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Contents | 1. Introduction to structural design | 2. Loads |

Introduction to wood | Material properties | Sectional properties | Design approaches | Construction systems | Tension elements | Columns |

Wood beams are generally designed for bending stress and then checked for shear and deflection.
Using allowable stress design, the required section modulus is found by dividing the maximum bending
moment by the adjusted allowable bending stress, *F _{b}*' , as shown in Equation 1.24. This adjusted value is found by multiplying the tabular value,

The flat use factor, *C _{fu}*, accounts for the apparent increase in bending strength when beams are stressed about their weak axes. The repetitive member factor,

For glued-laminated (glulam) beams only, the size factor is replaced by a "volume" factor, *C _{V}*. Like the size factor, the volume factor is designed to account for the increased probability of brittle tensile failure in larger structural elements. Because the beam stability factor,

Because some adjustment factors cannot be determined until the cross-sectional dimensions of
the beam are known, the design process may become an iterative one, based on the analysis of trial
sections. In this process, tabular values of allowable bending stress and modulus of elasticity are
found in Appendix Tables A-3.5 and A-3.9; values for allowable shear stress, *F _{v}*, are found in Appendix Table A-3.7. Shear stress is only adjusted for duration of load and wet service conditions (Appendix Table A-3.8). When computing deflections, the only adjustment to modulus of elasticity,

In the examples that follow, the maximum shear force, *V*, could have been reduced by considering
the value at a distance, *d*, from the face of the supports, as illustrated in Figure 1.67. Where shear does not appear to be a critical factor in the design of the beam, this reduction is usually unnecessary; however, if shear appears to govern the beam design, it may be beneficial to use the reduced
value of *V* in the calculation of actual shear stress.

** Problem definition.** Can a 2 × 8 Hem-Fir No. 2 joist, spaced 16 in. on center, be used in a residential application, spanning 12 ft? Assume a dead load corresponding to that listed in Appendix Table A-2.1 for wood floor systems with 2 × 10 joists.

** Solution overview.** Find loads; check bending stress (or required section modulus); check shear stress (or required cross-sectional area); check deflection.

*Problem solution*

**1.** Find loads:

a. From Appendix Table A-2.2, the live load, *L* = 40 psf; the live load distributed on 1 linear foot of the joist is *L* = 40(16/12) = 53.33 lb/ft. Live load reduction does not apply since *A _{T}*

b. From Appendix Table A-2.1, the dead load, *D* = 10.5 psf; the dead load distributed on 1 linear foot of the joist is *D* = 10.5(16/12) = 14 lb/ft.

c. The total distributed load, *w* = 53.33 + 14.0 = 67.33 lb/ft.

**2.** Create load, shear and moment diagrams as shown in Figure 3.18 to determine critical (i.e., maximum) shear force and bending moment.

Figure 3.18: Load, shear, and moment diagrams for Example 3.5

**3.** Find adjusted allowable bending stress:

a. From Appendix Table A-3.5, find the tabular allowable bending stress: *F _{b}* = 850 psi.

b. From Appendix Table A-3.6, find all relevant adjustments: *C _{F}* = 1.2;

c. Multiply the tabular stress value by the adjustments to get the adjusted allowable stress: *F _{b}*' = 850(1.2)(1.15) = 1173 psi.

**4.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**5.** From Appendix Table A-3.12, check the actual section modulus for a 2 × 8, bent about its strong (*x*) axis: *S _{x}* = 13.14 in

**6.** Find adjusted allowable shear stress:

a. From Appendix Table A-3.7, the tabular allowable shear stress, *F _{v}* = 150 psi.

b. From Appendix Table A-3.8, find all relevant adjustments: *C _{M}* = 1.0;

c. Multiply the tabular stress value by the adjustments to get the adjusted allowable stress: *F _{v}*' = 150(1.0)(1.0) = 150 psi.

**7.** From Equation 1.29, compute the required area, *A _{req}* = 1.5

**8.** From Appendix Table A-3.12, check the actual area of the cross section: *A _{act}* = 10.88 in

**9.** From Appendix Table A-1.3, find the allowable total-load deflection for a floor joist: Δ^{T}* _{allow}* = span/240 = (12 × 12)/240 = 0.6 in.; and the allowable live-load deflection for a floor joist: Δ

**10.** Using Appendix Table A-3.15, check the actual total-load deflection. Δ^{T}* _{act}* =

*C* = 22.46.

*L* = 12 × 12 = 144 in. (We are using the same symbol, *L*, for span and "live load"; the meaning should be clear from the context).

