Basis of the design of timber and masonry structures according to Eurocode

BUDAPEST UN IVERS ITY OF TECHNOLOGY AND ECONOMICS FACULTY OF C IV I L ENG INEER ING Depar tment o f S t ruc tu ra l Eng ineer ing

BASIS OF THE DESIGN

OF TIMBER AND MASONRY STRUCTURES

ACCORDING TO EUROCODE

Lectures notes for the BSc subject

Timber and Masonry Structures (BMEEOHSAT19)

v1.0

Edited by: Dr. Kálmán Koris

Budapest, 15th September 2014.

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TABLE OF CONTENTS

I. Design of timber structures according to EC5 (EN 1995-1-1) 3 1. Basis of design 3

1.1. Principles of limit state design 3 1.1.1. Ultimate limit states 3 1.1.2. Serviceability limit states 3

1.2. Basic variables 3 1.3. Material properties 4

1.3.1. The effect of member size on strength for solid timber 7 1.3.2. The effect of member size on strength for glued laminated timber 8 1.3.3. The effect of member size on strength for Laminated veneer lumber (LVL) 8

1.4. Basis of structural analysis 8 2. Ultimate limit states 10

2.1. Design of cross-sections subjected to stress in one principal direction 10 2.1.1. Tension parallel to the grain 10 2.1.2. Tension perpendicular to the grain 10 2.1.3. Compression parallel to the grain 10 2.1.4. Compression perpendicular to the grain 11 2.1.5. Bending 11 2.1.6. Shear 12 2.1.7. Torsion 14

2.2. Design of cross-sections subjected to combined stresses 14 2.2.1. Compression stresses at an angle to the grain 14 2.2.2. Combined bending and axial tension 14 2.2.3. Combined bending and axial compression 15

2.3. Stability of timber members 15 2.3.1. Columns subjected to either compression or combined compression and bending 15 2.3.2. Beams subjected to either bending or combined bending and compression 16

3. Limiting values for deflections of beams 17 4. Bracing 18

4.1. General aspects 18 4.2. Single members in compression 18 4.3. Bracing of beam or truss systems 19

5. Numerical examples 20 5.1. Member subjected to compression and bending 20 5.2. Column and tie beam connection 22 5.3. Dovetail halving connection 25

II. Design of masonry structures according to EC6 (EN 1996-1-1) 30 6. Terms and Definitions 30

6.1. Terms relating to masonry 30 6.2. Terms relating to masonry units 30 6.3. Terms relating to mortar 30

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6.4. Terms relating to wall types 31 7. Basis of design 32

7.1. Ultimate limit states 32 7.2. Serviceability limit states 33

8. Materials 33 8.1. Masonry units 33

8.1.1. Types and grouping of masonry units 33 8.1.2. Compressive strength of masonry units 34

8.2. Mortar 35 8.3. Mechanical properties of masonry 35

8.3.1. Characteristic compressive strength of masonry 35 8.3.2. Characteristic shear strength of masonry 36 8.3.3. Characteristic flexural strength of masonry 37

8.4. Deformation properties of masonry 39 8.4.1. Stress-strain relationship 39 8.4.2. Modulus of elasticity 39 8.4.3. Shear modulus 39

9. Analysis of structural members 40 9.1. Masonry walls subjected to vertical loading 40

9.1.1. The initial eccentricity 40 9.1.2. Effective height of masonry walls 40 9.1.3. Effective thickness of masonry walls 42 9.1.4. Slenderness ratio of masonry walls 42 9.1.5. Verification of unreinforced masonry walls subjected to mainly vertical loading 42 9.1.6. Walls subjected to concentrated loads 45 9.1.7. Unreinforced masonry walls subjected to shear loading 47

10. Structural behaviour and application of reinforced masonry 48 10.1. Advantages of the application of reinforced masonry 48 10.2. Typical properties of the reinforcement 49 10.3. Application of horizontal reinforcement for masonry 51

11. Numerical examples 55 11.1. Verification of masonry wall subjected to vertical loading 55 11.2. Verification of shear resistance 58

12. References 60

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I. Design of timber structures according to EC5 (EN 1995-1-1)

1. Basis of design

1.1. Principles of limit state design

1.1.1. Ultimate limit states

For a first order linear elastic analysis of a timber structure, whose distribution of in-ternal forces is not affected by the stiffness distribution within the structure (e.g. all members have the same time-dependent properties), mean values shall be used.

For a first order linear elastic analysis of a structure, whose distribution of internal forces is affected by the stiffness distribution within the structure (e.g. composite members containing materials having different time-dependent properties), final mean values adjusted to the load component causing the largest stress in relation to strength shall be used.

For a second order linear elastic analysis of a structure, design values, not adjusted for duration of load, shall be used.

1.1.2. Serviceability limit states

The deformation of a timber structure which results from the effects of actions (such as axial and shear forces, bending moments and joint slip) and from moisture shall remain within appropriate limits, having regard to the possibility of damage to surfac-ing materials, ceilings, floors, partitions and finishes, and to the functional needs as well as any appearance requirements.

The instantaneous deformation, uinst should be calculated for the characteristic com-bination of actions, using mean values of the appropriate moduli of elasticity, shear moduli and slip moduli.

The final deformation, ufin should be calculated for the quasi-permanent combination of actions.

1.2. Basic variables

Duration of load and moisture content affect the strength and stiffness properties of timber and wood-based elements and shall be taken into account in the design for mechanical resistance and serviceability. Actions caused by the effects of moisture content changes in the timber shall be also taken into account. The load-duration classes are given in Table 1.

Table 1: Load-duration classes

Load-duration class

Order of accumulated dura-tion of characteristic load

Examples of loading

Permanent more than 10 years self-weight

Long-term 6 months - 10 years storage

Medium-term 1 week - 6 months imposed floor load, snow

Short-term less than one week snow, wind

Instantaneous wind, accidental load

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Based on the moisture content, structures shall be assigned to one of the service classes given below:

• Service class 1 is characterised by a moisture content in the materials correspond-ing to a temperature of 20 °C and the relative humidity of the surrounding air only exceeding 65% for a few weeks per year. In service class 1 the average moisture content in most softwoods will not exceed 12%. This service class typically refers to a situation where the timber material is not exposed to rain or ground moisture and wind blown corrosive salts (see Fig. 1 a).

• Service class 2 is characterised by a moisture content in the materials correspond-ing to a temperature of 20 °C and the relative humidity of the surrounding air only exceeding 85% for a few weeks per year. In service class 2 the average moisture content in most softwoods will not exceed 20%. This service class typically refers to a situation where the timber material is not washed by direct or wind blown rain but may be subject to wind blown salts (see Figure 1 b).

• Service class 3 is characterised by climatic conditions leading to higher moisture contents than in service class 2. This service class typically refers to a situation where the material is washed by direct or wind blown rain (see Fig. 1 c).

Figure 1: Illustration of Service classes. a) Service class 1 (closed); b) Service class 2

(sheltered); a) Service class 3 (exposed)

1.3. Material properties

Modification factors should be considered for the influence of load-duration and mois-ture content on strength. For serviceability limit states, if the structure consists of members or components having different time-dependent properties, the final mean value of modulus of elasticity, Emean,fin, shear modulus Gmean,fin, and slip modulus, Kser,fin, which are used to calculate the final deformation should be taken from the fol-lowing expressions:

a)

b)

c)

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)1( def

meanmean,fin k

EE

+=

)1( def

meanmean,fin k

GG

+=

)1( def

serser,fin k

KK

+=

For ultimate limit states, where the distribution of member forces and moments is af-fected by the stiffness distribution in the structure, the final mean value of modulus of elasticity, Emean,fin, shear modulus Gmean,fin and slip modulus, Kser,fin should be calcu-lated from the following expressions:

)1( def2

meanmean,fin k

EE

ψ+=

)1( def2

meanmean,fin k

GG

ψ+=

)1( def2

serser,fin k

KK

ψ+=

where: Emean is the mean value of modulus of elasticity; Gmean is the mean value of shear modulus; Kser is the mean value of slip modulus; kdef is a factor for the evaluation of creep deformation taking into account

the relevant service class (see Table 2); ψ2 is the factor for the quasi-permanent value of the action causing the

largest stress in relation to the strength (if this action is a permanent action, ψ2 should be replaced by 1.0).

Where a connection is constituted of timber elements with the same time-dependent behaviour, the value of kdef should be doubled.

The design member stiffness property Ed or Gd shall be calculated as:

M

meand γ

=E

E M

meand γ

=G

G

where γM is the partial factor for a material property (see Table 3).

The design value fd of a strength property shall be calculated as:

M

kmodd γ

=f

kf

where: fk is the characteristic value of a strength property; γM is the partial factor for a material property (see Table 3); kmod is a modification factor taking into account the effect of the duration

of load and moisture content (see Table 4). lf a load combination consists of actions belonging to different load-duration classes, a kmod value should be chosen which corresponds to the action with the shortest duration (e.g. for a combination of dead load and a short-term load, a value of kmod corresponding to the short-term load should be used).

The design value Rd of a resistance (load-carrying capacity of a member or a con-nection) shall be calculated as:

M

kmodd γ

=R

kR

where: Rk is the characteristic value of load-carrying capacity; γM is the partial factor for a material property (see Table 3); kmod is a modification factor taking into account the effect of the duration

of load and moisture content (see Table 4).

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Table 2: Values of kdef for timber and wood-based materials Service class

Material Standard 1 2 3

Solid timber EN 14081-1 0.60 0.80 2.00

Glued Laminated

Timber EN 14080 0.60 0.80 2.00

LVL EN 14374, EN 14279 0.60 0.80 2.00

Plywood

EN 636 Type EN 636-1 Type EN 636-2 Type EN 636-3

0.80 0.80 0.80

-

1.00 1.00

- -

2.50

OSB EN 300 OSB/2 OSB/3, OSB/4

2.25 1.50

-

2.25

- -

Particleboard

EN 312 Type P4 Type P5 Type P6 Type P7

2.25 2.25 1.50 1.50

-

3.00 -

2.25

- - - -

Fibreboard, hard EN 622-2 HB.LA HB.HLA1, HB.HLA2

2.25 2.25

-

3.00

- -

Fibreboard, medium EN 622-3 MBH.LA1, MBH.LA2 MBH.HLS1, MBH.HLS2

3.00 3.00

-

4.00

- -

Fibreboard, MDF EN 622-5 MDF.LA MDF.HLS

2.25 2.25

-

3.00

- -

Table 3: Recommended partial factors γM for material properties and resistances

Fundamental combinations: Solid timber Glued laminated timber LVL, plywood, OSB Particleboards Fibreboards, hard Fibreboards, medium Fibreboards, MDF Fibreboards, soft Connections Punched metal plate fasteners

1.3 1.25 1.2 1.3 1.3 1.3 1.3 1.3 1.3 1.25

Accidental combinations 1.0

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Table 4: Values of kmod modification factor

