DIN en 15512 Worked Example1

It is assumed that: -

Aeff   has been determined by stub column tests (EN 15512, A.2.1) and compression tests on uprights (EM 15512, A.2.2) and therefore includes the effects of local buckling, distortional buckling.

-

W eff,y, W eff,z  have been determined by bending tests (EN 15512, A,2.9), so that they contain the effects of local buckling, distorsional buckling, lateral torsional buckling and flange curling.

However it has to be noted, that e.g. the resistance against distorsional bucking depends on the bracing node distance and the resistance against lateral torsional buckling on the effective torsional length. So care should be taken that the situation in the tests corresponds sufficiently with the situation in the real structure. For a frame with irregular bracing it might be useful/necessary to have different effective cross section properties e.g. depending on the distance of the bracing nodes. -

+

The values W eff,z and W eff,z refer to compression due to bending at the front side or at the rear  flanges. Usually these values are not identical. However, often only the more unfavourable situation is tested and the result of these tests is then used for bending in both directions.

4.3

Bracing member 

The bracing member is a cold formed U-profile with the main dimension 50/30/1,5 mm. The effective cross section values with respect to local buckling have been determined by ERF analytically according to EN 1993-1-3. The following values have been given: Table 3 Cross section

Bracing

properties

U 50/30/1,5

Steel f y

S235 JR  

kN/cm

2

23,5

M

1,00

 Aeff 

cm

2

Iy,g

cm

4

6,00

Iz,g

cm

4

1,20

It,g

cm

4

0,0110

Iw,g

cm

6

5,00

y0,eff 

cm

1,80

yS,eff 

cm

0,60

ey

cm

+ W z,eff 

1,40

Figure 3 M y0 ey

yS

z

y

0,90

cm

3

+ MRd,z,eff 

cm

4

g

kN/m

0,553 13,0 0,011

In contrast to the upright the shift of the center of gravity of the effective cross section is here taken into account by the values y S,eff¸ y0,eff .

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4.4

Beam

Typical composed box beams used in pallet racking are asymmetric with respect to the horizontal and vertical axes. So the main axes are often slightly inclined (< 5°). In the static calculations this effect usually can be neglected. The cross section values in Table 4 refer to the horizontal and vertical axes. Table 4 Cross section

Beam

properties

Box 150/50/1,5

Steel f y

S235 JR  

kN/cm

2

23,5

M  A Iy Iz W y,eff  W z,eff 

1,00 cm

8,00

cm

4

260

cm

4

24,0

cm

3

39,6

cm

3

6,38

MRd,y

 

930,0

MRd,z

 

150,0

kN/m

0,063

g

Figure 4

y

z

It is assumed, that W y,eff  and W z,eff  have been determined by testing, so that possible effects of local buckling, distorsional buckling not to be considered in the member checks. It is further assumed 1

that tests showed that the beam is not subject to effects of lateral torsional buckling .

1

Usually lateral torsional buckling is not critical for this type of box beam. However, depending on the height to width ratio of the profile, the number of welds which connect the channel sections the over the length of  the beam, and other parameters lateral torsional buckling might not be negligible.

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4.5

Beam end connector 

The following values for the bending strength M Rd, rotational stiffness k d, shear strength S Rd and

looseness ϕℓ  are given: Table 5  Connector

H4

Figure 5 

properties MRd,1

kNcm

230

kd,1

kN/rad

10000

MRd,2

kNcm

200

kd,2

kN/rad

12500

MRd,3

kNcm

170

kd,3

kN/rad

13000

VRd,up

kN

5,50

VRd,down

kN

18,0

ϕℓ

rad

0,00

upwards bending

downwards bending

 All the values are determined by tests according to EN 15512 A.2.4 to A.2.6. For the bending strength and stiffness three pairs of values are specified. That means that the stiffness has been evaluated from the test curves not only for the maximum bending strength MRd,1   but also for  reduced values MRd,i = ηi ⋅ MRd,1 (ηi  < 1, see EN 15512, A.2.4.5.1). The values MRd,i  and kd,i  relate to downwards bending. For upwards bending no values are given, since in this example upwards directed moments only appear at unloaded beams and are relatively small. According to EN 15512, 9.5.1 for upwards bending separate values for M Rd  and kd  need not to be used in the analysis as long as these values are at least 50% of the values under downwards bending. VRd,down is the resistance against failure of the connector under a downwards directed shear force. VRd,up   is the resistance against disengaging of the connector under an upwards directed shear  force. VRd,up   is determined by tests according to EN 15512, A.2.6. This value is usually limited by the shear strength or the fit of the safety lock in upright and connector.

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4.6

Partial restraint of the upright base

The partial restraint of the upright base usually depends on the contact pressure between upright base and slab. Rotational stiffness and bending strength are determined for different compressive forces in the upright by tests according EN 15512, A.2.7.2. The results are given in Table 6 and in the chart below. The values relate to bending about the y-axis of the upright. Table 6  N

kd

MRd

kN

kNcm/rad

kNcm

30

7800

159,6

40

10400

208,0

50

13000

252,5

60

15600

292,1

70

20400

326,2

80

25100

354,0

90

29900

375,1

100

34600

389,0

110

39400

395,5

120

44100

394,5

130

48900

385,9

140

53600

369,9

60000

450 400

50000 350      ] 40000      d     a     r      /     m     c 30000      N      k      [

300

     k 20000

150

     ]     m 250     c      N      k      [

200

     d

     d      R

     M

100 10000 50 0

0 0

20

40

60

80

100

120

140

160

N [kN]

For intermediate values of N, the values M Rd  and kd  can be determined by linear interpolation.

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4.7

Shear stiffness of the upright frames

The shear stiffness SD  is the ratio of the shear force in the frame V and

the angle of the shear deformation θ resulting from V, i.e. S D = V / θ. The

Figure 6  V

shear stiffness is determined by tests according to EN 15512, A.2.8. It is used in the global analysis to determine the second order effects on the inner forces due to global imperfection in cross-aisle direction.

θ

In the common software used for structural analyses, the shear stiffness SD  of a frame cannot directly be defined. In this case the axial stiffness of  the bracing members must be reduced. This can be done e.g. by using a reduced cross section area of the bracing members or by applying longitudinal springs at the ends of the bracing members. For this example

V

the latter case was chosen. The table below contains the values of the spring stiffness depending for two angles of inclination of the bracing.

α [°]

Cv  [kN/cm]

30

57,5

60

47,5

In the present example the spring only needs to be applied at one end of 

Figure 7 

α

bracing members as shown in Figure 7.

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5. Actions 5.1

Dead load

Self-weight of frame bracing per upright (see annex A): Gbr 

= 0,205 kN/2

= 0,10 kN

Self-weight of one upright Gupr 

= H ⋅ g = 10,164 m ⋅ 0,055 kN/m

= 0,56 kN

Self-weight of one beam: Gbeam = L ⋅ g = 3,60 m ⋅ 0,063 kN/m ⇒

= 0,23 kN

Dead load per inner upright inner frames G=

0,1 + 0,56 + 5 ⋅ 0,23

= 1,81 kN

In the global analysis the dead load of the members being part of the static model is generated automatically by the software.

5.2

Unit loads

5.2.1

Loading of the beams

In the present example the loading on the beams can be distributed uniformly over the length L of  the beam (see EN 15512, 9.4.2). It is assumed that -

the centre of gravity is in the middle of the base area of the pallet

-

the placement imperfection in cross-aisle direction is not

Figure 8 

systematic -

2Q

the placement tolerance is max. 50 mm.

So the most unfavourable loading situation would be as shown in Figure 8. According to EN 15512, 6.3.2 the influence of the

50

placement tolerance on the actual beam load can be neglected, if  Qe

Qe / Q ≤ 1,12 Here:

1100 1200

Qe / Q = (1200 - 50) / (1100 - 50) = 1,095 ≤ 1,12 ⇒

Placement imperfection can be neglected.

So the load to be applied on the beam is Qbeam  = 4 x 850 kg ⋅ 9,81 m/s

2

⋅

-3

10 / 2 = 16,67 kN

Uniformly distributed over the beam length this becomes: qbeam  = 16,67 / 3,60 m = 4,63 kN/m

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5.2.2

Bay load for check of down-aisle stability

For this example is specified, that for the check of the global down-aisle stability the average compartment load can be assumed as 80% of the maximum compartment load. The single upright frame and the beams however have to be calculated for the full load.

Figure 9

That means that the bending moments in uprights and the moment at the beam end connectors arising from the global

4,63 kN/m

imperfection in down-aisle direction are calculated for the situation, that the beams are 80% loaded. But the moments in the uprights due to pattern loading and the moment at the beam

3,34 kN

end connectors due to bending of the beam and all other inner 

3,34 kN

forces have to be determined for the full beam load. In order to combine these requirements in the global analysis, the loads on the beams are applied with the maximum value, but the axial force in the uprights is reduced by 20% by means of upwards directed single loads at each beam level with the value of  0,2 ⋅ 16,67 kN = 3,34 kN (see Figure 9). The axial forces of the uprights have to be corrected later “manually” to 100% loading for the member  checks.