*P* = *w*(*L*/12) = 67.33(144/12) = 808 lb.

*E* = 1,300,000 psi (Appendix Table A-3.9).

*I _{x}* = 47.63 in

Δ^{T}* _{act}* = 22.46(808)(144/12)3/(1,300,000 × 47.63) = 0.5 in.

Since Δ^{T}* _{act}* = 0.5 in. ≤ Δ

**11.** Using Appendix Table A-3.15, check the actual live-load deflection. Δ^{L}* _{act}* =

*C* = 22.46.

*L* = 12 × 12 = 144 in. (We are using the same symbol, *L*, for span and "live load"; the meaning should be clear from context).

*P* = *w*(*L*/12) = 53.33(144/12) = 640 lb (Use live load only!).

*E* = 1,300,000 psi (Appendix Table A-3.9).

*I _{x}* = 47.63 in

Δ^{L}* _{act}* = 22.46(640)(144/12)

Since Δ^{L}* _{act}* = 0.4 in. ≤ Δ

**12.** Conclusion: The 2 × 8 is OK for bending, shear and deflection. Therefore it is acceptable.

** Problem definition.** Can a 14 × 20 Hem-Fir No. 2 girder be used in a "heavy timber" office building application, as shown in Figure 3.19? Assume that only the beams framing into the girder provide lateral bracing at the third points. Assume a total dead load of 18 psf and a live load corresponding to office occupancy.

Figure 3.19: Framing plan and view of girder for Example 3.6

** Solution overview.** Find loads; check bending stress (or required section modulus); check shear stress (or required cross-sectional area); check deflection.

*Problem solution*

**1.** Find loads:

a. From Appendix Table A-2.2, the live load for office occupancy, *L* = 50 psf; with live load reduction, we get: *L* = 50(0.25 + 15/) = 50(0.935) = 46.7 psf.

b. The dead load, *D* = 18 psf (given).

c. A total concentrated load, *PL*, acts on a tributary area of 10 × 8 = 80 ft^{2}, so *P* = (*D* + *L*)(80) = (18 + 46.7)(80) = 5176 lb.

**2.** Create load, shear and moment diagrams as shown in Figure 3.20 to determine critical (i.e., maximum) shear force and bending moment.

Figure 3.20: Load, shear, and moment diagrams for
Example 3.6

**3.** From Appendix Table A-3.5, the tabular value is *F _{b}* = 675 psi.

**4.** Find the adjustments to the allowable bending stress:

a. From Appendix Table A-3.6: *C _{F}* = (12/19.5)

b. From Appendix Table A-3.6: *C _{r}* = 1.0.

c. From Appendix Table A-3.6: *C _{M}* = 1.0.

d. From Appendix Table A-3.6: *C _{D}* = 1.0.

e. In addition, the beam stability factor must be computed:

*C _{L}* = A – where:

*l _{e}* = 1.68

*E*'_{min} = 400,000 psi (from Appendix Table A-3.9)

*b* = 13.5 in.; *d* = 19.5 in. ("green" dimensions of a 14 × 20 from Appendix Table A-3.12)

*F _{b}** =

*F _{bE}* = 1.20(13.5

*A* = (1 + 27,864/640)/1.9 = 23.44

*B* = (27,864/640)/0.95 = 45.83

f. *C _{L}* =

**5.** The adjusted allowable stress, *F _{b}*' =

**6.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**7.** From Appendix Table A-3.12, check the actual section modulus about the strong (*x*) axis: *S _{x}* = 855.6 in

**8.** Find the adjusted allowable shear stress:

a. From Appendix Table A-3.7, the tabular allowable shear stress, *F _{v}* = 140 psi.

b. From Appendix Table A-3.8, find all relevant adjustments: *C _{M}* = 1.0;

c. The adjusted allowable shear stress, *F _{v}*' = 140(1.0)(1.0) = 140 psi.

**9.** From Equation 1.29, compute the required area, *A _{req}* = 1.5

**10.** From Appendix Table A-3.12, check the actual area of a 14 × 20 cross section: *A _{act}* = 263.3 in

**11.** From Appendix Table A-1.3, find the allowable total-load deflection for a floor beam: Δ^{T}* _{allow}* =
span/240 = (24 × 12)/240 = 1.2 in.; and the allowable live-load deflection for a floor joist: Δ

**12.** From Appendix Table A-1.3, check the actual total-load deflection: Δ^{T}* _{act}* =

*C* = 61.34.

*L* = 24 × 12 = 288 in.

