Wood Design for Architects: Engineering for the Non-engineer. AIA Statement

Transcription

1 Wood Design for Architects: Engineering for the Non-engineer AIA Statement The Wood Products Council is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be reported to AIA/CES for AIA members. Certificates of Completion for both AIA members and non-aia members are available upon request. This program is registered with AIA/CES for continuing gprofessional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Karyn A. Beebe, P.E., LEED AP (858) Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation. Copyright Materials This presentation is protected by US and International Copyright laws. Reproduction, distribution, display and use of the presentation without written permission of the speaker is prohibited. The Wood Products Council 2012 Learning Objectives At the end of this program, participants will be able to: 1. Define the design criteria such as lateral loads in the region, understand how they impact buildings, and consequently be better prepared to design for them. 2. Identify the latest changes to the International Codes with respect to engineered wood provisions 3. Through h working design examples, apply their new knowledge to building design. 4. Utilize design resources (APA literature and a list of websites and g ( publications) addressing the challenges facing today s wood building designer

2 Presentation Agenda Wood as a Structural Material Wood as a structural t material Design Criteria Building Elements The Unified Structure Design Examples Wood has a strength direction Wood as a structural material Load parallel to grain Load perpendicular to grain Compression Parallel columns, posts, truss chords Perpendicular deformation of member Tension Parallel Highest strength beams, panels Perpendicular Weakest capacity - connections Stronger Weaker

3 Mechanical Properties of Wood Mechanical Properties of Wood Bending Load, w: Develop strength in extreme fiber High strength to weight ratio d x L Calculate maximum moment Resisting Moment = M = S x F b S = Section Modulus = bd 2 /6 F b = Allowable Bending Stress b Moment, M: M=w*x/2*(L-x) M max =wl 2 /8 Typical Units are pound-feet (lb-ft) Mechanical Properties of Wood 2 Failure Modes Strength Have fibers crushed, split, or otherwise destructed? Stiffness How much has the beam moved under a given load? Mechanical Properties of Wood Deflection Modulus of Elasticity Calculate maximum deflection Deflection limitsit Building code vs. Manufacturers Recommendations Typical units are inches (in)

4 Mechanical Properties of Wood Maximum Deflection Load, w: Deflection, : x L Limited to reduce floor bounce and prevent cracking of finish materials such as drywall and tile (CBC Table ) Construction Roof members w/drywall clg. Live load deflection Total load deflection L/240 L/180 max =5w*L 4 /(384*E*I) L Floor L/360 L/240 members I joists* L/480 *Per APA form E30, pg 26 Mechanical Properties of Wood Mechanical Properties of Wood Shear stress Critical at connections, reactions, point loads Typically not failure mode in flexural members May control in short spans with heavy loading, cantilever, or continuous spans Avoid stress concentrations at notches or changes in cross section Calculate maximum shear force Allowable Shear Stress = Fv >= 1.5V/A V = maximum shear A = Cross Sectional Area Typical units for Shear are pounds (lbs) Load, w: Shear, V: x V=w(L/2-x) L V max =wl/2

5 Tension Perpendicular to Grain Load Path Continuity Wood splits from: notches hanging loads restraint by connector Spread out loads from fasteners Consider alternative (large single fastener) Best (multiple small fasteners) Load Path Continuity Notching Tension perpendicular to grain

6 Connecting Wood Do Not Mix I-joists with Dimension i Lumber Wood, like other materials, moves in varying environments Dispersal of Strength Reducing Characteristics Wood as a Structural Material I-Joist vs. Lumber Both at 16" o.c. 36% less wood fiber I-Joist at 19.2" o.c & Lumber at 16" o.c. 46% less wood fiber VS. I-Joist Lumber

7 Design Criteria: Loads Vertical Load Dead Loads (permanent) structure, partitions, finishes Live Loads people, furniture, snow Wind and Seismic Impact loads (The effect of loads are lessened with shorter duration) Loads (2010 CBC Chapter 16 & CRC Section R301) Dead Loads (Section 1606) Weight of permanent loads: construction materials, fixed equipment Increases from joists to beams Dead Loads? Live Loads (Table ) Live Load Reduction (1607.9) Reduction in Roof Live Loads ( ) Based on supported area, slope roof Decreases from joists to beams