5.2.3

Pattern loading

For the upright design: For a regular rack layout, it is usually sufficient, to consider the fully loaded rack structure with the exception a single unloaded compartment near the middle of the structure at the lowest beam level (EN 15512, 10.2.2.2, figure 27 a)). However for the present example an irregular braced upright frame is specified. In this frame the buckling lengths of the upright are at the top of the frame are significantly greater than at the bottom, which means in reverse, that the compressive strength of  the upright is significantly smaller at the top of the frame than at the bottom. So the unloaded compartment in the lowest beam level is not inevitably the most critical situation. It might be necessary to consider pattern loading also in the higher beam levels. Never the less in the global analysis here only the case of an unloaded compartment in the lowest beam level will be considered. The problem of pattern loading at the higher beam levels will be treated in conjunction with the member checks of the uprights in section 10. For the beam design: For a regular rack layout, pattern loading at the top beam level with one loaded beam between to unloaded beams can be deemed to be most critical. Pattern loading for the beam design will be treated separately from the global analysis in conjunction with the member checks.

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5.2.4

Centre of gravity of the unit loads

The forces resulting from unit loads are applied in the static model directly at the beam levels. The vertical distance between the beam level and the centre of gravity of the unit loads usually can be neglected. For the design under seismic action an exacter approach can be necessary.

5.3

Placement load

5.3.1

Horizontal placement load in down-aisle direction

The horizontal placement load shall be applied in down-aisle and cross-aisle direction but not simultaneously in both directions (EN 15512, 6.3.4.1).  At racks over 6 m in height a horizontal placement load of either  -

0,25 kN at the top of the rack or 

-

0,50 kN at any height up to 3 m

has to be applied (see EN 15512, 6.3.4.2).  According to section 6.3.4.3 of EN 15512 the horizontal placement load in down-aisle direction only needs to be applied at beam levels. 5.3.2

Horizontal placement load in cross-aisle direction

In the cross-aisle direction the most unfavourable position for the placement load is considered as a) the top of the upright frame b) midway between to bracing nodes of the upright frame lattice c) the mid-span of a beam, where the placement load can be distrusted equally on both beams of  a compartment. The load can be assumed acting in the neutral axis of the beam. (see EN 15512 , 6.3.4.4)  As already set forth above, for the rack structure in this example section 6.3.4.2. of EN 15512 only requires placement loads at the top of the frame and below 3 m. However, because of the nonuniform spacing of the bracing, it seems to be rational, that the effect of horizontal placement load acting midway between the bracing nodes should be considered at any height. For positions between 3 m and 6 m the horizontal placement load will be determined by linear interpolation between 0,50 kN and 0,25 kN. The bending moments in the uprights resulting from the placement loads will be estimated on the safe side in conjunction with the member checks. 5.3.3

Vertical placement load

The compartments are designed for 4 unit loads. It is specified that the fork lift trucks are handling maximum one unit load simultaneously. So according to EN 15512, 6.3.3 vertical placement loads need not be taken into account.

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5.4

Actions due to impact

5.4.1

Horizontal loads

 According to EN 15512, 6.4.3 a horizontal load shall be applied on the aisle side upright at the most unfavourable height between floor and 400 mm of either: -

1,25 kN in down-aisle direction or 

-

2,50 kN in the cross-aisle front to back direction

The horizontal load is assumed to occur not simultaneously in both directions. 5.4.2

Vertical loads

If goods are placed with manually operated mechanical equipment (e.g. fork lift trucks), rack components above a unit load shall be able to absorb an accidental upwards directed vertical force of 5,0 kN (EN 15512, 6.4.2). This force will usually be applied at the beam end, to verify that connector does not disengage. Following the standard, the force would consequently also have to be applied at mid-span of the beam. However in comparison with the compartment load this will not be critical here.

5.5

Actions due to installation

Not specified here.

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6. Imperfections 6.1

Global Imperfections

6.1.1

Down-aisle direction

 According to EN 15512, 5.3.2 an imperfection in the form of an inclination of the upright from vertical shall be taken into account. The angle of the inclination is:

ϕ = ϕs + ϕℓ ≥ 1/500 In this is:

ϕs

the maximum specified out-of-plumb divided by the frame height. The maximum out-of-plumb of any upright in any direction shall not exceed H/350 measured in the unloaded condition 2

immediately after installation (EN 15512, 8.5.7.2 ). However, depending on the operational method smaller values might be required. Maximum permissible values for the out-of-plumb are also defined in EN 15620, tables 2, 7 and 11.

The designer may specify smaller values of ϕ s, if correspondingly smaller installation tolerances can be achieved in practice.

ϕℓ

looseness of the connector, determined by tests according to A.2.5 of EN 15512.

So with the values given in section 2 of this document:

ϕ = 1/350 + 0 = 1/350 6.1.2

Cross-aisle direction

 As in down-aisle direction an imperfection in the form of an inclination of the upright from vertical shall be taken into account. However, for upright stabilized by bracing systems the angle of the inclination is given in to EN 15512, 5.3.3.2 as:

ϕ = √(0,5 + 1/nf ) ⋅ 2 ϕs but

ϕ ≤ 2 ϕs ϕ ≥ 1/500 Herein is

ϕs

the specified out-of-plumb divided by the frame height as described in section 7.1.

nf 

the number of upright frames connected together 

 A single run of racking is assumed. Thus with the values given in section 2 of this document:

ϕ = √(0,5 + 1/nf ) ⋅ 2 ϕs = √(0,5 + 1/1) ⋅ 2 1/350 = 1/175

2

It is not quite clear, what is exactly meant by „any direction“: The limits for the out-of-plumb for the crossaisle and down-aisle direction Cx  and Cz  in EN 15620, table 2 are defined independently of each other. If  any direction means “cross-aisle, down-aisle and any direction in-between”, this would be a stricter limit for  the out-of-plumb than in EN 15620!

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Since the upright frames are usually asymmetric, it might be necessary to consider the effect of the imperfection in both directions.

6.2

Local imperfections

6.2.1

Bracing imperfections

Local bracing imperfections according to EN 15512, 5.3.3.3 usually need not to be applied at typical upright frames in cross-aisle direction. Local bracing imperfections become more critical, the more uprights have to be stabilized by one bracing system. This might be e.g. in racking with bracing in down-aisle direction, especially for the design of the horizontal bracing. 6.2.2

Member imperfections

Member imperfections are usually applied as bow imperfection and increase the inner moments of  an element under compression at mid-span and/or at its ends (depending on the support conditions). The member imperfections usually are applied for two reasons. -

as a substitute for the check of a member with respect to flexural buckling. In conjunction with a second order analysis, only needs to be checked that the stresses do not exceed the design value for the yield stress.

-

for structures with moment resisting joints, where the moments at the joints are sensible to local imperfections of the connected members under compression. 3

 According to EN 15512, 5.3.5 member imperfections are generally not required . However for very unusual rack configurations e.g. with extreme slenderness of the uprights and small number of  connectors stabilizing the system, it might be useful to take account of member imperfections.

6.3

Application of global imperfections in the analysis

Global imperfections can be taken into account -

geometrically (horizontal shift of the nodes of the static model)

-

by substitute forces (see e.g. EN 1993-1-1, figure 5.3 and 5.4) perpendicular to an element with axial load

In case of imperfections as described in section 7.1 and 7.2.1 both methods lead to the same result. The software used for this example allows to define imperfections as an individual load case. For  each load combination the axial forces are calculated first and then the substitute forces are determined and adjusted for each further iteration step.

3

In this point EN 15512 deviates from other standards dealing with the design of steel structures (see e.g. EN 1993-1-1, 5.3.2 (6))

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7. Load combinations 7.1

General

The required load factors are given in table 2 of EN 15512. In the following table the loads are assigned to the relevant load factors. The values in brackets refer to a favourable effect of the loads. Table 7   Action

type of action

Ultimate limit state

Serviceability

permanent or 

accidental load

variable

situation

limit state

load situation

Dead load Unit load Placement load Impact

permanent variable variable accidental

γG = 1,3 [1,0] γQ = 1,4 [0,0] γQ = 1,4 [0,0] --

γGA = 1,0 γQA = 1,0

1,0

--

--

γ A = 1,0

--

1,0

The rules for load combinations are given in EN 15512, 7.2:

ΣγG  G k + γQ Q k,1 ΣγG  G k + 0,9 Σ γQ,i  Q k,i ΣγGA  G k + Σ γQA,i  Q k,I + γ A A k where Gk

characteristic value of permanent action

Qk,1 characteristic value one of the variable loads Qk,i characteristic value of a variable loads  Ak

characteristic value of an accidental load

7.2

Global analysis in down-aisle direction

The combination rules and the general guidelines in section 10.2.2.2 of EN 15512 lead to the 4

following load combinations (LG ): LG1:

1,3 D + 1,40 P1 + ImpX

LG2:

1,3 D + 1,26 P1 + 1,26 H1 + ImpX

LG3:

1,3 D + 1,26 P1 + 1,26 H2 + ImpX

LG11: 1,3 D + 1,40 P2 + ImpX LG12: 1,3 D + 1,26 P2 + 1,26 H1 + ImpX LG13: 1,3 D + 1,26 P2 + 1,26 H2 + ImpX LG21: 1,0 D + 1,00 P1 + Impx in which is: D

Dead load

P1

Unit loads, fully loaded rack

4

The abbreviation “LG” refers to the term “load group“ which is used in the software RStab for load nd

combinations in 2

order analyses.

