Design of sawn timber beams or joists.

Design requirement

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

F’_b = F_b*C_D*C_M*C_t*C_F*C_V*C_fu*C_r*C_c*C_f

Where

F_b is allowable bending stress in NDS supplement.

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

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

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

C_L is beam stability factor, (See below)

C_F 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)

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

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

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

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

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

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

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

F’_v = F_v*C_D*C_M*C_t*C_H

Where F_v is allowable shear stress in NDS supplement and,

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

Adjustment factors

Load duration factor, C_D

Beam stability factor, C_L.

C_L = 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, C_L 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, C_F

The size, C_F, 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, C_F = 1.1.  For dimension lumber, wider than 12”, C_F = 0.9 except Dense structural 86. 72, and 65. in which, C_F =0.9.  When the depth of Dense structural 86, 72, and 65, dimension lumber exceeds 12”, C_F =(12/d)1/9.

Repetitive member factor, C_r

C_r applies to dimension lumber 2” to 4” thick that subjected to bending.  C_r =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, C_M

When the moisture of dimension lumber exceeds 19%, the design value Fb shall be multiplied by C_M = 0.85 except that when F_b * C_M  £ 1500 psi, C_M =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 = F_b*C_D*C_M*C_t*C_F*C_V*C_fu*C_r*C_c*C_f

4. Calculate elastic modulus

E’=E*C_M*C_t

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

6. 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.

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

F”_v = F_v*C_D*C_M*C_t*C_H

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: W_L = 40 psf

Floor dead load: W_D = 10 psf

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

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

Solution:

Calculate Design load: W = [W_D + W_SD+ W_L]*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: f_b =M/S = 1388 psi

Try Southern pine No. 2, F_b = 1500 psi  

Load duration factor for dead load: C_D = 0.9

Load duration factors for live load: C_D = 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, C_L = 1

Repetition factor for joist: C_r = 1.15

Wet service factor: C_M = 1

Temperature factor: C_t = 1

Other factors not applicable

Allowable stress, F’_b = F_b*C_D* C_L* C_r* C_M* C_t = 1725 psi       O.K.

Check deflection:

Elastic modulus: E = 1600000 psi*C_M* C_t = 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, f_v = V/bd = 46 psi

Conservatively assume shear stress factor, C_H = 1

Allowable shear stress, F_v = 90 psi * C_D* C_M* C_t *C_H = 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: W_L = 40 psf

Floor dead load: W_D = 10 psf

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

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

Solution:

Calculate Design load: W = [(W_D + W_SD+ W_L]*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: f_b =M/S = 1126 psi

Try Southern pine No. 2, F_b = 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., C_L = 1

Wet service factor: C_M = 1

Temperature factor: C_t = 1

Load duration factor for dead load: C_D = 0.9

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

Other factors not applicable

Allowable stress, F’_b = F_b* C_D*C_L* C_M* C_t = 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, C_H = 1

Allowable shear stress, Fv = 90 psi * C_D*C_M* C_t *C_H = 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: W_L = 40 psf

Floor dead load: W_D = 10 psf

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

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

Solution:

Calculate Design load: W = [W_D + W_SD+ W_L]*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: f_b =M/S = 938.5 psi

Try Douglas Fir-Larch No. 1, F_b = 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, C_L = 1

Repetition factor for joist: C_r = 1.15

Wet service factor: C_M = 1

Temperature factor: C_t = 1

From NDS Table, size factor, C_F = 1

Load duration factor for dead load: C_D = 0.9

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

Other factors not applicable

Allowable stress, F’_b = F_b* C_D*C_L* C_r* C_M* C_t*C_F = 1150 psi       O.K.

Check deflection:

Elastic modulus: E = 1700000 psi*C_M* C_t = 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, C_H = 1

Allowable shear stress, Fv = 95 psi * C_D*C_M* C_t *C_H = 95 psi    O.K.