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

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(1)BUDAPEST UNIVERSITY OF TECHNOLOGY AND ECONOMICS FACULTY OF CIVIL ENGINEERING Department of Structural Engineering. 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..

(2) -1TABLE OF CONTENTS 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 1.1.2. Serviceability limit states 1.2. Basic variables 1.3. Material properties 1.3.1. The effect of member size on strength for solid timber 1.3.2. The effect of member size on strength for glued laminated timber 1.3.3. The effect of member size on strength for Laminated veneer lumber (LVL) 1.4. Basis of structural analysis 2. Ultimate limit states 2.1. Design of cross-sections subjected to stress in one principal direction 2.1.1. Tension parallel to the grain 2.1.2. Tension perpendicular to the grain 2.1.3. Compression parallel to the grain 2.1.4. Compression perpendicular to the grain 2.1.5. Bending 2.1.6. Shear 2.1.7. Torsion 2.2. Design of cross-sections subjected to combined stresses 2.2.1. Compression stresses at an angle to the grain 2.2.2. Combined bending and axial tension 2.2.3. Combined bending and axial compression 2.3. Stability of timber members 2.3.1. Columns subjected to either compression or combined compression and bending 2.3.2. Beams subjected to either bending or combined bending and compression 3. Limiting values for deflections of beams 4. Bracing 4.1. General aspects 4.2. Single members in compression 4.3. Bracing of beam or truss systems 5. Numerical examples 5.1. Member subjected to compression and bending 5.2. Column and tie beam connection 5.3. Dovetail halving connection II. Design of masonry structures according to EC6 (EN 1996-1-1) 6. Terms and Definitions 6.1. Terms relating to masonry 6.2. Terms relating to masonry units 6.3. Terms relating to mortar. 3 3 3 3 3 3 4 7 8 8 8 10 10 10 10 10 11 11 12 14 14 14 14 15 15 15 16 17 18 18 18 19 20 20 22 25 30 30 30 30 30.

(3) -26.4. Terms relating to wall types 7. Basis of design 7.1. Ultimate limit states 7.2. Serviceability limit states 8. Materials 8.1. Masonry units 8.1.1. Types and grouping of masonry units 8.1.2. Compressive strength of masonry units 8.2. Mortar 8.3. Mechanical properties of masonry 8.3.1. Characteristic compressive strength of masonry 8.3.2. Characteristic shear strength of masonry 8.3.3. Characteristic flexural strength of masonry 8.4. Deformation properties of masonry 8.4.1. Stress-strain relationship 8.4.2. Modulus of elasticity 8.4.3. Shear modulus 9. Analysis of structural members 9.1. Masonry walls subjected to vertical loading 9.1.1. The initial eccentricity 9.1.2. Effective height of masonry walls 9.1.3. Effective thickness of masonry walls 9.1.4. Slenderness ratio of masonry walls 9.1.5. Verification of unreinforced masonry walls subjected to mainly vertical loading 9.1.6. Walls subjected to concentrated loads 9.1.7. Unreinforced masonry walls subjected to shear loading 10. Structural behaviour and application of reinforced masonry 10.1. Advantages of the application of reinforced masonry 10.2. Typical properties of the reinforcement 10.3. Application of horizontal reinforcement for masonry 11. Numerical examples 11.1. Verification of masonry wall subjected to vertical loading 11.2. Verification of shear resistance 12. References. 31 32 32 33 33 33 33 34 35 35 35 36 37 39 39 39 39 40 40 40 40 42 42 42 45 47 48 48 49 51 55 55 58 60.

(4) -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 internal 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 surfacing 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 combination 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 duration 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.

(5) -4Based 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 corresponding 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 corresponding 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).. b). a). 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 moisture 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 following expressions:.

(6) -5E mean,fin =. E mean (1 + k def ). Gmean,fin =. Gmean (1 + k def ). K ser,fin =. K ser (1 + k def ). For ultimate limit states, where the distribution of member forces and moments is affected 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 calculated from the following expressions: E mean,fin =. E mean (1 + ψ 2 k def ). where: Emean Gmean Kser kdef ψ2. Gmean,fin =. Gmean (1 + ψ 2 k def ). K ser,fin =. K ser (1 + ψ 2 k def ). is the mean value of modulus of elasticity; is the mean value of shear modulus; is the mean value of slip modulus; is a factor for the evaluation of creep deformation taking into account the relevant service class (see Table 2); 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: Ed =. E mean γM. Gd =. Gmean γM. 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: f d = k mod. fk γM. where: fk. γM. kmod. is the characteristic value of a strength property; is the partial factor for a material property (see Table 3); 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 shortterm 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 connection) shall be calculated as: R d = k mod. Rk γM. where: Rk. γM. kmod. is the characteristic value of load-carrying capacity; is the partial factor for a material property (see Table 3); is a modification factor taking into account the effect of the duration of load and moisture content (see Table 4)..

