Structural Analysis & Design - Timber design
- 1. STR ENG. 2 TIMBER DESIGN TO EC5
- 2. APPLICATIONS ▪ STRUCTURAL MEMBERS ▪ NON-STRUCTURAL COMPONENTS ▪ INTERIOR AND EXTERIOR USE ▪ CLADDING AND INSULATION ▪ REUSABLE - RECYCLABLE ▪ ECO-FRIENDLY
- 3. BASIS FOR DESIGN Design models for the different limit states shall, as appropriate, consider: ▪ different material properties (e.g., strength and stiffness). ▪ different time-dependent behaviour of the materials (duration of load, creep); ▪ different climatic conditions (temperature, moisture variations); ▪ different design situations (stages of construction, change of support conditions). Basic Variables − EN 1991-1-1 Densities, self-weight and imposed loads − EN 1991-1-3 Snow loads − EN 1991-1-4 Wind actions − EN 1991-1-5 Thermal actions − EN 1991-1-6 Actions during execution − EN 1991-1-7 Accidental actions
- 4. The load-duration classes are characterised by the effect of a constant load acting for a certain period in the life of the structure. For a variable action, the appropriate class shall be determined according to the typical variation of the load with time LOAD – DURATION CLASS
- 5. ▪ 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. NOTE: In service class 1 the average moisture content in most softwoods will not exceed 12 %. ▪ 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. NOTE: In service class 2 the average moisture content in most softwoods will not exceed 20 %. ▪ Service class 3 is characterised by climatic conditions leading to higher moisture contents than in service class 2. SERVICE CLASS
- 6. ▪ 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. NOTE: In service class 1 the average moisture content in most softwoods will not exceed 12 %. ▪ 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. NOTE: In service class 2 the average moisture content in most softwoods will not exceed 20 %. ▪ Service class 3 is characterised by climatic conditions leading to higher moisture contents than in service class 2. SERVICE CLASS
- 7. VERIFICATION OF STRENGTH BY THE PARTIAL FACTOR METHOD 𝐾𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑋𝑘 𝛾𝑀 − Xk: is the characteristic value of a strength property. − γM: is the partial factor of safety for a material property. − kmod: is a modification factor considering the effect of duration of the load and moisture content. − kh: is a modification factor considering member size. 𝑓𝑚,𝑦,𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑓𝑚,𝑘 𝛾𝑀 𝑓𝑐,0,𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑓𝑐,𝑘 𝛾𝑀 e.g., 𝑓𝑚,𝑦,𝑑; 𝑓𝑐,0,𝑑 are the design bending and compression strength, respectively. Recommended values of kmod
- 8. MODIFICATION FACTOR FOR SERVICE CLASS For rectangular solid timber with a characteristic timber density ρk ≤ 700 kg/m3 the reference depth in bending or width (maximum cross-sectional dimension) in tension is 150 mm. For depths in bending or widths in tension of solid timber less than 150 mm, the characteristic strength may be increased by the factor kh, given by: 𝑘ℎ = 𝑚𝑖𝑛 150 ℎ 0.2 1.3 e.g., for h=3 ½”=88.9 mm 𝑘ℎ = 𝑚𝑖𝑛 1.11 1.3
- 9. VERIFICATION OF STRENGTH BY THE PARTIAL FACTOR METHOD 𝐾𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑋𝑘 𝛾𝑀 𝑓𝑚,𝑦,𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑓𝑚,𝑘 𝛾𝑀 𝑓𝑐,0,𝑑 = 𝑘ℎ ∙ 𝑘𝑚𝑜𝑑 𝑓𝑐,𝑘 𝛾𝑀 e.g., 𝑓𝑚,𝑦,𝑑; 𝑓𝑐,0,𝑑 are the design bending and compression strength, respectively. Recommended values of kmod
- 10. ULTIMATE LIMIT STATE In this course we will focus on straight solid timber. Notwithstanding, the principles for design apply to glued laminated timber or wood-based structural products of constant cross-section, whose grain runs essentially parallel to the length of the member as indicated in the figure below. The grain is the longitudinal arrangement of wood fibers. The two basic categories of grain are straight grain and cross grain.