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P2

Unit loads, pattern loading in the bottom beam level

H1

Horizontal placement load of 0,25 kN in down-aisle direction at the top beam level

H2

Horizontal placement load of 0,50 kN in down-aisle direction at the bottom beam level

ImpX

Global imperfection in down-aisle direction

The effects of  -

the accidental load and

-

pattern loading at the higher beam levels

will be treated in conjunction with the member checks of the uprights.

7.3

Global analysis in cross-aisle direction

The combination rules and the general guidelines in section 10.2.2.3 in EN 15512 lead to the following load combinations: LG1:

1,3 D + 1,40 P + ImpZ

LG2:

1,3 D + 1,26 P + 1,26 H + ImpZ

LG11: 1,0 D + 1,00 P + ImpZ D

Dead load

P

Unit loads, fully loaded rack

H

Horizontal placement load of 0,25 kN in cross-aisle direction at the top of the frame

ImpZ

Global imperfection in cross-aisle direction

 According to EN 15512, 10.2.2.3, Note 1 pattern loading needs not be considered in the load combinations for the cross-aisle directions. For the present example this is acceptable. However it is recommended to take account of pattern loading, if it together with the effect of the imperfection in cross-aisle direction becomes clearly critical for the member check of the uprights. The effects of  -

the accidental load and

-

the horizontal placement loads in cross-aisle direction between at mid-span between to bracing nodes

will be treated in conjunction with the member checks of the uprights.

7.4

Beam design

For the beam design the following load combinations are considered: 1,40 P 1,26 P + 1,26 H with P

Unit loads, pattern loading

H

Horizontal placement load of 0,25 kN in cross-aisle direction at mid-span of the beam

For simplification the dead load is neglected here, since it is only (0,23 kN / 16,67 kN) ⋅ 100% = 1,4% of the unit loads.

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8. Modeling 8.1

General

In the present example the global analysis is carried out on

Figure 10 

two plane models, which represent the behaviour of the structure in cross-aisle and down-aisle direction. The system

Upright

lines of the models correspond widely to the centroidal axes of  the gross cross section of the members (EN 15512, 10.1.1).

Bracing member 

However, in case of the frame bracing the system line is

Centroidal axis

principally assumed to go through the bolts of the joints to the uprights (see figure Figure 10). The eccentricity between

System line

system line and centroidal axis of the bracing member is later  taken into account in the member checks (EN 15512, 9.7.4.3 d). 8.2

Bracing eccentricities

 According to EN 15512, 8.6 eccentricities of the frame bracing can be neglected, if they do not exceed the limits specified in figure 11b of EN 15512. In Figure 11 shows different types of 

Figure 11 a)

du  = 75 mm

bracing nodes and the modification of the system lines doe to the neglect of the eccentricities. The blue dash-dotted lines are then the system lines that would be used as static model. Being consequent, the nodes of the static model are at the

100 mm

height of the intersection points of the original system lines as

≤ 1,5 du

shown e.g. in Figure 11 a). However, in most cases it will be accurate enough, to assume the nodes of the static model in the middle between the joints of the bracing. The latter  b)

approach is used in these calculation. The “exact” positions of  the nodes are determined in annex A. For the decision, if an eccentricity can be neglected, bracing members, which do not participate in transmitting the shear  force in the frame, should be considered as being not existent

≈ 0 kN

(see Figure 11 b)). In the given frame the second horizontal

2 x 50 mm

≤ 1,5 du

would be such a member. Following EN 15512 the bracing eccentricities can be neglected for the given frame configuration. 8.3

c)

Eccentricity between beam and upright

 According EN 15512, 8.7 the horizontal distance e between the centroidal axes of upright and beam can be neglected, if e

50 mm

is smaller than 0,25 d u. It is assumed, that the front of the box

≤ du

beam is in one plane with the front of the uprights. So with:

64 mm

e = ys,upright - dbeam/2 = 30 - 50/2 = 5 mm

≤ 0,25 du

≤ 1,5 du

= 0,25 ⋅ 75 = 18,8 mm

follows, that this eccentricity can be neglected. System line with eccentricity System line without eccentricity

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8.4

Beam end connector 

In the static model the behaviour of the beam end connector is

Figure 12 

represented by a rotational spring. In the present example the F

spring is assumed to be linear elastic.

M

In section 4.5 different sets of values for the bending strength MRd,i  and stiffness k d,i  are given. With increasing values for the stiffness, the values for the bending strength are decreasing. For the global analysis any value for k d   can be chosen. In order to decide, which values to take, a pre-calculation considering the down-aisle direction under full load has been carried out. It showed, that with the higher values of the stiffness (k d,2   = 12500 kNcm/rad, k d,3  = 13000 kN/cm/rad) the corresponding

bending

strengths

MRd,2   and

MRd,3

are

b For evaluation of beam end connector tests: M= b ⋅ F

significantly exceeded in all beam levels. So the global analysis is carried out with k d,1 = 10000 kNcm/rad. Figure 13  According to EN 15512, 10.1.1 the system lines of the models

kd

shall correspond to the centroidal axes of the members. On the other hand according to EN 15512, 9.4.1 for beam check analysis the beam length may be taken as the distance between the faces of the two adjacent uprights, i.e. as the clear compartment width. Further bending strength and stiffness of the beam end connector are determined according

EIbeam

to EN 15512, A.2.4 related to the edge of the upright front as

EI = ∞

shown in Figure 12. In order to be in line with these requirements/assumptions, in the static model the beam is not directly connected to the upright but by an rigid arm with a length of half the upright width (see Figure 13) as proposed in section 9.4.1 of EN 15512.

8.5

Partial restraint of the upright base

The stiffness of the partial restraint of the upright base depends on the axial design force N Sd  in the upright. So for the global analysis of the down-.aisle analysis in a first step the axial forces in the uprights have to be determined. They depend mainly on the load situation “fully loaded” or “pattern loading” and on the load factors. It will be sufficient exact, if effects of the placement load, loads due to impact and imperfections are neglected. So in this example the properties of the partial restraint would have to be determined the for the following load combinations of down-aisle direction (see section 7) Table 8  Load combination

Loading situation

Load factor for unit loads

LG1

fully loaded (80%)

1,40

LG2, LG3

fully loaded (80%)

1,26

LG11

pattern loading (80%)

1,40

LG12, LG13

pattern loading (80%)

1,26

LG21

fully loaded (80%)

1,00

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So as an example for an inner frame and for the load combination LG1 the stiffness and bending strength would be: NSd

= 1,3 ⋅ 1,81 + 5 x 1,4 ⋅ 0,8⋅ 16,67

= 95,7 kN

kd(NSd)

= (119-110) / (120 - 110) ⋅ (44100 - 39400) + 39400

= 43630 kN

MRd(NSd) = (119-110) / (120 - 110) ⋅ (394,5 - 395,5) + 395,5

= 394,6 kNcm

The software used for the present example allows to define the rotational stiffness at the upright base depending on the axial force in the upright. For this example the following settings were made: Figure 14

In the settings positive values of P-z’ correspond to compressive forces in the upright.

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9. Global analysis The global analysis is documented in annex B, where the file name “DA-01” refers to down-aisle direction and “CA-01” to the cross-aisle direction. The documentation contains -

the input for system, load cases and load combinations,

-

the settings for the calculation, and

-

the inner forces and deflection for the critical elements.

The models used for the global analysis are shown below: Figure 15 - DA-01

Figure 16 - CA-01

Shear deformations of the members are neglected in the calculation. So the relevant cross section area for shear is set A z  = 0.

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 As described in section 8.4, in the static model the beam is not directly connected to the upright but by a rigid arm with a length of half the upright width. In the software this rigid arm is not modeled as a beam element with high rigidity, but as a so called eccentricity. This is practically the same as an infinitely rigid element, but the internal forces in this element are not shown as a result of the analyses. As a consequence, the moments at the beam ends and at the upright are not in equilibrium. In order to facilitate the member checks, in the model for the down-aisle direction additional nodes have been introduced at the two critical uprights, which coincide with the bracing nodes.