Load-duration class

Material Standard Service class

Perma-nent

action

Long term

action

Medium term

action

Short term

action

lnstanta-neous action

Solid timber EN 14081-1 1 2 3

0.60 0.60 0.50

0.70 0.70 0.55

0.80 0.80 0.65

0.90 0.90 0.70

1.10 1.10 0.90

Glued laminated timber

EN 14080 1 2 3

0.60 0.60 0.50

0.70 0.70 0.55

0.80 0.80 0.65

0.90 0.90 0.70

1.10 1.10 0.90

LVL EN 14374, EN 14279

1 2 3

0.60 0.60 0.50

0.70 0.70 0.55

0.80 0.80 0.65

0.90 0.90 0.70

1.10 1.10 0.90

Plywood

EN 636 Type EN 636-1 Type EN 636-2 Type EN 636-3

1 2 3

0.60 0.60 0.50

0.70 0.70 0.55

0.80 0.80 0.65

0.90 0.90 0.70

1.10 1.10 0.90

OSB

EN 300 OSB/2 OSB/3, OSB/4 OSB/3, OSB/4

1 1 2

0.30 0.40 0.30

0.45 0.50 0.40

0.65 0.70 0.55

0.85 0.90 0.70

1.10 1.10 0.90

Particle- board

EN 312 Type P4, P5 Type P5 Type P6, P7 Type P7

1 2 1 2

0.30 0.20 0.40 0.30

0.45 0.30 0.50 0.40

0.65 0.45 0.70 0.55

0.85 0.60 0.90 0.70

1.10 0.80 1.10 0.90

Fibreboard, hard

EN 622-2 HB.LA, HB.HLA 1, 2 HB.HLA 1,2

1 2

0.30 0.20

0.45 0.30

0.65 0.45

0.85 0.60

1.10 0.80

Fibreboard, medium

EN 622-3 MBH.LA1,2 MBH.HLS1,2 MBH.HLS1,2

1 1 2

0.20 0.20

-

0.40 0.40

-

0.60 0.60

-

0.80 0.80 0.45

1.10 1.10 0.80

Fibreboard, MDF

EN 622-5 MDF.LA, MDF.HLS MDF.HLS

1 2

0.20

-

0.40

-

0.60

-

0.80 0.45

1.10 0.80

1.3.1. The effect of member size on strength for solid timber

For rectangular solid timber (with a characteristic timber density ρk ≤ 700 kg/m3) the reference depth in bending or width in tension is 150 mm. For depths in bending or widths in tension of solid timber less than 150 mm the characteristic values for fm,k or ft,0,k may be increased by the factor kh, given by:

=

3.1

150min

2.0

h hk

where h is the depth for bending members or width for tension members, in [mm].

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1.3.2. The effect of member size on strength for glued laminated timber

For rectangular glued laminated timber, the reference depth in bending or width in tension is 600 mm. For depths in bending or widths in tension of glued laminated timber less than 600 mm the characteristic values for fm,k or ft,0,k may be increased by the factor kh given by:

=

1.1

600min

1.0

h hk

where h is the depth for bending members or width for tension members, in [mm].

1.3.3. The effect of member size on strength for Laminated veneer lumber (LVL)

The reference depth in bending is 300 mm. For depths in bending not equal to 300 mm the characteristic value for fm,k should be multiplied by the factor kh, given by:

s

hk

=2.1

300minh

where h is the depth of the member, in [mm]; s is the size effect exponent (s=2ν–0.05 where n is the coefficient of variation of

the test results. ν may be taken less than 0.10 only if it is verified from at least two years of documented experience).

The reference length in tension is 3000 mm. For lengths in tension not equal to 3000 mm the characteristic value for ft,0,k should be multiplied by the factor kℓ given by:

2/

1.1

0003min

s

k

=

where ℓ is the length of the member, in [mm].

1.4. Basis of structural analysis

The global structural behaviour should be assessed by calculating the action effects with a linear material model (elastic behaviour).

For structures able to redistribute the internal forces via connections of adequate ductility, elastic-plastic methods may be used for the calculation of the internal forces in the members.

The model for the calculation of internal forces in the structure or in part of it shall take into account the effects of deformations of the connections. In general, the influ-ence of deformations in the connections should be taken into account through their stiffness (rotational or translational for instance) or through prescribed slip values as a function of the load level in the connection.

Deviations from straightness and inhomogeneities of the material shall be taken into account by the structural analysis. These effects are taken into account implicitly by the design methods given in EC5 (EN 1995-1-1) standard.

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Reductions in the cross-sectional area shall be taken into account in the member strength verification. Reductions in the cross-sectional area may be ignored for the following cases:

– nails and screws with a diameter of 6 mm or less, driven without pre-drilling; – holes in the compression area of members, if the holes are filled with a material

of higher stiffness than the wood.

When assessing the effective cross-section at a joint with multiple fasteners, all holes within a distance of half the minimum fastener spacing measured parallel to the grain from a given cross-section should be considered as occurring at that cross-section.

In case of timber frames and arches, the effects of deflection (induced by external loads) on internal forces and moments shall be taken into account (e.g. by the appli-cation of second order linear analysis, see Figure 2).

Figure 2: Examples of assumed initial deviations in the geometry for a frame for a second order linear analysis (a), corresponding to a symmetrical load (b) and non-

symmetrical load (c)

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2. Ultimate limit states

2.1. Design of cross-sections subjected to stress in one principal direction

Following design methods apply to straight solid timber, glued laminated timber or wood-based structural products of constant cross-section, whose grain runs essen-tially parallel to the length of the member. The member is assumed to be subjected to stresses in the direction of only one of its principal axes (see Figure 3).

Figure 3: Member axes

2.1.1. Tension parallel to the grain

The following expression shall be satisfied:

σt,0,d ≤ ft,0,d

where σt,0,d is the design tensile stress along the grain; ft,0,d is the design tensile strength along the grain.

2.1.2. Tension perpendicular to the grain


σt,90,d ≤ kvol·ft,90,d

where σt,90,d is the design tensile stress perpendicular to the grain; ft,90,d is the design tensile strength perpendicular to the grain; kvol=1 in case of solid timber; kvol=(V/V0)

0.2 in case of glued laminated timber, where V0=0.01 m3 is the ref-erence volume, and V is the tensioned volume of the member in [m3]. V should not be taken greater than the two-thirds of the to-tal volume of the beam.

2.1.3. Compression parallel to the grain


σc,0,d ≤ fc,0,d

where σc,0,d is the design compressive stress along the grain; fc,0,d is the design compressive strength along the grain.

The above expression is used only for strength control of timber members. The sta-bility analysis of compressed members is discussed in chapter 2.3.1.

Key:

(1) direction of grain

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2.1.4. Compression perpendicular to the grain


σc,90,d ≤ kc,90 fc,90,d with σc,90,d = ef

dc,90,

A

F

where σc,90,d is the design compressive stress in the effective contact area per-pendicular to the grain;

Fc,90,d is the design compressive load perpendicular to the grain; Aef is the effective contact area in compression perpendicular to the grain; fc,90,d is the design compressive strength perpendicular to the grain; kc,90 is a factor taking into account the load configuration, the possibility of

splitting and the degree of compressive deformation.

The effective contact area perpendicular to the grain, Aef should be determined taking into account an effective contact length parallel to the grain, where the actual contact length, ℓ at each side is increased by 30 mm, but not more than ℓ or ℓ1/2 (see Figure 4).

Figure 4: Timber structural member on (a) continuous and (b) discrete supports

For members on continuous supports, provided that ℓ1 ≥ 2h (see Figure 4 a), the value of kc,90 should be taken as:

kc,90 = 1.25 for solid softwood timber, kc,90 = 1.5 for glued laminated softwood timber,

where h is the depth of the member and ℓ is the contact length.

For members on discrete supports, provided that ℓ1 ≥ 2h (see Figure 4 b), the value of kc,90 should be taken as:

kc,90 = 1.5 for solid softwood timber, kc,90 = 1.75 for glued laminated softwood timber, if ℓ ≤ 400 mm.

where h is the depth of the member and ℓ is the contact length.

If the above conditions are not fulfilled, then the value of kc,90 should be taken as 1.0.

2.1.5. Bending

The following expressions shall be satisfied:

0.1dm,z,

dm,z,m

dy,m,

dy,m, ≤σ

+σ

fk

f and 0.1

dm,z,

dm,z,

dy,m,

dy,m,m ≤

σ+

σff

k

(a) (b)

- 12 -

where σm,y,d and σm,z,d are the design bending stresses about the principal axes as shown in Figure 5;

fm,y,d and fm,z,d are the corresponding design bending strengths.

Figure 5: Distribution of bending stresses in a member

The factor km makes allowance for re-distribution of stresses and the effect of inho-mogeneities of the material in a cross-section. Its value should be taken as follows:

• For solid timber, glued laminated timber and LVL: – for rectangular sections: km = 0.7 (because of the stress-concentration

in the corners of the cross section) – for other cross-sections: km = 1.0

• For other wood-based structural products, for all cross-sections: km = 1.0

The above expressions are used only for strength control of timber members. The stability analysis of members subjected to bending is discussed in chapter 2.3.2.

2.1.6. Shear

For shear with a stress component parallel to the grain (see Figure 6 a), as well as for rolling shear with both stress components perpendicular to the grain (see Fig-ure 6 b), the following expression shall be satisfied:

τd ≤ fv,d

where τd is the design shear stress;

fv,d is the design shear strength for the actual condition. The shear strength for rolling shear is approximately equal to twice the tensile strength per-pendicular to grain.

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Figure 6: (a) Member with a shear stress component parallel to the grain (b) Member

with both stress components perpendicular to the grain (rolling shear)

For the verification of shear resistance of members in bending, the influence of cracks should be taken into account using an effective width of the member (see Fig-ure 7) given as:

bef = kcr b

where b is the width of the relevant section of the member, and the recommended value of kcr is given as:

kcr = 0.67 for solid and glued laminated timber; kcr = 1.0 for other wood-based products.

Figure 7: Effective width of a member subjected to shear

At supports, the contribution to the total shear force of a concentrated force F act-ing on the top side of the beam and within a distance h or hef from the edge of the support may be disregarded (see Figure 8). For beams with a notch at the support this reduction in the shear force applies only when the notch is on the opposite side to the support.

Figure 8: Conditions at a support, for which the concentrated force F may be disre-

garded in the calculation of the shear force

bef

b

a) b)

- 14 -

2.1.7. Torsion


τtor,d ≤ kshape fv,d

where

+=

sectioncrossrrectangulaafor0.2

15.01min

sectioncrosscircularafor2.1

shapeb

hk

τtor,d is the design torsional stress; fv,d is the design shear strength; kshape is a factor depending on the shape of the cross-section; h is the larger cross-sectional dimension; b is the smaller cross-sectional dimension.

2.2. Design of cross-sections subjected to combined stresses

Following design methods apply to straight solid timber, glued laminated timber or wood-based structural products of constant cross-section, whose grain runs essen-tially parallel to the length of the member. The member is assumed to be subjected to stresses from combined actions or to stresses acting in two or three of its principal axes.

2.2.1. Compression stresses at an angle to the grain

Interaction of compressive stresses in two or more directions shall be taken into ac-count. The compressive stresses at an angle α to the grain, (see Figure 9), should satisfy the following expression:

α+α≤σ

22

d,90c,90c,

d,0c,

d,0c,dα,c,

cossinfk

f

f

where σc,α,d is the compressive stress at an angle α to the grain; kc,90 is a factor taking into account the effect of any of stresses perpen-

dicular to the grain (see chapter 2.1.4).