- 11. STRENGTH PROPERTIES FOR STRENGTH CLASS (SOFTWOOD CASE)
- 12. DESIGN OF CROSS SECTIONS Tension parallel to the grain Compression parallel to the grain 𝜎𝑡,0,𝑑 ≤ 𝑓𝑡,0,𝑑 𝜎𝑐,0,𝑑 ≤ 𝑓𝑐,0,𝑑 𝜎𝑡,0,𝑑 𝑓𝑡,0,𝑑 Design tensile stress along the grain Design tensile strength along the grain 𝜎𝑐,0,𝑑 𝑓𝑐,0,𝑑 Design compressive stress along the grain Design compressive strength along the grain NOTE: Rules for the instability of members are given in section 4. To check for instability of members in tension assume σ_(t,0,d)=0 The stress at a cross-section from an axial force may be calculated from the general expression, 𝜎𝑁 = 𝑁 𝐴 𝜎𝑁 N A Axial stress, e.g., 𝜎𝑡,0,𝑑 or 𝜎𝑐,0,𝑑 Axial force Area of the cross section
- 13. DESIGN FOR BENDING Bending Condition 1 Condition 2 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝑚,𝑦,𝑑 , 𝜎𝑚,𝑧,𝑑 𝑚,𝑦,𝑑 , 𝑓𝑚,𝑧,𝑑 Design bending stress about the principal axes y, and z Design bending strength about the principal axes y, and z 𝑘𝑚 For solid timber, glued laminated timber, and LVL: For rectangular sections 𝑘𝑚 = 0.7 For other sections 𝑘𝑚 = 1.0 Bending Condition 1 Condition 2 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝜎𝑚,𝑦,𝑑 , 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑦,𝑑 , 𝑓𝑚,𝑧,𝑑 Design bending stress about the principal axes y, and z Design bending strength about the principal axes y, and z 𝑘𝑚 For solid timber, glued laminated timber, and LVL: For rectangular sections 𝑘𝑚 = 0.7 For other sections 𝑘𝑚 = 1.0 Bending Condition 2 ≤ 1 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 g stress about the principal axes y, and z g strength about the principal axes y, and z er, glued laminated timber, and LVL: r sections 𝑘𝑚 = 0.7 ons 𝑘𝑚 = 1.0 y y
- 14. DESIGN FOR SHEAR Shear General criteria For shear acting simulta influence of cracks shou effective width of the m 𝜏𝑑 ≤ 𝑓𝑣,𝑑 𝑏𝑒𝑓 𝜏𝑑 𝑓𝑣,𝑑 Design shear stress Design shear strength 𝑘𝑐𝑟 For solid and glued lamina For other sections 𝑘𝑐𝑟 = 1 Shear General criteria For shear acting simultaneously with bending, t influence of cracks should be considered using effective width of the member given as, 𝜏𝑑 ≤ 𝑓𝑣,𝑑 𝑏𝑒𝑓 = 𝑘𝑐𝑟𝑏 sign shear stress sign shear strength 𝑘𝑐𝑟 For solid and glued laminated timber𝑘𝑐𝑟 = 0.67 For other sections 𝑘𝑐𝑟 = 1.0 Shear General criteria For shear acting simultaneously with bending, t influence of cracks should be considered using effective width of the member given as, 𝜏𝑑 ≤ 𝑓𝑣,𝑑 𝑏𝑒𝑓 = 𝑘𝑐𝑟𝑏 sign shear stress sign shear strength 𝑘𝑐𝑟 For solid and glued laminated timber𝑘𝑐𝑟 = 0.67 For other sections 𝑘𝑐𝑟 = 1.0 Shear For shear acting simultaneously with bending, the influence of cracks should be considered using an effective width of the member given as, 𝑏𝑒𝑓 = 𝑘𝑐𝑟𝑏 𝑘 For solid and glued laminated timber𝑘𝑐𝑟 = 0.67 Shear For shear acting simultaneously with bending, the influence of cracks should be considered using an effective width of the member given as, 𝑏𝑒𝑓 = 𝑘𝑐𝑟𝑏 𝑘𝑐𝑟 For solid and glued laminated timber 𝑘𝑐𝑟 = 0.67 For other sections 𝑘𝑐𝑟 = 1.0 The maximum elastic shear stress (τd) in a rectangular section occurs at the N.A. 𝜏𝑚𝑎𝑥 = 𝜏𝑑 = 1.5 ∙ 𝑉𝑑 𝑏 ∙ ℎ