*P* = (46.7 + 18)(10 × 8) = 5176 lb.

*E* = 1,100,000 psi (from Appendix Table A-3.9).

*I* = 8342 in^{4} (from Appendix Table A-3.12).

Δ^{T}* _{act}* = 61.34(5176)(288/12)

**13.** From Appendix Table A-3.15, check the actual live-load deflection: Δ^{L}* _{act}* =

*C* = 61.34.

*L* = 24 × 12 = 288 in.

*P* = 46.7(10 × 8) = 3736 lb (Use live load only!)

*E* = 1,100,000 psi (from Appendix Table A-3.9).

*I* = 8342 in^{4} (from Appendix Table A-3.12).

Δ^{L}* _{act}* = 61.34(3736)(288/12)

**14.** Conclusion: The 14 × 20 is OK for bending, shear and deflection. Therefore it is acceptable.

** Problem definition.** Design a 32 ft-long glulam roof girder of stress class 20F-1.5E for the one-story industrial building shown in the framing plan (Figure 3.21). Assume a snow load,

Figure 3.21: Framing plan for Example 3.7

** Solution overview.** Find loads; begin iterative design process by assuming unknown adjustments to allowable stresses; then check bending stress (required section modulus), shear stress (required cross-sectional area) and deflection, as in analysis examples. Recompute if necessary with bigger (or smaller) cross section until bending, shear and deflection are OK.

*Problem solution*

**1.** Find loads:

*S* = 30 psf (given).

*D* = 20 psf (given).

From Appendix Table A-2.7, it can be seen by examining the various load combinations that the most severe effects occur with the combination: dead load plus snow load, or *D* + *S*.

Using *D* + *S*, the total concentrated load, *P*, acts on tributary area of 28 × 8 = 224 ft^{2}, so: *P* = (*D* + *S*) × (tributary area) = (20 + 30)(224) = 11,200 lb.

**2.** Create load, shear and moment diagrams as shown in Figure 3.22 to determine critical (i.e., maximum) shear force and bending moment.

Figure 3.22: Load, shear, and moment diagrams for Example 3.7

**3.** Find provisional adjusted allowable bending stress:

a. From Appendix Table A-3.5 part *D*, the design (tabular) value for bending is: *F _{b}* = 2000 psi.

b. From Appendix Table A-3.6, the relevant adjustments are as follows: *C _{r}* = 1.0;

c. The adjusted value for allowable bending stress, *F _{b}*' = 2000(1.15)(0.9) = 2070 psi.

**4.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**5.** Compute the required depth, *d*, based on the section modulus for a rectangular cross section, *S* = *bd*^{2}/6 = 1039 in^{3} and *b* = 8.75 in. (given). In this case, 8.75*d*^{2}/6 = 1039, from which *d* = 26.7 in. Rounding up to the first multiple of 1.5 in. (the depth of an individual lamination), we get: *d* = 27 in.

*Trial 1: 8-3/4 in. × 27 in. cross section*

**1.** Find allowable bending stress: as before, *F _{b}* = 2000 psi.

**2.** Find adjustments to allowable bending stress (Appendix Table A-3.6):

*C _{r}* = 1.0.

*C _{M}* = 1.0.

*C _{D}* = 1.15.

*C _{L}* or

*C _{V}* = (21/32)

*C _{L}* =

*l _{e}* = 1.54

*E*'_{min} = 780,000 psi, from Appendix Table A-3.9 parts *B* and *C*.

*b* = 8.75 in.; *d* = 27 in.

*F _{b}** =

*F _{bE}* = 1.20(8.75

*A* = (1 + 17,934/2300)/1.9 = 4.63.

*B* = (17,934/2300)/0.95 = 8.21.

*C _{L}* =

Since *C _{V}* = 0.84 <

**3.** The adjusted design value for bending is *F _{b}*' =

**4.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**5.** Check that actual section modulus is greater or equal to required section modulus: actual *S _{x}* =

*Trial 2: 8-3/4 in. × 28-1/2 in. cross section (Figure 3.23)*

Figure 3.23: Glued laminated cross section for trial 2, Example 3.7

**1.** Find allowable bending stress: as before, *F _{b}* = 2000 psi.

**2.** Find adjustments to allowable bending stress (Appendix Table A-3.6):

*C _{r}* = 1.0.

*C _{M}* = 1.0.

*C _{D}* = 1.15.

*C _{L}* or

*C _{V}* = (21/32)

*C _{L}* = A – where:

*l _{e}* = 1.54

*E*'_{min} = 780,000 psi, from Appendix Table A-3.9 parts *B* and *C*.

