同济大学：《木结构 Timber Engineering》课程教学资源（电子教案）09 Earthquake Resistance Design

Review od as bu materia Earthquake Resistance Design --Xiang'E Elementary Schoo What advantages and problems are there in timber Review Review 4.Whal kind ol tm Words and Expressions 园 Earthquake Resistance Design 1.Project General Information 2.Seismic Resistance Design h话因 3.Project photos 1

1 Earthquake Resistance Design ------Xiang’E Elementary School HE Minjuan Professor, Civil Engineering Tongji University Timber Engineering Review 1. Why do we use wood as building material? Wood: renewable building material The key: Sustainable Forest Management 2. What advantages and problems are there in timber constructions?. 3. How can we use wood as building material? Log, Square Lumber Wood products: Dimension lumber, Glued Laminated timber (Glulam) OSB, PSL Review 4. What kind of timber construction do we build? Post-beam construction Light wood frame construction Review 5. How can we design timber construction? Properties of wood material Strength of wood and wood products Design of timber members Design of connections Shearwall and Diaphragm Words and Expressions Concrete foundation Boundary member Bottom plate Embedded bolt Anchorage Overturning resistance anchorage design Hold-down Seismic fortification intensity 抗震设防烈度 Characteristic period of site 场地特征周期 Horizontal seismic action 水平地震作用 Radius of gyration 回转半径 Overturning resistance 抗倾覆 Anchorage 锚固 Earthquake Resistance Design 1. Project General Information 2.Seismic Resistance Design 3.Project photos

1.Project General Information 1.Project General Information Area(m) 1.Project General Information 2.Seismic Design Take the Dormitory Building as anexamp a.Seismic Fortification Intensity:8 b.Design basic a ation of ground-motion:0.2g 名 2

2 1. Project General Information Dormitory Building (DB) Dining Hall (DH) Teaching Building (TB) Building Area (m2 ) Story Story-height (m) Ridge-height (m) TB 3400 2 3.6 11.050 DB 1210 3 3.6 11.390 DH 680 1/2(part) — 11.000 Building Plan dimension (m × m) TB 9.9×52.8/16.8×44 /10.4 × 32.8 DB 16×26.6 DH 16×24 Dormitory Building (DB) Dining Hall (DH) Teaching Building (TB) 1. Project General Information a. Seismic Fortification Intensity: 8 b. Design basic acceleration of ground-motion: 0.2g c. Characteristic period of site: 0.4s d. Fortification category of Building: B 1. Project General Information
Take the Dormitory Building as an example 2. Seismic Design The First Floor The Second Floor

2.Seismic Design 胞 Main Codes used for this prejet 2.Seismic Design 2.Seismic Design General design procedure General design procedure Step 1.Natural period e wmDT592.2) Fu=aG 6- a=0.16 2.Seismic Design General design procedure 3

3 The Third Floor
Main Codes used for this project Code for Seismic Design of Buildings GB50011-2001. (CSDB) Code for design of timber structures GB50005-2003. (CDTS) 2. Seismic Design
General design procedure Step 1. Natural period calculation (CDTS: 9.2.2) Empirical equation: T=0.05H0.75 Result: T=0.05H0.75=0.05×14.390.75=0.369s Step 2. Seismic influence coefficient definition (CSDB: 5.1.4, 5.1.5) 1 α = 0.16 —— Natural period calculation 2. Seismic Design Step 3. Horizontal seismic action calculation (CSDB: 5.2.1) FEK =α1Geq ( ) n EK n j jj ii i F HG G H F = −δ ∑ = 1 1 H3=5.395m H1=3.600m G3=665.984kN G2=1449.349kN G1=1440.476kN H2=3.600m F1=104.450kN F2=210.188kN F3=168.952kN 2. Seismic Design
General design procedure —— Seismic action calculation Step 4. Horizontal seismic action distribution between the wood-framed shear walls at each storey. • Action distribution according to the flexible diaphragms principle. (CSDB: 5.2.6) • It is assumed that walls with height-to-width ratio more than 3.5 do not resist the horizontal force. (CDTS: Appendix Q)
General design procedure —— seismic action distribution 2. Seismic Design First Plane

2.Seismic Design 2.Seismic Design -176 Seismic base shear on first storey L306892+210188+044线-6☒667V 2.Seismic Design 2.Seismic Design Design of Shear Walls General design procedure Design of Shear Walls ofthe 126nuth wulle nm二5y 2.Seismic Design 2.Seismic Design General deig procedureDeier Wall General 4