(7) -6Table 2: Values of kdef for timber and wood-based materials Service class Material Standard 1 2. Solid timber Glued Laminated Timber LVL Plywood. OSB. Particleboard. Fibreboard, hard Fibreboard, medium Fibreboard, MDF. 3. EN 14081-1. 0.60. 0.80. 2.00. EN 14080. 0.60. 0.80. 2.00. EN 14374, EN 14279. 0.60. 0.80. 2.00. 0.80 0.80 0.80. 1.00 1.00. 2.50. 2.25 1.50. 2.25. -. 2.25 2.25 1.50 1.50. 3.00 2.25. -. 2.25 2.25. 3.00. -. 3.00 3.00. 4.00. -. 2.25 2.25. 3.00. -. EN 636 Type EN 636-1 Type EN 636-2 Type EN 636-3 EN 300 OSB/2 OSB/3, OSB/4 EN 312 Type P4 Type P5 Type P6 Type P7 EN 622-2 HB.LA HB.HLA1, HB.HLA2 EN 622-3 MBH.LA1, MBH.LA2 MBH.HLS1, MBH.HLS2 EN 622-5 MDF.LA MDF.HLS. Table 3: Recommended partial factors γM for material properties and resistances Fundamental combinations: 1.3 Solid timber 1.25 Glued laminated timber 1.2 LVL, plywood, OSB 1.3 Particleboards 1.3 Fibreboards, hard 1.3 Fibreboards, medium 1.3 Fibreboards, MDF 1.3 Fibreboards, soft 1.3 Connections 1.25 Punched metal plate fasteners 1.0 Accidental combinations.

(8) -7-. Material. Solid timber Glued laminated timber LVL. Plywood. OSB. Particleboard Fibreboard, hard Fibreboard, medium Fibreboard, MDF. Table 4: Values of kmod modification factor Load-duration class Service PermaLong Medium Short Standard class nent term term term action action action action 1 0.60 0.70 0.80 0.90 EN 14081-1 2 0.60 0.70 0.80 0.90 3 0.50 0.55 0.65 0.70 1 0.60 0.70 0.80 0.90 EN 14080 2 0.60 0.70 0.80 0.90 3 0.50 0.55 0.65 0.70 1 0.60 0.70 0.80 0.90 EN 14374, 2 0.60 0.70 0.80 0.90 EN 14279 3 0.50 0.55 0.65 0.70 EN 636 0.90 0.80 0.70 0.60 1 Type EN 636-1 0.90 0.80 0.70 0.60 2 Type EN 636-2 0.70 0.65 0.55 0.50 3 Type EN 636-3 EN 300 OSB/2 1 0.30 0.45 0.65 0.85 OSB/3, OSB/4 1 0.40 0.50 0.70 0.90 OSB/3, OSB/4 2 0.30 0.40 0.55 0.70 EN 312 0.85 0.65 0.45 0.30 1 Type P4, P5 0.60 0.45 0.30 0.20 2 Type P5 0.90 0.70 0.50 0.40 1 Type P6, P7 0.70 0.55 0.40 0.30 2 Type P7 EN 622-2 HB.LA, HB.HLA 1, 2 1 0.30 0.45 0.65 0.85 HB.HLA 1,2 2 0.20 0.30 0.45 0.60 EN 622-3 MBH.LA1,2 1 0.20 0.40 0.60 0.80 MBH.HLS1,2 1 0.20 0.40 0.60 0.80 MBH.HLS1,2 2 0.45 EN 622-5 MDF.LA, MDF.HLS 1 0.20 0.40 0.60 0.80 MDF.HLS 2 0.45. lnstantaneous action 1.10 1.10 0.90 1.10 1.10 0.90 1.10 1.10 0.90 1.10 1.10 0.90 1.10 1.10 0.90 1.10 0.80 1.10 0.90 1.10 0.80 1.10 1.10 0.80 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:.  150  0.2   k h = min h   1.3 where h is the depth for bending members or width for tension members, in [mm]..

(9) -81.3.2. 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:  600  0.1   k h = min h   1.1 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:  300    k h = min  h   1.2. s. 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:  3 000    k  = min     1.1. s/2. 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 influence 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..

(10) -9Reductions 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 application 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 nonsymmetrical load (c).

(11) - 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 essentially 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).. Key: (1) direction of grain. 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 ft,0,d. is the design tensile stress along the grain; 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 ft,90,d kvol=1 kvol=(V/V0)0.2. is the design tensile stress perpendicular to the grain; is the design tensile strength perpendicular to the grain; in case of solid timber; in case of glued laminated timber, where V0=0.01 m3 is the reference volume, and V is the tensioned volume of the member in [m3]. V should not be taken greater than the two-thirds of the total 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 fc,0,d. is the design compressive stress along the grain; is the design compressive strength along the grain.. The above expression is used only for strength control of timber members. The stability analysis of compressed members is discussed in chapter 2.3.1..