Figure 9: Compressive stresses at an angle to the grain

2.2.2. Combined bending and axial tension


0.1dm,z,

dm,z,m

dy,m,

dy,m,

d,0t,

d,0t, ≤σ

+σ

+σ

fk

ff and 0.1

dm,z,

dm,z,

dy,m,

dy,m,m

d,0t,

d,0t, ≤σ

+σ

+σ

ffk

f

- 15 -

2.2.3. Combined bending and axial compression


0.1dm,z,

dm,z,m

dy,m,

dy,m,

2

d,0c,

d,0c, ≤σ

+σ

+

σf

kff

and 0.1dm,z,

dm,z,

dy,m,

dy,m,m

2

d,0c,

d,0c, ≤σ

+σ

+

σff

kf

The above expressions are used only for strength control of timber members. The sta-bility analysis of members subjected to bending is discussed in the following chapter.

2.3. Stability of timber members

The bending stresses due to initial curvature, eccentricities and induced deflection shall be taken into account, in addition to those due to any lateral load. Column sta-bility and lateral torsional stability shall be verified using the characteristic material properties (e.g. E0,05).

2.3.1. Columns subjected to either compression or combined compression and bending

The relative slenderness ratios of the column should be taken as:

05,0

k,0c,yrel,y E

f

πλ

=λ and 05,0

k,0c,zzrel, E

f

πλ

=λ

where λy and λrel,y are .y (deflection in the z direction); λz and λrel,z are slenderness ratios corresponding to bending about the

z axis (deflection in the y direction); E0,05 is the 5% value of the modulus of elasticity parallel to the grain.

For cases, where both λrel,z ≤ 0.3 and λrel,y ≤ 0.3 the stresses should satisfy the ex-pressions presented in chapter 2.2.3. In all other cases the stresses, which will be increased due to deflection, should satisfy the following expressions:

0.1f dm,z,

dm,z,m

dy,m,

dy,m,

d,0c,yc,

d,0c, ≤σ

+σ

+σ

fk

fk

0.1dm,z,

dm,z,

dy,m,

dy,m,m

d,0c,c,z

d,0c, ≤σ

+σ

+σ

ffk

fk

where the symbols are defined as follows:

2rel,y

2yy

c,y1

λ−+=

kkk with ( )( )2

rel,yrel,ycy 3,015,0 λ+−λβ+=k

2

zrel,2zz

zc,1

λ−+=

kkk with ( )( )2

zrel,zrel,cz 3,015,0 λ+−λβ+=k

- 16 -

βc is a factor for members within the straightness limits defined as:

=βLVL and timber laminated glued for1.0

timber solid for2.0c

2.3.2. Beams subjected to either bending or combined bending and compression

Lateral torsional stability of beams shall be verified both in the case (a) where only a moment My exists about the „strong” axis y, and (b) where a combination of moment My and compressive force Nc exists.

In case of bending the relative (torsional) slenderness should be taken as:

critm,

km,mrel, σ

=λf

where σm,crit is the critical bending stress calculated according to the classical theory of stability, using 5% stiffness values. The critical bending stress should be taken as:

yef

tor05,0z05,0

y

crity,critm, W

IGIE

W

M

π==σ

where E0,05 is the 5% value of modulus of elasticity parallel to grain; G0,05 is the 5% value of shear modulus parallel to grain; Iz is the second moment of area about the „weak” axis z; Itor is the torsional moment of inertia; ℓef is the effective length of the beam, depending on the support condi-

tions and the load configuration, according to Table 5; Wy is the section modulus about the „strong” axis y.

For softwood with solid rectangular cross-section, σm,crit should be taken as:

05,0ef

2

critm,78,0

Eh

b

=σ

where b is the width of the beam, and h is the depth of the beam.

(a) In the case where only a moment My exists about the „strong” axis y, the stresses should satisfy the following expression:

σm,d ≤ kcrit fm,d

where σm,d is the design bending stress; fm,d is the design bending strength; kcrit is a factor which takes into account the reduced bending strength due to

lateral buckling.

The deviation from straightness measured midway between the supports should, for columns and beams where lateral instability can occur, or members in frames, be limited to 1/500 times the length of glued laminated timber or LVL members and to 1/300 times the length of solid timber. If this condition is met, the factor kcrit may be determined from the following expression:

- 17 -

λ<λ

≤λ<λ−

≤λ

=

mrel,2mrel,

mrel,mrel,

mrel,

crit

4.1if1

4.175.0if75.056.1

75.0if1

k

The value of factor kcrit may be taken as 1.0 for a beam where lateral displacement of its compressive edge is prevented throughout its length and where torsional rota-tion is prevented at its supports.

(b) In the case where a combination of moment My about the „strong” axis y and a compressive force Nc exists, the stresses should satisfy the following expression:

0.1d,0c,c,z

dc,

2

dm,crit

dm, ≤σ

+

σfkfk

where σm,d is the design bending stress; σc,0,d is the design compressive stress parallel to grain; fm,d is the design bending strength; fc,0,d is the design compressive strength parallel to grain; kc,z is a factor described in chapter 2.3.1.

Table 5: Effective length as a ratio of the span

Beam type Loading type ef / *

Simply supported Constant moment Uniformly distributed load Concentrated force at the middle of the span

1.0 0.9 0.8

Cantilever Uniformly distributed load Concentrated force at the free end

0.5 0.8

* The ratio between the effective (buckling) length ef and the span ℓ is valid for a

beam with torsionally restrained supports and loaded at the centre of gravity. lf the load

is applied at the compression edge of the beam, ef should be increased by 2h and

may be decreased by 0.5h for a load at the tension edge of the beam.

3. Limiting values for deflections of beams

The components of deflection resulting from a combination of actions are shown in Figure 9, where the symbols are defined as follows:

- wc is the precamber (if applied); - winst is the instantaneous deflection; - wcreep is the creep deflection; - wfin is the final deflection; - wnet,fin is the net final deflection.

- 18 -

Figure 9: Components of deflection

The net deflection below a straight line between the supports, wnet,fin should be taken as:

wnet,fin = winst + wcreep - wc = wfin - wc

The recommended range of limiting values of deflections for beams with span ℓ is given in Table 6 depending upon the level of deformation deemed to be acceptable.

Table 6: Examples of limiting values for deflections of beams

winst wnet,fin wfin

Beam on two supports ℓ/300 to ℓ/500 ℓ/250 to ℓ/350 ℓ/150 to ℓ/300

Cantilevering beams ℓ/150 to ℓ/300 ℓ/125 to ℓ/175 ℓ/75 to ℓ/150

4. Bracing

4.1. General aspects

Timber structures which are not otherwise adequately stiff shall be braced to prevent instability or excessive deflection. The stress caused by geometrical and structural im-perfections, and by induced deflections (including the contribution of any joint slip) shall be taken into account. The bracing forces shall be determined on the basis of the most unfavourable combination of structural imperfections and induced deflections.

4.2. Single members in compression

For single timber elements in compression, requiring lateral support at intervals a (see Figure 10), the initial deviations from straightness between supports should be within a/500 for glued laminated or LVL members, and a/300 for other members. Each intermediate support should have a minimum spring stiffness:

a

NkC d

s=

where ks is a modification factor (see Table 7); Nd is the mean design compressive force in the element; a is the bay length (see Figure 10).

- 19 -

The design stabilizing force Fd at each support should be calculated as:

LVL and timber laminated glued for

timber solid for

2f,

d

1f,

d

d

=

k

Nk

N

F

where kf,1 and kf,2 are modification factors (the recommended values can be found in Table 7).

Figure 10: Examples of single members in compression braced by lateral supports

The design stabilizing force Fd for the compressive edge of a rectangular beam should be also determined from the above expression. In this case the value of the mean design compressive force can be calculated as:

( )h

MkN d

critd 1−=

The value of kcrit should be determined for the unbraced beam according to chapter 2.3.2., and Md is the maximum design moment acting on the beam of depth h.

4.3. Bracing of beam or truss systems

For a series of n parallel timber members which require lateral supports at intermedi-ate nodes A, B, etc. (see Figure 11) a bracing system should be provided, which, in addition to the effects of external horizontal load (e.g. wind), should be capable of resisting an internal stability load per unit length q, as follows:

3f,

dd k

Nnkq =

where

=

15

0,1

mink

Nd is the mean design compressive force in the members; ℓ is the overall span of the stabilizing system, in [m]; kf,3 is a modification factor (see Table 7).

- 20 -

Figure 11: Beam or truss system requiring lateral supports

The values of the modification factors ks, kf,1, kf,2 and kf,3 depend on influences such as workmanship, span of the girders, etc. Ranges of values are given in Table 7 where the recommended values are underlined.

Table 7: Recommended values of modification factors

Modification factor Range

ks 4 to 1

kf,1 50 to 80

kf,2 80 to 100

kf,3 30 to 80

The horizontal deflection of the bracing system due to force qd and any other external load (e.g. wind), should not exceed ℓ/500.

5. Numerical examples

5.1. Member subjected to compression and bending

We have a three-hinged timber frame as illustrated in Figure 12. The frame is loaded by the PEd concentrated and the qEd uniformly distributed design forces. The task is to check the stability of member "a" in the plane of the frame.

Cross section

of bar “a”

b

h

PEd

qEd

a

α α

LL

Figure 12: Three-hinged frame subjected to compression and bending

(1) n members of truss system (2) Bracing system (3) Deflection of truss system due to

imperfections and second order effects

(4) Stabilizing forces (5) External load on bracing (6) Reaction forces of bracing due to

external loads (7) Reaction forces of truss system due

to stabilizing forces

- 21 -

Parameters for the calculation:

Strength class of timber: C22 (characteristic strength values are given in Table 8) Service class: 1 L = 4.00 m qEd = 2.00 kN/m (short term action) PEd = 20.00 kN (short term action) h = 150 mm b = 100 mm α = 45º γM = 1.3 (for solid timber, see Table 3) kmod = 0.9 (solid timber, service class 1, short term action, see Table 4)

Calculation:

A) Calculation of design forces in bar " a"

NEd PEd cos α( )⋅ 14.14 kN⋅==

MEd

qEd L2⋅

84 kNm⋅==

B) Calculation of section properties

A b h⋅ 15000 mm2⋅==

Iyb h

3⋅12

2.813 107× mm

4⋅==

C) Calculation of design stresses

σc.0.d

NEd

A0.94

N

mm2

⋅==

σm.d

MEd

Iy

h

2⋅ 10.67

N

mm2

⋅==

D) Calculation of neccesary design strength values

fc.0.k 20N

mm2

= (see Table 8) fc.0.d

fc.0.k

γMkmod⋅ 13.85

N

mm2

⋅==

fm.k 22N

mm2

= (see Table 8) fm.d

fm.k

γMkmod⋅ 15.23

N

mm2

⋅==

Remark: The size effect factor kh can be neglected for the calculation of the design bending

strength fm,d, because the depth of the cross section is not smaller than the reference

depth (150 mm).

PEd

qEd

NEd = PEd·cosα

α

.