4 2. Seismic Design 1.3 168.952 210.188 104.45 628.667 × + + = ( ) kN 4 (3.3 / 2 3.3 / 2) 16 628.667 78.58 26.4 16 F kN + × = × = × -axis Seismic base shear on first storey: The force resisted by walls arranged along 4-axis: 2. Seismic Design l1=1.76m l2=4.66m v1 v2 v3 1 v4 1 4 0 1 2 3 4 1.76 78.58 10.77 1.76 4.66 1.76 4.66 l V V V kN l l l l = = × = × = + + + + + + 2 2 3 0 1 2 3 4 4.66 78.58 28.52 1.76 4.66 1.76 4.66 l V V V kN l l l l = = × = × = + + + + + + Along 4-axis, there are two 1.76 m-length walls and two 4.66 m-length walls, it is assumed that force resisted by walls is linear to the wall length. Action distribution according to the flexible diaphragms principle 4 (3.3 / 2 3.3 / 2) 16 628.667 78.58 26.4 16 F kN + × = × = × -axis Step 5. The wood-framed shear wall configurations determination (CDTS: 9.2.5, Appendix Q) • Assumed a piece of wood-framed shear wall subjected to horizontal force F. V = ∑ lf d d vd 1 2 3 f f k k k = × × × Design value of shear resistance strength of unit length wood-framed shear wall. (Table Q.0.1-1) • Wood-framed shear wall resistance V calculation. • Wood-framed shear wall resistance should larger than horizontal force F, namely, V >F.
General design procedure —— Design of Shear Walls 2. Seismic Design Double sides sheathed walls with 9.5 mm-thickness sheathings, 3.1 mm-diameter nails, 150 mm-nail spacing on sheathing sides are chosen. V f k k k 1.76 vd 1 2 3 2 / =2 4.3 1.0 0.8 1.0 1.76 / 0.8 15.125 RE = × × × × × × × × × = γ kN 4.66 vd 1 2 3 2 / =2 4.3 1.0 0.8 1.0 4.66 / 0.8 40.075 V f k k k RE = × × × × × × × × × = γ kN Shear capacity of the 1.76 m-length walls: Shear capacity of the 4.66 m-length walls: 4.66 4.66 V kN F kN = > = 40.075 28.52 1.76 1.76 V kN F kN = > = 15.125 10.77 Shear capacity and seismic force comparison:
General design procedure —— Design of Shear Walls 2. Seismic Design Results Storey Axis Nail Spacing (mm) Remarks First storey 6~8 75 1. Diameter of common wire nails: 3.3mm; 2. Min. penetration into frame members by nails: 35mm; 3. Thickness of wall sheathings: 9.5mm; 4. Frame member section dimension: 38mm × 140mm; 5. Frame member section spacing: 406mm; 1~5, 9 150 A~F 75 Second storey 6~8 100 1~5, 9 150 B/E 75 A/C/D/F 150 Third storey 1~9 150 A~F 150 Wood-framed shear walls configurations
General design procedure —— Design of Shear Walls 2. Seismic Design Step 6. Design of boundary members of the wood-framed shear wall. (CDTS: 5.1, 9.2.5, Appendix Q) • Axial force Nf calculation of the boundary member. 0 f M N B • Boundary member design (capacity and stability). = • Overturning resistance（抗倾覆） anchorage design. Boundary member Boundary member Anchorage Bolt Top plates Floor diaphragm Bottom plate Concrete foundation Boundary member Bottom plate Embedded bolt Anchorage
General design procedure —— Design of boundary members 2. Seismic Design

2.Seismic Design 园 2.Seismic Design 胞 N-2x3x40x4.x10X1.36.39 N=04w<文-给”=a9 Compression resistance cakculation. N=2x38x140x12x10×1.1=140.48y 2.Seismic Design 2.Seismic Design 胞 2.Seismic Design 2.Seismic Design 胞 Gemeral desig proced e 13x168.952+2101s+101.45)-62s.67w rma 5

