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Timber Design

Design of sawn timber beams or joists.

Design requirement

  1. Maximum bending stress, fb must not exceed allowable stress parallel to grain,

F’b = Fb*CD*CM*Ct*CF*CV*Cfu*Cr*Cc*Cf

Where

Fb is allowable bending stress in NDS supplement.

CD is load duration factor, (see NDS Table 2.3.2 reproduced below)

CM is wet service factor, (use when moisture of timber is higher than 19%)

Ct is temperature factor, (when timber is used in temperature higher than 150°F)

CL is beam stability factor, (See below)

CF is size factor, (apply only to visually graded sawn lumber members, and to round timber bending members, not apply simultaneously with Cv for glued laminated timber)

CV is volume factor, (apply only to glued laminated timber bending member)

Cfu is flat use factor, (when 2”-4” timber is loaded at wide face)

Cr is repetitive member factor, (apply to dimension bending member 2”-4” thick)

Cc is curvature factor (apply to curved glued laminated bending member)

Cf is form factor. (for round or diamond section)

  1. Deflection should not exceed allowable limit. The elastic modulus shall be calculated as E’=E*CM*Ct, Where E is modulus of elasticity in NDS supplement

  2. Maximum shear stress, fv shall not exceed allowable shear stress,

F’v = Fv*CD*CM*Ct*CH

Where Fv is allowable shear stress in NDS supplement and,

CH is shear stress factor depends on length of split and shake.  Value of CH varies from 2 for no split to 1 with 1-1/2 split.

Adjustment factors

Load duration factor, CD

Load duration

CD

Design load

Permanent

0.9

Dead load

Ten years

1.0

Occupancy live load

Two months

1.15

Snow load

Seven days

1.25

Construction load

Ten minutes

1.6

Wind/Earthquake load

Impact

2.0

Impact load

Beam stability factor, CL.

CL = 1 for the following condition for member, with nominal depth, B and width, D.

  1. D/B £ 2

  2. 2 < D/B £ 4 – solid blocking is provided at both ends of member.

  3. D/B = 5, one edge (tension or compression) is fully supported.

  4. D/B = 6, bridge, full depth blocking, cross bracing at 8 ft maximum, and both edges are fully supported or compressive edge is fully supported to prevent lateral displacement, and the ends at the point of bearing are laterally supported to prevent rotation;

  5. D/B = 7, both edge fully supported.

When the conditions were not met, CL is calculated based on a complicated equation in NDS section 3.3.3.7.  Normally, it is easier to meet the requirement then to go through the complicated equation.

Size factor, CF

The size, CF, for timber species other than southern pine are listed in Table 4-A.  For southern pine 2” to 4” thick, size factor needs not be applied.  For southern pine 4” thick, 8” and wider, CF = 1.1.  For dimension lumber, wider than 12”, CF = 0.9 except Dense structural 86. 72, and 65. in which, CF =0.9.  When the depth of Dense structural 86, 72, and 65, dimension lumber exceeds 12”, CF =(12/d)1/9.

Repetitive member factor, Cr

Cr applies to dimension lumber 2” to 4” thick that subjected to bending.  Cr =1.15 when members are used as joist, truss chords, rafters, etc and spacing is not exceed 24” and not less than 3”. 

Wet service factor, CM

When the moisture of dimension lumber exceeds 19%, the design value Fb shall be multiplied by CM = 0.85 except that when Fb * CM  £ 1500 psi, CM =1.

Design procedure for Timber Beam and Joist

  1. Calculate design load and moment

  2. Select timber species and cross section. Determine maximum bending stress, f”b=M/S, where M is design moment with load duration factor, S is section modulus.

  3. Determine allowable bending stress, with the rest of multiplication factors

F”b = Fb*CD*CM*Ct*CF*CV*Cfu*Cr*Cc*Cf

  1. Calculate elastic modulus

E’=E*CM*Ct

  1. Calculate deflection of beam with load without load duration factor.

  2. Calculate shear stress, f”v = VQ/Ib or for rectangular member, f”v = 3V/2b

Where, V is shear force with load duration factor, Q is first moment of inertia, I is second moment of inertia, b is width of the member, d is depth of the member.

  1. Calculate allowable shear stress with the rest of multiplication factors

F”v = Fv*CD*CM*Ct*CH

Examples

Example 1: Design of 2x 10 floor joist with southern pine

Design data:

Length of floor joist: L = 16 ft

Spacing of floor joist: s = 16 in.

Top of joist supported by plywood sheathing.

Design load:

Floor live load: WL = 40 psf

Floor dead load: WD = 10 psf

Superimposed dead load including mechanical and electric load, WSD = 8 psf

Timber: Southern pine, moisture less than 19%, used in normal room temperature.

Solution:

Calculate Design load: W = [WD + WSD+ WL]*s = 77.3 lb/ft

Design moment: M = W*L2/8 = 2475 lb-ft

Try 2x10 joist

Nominal dimension, B = 2 in, D = 10 in

Actual dimension, b = 1.5 in, d = 9.25 in

Section modulus: S = 21.39 in3, Modulus of inertia, I = 98.93 in4.