MEd = qEd·L2/8

Figure 13: Determination of internal forces in bar “a”

- 22 -

E) Determination of buckling reduction factor kc,y

Buckling length: L0 L 4 m== (because both ends of bar " a" are hinged)

Slenderness ratio: λy

L0

Iy

A

92.38== (bending about the y axis)

Modulus of elasticity: E0.05 6.7kN

mm2

= (see Table 8)

Relative slenderness: λrel.y

λy

π

fc.0.k

E0.05⋅ 1.61== > 0.3 so stability should be checked

βc 0.2= (for solid timber)

ky 0.5 1 βc λrel.y 0.3−( )⋅+ λrel.y2+

⋅ 1.92==

kc.y1

ky ky2

λrel.y2−+

0.336==

F) Stability control of bar " a" in the plane of the frame

km 0.7= (rectangular cross section)

σc.0.d

kc.y fc.0.d⋅

σm.d

fm.d+ 0.903= < 1.00 satisfactory

5.2. Column and tie beam connection

We have a wedge locked column and tie beam connection as illustrated in Figure 14. The tie beam is subjected to a PEd concentrated force. The task is to determine the maximum value of the PEd force that can be resisted by the given connection. Parameters for the calculation:

Strength class of timber for column and beam: C16 (see Table 8) Strength class of timber for wedge: D30 (see Table 8) Service class: 1 PEd is long term action Cross section of the column is 240×240 mm Cross section of the tie beam is 100×120 mm Cross section of the wedge is 50×50 mm γM = 1.3 (for solid timber, see Table 3) kmod = 0.7 (solid timber, service class 1, long term action, see Table 4)

- 23 -

Figure 14: Wedge locked column and tie beam connection

Calculation:

A) Calculation of the maximum allowed tensile force at the connection

The beam will be examined at the wedge (weakened cross section, see Figure 15).

Critical cross-sectional area: At hbeam bbeam bwedge−( )⋅ 7000 mm2⋅==

Characteristic tensile strength: ft.0.k 10N

mm2

= (see Table 8)

Size effect factor: kh min 1.3150mm

bbeam

0.2,

1.046==

Design tensile strength: ft.0.d

ft.0.k

γMkmod⋅ kh⋅ 5.63

N

mm2

⋅==

Design tensile resistance: Rt.d ft.0.d At⋅ 39.41 kN⋅==

B) Calculation of the maximum allowed compressive force at the connection

The connection of the wedge and the column will be examined for compressionperpendicular to grain (see Figure 15).

Critical cross-sectional area: Ac hbeam bwedge⋅ 5000 mm2⋅==

Characteristic compressive strength: fc.90.k 2.2N

mm2

= (see Table 8)

FEd

Side view

240

10050

Front view

50

50

100

50 50 50

45 45

120

FEd

240

60

60

Plan view

Key:

Column Tie Beam Wedge

Grain direction

- 24 -

At Ac

Av

Avr

Figure 15: Critical cross sections for tension, compression and shear

Design compressive strength: fc.90.d

fc.90.k

γMkmod⋅ 1.18

N

mm2

⋅==

Design compressive resistance: Rc.d fc.90.d Ac⋅ 5.92 kN⋅==

C) Calculation of the maximum allowed shear force at the connection

The free end of the beam is examined for shear parallel to the grain (see Figure 15).

Critical cross-sectional area: Av 2hbeam 100⋅ mm 20000 mm2⋅==

Characteristic shear strength: fv.k 1.8N

mm2

= (see Table 8)

Design shear strength: fv.d

fv.k

γMkmod⋅ 0.97

N

mm2

⋅==

Design shear resistance: Rv.d fv.d Av⋅ 19.38 kN⋅==

The wedge is examined for rolling shear (see Figure 15).

Critical cross-sectional area: Avr 2bwedge hwedge⋅ 5000 mm2⋅==

Characteristic tensile strength of the wedge (timber class D30):

ft.90.k 3N

mm2

= (see Table 8)

Characteristic rolling shear strength: fvr.k 2 ft.90.k⋅ 6N

mm2

⋅==

Design rolling shear strength: fvr.d

fvr.k

γMkmod⋅ 3.23

N

mm2

⋅==

Design rolling shear resistance: Rvr.d fvr.d Avr⋅ 16.15 kN⋅==

- 25 -

D) Calculation of the maximum allowed PEd force

The maximum value of the PEd force that can be resisted by the given connection:

PEd.max min Rt.d Rc.d, Rv.d, Rvr.d, ( ) 5.92 kN⋅==

The critical failure mode is compressive failure perpendicular to grain at the connectionof the wedge and the column.

5.3. Dovetail halving connection

We have a dovetail halving connection as illustrated in Figure 16. The connection is subjected to a PEd concentrated force. The task is to verify the given connection.

160

PEd

150

75

PEd/2

50

110

PEd

150

70

PEd/2

Side view Cross section

Figure 16: Dovetail halving connection


Strength class of timber: C24 (characteristic strength values are given in Table 8) Service class: 1 PEd = 10.00 kN (medium term action) γM = 1.3 (for solid timber, see Table 3) kmod = 0.8 (solid timber, service class 1, medium term action, see Table 4)

Calculation: A) Calculation of the tensile resistance of the connection

The critical cross section for tension is displayed in Figure 17.

Critical cross-sectional area: At 70mm 75⋅ mm 5250 mm2⋅==

Characteristic tensile strength: ft.0.k 14N

mm2

= (see Table 8)

- 26 -

At Avr Av Ac

Figure 17: Critical cross sections for tension, shear and compression

Design tensile strength: ft.0.d

ft.0.k

γMkmod⋅ 8.62

N

mm2

⋅==

Remark: The size effect factor kh can be neglected for the calculation of the design tensile

strength ft,0,d, because the depth of the cross section is not smaller than the reference

depth (150 mm).

Design tensile resistance: Rt.d ft.0.d At⋅ 45.23 kN⋅==

B) Calculation of the shear resistance of the connection

The end of the column is examined for shear parallel to the grain (see Figure 17).

Critical cross-sectional area: Av 2 110× mm 75⋅ mm 16500 mm2⋅==

Characteristic shear strength: fv.k 2.5N

mm2

= (see Table 8)

Design shear strength: fv.d

fv.k

γMkmod⋅ 1.54

N

mm2

⋅==

Design shear resistance: Rv.d fv.d Av⋅ 25.38 kN⋅==

The beam is examined for rolling shear (see Figure 17).

Critical cross-sectional area: Avr 2 110× mm 75⋅ mm 16500 mm2⋅==

Characteristic tensile strength perpendicular to grain:

ft.90.k 0.5N

mm2

= (see Table 8)

Characteristic rolling shear strength: fvr.k 2 ft.90.k⋅ 1N

mm2

⋅==

Design rolling shear strength: fvr.d

fvr.k

γMkmod⋅ 0.62

N

mm2

⋅==

Design rolling shear resistance: Rvr.d fvr.d Avr⋅ 10.15 kN⋅==

- 27 -

C) Calculation of the compressive resistance of the connection

The connection surface of the column and the beam will be examined for compression atan angle to the grain (see Figure 17).

Length of the connection surface: h 110mm( )2

40mm( )2+ 117 mm==

h α

β Rc,α,d

Figure 18: Compression at the connection surface

The compressive force will be perpendicular to the connection surface (see Figure 18).The angle of the force to the grain will be:

α atan110mm

40mm

70.02 °== for the column (see Figure 18)

β 90° α− 19.98 °== for the beam (see Figure 18)

Critical cross-sectional area: Ac h 75⋅ mm 8778.52 mm2⋅==

Calculation of design compressive strength at an angle to the grain:

fc.0.k 21N

mm2

= fc.0.d

fc.0.k

γMkmod⋅ 12.92

N

mm2

==

fc.90.k 2.5N

mm2

= fc.90.d

fc.90.k

γMkmod⋅ 1.54

N

mm2

==

kc.90 1= (see chapter 2.1.4)

fc.α.d

fc.0.d

fc.0.d

kc.90 fc.90.d⋅sin α( )

2⋅ cos α( )2+

1.71N

mm2

==

fc.β.d

fc.0.d

fc.0.d

kc.90 fc.90.d⋅sin β( )

2⋅ cos β( )2+

6.93N

mm2

==

Critical is the compression on the column (fc,α,d < fc,β,d).

- 28 -

Design compressive resistance of one connection surface:

Rc.α.d fc.α.d Ac⋅ 15.05 kN⋅==

Total design compressive resistance parallel to the given PEd tensile force:

Rc.d 2 Rc.α.d⋅ cos α( )⋅ 10.29 kN⋅==

D) Verification of the connection

The resistance of the given connection:

Rd min Rt.d Rc.d, Rv.d, Rvr.d, ( ) 10.15 kN⋅== > PEd = 10 kN satisfactory

(The critical failure mode is the rolling shear of the beam.)

- 29 -

Table 8: Strength classes and characteristic values according to EN 338

C1

4C

16

C1

8C

20

C2

2C

24

C2

7C

30

C3

5C

40

C4

5C

50

D3

0D

35

D4

0D

50

D6

0D

70

Be

nd

ing

f m,k

14

16

18

20

22

24

27

30

35

40

45

50

30

35

40

50

60

70

Te

nsi

on

pa

ralle

l to

gra

inf t,

0,k

81

01

11

21

31

41

61

82

12

42

73

01

82

12

43

03

64

2

Te

nsi

on

pe

rpe

nd

icu

lar

to g

rain

f t,90

,k0

,40

,50

,50

,50

,50

,50

,60

,60

,60

,60

,60

,60

,60

,60

,60

,60

,60

,6

Co

mp

ress

ion

pa

ralle

l to

gra

inf c

,0,k

16

17

18

19

20

21

22

23

25

26

27

29

23

25

26

29

32

34

Co

mp

ress

ion

pe

rpe

nd

icu

lar

to g

rain

f c,9

0,k

22

,22

,22

,32

,42

,52

,62

,72

,82

,93

,13

,28

8,4

8,8

9,7

10

,51

3,5

Sh

ea

rf v

,k1

,71

,82

2,2

2,4

2,5

2,8

33

,43

,83

,83

,83

3,4

3,8

4,6

5,3

6

Me

an

va

lue

of

mo

du

lus

of

ela

stic

ity p

ara

llel t

o

gra

inE

0,m

ean

78

99

,51

01

11

1,5

12

13

14

15

16

10

10

11

14

17

20

5%

va

lue

of

mo

du

lus

of

ela

stic

ity p

ara

llel t

o

gra

inE

0,05

4,7

5,4

66

,46

,77

,47

,78

8,7

9,4

10

10

,78

8,7

9,4

11

,81

4,3

16

,8

Me

an

va

lue

of

mo

du

lus

of

ela

stic

ity

pe

pe

nd

icu

lar

to g

rain

E90

,mea

n0

,23

0,2

70

,30

,32

0,3

30

,37

0,3

80

,40

,43

0,4

70

,50

,53

0,6

40

,69

0,7

50

,93

1,1

31

,33

Me

an

va

lue

of

she

ar

mo

du

lus

Gm

ean

0,4

40

,50

,56

0,5

90

,63

0,6

90

,72

0,7

50

,81

0,8

80

,94

10

,60

,65

0,7

0,8

81

,06

1,2

5

De

nsi

tyρ k

29

03

10

32

03

30

34

03

50

37

03

80

40

04

20

44

04

60

53

05

60

59

06

50

70

09

00

Me

an

va

lue

of

de

nsi

tyρ m

ean

35

03

70

38

03

90

41

04

20

45

04

60

48

05

00

52

05

50

64

06

70

70

07

80

84

010

80

Str

en

gth

pro

pe

rtie

s [

N/m

m2]

