STRC09 Timber Beams Columns Connections 0716

Structural Engineer Exam Review Course

Timber and Wood (Beams, Columns, and Connections)

Timber and Wood: Beams, Columns, and Connections Structural Engineering Review Course

STRC ©2015 Professional Publications, Inc.

© 2015 Professional Publications, Inc.

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Timber and Wood: Beams, Columns, and Connections

Overview • adjustment factors • design for flexure • design for shear • design for compression • design for tension • design of connections



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Learning Objectives You will learn • how to determine strength of lumber components using the National Design Standard • how to determine allowable strength of lumber components using the National Design Standard

• how to use allowable stress design method • terminology used in wood and timber construction

• how to calculate strength of a nail or bolt for wood construction



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Prerequisite Knowledge and Skills You should already be familiar with • stress mechanics – bending stress, shear stress, axial stress 

P A



M S



3V 2A

• design standards • IBC load combinations • ASCE7 loading criteria (wind and seismic loads)

• structural analysis • shear and bending moment diagrams • deflection of beams STRC ©2015 Professional Publications, Inc.


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Referenced Codes and Standards • International Building Code (IBC, 2012) • Minimum Design Loads for Buildings and Other Structures (ASCE/SEI7, 2010) • National Design Specification for Wood Construction ASD/LRFD, with Commentary and Supplement (NDS, 2012) • Special Design Provisions for Wind and Seismic (SDPWS, 2008)



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Sizes for Structural Classifications dressed lumber • dressed on a planing machine to create smooth surfaces and uniform sizes

CSTB Table 3.1 Examples of Nominal and Net Dimensions

• net (actual) dimensions are less than nominal dimensions • rough sawn lumber about 1/8 in larger than dressed lumber



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Wood Products • sawn timber: 5 in × 5 in and larger,

CSTB Fig. 3.1 Use and Size Categories

• dimension lumber: 2 in to 4 in thick, 2 in or more wide • glulam timber: thin lams of sawn lumber glued together in a factory • decking: 2 in to 4 in thick, 4 in or more wide, typically tongue in groove



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Allowable Stress Design use internal stress of element, calculated from applied external loads • elementary mechanics of materials equations • linear elastic structural analysis compare to strength of element, reduced by factor of safety • material graded to similar levels of quality • strength from sample testing of materials • factor of safety determined by code writers



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Applied Stress applied stress • calculated stress in a member due to loading applied to the structure and translated to the member • represented by lower‐case f • wood design based upon fundamental stress mechanics



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Reference Design Values • Wood is sorted into grades of lumber based upon species of tree and quality of individual pieces. • lower quality = lower strength = cheaper wood • reference design value of strength designated with upper case F, obtained from NDS Supplement



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Tabulated Wood Strength NDS Supplement Table 4A: Reference Design Values for Visually Graded Dimension Lumber



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Adjustment Factors adjustment factors • change reference design value to allowable design value that matches real‐world applications of timbers and wood • strength adjusted because a single piece of wood will have different strengths, depending on use • assigned symbol C with a subscript • most common adjustment factors • CD = load duration factor • CM = wet service factor • CF = size factor



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Applicability of Adjustment Factors • some factors applicable to wide variety of modes of loading (flexure, compression, bending, etc…)

Table 5.1: Applicability of Adjustment Factors

• some factors applicable to specific modes of loading • Most likely conditions have factors of 1.0. • Unique and unlikely situations cause factors to be greater or less than 1.0.



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Allowable Design Value allowable design value • product of reference design value and adjustment factors • represented by symbol F’ in NDS



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Fundamental Design Equation Applied stress must be less than or equal to allowable design value. f  F' F'  FCDCM ...



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Load Duration Factor • ood stronger if load applied for short amount of time • For a given load combination, choose shortest duration load (largest load duration factor) and use for entire load combination.

Table 5.2 Load Duration Factors

If load combination is dead load alone (CD = 0.9), CD is 0.9 If load combination is dead load (CD = 0.9), live load (CD = 1.00), and earthquake load (CD = 1.60), CD is 1.6 STRC ©2015 Professional Publications, Inc.


