Wood Design. fv = shear stress fv-max = maximum shear stress Fallow = allowable stress Fb = tabular bending strength = allowable bending stress

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1 Wood Design Notation: a = name for width dimension A = name for area Areq d-adj = area required at allowable stress when shear is adjusted to include self weight b = width of a rectangle = name for height dimension c = largest distance from the neutral axis to the top or bottom edge of a beam = constant in Cp expression c1 = coefficient for shear stress for a rectangular bar in torsion CC = curvature factor for laminated arches CD = load duration factor Cfu = flat use factor for other than decks C = size factor CH = shear stress factor Ci = incising factor CL = beam stability factor CM = wet service factor Cp = column stability factor for wood design Cr = repetitive member factor for wood design CV = volume factor for glue laminated timber design Ct = temperature factor for wood design d = name for depth = calculus symbol for differentiation dmin = dimension of timber critical for buckling D = shorthand for dead load = name for diameter DL = shorthand for dead load E = modulus of elasticity Emin = reference modulus of elasticity for stability E min = adjusted modulus of elasticity for stability f = stress (strength is a stress limit) fb = bending stress ffrom table = tabular strength (from table) = bearing stress fp fv = shear stress fv-max = maximum shear stress allow = allowable stress b = tabular bending strength = allowable bending stress b = allowable bending stress (adjusted) c = tabular compression strength parallel to the grain c = allowable compressive stress (adjusted) c = intermediate compressive stress dependent on load duration = theoretical allowed buckling stress c = tabular compression strength perpendicular to the grain connector = shear force capacity per connector p = tabular bearing strength parallel to the grain = allowable bearing stress t = tabular tensile strength u = ultimate strength v = tabular bending strength = allowable shear stress y = yield strength h = height of a rectangle H = name for a horizontal force I = moment of inertia with respect to neutral axis bending Itrial = moment of inertia of trial section Ireq d = moment of inertia required at limiting deflection Iy = moment of inertia with respect to an y-axis J = polar moment of inertia K = effective length factor for columns Le = effective length that can buckle for column design, as is l e L = name for length or span length LL = shorthand for live load LRD = load and resistance factor design M = internal bending moment Mmax = maximum internal bending moment 75

2 n = number of connectors across a joint, as is N Mmax-adj = maximum bending moment adjusted to include self weight p = pitch of connector spacing = safe connector load parallel to the grain P = name for axial force vector Pallowable = allowable axial force q = safe connector load perpendicular to the grain Qconnected = first moment area about a neutral r R axis for the connected part = radius of gyration = interior radius of a laminated arch = radius of curvature of a deformed beam = radius of curvature of a laminated arch = name for a reaction force S = section modulus Sreq d = section modulus required at allowable stress Sreq d-adj = section modulus required at allowable stress when moment is adjusted to include self weight T = torque (axial moment) V = internal shear force Vmax = maximum internal shear force Vmax-adj = maximum internal shear force adjusted to include self weight w = name for distributed load wself wt = name for distributed load from self weight of member W = shorthand for wind load x = horizontal distance y = vertical distance Z = force capacity of a connector actual = actual beam deflection allowable = allowable beam deflection limit = allowable beam deflection limit max = maximum beam deflection γ = density or unit weight θ ρ Σ = slope of the beam deflection curve = radial distance = symbol for integration = summation symbol Wood or Timber Design Structural design standards for wood are established by the National Design Specification (NDS) published by the National orest Products Association. There is a combined specification (from 005) for Allowable Stress Design and limit state design (LRD). Tabulated wood strength values are used as the base allowable strength and modified by appropriate adjustment factors: = C C C... Size and Use Categories Boards: 1 to 1½ in. thick in. and wider Dimension lumber to 4 in. thick in. and wider Timbers 5 in. and thicker 5 in. and wider D M from table 76

3 Adjustment actors (partial list) CD CM C Cfu Ci Ct Cr CH CV CL CC load duration factor wet service factor (1.0 dry < 16% moisture content) size factor for visually graded sawn lumber and round timber > 1 depth 1 C 1 9 = ( ) 1.0 d flat use factor (excluding decking) incising factor (from increasing the depth of pressure treatment) temperature factor (at high temperatures strength decreases) repetitive member factor shear stress factor (amount of splitting) volume factor for glued laminated timber (similar to C) beam stability factor (for beams without full lateral support) curvature factor for laminated arches Tabular Design Values b : bending stress t: tensile stress v: horizontal shear stress c : compression stress (perpendicular to grain) c: compression stress (parallel to grain) E: modulus of elasticity p: bearing stress (parallel to grain) Wood is significantly weakest in shear and strongest along the direction of the grain (tension and compression). Load Combinations and Deflection The critical load combination is determined by the largest of either: dead load ( dead load + any combination of live load or 0. 9 C D ) The deflection limits may be increased for less stiffness with total load: LL + 0.5(DL) 77