8 Loads (2010 CBC Chapter 16 & CRC Section R301.2) Lateral Load Climatic Loads Snow (Section 1608) Rain (Section 1611) Lateral Loads Wind (Section 1609) Seismic (Section 1613) Lateral a Loads: National a Issue Loads (2010 CBC Chapter 16 & CRC Section R301) Wind Hazard Earthquake Hazard Load Combinations (Section 1605) 21 Equations M t h k ll bi ti f i l di Must check all combinations for maximum loading Examples: D + L + (L r or S or R) D + L + ωw 0.9D + E/1.4

9 Adjustment Factors Capacity +/- based on: Duration of Load Moisture Temperature Chemical Treatments (Found in the National Design Specification for Wood Construction (NDS) published by the American Forest & Paper Association) Load Duration Factor Wood capacity greater for short time loading LOAD DURATION Load Duration Typical Loads Factor - CD Permanent 0.9 Dead Load Ten years 1.0 Floor live load Two months 1.15 Snow load Seven days 1.25 Construction load Ten minutes Wind/Earthquake Impact 2.0 Vehicles These factors are applied to member capacity Design Considerations End restraint conditions: Simple span has 2 supports Continuous has 3 or more supports Cantilevered has 1 support Supports may be beams, columns, walls Design Considerations Loading Conditions: Uniform: Dead load, Live loads, pounds per lineal feet (plf) x L Simple Continuous Point: Interior Columns, walls, pounds (lbs) Cantilevered x L

10 Building Elements Why Engineer? When a building, or portion, doesn t meet conventional requirements it must be engineered (CBC , CRC R ) Gravity Design Horizontal members Panels Joists Beams Vertical members Studs Columns Wood Structural Panels Building Elements: Panels Face Core Center Core Back

11 Building Elements: Panels Building Elements: Panels OSB layers are engineered for strength. Roof Span Deflection = L/240 Live load = 30 psf Dead load =10psf Floor Span Deflection = L/360 Live load = 100 psf Dead load = 10 psf Building Elements: Panels Building Elements: Panels Rated Sheathing Floor, wall or roof Plywood or OSB A PA RATED SHEATHING 32/16 SIZED FOR SPACING EXPOSURE 1 THICKNESS IN. 000 PS 2-10 SHEATHING PRP-108 HUD-UM-40 15/32 CATEGORY Roof Covering Rated Sheathing Floor, wall or roof Plywood or OSB A PA RATED SHEATHING 32/16 SIZED FOR SPACING EXPOSURE 1 THICKNESS IN. 000 PS 2-10 SHEATHING PRP-108 HUD-UM-40 PRP-108 HUD-UM-40 15/32 CATEGORY

12 Building Elements: Panels Sturd-I-Floor Combined subfloor & underlayment Resistant to concentrated & impact loads Plywood or OSB A PA RATED STURD-I-FLOOR 20 oc SIZED FOR SPACING T & G NET WIDTH 47-1/2 EXPOSURE 1 THICKNESS IN. 000 PS 2-10 PRP /32 CATEGORY SINGLE FLOOR HUD-UM-40 Carpet & pad APA Form E30 Table 30 Span Rating Conditions APA Form E30 Table 33 Strength th axis perpendicular to supports Continuous across 2 or more spans

13 Building Elements: Joists Building Elements: Joists I-joist Used for floor & roof framing Long lengths available Uniform Load Flange (LVL or lumber) Compression Tension Web (OSB) LC Forces are Max. at L C Building Elements: Joists Uniform Load Shear Force C B B C Rule of Thumb: Hole size inversely proportional to shear force

14 Building Elements: Rim Board Building Elements: Beams Laminated Veneer Lumber (LVL) Veneers bonded together Beams, headers, rafters & scaffold planking Constructability Field Notching and Drilling of LVL (Form G535) Horizontal Hole Drilling All grain parallel to length

15 Vertical Holes Strength reduction = 1.5 x Hole diameter/beam width (Forms S560 and G535) Example: 6 Beam width 1 diameter vertical hole Reduction = 1 x 1.5/6 Reduction = Beam is 75% of original strength Side-loaded Multi-ply Beams Connection of plies is specified in the NDS and individual LVL manufacturer literature Pre-engineered engineered Connectors Glulam Joist and beam hangers Top and face mount Product specific Use correct nail Fill all holes Ensure proper p fastener penetration Glulam