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10. Checks 10.1 Beams 10.1.1 Strength, unit loads only In section 4.4 is assumed that the bending strength has been determined by testing. So effects of  local buckling, distorsional buckling, lateral torsional buckling, concentrated transverse forces, the interaction of bending and shear are already covered by the effective section moduli. So the values MRd,y  and MRd,z  as given in section 4.4 can be used directly for the member check. For the determination of the design moments M Sd,y   at mid-span of the beam, in this example pattern loading at the top beam level will be critical. For this case in 9.4.3.3 of EN 15512 a formula (13) is given:

MSd,y

        Wd  L 2 / 3    m   1  =  8 2EIb           m 1     k e L    

where: ke

= 1 / (1/kb  + h/3EIc)

the effective stiffness, i.e. stiffness of the beam end connector  reduced by the effect of the flexibility of the upright

γQ

= 1,4

Qbeam

= 16,67 kN

total load on the beam

L

= 360 cm

clear compartment width

βm, βθ

= 1

values according annex F of EN 15512 for adjusting the formula to non-uniform loading, see also FEM 10.2.02, 4.4.2

Ib

4

= 260 cm 2

second moment of area of the beam about the y-axis

Ic

= 88 cm

second moment of area of the upright about the y-axis

h

= 200 cm

storey height, here between 4 and 5 beam level

kb

= 10000 kNcm/rad

stiffness of the beam end connector 

th

th

with: ke

=

10000 10000  360 1 3  21000  88

= 7349 kNcm/rad follows:

MSd,y

      1,4  16,67  360 2 / 3  1,0   =  1,0  1  8   2  21000  260    1,0  1   7349  360        = 913,7 kNcm

≤ 930,0 kNcm = MRd,y

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It should be noted, that the static model assumed to determine ke   does not comply with the assumptions in global analysis, since: 1)

The restraining effect due to the adjacent beam(s) is neglected, which leads to a larger moment at mid-span

Figure 17 

compared to the global analysis. 2)

The formula for ke   based on the assumption that distance of  the uprights coincides with the clear compartment width. Taking account of the real distance of the upright axes, e.g. by rigid arm as in the global analysis, would result in a greater  bending moment in the upright than at the connector (see Figure 17). That means in reverse, that the bending stiffness of the upright h/3EIc   in the formula of k e   is overestimated,

model assumed for  formula in EN 15512

which leads to a smaller moment at mid-span compared to the global analysis. However, usually the formula for MSd   will give sufficient exact results.  As counter check the bending moment in the beam is calculated with the model of the entire structure in down-aisle direction in annex B. The load case “Pallet P3” results in MSd,y

model complying with the global analysis

= 1,4 ⋅ 651,75 = 912,5 kNcm

≈ 913,7 kNcm. So in this example both methods lead to nearly the same result.

10.1.2 Strength, unit loads + placement loads The maximum horizontal placement load occurs at the bottom beam level. The formula (13) of  EN 15512 for MSd. In order to receive exacter results k e  can be modified as follows: ke

= 1 / (1/kb  + 1/kc)

with kc

= 3EIc  / h1  + 3EIc  / h2

in which h1

= 192,5 cm is the storey height below the 1st beam level

h2

= 200,0 cm is the storey height between 1 and 2

st

nd

beam level

With this modification follows: ke

= 8497

The load factor for the combined action of unit load and placement load is 0,9 ⋅ γ Q. So with formula (13) the design moment is: MSd,y

= 807,3 kNcm

 About the vertical axis the connection of upright and beam is practically fully hinged. So the moment at mid-span due to the horizontal placement load is:

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MSd,z

= 0,9 ⋅ γQ ⋅ H ⋅ L / 4 = 0,9 ⋅ 1,4 ⋅ 0,25 kN ⋅ 360 cm / 4 = 28,4 kNcm

The member check for biaxial bending is then

1

≥ MSd,y  / MRd,y + MSd,z / MRd,z = 807,3 / 930 + 28,4 / 150 not o.k.‼

= 1,057

Thus, formally the beam is not designed sufficiently for the combined action of unit loads and horizontal placement load. However the yield stress is only exceeded in two opposite edges of the profile. So depending on the design of the beam it might be assessed, if this can be accepted.

10.1.3 Deflection

 According to In section 9.4.5 of EN 15512 The maximum deflection at mid-span Δmax   can be determined as

Δmax

        5  Qbeam  L 0,8      1  =  384  EIb 2EIb           1  k L    e       

βΔ is again a value to adjust the formula to non-uniform loading. For uniform loading β Δ  equals 1. The other parameters are as described in the section before. Thus:

      5  16,67  360 0,8  1,0   =  1,0  1  384  21000  260   2  21000  260    1,0  1   7349  360        3

Δmax

= 1,56 cm The counter check with the global analysis in annex B results in:

Δmax

= 17,6 - (1,5 + 2,4) / 2 = 15,7 mm

The maximum permissible deflection is L / 200 = 360 cm / 200 = 1,80 cm. So the beam is designed sufficiently with respect to the deflection.

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10.2 Beam end connectors 10.2.1 Downwards bending + downwards shear  The design moments at the connectors are determined in the global analysis in (DA-01, annex B). The maximum moments occur at the bottom beam level for the load combination LG1: 1,3 D + 1,40 P1 + ImpX as MSd

≈ 244 kNcm

The corresponding design shear force is: VSd

= 12,2 kN

The global analysis has been carried out with the connector stiffness k d,1. So the bending strength is: MRd,1

= 230 kNcm

The shear strength is VRd

= 18,0 kN

 According to EN 15512, 9.5.4 is has to be verified, that

1

5

≥ MSd / MRd  + (VSd  - MRd / a) / VRd

Herein the value “a” corresponds to the length of the cantilever arm in the beam end connector  tests (“b” in figure A.2.5), which is defined as 400 mm. So for this example follows: 244 / 230 + (12,2 - 230 / 40) / 18 = 1,41 not o.k. ‼

≫   1,00

So following EN 15512 the beam end connector would be significantly too weak! MRd   would have to be increased by 39%, to satisfy the check. So a redistribution of 15% of the 6

moments at the beam ends, as shown in figure 17 of EN 15512, will not solve the problem here . However, it must be pointed out, that in the past (see FEM 10.2.02) the strength of the connectors was checked separately for bending and shear, i.e. MSd / MRd ≤ 1 VSd / VRd

5 6

≤ 1

In the formula the letter “V” instead of “S” is used in accordance with EN 1993.  Also, the try to reduce the connector stiffness, in order to take account of plastic reserves and show that at least MSd ≈ MRd   can be achieved, was not successful. Reducing the stiffness at the connectors iteratively, where MSd  > MRd  only lead to higher values of MSd. So in a global analysis, taking account of a non-linear 

M-θ-curve for the connector, the system would be instable.

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This approach was justified by the experience from

Figure 18 

beam end connector tests, which showed, that the bending strength of connectors usually is not affected

too much by the shear force, at least for a ratio M/V ≥

M {MSd,2  ; VSd,2}

ca. 20 cm. But of course this cannot be seen as a {MRd,1  ; VRd,1}

universal rule, which can be applied on any type of 

{MRd,2  ; VRd,2}

connector.  An alternative way of proving a sufficient load bearing

{MSd,1  ; VSd,1}

capacity might be, to determine the bending strength by testing the connector with a short cantilever arm ( ≪ 400 mm), instead of carrying out shear tests. In this way, one would achieve different pairs of {M Rd,i  ; VRd,i }. Figure 18 illustrates the check resulting from this procedure, in

V

which {MRd,i ; VRd,i } are determined for two different cantilever lengths. The connector would then be designed sufficiently for the combined action of {M Sd,1 ; VSd,1} but not for {M Sd,2  ; VSd,2}.

10.2.2 Upwards bending + upwards shear  In the present example the moments and forces due to upwards bending and the upwards shear at the unloaded beams are small and need therefore not to be considered any further.

10.2.3 Upwards shear due to impact  Accidental vertical load due to impact :

VSd,up

= 5,0 kN

Shear resistance

VRd,up

= 5,5 kN

(see 5.4.2)

≥ 5,0 kN

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10.3 Uprights 10.3.1 General Principally two checks have to be considered: (1) Bending and axial compression without lateral-torsional buckling (EN 15512, 9.7.6.3)

1

≥

=

N Sd

 min  A eff   f y /  M N Sd Nb,Rd,min





k y  MSd,y MRd,y

k y  M Sd, y Weff ,y  f y /  M





k z  M Sd,z Weff ,z  f y /  M

k z  M Sd,z MRd,z

χmin is defined as “the lesser of χ db, χy and χz, where χdb   is the reduction factor calculated in 9.2.7.c and χy  and χz   are the reduction factors from 9.7.4 for the y-y and z-z axes respectively.” (2) Bending and axial compression with lateral-torsional buckling (EN 15512, 9.7.6.4)

1

≥

=

N Sd

 min  A eff   f y /  M NSd Nb,Rd,min





k LT  MSd,y Mb,Rd,LT

k LT  MSd,y

 LT  W eff ,y  f y /  M 



k z  MSd,z W eff ,z  f y /  M

k z  MSd,z MRd,z

χmin is defined as “the smallest of χ db (from 9.2.7c). χy and χz, (from 9.7.4) and of the reduction factors corresponding to the distorsional and flexural torsional buckling modes.” χ LT   is the reduction factor for lateral-torsional buckling.

χy  and χz  are the reduction factors for flexural buckling. For easier understanding the reduction factor for flexural-torsional buckling will be named in this document χ FT. As specified in section 4.2 the beam is not subject to effects of lateral-torsional buckling. Thus χ LT  equals 1. χdb obviously relates to distorsional buckling. Concerning the calculation of χ db  the standard refers to clause 9.7.2 c). However, this clause describes, how to modify the effective area of the upright to

take account of distorsional buckling. A value χ db  is neither defined there nor anywhere else in the standard. As set forth in section 4.2 of this document, the effects of local buckling and distorsional buckling are already taken into account by the effective cross section properties A eff   and W eff . A

further reduction by a value χ db is therefore not necessary. For comparison, in EN 1993-1-3 the material thickness of those parts of the cross section that are

subject to distorsional buckling are reduced by a factor χ d. The checks with respect to flexural buckling and flexural-torsional buckling are then calculated with this reduced area. That means, that distorsional buckling is considered in interaction with flexural or flexural-torsional buckling and not as an completely separate failure mode, as suggested by the clauses 9.7.6.3 and 9.7.6.4 of EN 15512. NSd, MSd,y and MSd,z are the maximum values of the inner design forces regardless at which location of the considered member length the maximum of each single value occurs.

