediate between the uncracked and fully cracked conditions and an adequate prediction of this behavior is given by:

α = ζ αII + (1 - ζ) αI

where:

α is in this case the curvature calculated for the uncracked and fully cracked conditions,

ζ is a distribution coefficient given by ζ = 1 - β (σsr / σs)2

ζ is zero for uncracked sections,

β is a coefficient taking account of the influence of the duration of the loading or of repeated loading on the average strain,

σs is the stress in the tension steel calculated on the basis of a crack- ed section,

σsr is the stress in the tension steel calculated on the basis of a cracked section under the loading which will just cause cracking at the section being considered.

Note that stresses and moment of inertia are calculated with applied reinforcement if it is selected, otherwise with required reinforcement.


Design calculations for bar structures

Material properties

Concrete

1557394141236-821.png

  • Ultimate limit states:
    Continuous line is used.
     
  • Servicibility limit states:
    Stage II is used (dashed line, without horizontal section).

Steel

1557394271242-563.png

  • Ultimate limit states:
    B graph with horizontal line is used.
     
  • Servicibility limit states:
    The same as ultimate but without safety factor.

Longitudinal reinforcement

Analysis of second order effects with axial load

According to EC2 5.8.

  • For calculation of 2nd order effect Nominal curvature method (5.8.8) is used.
  • If there is no compression force in the section the eccentricity is equal to 0,0.
  • Buckling lengths l0x and l0y are specified by the user.
  • Curvature:

1 / r = kr kϕ 1 / r0

where:

kr is a correction factor depending on axial load,

κϕ is a factor for taking account of creep,

1 / r0 = εyd / (0,45 d),

d is the effective depth,

d = (h / 2) + is

where is is the radius of gyration of the total reinforcement area.

kr = (nu - n) / (nu - nbal) ≤ 1

where:

n = NEd / (Ac fcd), relative axial force,

NEd is the design value of axial force, nu = 1 + ω,

nbal is the value of n at maximum moment resistance; the value 0,4 is used,

ω = As fyd / (Ac fcd),

As is the total area of reinforcement,

Ac is the area of concrete cross section,

kϕ = 1 + β ϕef ≥ 1

where:

ϕef is effective creep ratio, defined by the user,

β = 0,35 + fck /200 - λ / 150,

λ is the slenderness ratio.

  • 2nd order effect is ignored, if:

λ ≤ λlim

λlim = 20 A B C / √n

where: A = 1 / (1 + 0,2 ϕef),

B = √1 + 2 ω,

C = 0,7

ϕef is effective creep ratio,

ω = As fyd / (Ac fcd), mechanical reinforcement ratio,

As is the total area of longitudinal reinforcement,

n = NEd / Ac fcd), relative normal force,

rm = M01 / M02, moment ratio,

M01, M02 are the first order en moments |M02| ≥ |M01|.

  • Geometric imperfection (5.2 (7) a):

ei = l0 / 400

  • The minimum of all eccentricities (1st order + imperfection + 2nd order effect): max (20,0; h / 30,0).
  • Imperfection and 2nd order effect considered in both directions.
  • The eccentricity is calculated in four possible positions:
    • Stiff direction+, weak direction+
    • Stiff -, weak+
    • Stiff+, weak-
    • Stiff-, weak-

Torsion

  • Necessary longitudinal reinforcement area (Asl):
    TEd is the applied design torsion (see Figure 6.11):

1557398781245-600.png

The required cross-sectional area of the longitudinal reinforcement for torsion &Sigma;Asl may be calculated from:

1557398836470-691.png

where:

uk is the perimeter of the area Ak,

fyd is the design yield stress of the longitudinal reinforcement Asl,

θ is the angle of compression struts, θ = 45 deg.

Considering torsion in calculation of longitudinal bars:

Calculation of torsional capacity by edges, considering all bars placed in tef strip. The minimum of capacities gives the torsional capacity of the section. Utilization for torsion calculated for all bars placed in the strip one by one.