4 Criteria for Design of Beams Allowable normal stress or normal stress from LRD should not be exceeded: Knowing M and b, the minimum section modulus fitting the limit is: b fb = S req' d Mc I M b Besides strength, we also need to be concerned about serviceability. This involves things like limiting deflections & cracking, controlling noise and vibrations, preventing excessive settlements of foundations and durability. When we know about a beam section and its material, we can determine beam deformations. Determining Maximum Bending Moment Drawing V and M diagrams will show us the maximum values for design. Remember: V = Σ( w) dx M = Σ( V ) dx dv dx dm = w = V dx Determining Maximum Bending Stress or a prismatic member (constant cross section), the maximum normal stress will occur at the maximum moment. or a non-prismatic member, the stress varies with the cross section AND the moment. Deflections If the bending moment changes, M(x) across a beam of constant material and cross section then the curvature will change: The slope of the n.a. of a beam, θ, will be tangent to the radius of curvature, R: 1 θ = slope = M x dx EI ( ) The equation for deflection, y, along a beam is: y = 1 1 θdx = M ( x) EI EI dx 1 = R M ( x) EI Elastic curve equations can be found in handbooks, textbooks, design manuals, etc.. Computer programs can be used as well (like Multiframe). 78

5 Elastic curve equations can be superpositioned ONLY if the stresses are in the elastic range. The deflected shape is roughly the same shape flipped as the bending moment diagram but is constrained by supports and geometry. Boundary Conditions The boundary conditions are geometrical values that we know slope or deflection which may be restrained by supports or symmetry. At Pins, Rollers, ixed Supports: y = 0 At ixed Supports: θ = 0 At Inflection Points rom Symmetry: θ = 0 The Slope Is Zero At The Maximum Deflection ymax: dy θ = = slope = 0 dx Allowable Deflection Limits All building codes and design codes limit deflection for beam types and damage that could happen based on service condition and severity. y x L max ( ) = actual allowable = value Use LL only DL+LL Roof beams: Industrial (no ceiling) L/180 L/10 Commercial plaster ceiling L/40 L/180 no plaster L/360 L/40 loor beams: Ordinary Usage L/360 L/40 Roof or floor (damageable elements) L/480 79

6 Lateral Buckling With compression stresses in the top of a beam, a sudden popping or buckling can happen even at low stresses. In order to prevent it, we need to brace it along the top, or laterally brace it, or provide a bigger Iy. Beam Loads & Load Tracing In order to determine the loads on a beam (or girder, joist, column, frame, foundation...) we can start at the top of a structure and determine the tributary area that a load acts over and the beam needs to support. Loads come from material weights, people, and the environment. This area is assumed to be from half the distance to the next beam over to halfway to the next beam. The reactions must be supported by the next lower structural element ad infinitum, to the ground. Design Procedure The intent is to find the most light weight member satisfying the section modulus size. 1. Know for the material or U for LRD.. Draw V & M, finding Mmax. M max 3. Calculate Sreq d. This step is equivalent to determining f b = b bh S 4. or rectangular beams - or timber: use the section charts to find S that will work and remember that the beam self weight will increase Sreq d. Determine the updated Vmax and Mmax including the beam self weight, and verify that the updated Sreq d has been met. 5. Consider lateral stability. 6. Evaluate horizontal shear stresses using Vmax to determine if f v v or find Areq d or rectangular beams S = 7. Provide adequate bearing area at supports: 3V fv max = = 1. 5 A Tρ T 8. Evaluate shear due to torsion f v = or v J c1ab (circular section or rectangular) 80 V A A req'd 3V 9. Evaluate the deflection to determine if maxll LL allowed and/or max Total Total allowed note: when calculated > limit, Irequired can be found with: and Sreq d will be satisfied for similar self weight 6 w self wt = γa f P A p = c or c too big I req ' d I lim it trial v