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The values k y, kz  and kLT  are defined in the sections 9.7.6.3 and 9.7.6.4. k y  and kz  can be assumed

to be ≤ 1, if the inner forces (NSd, MSd,y, MSd,z) are result of a second order analysis. k LT  is generally ≤ 1 (see also 10.3.5 in this document). So setting k y, kz  and kLT  as 1 a simplified design rule may be used in this example, which is slightly on the safe side:

(3) 1 ≥

=

N Sd

 min  A eff   f y /  M NSd Nb,Rd,min



MSd,y MRd,y





M Sd,y W eff ,y  f y /  M



M Sd,z Weff ,z  f y /  M

MSd,z MRd,z

in which χmin = min{χy ; χz ; χFT}

10.3.2 Buckling lengths The definition of the buckling length is connected to the term of the “system length”. However the “system length” is not clearly defined in the standard, which is problematic especially in conjunction with bracing eccentricities. So following an own interpretation, the system lengths for the uprights in this example are assumed as: a) for flexural buckling in down-aisle direction - Ly:

Figure 19

the distance from the upright base to the system line (= centroidal axis) of the bottom beam or the vertical distance of the system lines of two beam

Lz

levels located above each other  b) for flexural buckling in cross-aisle direction - Lz:

Ly

the distance of the bracing nodes in the static model. (Note: the system lengths are greater, if  bracing eccentricities are neglected)

Lz

7

c) for torsional buckling - LT : the distance of the bracing nodes in the real

LT

structure Exemplarily the system lengths are shown in Figure 19. For the determination of the buckling lengths section 9.7.4.3 of EN 15512, gives the following general rule: “If the axial forces and bending moment in the plane of buckling of a member have been determined on the basis of a second-order analysis, they are already enhanced by second order  effects and the buckling length may be considered as equal to the system length. When second  order global analysis is used it is permissible to use in-plane buckling lengths for the non-sway  mode for member design.”

7

It is discussed controversially, if the connection of the beam to the upright can be considered as a torsional restraint for the upright. Since this surely depends on the type of connector and no information was provided by ERF concerning this point, no torsional restraint is assumed.

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Decisive for the fact, if axial forces and bending moment are determined in the plane of buckling, is the direction, in which the global imperfection is applied. Precondition is, that the global imperfection is sufficient similar to the global buckling mode of the structure in the considered plane. Concerning the case of buckling out of the plane of the applied global imperfection the paragraph could be interpreted so, as if the real buckling lengths of global buckling of the structure have to be used. However, it is assumed, that this is not intended and that the system lengths can be used. This would correspond to the procedure, which is nowadays usually used for design calculations of  pallet racking. Concerning the buckling lengths of braced frames the clauses a) to e) in section 9.7.4.3 of EN 15512 give additional information. The buckling lengths with respect to torsional buckling are defined in 9.7.5.2 of EN 15512 as “ - 1,0 × distance between bracing points when connections provide full torsional restraint; - 0,5 × distance between bracing points when connections provide full torsional restraint and full  warping restraint.”  For typical bracing member connections additional information is given in figure 24 of EN 15512. For this example the following buckling lengths in-plane buckling length are not taken into account. So the buckling length used are Table 9 Direction of 

Buckling mode

Buckling length

global imperfection

and direction

Le..

flexural buckling

Ley = βy  × Ly

down-aisle

βy  = 1,0 Lez = βz × Lz

down-aisle + cross-aisle

flexural buckling cross-aisle

βz  = 0,9 Lez = βz × Lz

βz  = 1,0 torsional buckling flexural-torsional buckling

LeT = βT  × LT

βT = 0,7 Ley = βy  × Ly βy  = 1,0

Comment on the safe side in conjunction with global imperfection in downaisle direction Lz  = length from upright base to st

the 1 bracing node in static model (EN 15512, 9.7.4.3 b) Lz  = length between the bracing st

nodes above the 1 bracing node in static model EN 15512, figure 24 a)

The buckling length Lez  could be reduced further for those parts of the upright, where the spacing between two bracing nodes crosses a beam level, so that the axial force is not constant over L z.

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10.3.3 Reduction factors χ

The determination of the reduction factors χ y, χz  for flexural buckling is given in 9.7.4.1 of EN 15512 as follows:

1

χ

=

ϕ

= 0,5 [1+ α (   -0,2) ⋅ 2 ]



=

Ncr 

=

   2  2

 A eff   f y N cr 

 2  E  Ig L2e..

herein is:  Aeff 

= effective area of the upright

Ig

= second moment of area of the gross cross section area for the relevant axis

Le..

= buckling length for the relevant axis

α

= imperfection factor according to table 9 of EN 15512 (= 0,34 for a Ω-shaded cross section)

With 9.7.5.1 and 9.7.4.2 of EN 15512 the reduction factor with respect to flexural-torsional buckling can be determined as:

1

χFT

=

ϕ

= 0,5 [1+ α (   -0,2) ⋅ 2 ]



=

Ncr,FT

=

   2  2

 A eff   f y Ncr ,FT 2 2          y N N 0 , g cr  , T cr  , T   4   Ncr ,T  1   1    i 0,g  Ncr ,y 2 Ncr ,y   Ncr ,y       

N cr ,y

1

   2  E  I w,g     G  IT,g    L2eT    

Ncr,T

=

β

 y   = 1   0,g    i 0,g   2

i0,g

i 02,g

2

   

2

2

2

= iy,g + iz,g + y0,g

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iy,g

=

iz,g

=

I y ,g  A g

I z, g  A g

with: IT,g

= Torsional constant of the gross cross section

Iw,g

= warping constant of the gross cross section

y0,g

= distance between shear centre and centre of gravity of the gross cross section along y-axis

Iy,g, Iz,g

= second moment of area of the gross cross section

Ncr,y

= critical force for flexural buckling about the y-axis, see 10.3.3

Ncr,T

= critical force for torsional buckling

Ncr,FT

= critical force for flexural-torsional buckling

α

= imperfection factor

α is not specified explicitly for flexural torsional buckling in EN 15512! For this example is assumed, that the Ω-section behaves similar to a channel section. So, in accordance with clause 6.2.3 (4) of EN 1993-1-3 buckling curve b with α = 0,34 will be used.

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10.3.4 Design buckling resistance In the following table the design buckling resistance N b,Rd = χ ⋅  Aeff  ⋅ f y / γM  of the upright for flexural and flexural-torsional buckling are determined: Table 10  Profile : Position : Beam level :

U1 0-1

=

35,5

y

M = E= G=  Ag =  A = Iy = Iz = IT = Iw = y0 = Ly = Lz = LT =

y = z = FT =

1,00 21.000 8.077 7,00 6,40 88,0 33,0 0,120 1.300 6,00 192,5 106,4 95,0

U2 0-2  

35,5

  1,00   21.000   8.077   7,00   6,40   88,0   33,0   0,120   1.300   6,00 192,5 110,0 100,0

Upright 100/75/2,5 U3 U4 1-2 2-3

Unit U5 3-4

U6 4-5

35,5

35,5

35,5

35,5   [kN/cm2]

1,00 21.000 8.077 7,00 6,40 88,0 33,0 0,120 1.300 6,00 200,0 140,0 130,0

1,00 21.000 8.077 7,00 6,40 88,0 33,0 0,120 1.300 6,00 200,0 200,0 190,0

1,00 21.000 8.077 7,00 6,40 88,0 33,0 0,120 1.300 6,00 200,0 230,0 200,0

1,00 21.000   [kN/cm2] 8.077   [kN/cm2] 7,00 [cm2] 6,40 [cm2] 88,0 [cm4] 33,0 [cm4] 0,120 [cm4] 1.300 [cm6] 6,00   [cm] 200,0 [cm] 235,0 [cm] 220,0 [cm]

1,00

1,00

1,00

1,00

1,00

1,00

0,90

1,00

1,00

1,00

1,00

1,00

0,70 227,2

0,70 227,2

0,70 227,2

0,70 227,2

0,70 227,2

0,70 227,2

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34

0,34 492,2 745,9

0,34 492,2 565,3

0,34 456,0 349,0

0,34 456,0 171,0

0,34 456,0 129,3

0,34 456,0 123,9

y,rel = z,rel = y = z = y = z =

0,679

0,679

0,706

0,706

0,706

0,706

0,552

0,634

0,807

1,153

1,326

1,354

0,812

0,812

0,835

0,835

0,835

0,835

0,712

0,775

0,929

1,326

1,570

1,613

0,795

0,795

0,780

0,780

0,780

0,780

0,860

0,820

0,720

0,504

0,415

0,402

Nb,Rd,y = Nb,Rd,z = i0 =

180,7 195,5 7,300

180,7 186,2 7,300

177,3 163,6 7,300

177,3 114,6 7,300

177,3 94,2 7,300

177,3   [kN] 91,2   [kN] 7,300 [cm]

 =

0,324 1162 373,0

0,324 1050 362,8

0,324 629 289,3

0,324 304 199,4

0,324 276 187,6

0,324 231 166,6

FT,rel = FT = FT =

0,780

0,791

0,886

1,067

1,101

1,168

0,903

0,914

1,009

1,217

1,259

1,346

0,736

0,730

0,670

0,555

0,535

0,496

Nb,Rd,FT =

167,3

165,8

152,2

126,1

121,5

112,7   [kN]

NRk =

y = z = FT = Ncr,y = Ncr,z =

Ncr,T = Ncr,FT =

[kN]

[kN] [kN]

[kN] [kN]

The upright positions U1, U2, .. are shown in the report for the model CA-01 in annex B.