Area of these bars decreased in the calculation of axial effects (N, My, Mz) in proportion of utilization (see formula below):

1557399352673-816.png

where:

A is area of the bar,

A’ is decreased area used in calculation

ULS checking

1557399421450-202.png

SLS checking

Crack width calculated according to EC2 7.3.

• Crack width calculated as:

wk = sr,maxsm - εcm)

where:

sr,max is the maximum crack spacing,

εsm is the mean strain in the reinforcement under the relevant combination of loads, including the effect of imposed derforma- tions and taking into account the effects of tension stiffening. Only the additional tensile strain beyond the state of zero strain of the concrete at the same level is considered,

εcm is the mean strain in the concrete between cracks.

εsm - εcm may be calculated from the expression:

1557399634681-893.png

where:

σs is the tension reinforcement assuming a cracked section,

αe is the ratio Es / Ecm

ρp,eff = As / Ac,eff,

Ac,eff is calculated as below,

kt is a factor dependent on the duration of the load,

kt = 0,6 for short term loading,

kt = 0,4 for long term loading (always supposed by the program),

Ac,eff:

1557400792364-457.png

hc,ef = min (2,5 (h - d), (h - x) / 3, h / 2)

sr,max = k3 c + k1 k2 k4 φ / ρp,eff

where:

φ is the bar diameter. Where a mixture of bar diameters is used in a section, an equivalent diameter, φeq, should be used. For a section with n1 bars of diameter φ1 and n2 bars of diameter φ2, the following expression should be used,

1557402498643-546.png

c is the cover to the longitudinal reinforcement,

k1 is a coefficient which takes account of the bond properties of the bonded reinforcement:

k1 = 0,8 for high bond bars,

k1 = 1,6 for bars with an effectively plain surface (e.g. prestressing tendons),

k2 is a coefficient which takes account of the distribution of strain:

k2 = 0,5 for bending,

k2 = 1,0 for pure tension

For cases of eccentric tension or for local areas, intermediate values of k2 should be used which may be calculated from the relation:

k2 = (ε1 + ε2) / 2 ε1,

where: ε1 is the greater and ε2 is the lesser tensile strain at the boundaries of the section considered, assessed on the basis of a cracked section.

Recommended values of k3 = 3,4 and k4 = 0,425 are used.

• Maximum crack spacing:

sr,max = 1,3 (h - x)

Space between bars

  • Minimum distance:
    The clear distance (horizontal and vertical) between individual parallel bars or horizontal layers of parallel bars should be not less than the maximum of k1 bar diameter, (dg + k2 mm) or 20 mm where dg is the maximum size of aggregate.
  • Maximum distance:
    The longitudinal bars should be so arranged that there is at least one bar at each corner, the others being distributed uniformly around the inner periphery of the links, with a spacing not greater than 350 mm.

Lengthening and anchorage

  • Because of shear effect (shift rule):

ai = 0,9 max (h, b)

The code prescribes d instead of h, but the difference can be ignored.

  • Anchorage:

fbd = 2,25 η1 η2 fctd 

where:

fctd is design value of concrete tensile strength. Due to the increasing brittleness of higher strength concrete, fctk,0,05 should be limited here to the value for C60/75, unless it can be verified that the average bond strength increases above this limit

η1 is a coefficient related to the quality of the bond condition and the position of the bar during concreting:

η1 = 0,7

η2 is related to the bar diameter:

η2 = 1,0 for φ ≤ 32 mm,

η2 = (132 - φ) / 100 for φ > 32 mm

lb,rqd = (φ / 4) (σsd / fbd)

where:

σsd = fyd (fully utilized bar supposed),

lbd = α1 α2 α3 α4 α5 lb,rqd ≥ lb,min,

αi = 1,0

lb,min is the minimum anchorage length if no other limitation is applied:

  • for anchorages in tension:
    lb,min > max (0,3 lb,rqd; 10 φ; 100 mm),
     
  • for anchorage in compression:
    lb,min > max (0,6 lb,rqd; 10 φ; 100 mm),
    Rule given for compression is used.