7 OR ANY EVALUATION: Redesign (with a new section) at any point that a stress or serviceability criteria is NOT satisfied and re-evaluate each condition until it is satisfactory. Load Tables for Uniformly Loaded Joists & Rafters Tables exists for the common loading situation for joists and rafters that of uniformly distributed load. The tables either provide the safe distributed load based on bending and deflection limits, they give the allowable span for specific live and dead loads. If the load is not uniform, an equivalent distributed load can be calculated from the maximum moment equation. Decking lat panels or planks that span several joists or evenly spaced support behave as continuous beams. Design tables consider a 1 unit wide strip across the supports and determine maximum bending moment and deflections in order to provide allowable loads depending on the depth of the material. The other structural use of decking is to construct what is called a diaphragm, which is a horizontal or vertical (if the panels are used in a shear wall) unit tying the sheathing to the joists or studs that resists forces parallel to the surface of the diaphragm. Criteria for Design of Columns If we know the loads, we can select a section that is adequate for strength & buckling. If we know the length, we can find the limiting load satisfying strength & buckling. Any slenderness ratio, Le/d 50: P f c c A = = ( C )( C )( C )( C )( C ) The allowable stress equation uses factors to replicate the combination crushing-buckling curve: c c D M t p where: c = allowable compressive stress parallel to the grain c = compressive strength parallel to the grain CD = load duration factor CM = wet service factor (1.0 for dry) Ct = temperature factor 1+ ( / c ) C p = C = size factor c Cp = column stability factor off chart or equation: c = 0.9 for glue-lam and 0.8 for sawed lumber 1+ / c c / c c 81

8 or preliminary column design: c = c C = p ( cc D ) C p Procedure for Analysis 1. Calculate Le/dmin (KL/d for each axis and chose largest). Obtain c compute 0.8 = with E min = Emin ( CM )( Ct )( CT )( Ci ) Emin L d e ( ) where E min the modulus of elasticity for stability K E (or by outdated text: = with K =0.3 for sawn, = for glu-lam) l ( ) e 3. Compute c cc D with CD = 1, normal, CD =1.5 for 7 day roof, etc Calculate c and get Cp from table or calculation 5. Calculate c = c C p 6. Compute Pallowable = c A or alternatively compute factual = P/A 7. Is the design satisfactory? Is P Pallowable? yes, it is; no, it is no good or Is factual c? yes, it is; no, it is no good d Procedure for Design 1. Guess a size by picking a section. Calculate Le/dmin (KL/d for each axis and choose largest) 3. Obtain c compute 0.8 = with E min = Emin ( CM )( Ct )( CT )( Ci ) Emin L d e ( ) where E min the modulus of elasticity for stability K E (or by outdated text: = with K =0.3 for sawn, = for glu-lam) l ( ) e 4. Compute c cc D with CD = 1, normal, CD =1.5 for 7 day roof Calculate c and get Cp from table or calculation 6. Calculate c = c C p 7. Compute Pallowable = c A or alternatively compute factual = P/A 8. Is the design satisfactory? d Is P Pallowable? yes, it is; no, pick a bigger section and go back to step. or Is factual c? yes, it is; no, pick a bigger section and go back to step. 8

9 Trusses Timber trusses are commonly manufactured with continuous top or bottom chords, but the members are still design as compression and tension members (without the effect of bending.) Stud Walls Stud wall construction is often used in light frame construction together with joist and rafters. Studs are typically -in. nominal thickness and must be braced in the weak axis. Most wall coverings provide this function. Stud spacing is determined by the width of the panel material, and is usually 16 in. The lumber grade can be relatively low. The walls must be designed for a combination of wind load and bending, which means beam-column analysis. Columns with Bending (Beam-Columns) f c fbx The modification factors are included in the form: + where: 1 f c c bx fc 1 = magnification factor accounting for P- bx x = allowable bending stress f = working stress from bending about x-x axis bx x 1.0 In order to design an adequate section for allowable stress, we have to start somewhere: 1. Make assumptions about the limiting stress from: - buckling - axial stress - combined stress. See if we can find values for r or A or S (=I/cmax) 3. Pick a trial section based on if we think r or A is going to govern the section size. 4. Analyze the stresses and compare to allowable using the allowable stress method or interaction formula for eccentric columns. 5. Did the section pass the stress test? - If not, do you increase r or A or S? - If so, is the difference really big so that you could decrease r or A or S to make it more efficient (economical)? 6. Change the section choice and go back to step 4. Repeat until the section meets the stress criteria. 83