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Wet Service Factor • wood weaker if moisture content is high • wood usually in dry (low moisture content) condition (CM = 1.0) • when wet, CM varies, but usually = 0.80 or 0.85 • moisture content required to be considered “wet” varies depending on type of wood product • sawn lumber: moisture content greater than 19% to be considered wet • glulam lumber: moisture content greater than 16% to be considered wet



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Example: Adjustment Factor Example 5.1



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Example: Adjustment Factor Example 5.1



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Example: Adjustment Factor



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Poll: Allowable Strength Adjustment Factors Which statement about the adjustment factors used for wood design is true? (A) They are part of determining the factor of safety for wood design. (B) The higher the value of the load duration factor, CD, the more difficult it is to determine the precise magnitude of the design load. (C) The adjustment factors reflect that the strength of wood, varies based on installation conditions, loading, and wood quality. (D) The adjustment factors depend upon the species of the wood (i.e. Douglas fir‐larch vs. southern pine).



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Poll: Allowable Strength Adjustment Factors Which statement about the adjustment factors used for wood design is true? (C) The adjustment factors reflect that the strength of wood, varies based on installation conditions, loading, and wood quality. Wood, as an organic material, changes depending upon what environment and situation the it is exposed to; e.g., wood is actually stronger if loaded quickly while it is in a dry low temperature situation.



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Design for Flexure • design for flexure = primary task of structural engineers • flexure (bending) is dominant role for wood structural elements (ex., floor joists, roof rafters, girders) • construct beam bending moment diagram • locate maximum moment in beam • determine maximum applied stress 

Mc l



M S



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Size/Volume Factor • grade of wood has same reference strength, independent of cross‐section dimensions • if cross‐section is large, more likely that defect exists • sawn lumber: size factor, CF (applied when depth greater than 12 in)

• glulam beam: volume factor, CV



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Repetitive Member Factor • represented by symbol Cr • Closely spaced wood beams can share concentrated loads to adjoining beams. • NDS allows a Cr value of 1.15 to be applied if: • at least three beams lined up side by side • spacing of beams not more than 24 in apart • sheathing on at least one face of beams



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Flat Use Factor • represented by symbol Cfu • wood members expected to bend around strong axis • when lumber used flatwise (load applied to wide face), members bend around weak axis, and Cfu needed • material may be used as flooring • beam may be bent horizontally (wind load on side of beam) • magnitude of factor varies by type of material • does not apply to material designated as ‘decking’ (decking assumed to be bent about weak axis) STRC ©2015 Professional Publications, Inc.


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Lateral Stability Factor • needed to prevent compression edge of beam from buckling • sheathing on top of beam usually means CL = 1.0 • complicated equation based on ratio of Euler buckling strength, FbE , and strength of beam without buckling considered, Fbx*



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Example: Beam Design Example 5.4



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Example: Beam Design Example 5.4



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Example: Beam Design



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Example: Demand/Capacity Ratio A beam is subjected to a dead load stress of 800 lbf/in2, a live load stress of 600 lbf/in2, and a wind load stress of 500 lbf/in2. The beam is made from lumber with a reference design strength value of 1500 lbf/in2, and is used in a situation where all adjustment factors except the load duration factor are 1.0. Find the demand/capacity ratio for all applicable IBC load combinations.



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Example: Demand/Capacity Ratio A beam is subjected to a dead load stress of 800 lbf/in2, a live load stress of 600 lbf/in2, and a wind load stress of 500 lbf/in2. The beam is made from lumber with a reference design strength value of 1500 lbf/in2, and is used in a situation where all adjustment factors except the load duration factor are 1.0. Find the demand/capacity ratio for all applicable IBC load combinations.

Solution From Table 5.1, the load duration factor for dead load is 0.9. The allowable strength design value for the dead load loading condition is lbf   Fb'  0.90 fb   0.90   1500 2  in    1350 lbf/in 2



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Example: Demand/Capacity Ratio The demand/capacity ratio for the dead load loading condition is lbf fb in 2  0.593  Fb' 1350 lbf in 2 800

For dead load plus live load loading condition, the total applied stress is fb  800

The load duration factor for an occupancy live load is 1.0, and the largest load duration factor governs, so the allowable design value for the dead load plus live load loading condition is lbf   Fb'  1.0 fb  1.0  1500 2  in    1500 lbf/in 2

lbf lbf  600 2  1400 lbf/ft 2 2 in in



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Example: Demand/Capacity Ratio The demand/capacity ratio for the dead load plus live load loading condition is lbf fb in 2  0.933  Fb' 1500 lbf in 2 1400