16 Engineered Lay-ups TOP Stamp Compression zone Inner zone Tension zone Critical Tension Zone Building Elements: Beams Stock Beams Camber is not an issue Camber in stock beams is usually zero or based on a 3500 or 5000 radius where a 20 beam has a curvature of 1/8 or less Constructability Field Notching and Drilling of Glulam (Form S560) Horizontal Hole Drilling 3500 radius Zero camber

17 Building Elements: Beams Building Elements: Beams Preservative treatment Naturally durable wood species Alaskan Yellow Cedar Port Orford Cedar Exposed Conditions Treated Beams and Columns for Decks LVL Hybrid Glulam with LVL Outer Laminations Full length with no finger joints required LVL has greater tensile strength compared to lumber 30F-2.1E stress level achieved Direct substitute t for many SCL products LVL Laminations Building Elements: Beams Architectural Appearance + Full Framing Width + IJC Depths Maximize versatility exposed or not Ease of construction no shimming required Stud Capacity Buckling capacity usually controls Buckling controlled by: Stud length Buckled shape Stud size (2x4 vs. 2x6) Stud grade (No.1, No. 2, Stud) Bracing in weak direction (blocking, drywall) Strong direction Weak direction

18 Stud Bending Built-up up Lumber Columns Lateral force on studs further reduces the buckling capacity. This controls the design of exterior studs subjected to lateral wind or seismic forces. Buckled shape Multi-ply columns Guidance provided in NDS for: Nailed or bolted laminated columns Nailed K f = 0.60 Bolted K f = 0.75 Built-up up Lumber Columns Building Elements Nail spacing dictated by NDS for reduced Kf Lateral Design Horizontal members Diaphragms Vertical members Shear Walls

19 Lateral Load Path Gravity Load Path Lateral Load Path Designing Wood Structures to Resist Lateral Loads Conventional Light Frame Construction Prescriptive, uses bracing Limited as defined by provisions Engineered Lateral Force Resisting System Uses shear walls, diaphragms, collectors, etc.

20 Blocked Diaphragm Unblocked Diaphragm Engineered Shear Walls Specific stud species Wood structural panels of specific grade and thickness Height to width ratio (SDPWS Table 4.3.4) For F shear walls and perforated shear walls h:w must not exceed 2:1 or 3.5:1 ratio Hold-down anchors Specific nail size and spacing requirements Base shear anchor bolts

21 Max. Shear Wall Aspect Ratios ( ) Aspect ratio = height-to-width ht t idth ratio Height = bottom of bottom plate to top of top plate Width = sheathed width of wall Design 1997 UBC 2000 IBC IBC Wind 3.5:1 3.5:1 3.5:1 Zone 4 2: Seismic Zone : SDC D-F -- 2:1 2:1 a SDC A-C :1 2:1 a a. May be reduced to 3.5:1 if allowable shear is reduced by 2w/h Shear Wall Design (SDPWS 4.3) Segmented Force Transfer Perforated 1. Aspect Ratio for seismic 2:1 2. Aspect ratio up to 3.5:1, if allowable shear is reduced by 2w/h 1. Code does not provide guidance for this method 2. Different approaches using rational analysis could be used 1. Code provides specific requirements 2. The capacity is determined based on empirical equations and tables Hold-Down Placement Traditional Hold-Down Placement Perforated

22 The Unified Structure Lateral Loads(Wind) F = PA Effort is devoted d to determining: P wind pressure Lateral Loads(Seismic) General Modes of Failure F = ma Effort is devoted d to determining: a acceleration Uplift Base Shear Racking Overturning

23 Breached Building Envelope - F-2 Tornado Easy Upgrade! Reference: APA Report Midwest Tornados 2003 Bottom Plate to Foundation