Stirrups

Shear

In Figure 6.5 below the following notations are shown:

α is the angle between shear reinforcement and beam axis perpendicular to the shear force (measured positive as shown in Figure 6.5),

θ is the angle between the concrete compression strut and the beam axis per- pendicular to the shear force,

Ftd is the dessign value or the tensile force in the longitudinal reinforcement,

Fcd is the design value of the concrete compression force in the direction of the longitudinal member axis,

bw is the minimum width between tension and compression chords,

z is the inner lever arm, for a member with constant depth, corresponding to the bending moment in the element under consideration. In the shear analy- sis of reinforced concrete without axial force, the approximate value z = 0,9 d may normally be used.

1557404124068-157.png

  • Member do not require shear reinforcement, if:

The design value for the shear resistance VRd,c is given by:

VRd,c = [CRd,c k (100 ρl fck)1/3 + k1 σcp] bw

with a minimum of:

VRd,c = (vmin + kσcp) bw

where: fck is in MPa

1557404308626-814.png

Asl is the area of the tensile reinforcement, which extends: ≥ (lbd + d) beyond the section aonsidered (see Figure 6.3), bw is the smallest width of the cross-section in the tensile area [mm],

σcp = NEd / Ac < 0,2 fcd [MPa],

NEd is the axial force in the cross-section due to loading or prestressing [in N] (NEd > 0 for compression). The influence of imposed deformations on NE may be ignored,

Ac is the area of concrete cross section [mm2],

VRd,c is [N]

The recommended value for CRd,c is 0,18 / γc, that for vmin is given by the expression below and that for k1 is 0,15.

1557404927126-189.png

1557404943122-538.png

  • Upper limit of shear:

VRd,max = αcw bw z ν1 fcd / (cotθ + tanθ) 

where:

Asw is the cross-sectional area of the shear reinfocement,

s is the spacing of the stirrups,

fywd is the design yield strength of the shear reinforcement,

ν1 is a strength reduction factor for concrete cracked in shear,

αcw is a coefficient taking account of the state of the stress in the compression chord.

The recommended value of ν1 is ν (see expression below). The recommended value of αcw is as follows:1 for non-prestressed structures,

1557405487679-357.png

Capacity of stirrups:

1557405540899-916.png

where:

Asw is the cross-sectional area of the shear reinforcement,

s is the spacing of the stirrups,

fywd is the design yield strength of the shear reinforcement.

Torsion

TEd is the applied design torsion (see Figure 6.11)

1557405627044-695.png

Ak is the area enclosed by the centre-lines of the connecting walls, including inner hollow areas,

τt,i is the torsional shear stress in wall i,

tef,i is the effective wall thickness. It may be taken as A/u, but should not be taken as less than twice the distance between edge and center of the longitudinal reinforcement. For hollow sections the real thickness is an upper limit,

A is the total area of the cross-section within the outer circumference, including inner hollow areas,

u is the outer circumference of the cross-section,

zi is the side length of wall i defined by the distance between the intersection points with the adjacent walls,

θ = 45 deg, in all calculations.

  • Member do not require torsional reinforcement, if:

TRd,c = fcd tef 2 Ak ≤ TEd

  • Upper limit of torsion:

TRd,max = 2 ν αcw fcd Ak tef,i sinθ cosθ 

where ν and αcw are as above.

  • Force in stirrups:

TRd,max=2ναcwfcdAktef,isinθcosθ

The shear force VEd,i in a wall i due to torsion is given by:

VEd,i = τt,i tef,i zi

zi is section height used to be able to sum with shear.

  • Capacity of stirrups:

See Shear.

Shear and torsion

  • Forces in stirrups:

VEd = VEd(shear) / 2 + VEd(torsion)

  • No stirrup required:

TEd / TRd,c + VEd / VRd,c ≤ 1,0 

  • Upper limit of the effects:

TEd / TRd,max + VEd / VRd,max ≤ 1,0 

  • Calculation is done in two directions y' and z' independently.

 

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Created by IwonaBudny on 2018/11/29 15:51
   
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