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Basis of the design of timber structures according to EC5 and basis of the design of masonry structures according to EC6 with worked examples.
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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.
- 1 -
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
- 2 -
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
- 3 -
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
- 4 -
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)
- 5 -
)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).
- 6 -
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
- 7 -
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].
- 8 -
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.
- 9 -
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)
- 10 -
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
The following expression shall be satisfied:
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
The following expression shall be satisfied:
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
- 11 -
2.1.4. Compression perpendicular to the grain
The following expression shall be satisfied:
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.
- 13 -
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
The following expression shall be satisfied:
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
The following expressions shall be satisfied:
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
The following expressions shall be satisfied:
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
Parameters for the calculation:
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
each of multiple holes ≤ 2
gripholes up to a total of 12,5
each of multiple holes ≤ 30
calcium silicate
each of multiple holes ≤ 15
gripholes up to a total of 30
not used not used Volume of any hole (% of the gross volume)
≤ 12.5
concreteb
each of multiple holes ≤ 30
gripholes up to a total of 30
each of multiple holes ≤ 30
gripholes up to a total of 30
each of multiple holes ≤ 25
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
Side view Cross section
AA
BB h
e0
Figure 48: Masonry wall subjected to vertical compression
- 56 -
Parameters for the calculation:
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)
First order eccentricity: e0 15mm=
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)
First order eccentricity: e0 15mm=
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==
- 58 -
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
Side view Cross section
K K
L
Figure 49: Masonry wall subjected to compression and shear
Parameters for the calculation:
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.