24 Lateral Force Resisting Systems Hold-down down hardware Lateral connection strength depends on: The Unified Structure Crushing g( (bearing) strength of wood Size of wood pieces Fastener size and strength Plus appropriate end use adjustment factors (i.e. Wet service, edge distance, end grain, etc.) The Unified Structure Consistency Counts Withdrawal Connection Strength Depends On: Depth of penetration Wood density Fastener size and type Plus appropriate end use adjustment factors i.e. wet service, edge distance, end grain, etc. Nail sizes Are you using the right nail? Specify pennyweight, type, diameter and length Ex: 8d common = x 2-1/2 Type 8d Nail Sizes Length (in.) Wire Dia. (in.) Finish 2-1/2" Box & casing 2-1/2" Siding 2-3/8" Cooler 2-3/8" Common 2-1/2" Ring- or or 2-1/2" screw-shank 0.131

25 Consistency Counts Consistency Counts Overdriven fasteners Overdriven Fasteners Overdriven Not Overdriven Overdriven Fasteners Overdriven Distance Action < 20% <1/8" None > 20% < 1/8" Any > 1/8" Add 1 for every two overdriven APA Publication TT-012 Consistency Counts Pre-engineered engineered Connectors Overdriven Fasteners Joist and beam hangers Overdriven Fasteners Any Overdriven Distance Due to Thickness Swelling Action None Top and face mount Product specific Use correct nail Fill all holes Ensure proper fastener penetration APA Publication TT-012

26 Consistency Counts Inconsistent Spacing & Span Variable Spacing Be Careful! Inconsistent feel & performance Consistency Counts Consistent Spacing & Span Floor Sheathing Example Answer: From Table 12, APA form E30 (pg 33): For 16 oc spacing = 7/16 32/16 wood structural panel (WSP) For 24 oc spacing = 23/32 or ¾ 48/24 WSP Given: Span = 16 and 24 Live load = 40 psf Dead load = 10 psf Design reference: APA form E30

27 APA Form E30 Table 12 Joist Example Answer: From Table 8, APA form E30 (pg 26): For 16 oc spacing = 9-1/2 PRI-20 For 24 oc spacing = 9-1/2 PRI-60 or 11-7/8 PRI-20 Given: Spacing = 16 and 24 Live load = 40 psf Dead load = 10 psf Simple Span = 15 Design reference: APA form E30 APA Form E30 Table 8 Beam Example Answer: From Table 3A, APA form EWS X440B (pg 18): 3-1/8 x 16-1/2 3-1/2 x /8 x 12 Or 5-1/2 x12 Given: Span = 16-3 Span roof trusses = 24 Live load = 40 psf Dead load = 10 psf Design reference: APA form EWS X440B

28 Beam Sizing Given: Span = 22' Floor live load = 40 psf Floor dead load = 15 psf Tributary Width = 18 Find: Beam size for l/360 deflection Answer: From Glulam Floor Beam, APA form C415 For 24F-1.8E beams, see Table 1a (pg 3): 5-1/8 x 22-1/2, 5-1/2 x 22-1/2, or 6-3/4 x 21 For IJC 24F-1.8E beams, see Table 2a (pg 5): 3-1/2x24, 5-1/2 x 20 or 7 x 18 For 30F-1.8E beams, see Table 3a (pg 8): 3-1/2x22, 5-1/2 x 18 or 7 x 18

29 Beam Sizing Structural calculations: 1. Define design criteria 2. Check maximum bending 3. Check Shear 4. Check deflection 1. Span = 22 Floor live load = 40 psf Floor dead load = 15 psf Tributary Width = 18 Max. Deflection = l/360 Uniform load = W = (D + L)*Tributary Width = 18 *( )psf = 990plf Select 6-3/4 x 21 beam to begin design APA Form Y117 Table 5 Structural calculations: 2. Check maximum bending Beam Sizing M max = wl 2 /8 = 990*(22) 2 /8 = 59,895 lb-ft From APA Form Y117 (pg 12), Moment Capacity = M = 99,225 lb-ft Since M max < M, OK

30 Structural calculations: 3. Check maximum shear Beam Sizing V max = wl/2 = 990*(22)/2 = 10,890 lb From APAForm Y117 (pg 12), Shear Capacity = V = 25,043 lb Beam Sizing Structural calculations: 4. Check maximum deflection max = 5wl 4 /(384EI) = 5*990plf*(22 ) 4 /(384*9377x10 6 lb-in 2 )*(12 /1 ) 3 = 0.56 From APA Form Y117(pg 12), EI = 9377x10 6 lb-in 2 = l/360 = 22 /360*(12 /1 ) = 0.73 Since max <, OK Since V max <V V, OK Therefore, 6-3/4 x 21 beam works. If not, select new beam and repeat steps 2-4. What is new? From Steel APA Form C415 Table 4A Given: Span = 10 W10x12 Design reference: APA form EWS C415