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10.3.5 Moment coefficients k.. The in the member checks the design moments MSd,y  and MSd,z  can be reduced by moment coefficients k y, kz and kLT. ky  and kz  are defined in section 9.7.6.3 of EN 15512 as follows

 y  NSd  y  A eff   f y

kz

= 1-

 y ⋅ (2 βM,y - 4)

μz

=

ky

= 1-

μy

=

≤ 0,9

 z  NSd  z  A eff   f y

 z ⋅ (2 βM,z - 4)

≤ 0,9

ky  and kz can be assumed to be ≤ 1, if the inner forces (N Sd, MSd,y, MSd,z) are result of a second order analysis. For a linear moment diagram (M Q  = 0 in figure 25 of EN 15512) k y  and kz are

smaller than 1 only, if the ratio of the end moments ψ is smaller than -0,286. kLT  shall be determined according to 9.7.6.4 of EN 15512 as: kLT

= 1-

 LT  NSd  Z  A eff   f y

≤ 1 μLT

= 0,15 ⋅  z ⋅ (2 βM,LT  - 4)

≤ 0,9  According to EN 15512  z  in the formula for μ LT may be determined for a flexural buckling length equal to the maximum vertical spacing of the beams. This seems not to be logical, since the upright is held against deflection in z-direction and torsion at the bracing nodes. So in this example the value  z is used as for the determination of the buckling reduction factor χ z.

The values βM,..  are the equivalent uniform moment factors depending on the moment diagram over  the considered system length of the upright. The values shall be determined with the formulae given in figure 25 of EN 15512. In the following table the bending moments and relevant system

lengths are assigned to the values β M,.. Table 11 Factor

Moment diagram of

βM,y βM,z βM,LT

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MSd,y MSd,z MSd,y

buckling reduction

System length for 

factor 

moment diagram

χy χz χz

Ly

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Lz Lz

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10.3.6 Checks (1) Pre-calculation For the check of the uprights the simplified design rule (3) given in section 10.3.1 is used. Since the frame bracing is irregular, first a pre-calculation is carried out in the following table, in order to assess which part of the upright might be critical for the member check: Table 12  Beam

nb

level

NSd

Nb,Rd,min

kN

kN

NSd / N b,Rd

 

Reserve 1 - NSd / N b,Rd

4-5

1

23,9

91,2

0,261

0,739

3-4

2

47,7

94,2

0,506

0,494

2-3

3

71,6

114,6

0,624

0,376

1-2

4

95,4

152,2

0,627

0,373

0-1

5

119,3

165,8

0,719

0,281

with: NSd

= nb ⋅ (γG ⋅ G/5 + γQ ⋅ Qbeam)

nb

= number of beam levels above the considered part of the upright

Qbeam

= 16,67 kN

G

= 1,81 KN

The term 1 - NSd  / Nb,Rd,min is the reserve of the upright, that can be used to carry bending moments. The bending moments in the uprights due to imperfection and placement loads decrease in the upper beam levels. The bending moments due to pattern loading are in a rack with constant compartment heights similar for all beam levels, with exception of the top beam level, where the bending moments are approximately twice as high. That shows that it will be sufficient to consider  the uprights below the second beam level for the member checks. (2) Checks, down-aisle direction The axial forces of the uprights in the down-aisle model DA-01 are calculated for average frame load resulting from 80% of the maximum compartment loads. However for the single frame the full load has to be considered. So to the axial loads N Sd,80%  determined in global analysis, N Sd,20% for  the missing 20% has to be added, so that N Sd  = NSd,80% + NSd,20%. Table 13 contains the results of the global analysis, the determination of N Sd, the relevant buckling resistance from section 10.3.4 and the member check.

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Table 13 Beam

Pos.

level 0-1

1-2 *)

Load

nb

comb.

Load

NSd,80%

NSd,20%

NSd

MSd,y

Nb,Rd,min   Check

factor

kN

kN

kN

kN

kN

-

U1

LG1

5,0

1,40

95,6

23,3

118,9

122,8

167,3

0,897

U1

LG2

5,0

1,26

86,3

21,0

107,3

107,9

167,3

0,805

U1

LG3

5,0

1,26

86,3

21,0

107,3

107,6

167,3

0,804

U2

LG11

4,5

1,40

86,3

21,0

107,3

150,9

165,8

0,876

U2

LG12

4,5

1,26

77,9

18,9

96,8

138,1

165,8

0,793

U2

LG13

4,5

1,26

77,9

18,9

96,8

143,0

165,8

0,801

U2

LG11

4,0

1,40

76,5

18,7

95,2

168,8

165,8

0,830

U2

LG12

4,0

1,26

69,1

16,8

85,9

152,2

165,8

0,749

on the safe side taken from the upright base. Since Nb,Rd,min  = Nb,Rd,FT  , MSd,y could be taken from x = 64 mm = position st of 1 bracing node of the real structure.

Check

= NSd / Nb,Rd,min + MSd,y / MRd,y  + 0 kNcm / M Rd,z

NSd,20%

= nb  × (Load factor × Qbeam)

(3) Checks, cross-aisle direction Table 14 Beam level

Pos.

Load comb.

NSd kN

MSd,z kNcm

0-1

U1

LG1

128,2

5,1

Check ⇒

Nb,Rd,min   Check kN 167,3

0,774

= NSd / Nb,Rd,min + MSd,z  / MRd,z  + 0 kNcm / M Rd,y

The upright is designed sufficiently

10.3.7 Effect of horizontal placement load between bracing nodes The inner forces for the load combination with horizontal placement load are determined for a safety factor that is approximately 10% smaller than for the case of without horizontal placement load. So, if it can be proved that the moments due to the horizontal placement loads are not bigger  than 10% of the bending strength of the upright, the load combinations with horizontal placement load will not be decisive for the member check. The table below contains this proof: Table 15  H

Lz

Lez

Ncr,z

mm

kN

mm

mm

kN

532

0,50

1064

958

745,9

5,0

107,4

1,17

19,6

0,061

1614

0,50

1100

1100

565,3

5,0

107,4

1,23

21,4

0,067

3964

0,42

1400

1400

349,0

4,0

85,9

1,33

24,6

0,077

4664

0,36

1700

1700

236,7

3,0

64,4

1,37

26,6

0,083

5664

0,28

2000

2000

171,0

3,0

64,4

1,60

28,1

0,088

6664

0,25

2300

2300

129,3

2,0

42,9

1,50

27,1

0,085

8989

0,25

2350

2350

123,9

1,0

21,5

1,21

22,4

0,070

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nb

NSd



h(H)

kN

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with: h(H)

= position of the placement load in height

H

= placement load

Ncr,z

= π2 E Iz,g / Lez2

nb

= number of beam levels above h(H)

NSd

= nb × (γG  G/5 + 0,9 ⋅ γQ ⋅ Qbeam)

G

= 1,81 kN

Qbeam

= 16,67 kN

η MSd,z

= 1 / (1 - NSd  / Ncr,z) = η ⋅ 0,9 ⋅ γQ ⋅ H ⋅ Lz / 4

Check

= MSd,z / MRd,z

nd

(amplification factor 2 order) (on the safe side pinned ends assumed)

 All checks ≤ 0,1 ⇒ Effect of horizontal placement load between bracing nodes is not critical here.