(12) - 11 2.1.4. Compression perpendicular to the grain The following expression shall be satisfied: σc,90,d ≤ kc,90 fc,90,d where σc,90,d Fc,90,d Aef fc,90,d kc,90. with. σc,90,d =. Fc,90,d Aef. is the design compressive stress in the effective contact area perpendicular to the grain; is the design compressive load perpendicular to the grain; is the effective contact area in compression perpendicular to the grain; is the design compressive strength perpendicular to the grain; 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).. (a). (b). 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 kc,90 = 1.5. for solid softwood timber, 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 kc,90 = 1.75. for solid softwood timber, 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:. σ m,y,d fm,y,d. + km. σ m,z,d fm,z,d. ≤ 1 .0. and. km. σ m,y,d fm,y,d. +. σ m,z,d fm,z,d. ≤ 1 .0.

(13) - 12 where σm,y,d and σm,z,d fm,y,d and fm,z,d. are the design bending stresses about the principal axes as shown in Figure 5; 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 inhomogeneities 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 Figure 6 b), the following expression shall be satisfied: τd ≤ fv,d where τd fv,d. is the design shear stress; is the design shear strength for the actual condition. The shear strength for rolling shear is approximately equal to twice the tensile strength perpendicular to grain..

(14) - 13 -. a). b). 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 Figure 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 kcr = 1.0. for solid and glued laminated timber; for other wood-based products.. bef b Figure 7: Effective width of a member subjected to shear. At supports, the contribution to the total shear force of a concentrated force F acting 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 disregarded in the calculation of the shear force.

(15) - 14 2.1.7. Torsion The following expression shall be satisfied: τtor,d ≤ kshape fv,d. where. k shape. τtor,d fv,d kshape h b. for a circular cross section 1.2  h  = 1 + 0.15 b for a rectangula r cross section min  2.0  is the design torsional stress; is the design shear strength; is a factor depending on the shape of the cross-section; is the larger cross-sectional dimension; 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 essentially 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 account. The compressive stresses at an angle α to the grain, (see Figure 9), should satisfy the following expression: σ c,α,d ≤. fc,0,d fc,0,d k c,90 fc,90,d. where σc,α,d kc,90. sin 2 α + cos 2 α. is the compressive stress at an angle α to the grain; is a factor taking into account the effect of any of stresses perpendicular 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: σ t,0,d f t,0,d. +. σ m,y,d fm,y,d. + km. σ m,z,d fm,z,d. ≤ 1 .0. and. σ t,0,d f t,0,d. + km. σ m,y,d fm,y,d. +. σ m,z,d fm,z,d. ≤ 1 .0.

(16) - 15 2.2.3. Combined bending and axial compression The following expressions shall be satisfied: 2. 2. σ  σ c,0,d  σ   + m,y,d + k m m,z,d ≤ 1.0 f  fm,y,d fm,z,d  c,0,d . and. σ  σ c,0,d  σ   + k m m,y,d + m,z,d ≤ 1.0 f  fm,y,d fm,z,d  c,0,d . The above expressions are used only for strength control of timber members. The stability 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 stability 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: λ rel,y =. λy. fc,0,k. π. E 0,05. and. λ rel,z =. λz π. fc,0,k E 0,05. 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 expressions presented in chapter 2.2.3. In all other cases the stresses, which will be increased due to deflection, should satisfy the following expressions: σ c,0,d k c,y fc,0,d. σ c,0,d k c,z fc,0,d. +. σ m,y,d. fm,y,d. + km. + km. σ m,y,d fm,y,d. +. σ m,z,d fm,z,d. σ m,z,d fm,z,d. ≤ 1 .0. ≤ 1. 0. where the symbols are defined as follows: k c,y = k c,z =. 1. ky +. k y2. − λ2rel,y. 1. kz +. k z2. − λ2rel,z. (. (. ). with. k y = 0,5 1 + β c λ rel,y − 0,3 + λ2rel,y. with. k z = 0,5 1 + β c (λ rel,z − 0,3 ) + λ2rel,z. (. ) ).

(17) - 16 -. βc is a factor for members within the straightness limits defined as: 0.2 for solid timber βc =  0.1 for glued laminated timber and LVL. 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:. λ rel,m =. fm,k σ m,crit. 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: σ m,crit =. M y,crit Wy. where E0,05 G0,05 Iz Itor ℓef. Wy. =. π E 0,05 I z G0,05 I tor  ef W y. is the 5% value of modulus of elasticity parallel to grain; is the 5% value of shear modulus parallel to grain; is the second moment of area about the „weak” axis z; is the torsional moment of inertia; is the effective length of the beam, depending on the support conditions and the load configuration, according to Table 5; is the section modulus about the „strong” axis y.. For softwood with solid rectangular cross-section, σm,crit should be taken as:. σ m,crit =. 0,78 b 2 E 0,05 h  ef. 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:.