Bending stress: fb =M/S = 1388 psi

Try Southern pine No. 2, Fb = 1500 psi  

Load duration factor for dead load: CD = 0.9

Load duration factors for live load: CD = 1.0 (Use 1 per NDS 2001)

The depth to width ratio based on nominal dimension, D/B = 5

Since compressive edge is fully supported by plywood floor, CL = 1

Repetition factor for joist: Cr = 1.15

Wet service factor: CM = 1

Temperature factor: Ct = 1

Other factors not applicable

Allowable stress, F’b = Fb*CD* CL* Cr* CM* Ct = 1725 psi       O.K.

Check deflection:

Elastic modulus: E = 1600000 psi*CM* Ct = 1600000 psi

Deflection: D = 5*W*L4/(384*E*I) = 0.75 in          < L/240 O.K.

Check shear stress

Maximum shear force. V = W*L/2 = 640 lb

Shear stress, fv = V/bd = 46 psi

Conservatively assume shear stress factor, CH = 1

Allowable shear stress, Fv = 90 psi * CD* CM* Ct *CH = 90 psi    O.K.

Example 2: Design of 3-3x12 beam with Southern pine.

Design data:

Length of beam: L = 16 ft

Tributary width: s = 8 ft

Top of beam supported by floor joists at 16 in O.C.

Design load:

Floor live load: WL = 40 psf

Floor dead load: WD = 10 psf

Superimposed dead load including mechanical and electric load, WSD = 8 psf

Timber: Southern pine, moisture less than 19%, used in normal room temperature.

Solution:

Calculate Design load: W = [(WD + WSD+ WL]*s = 464 lb/ft

Design moment: M = W*L2/8 = 14850 lb-ft

Use 3-3x12 nailed together with 12d nails at 12 in O.C. from both sides staggered.

Nominal dimension, B = 9 in, D = 12 in

Actual dimension, b = 7.5 in, d = 11.25 in

Section modulus: S = 158.2 in3, Modulus of inertia, I = 890 in4.

Bending stress: fb =M/S = 1126 psi

Try Southern pine No. 2, Fb = 1500 psi

The depth to width ratio based on nominal dimension, D/B = 1.33

Since compressive edge is supported by floor joist at 16 in O.C., CL = 1

Wet service factor: CM = 1

Temperature factor: Ct = 1

Load duration factor for dead load: CD = 0.9

Load duration factors for live load: CD = 1.0 (use 1 per NDS 2001)

Other factors not applicable

Allowable stress, F’b = Fb* CD*CL* CM* Ct = 1500 psi       O.K.

Check deflection:

Elastic modulus: E = 1600000 psi*CM* Ct = 1600000 psi

Deflection: D = 5*W*L4/(384*E*I) = 0.48 in          < L/240 O.K.

Check shear stress

Maximum shear force. V = W*L/2 = 3840 lb

Shear stress, fv = V/bd = 46 psi

Conservatively assume shear stress factor, CH = 1

Allowable shear stress, Fv = 90 psi * CD*CM* Ct *CH = 90 psi    O.K.

Example 3: Design of 2x12 floor joist with Douglas Fir-Larch

Design data:

Length of floor joist: L = 16 ft

Spacing of floor joist: s = 16 in.

Top of joist supported by plywood sheathing.

Design load:

Floor live load: WL = 40 psf

Floor dead load: WD = 10 psf

Superimposed dead load including mechanical and electric load, WSD = 8 psf

Timber: Southern pine, moisture less than 19%, used in normal room temperature.

Solution:

Calculate Design load: W = [WD + WSD+ WL]*s = 77.3 lb/ft

Design moment: M = W*L2/8 = 2475 lb-ft

Try 2x12 joist

Nominal dimension, B = 2 in, D = 12 in

Actual dimension, b = 1.5 in, d = 11.25 in

Section modulus: S = 31.64 in3, Modulus of inertia, I = 178 in4.

Bending stress: fb =M/S = 938.5 psi

Try Douglas Fir-Larch No. 1, Fb = 1000 psi

The depth to width ratio based on nominal dimension, D/B = 6

Since compressive edge is fully supported by plywood floor,

Provide solid blocking at both ends, and cross bracing at mid-span,

Maximum spacing = 8 ft, CL = 1

Repetition factor for joist: Cr = 1.15

Wet service factor: CM = 1

Temperature factor: Ct = 1

From NDS Table, size factor, CF = 1

Load duration factor for dead load: CD = 0.9

Load duration factors for live load: CD = 1.0  (Use 1 per NDS 2001)

Other factors not applicable

Allowable stress, F’b = Fb* CD*CL* Cr* CM* Ct*CF = 1150 psi       O.K.

Check deflection:

Elastic modulus: E = 1700000 psi*CM* Ct = 1700000 psi

Deflection: D = 5*W*L4/(384*E*I) = 0.38 in          < L/240 O.K.

Check shear stress

Maximum shear force. V = W*L/2 = 640 lb

Shear stress, fv = V/bd = 38 psi

Conservatively assume shear stress factor, CH = 1

Allowable shear stress, Fv = 95 psi * CD*CM* Ct *CH = 95 psi    O.K.