10.3.8 Effect of horizontal accidental loads The inner forces for the load combination with horizontal accidental load are determined for a safety factor that is approximately 1 - 1/1,4 = 28,5% smaller than for the case of without accidental load. So, if it can be proved that the moments due to the horizontal placement loads are not bigger  than 28,5% of the bending strength of the upright, the load combinations with horizontal accidental load will not be decisive for the member check: (1) Accidental load A in down-aisle direction, 400 mm above the floor:  A

= 1,25 kN

Ly

= 192,5 cm

Ley

= 192,5 cm

Ncr,y

= π2 E Iy,g / Ley2 = 492,2 kN

NSd

= 107,4 kN

(see 10.3.5)

η

= 1 / (1 - NSd  / Ncr,z)

(amplification factor 2 order)

nd

= 1,28 MSd,y

= η ⋅  A ⋅ (Ly - 40 cm) / Ly ⋅ 40 cm

(on the safe side pinned ends assumed)

= 50,7 kNcm Check

= MSd,y  / MRd,y = 50,7 kNcm / 660 kNcm = 0,077 < 0,285

⇒  Accidental

load in down-aisle direction is not critical here

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(2) Accidental Accidental load A in cross-aisle cross-aisle direction, direction, 400 mm above the floor:  A

= 2,5 kN

Lz

= 106 106,4 ,4 cm

Lez

= 0,9 ⋅ 106,4 cm = 95,8 95,8 cm

Ncr,z

= π2 E Iz,g / Lez2 = 745, 745,2 2 kN

NSd

= 107, 107,4 4 kN

(see 10.3.5)

η

= 1 / (1 - NSd  / Ncr,z)

(amplification factor 2 order)

nd

= 1,17 MSd,y

= η ⋅  A ⋅ (Lz  - 40 cm) / Lz ⋅ 40 cm

(on the safe side pinned ends assumed)

= 73,0 73,0 kNcm kNcm Check

= MSd,y  / MRd,y = 73,0 73,0 kNcm kNcm / 320 kNcm kNcm = 0,22 0,228 8 < 0,28 0,285 5

⇒  Accidental

load in cross-aisle direction is not critical here

10.4 10.4 Frame Frame bracing bracing In the global analysis in model CA-01 the imperfection and horizontal placement load was only applied in one direction. It is assumed, that the tensile forces in the bracing members will occur as compression forces with the same magnitude, if imperfection and horizontal placement load are applied reversely. The determination of the buckling resistance is based on the following assumptions

 αy = αz = αFT  = 0,49

-

the bracing bracing member is is a U-profil U-profile. e. So itit is assigned assigned to buckling buckling curve curve “c” “c”

-

The buckling buckling lengths lengths for flexural flexural buckling buckling correspo correspond nd to the system system length (on the safe side) side)

-

the connect connection ion to the uprigh uprightt provide provides s full full torsional torsional but no warpin warping g restrai restraint nt (axis of the bolt

-

 βT  = 1,0.

close to the zero crossing of the warping ordinate)

⇒

Ag  has not been specified for this example

Aeff  is assumed

⇒  Ag  =

⇒

So the buckling resistance is determined for the critical members as follows:

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Table 16  Profile : Position :

1

= M = E= G=  Ag =  Aeff  = Iy = Iz = IT = Iw = y0 = Ly = Lz = LT = y = z = T = NRk = y = z = FT = Ncr,y = Ncr,z = y,rel = z,rel = y = z = y = z = Nb,Rd,y = Nb,Rd,z = i0 =  = Ncr,T = Ncr,FT = FT,rel = FT = FT = Nb,Rd,FT =

23,5 1,00 21.000 8.077 1,40 1,40 6,00 1,20 0,01 5,00 1,80 148,8 148,8 148,8 1,00 1,00 1,00 32,9 0,49 0,49 0,49 56,2 11,2 0,765 1,711 0,931 2,335 0,684 0,255 22,5 8,39 2,895 0,613 16,2 14,3 1,517 1,973 0,309 10,2

y

Bracing U 50/30/1,5 10                      

2233,5 1,00 1, 21.000 21 8.077 8. 1,40 1, 1,40 1, 6,00 6, 1,20 1, 0,01 0, 5,00 5, 1,80 1, 155,1 155,1 155,1 1,00 1,00 1,00 32,9 0,49 0,49 0,49 51,7 10,3 0,798 1,784 0,965 2,479 0,664 0,238 21,8 7,83 2,895 0,613 15,7 13,8 1,544 2,022 0,301 9,89

Unit 13 23,5 1,00 21.000 8.077 1,40 1,40 6,00 1,20 0,01 5,00 1,80 158,8 158,8 158,8 1,00 1,00 1,00 32,9 0,49 0,49 0,49 49,3 9,9 0,817 1,826 0,985 2,566 0,652 0,229 21,4 7,53 2,895 0,613 15,5 13,5 1,560 2,049 0,296 9,74

 

[kN/cm2]

  [kN/cm2]   [kN/cm2] [cm2] [cm2] [cm4] [cm4] [cm4] [cm6]   [cm] [cm] [cm] [cm]

[kN]

[kN] [kN]

   

[kN] [kN] [cm] [kN] [kN]

 

[kN]

The effe effect ct of the the acci accide dent ntal al load load “A” “A” (400 (400 mm abov above e the the floo floor) r) on the the bott bottom om frame frame brac bracin ing g member is determined here as follows:  A

= 2,5 kN

Lz,upr 

= 106, 106,4 4 cm

(system length of upright)

Lz,br 

= 148,8 148,8 cm

(system length of bottom bracing member)

Dsys

= 104 104 cm

(width c.t.c. of upright frame)

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NSd,A

2

= A ⋅ (3 - (40cm / Lz,upr ) / 2) ⋅ 40 cm / Lz,upr  ⋅ Lz,br  / Dsys

Figure 20 

8

= 1,92 1,92 kN

With the axial force from load combination LG11 follows: NSd

NSd,A

= 1,92 1,92 kN + 1,56 1,56 kN = 3,48 3,48 kN

A

The bending moments due to the eccentricity e of the bolted connection and the centroidal axis of the bracing members is calculated as follows: e

= ey  - yS,eff  = 9 mm - 6 mm = 3 mm

Ncr,z

η MSd,z

= π2 E Iz,g / Lz2 = 1 / (1 - NSd  / Ncr,z) = η ⋅ e ⋅ NSd

(amplification factor 2

nd

order)

For the check the simplified design rule (3) of 10.3.1 will be used. With M Sd,y = 0 follows: Check

= NSd / Nb,Rd,min  + MSd,z  / MRd,z 4

With Iz,g = 1,20 cm and MRd,z  = 13,0 kNcm the checks are given in the table below: Table 17  Pos.

Load

NSd

Lz

Ncr,z

comb.

kN

mm

kN



MSd,z

Nb,Rd,min   Check

kNcm

kN

1

LG2

2,65

1488

11,2

1,309

1,040

8,39

0,396

1

LG11+A

3,48

1488

11,2

1,4449

1,513

8,39

0,531

10

LG1

1,56

1551

10,3

1,178

0,551

7,83

0,242

13

LG1

0,81

1588

9,9

1,089

0,265

7,53

0,128

 Actually also the effect of a horizontal placement load of 0,5 kN in a height of ≤ 3000 mm would have to be considered for the check of the bracing. However, it can be expected, that this does not lead here to a more critical situation as the accidental load. ⇒

The bracing bracing members are designed designed sufficiently sufficiently..

The design of the bolted connections between bracing member and upright usually can checked using the design rules of EN 1993-1-8, table 3.4. Where the distances e i  or pi  do not fall within the limits of table 3.3 in EN 1993-1-8, testing might be necessary.

8

In cases where the shear stiffness S D  of the frame is low compared to the maximum frame load Qframe , NSd,A should be multiplied by an amplification factor η for 2

nd

order effects, as e.g.

η = 1/(1 - QFrame/SD). In the present example, it is not necessary.

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10.5 Upright to floor connection 10.5.1 Partial restraint It has to be showed, that for a given NSd  at the upright base, M Rd(NSd) is not exceeded, i.e. Check

= MSd,y  / MRd(NSd)

≤ 1 With the inner forces in DA-01 follows: Table 18  Position

Load

NSd

MSd,y

MRd(NSd)

Check

comb.

kN

kNcm

kNcm

LG1

95,6

122,8

382,9

0,321

inner 

LG2

86,3

108,1

367,3

0,294

upright

LG11

86,3

138,0

367,3

0,376

LG12

77,9

119,5

348,2

0,343

outer 

LG1

49,1

80,7

248,5

0,325

upright

LG2

44,4

70,4

227,6

0,309

The check of the partial restraint does usually not include the check of the contact pressure on the concrete floor. The contact pressure has to be considered separately (see 10.5.4).

10.5.2 Anchorage  According to section 9.10.4 of EN 15512 each “upright to floor connection shall be able to transfer  a minimum un-factored force of 3 kN in tension and 5 kN in shear ”. Is assumed that the 3 kN and 5 kN are two components of one force and therefore have to be taken into account simultaneously. The wording “un-factored force” leaves space for interpretation, if the force has to be multiplied by a load factor or not. The value of the load factor, would also not be clear: 1,4 as for a placement load, 1,5 as for a live load or 1,0 as for a accidental load? Since the force is not intended to be combined with other forces, it would have been much clearer, if a design load (including the intended load factor) had been specified. The term “upright to floor connection” allows the interpretation, that also the upright to base plate connection and the base plate itself shall be designed for the given force. Probably this is not intended, since the headline of section 9.10.4 is “Design of anchorages”.  According to section 7.6 of EN 15512 the stability against overturning shall be verified. The empty rack shall remain stable under the action of single horizontal placement load in the most unfavourable position and direction, which is usually the top of the end frame and the cross-aisle direction. The anchors shall be designed to resist the uplift force FT,Sd   resulting from this consideration. In the present example the uplift force can be determined as follows: Self-weight of the end frame: G

= 2 ⋅ (Gupr  + Gbr ) + 10/2 ⋅ Gbeam

(see 5.1)

= 2 ⋅ (0,53 + 0,10) + 10/2 ⋅ 0,23 = 2,41 kN

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Page 45

Uplift force: FT,Sd

= γQ ⋅ QHp ⋅ H / Dsys - γG ⋅ G / 2 = 1,4 ⋅ 0,25 kN ⋅ 10164 mm / 1040 mm - 1,0 ⋅ 2,41 kN / 2 = 2,22 kN

⇒

The force of 3 kN in tension + 5 kN in shear is decisive for the design of the anchors.