(18) - 17 -. k crit.  1   = 1.56 − 0.75λ rel,m  1  2  λ rel,m. if if if. λ rel,m ≤ 0.75 0.75 < λ rel,m ≤ 1.4 1.4 < λ rel,m. 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 rotation 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: 2.  σ m,d  σ c,d   + ≤ 1. 0 k f  k c,z fc,0,d  crit m,d . where σm,d σc,0,d fm,d fc,0,d kc,z. is the design bending stress; is the design compressive stress parallel to grain; is the design bending strength; is the design compressive strength parallel to grain; is a factor described in chapter 2.3.1. Table 5: Effective length as a ratio of the span. Beam type. Simply supported Cantilever *. Loading type.  ef / . Constant moment Uniformly distributed load Concentrated force at the middle of the span Uniformly distributed load Concentrated force at the free end. 1.0 0.9 0.8 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 - winst - wcreep - wfin - wnet,fin. is the precamber (if applied); is the instantaneous deflection; is the creep deflection; is the final deflection; is the net final deflection..

(19) - 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 imperfections, 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: C = ks. Nd a. where ks Nd a. is a modification factor (see Table 7); is the mean design compressive force in the element; is the bay length (see Figure 10)..

(20) - 19 The design stabilizing force Fd at each support should be calculated as:.  Nd k  Fd =  f,1 N  d  k f,2. for solid timber for glued laminated timber and LVL. 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: N d = (1− k crit ). Md h. 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 intermediate 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: qd = k . n Nd k f,3 . 1,0  where k  = min 15    Nd is the mean design compressive force in the members; ℓ is the overall span of the stabilizing system, in [m]; is a modification factor (see Table 7). kf,3.

(21) - 20 -. (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. 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. PEd. L. Cross section of bar “a” L. qEd. h. a α. α. b. Figure 12: Three-hinged frame subjected to compression and bending.

(22) - 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". PEd. NEd = PEd⋅ cos ( α) = 14.14⋅ kN MEd =. qEd⋅ L. NEd = PEd·cosα α. 2. qEd. = 4⋅ kNm. 8. B) Calculation of section properties A = b⋅ h = 15000⋅ mm Iy =. b⋅ h. MEd = qEd·L2/8. 2. 3. 7. = 2.813 × 10 ⋅ mm. 12. .. 4. Figure 13: Determination of internal forces in bar “a”. C) Calculation of design stresses σc.0.d = σm.d =. NEd. N. = 0.94⋅. A. mm. 2. MEd h N ⋅ = 10.67⋅ Iy 2 2 mm. D) Calculation of neccesary design strength values fc.0.k = 20 fm.k = 22. N mm. 2. N mm. 2. (see Table 8). (see Table 8). fc.0.d = fm.d =. fc.0.k γM fm.k γM. ⋅ kmod = 13.85⋅. ⋅ kmod = 15.23⋅. N mm. 2. N mm. 2. Remark: The size effect factor k h 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)..

(23) - 22 E) Determination of buckling reduction factor kc,y Buckling length:. L0 = L = 4 m. Slenderness ratio:. λy =. L0 Iy. (because both ends of bar " a" are hinged). = 92.38. (bending about the y axis). A Modulus of elasticity:. E0.05 = 6.7. λy Relative slenderness: λrel.y = ⋅ π. kN mm. 2. fc.0.k E0.05. (see Table 8). = 1.61. > 0.3 so stability should be checked. (for solid timber). βc = 0.2. ky = 0.5⋅ 1 + βc⋅ λrel.y − 0.3 + λrel.y  = 1.92. (. kc.y =. ). 1 ky +. 2. 2. 2. = 0.336. ky − λrel.y. F) Stability control of bar " a" in the plane of the frame (rectangular cross section). km = 0.7 σc.0.d kc.y⋅ fc.0.d. +. σm.d = 0.903 fm.d. < 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).

(24) - 23 Side view. Front view. .  . . 240. 50 100 50. FEd. . . 50 100. 50. 50 45. .  Column  Tie Beam  Wedge. 240. . 60 120 60. . 45. Key:. Plan view FEd. 50. Grain direction. 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⋅ mm. Characteristic tensile strength:. ft.0.k = 10. N mm. (see Table 8). 2. Size effect factor:.  150mm  kh = min 1.3 ,      bbeam . Design tensile strength:. ft.0.d =. Design tensile resistance:. ft.0.k γM. 0.2.  = 1.046  . ⋅ kmod⋅ kh = 5.63⋅. N mm. 2. 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 compression perpendicular to grain (see Figure 15). Critical cross-sectional area:. Ac = hbeam⋅ bwedge = 5000⋅ mm. Characteristic compressive strength:. fc.90.k = 2.2. N mm. 2. 2. (see Table 8). 2.