The load duration factor for a wind load is 1.60, and the largest load duration factor governs, so the allowable design value for the dead load plus live load plus wind load loading condition is

For the dead load plus live load plus wind load loading condition, the total applied stress is

lbf   Fb'  1.60 fb  1.60   1500 2  in    2400 lbf/in 2

fb  D  0.45W  0.75 L lbf lbf  lbf     0.45 500  0.75 600         in 2 in 2  in 2     1475 lbf/in 2  800



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Example: Demand/Capacity Ratio The demand/capacity ratio for the dead load plus live load plus wind load loading condition is lbf fb in 2  0.61  Fb' 2400 lbf in 2 1475



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Beams in Shear: General Requirements • shear is often controlling issue for wood design

Fig. 5.2 Shear Determination in a Beam

• beam usually chosen based on bending, but needs to have shear stress checked



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Beams in Shear: General Requirements • shear stress for any cross‐section and for any location


• maximum shear stress for a rectangular beam



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Critical Location for Shear • calculate shear stress at critical location of beam


• critical location usually distance d away from face of support • add any point loads located near face of support



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Notched Beams • notch reduces shear strength of beam; more severe than simple reduction in cross‐section • NDS restricts location and size of notches



Fig. 5.3 Notched Beams

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Notched Beams Fig. 5.3 Notched Beams

• For compression flange notches • when length e less than depth at notch,

• when e longer than depth of notch,

• For notches on the tension face of a rectangular member 2  d  Vr   Fvbd n  n  3  d 



2

 NDS 3.4-3

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Example: Maximum Shear Load Example 5.5



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Example: Maximum Shear Load Example 5.5



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Example: Maximum Shear Load



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Example: Maximum Shear Load



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Compression Members – Axial Compression • Columns fail in one of two ways. • crushing – usually at support • buckling – usually at mid‐length of member

• crushing strength • considers all adjustment factors except column stability factor • crushing strength designated as

• compression stress • use gross area unless opening exists in middle portion of length •

fc 

P A



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Compression Members – Axial Compression • length, l, is distance between lateral supports • length can be different between two directions of buckling • brace at mid‐height restricts buckling about short dimension of cross‐section



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Compression Members – Effective Length effective length accounts for column end connections • found by multiplying length, l, by effective length factor, K • uses design values for K (not theoretical) • can be different for two directions of buckling • used in buckling calculations



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Compression Members – Effective Length

when end connections are not clearly known, usually assume pin connections



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Compression Members – Slenderness Limits • slenderness ratio: ratio of effective length and depth, d, for each of the two directions of buckling Kl / d

d may be equal to dx or dy. • NDS does not allow compression members with slenderness ratio greater than 50 (longest length allowed for a 2x truss member is 1.5 in × 50 = 75 in ) • rare for a post to be smaller than 4 in × 4 in



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Column Stability Factor • column stability factor, CP, accounts for • FcE = critical buckling design value buckling of cross‐section 0.822 E ' FcE 

• c = 0.8 for sawn lumber • c = 0.9 for glulam • F’ = ratio of FcE and Fc*

 le    d 

min 2

• le = effective length of compression member • Fc* = reference compression design value, Fc, multiplied by all applicable adjustment factors except CP Fc*  FcCDCM CT CiCF



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Example: Compression Strength Check Example 5.7



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Example: Compression Strength Check Example 5.7



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Example: Compression Strength Check



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Tension Members – Axial Tension Only • applied tension stress ft 

T Anet

• net area, Anet, must account for loss of wood due from bolt holes or openings in cross‐section • NDS only allows tension stress parallel to the grain of the wood (do not load member perpendicular to grain).



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Combined Axial and Bending Moments biaxial bending moment • tension combined with bending moment • compression combined with bending moment • NDS procedure is to determine applied stress for each type of loading independently.



