Designing Glulam Columns
Glulam Column Example
Note that for all variable defintions, tables, or clause references please refer to the Wood Design Manual 2010. ^{[1]}
Question
Design a compression member out of glulam. The species is Spruce-Pine-Jack 12C-E. Assume dry service conditions. The column height is 8m with pinned ends. The column is supported at mid span in the weak axis direction. The unfactored loads are:
- D.L.= 200 KN (Compression)
- L.L.= 100 KN (Compression)
Solution:
The first step is to compute the maximum factored load. This is done by calculating the load combinations and then calculating the duration factors. Note that only the first 3 load combinations will be computed because load cases 4 and 5 will not govern.
Load Combinations from table 4.2.4.1:
- $ P_{f}= 1.4D=1.4(200)= 280 KN $
- $ P_{f}= 1.25D+1.5L=1.25(200)+1.5(100)= 400 KN $
- $ P_{f}= 1.25D+0.5L=1.25(200)+0.5(100)= 300 KN $
- $ P_{f}= 1.25D =1.25(200)= 250 KN $ (this is similar to number 3, but it does not include the companion load)
- For case 1 and 4, $ K_{d}=0.65 $. This value if from table 4.3.2.2.
- For the other cases, the duration factor must be computed manually using the equation in clause 4.3.2.3 because the specified long term load is greater than the specified standard term.
$ K_{d}=1.0-0.50log(P_{L}/P_{S}) $
- $ K_{d}=1.0-0.50log(200/100)=0.85 $
To design the column, we must now select the load that governs for each duration factor. For the duration factor of 0.65, a load of 280 KN will govern. For a duraion factor of 0.85, a load of 400 KN will govern.
The second step is to select a member size from the column selection tables in the handbook. After selecting a size, the next step is to calculate the compresseive resistance for each available duration factor and compare it to the factored loads calculated earlier.
To calculate the compressive resistance we will use clause 6.5.8.4:
$ P_{r}=\phi AK_{Zcg}K_{c} $ where:
- $ F_{c}= f_{c}[K_{D}K_{H}K_{Sc}K_{T}] $
- $ K_{Zcg}=0.68(Z)^{-0.13} \leq1 $
Where:
- Z is the member volume in $ m^{3} $
- $ \phi=0.8 $
- A is the cross-sectional area in $ mm^{2} $
From clause 6.5.8.5, the slenderness factor is:
- $ K_{C}=[1.0+\frac{F_{c}K_{Zcg}C^3_{c}}{35E_{05}K_{SE}K_{T}} ]^{-1} $
Where:
- $ E_{05}=0.87E $
From clause 6.5.8.2, the slenderness ratio is:
- $ C_{c}=\frac{L_{e}}{w}\leq50 $
- $ C_{c}=\frac{L_{e}}{d}\leq50 $
After stating all the necessary equations, we will move on to the calculations. We will start calculating for a duration factor of 0.65 and then 0.85.
From the column tables on page 137 the selected member is Spruce-Pine-Jack 12C-E 265x304. Also, the effective length factorfor pinned end supports is k=1 for the strong and weak axis. This results in an effective length of 8m for the strong axis, and 4m for the weak axis.
- $ C_{c}=\frac{8000}{304}=26.32 $
- $ C_{c}=\frac{4000}{265}=15.1 $
The slenderness ratio in the strong and weak axis is less than 50, so we can use these numbers. Since we are dealing with a glulam member, we can ignore checking the compressive resistance of the smaller slenderness ratio value. So, we can use the governing value of 26.32 in our calculations.
For a duration factor of 0.65:
- $ F_{c}= (25.2)[(0.65)(1)(1)1(1)]= 16.38 Mpa $
where:
- $ f_{c}=25.2Mpa $ from table 6.3
- $ K_{H}=1 $ from clause 6.4.3
- $ K_{Sc}=1 $ from table 6.4.2
- $ K_{T}=1 $ from clause 6.4.4
- $ K_{Zcg}=0.68[(8)(0.265)(0.304)]^{-0.13} \leq1 $
- $ K_{Zcg}=0.72m^{-3}\leq1 $
- $ K_{C}=[1.0+\frac{(16.38)(0.72)(26.32)^3}{35(8439)(1)(1)}} ]^{-1}= 0.58 $
where:
- $ F_{c}=16.38Mpa $
- $ K_{Zcg}=0.72 $
- $ C_{c}=26.32 $
- $ K_{Se}=1 $ from table 6.4.2
- $ K_{T}=1 $ from clause 6.4.4
- $ E=9700 $ from table 6.3
- $ E_{05}=0.87(9700)=8439 Mpa $
After computing all the variables, the total compressive resistance of the member is:
- $ P_{r}=(0.8)(16.38)(265*304)(0.72)(0.58)(10^{-3})=440KN $
- $ P_{r} \geq P_{f} $
- $ 440 \geq 280 $
Since the factored resistance is larger than the factored load, the member is adequate when the duration factor is 0.65.
For a duration factor of 0.85:
- $ F_{c}= (25.2)[(0.85)(1)(1)1(1)]= 21.42Mpa $
where:
- $ f_{c}=25.2Mpa $ from table 6.3
- $ K_{H}=1 $ from clause 6.4.3
- $ K_{Sc}=1 $ from table 6.4.2
- $ K_{T}=1 $ from clause 6.4.4
- $ K_{Zcg}=0.68[(8)(0.265)(0.304)]^{-0.13} \leq1 $
- $ K_{Zcg}=0.72m^{-3}\leq1 $
- $ K_{C}=[1.0+\frac{(21.42)(0.72)(26.32)^3}{35(8439)(1)(1)}} ]^{-1}= 0.51 $
where:
- $ F_{c}=21.42Mpa $
- $ K_{Zcg}=0.72 $
- $ C_{c}=26.32 $
- $ K_{Se}=1 $ from table 6.4.2
- $ K_{T}=1 $ from clause 6.4.4
- $ E=9700 $ from table 6.3
- $ E_{05}=0.87(9700)=8439 Mpa $
After computing all the variables, the total compressive resistance of the member is:
- $ P_{r}=(0.8)(21.42)(265*304)(0.72)(0.51)(10^{-3})=507KN $
- $ P_{r} \geq P_{f} $
- $ 507 \geq 400 $
Since the factored resistance is larger than the factored load, the member is adequate when the duration factor is 0.85.
To conclude, a member size of 265x304 is able to support the compressive applied loads of 200 KN for the specified dead load and 100 KN for the specified live load.
References
- ↑ Wood Design Manual 2010. Ottawa, ON: Canadian Wood Council (2010).