Sti

ffn

es

s p

rop

ert

ies

[k

N/m

m2 ]

De

ns

ity

[kg

/m3 ]

De

cid

uo

us

sp

ec

ies

Po

pla

r a

nd

so

ftw

oo

d s

pe

cie

s

- 30 -

II. Design of masonry structures according to EC6 (EN 1996-1-1)

6. Terms and Definitions

6.1. Terms relating to masonry

masonry: an assemblage of masonry units laid in a specified pattern and joined to-gether with mortar

unreinforced masonry: masonry not containing sufficient reinforcement so as to be considered as reinforced masonry

reinforced masonry: masonry in which bars or mesh are embedded in mortar or con-crete so that all the materials act together in resisting action effects

masonry bond: disposition of units in masonry in a regular pattern to achieve com-mon action

characteristic strength of masonry: value of the strength of masonry having a pre-scribed probability of 5% of not being attained in a hypothetically unlimited test se-ries. This value generally corresponds to a specified fractile of the assumed statisti-cal distribution of the particular property of the material or product in a test series. A nominal value is used as the characteristic value in some circumstances.

compressive strength of masonry: the strength of masonry in compression without the effects of platen restraint, slenderness or eccentricity of loading

shear strength of masonry: the strength of masonry shear subjected to shear forces

flexural strength of masonry: the strength of masonry in bending

6.2. Terms relating to masonry units

masonry unit: a preformed component, intended for use in masonry construction

groups 1, 2, 3 and 4 masonry units: group designations for masonry units, according to the percentage units when laid

gross area: the area of a cross-section through the unit without reduction for the area of holes, voids and re-entrants

compressive strength of masonry units: the mean compressive strength of a specified number of masonry units

normalized compressive strength of masonry units: the compressive strength of ma-sonry units converted to the air dried compressive strength of an equivalent 100 mm wide x 100 mm high masonry unit

6.3. Terms relating to mortar

masonry mortar: mixture of one or more inorganic binders, aggregates and wa-ter, and sometimes additions and/or admixtures, for bedding, jointing and pointing of masonry

general purpose masonry mortar: masonry mortar without special characteristics

thin layer masonry mortar: designed masonry mortar with a maximum aggregate size less than or equal to a prescribed figure (see Figure 19)

- 31 -

thin mortarlayer

Figure 19: Masonry laid on thin mortar layer

lightweight masonry mortar: designed masonry mortar with a dry hardened density equal to or below 1300 kg/m3

designed masonry mortar: a mortar whose composition and manufacturing method is chosen in order to achieve specified properties (performance concept)

prescribed masonry mortar: mortar made in predetermined proportions, the proper-ties of which are assumed from the stated proportions of the constituents (recipe con-cept)

factory made masonry mortar: mortar batched and mixed in a factory

compressive strength of mortar: the mean compressive strength of a specified num-ber of mortar specimens after curing for 28 days

bed joint: a mortar layer between the bed faces of masonry units

perpend joint (head joint): a mortar joint perpendicular to the bed joint and to the face of wall

longitudinal joint: a vertical mortar joint within the thickness of a wall, parallel to the face of the wall

6.4. Terms relating to wall types

load-bearing wall: a wall primarily designed to carry an imposed load in addition to its own weight

single-leaf wall: a wall without a cavity or continuous vertical joint in its plane (see Figure 20)

double-leaf wall: a wall consisting of two parallel leaves with the longitudinal joint be-tween filled solidly with mortar and securely tied together with wall ties so as to result in common action under load (see Figure 21)

cavity wall: a wall consisting of two parallel single-leaf walls, effectively tied to-gether with wall ties or bed joint reinforcement. The space between the leaves is left as a continuous cavity or filled with non-loadbearing thermal insulating mate-rial (see Figure 22)

shear wall: a wall to resist lateral forces in its plane

stiffening wall: a wall set perpendicular to another wall to give it support against lat-eral forces or to resist buckling and so to provide stability to the building

- 32 -

Figure 20: Sample cross sections of single-leaf wall (a) without vertical joint; (b) with vertical joint

continuous longitudinal joint

Figure 21: Sample cross section of double-leaf wall

Figure 22: Sample cross sections of cavity wall

7. Basis of design

7.1. Ultimate limit states

The relevant values of the partial factor for materials γM shall be used for the ulti-mate limit state for ordinary and accidental situations. When analysing the struc-ture for accidental actions, the probability of the accidental action being present shall be taken into account. The value of partial factor depends on the class of the masonry (quality of the erection, see Table 9), the category of the masonry unit during the manufacturing process (quality control), and the type of the mortar. Recommended values of the partial factor, given as classes that may be related to execution control, are given in the table 9.

vertical joints

a) b)

- 33 -

Table 9: Recommended values of the partial factor γM

Class Requirements during the construction

1 2 3 4 5

Construction is supervised by a suitably qualified and experienced person employed by the contractor.

X X X X X

Construction is supervised by a suitably qualified and ex-perienced person who is independent from the contractor.

X X X

Strength of mortar is tested and controlled in the laboratory on specimens prepared on site.

X X

Factory made designed masonry mortar is used for the construction of the wall.

X

Prescribed mortar mixed on site is used for the con-struction of the wall.

X X X X

The mortar saturation of joints 100% 100% 100% 90% 80% There is no smaller unit than half unit in the masonry

There is no smaller unit than quarter unit in the masonry

Way of bricklaying Cutting of masonry units is done by me-chanical or manual sawing

Material γM

A Units of Category I., designed mortar 1.5 1.7 2.0 2.2 2.5 B Units of Category I., prescribed mortar 1.7 2.0 2.2 2.5 2.7 C Units of Category II., any mortar 2.0 2.2 2.5 2.7 3.0

7.2. Serviceability limit states

Where simplified rules are given in the relevant clauses dealing with serviceability limit states, detailed calculations using combinations of actions are not required. When needed, the partial factor for materials, for the serviceability limit state, is γM = 1.0.

8. Materials

8.1. Masonry units

8.1.1. Types and grouping of masonry units

Masonry units shall comply with any of the following types:

− clay units − calcium silicate units − aggregate concrete units (dense or lightweight aggregate) − autoclaved aerated concrete units − manufactured stone units − dimensioned natural stone units

Masonry units may be Category I. or Category II. depending on their quality control during the manufacturing process (normally the manufacturer will state the grouping of his units). Masonry units should be grouped as Group 1, Group 2 , Group 3 or Group 4. Autoclaved aerated concrete, manufactured stone and dimensioned natural stone units are considered to be Group 1 . The geometrical requirements for group-ing of clay, calcium silicate and aggregate concrete units are given in table 10.

- 34 -

Table 10: Geometrical requirements for grouping of masonry units

Materials and limits for masonry units

Group 2. Group 3. Group 4.

Group 1. (all mate-

rials) Unit

Vertical holes [%] Horizontal holes [%]

clay > 25; ≤ 55 ≥ 25; ≤ 70 > 25; ≤ 70

calcium silicate

> 25; ≤ 55 not used not used Volume of all holes (% of the gross volume)

≤ 25

concreteb > 25; ≤ 60 > 25; ≤ 70 > 25; ≤ 50

clay

each of multiple holes ≤ 2

gripholes up to a total of 12,5


gripholes up to a total of 12,5


calcium silicate


gripholes up to a total of 30

not used not used Volume of any hole (% of the gross volume)

≤ 12.5

concreteb






web shell web shell web shell

Clay ≥ 5 ≥ 8 ≥ 3 ≥ 6 ≥ 5 ≥ 6

calcium silicate ≥ 5 ≥ 10 not used not used

Declared values of thickness of webs and shells [mm]

No re-quirement

concreteb ≥ 15 ≥ 18 ≥ 15 ≥ 15 ≥ 20 ≥ 20

Clay ≥ 16 ≥ 12 ≥ 12

calcium silicate ≥ 20 not used not used

Declared value of combined thicknessa of webs and shells (% of the overall width)

No re-quirement

concreteb ≥ 18 ≥ 15 ≥ 45

a The combined thickness is the thickness of the webs and shells, measured horizontally in the relevant direc-tion. The check is to be seen as a qualification test and need only be repeated in the case of principal changes to the design dimensions of units. b In the case of conical holes, or cellular holes, use the mean value of the thickness of the webs and theshells.

8.1.2. Compressive strength of masonry units

The mean compressive strength of the masonry units is usually declared by the manu-facturer. The compressive strength of masonry units, to be used in design, shall be the normalised mean compressive strength, fb which value also considers the width and height of the unit. The normalised mean compressive strength can be calculated as:

fb = δ·f

where f is the mean compressive strength of the masonry unit; δ is a coefficient depending on the dimensions of the unit, see Table 11.

- 35 -

Table 11: Values of the δ coefficient

Width of the masonry unit [mm] Height of the masonry unit

[mm] 50 100 150 200 250 or greater50 0.85 0.75 0.70 - - 65 0.95 0.85 0.75 0.70 0.65

100 1.15 100 0.90 0.80 0.75 150 1.30 1.20 1.10 1.00 0.95 200 1.45 1.35 1.25 1.15 1.10

250 or greater 1.55 1.45 1.35 1.25 1.15

8.2. Mortar

Masonry mortars are defined as general purpose, thin layer or lightweight mortar ac-cording to their constituents. Masonry mortars are considered as designed or pre-scribed mortars according to the method of defining their composition. Masonry mor-tars may be factory made (pre-batched or pre-mixed), semi-finished factory made or site-made, according to the method of manufacture.

Mortars should be classified by their compressive strength, expressed as the letter M followed by the compressive strength in N/mm2, for example, M5. Prescribed ma-sonry mortars, additionally to the M number, will be described by their prescribed constituents, e.g. 1:1:5 is the cement:lime:sand by volume.

8.3. Mechanical properties of masonry

8.3.1. Characteristic compressive strength of masonry

The characteristic compressive strength of masonry, fk shall be determined from re-sults of tests. If there are no test results available, the characteristic compressive strength for masonry made with general purpose mortar or lightweight mortar may be determined from the following equation:

0.3m

0.7bk ffKf ⋅⋅=

where fk is the characteristic compressive strength of the masonry, in [N/mm2]; K is a constant (see Table 12); fb is the normalised mean compressive strength of the units, in the direc-

tion of the applied action, in [N/mm2]; fm is the compressive strength of the mortar, in [N/mm2].