5
General design procedure • Axial force Nf calculation of the boundary member. H3=5.395m H1=3.600m G3=665.984kN G2=1449.349kN G1=1440.476kN H2=3.600m F1=104.450kN F2=210.188kN F3=168.952kN H3=5.395m H1=3.600m H2=3.600m f1=16.97kN f2=34.16kN f3=27.45kN Structure Walls arranged along 4-axis 27.45 12.595 34.16 7.2 16.97 3.6 50.84 6.42 2 Nf kN × + × + × = = ×
General design procedure —— Design of boundary members 2. Seismic Design • Tension resistance calculation, two dimension lumbers are adopted to built up the boundary member. 3 N kN t 2 38 140 4.8 10 1.3 66.39 − = × × × × × = 66.39 50.84 82.99 0.8 t f RE N N kN kN γ = < = = • Compression resistance calculation. 140.448 50.84 175.56 0.8 c f RE N N kN kN γ = < = = 3 2 38 140 12 10 1.1 140.448 N kN c − = × × × × × = 2. Seismic Design • Stability resistance calculation. 1 3 4 140 76 5121386.667 12 I mm = × × = 2 A mm = × = 140 76 10640 5121386.667 21.94 10640 I i mm A = = = 0 1220 55.6 21.94 l i λ = = = Inertia of gross cross-section Gross cross-section area Radius of gyration Slenderness ratio ( ) ( ) 2 2 1 1 0.674 1 / 80 1 55.6 / 80 ϕ λ = = = + + 3 2 2 0 50.84 10 7.09 / 1.1 12 13.2 / 0.674 38 140 2 f c N N mm kf N mm ϕA × = = < = × = × × × Stability factor 2. Seismic Design • Local compression resistance calculation. 3 50.84 10 2 2 ,90 4.78 / 6.125 / 10640 0.8 f c c N f N mm N mm A × = = ≤ = 2. Seismic Design • Overturning resistance anchorage design. C C Concrete foundation Boundary member Bottom plate Embedded bolt Anchorage 2. Seismic Design Step 7. Nail connections design between walls and floors. (CDTS: 6.2.2, 9.2.5) • Nail connections between the first-story walls and second floor diaphragms in transverse direction are taken as an example. 1.3 168.952 210.188 104.45 628.667 × + + = ( ) kN Seismic force acts on first storey: 2 2 3 10.2 3.8 12 10 0.51 N k d f kN v v c − = = × × × = Single-nail connection capacity (diameter: 3.8mm):
General design procedure —— Nail connections design 2. Seismic Design

2.Seismic Design 园 2.Seismic Design 腕 kny7+46=094 13x163.952+210.1第+04.45)=628667W So the 2.Seismic Design 3.Construction Photos General design procedure Tlry7x为+46个=9hm 3.Construction Photos 园 3.Construction Photos ■ 6

6 Nails needed: Total length of shear walls in first storey: 628.667 1232.68 0.51 = 17 1.76 4.66 109.14 × + = ( ) m Nail spacing: 3 109.14 10 88.5 1232.68 mm × = So, the designed nail spacing is 75mm 2. Seismic Design Step 8. Bolts connections design between walls and foundation. (CDTS: 6.2.2, 9.2.5) 1.3 168.952 210.188 104.45 628.667 × + + = ( ) kN Seismic force acts on first storey: Single-bolt connection capacity (M20): 2 2 3 5.5 20 12 10 7.62 N k d f kN v v c − = = × × × =
General design procedure —— Bolts connections design 2. Seismic Design
General design procedure Step 8. Bolt connections design between walls and foundation. (CDTS: 6.2.2, 9.2.5) Bolts needed: Total length of shear walls in first storey: 628.667 82.5 7.62 = 17 1.76 4.66 109.14 × + = ( ) m Bolt spacing: 3 109.14 10 1323 82.5 mm × = So, the designed bolts spacing is 1200mm 2. Seismic Design worse Wood-framed shear walls under construction 3. Construction Photos worse Wood-framed shear walls and floor diaphragm under construction 3. Construction Photos worse Wood-framed shear walls under construction on the finished floor diaphragm Roof diaphragm under construction 3. Construction Photos

3.Construction Photos 3.Construction Photos Photos of the completed school Photos of the completed dormitory building 3.Construction Photos 3.Construction Photos 百年凡海 Photos of the completed teaching building Photos of the completed dining room Timber Engineering Thanks! HE Minjuan hemj@tongji.edu.cn 7

7 worse Photos of the completed school 3. Construction Photos worse Photos of the completed dormitory building 3. Construction Photos worse Photos of the completed teaching building 3. Construction Photos worse Photos of the completed dining room 3. Construction Photos Thanks! HE Minjuan hemj@tongji.edu.cn Timber Engineering