31 APA Form C415 Table 5A To Wood For 24F-1.8E Glulams: See Table 4a, APA form EWS C415 (pg 11): 3-1/2x15, 5-1/2x13-1/2, or 7x10-1/2 1/2 For 30F-2.1E Glulams: See Table 5a, APA form EWS C415 (pg 13): 3-1/2x14, 5-1/8x11-7/8, or 7x9-1/2 Given: ShearWalls:Windv Wind v. Seismic 7/16 OSB 8d common V / 6 edge/field 8 nail spacing Gypsum on opposite face H v H ShearWalls:Windv Wind v. Seismic Wind d Capacity: V=(450 plf x plf) x 5.33 = 3891 lb Length of wall From table For wind S i i C it Seismic Capacity: V=450 plf x 5.33 = 2399 lb For gypsum from table

32 Shear Wall Design Examples Segmented Shear Wall Approach Force Transfer Around Opening Approach Perforated Shear Wall Approach V Design Example V = 3,750 lbs Segmented Approach Segmented Approach V V Do not consider contribution of wall below and above openings H v H H v H H v H H v H Code Limitation V = 3,750 lbs Height/width Ratio = 8:3.5 2w/h = (2)(3.5)/8 = 0.875

33 Segmented Approach Segmented Approach 1. Unit Shear V = V/L = 3,750/15 = 250 lbs/ft 2. Allowable Shear 3-6 walls v allowable = 380 (0.875)=332 lbs/ft > 250 lbs/ft 3. Allowable Shear 4 walls (2:1 h:w) v allowable = 260lb/ft > 250 lbs/ft 4. Hold-down forces H = vh = 250 x 8 = 2,000 lbs Note: For simplicity Dead Load contribution and various footnote adjustments are omitted 15/32 Rated Sheathing 4 o.c. at 3.5 walls 15/32 Rated Sheathing 6 4 walls 8 hold lb capacity V 15/32 Rated Sheathing 4 o.c V = 3,750 lbs v = 250 lbs/ft H = 2,000 lbs H v H H v H H v H H v H 8 hold lb capacity /32 Rated Sheathing 6 o.c. Segmented Shear Wall Approach Force Transfer Around Opening Approach Perforated Shear Wall Approach Perforated Shear Wall Approach V H v, t v, t v, t v, t H V = 3,750 lbs Height/width Ratio = 8:3.5 2w/h = (2)(3.5)/8 = 0.875

34 Perforated Shear Wall Approach Perforated Shear Wall Approach 1 Unit shear in the wall v = 3,750/15 = 250 lb/ft SDPWSTable4335ShearResistanceAdjustmentFactor Shear Factor, C O 2 Percent of Full-Height Sheathed 15/26 = 0.57 (57%) 3 Maximum opening height 2H/3 = % 0.61 Perforated Shear Wall Approach Perforated Shear Wall Approach 4 Co Shear Resistance Adjustment t Factor Co = say Adjusted Shear Resistance v allowable = 490 x x 0.61 = 262 lbs/ft > 250 lbs/ft 6. Uplift at Perforated Shear Wall ends (hold downs) H = (250/0.61) x 8 = 3,280 lbs 7. In-plane Shear Anchorage H = 250/0.61 = 410plf 15/32 Rated Sheathing 3 o.c. 8. Uplift anchorage between shear wall ends t = 250/0.61 = 410 plf (at full segments only) 9. Deflection is determined based on the deflection of any segment of the wall divided by C o

35 Segmented Approach Force Transfer Perforated 15/32 Rated sheathing 4 o.c. (3-6 6 o.c. (4 walls) 8 hold lb capacity 15/32 Rated Sheathing 4 o.c. 2 hold 1,550 lb capacity 2 Straps 1,250 lb 15/32 Rated Sheathing 3 o.c. 2 hold 3280 lb capacity Questions/ Comments? This concludes The American Institute of Architects Continuing Education Systems Course H v, t v, t v, t v, t H extensive plate anchorage Karyn A. Beebe, P.E., LEED AP (858) Wood Products Council