For masonry made with thin layer mortar in bed joints of thickness 0.5 to 3 mm, and clay units of Group 1 and 4, calcium silicate, aggregate concrete or autoclaved aer-ated concrete units, the characteristic compressive strength can be determined from:

0,85bk fKf ⋅=

For masonry made with thin layer mortar, in bed joints of thickness 0.5 to 3 mm, and clay units of Group 2 and 3, the characteristic compressive strength can be deter-mined from:

0,7bk fKf ⋅=

- 36 -

The above equations can be used if the following requirements are satisfied:

− the masonry is detailed in accordance with EN 1996-1-1 Standard; − fb is not taken to be greater than 75 N/mm2 when units are laid in general pur-

pose mortar, and fb is not taken to be greater than 50 N/mm2 when units are laid in thin layer mortar;

− fm is not taken to be greater than 20 N/mm2 nor greater than 2fb when units are laid in general purpose mortar, and fm is not taken to be greater than 10 N/mm2 when units are laid in lightweight mortar;

− the thickness of the masonry is equal to the width or length of the unit, so that there is no mortar joint parallel to the face of the wall through all or any part of the length of the wall;

− the coefficient of variation of the strength of the masonry units is not more than 25%.

Table 12: Values of K for use with general purpose, thin layer and lightweight mortars

Lightweight mortar of density

Masonry unit General purpose mortar

Thin layer mortar

(bed joint ≥ 0.5 mm and ≤ 3 mm )

600 ≤ ρd

≤ 800 kg/m3

800 < ρd

≤ 1 300kg/m3

Group 1. 0.55 0.75 0.30 0.40 Group 2. 0.45 0.70 0.25 0.30 Group 3. 0.35 0.50 0.20 0.25

Clay

Group 4. 0.35 0.35 0.20 0.25 Group 1. 0.55 0.80 ‡ ‡ Calcium

silicate Group 2. 0.45 0.65 ‡ ‡ Group 1. 0.55 0.80 0.45 0.45 Group 2. 0.45 0.65 0.45 0.45 Group 3. 0.40 0.50 ‡ ‡

Aggregate concrete

Group 4. 0.35 ‡ ‡ ‡ Autoclaved aerated concrete

Group 1. 0.55 0.80 0.45 0.45

Manufactured stone

Group 1. 0.45 0.75 ‡ ‡

Dimensioned natural stone

Group 1. 0.45 ‡ ‡ ‡

‡ Combination of mortar/unit not normally used, so no value given.

8.3.2. Characteristic shear strength of masonry

The characteristic shear strength, fvk shall be determined from the results of tests on masonry. If there are no test results available, the characteristic shear strength of masonry, using general purpose mortar, thin layer mortar in beds of thickness 0.5 to 3.0 mm, or lightweight mortar with all joints satisfying the detailing requirements, may be taken from the equation:

dvk0vk 4.0 σ⋅+= ff ≤ 0.065·fb

where fvk0 is the characteristic initial shear strength, under zero compressive stress (see Table 13);

- 37 -

σd is the design compressive stress perpendicular to the shear in the member at the level under consideration, using the appropriate load combination based on the average vertical stress over the compressed part of the wall that is providing shear resistance;

fb is the normalised compressive strength of the masonry units, for the di-rection of application of the load on the test specimens being perpen-dicular to the bed face.

The characteristic shear strength of masonry using general purpose mortar, thin layer mortar in beds of thickness 0.5 to 3.0 mm, or lightweight mortar, and having the per-pend joints unfilled, but with adjacent faces of the masonry units closely abutted to-gether, may be taken from equation:

dvk0vk 4.05.0 σ⋅+⋅= ff ≤ 0.045·fb

The initial shear strength of the masonry, fvk0 may be determined from either test re-sults, or from the values given in Table 13, provided that general purpose mortars do not contain admixtures or additives.

Table 13: Values of the initial shear strength of masonry fvk0

fvk0 [N/mm2]

Masonry unit General purpose mortar of the Strength class given

Thin layer mor-tar (bed joint ≥ 0.5 mm and

≤ 3 mm)

Lightweight

mortar

M10 – M20 0.30 M2,5 – M9 0.20 Clay M1 – M2 0.10

0.30 0.15

M10 – M20 0.20 M2,5 – M9 0.15 Calcium silicate M1 – M2 0.10

0.40 0.15

Aggregate concrete

M10 – M20 0.20

Autoclaved Aerated Concrete

M2,5 – M9 0.15

Manufactured stone and Dimen-sioned natural stone

M1 – M2 0.10

0.30 0.15

8.3.3. Characteristic flexural strength of masonry

In relation to out-of plane bending, the following situations should be considered: flex-ural strength having a plane of failure parallel to the bed joints, fxk1; flexural strength having a plane of failure perpendicular to the bed joints, fxk2 (see Figure 23).

The characteristic flexural strength of masonry, fxk1 and fxk2, shall be determined from the results of tests on masonry. Where test data are not available values of the char-acteristic flexural strength of masonry made with general purpose mortar, thin layer mortar or lightweight mortar, may be taken from Tables 14 and 15, provided that thin layer mortar and lightweight mortars are M5, or stronger.

- 38 -

Figure 23: Planes of failure of masonry in bending

Table 14. Values of fxk1, for plane of failure parallel to bed joints

fxk1 [N/mm2]

General purpose mortar Masonry unit

fm < 5 N/mm2 fm ≥ 5 N/mm2 Thin layer mortar Lightweight

mortar

Clay 0.10 0.10 0.15 0.10

Calcium silicate 0.05 0.10 0.20 not used

Aggregate concrete 0.05 0.10 0.20 not used

Autoclaved aerated

concrete 0.05 0.10 0.15 0.10

Manufactured stone 0.05 0.10 not used not used

Dimensioned natural stone

0.05 0.10 0.15 not used

Table 15. Values of fxk2, for plane of failure perpendicular to bed joints

fxk2 [N/mm2]

General purpose mortar Masonry unit

fm < 5 N/mm2 fm ≥ 5 N/mm2 Thin layer mortar Lightweight

mortar

Clay 0.20 0.40 0.15 0.10

Calcium silicate 0.20 0.40 0.30 not used

Aggregate concrete 0.20 0.40 0.30 not used

ρ < 400 kg/m3 0.20 0.20 0.20 0.15 Autoclaved

aerated

concrete ρ ≥ 400 kg/m3 0.20 0.40 0.30 0.15

Manufactured stone 0.20 0.40 not used not used

Dimensioned natural stone 0.20 0.40 0.15 not used

fxk2 should not be taken to be greater than the flexural strength of the unit.

a) plane of failure parallel to bed joints, fxk1

a) plane of failure perpen-dicular to bed joints, fxk2

- 39 -

8.4. Deformation properties of masonry

8.4.1. Stress-strain relationship

The stress-strain relationship of masonry in compression is non-linear. For the pur-poses of designing a masonry section the stress-strain relation is may be taken as linear, parabolic, parabolic rectangular (see Figure 24) or as rectangular.

fk

σ

ε

1)

arctan(E)

εm1

f

2)

fd

⅓f

εmu

3)

Key:

1) Typical diagram 2) Idealised diagram (parabolic-

rectangular) 3) Design diagram

Figure 24: Stress-strain relationship for masonry in compression

8.4.2. Modulus of elasticity

The short term secant modulus of elasticity, E shall be determined by tests. In the absence of a value determined by tests, the short term secant modulus of elasticity of masonry for use in structural analysis, may be taken to be:

kfKE e ⋅=

where fk is the characteristic compressive strength of masonry and the value of Ke depends on the type of the masonry unit. For clay units Ke = 1000, and for calcium silicate units Ke = 700 can be used. The long term modulus should be based on the short term secant value, reduced to allow for creep effects, such that:

∞φ+=

1term long

EE

where φ∞ is the final creep coefficient. The value of final creep coefficient is about 0.5 – 1.5 for clay and autoclaved aerated concrete units, 1.0 – 2.0 for calcium sili-cate, aggregate concrete and manufactured stone units, and ~0.0 for natural stone units.

8.4.3. Shear modulus

The shear modulus, G may be taken as 40% of the elastic modulus, E.

- 40 -

9. Analysis of structural members

9.1. Masonry walls subjected to vertical loading

9.1.1. The initial eccentricity

An initial eccentricity, einit shall be assumed for the full height of a wall to allow for construction imperfections. The initial eccentricity may be assumed to be

einit = hef/450

where hef is the effective height of the wall according to the next chapter.

9.1.2. Effective height of masonry walls

The effective height of a load bearing wall shall be assessed taking account of the rela-tive stiffness of the elements of structure connected to the wall and the efficiency of the connections. A wall may be stiffened by floors or roofs, suitably placed cross walls, or any other similarly rigid structural elements to which the wall is connected. Stiffening walls should have a length of at least 1/5 of the clear height and have a thickness of at least 0.3 times the effective thickness of the wall to be stiffened. lf the stiffening wall is interrupted by openings, the minimum length of the wall between openings, encompass-ing the stiffened wall, should be as shown in Figure 25, and the stiffening wall should extend a distance of at least 1/5 of the storey height beyond each opening.

t

4)

> h/5 h

h1

thh

≥+2

)(

5

1 21

1) 2)

h2

h2

3)

Key:

1) stiffened wall 2) stiffening wall 3) h2 (window) 4) h2 (door)

Figure 25: Minimum length of stiffening wall with openings

The effective height of a wall should be taken as:

hh nef ρ=

where hef is the effective height of the wall; h is the clear storey height of the wall;

- 41 -

ρn is a reduction factor where n = 2, 3 or 4 depending on the edge restraint or stiffening of the wall.

The reduction factor, ρn may be assumed to be:

(a) For walls restrained at the top and bottom by reinforced concrete floors or roofs spanning from both sides at the same level or by a reinforced concrete floor spanning from one side only and having a bearing of at least 2/3 of the thickness of the wall but not less than 85 mm:

ρ2 = 0.75

unless the eccentricity of the load at the top of the wall is greater than 0.25 times the thickness of wall, in which case:

ρ2 = 1.0

(b) For walls restrained at the top and bottom by timber floors or roofs spanning from both sides at the same level or by a timber floor spanning from one side having a bearing of at least 2/3 the thickness of the wall but not less than 85 mm:

ρ2 = 1.0

(c) For walls restrained at the top and bottom and stiffened on one vertical edge (with one free vertical edge):

– if h ≤ 3.5·ℓ

22

2

3

31

1 ρ

⋅⋅ρ

+

=ρ

h

with ρ2 = 0.75 or 1.0 according to paragraph (a) or (b);

– if h > 3.5·ℓ

3.05.1

3 ≥⋅=ρh

where ℓ is the length of the wall.

(d) For walls restrained at the top and bottom and stiffened on two vertical edges:

– if h ≤ 1.15·ℓ

22

2

4

1

1 ρ

⋅ρ+

=ρ

h

– if h > 1.15·ℓ

h

⋅=ρ 5.04

where ℓ is the length of the wall.

Values for ρ3 and ρ4 are shown in graphical in Figure 26.

- 42 -

Figure 26: Graphs showing values of ρ3 and ρ4

9.1.3. Effective thickness of masonry walls

The effective thickness, tef of a single-leaf wall, a double-leaf wall, a faced wall, a shell bedded wall or a grouted cavity wall should be taken as the actual thickness of the wall, t.