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Combined Axial Tension and Bending tension combined with bending moment • check edge of cross‐section with largest • check other edge for potential buckling total tension stress • Fb** = allowable bending stress (not including volume factor) • Fb* = allowable bending stress (not including beam stability factor)



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Example: Axial Tension and Bending Example 5.9



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Example: Axial Tension and Bending Example 5.9



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Example: Axial Tension and Bending



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Example: Axial Tension and Bending



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Combined Axial Compression and Flexure combined axial compression and flexure

• fb2 = bending force where biaxial bending occurs • Cm1 = moment magnification factor for biaxial bending and axial compression

• bending magnifies potential instability from compression force • determine applied stresses and allowable design values for each type of stress independently • fb1 = bending force where bending occurs about only one axis

Cm1  1.0  f c / FcE1

• Cm2 = moment magnification factor for biaxial bending and axial compression Cm 2  1.0  f c / FcE 2   f b1 / FbE 

• factors can also be determined from NDS



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Example: Axial Compression and Flexure A short post is subject to an axial wind load and a compressive dead load, as shown. The post is made from select structural 4×10 Douglas fir‐larch members (actual dimensions: 3.5 in × 9.25 in) and is fixed at the base and free at the top and. The post is used such that all adjustment factors except the load duration factor and column stability factor are 1.2. Does the post satisfy the NDS interaction equations for combined axial compression and flexure?



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Example: Axial Compression and Flexure A short post is subject to an axial wind load and a compressive dead load, as shown. The post is made from select structural 4×10 Douglas fir‐larch members (actual dimensions: 3.5 in × 9.25 in) and is fixed at the base and free at the top and. The post is used such that all adjustment factors except the load duration factor and column stability factor are 1.0. Does the post satisfy the NDS interaction equations for combined axial compression and flexure?

Determine the applied stresses. fc 

P 20,000 lbf  A  3.5 in  9.25 in 

 618

25,000

lbf in

M  S 1    3.5 in  3.5 in  9.25 in  6 lbf  1323 2 in

fb 



lbf in 2

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Example: Axial Compression and Flexure The reference design values for select structural Douglas fir‐larch members are Fb  1500 lbf/ft 2 Fc  1700 lbf/ft 2

Find the compressive design value of the member. The load duration factor for a dead load plus wind load loading condition is 1.60. Fc*  FcCDCM

Emin  690,000 lbf/ft 2

Find the effective length. The post is a cantilever from the ground, so K = 2.1 in.

lbf     1700 2  1.60 1.0  in    2720 lbf/in 2

le  Kl   2.1 in  30 in   63 in



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Example: Axial Compression and Flexure Find the critical buckling design value. The shortest dimension of the cross‐ section will buckle, so d = 3.5 in. FcE 



' min 2

0.822 E  le    d 

The ratio of the compressive design value and the critical buckling design value is F' 

FcE Fc*

lbf in 2  lbf 2720 2 in  0.644 1751

 0.822   690,000  2  63 in     3.5 in 

lbf   in 2 

 1751 lbf/in 2



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Example: Axial Compression and Flexure Find the column stability factor. Select structural Douglas fir‐larch is a sawn lumber, so c = 0.8. 2

 1.0  F '  F ' 1.0  F ' CP      c 2c  2c  2

 1.0  0.644  0.644 1.0  0.644      2 0.8 0.8  2  0.8       0.526

The allowable compression design value is F  Fc*CP lbf     2720 2   0.526  in    1431 lbf/in 2



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Example: Axial Compression and Flexure Find the moment magnification factor. Cm1  1.0 

fc FcE1

lbf in 2  0.647  1.0  lbf 1751 2 in 618

Check the interaction equation. There is no biaxial bending, so disregard the biaxial bending ratio. 2

 fc  fb1 f  ' b 2  1.00  '  '  Fc  Fb1Cm1 Fb 2Cm 2 2

 fc  fb1  '  '  Fc  Fb1Cm1

lbf   617 in 2  lbf  1431 2 in 

2

lbf  1323 2  in  1.04   lbf    1.6  1500 2   0.647   in   STRC ©2015 Professional Publications, Inc.