(25) - 24 -. At. Avr. Ac. Av. Figure 15: Critical cross sections for tension, compression and shear Design compressive strength: Design compressive resistance:. fc.90.d =. fc.90.k γM. ⋅ kmod = 1.18⋅. N mm. 2. 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⋅ 100mm = 20000⋅ mm. Characteristic shear strength:. fv.k = 1.8. Design shear strength: Design shear resistance:. fv.d =. N mm. fv.k γM. 2. 2. (see Table 8). ⋅ kmod = 0.97⋅. N mm. 2. 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⋅ mm. Characteristic tensile strength of the wedge (timber class D30): ft.90.k = 3 Characteristic rolling shear strength:. Design rolling shear strength: Design rolling shear resistance:. N mm. 2. fvr.k = 2⋅ ft.90.k = 6⋅ fvr.d =. fvr.k γM. (see Table 8) N mm. 2. ⋅ kmod = 3.23⋅. N mm. Rvr.d = fvr.d⋅ Avr = 16.15⋅ kN. 2. 2.

(26) - 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 connection of 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. Side view. Cross section. 150. 150 75 50. 70. PEd/2. 160. 110. PEd/2. PEd. PEd. 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⋅ 75mm = 5250⋅ mm. Characteristic tensile strength:. ft.0.k = 14. N mm. 2. 2. (see Table 8).

(27) - 26 -. Avr. At. Ac. Av. Figure 17: Critical cross sections for tension, shear and compression. Design tensile strength:. ft.0.d =. ft.0.k γM. ⋅ kmod = 8.62⋅. N mm. 2. Remark: The size effect factor k h 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 × 110mm⋅ 75mm = 16500⋅ mm. Characteristic shear strength:. fv.k = 2.5. Design shear strength: Design shear resistance:. fv.d =. N mm. fv.k γM. 2. (see Table 8). 2. ⋅ kmod = 1.54⋅. N mm. 2. Rv.d = fv.d⋅ Av = 25.38⋅ kN. The beam is examined for rolling shear (see Figure 17). Critical cross-sectional area:. Avr = 2 × 110mm⋅ 75mm = 16500⋅ mm. Characteristic tensile strength perpendicular to grain: ft.90.k = 0.5 Characteristic rolling shear strength:. Design rolling shear strength: Design rolling shear resistance:. N mm. 2. fvr.k = 2⋅ ft.90.k = 1⋅ fvr.d =. fvr.k γM. (see Table 8) N mm. 2. ⋅ kmod = 0.62⋅. N mm. Rvr.d = fvr.d⋅ Avr = 10.15⋅ kN. 2. 2.

(28) - 27 C) Calculation of the compressive resistance of the connection The connection surface of the column and the beam will be examined for compression at an angle to the grain (see Figure 17). Length of the connection surface:. h =. 2. Rc,α,d. β h. 2. ( 110mm) + ( 40mm) = 117 mm. α. 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:.  110mm  = 70.02 °   40mm . α = atan. for the column (see Figure 18). β = 90° − α = 19.98 °. for the beam (see Figure 18). Critical cross-sectional area:. Ac = h⋅ 75mm = 8778.52⋅ mm. 2. Calculation of design compressive strength at an angle to the grain: fc.0.k = 21. N mm. fc.90.k = 2.5 kc.90 = 1 fc.α.d =. fc.0.d =. 2. N mm. 2. γM. ⋅ kmod = 12.92. fc.90.k γM. (see chapter 2.1.4) fc.0.d fc.0.d. kc.90⋅ fc.90.d fc.β.d =. fc.90.d =. fc.0.k. = 1.71 2. ⋅ sin ( α) + cos ( α). 2. fc.0.d fc.0.d kc.90⋅ fc.90.d. mm. = 6.93 2. ⋅ sin ( β) + cos ( β). 2. N 2. N mm. 2. Critical is the compression on the column (fc,α,d < fc,β,d).. N mm. ⋅ kmod = 1.54. 2. N mm. 2.

(29) - 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.).