- 15. EXAMPLE 1: BENDING AND SHEAR A 100x200 (bxh) rectangular solid section in grade C20 structural timber is used as a 4m long floor joist. It is subjected to a permanent load of 1.5 kN/m and an imposed load of 1 kN/m. Check whether the section can safely carry the applied loading at ULS assuming Service Class 1. (Full solution in Lecture Notes page 11) Values of kmod
- 16. COMBINED BENDING AND AXIAL FORCE Bending and Tension Condition 1 Condition 2 𝜎𝑡,0,𝑑 𝑓𝑡,0,𝑑 2 + 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝜎𝑡,0,𝑑 𝑓𝑡,0,𝑑 2 + 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 The definitions of parameters given above for separate effects, still apply. Bending and Compression Condition 1 Condition 2 𝜎𝑐,0,𝑑 𝑓𝑐,0,𝑑 2 + 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝜎𝑐,0,𝑑 𝑓𝑐,0,𝑑 2 + 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 The definitions of parameters given above for separate effects, still apply. NOTE: Check the instability condition following the criteria given in section 4 NOTE: The method given in section 4 can be used to check the instability condition, assuming 𝝈𝒕,𝟎,𝒅 = 𝟎
- 17. EXAMPLE 2: BENDING AND COMPRESSION A short column of 750 mm length is pin jointed at each end and subject to a permanent axial compression Nd = 20.25 kN and medium-term variable moments equal to My,d = 2.49 kNm and Mz,d = 1.56 kNm. The section is class C20, has 150 mm width, 200 mm height, and is located on the ground floor. Check the suitability of the section to withstand the applied loads at the ultimate limit state for the combined permanent and variable load. (Full solution in Lecture Notes page 13) Bending and Compression Condition 1 Condition 2 𝜎𝑐,0,𝑑 𝑓𝑐,0,𝑑 2 + 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝜎𝑐,0,𝑑 𝑓𝑐,0,𝑑 2 + 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 The definitions of parameters given above for separate effects, still apply. NOTE: Check the instability condition following the criteria given in section 4
- 18. STABILITY OF MEMBERS: COLUMNS Relative slenderness ratios About y-axis About z-axis 𝜆𝑟𝑒𝑙,𝑦 = 𝜆𝑦 𝜋 𝑓𝑐,0,𝑘 𝐸0.05 𝜆𝑟𝑒𝑙,𝑧 = 𝜆𝑧 𝜋 𝑓𝑐,0,𝑘 𝐸0.05 𝜆𝑦, 𝜆𝑟𝑒𝑙,𝑦 𝜆𝑧, 𝜆𝑟𝑒𝑙,𝑧 Slenderness ratios corresponding to bending about y-y i.e., deflection in the z-direction. Slenderness ratios corresponding to bending about z-z i.e., deflection in the y-direction. 𝐸0.05 Fifth percentile value of the modulus of elasticity parallel to the grain The member can be considered stable if both 𝝀𝒓𝒆𝒍,𝒚 ≤ 𝟎. 𝟑 and 𝝀𝒓𝒆𝒍,𝒛 ≤ 0.3. In which case stress levels should only satisfy the conditions given in Section 3. To define the value of 𝜆𝑦 or 𝜆𝑧 you may use the following effective length values.