10 Laminated Arches The radius of curvature, R, is limited because of residual bending stresses between lams of thickness t to 100t for Southern pine and hardwoods and 50t for softwoods. The allowable bending stress for combined stresses is ( C C ) b = b C where t C C = r and r is the radius to the inside of the lamination. Criteria for Design of Connections Connections for wood are typically mechanical fasteners. Shear plates and split ring connectors are common in trusses. Bolts of metal bear on holes in wood, and nails rely on shear resistance transverse and parallel to the nail shaft. Timber rivets with steel side plates are allowed with glue laminated timber. Connections must be able to transfer any axial force, shear, or moment from member to member or from beam to column. Bolted Joints Stress must be evaluated in the member being connected using the load being transferred and the reduced cross section area called net area. Bolt capacities are usually provided in tables and take into account the allowable shearing stress across the diameter for single and double shear, and the allowable bearing stress of the connected material based on the direction of the load with respect to the grain. Problems, such as ripping of the bolt hole at the end of the member, are avoided by following code guidelines on minimum edge distances and spacing. Nailed Joints Because nails rely on shear resistance, a common problem when nailing is splitting of the wood at the end of the member, which is a shear failure. Tables list the shear force capacity per unit length of embedment per nail. Jointed members used for beams will have shear stress across the connector, and the pitch spacing, p, can be determined from VQconnected area the shear stress equation when the capacity,, is known: nconnector p I 84

11 Example 1 (pg 38) c = 440 psi, γ = 36.3 lb/ft 3 85

12 Example 1 (continued) 86

13 Example (pg 36) Using D-L No. 1, design the simply supported floor beam shown to meet bending, shear, and deflection criteria. b = 1300 psi; v = 85 psi; E = 1.6 x 10 6 psi; γ = 35 lb/ft 3. PLL = 1,500 lb. 9 ft SOLUTION: Ordinarily, floor joist are closely spaced together, which allows for the use of the repetitive use factor, Cr. loor joist also will not have very large uniformly distributed loads because the spacing is typically feet apart or less. We will assume this beam is not a joist. But it does have live load from occupancy which is considered normal, so the load duration factor, CD, is 1. The other conditions (like temperature and moisture) must be assumed to be normal (and have values of 1.0). The allowable stresses can be determined from: 'b = CDb = (1.0)(1300) = 1300 psi 'v = CDv = (1.0)(85) = 85 psi,675 lb 17,35 lb-ft 475 lb Bending: Shear: lb ft 17, 35 (1 in M ft ) 3 Sreq ' d = = 159.9in A 1300 psi b req ' d 3V 3(,85 lb) 49.8in = (85 psi) = Try a 6 x 14. This satisfies both requirements with the least amount of area. (See the 8 x 1.) (A = 74.5 in, S = in 3, Ix = in 4 ) 35 lb 3 (74.5 in ) ft w lb self wt = γ A = = ft (1 in ft ) which is additional dead load! Because the maximum moment from the additional distributed load is not at the same location as the maximum moment from the diagram, we have to find Ms.w. at 11 feet from the end in order to add them: lb lb ft ft (11 )(0 11 ) 17, 35 lb ft ft ft M 18, 18.5 lb ft 18, 18.5 (1 in ft ) 3 adjusted = + = and Sreq ' d = 168.in 1300 psi v -1,05 lb -,85 lb The same holds true for the contribution to the shear: lb ft (0 ft) Vadjusted =,85lb + = 3, 005.5lb and A Check that the section chosen satisfies the new required section modulus and area: Is Sthat I have Sthat I need? Is Athat I have Athat I need? Is in in 3? No, NOT OK. Is 74.5 in in? Yes, OK. 3( lb) (85 psi) req ' d = 53.04in Because the section that I have is not adequate, I need to choose one that is. This will have a larger self weight that must be determined and included in the maximum moment (with the initial maximum). It will make Sreq d and Areq d bigger as well, and the new section properties must be evaluated with respect to these new values. Try a 8 x 14. This satisfies both requirements with the least amount of area. (A = in, S = 7.8 in 3, Ix = in 4 ). w lb 3 in ft self wt. A in ft M adjusted 35 (101.5 ) = γ = = 4.6 (1 ) lb ft 4.6 lb ft (11 ft)(0 11 ft) = 17,35 + = 18,54.7 which is more additional dead load! lb ft lb ft and S lb ft 18, 54.7 (1 in ) ft 3 req ' d = 171.in 1300 psi The same holds true for the contribution to the shear: 4.6 lb ft (0 ft) Vadjusted =,85lb + = 3, 071lb and A 3(3071 lb) (85 psi) req ' d = 54.in 87