(30) 0,4 16 2 1,7. ft,90,k fc,0,k fc,90,k fv,k. Tension perpendicular to grain. Compression parallel to grain. Compression perpendicular to grain. Shear. 0,44. E90,mean Gmean. Mean value of modulus of elasticity pependicular to grain. Mean value of shear modulus. 290 350. ρk ρmean. Density. Mean value of density. Density [kg/m ]. 3. 0,23. E0,05. 5% value of modulus of elasticity parallel to grain 4,7. E0,mean. Mean value of modulus of elasticity parallel to grain. Stiffness properties [kN/mm ] 7. 8. ft,0,k. Tension parallel to grain. 2. 14. fm,k. C14. Bending. Strength properties [N/mm ]. 2. 370. 310. 0,5. 0,27. 5,4. 8. 1,8. 2,2. 17. 0,5. 10. 16. C16. 380. 320. 0,56. 0,3. 6. 9. 2. 2,2. 18. 0,5. 11. 18. C18. 390. 330. 0,59. 0,32. 6,4. 9,5. 2,2. 2,3. 19. 0,5. 12. 20. C20. 410. 340. 0,63. 0,33. 6,7. 10. 2,4. 2,4. 20. 0,5. 13. 22. C22. 420. 350. 0,69. 0,37. 7,4. 11. 2,5. 2,5. 21. 0,5. 14. 24. C24. 450. 370. 0,72. 0,38. 7,7. 11,5. 2,8. 2,6. 22. 0,6. 16. 27. C27. 460. 380. 0,75. 0,4. 8. 12. 3. 2,7. 23. 0,6. 18. 30. C30. 480. 400. 0,81. 0,43. 8,7. 13. 3,4. 2,8. 25. 0,6. 21. 35. C35. Poplar and softwood species. 500. 420. 0,88. 0,47. 9,4. 14. 3,8. 2,9. 26. 0,6. 24. 40. C40. 520. 440. 0,94. 0,5. 10. 15. 3,8. 3,1. 27. 0,6. 27. 45. C45. 550. 460. 1. 0,53. 10,7. 16. 3,8. 3,2. 29. 0,6. 30. 50. C50. 640. 530. 0,6. 0,64. 8. 10. 3. 8. 23. 0,6. 18. 30. D30. 670. 560. 0,65. 0,69. 8,7. 10. 3,4. 8,4. 25. 0,6. 21. 35. D35. 700. 590. 0,7. 0,75. 9,4. 11. 3,8. 8,8. 26. 0,6. 24. 40. D40. 780. 650. 0,88. 0,93. 11,8. 14. 4,6. 9,7. 29. 0,6. 30. 50. D50. 840. 700. 1,06. 1,13. 14,3. 17. 5,3. 10,5. 32. 0,6. 36. 60. D60. Deciduous species. 1080. 900. 1,25. 1,33. 16,8. 20. 6. 13,5. 34. 0,6. 42. 70. D70. - 29 -. Table 8: Strength classes and characteristic values according to EN 338.

(31) - 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 together 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 concrete so that all the materials act together in resisting action effects masonry bond: disposition of units in masonry in a regular pattern to achieve common action characteristic strength of masonry: value of the strength of masonry having a prescribed probability of 5% of not being attained in a hypothetically unlimited test series. This value generally corresponds to a specified fractile of the assumed statistical 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 masonry 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 water, 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).

(32) - 31 -. thin mortar layer. 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 properties of which are assumed from the stated proportions of the constituents (recipe concept) factory made masonry mortar: mortar batched and mixed in a factory compressive strength of mortar: the mean compressive strength of a specified number 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 between 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 together 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 material (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 lateral forces or to resist buckling and so to provide stability to the building.

(33) - 32 a). b) vertical joints. 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 ultimate limit state for ordinary and accidental situations. When analysing the structure 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..

(34) - 33 Table 9: Recommended values of the partial factor γM Requirements during the construction Construction is supervised by a suitably qualified and experienced person employed by the contractor. Construction is supervised by a suitably qualified and experienced person who is independent from the contractor. Strength of mortar is tested and controlled in the laboratory on specimens prepared on site. Factory made designed masonry mortar is used for the construction of the wall. Prescribed mortar mixed on site is used for the construction of the wall. The mortar saturation of joints. Class 1. 2. 3. 4. 5. X. X. X. X. X. X. X. X. X. X. X. 100%. X. X. X. X. 100%. 100%. 90%. 80%. There is no smaller There is no smaller unit than unit than half unit in quarter unit in the masonry the masonry Cutting of masonry units is done by mechanical or manual sawing. Way of bricklaying. γM. Material A B C. Units of Category I., designed mortar Units of Category I., prescribed mortar Units of Category II., any mortar. 1.5 1.7 2.0. 1.7 2.0 2.2. 2.0 2.2 2.5. 2.2 2.5 2.7. 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 grouping of clay, calcium silicate and aggregate concrete units are given in table 10..