  1.00, not acceptable 71

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Design of Connections • types of connectors: bolts, lag screws, nails, screws, shear plates, split rings • failure modes: shear and withdrawal • find reference design values and apply relevant adjustment factors



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Design of Connections Table 5.9 Adjustment Factors for Connections



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Connection Adjustment Factors group action factor, Cg when multiple connectors resist load



Fig. 5.6 Staggered Fasteners

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Connection Adjustment Factors geometry factor, CΔ when connectors closely spaced or near edge of material penetration depth factor, Cd connector not as deep into wood as required for full strength end grain factor, Ceg connector installed parallel to wood grain metal side plate factor, Cst when metal cover plates used on connection STRC ©2015 Professional Publications, Inc.


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Bolted Connection Capacity Bolt strength varies depending on orientation of load to grain of wood. When one member is thinner than the other, the thin member is called a side member. NDS Table 11A

Reproduced from National Design Specification for Wood Construction ASD/LRFD, 2012 ed. American Forest & Paper Association, Washington, DC.



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Bolted Connection Capacity • tabulated strengths from NDS for loading parallel to grain and perpendicular to grain • Use Hankinson formula for other directions. NDS Eq. 11.4‐1

• Z = allowable design value for lag screw with load applied at angle α to wood surface • Z′ = adjusted lateral design value for single fastener connection • W′ = adjusted withdrawal design value for fastener • p = depth of fastener penetration into wood member



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Example: Direction of Loading on Bolt Bolts in a wood structure are subjected to a load, T, as shown. In what direction is the load applied to the bolts? (A) only perpendicular to the grain (B) only parallel to the grain (C) both parallel and perpendicular to the grain (D) Wood grain direction does not apply to glulam beams.



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Example: Direction of Loading on Bolt Bolts in a wood structure are subjected to a load, T, as shown. In what direction is the load applied to the bolts? (A) only perpendicular to the grain (B) only parallel to the grain (C) both parallel and perpendicular to the grain

The grain orientation always runs along the long dimension of the wood, so in this case it runs from one end of the glulam member to the other. The force is in the same direction, so the load is applied parallel to the grain. The answer is (B).

(D) Wood grain direction does not apply to glulam beams.



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Example: Bolt Connection Example 5.10



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Example: Bolt Connection Example 5.10



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Example: Bolt Connection



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Split Ring and Shear Plate Connectors • bolt strength limited by strength of wood bolt bears against • when large forces transferred between tension members, installation of metal ring into faying surface between pieces greatly increases shear strength



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Example: Split Ring Connection Example 5.13



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Example: Split Ring Connection Example 5.13



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Example: Split Ring Connection



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Nail Strength • manufactured from high‐strength steel wire • nail made up of head and shank (part of the nail that is not the head) • all nail measurements based upon shank diameter. • reference design values taken from NDS Tables • for full strength, nails must penetrate the second piece of lumber to a distance at least 10 times shank diameter • nails with shank diameter less than 0.25 in considered small dowel connectors; grain direction not considered when determining strength of small dowel connectors • nails often toe‐nailed into pieces, so adjustment factors used to reduce strength of nail STRC ©2015 Professional Publications, Inc.


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Nail Withdrawal endgrain withdrawal nail pulling out from end of member parallel to grain of lower piece; unacceptable sidegrain withdrawal nail pulling out from end of member perpendicular to grain of lower piece; acceptable



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Example: Nail Connection Example 5.15



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Example: Nail Connection Example 5.15



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Nail Connection Example



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Summary • applied stresses determined from fundamental stress mechanics • reference design values for wood and most connectors provided by NDS • reference design values adjusted for various applications using adjustment factors • load duration factor – applied based on shortest duration load of combination • wet service factor – reduced strength when moisture content is high • size factor – accounts for reduced strength due to defects in large beams • stability factors – account for buckling of beams or columns • adjusted reference design value called allowable design value



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Summary • determine applied stress and allowable design value for each type of stress (bending stress, shear stress, compression stress, tension stress) • combine individual stresses for members with complex stress patterns as necessary • determine connector strength for individual connector (shear strength, withdrawal strength) • combine connectors into connection assemble



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Learning Objectives You have learned • how to determine strength of lumber components using the National Design Standard • how to determine allowable strength of lumber components using the National Design Standard

• how to use allowable stress design method • terminology used in wood and timber construction

• how to calculate strength of a nail or bolt for wood construction



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Overview • adjustment factors • design for flexure • design for shear • design for compression • design for tension • design of connections



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STRC09 Timber Beams Columns Connections 0716

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