# Timber Column Design per the CSA O86-19 Standard

### Technical Article

Using the RF-TIMBER CSA module, timber column design is possible according to the CSA O86-19 standard. Accurately calculating timber member compressive resistance and adjustment factors is important for safety considerations and design. The following article will verify the factored compressive resistance in the RFEM add-on module RF-TIMBER CSA using step-by-step analytical equations per the CSA O86-19 standard including the column modification factors, factored compressive resistance, and final design ratio.

#### Timber Column Analysis

A simply supported 10 ft. long, nominal 89 mm ⋅ 89 mm Douglas Fir-Larch Structural (DF-L SS) column with an axial load of 5.00 kips will be designed. The goal of this analysis is to determine the adjusted compressive factors and compressive resistance of the column. A standard term load duration is assumed. Loading criteria are simplified for this example. Typical load combinations can be referenced in Sect. 5.2.4 [1]. In Figure 01, a diagram of the simple column with loads and dimensions is shown.

Image 01 - Knowledge Base Article | Timber CSA Column and Cross-section Dimensions

#### Column Properties

The cross-section used in this example is a 89 mm ⋅ 89 mm nominal dimension lumber. The actual cross-section property calculations of the sawn timber column can be viewed below:

b = 3.50 in., d = 3.50 in., L = 10 ft.

Gross cross-section area:

A_{g} = b ⋅ d = (3.50 in.) ⋅ (3.50 in.) = 12.25 in.²

Section modulus:

${\mathrm{S}}_{\mathrm{x}}=\frac{\mathrm{b}\xb7{\mathrm{d}}^{2}}{6}=\frac{(3.50\mathrm{in}.)\xb7{(3.50\mathrm{in}.)}^{2}}{6}=7.15\mathrm{in}{.}^{3}$

Moment of inertia:

${\mathrm{I}}_{\mathrm{x}}=\frac{\mathrm{b}\xb7{\mathrm{d}}^{3}}{12}=\frac{(3.50\mathrm{in}.)\xb7{(3.50\mathrm{in}.)}^{3}}{12}=12.50\mathrm{in}{.}^{4}$

The material that will be used for this example is DF-L SS. The material properties are as follows.

Reference compressive design value:

f_{c} = 2,001.52 psi

Modulus of elasticity:

E = 1,740,450.00 psi

#### Column Modification Factors

For the design of timber members per the CSA O86 - 19 standard, modification factors must be applied to the reference compressive design value (f_{c}). This will ultimately provide the adjusted compressive design value (F_{c}).

F_{c} = f_{c} ⋅ (K_{D} ⋅ K_{H} ⋅ K_{sc} ⋅ K_{T})

Below, each modification factor is further explained and determined for this example.

K_{D} - The load duration factor accounts for different load periods. Snow, wind, and earthquakes loads are considered with K_{D}. This means K_{D} is dependent on the load case. In this case, K_{D} is set to 0.65 per Table 5.3.2.2 [1] assuming a long-term load duration.

K_{SE} - The wet service factor considers dry or wet service conditions on sawn lumber as well as cross-section dimensions. For this example, we are assuming compression at the extreme fiber and wet service conditions. Based on Table 6.4.2 [1] K_{s} is equal to 0.84.

K_{T} - The treatment adjustment factor considers wood that has been treated with fire-retardant or other strength-reducing chemicals. This factor is determined from strength and stiffness capacities based on documented time, temperature, and moisture test. For this factor, Sect. 6.4.3 [1] is referenced. For this example, 0.95 is multiplied by the modulus of elasticity and 0.85 for all other properties when assuming wet service conditions.

K_{Zc} - The size factor considers varying sizes of lumber and how the loading is applied to the column. More info on this factor can be found in Sect. 6.4.5 [1]. For this example, K_{Z} is equal to 1.30 based on dimensions, compression and shear, and Table 6.4.5 [1].

K_{H} - The system factor takes into account sawn lumber members that consists of three or more essentially parallel members. These members cannot be spaced more than 610 mm apart and mutually support the load. This criteria is defined as case 1 in Sect. 6.4.4 [1]. For this example K_{H} is equal to 1.10 using Table 6.4.4 because we assume it as a compression member and case 1.

K_{L} - The lateral stability factor considers lateral supports provided along the member length which help prevent lateral displacement and rotation. The lateral stability factor (K_{L}) is calculated below.