10.5.3 Contact pressure on the floor   According to EN 15512, 9.9.1 the contact pressure is determined neglecting the bending moments resulting from to the partial restraint of the uprights. The compression force in the upright is distributed uniformly over the effective contact area A bas. According to section 9.9.2 of EN 15512  Abas   follows the contour of the cross section of the upright. The stress in the upright is distributed over a width of maximum 2 ⋅ e, where e

= tb ⋅

f y 3  f  j

in which: tb

= thickness of the base plate

f y

= “design strength of the base plate”

f  j

= design strength of the floor material

f y  is defined is defined in section 4 of EN 15512 as “ yield strength”. In conjunction with a base plate the term “design strength” could be interpreted as f y,d = f y / γM. However, since γ M  is given as 1,0 the result for e will be the same regardless if f y  or f y,d  will be used. f  j  is given for concrete in 9.10.1 of EN 15512 as f  j

= 2,5 f ck / γM

with: f ck

= characteristic compressive cylinder strength for concrete

γM

= 1,5

If the distance from the upright face to the edge of the base plate is less than e, a reduced value of  e equal to the distance from the upright face to the edge of the base plate shall be used. Figure 21 illustrates this issue exemplarily.

Institut für Schweißtechnik u. Ingenieurbüro Dr. Möll GmbH

64289 Darmstadt  An der Schleifmühle 6 Tel.: (06151) 74097 u. 713051 Fax.: (06151) 74140

Project StB-27-2011-I

Page 46

Figure 21 a)

b)

e

e

 Abas

The effect of a possible eccentricity between the centres of gravity of A bas  and of the cross section of the upright is usually neglected. But it might be useful to avoid too extreme disproportions. The design upright compression force N Sd  shall satisfy the following relationship: NSd

≤ f  j ⋅  Abas

10.5.4 Base plate The base plate is designed sufficiently for the compression forces, if the check for the contact pressure is fulfilled. The base plate and its connection to the upright should also be able to carry the

Figure 22  a)

b)

uplift force determined in conjunction with the stability against overturning. If the supporting effect of the contact with the

Ft,Sd

Ft,Sd

concrete is taken into account the increase of the anchor force should be taken in to account into account in the check for the anchor (see Figure 22 b).

Fc,Sd

Ft,Sd

Ft,Sd  + Fc,Sd

10.5.5 Effect of horizontal loads due to impact The effect of the horizontal load due to impact is usually not taken into consideration for the design of the upright to floor connection. A reason for this might be, that in the loaded condition the friction is usually big enough to transmit the specified horizontal force of max. 2,50 kN from the upright to the floor. However, for the unloaded condition major deformations can occur. This applies especially, if the base plate is fixed to the floor with only one anchor, which is located away from the system line of the upright.

Institut für Schweißtechnik u. Ingenieurbüro Dr. Möll GmbH

64289 Darmstadt  An der Schleifmühle 6 Tel.: (06151) 74097 u. 713051 Fax.: (06151) 74140

Project StB-27-2011-I

Page 47

10.6 Lateral deflection of the entire structure From global analysis in down-aisle direction (DA-01)

δh

= 26,6 mm

≤ H/200 = 9925 / 200 = 49,6 mm

⇒

o.k.

⇒

o.k.

From global analysis in cross-aisle direction (CA-01)

δh

= 6,6 mm

≤ H/200 = 10164 / 200 = 50,8 mm

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64289 Darmstadt  An der Schleifmühle 6 Tel.: (06151) 74097 u. 713051 Fax.: (06151) 74140

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Page 48

A. Frame bracing Dbr  = ybr  =

980 60,0

No Alignment

hl [mm]

hr  [mm]

Lbr  [mm]

hbr  [mm]

[kN]

64

64

980

0

0,0

1 2 3

horizontal diagonal horizontal

114 1064

mm mm

1014 1064

Abr  = e1 =

1331 980

900 0

1,40 25



cm2 mm abr  [mm]

e [mm]

50

0,0

64,0

50

50,0

1064,0

42,6

0,015

0,0

4

diagonal

1564

1114

1078

450

24,7

5

diagonal

1664

2114

1078

450

24,7

0,011 0,0 50,0

2664

2214

1078

450

24,7

7

diagonal

2764

3214

1078

450

24,7

50,0 50,0

3764

3314

1078

450

24,7

9

diagonal

3864

4614

1234

750

37,4

50,0 37,5

4714

1234

750

37,4

11 diagonal

5564

6614

1436

1050

47,0

50,0 41,7

13 diagonal 14 diagonal

7864 10064

8914 9014

1436 1436 1436

1050 1050 1050

10164

10164

980

0

50,0

6664,0

47,0

0,016 100

50,0

7814,0

100

50,0

8964,0

47,0

0,016

47,0

0,016 100

15 horizontal

5505,7 0,016

100 6714

4664,0 0,014

100

7764

3801,5 0,014

100 5464

3264,0 0,012

100

10 diagonal

2714,0 0,012

100 diagonal

2164,0 0,012

100

8

1614,0 0,012

100 diagonal

1064,0 0,012

100

6

G kN 0,011

50

12 diagonal

hsys [mm]

100,0

10164,0

0,0

0,011 (36)

G = Dbr  ybr  yS hl hr  Lbr  abr,Y e hsys G

0,205

= horizontal distance of the bracing nodes = distance front of upright to bracing joint = distance front of upright to axis of upright = position of bracing node in height at left upright = position of bracing node in height at right upright = length of bracing member  = [D br 2 + (h r -hl)2]0,5 = distance of the joints at a bracing node in Y-direction = distance of intersection point of the bracing axes from the upright axis = [e br,Y / (tan ai + tan ai+1)] - (ybr  - y S) = position of bracing nodes of the system lines = max{hl,i ; hr,i} + e = A br  (Lbr  + 2 e1) 78,5 kN/m3

see also Figure A.1

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Page A.1

Figure A.1 hr,i+1

hbr,i+1

αi+1

αi

hl,i+1 hl,i

abr 

e

αi

hbr,i

hr,i

ybr 

Institut für Schweißtechnik u. Ingenieurbüro Dr. Möll GmbH

Dbr 

64289 Darmstadt  An der Schleifmühle 6 Tel.: (06151) 74097 u. 713051 Fax.: (06151) 74140

Project StB-27-2011-I

Page A.2

ISIB Dr. Möll GmbH  An der Schleifmühle 6, 64289 DARMSTADT Tel: 06151/713051 - Fax: 06151/74140

Project:  StB-11-27 Worked examples

 

Structure: DA-01 Unbraced rack, down-aisle direction

CONTENTS Graph. Graph. 1.1 1.2 1.3 1.4 1.5 1.7 1.8 1.8.5

Graph. Graph. Graph. Graph. Graph.

Page: Sheet:

1/29 1

STRUCTURE Date:

07.08.2011

CONTENTS

General Data Structure Structure Structure - Beam to upright connection Nodes Materials Cross-sections Member Releases Member Eccentricities Members Nodal Supports Nodal Supports - Non-linearities Stiffness Diagram Loads Load Cases LC 1 - Dead load D LC1: Dead load D LC 11 - Pallets P1 LC11: Pallets P1 LC 12 - Pallets P2 LC12: Pallets P2 LC 13 - Pallets P3 LC13: Pallets P3 LC 14 - Pallets P4 LC14: Pallets P4 LC 21 - Placement load H1

1 2 2 3 3 4 4 4 4 4 6 6 6 7 7 7 7 8 8 9 9 10 10 11 11 12

Graph. Graph. Graph.

Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph. Graph.

LC21: Placement load H1 LC 22 - Placement load H2 LC22: Placement load H2 LC 31 - Imperfections ImpX LC31: Imperfections ImpX Load groups Settings for non-linear analysis Results - Load Cases, Load Groups Uprights Uprights Uprights Uprights Uprights Uprights Beams Beams Beams Beams Beams Beams Deformations u-X, LG21: 1,0*G + 1,0*P1 + ImpX Vertical deflection of the beam (Member 10) Vertical deflection of the beam (Member 6)

12 13 13 14 14 15 16 17 17 18 19 20 21 22 23 24 25 26 27 28 29 29 29 29 29 29

GENERAL DATA COMPUTING METHOD Structural Analysis Design Dynamic Analysis

Linear Stati c Analysi s Second-Order Analysis (Non-linear, Timoshenko) Large Deformation Analysis (Non-linear, Newton-Raphson) Postcritical Analysis (Non-linear, Newton-Raphson)

Load Cases Load Groups Load Combinations

Design Cases Dynamic Cases Buckling Curves

STRUCTURAL DATA PARAMETERS 1D Continuous Beam 2D Construction Type 3D Construction Type Grid

72 2 2 1 0

Nodes Materials Sections Element Hinges Element Partitions

103 0 0 0 0

Elements Cables Tapered Elements Elastic Foundations Sets of Elements

RSTAB 7.04.0940 - Spatial Framed Structures

www.dlubal.com

StB-27/2011-I

B.1