*b* = 8.75 in.; *d* = 28.5 in.

*F _{b}** =

*F _{b}*

*A* = (1 + 16,990/2300)/1.9 = 4.41.

*B* = (16,990/2300)/0.95 = 7.78.

*C _{L}* = A – = 4.41 – = 0.99.

Since *C _{V}* = 0.83 <

**3.** The adjusted design value for bending is *F _{b}*' =

**4.** From Equation 1.24, compute the required section modulus: *S _{req}* = M/

**5.** Check that actual section modulus is greater or equal to required section modulus: actual *S _{x}* =

**6.** Find adjusted allowable shear stress:

a. From Appendix Table A-3.7 part *C*, the design value for shear, *F _{v}* = 195 psi.

b. From Appendix Table A-3.8, the relevant adjustments are as follows:
*C _{M}* = 1.0;

c. The adjusted allowable stress for shear, *F _{v}*' = 195(1.15) = 224.25 psi.

**7.** Based on Equation 1.29, the required cross-sectional area to resist shear, *A _{req}* = 1.5V/

**8.** Check actual cross-sectional area = 8.75 × 28.5 = 249.4 in^{2}; since the actual area, *A _{act}* = 249.4 in

**9.** From Appendix Table A-1.3, the allowable total load deflection for a roof with no ceiling, Δ^{T}* _{allow}* =
span/120 = 32(12)/120 = 3.20 in.; and the allowable live load (actually snow load in this case)
deflection for a roof with no ceiling, Δ

**10.** From Appendix Table A-3.15, the actual total load deflection is Δ^{T}* _{act}* =

*C* = 85.54.

*P* = (S + D)(tributary area) = (30 + 20)(28 × 8) = 11,200 lb.

*L* = 32 × 12 = 384 in.

*E*' = 1,500,000 psi, from Appendix Table A-3.9 parts A and C. The "average" adjusted modulus of elasticity, *E*', is used for deflection calculations, whereas the adjusted minimum modulus of elasticity, *E*'_{min}, is used in buckling or stability calculations.

*I* = *bd*^{3}/12 = (8.75)(28.53)/12 = 16,879.6 in^{4} (Equation 1.8).

Δ^{T}* _{act}* = 85.54(11,200)(384/12)

**11.** From Appendix Table A-3.15, the actual live (snow) load deflection is Δ_{act} = *CP*(*L*/12)^{3}/(*EI*) where:

*C* = 85.54.

*P* = (S)(tributary area) = (30)(28 × 8) = 6720 lb (Use snow load only!).

*L* = 32 × 12 = 384 in.

*E*' = 1,500,000 psi, from Appendix Table A-3.9 parts A and C. The "average" adjusted modulus of elasticity, *E*', is used for deflection calculations, whereas the adjusted minimum modulus of elasticity, *E*'_{min}, is used in buckling or stability calculations.

*I* = *bd*^{3}/12 = (8.75)(28.5^{3})/12 = 16,879.6 in^{4}.

Δ^{L}* _{act}* = 85.54(6720)(384/12)3/(1,500,000 × 16,880) = 0.74 in. Since Δ

**12.** Conclusion: The 8-3/4 in. × 28-1/2 in. section is OK for bending, shear and deflection. Therefore it is acceptable.

** Problem definition.** Design a Douglas Fir-Larch (North) No.1/No.2 girder using 4× lumber to support a residential live load as shown in Figure 3.24. Assume 10.5 psf for dead load. Loads on the girder can be modeled as being uniformly distributed since joists are spaced closely together.

Figure 3.24: Framing plan for Example 3.8

** Solution overview.** Find loads; find known adjustments to allowable bending stress; use Appendix Table A-3.16 to directly compute lightest cross section for bending; check for shear and deflection. Alternatively, begin iterative design process by assuming unknown adjustments to allowable stresses; then check bending stress (required section modulus), shear stress (required cross-sectional area) and deflection, as in analysis examples. Recompute if necessary with bigger (or smaller) cross section until bending, shear and deflection are OK.

*Problem solution*

**1.** Find loads:

a. From Appendix Table A-2.2, the live load for a residential occupancy, *L* = 40 psf.

b. The dead load, *D* = 10.5 psf (given).

c. The total distributed load, *w* = (*D* + *L*)(tributary area) = (10.5 + 40)(6) = 303 lb/ft. Live load reduction does not apply since *K _{LL}* times the tributary area is less than 400 ft

**2.** Create load, shear and moment diagrams as shown in Figure 3.25 to determine critical (i.e., maximum) shear force and bending moment.