(35) - 34 Table 10: Geometrical requirements for grouping of masonry units Materials and limits for masonry units Group 1. (all materials) Volume of all holes (% of the gross volume). Volume of any hole (% of the gross volume). ≤ 25. ≤ 12.5. Declared values of thickness of No rewebs and shells quirement [mm] Declared value of combined a No rethickness of quirement webs and shells (% of the overall width). Group 2. Unit. Group 3.. Group 4. Horizontal holes [%]. Vertical holes [%]. clay. > 25; ≤ 55. ≥ 25; ≤ 70. > 25; ≤ 70. calcium silicate. > 25; ≤ 55. not used. not used. 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. 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. concreteb. ≥ 15. ≥ 18. not used ≥ 15. ≥ 15. not used ≥ 20. ≥ 20. Clay. ≥ 16. ≥ 12. ≥ 12. calcium silicate. ≥ 20. not used. not used. concreteb. ≥ 18. ≥ 15. ≥ 45. a The combined thickness is the thickness of the webs and shells, measured horizontally in the relevant direction. 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 the shells.. 8.1.2. Compressive strength of masonry units The mean compressive strength of the masonry units is usually declared by the manufacturer. 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..

(36) - 35 Table 11: Values of the δ coefficient Height of the masonry unit [mm] 50 65 100 150 200 250 or greater. Width of the masonry unit [mm] 50 0.85 0.95 1.15 1.30 1.45 1.55. 100 0.75 0.85 100 1.20 1.35 1.45. 150 0.70 0.75 0.90 1.10 1.25 1.35. 200 0.70 0.80 1.00 1.15 1.25. 250 or greater 0.65 0.75 0.95 1.10 1.15. 8.2. Mortar. Masonry mortars are defined as general purpose, thin layer or lightweight mortar according to their constituents. Masonry mortars are considered as designed or prescribed mortars according to the method of defining their composition. Masonry mortars 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 masonry 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 results 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: fk = K ⋅ fb0.7 ⋅ fm0.3 is the characteristic compressive strength of the masonry, in [N/mm2]; where fk K is a constant (see Table 12); fb is the normalised mean compressive strength of the units, in the direction 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 aerated concrete units, the characteristic compressive strength can be determined from:. fk = K ⋅ fb0,85 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 determined from:. fk = K ⋅ fb0,7.

(37) - 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 purpose 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. Masonry unit. Clay Calcium silicate Aggregate concrete. Group 1. Group 2. Group 3. Group 4. Group 1. Group 2. Group 1. Group 2. Group 3. Group 4.. General purpose mortar 0.55 0.45 0.35 0.35 0.55 0.45 0.55 0.45 0.40 0.35. Thin layer mortar (bed joint ≥ 0.5 mm and ≤ 3 mm ) 0.75 0.70 0.50 0.35 0.80 0.65 0.80 0.65 0.50 ‡. Autoclaved aerated Group 1. 0.55 0.80 concrete Manufactured Group 1. 0.45 0.75 stone Dimensioned Group 1. 0.45 ‡ natural stone ‡ Combination of mortar/unit not normally used, so no value given.. Lightweight mortar of density 600 ≤ ρd ≤ 800 kg/m3. 800 < ρd ≤ 1 300kg/m3. 0.30 0.25 0.20 0.20 ‡ ‡ 0.45 0.45 ‡ ‡. 0.40 0.30 0.25 0.25 ‡ ‡ 0.45 0.45 ‡ ‡. 0.45. 0.45. ‡. ‡. ‡. ‡. 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:. f vk = f vk0 + 0.4 ⋅ σ d ≤ 0.065·fb where fvk0. is the characteristic initial shear strength, under zero compressive stress (see Table 13);.

(38) - 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 direction of application of the load on the test specimens being perpendicular 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 perpend joints unfilled, but with adjacent faces of the masonry units closely abutted together, may be taken from equation:. f vk = 0.5 ⋅ f vk0 + 0.4 ⋅ σ d ≤ 0.045·fb The initial shear strength of the masonry, fvk0 may be determined from either test results, 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. Masonry unit. Clay. Calcium silicate Aggregate concrete Autoclaved Aerated Concrete Manufactured stone and Dimensioned natural stone. fvk0 [N/mm2] Thin layer morGeneral purpose mortar of tar (bed joint the Strength class given ≥ 0.5 mm and ≤ 3 mm) M10 – M20 0.30 0.30 M2,5 – M9 0.20 M1 – M2 0.10 M10 – M20 0.20 0.40 M2,5 – M9 0.15 M1 – M2 0.10 M10 – M20. 0.20. M2,5 – M9. 0.15 0.30. M1 – M2. Lightweight mortar 0.15. 0.15. 0.15. 0.10. 8.3.3. Characteristic flexural strength of masonry In relation to out-of plane bending, the following situations should be considered: flexural 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 characteristic 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..