The effective thickness, tef of a cavity wall in which both leaves are connected with wall ties should be determined using equation:

3 32

31tefef ttkt +=

where t1, t2 are the actual thicknesses of the leaves, and t1 is the thickness of the outer or unloaded leaf and t2 is the thickness of the inner or loaded leaf;

kef = E1/E2 is a factor to allow for the relative E values of the leaves t1 and t2.

9.1.4. Slenderness ratio of masonry walls

The slenderness ratio of a masonry wall shall be obtained by dividing the value of the effective height, hef by the value of the effective thickness, tef. The slenderness ratio of the masonry wall should not be greater than 27 when subjected to mainly vertical loading.

9.1.5. Verification of unreinforced masonry walls subjected to mainly vertical loading

The resistance of masonry walls subjected to vertical loading shall be based on the geometry of the wall, the effect of the applied eccentricities and the material proper-

0 1 2 3 4 5

h/l

0.2

0.4

0.6

0.8

1.0

ρ2 = 1,0

ρ2 = 0,75

0 1 2 3 4 5

h/l

0.0

0.2

0.4

0.6

0.8

1.0

ρ2 = 0,75

ρ2 = 1,0ρ3 ρ4

- 43 -

ties of the masonry. In calculating the vertical resistance of masonry walls, it may be assumed that:

− plane sections remain plane; − the tensile strength of masonry perpendicular to bed joints is zero.

At the ultimate limit state, the design value of the vertical load applied to a masonry wall, NEd shall be less than or equal to the design value of the vertical resistance of the wall, NRd such that:

RdEd NN ≤

The design value of the vertical resistance of a single leaf wall per unit length, NRd is given by:

dRd ftN ⋅⋅Φ=

where Φ is the capacity reduction factor, Φi at the top or bottom of the wall, or Φm in the middle of the wall, as appropriate, allowing for the effects of slenderness and eccentricity of loading;

t is the thickness of the wall; Mkd / γ= ff is the design compressive strength of the masonry.

(a) The value of the reduction factor for slenderness and eccentricity, Φi at the top or bottom of the wall may be based on a rectangular stress block as follows:

t

eii 21−=Φ

where teeN

Me ⋅≥++= 05.0inithe

id

idi is the eccentricity at the top or the bottom of

the wall, as appropriate;

Mid is the design value of the bending moment at the top or bottom of the wall (see Figure 27);

Nid is the design value of the vertical load at the top or bottom of the wall;

ehe is the eccentricity at the top or bottom of the wall, if any, resulting from horizontal loads (e.g. wind or soil pressure);

einit is the initial eccentricity with a sign that increases the absolute value of the ei eccentricity (see Chapter 9.1.1);

t is the thickness of the wall.

(b) In the middle of the wall height, the reduction factor, Φm taking into account the slenderness of the wall and the eccentricity of loading, for any modulus of elasticity E and characteristic compressive strength of unreinforced masonry fk, may be esti-mated from:

21m

2

eu

AΦ−

⋅=

- 44 -

1)

2)

3)

h/2

h N1d

Nmd

N2d

h/2

Key:

1) M1d (at underside of floor)

2) Mmd (at mid height of wall)

3) M2d (at top of floor)

Figure 27: Moments from calculation of eccentricities

where e is the base of natural logarithms;

t

eA mk

1 21−=

t

eu

mk17.173.0

063.0

−

−λ=

with: E

f

t

h k

ef

ef=λ

teee ⋅⋅≥+= 050kmmk is the eccentricity at the middle height of the wall;

inithmmd

mdm ee

N

Me ±+= is the eccentricity due to loads;

Mmd is the design value of the greatest moment at the middle of the height of the wall resulting from the moments at the top and bottom of the wall (see Figure 27 ), including any load applied eccentrically to the face of the wall;

Nmd is the design value of the vertical load at the middle height of the wall, including any load applied eccentrically to the face of the wall;

ehm is the eccentricity at mid-height resulting from horizontal loads (for example wind);

einit is the initial eccentricity (see Chapter 9.1.1); hef is the effective height of the wall (see Chapter 9.1.2); tef is the effective thickness of the wall (see Chapter 9.1.3);

- 45 -

m

ef

efk 002.0 et

t

he ⋅φ= ∞ is the eccentricity due to creep. For walls having a

slenderness ratio of λc=15 or less, the creep ec-centricity, ek may be taken as zero.

φ∞ is the final creep coefficient (approximate values of final creep coeffi-cient can be found in Chapter 8.4.2).

The values of reduction factor Φm in case of clay masonry units (E = 1000·fk) are illustrated in Figure 28.

0 5 10 15 20 25 30

hef /tef

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Φm

Figure 28: Values of Φm reduction factor against slenderness ratio for different eccen-

tricities, based on an E = 1000·fk (clay units)

9.1.6. Walls subjected to concentrated loads

The design value of a concentrated vertical load, NEdc applied to a masonry wall, shall be less than or equal to the design value of the vertical concentrated load resis-tance of the wall, NRdc such that

RdcEdc NN ≤

When a wall, built with Group 1 masonry units (other than a shell bedded wall) is subjected to a concentrated load, the design value of the vertical load resistance of the wall is given by:

dbRdc fAN β=

where

−

+=β

ef

b

c

1 1,15,13,01A

A

h

a is an enhancement factor for concentrated

loads, considering that:

⋅+

≤β≤5.1

225.1

min00.1c

1

h

a

- 46 -

a1 is the distance from the end of the wall to the nearer edge of the loaded area (see Figure 29);

hc is the height of the wall to the level of the load;

Ab is the loaded area;

Aef is the effective area of bearing, i.e.: Aef = ℓ efm·t;

ℓ efm is the effective length of the bearing as determined at the mid height of the wall or pier (see Figure 29);

t is the thickness of the wall, taking into account the depth of recesses in joints greater than 5 mm;

ef

b

A

A is not to be taken greater as 0.45.

Values for the enhancement factor β are shown in graphical form in Figure 30.

h

NEdc

1) 2)

t

≤ t/4

ℓefm

60º

h c

NEdc NEdc

60º 60º60º

NEdc

ℓefm

ℓefm ℓefm

h c/2

a1

a1

Ab

NEdc

t

Key:

1) Plan view 2) Section

Figure 29: Walls subjected to concentrated load

The eccentricity of the concentrated load from the centre line of the wall should not be greater than t/4 (see Figure 29). In all cases, the requirements of Chapter 9.1.5. should be met at the middle height of the wall below the bearings, including the ef-fects of any other superimposed vertical loading, particularly for the case where con-centrated loads are sufficiently close together for their effective lengths to overlap. The concentrated load should bear on a Group 1 unit or other solid material of length equal to the required bearing length plus a length on each side of the bearing based on a 60° spread of load to the base of the solid material; for an end bearing the addi-tional length is required on one side only.

- 47 -

0 0.1 0.2 0.3 0.4 0.5

Ab / Aef

1.0

1.1

1.2

1.3

1.4

1.5

1.6β

0.45

2a1

h = 1

a1= 0

Figure 30: Graph showing the enhancement factor β for concentrated loads under bearings

9.1.7. Unreinforced masonry walls subjected to shear loading

When analysing masonry walls subjected to shear loading, the elastic stiffness of the walls, including any flanges, should be used as the stiffness of the wall. For walls higher than twice their length, the effect of shear deformations on the stiffness can be neglected. An intersecting wall, or a portion of such a wall, may be considered to act as a flange to a shear wall provided that the connection of the shear wall to the flange is able to resist the corresponding shearing actions, and provided the flange will not buckle within the length assumed.

The length of any intersecting wall, which may be considered to act as a flange (see Figure 31), is the thickness of the shear wall plus, on each side of it – where appropriate – the least of:

− htot/5, where htot is the overall height of the shear wall; − half the distance between shear walls (ℓs), when connected by the intersecting wall; − the distance to the end of the wall; − half the clear height (h); − six times the thickness of the intersecting wall, t.

In intersecting walls, openings with dimensions smaller than h/4 or ℓs/4 may be disre-garded. Openings with dimensions greater than h/4 or ℓs/4 should be regarded as marking the end of the wall.

- 48 -

1)

3) ℓs

2)

Key:

1) the lesser of

⋅ th

h

6

2/

2/

5/

s

tot

2) intersecting wall 3) shear wall

Figure 31: Flange widths that can be assumed for shear walls

lf the floors can be idealised as rigid diaphragms, the horizontal forces may be dis-tributed to the shear walls in proportion to their stiffness. The maximum horizontal load on a shear wall may be reduced by up to 15% provided that the load on the par-allel shear walls is correspondingly increased. The distribution of shear stress along the compressed part of a wall may be assumed to be constant.

At the ultimate limit state the design value of the shear load applied to the ma-sonry wall, VEd shall be less than or equal to the design value of the shear resis-tance of the wall, VRd such that:

RdEd VV ≤

The design value of the shear resistance is given by:

cvdRd AfV ⋅=

where Mvkvd / γ= ff is the design shear strength of masonry, based on the average

of the vertical stresses over the compressed part of the wall that is providing the shear resistance (see Chapter 8.3.2);

t is the thickness of the wall resisting the shear;

Ac is the area of the compressed part of the wall, ignoring any part of the wall that is in tension. In Ac the appropriate area of inter-secting wall can be considered according to Figure 31. If there are no flanges connected to the shear wall, the compressed area is calculated as: Ac = ℓc·t, where ℓc is the length of the com-pressed part of the wall.

10. Structural behaviour and application of reinforced masonry

10.1. Advantages of the application of reinforced masonry

The advantages of the use of reinforced masonry (see Figure 32):

− By use of a horizontal reinforcement, the tensile strength of the masonry wall increases in the plane of the wall, while the flexural strength increases in the perpendicular direction to the plane of the wall.

− Cracks in the masonry wall are significantly reduced. Cracks usually de-velop from slow deformation of the wall, uneven settlements, dynamic ef-

- 49 -

fects acting to the masonry (earthquakes, industrial vibration, etc.), and peak stresses around the openings.

− Horizontal reinforcement can be easily placed in the mortar layer (horizon-tal joints) without technical difficulties.

Figure 32: Masonry wall reinforced in the horizontal joint

10.2. Typical properties of the reinforcement

The reinforcement to be placed in the mortar layer (bed joint reinforcement) is factory welded and has a truss-like shape (see Figure 33). S500 grade carbon steel reinforce-ment with corrosion protection is usually used. The method of corrosion protection can be:

− application of zinc coating (Z); − application epoxy coating (E); − use of stainless steel (S)

The typical individual length of the masonry reinforcement is about 3.00 m. The char-acteristic yield strength of the reinforcement is fyk = 500 N/mm2. The cross section of the steel bars is circular for ordinary mortar layers (up to 10 mm in thickness). For thin mortar layers, flattened steel bars can also be used with thickness of about 1.5 mm (see Figure 34).

Figure 33: Welded masonry reinforcement with circular cross section

Figure 34: Welded masonry reinforcement with flattened cross section

- 50 -

The members of the reinforcement are connected by overlapping in the horizontal mortar layer (see Figure 35).

Figure 35: Overlapping of the steel bars in the horizontal joint

The reinforcement in different rows of masonry units below each other should be overlapped at different locations (see Figure 36).