Figure 3.25: Load, shear, and moment diagrams for Example 3.8

**3.** Find partially-adjusted allowable bending stress:

a. From Appendix Table A-3.5, the design (tabular) value for bending stress, *F _{b}* = 850 psi.

b. From Appendix Table A-3.6, the following adjustments can be determined: *C _{r}* = 1.0;

c. The adjusted value for bending stress, with all adjustments known except for *C _{F}*, is

**4.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**5.** Rather than doing several "trial" designs, it is possible to find the correct cross section for bending directly, by using a table of combined size factors (*C _{F}*) and section moduli (

**6.** Find adjusted allowable shear stress:

a. From Appendix Table A-3.7, the design (tabular) allowable shear stress *F _{v}* = 180 psi.

b. From Appendix Table A-3.8, there are no adjustments for shear stress; i.e.: *C _{M}* = 1.0;

c. The adjusted value for allowable shear stress, *F _{v}*' = 180 psi.

**7.** Based on Equation 1.29, the required cross-sectional area to resist shear, *A _{req}* = 1.5

**8.** From Appendix Table A-3.12, we can check the actual area of the cross section, *A _{act}* = 25.38 in

**9.** From Appendix Table A-1.3, find the allowable total-load deflection for a floor beam: Δ^{T}* _{allow}* = span/240 = (7.5 × 12)/240 = 0.375 in.; and the allowable live-load deflection for a floor joist: Δ

**10.** From Appendix Table A-3.15, we can check the actual total-load deflection: Δ^{T}* _{act}* =

*C* = 22.46.

*L* = 7.5 × 12 = 90 in.

*P* = w(L/12) = (40 + 10.5)(6)(90/12) = 2272.5 lb.

*E* = *E*' = 1,600,000 psi (from Appendix Table A-3.9).

*I* = 111.1 in^{4} (directly from Appendix Table A-3.12, or from the equation, *I* = *bd*^{3}/12).

Δ^{T}* _{act}* = 22.46(2272.5)(90/12)

**11.** From Appendix Table A-3.15, we can check the actual live-load deflection: Δ^{L}* _{act}* =

*C* = 22.46.

*L* = 7.5 × 12 = 90 in.

*P* = *w*(L/12) = (40 × 6)(90/12) = 1800 lb (Use live load only!).

*E* = *E*' = 1,600,000 psi (from Appendix Table A-3.9).

*I* = 111.1 in^{4} (directly from Appendix Table A-3.12, or from the equation, *I* = *bd*^{3}/12).

Δ^{L}* _{act}* = 22.46(1800)(90/12)

**12.** Conclusion: The 4 × 8 section is OK for bending, shear and deflection. Therefore it is acceptable.

It is also possible to find the lightest 4× section using an iterative design process without Appendix Table A-3.16. Using this method, the size factor, *C _{F}*, would need to be assumed, and then checked after a provisional cross section is found, as follows:

**1.** Assuming a size factor, *C _{F}* = 1.0, the adjusted allowable bending stress becomes

**2.** From Appendix Table A-3.12, we provisionally select a 4 ×10 with actual *S _{x}* = 32.38 in

*Trial 1: 4 × 10 cross section*

**1.** Find actual adjusted allowable bending stress: The design (tabular) value remains *F _{b}* = 850 psi; the actual size factor for a 4 × 10 is

*Trial 2: 4 × 8 cross section*

**1.** Find actual adjusted allowable bending stress for the 4 × 8: The size factor, *C _{F}* = 1.3, so the adjusted allowable bending stress,

**2.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**3.** From Appendix Table A-3.12, the actual section modulus for a 4 × 8, *S _{x}* = 30.66 in

**4.** Shear and deflection for the 4 × 8 are checked as shown above, using the first method, and are both OK.

**5.** Conclusion: The 4 × 8 section is OK for bending, shear and deflection. Therefore it is acceptable. But what about a 4 × 6, with a size factor just as large?

*Trial 3: 4 × 6 cross section*

**1.** Find actual adjusted allowable bending stress for the 4 × 6: The size factor, *C _{F}* = 1.3, so the adjusted allowable bending stress,

**2.** From Equation 1.24, compute the required section modulus: *S _{req}* =

**3.** From Appendix Table A-3.12, the actual section modulus for a 4 × 6, *S _{x}* = 17.65 in

**4.** Conclusion: Since the 4 × 6 is not OK, select the 4 × 8 section.

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.