- 19. STABILITY OF MEMBERS: COLUMNS The member can be considered stable if both 𝝀𝒓𝒆𝒍,𝒚 ≤ 𝟎. 𝟑 and 𝝀𝒓𝒆𝒍,𝒛 ≤ 0.3. In which case stress levels should only satisfy the conditions given in Section 3. To define the value of 𝜆𝑦 or 𝜆𝑧 you may use the following effective length values. where 𝜆𝑦 = ൗ 𝑙𝑒𝑓 𝑖𝑦 and 𝜆𝑧 = ൗ 𝑙𝑒𝑓 𝑖𝑧 𝜆𝑟𝑒𝑙,𝑦 = 𝜆𝑦 𝜋 𝑓𝑐,0,𝑘 𝐸0.05 ; 𝜆𝑟𝑒𝑙,𝑧 = 𝜆𝑧 𝜋 𝑓𝑐,0,𝑘 𝐸0.05 ; Effective length of columns Porteous J, Ross P, 2013. Designers’ guide to eurocode 5: Design of timber buildings EN 1995-1-1.
- 20. STABILITY OF MEMBERS: COLUMNS Columns in compression or combined compression and bending Condition 1 Condition 2 𝜎𝑐,0,𝑑 𝑘𝑐,𝑦 ∙ 𝑓𝑐,0,𝑑 + 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝑘𝑚 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝜎𝑐,0,𝑑 𝑘𝑐,𝑧 ∙ 𝑓𝑐,0,𝑑 + 𝑘𝑚 𝜎𝑚,𝑦,𝑑 𝑓𝑚,𝑦,𝑑 + 𝜎𝑚,𝑧,𝑑 𝑓𝑚,𝑧,𝑑 ≤ 1 𝑘𝑐,𝑦 = 1 𝑘𝑦 + 𝑘𝑦 2 − 𝜆𝑟𝑒𝑙,𝑦 2 𝑘𝑦 = 0.5 1 + 𝛽𝑐 𝜆𝑟𝑒𝑙,𝑦 − 0.3 + 𝜆𝑟𝑒𝑙,𝑦 2 𝑘𝑐,𝑧 = 1 𝑘𝑧 + 𝑘𝑧 2 − 𝜆𝑟𝑒𝑙,𝑧 2 𝑘𝑧 = 0.5 1 + 𝛽𝑐 𝜆𝑟𝑒𝑙,𝑧 − 0.3 + 𝜆𝑟𝑒𝑙,𝑧 2 𝛽𝑐 For solid timber 𝛽𝑐 = 0.2 For glued laminated timber and LVL 𝛽𝑐 = 0.1 𝑘𝑚 As given in 3.3.2 In all other cases the stresses, which will be increased due to deflection, should satisfy the following: Columns in compression or combined compression and bending
- 21. EXAMPLE 3: BENDING AND COMPRESSION ACTING ON A COLUMN A 150 mm (b) by 200 mm (h) column of strength class C20 is pin-jointed at each end and subjected to the design loading given below. The column is 4.50 m high (L), has its ends restrained in position but not in direction and functions in service class 2 conditions. Show that the column complies with the requirements of EN 1995-1-1 for the combined permanent and variable loading condition at the ultimate limit state. Timber properties are obtained from BS EN 338: 2009. Design loading Permanent axial action, Nd = 20.25 kN Design variable axial action, Qd = 42 kN Medium term variable moment along the column about the y–y axis, My,d = 2.49 kNm Medium-term variable moment along the column about the z–z axis, MZ,d = 1.56 kNm (Full solution in Lecture Notes page 15-16)