14 Example (continued) Check that the section chosen satisfies the new required section modulus and area: Deflection: Is Sthat I have Sthat I need? Is Athat I have Athat I need? Is 7.8 in in 3? Yes, OK. Is in 54. in? Yes, OK. The total deflection due to dead and live loads must not exceed a limit specified by the building code adopted (for example, the International Building Code) or recommended by construction manuals. or a floor beam, the usual limits are L/360 for live load only and L/40 for live and dead load. 0 ft(1 in ft ) 0 ft(1 in ft ) LL-limit = = 0.67in total-limit = = 1.0in Superpositioning (combining or superimposing) of several load conditions can be performed, but care must be taken that the deflections calculated for the separate cases to obtain the maximum must be deflections at the same location in order to be added together. In this case, the uniform distributed load maximum is at the center, while the concentrated load maximum is not. When you can t determine the actual location, you will need advice or tools. Multiframe indicates that the maximum is at the center with the combined loadings (see plot under the V & M diagrams). concentrated load (live load): (a is the distance from the supports to the load from the left) Pbx 1500 lb(9 ft)(10 ft) 3 x ( when x<a) = ( l b x ) = ((0 ft) (9 ft) (10 ft) )(1 in ) ft = in 6EI l 6(1.6 x10 psi)( in )(0 ft) distributed load (dead load: ( lb )(0 ) (1 in 5wl + ft ft ft) max at center 6 4 ( ) = = = 0.18in 384EI 384(1.6 x10 psi)( in ) distributed load (live load: (100 lb )(0 ) (1 in 5wl ft ft ft) max at center 6 4 ( ) = = = 0.146in 384EI 384(1.6 x10 psi)( in ) Is live that I have live-limit? Is (0.173 in in) =0.318 in 0.67 in? Yes, OK. Is total that I have total-limit? Is (0.173 in in in) = in 1.0 in? Yes, OK. USE an 8 x

15 Example 3 (pg 379) E min = 580 x 10 3 psi 89

16 Example 4 (pg 381) c = 1650 psi, E = 1.8 x 10 6 psi, E min = 915 x 10 3 psi Also verify with allowable load tables 90

17 Example 5 b = 15 psi c = 1350 psi E min = 580 x 10 3 psi A = 8.5 in S x = 7.56 in LB (D only) E min = E ( CM )( Ct )( Ci ) = 580,000(1.0)(1.0)(1.0)=580,000 psi 0.8E = 0.8(580,000) = = 909 psi min ( l / d ) (.9) e CD = 1.6 from wind loading c 909 = = = 376(0.35) = 83 psi 16in 1 in / ft

18 Example 6 Example 7 A nominal 4 x 6 in. redwood beam is to be supported by two x 6 in. members acting as a spaced column. The minimum spacing and edge distances for the ½ inch bolts are shown. How many ½ in. bolts will be required to safely carry a load of 1500 lb? Use the chart provided. 9

19 Example 8 93

20 Example 8 (continued) 94

21 ARCH 331 Note Set Su015abn

22 ARCH 331 Note Set Su015abn

23 ASD Beam Design low Chart Collect data: L, ω, γ, limits; find beam charts for load cases and actual equations Collect data: b & v and all adjustment factors (CD, etc.) ind Vmax & Mmax from constructing diagrams or using beam chart formulas ind Sreq d and pick a section from a table with Sx greater or equal to Sreq d Calculate ωself wt. using A found and γ. ind Mmax-adj & Vmax-adj. No Calculate Sreq d-adj using Mmax-adj. Is Sx(picked) Sreq d-adj? (OR calculate fb. Is fb b?) Yes Calculate Areq d-adj using Vmax-adj. Is A(picked) Areq d-adj? (OR calculate fv. Is fv v?) No pick a new section with a larger area Calculate max using superpositioning and beam chart equations with the Ix for the section is max limits? This may be both the limit for live load deflection and total load deflection.) too big I req ' d I lim it trial No pick a section with a larger Ix Yes (DONE) 97