K_{sc} - The specified strength of lumber shall be multiplied by a service-condition factor (K_{sc}). This factor is determined with reference to Table 6.10 [1].

#### Factored Specified Strength in Compression (F_{C})

The factored specified strength in compression (F_{c}) is determined in the section below. F_{c} is calculated by multiplying the specified strength for compression (f_{c}) by the following modification factors.

K_{D} = 1.00

K_{H} = 1.00

K_{SE} = 1.00

K_{T} = 1.00

We can now calculate F_{c} by using the following equation from Sect. 6.5.4.1 [1].

F_{c} = f_{c} ⋅ (K_{D} ⋅ K_{H} ⋅ K_{s} ⋅ K_{T})

F_{c} = 2001.52 psi

#### Lateral Stability Factor, K_{C}

The slenderness factor (K_{C}) is calculated from Sect. 6.5.5.2.5 [1]. Before K_{C} can be calculated, the factored modulus of elasticity for design of compression members (E_{05}) must be calculated. First, the size factor for compression for sawn lumber and for CLT (K_{Zc}) must be calculated with reference to Sect. 6.5.5.2.4 [1].

K_{Zc} = 6.3 ⋅ (d ⋅ L)^{-0.13}

K_{Zc} = 1.24

Then, the slenderness ratio for compression members (C_{c}) must be calculated based on Sect. 6.5.5.2.2 [1].

${\mathrm{C}}_{\mathrm{B}}=\sqrt{\frac{{\mathrm{L}}_{\mathrm{e}}\xb7\mathrm{d}}{{\mathrm{b}}^{2}}}$

C_{c} = 34.29

Next, the factored modulus of elasticity for compression members (E_{05}) needs to be determined based on Table 6.7 [1].

E_{05} = 8,000 MPa = 1,160,302 psi

Now that all of the variables required are calculated and determined, K_{C} can be calculated.

${\mathrm{C}}_{\mathrm{k}}=\sqrt{\frac{0.97\xb7\mathrm{E}\xb7{\mathrm{K}}_{\mathrm{SE}}\xb7{\mathrm{K}}_{\mathrm{T}}}{{\mathrm{F}}_{\mathrm{b}}}}$

K_{C} = 0.288

#### Column Design Ratio

The ultimate goal of this example is to obtain the design ratio for this simple column. This will determine if the member size is adequate under the given load or if it should be further optimized. Calculating the design ratio requires the factored compressive resistance parallel to grain (P_{r}) and factored axial load in compression (P_{f}).

The maximum axial compression load (P_{f}) applied is equal to 5.00 kips.

Next, the factored compressive resistance (P_{r}) can be calculated from Sect. 6.5.4.1 [1].

P_{r} = Φ ⋅ F_{C} ⋅ A ⋅ K_{Zc} ⋅ K_{C}

P_{r} = 7.00 kips

Finally, the design ratio (η) can now be calculated.

${\mathrm{K}}_{\mathrm{L}}=1-\frac{1}{3}\xb7{\left(\frac{{\mathrm{C}}_{\mathrm{B}}}{{\mathrm{C}}_{\mathrm{K}}}\right)}^{4}$

#### Application in RFEM

For timber design per the CSA O86-19 [1] standard in RFEM, the add-on module RF-TIMBER CSA analyzes and optimizes cross-sections based on loading criteria and member capacity for a single member or set of members. When modeling and designing the column example above in RF-TIMBER CSA, the results can be compared.

Image 02 - RFEM Model | CSA Timber Column Design

In the General Data table of the RF-TIMBER CSA add-on module, the member, loading conditions, and design methods are selected. The material and cross-sections are defined from RFEM and the load duration is set to standard term. The moisture service condition is set to dry and treatment is set to none or preservative (not incised). The slenderness factor K_{C} is calculated based on Sect. 6.5.5.2.5 [1]. The module calculations produce a factored axial load in compression (P_{f}) of 5.00 kip. and a factored compressive resistance parallel to grain (P_{r}) of 7.05 kips. A design ratio (η) of 0.71 is determined from these values aligning well with the analytical hand calculations shown above.

#### Author

#### Alex Bacon, EIT

Technical Support Engineer

Alex is responsible for customer training, technical support, and continued program development for the North American market.

#### Keywords

Timber Column Design CSA O86-19 Add-on Modules

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