- 22. STABILITY OF MEMBERS: BEAMS Relative slenderness and critical stress Relative slenderness Critical bending stress 𝜆𝑟𝑒𝑙,𝑚 = 𝑓𝑚,𝑘 𝜎𝑚,𝑐𝑟𝑖𝑡 𝜎𝑚,𝑐𝑟𝑖𝑡 = 𝑀𝑦,𝑐𝑟𝑖𝑡 𝑊 𝑦 = 𝜋 𝑙𝑒𝑓 ∙ 𝑊 𝑦 𝐸0.05𝐼𝑧𝐺0.05𝐼𝑡𝑜𝑟 𝜎𝑚,𝑐𝑟𝑖𝑡 Critical bending stress according to classical theory of stability using 5-percentile stiffness values. For softwood with solid rectangular cross-section: 𝜎𝑚,𝑐𝑟𝑖𝑡 = 0.78 ∙ 𝑏2 ℎ ∙ 𝑙𝑒𝑓 𝐸0.05 Where b and h are the width and depth of the beam, respectively. 𝐸0.05 𝐺0.05 Fifth percentile value of the modulus of elasticity parallel to the grain. Fifth percentile value of shear modulus parallel to the grain. 𝐼𝑧 𝐼𝑡𝑜𝑟 Second moment of area about the weak axis z. Torsional moment of inertia 𝑙𝑒𝑓 𝑊 𝑦 Effective length of beam depending on support conditions and load configurations – see table below. Section modulus about the strong axis y.
- 23. STABILITY OF MEMBERS: BEAMS Effective length of beams Porteous J, Ross P, 2013. Designers’ guide to eurocode 5: Design of timber buildings EN 1995-1-1.
- 24. STABILITY OF MEMBERS: BEAMS Effective length of beams Porteous J, Ross P, 2013. Designers’ guide to eurocode 5: Design of timber buildings EN 1995-1-1.
- 25. STABILITY OF MEMBERS: BEAMS Bending acting alone 𝜎𝑚,𝑑 ≤ 𝑘𝑐𝑟𝑖𝑡 ∙ 𝑓𝑚,𝑑 Where only My exists, the stress should satisfy the following condition:
- 26. STABILITY OF MEMBERS: BEAMS Combined bending and axial compression Where My and compressive force Nc exist, the stress should satisfy the follow 𝜎𝑚,𝑑 𝑘𝑐𝑟𝑖𝑡 ∙ 𝑓𝑚,𝑑 2 + 𝜎𝑐,0,𝑑 𝑘𝑐,𝑧 ∙ 𝑓𝑐,0,𝑑 ≤ 1 𝜎𝑚,𝑑 𝜎𝑐,0,𝑑 𝑓𝑚,𝑑 Design bending stress. Design compressive stress parallel to the grain. Design compressive strength parallel to the grain. 𝑘𝑐,𝑧 Given in section 4.1. 𝜎𝑚,𝑑 𝑘𝑐𝑟𝑖𝑡 ∙ 𝑓𝑚,𝑑 2 + 𝜎𝑐,0,𝑑 𝑘𝑐,𝑧 ∙ 𝑓𝑐,0,𝑑 ≤ 1 Where My and Nc exist, the stress should satisfy the following:
- 27. EXAMPLE 4: BENDING ACTING ALONE ON A BEAM A 100 mm wide (b) by 550 mm deep (h) (strength class D50) solid timber beam, AB, supports another beam CD at mid-span as shown in Figure 6.9. Beam CD provides lateral restraint to beam AB at C (mid span of beam AB) and applies a vertical design load of 34.8 kN at the compression surface of the beam. Beam AB has an effective span of 6.85 m (L) and is restrained torsionally and laterally against out-of-plane movement at the end supports. The design load is combined permanent and medium-term variable loading, and the beam functions in service class 1 conditions. Confirm that the bending strength of beam AB will be acceptable. (Full solution in Lecture Notes page 18-19)
- 28. Event Details: Structural Engineering II Event Code: 67292690 https://preview-intranet- uob.cloud.contensis.com/student/myuob/my -attendance.aspx