(39) - 38 -. a) plane of failure parallel to bed joints, fxk1. a) plane of failure perpendicular to bed joints, fxk2. 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] Masonry unit. General purpose mortar 2. 2. Thin layer mortar. Lightweight mortar. fm < 5 N/mm. fm ≥ 5 N/mm. 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] Masonry unit. General purpose mortar 2. 2. Thin layer mortar. Lightweight mortar. fm < 5 N/mm. fm ≥ 5 N/mm. 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. Autoclaved aerated concrete. ρ < 400 kg/m3. 0.20. 0.20. 0.20. 0.15. 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..

(40) - 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 purposes of designing a masonry section the stress-strain relation is may be taken as linear, parabolic, parabolic rectangular (see Figure 24) or as rectangular.. σ 1). f fk. 2). fd ⅓f. 3). Key: 1) Typical diagram 2) Idealised diagram (parabolicrectangular) 3) Design diagram. arctan(E) εm1. εmu. ε. 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:. E = K e ⋅ fk 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: E long term =. E 1 + φ∞. 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 silicate, 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..

(41) - 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 relative 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, encompassing 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.. 1). h. 2). t. 3) Key:. h2 > h/5. h1. h2 4). 1 (h1+ h2 ) 5. 2. 1) stiffened wall 2) stiffening wall 3) h2 (window) 4) h2 (door). ≥ t. Figure 25: Minimum length of stiffening wall with openings. The effective height of a wall should be taken as: hef = ρ n h where hef is the effective height of the wall; h is the clear storey height of the wall;.

(42) - 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 ρ3 =. 1. ρ ⋅ h  1+  2   3⋅ . 2. ρ2. with ρ2 = 0.75 or 1.0 according to paragraph (a) or (b); – if h > 3.5·ℓ ρ3 =. 1. 5 ⋅ . ≥ 0 .3. h. 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 ρ4 =. 1. ρ ⋅ h  1+  2    . 2. ρ2. – if h > 1.15·ℓ ρ4 =. 0 .5 ⋅  h. where ℓ is the length of the wall. Values for ρ3 and ρ4 are shown in graphical in Figure 26..

(43) - 42 -. 1.0. 1.0. ρ3. ρ4. ρ 2 = 1,0. ρ 2 = 1,0. 0.8 0.8. ρ 2 = 0,75. 0.6. 0.6. ρ 2 = 0,75 0.4. 0.4 0.2. 0.2. 0.0 0. 1. 2. 3. 4. 5. 0. 1. 2. h/l. 3. 4. 5. h/l. 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: t ef = 3 k tef t 13 + t 23. where t1, t2. kef = E1/E2. 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; 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-.

(44) - 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: NEd ≤ NRd The design value of the vertical resistance of a single leaf wall per unit length, NRd is given by: NRd = Φ ⋅ t ⋅ fd where Φ t fd = fk / γ M. 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; is the thickness of the wall; 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: Φi = 1 − 2. ei t. where ei =. M id Nid. + ehe + einit ≥ 0.05 ⋅ t. 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 estimated from: Φm = A1 ⋅ e. −. u2 2.

(45) - 44 -. N1d. h/2. 1). h. Nmd. h/2. 2). 3) N2d. Key: 1) 2) 3). M1d (at underside of floor) Mmd (at mid height of wall) M2d (at top of floor). Figure 27: Moments from calculation of eccentricities. where. with:. e is the base of natural logarithms; e A1 = 1 − 2 mk t λ − 0.063 u= e 0.73 − 1.17 mk t h fk λ = ef t ef E emk = em + ek ≥ 0 ⋅ 05 ⋅ t is the eccentricity at the middle height of the wall; M em = md + ehm ± einit is the eccentricity due to loads; N md 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);.

(46) - 45 ek = 0.002 φ ∞. hef t ef. t ⋅ em is the eccentricity due to creep. For walls having a slenderness ratio of λc=15 or less, the creep eccentricity, ek may be taken as zero.. φ∞. is the final creep coefficient (approximate values of final creep coefficient 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. Φm. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0. 5. 10. 15. 20. 25. 30. hef /tef Figure 28: Values of Φm reduction factor against slenderness ratio for different eccentricities, 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 resistance of the wall, NRdc such that NEdc ≤ NRdc 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: NRdc = β Ab fd  A a  where β = 1 + 0,3 1  1,5 − 1,1 b hc   Aef .   . is an enhancement factor for concentrated loads, considering that: a1  1.25 + 1.00 ≤ β ≤ min 2 ⋅ hc 1.5 .

(47) - 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;. Ab is not to be taken greater as 0.45. Aef. Values for the enhancement factor β are shown in graphical form in Figure 30. NEdc. NEdc. NEdc. NEdc a1. a1 60º. 60º. hc/2. 60º. hc. h. 60º. ℓefm ℓefm. ℓefm. ℓefm. NEdc. 1). 2). Key:. Ab t. ≤ t/4. 1) Plan view 2) Section. t 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 effects of any other superimposed vertical loading, particularly for the case where concentrated 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 additional length is required on one side only..

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