Figure 36: Overlapping of the steel bars at different locations

Reinforcement of special shapes also exist for different purposes, like construction of wall corners or straight arches (see Figure 37).

Figure 37: Special reinforcement for the construction of straight arches

- 51 -

10.3. Application of horizontal reinforcement for masonry

(a) Truss-like horizontal reinforcement or single steel bars with, or without additional ver-tical reinforcement may be used for the strengthening of wall corners (see Figure 38) and wall connections, as well as for the construction of wall bracings (see Figure 39).

Figure 38: Strengthening of wall corners

Figure 39: Reinforcement of wall connection and wall bracing

(b) Horizontal reinforcement can also be used for the construction of longer wall sections without dilatation (see Figure 40).

Figure 40: Long wall section (1) with dilatations; (2) with horizontal reinforcement with-

out dilatations

1)

2)

- 52 -

(c) Crack forming can be reduced and a more uniform stress distribution can be achieved near the doors and windows by the application of horizontal reinforcement (see Figure 41).

Figure 41: An opening in the wall: (1) without reinforcement; (2) with horizontal rein-

forcement

(d) More favourable stress distribution can be achieved and crack forming can be reduced in the nodes of wall connection by the application of horizontal reinforcement (see Figure 42).

Figure 42: Reinforcement of wall connections

1) 2)

- 53 -

(e) In case of buildings subjected to uneven soil settlements, stiffness of the wall can be increased and crack forming can be reduced by the application of horizontal rein-forcement (see Figure 43).

Figure 43: Reinforcement of wall in a building subjected to uneven settlements

Uneven settlements may caused by:

− essentially different soil types (with different compression modulus) under the building

− different loads acting to the ground (high building vs. low building) − different movement of buildings or building parts due to different foundation

solutions (pile foundation vs. spread footing) − differences due to local settlements (see Figure 44) caused by water washout,

burst pipe, etc.

The application of horizontal masonry reinforcement can be very important for limiting the impact of all these causes (primarily limiting the crack forming).

Figure 44: Uneven settlement under a reinforced masonry

(f) Horizontal masonry reinforcement may also be used to resist bending moments perpendicular to the plane of the wall. Figure 45 illustrates several examples for the application of horizontal reinforcement in walls subjected to bending perpendicularly to their planes.

- 54 -

Figure 45: Walls subjected to bending perpendicularly to their planes: (1) masonry

storage; (2) cellar wall subjected to soil pressure; (3) bounding wall (without vertical load) subjected to wind pressure

(g) In certain cases horizontal reinforcement is used to resist the bending moment acting in the plane of the wall. Such bending is may caused by a local effects, like concentrated reaction forces of floor beams (see Figure 46).

Figure 46: Concentrated compressive force acting to a wall

Infill masonry walls, or partition walls in concrete frame structures are subjected to bending in their plane too, due to the constraint movement of the wall. Crack forming

1) 2)

3)

- 55 -

can be significantly reduced in these walls by the application of horizontal reinforce-ment (see Figure 47).

Figure 47: (1) expected cracks without reinforcement; (2) solution with the application

of horizontal reinforcement (a – sliding layer; b - compressible foam)

11. Numerical examples

11.1. Verification of masonry wall subjected to vertical loading

Verify the load bearing capacity of the masonry wall given in Figure 48. The wall is loaded by the NEd uniformly distributed vertical load. The wall is made of Category I. and Group 2. clay units, and prescribed general purpose mortar of class M5. Class of the masonry wall is 3.

NEd

t

h/2

CC

NEd

h


AA

BB h

e0

Figure 48: Masonry wall subjected to vertical compression

- 56 -


Type and dimensions of masonry unit: Porotherm 38; ℓu=250 mm, wu=380 mm, hu= 238 mm Mean compressive strength of the applied masonry unit: f = 11 N/mm2

Volumetric weight of masonry: ρ = 8.21 kN/m3 (given by the manufacturer) NEd = 300 kN/m e0 = 15 mm h = 3.00 m t = 0.38 m ρ2 = 0.75 (reduction factor for the wall, see Chapter 9.1.2) φ∞ = 1,00 (final creep coefficient, see Chapter 8.4.2) γM = 2.2 (units of Category I., prescribed mortar, Class 3. masonry, see Table 9)

Calculation:

A) Calculation of the design compressive strength of masonry

Coefficient depending on the dimensions of the unit: δ 1.138= (see Table 11)

Normalised mean compressive strength: fb δ f⋅ 12.52N

mm2

⋅==

K 0.45= (Group 2. clay units, general purpose mortar, see Table 12)

Compressive strength of mortar: fm 5N

mm2

=

Characteristic compressive strength of masonry: fk K fm0.3⋅ fb

0.7⋅ 4.28N

mm2

⋅==

Design compressive strength of masonry: fdfk

γM1.94

N

mm2

⋅==

B) Calculation of the capacity reduction factor at the top of the wall (section A-A)

First order eccentricity: e0 15mm=

Effective height of the wall: hef ρ2 h⋅ 2.25m==

Initial eccentricity at the top of wall: einit

hef

4505 mm⋅==

Eccentricity at the top of wall: ei max e0 einit+ 0.05t, ( ) 20 mm⋅==

Capacity reduction factor at the top of the wall: Φi.A 12 ei⋅

t− 0.895==

C) Calculation of the capacity reduction factor at the bottom of the wall (section C-C)


Initial eccentricity at the top of wall: einit 0mm=

Remark: We asssume that the bottom of the wall is in the desired position, so no initialeccentricity is considered here.

- 57 -

Eccentricity at the top of wall: ei max e0 einit+ 0.05t, ( ) 19 mm⋅==

Capacity reduction factor at the top of the wall: Φi.C 12 ei⋅

t− 0.900==

D) Calculation of the capacity reduction factor in the middle of the wall (section B-B)


Initial eccentricity in the middle of the wall: einit 0.5hef

450⋅ 2.5 mm⋅==

Remark: We asssume that the initial eccentricity varies linearly between the bottom and thetop of the wall, thus between 0 and hef/450.

The eccentricity due to loads: em e0 einit+ 17.5 mm⋅==

Effective width of the wall (single-leaf wall): tef t 380 mm⋅==

The eccentricity due to creep: ek 0.002 ϕ∞⋅hef

tef⋅ t em⋅⋅ 1.0 mm⋅==

Eccentricity at the middle height of the wall: emk max em ek+ 0.05t, ( ) 19 mm⋅==

The short term secant modulus of elasticity: E 1000 fk⋅ 4277N

mm2

⋅== (clay units)

The long term modulus of elasticity: ElongtermE

1 ϕ∞+2139

N

mm2

⋅==

Slenderness of the wall: λhef

tef

fk

Elongterm⋅ 0.265==

uλ 0.063−

0.73 1.17emk

t⋅−

0.301==

A1 1 2emk

t⋅− 0.9==

Reduction factor in the middle of the wall: Φm A1 e

u2−

2⋅ 0.860==

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D) Verification of the appropriate cross sections of the wall

At the top of the wall (section A-A):

NRd.A Φi.A t⋅ fd⋅ 661.05kN

m== > NEd 300

kN

m= satisfactory

At the middle height of the wall (section B-B):

NRd.B Φm t⋅ fd⋅ 635.58kN

m== > NEd ρ t⋅

h

2⋅+ 304.68

kN

m= satisfactory

At the bottom of the wall (section C-C):

NRd.C Φi.C t⋅ fd⋅ 664.94kN

m== > NEd ρ t⋅ h⋅+ 309.36

kN

m= satisfactory

11.2. Verification of shear resistance

Verify the shear resistance at the bottom (section K-K) of the masonry wall given in Figure 49. The wall is loaded by the NEd vertical and HEd horizontal edge loads. The wall is made of Category I. and Group 2. clay units, and prescribed general purpose mortar of class M10. Class of the masonry wall is 2. Self weight of the wall can be neglected during the calculation.

NEdHEd

t

h

K K

NEd

h


K K

L

Figure 49: Masonry wall subjected to compression and shear


Type and dimensions of masonry unit: Porotherm 38; ℓu=250 mm, wu=380 mm, hu= 238 mm Mean compressive strength of the applied masonry unit: f = 11 N/mm2

Normalized mean compressive strength: fb = 12.52 N/mm2 (see previous example) NEd = 200 kN/m HEd = 100 kN h = 3.00 m L = 6.00 m t = 0.38 m γM = 2.0 (units of Category I., prescribed mortar, Class 2. masonry, see Table 9)

- 59 -

Calculation:

A) Calculation of internal forces in section K-K

Axial force: NEd NEd L⋅ 1200 kN⋅==

Bending moment: MEd HEd h⋅ 300 kNm⋅==

Shear force: VEd HEd 100 kN⋅==

B) Verification of the eccentricity of the axial force in section K-K

Cross sectional area: A t L⋅ 2.28m2==

Moment of inertia: It L

3⋅12

6.84m4==

Kern distance: dI

L

2A⋅

1000 mm⋅== > eN

MEd

NEd250 mm⋅==

The eccentricity of the axial force is smaller than the kern distance (i.e. the force remainswithin the core of the cross section) so the complete K-K section is under compression.The core of a rectangular cross section is illustrated in Figure 50.

C) Calculation of the characteristic shear strength

Average vertical stress in section K-K: σd

NEd

A0.53

N

mm2

⋅==

Initial shear strength of the masonry: fvk0 0.3N

mm2

= (see Table 13)

Characteristic shear strength: fvk min fvk0 0.4 σd⋅+ 0.065 fb⋅, ( ) 0.51N

mm2

⋅==

D) Verification of the shear resistance in section K-K

VRd

fvk

γMA⋅ 582 kN⋅== > VEd 100 kN⋅= satisfactory

h h/3 y

z

b/3

b

Figure 50: The core of a homogeneous rectangular cross section

- 60 -

12. References

[1] C. Hartsuijker, J. W. Welleman: “Engineering Mechanics: Volume 2: Stresses, Strains, Displacements”, Springer Science & Business Media, 2007.

[2] EN 1995-1-1:2010. Eurocode 5: Design of timber structures. Part 1-1: General. Common rules and rules for buildings.

[3] EN 1996-1-1:2005+A1:2013. Eurocode 6: Design of masonry structures. Part 1-1: General rules for reinforced and unreinforced masonry structures.

[4] EN 338:2010. Structural timber. Strength classes.

[5] I. Bódi, L. Erdődi, N. Friedman: “Verification of traditional timber connections” (in Hungarian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2011.

[6] I. Bódi: “Structural behavior and application of reinforced masonry”, (in Hungar-ian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2011.

[7] K. Koris, I. Bódi: “Basis of the design of timber structures according to MSZ EN 1995-1-1 (EC5) ”, (in Hungarian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2012.

[8] L. Varga: “Design of masonry structures according to MSZ EN 1996-1-1 (Euro-code 6)”, (in Hungarian), Electronic lecture note of the Department of Structural En-gineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2010.

[9] Structural fixings – On-site guide for building code compliance, MiTek New Zea-land Ltd, 2009.

[10] “Porotherm application and design guide”, (in Hungarian), Wienerberger Ltd. (www.wienerberger.hu), 2014.

Basis of the design of timber and masonry structures according to Eurocode

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