# Designing Tension Elements

Design of Lumber and Glulam in Tension based on CAN/CSA-O86-09

Zaid Barakat, Carleton University, Ottawa, Ontario, Canada

March,2015

## Contents

### 1.1 Designing for Lumber

Lumber and Timber products are Produced in accordance with CSA Standard O141 Softwood Lumber . The characteristics of structural grades are outlined in the National Lumber Grades Authority Standard Grading Rules for Canadian Lumber. [1]

Common tension members include webs and bottom chords of trusses, bottom flanges in wood I-joists, chord members in diaphragms and infrequently bracing members. Usually, the dimensions of tension members are governed by the connections rather than the tensile resistance of the section itself. The tension member must be large enough so that the spacing, in addition to end and edge distances for the 'fasteners' are adequate.

Factored tensile resistance at net section $T_{rn}\geq$ factored tensile force $T_{f}$

-For members connected with fasteners:

$T_{rN} =T_{r}\times (A_{N}/A_{G})$

• The following example illustrates the calculations required for designing a tension member out of lumber S.P.F of grade No.1 to resist the following conditions at the ultimate limit state:
Member to be 3m long
Assume dry conditions, no system effect.
Member is treated but unincised
The unfactored axial loads are: Dead - 100kN, Live - 70kN, Wind - 50kN

 Species Spruce-Pine-Fir Grade No.1 Service Condition Dry Treatment Treated but unincised

• Checking for load combinations for the factored load acting on member and finding the corresponding Duration ratio $K_{d}$ for each load combination according to the duration of loading :
 Load combination $K_{d}$ ratio 1 1.4D = 140 KN 0.65 2(a) 1.25D + 1.5L + 0.4 W = 250 KN 1.15 2(b) 1.25D + 1.5L + no companion load = 230 KN 0.923 3(a) 1.25D + 0.5L = 160 KN 0.923 3(b) 1.25D + 0.4W= 145 KN 1.15 4(a) 1.25D + 1.4W + 0.5L = 230 KN 1.15 4(b) 1.25D + 1.4W + no companion load = 195 KN 1.15
[2]

$K_{d}=1.0-0.5log({P_{L}}/{P_{D}})\geq0.65$

$K_{d}=1.0-0.5log(100/70)=0.9225>0.65$

-Since the $K_{d}$ ratio has a linear relationship with the factored tensile resistance parallel to grain, the governing load combination is to be divided by the $K_{d}$ ratio, with respect to limit state design:

$Q_{f} < Q_{R} \times K_{d} = Q_{f}/K_{d} < Q_{R}$

• Picking the load combinations that governs, where the combination corresponding to the highest factored load is the one that governs as to be extra cautious while applying the limit state design :

Cases that governs are '1, 2a , 2b,' therefore:

1) 140/0.65= 215.4 KN

2) a- 250/1.15= 217.39 KN

2) b- 230/0.923= 249.19 KN

- case 2(b) governs with the highest load.

Therefore, Design for $T_{f} = 230 KN and K_{d}= 0.923$

- Refer to Design tables for lumber S.P.F No.1 to select appropriate cross section for design with respect to limit state design, according to governing factored tension. [4]

• TRY S.P.F 191x241 where $T_{rn}$ would be found to be 220 KN < $T_{f}$=230 KN , therefore doesn't govern.
• TRY S.P.F 241x241 :

$T_{r}=\phi F_{t} A_{n} K_{zt}$[5]

where:

• $\phi$=0.9
• $F_{t} = f_{t} ( K_{D} K_{H} K_{st} K_{T} )$>[6]
• $f_{t}$=specified strength on tension parallel to grain
• For S.P.F No.1 we can find that $f_{t}$=5.6 MPA[7]
• $K_{D}=0.9225$[8]
• $K_{H}=1.1$ ( System factor: case 1 ) [9]
• $K_{st}=1.0$ ( Service condition factor: for Dry service condition, tension parallel to grain ) [10]
• $K_{T}=1.0$ ( Treatment factor : Treated but unincised )[11]

Therefore for $F_{t}$:

$F_{t}=5.6(0.9225\times1.1\times1.0\times1.0)=5.6826$MPA

• $K_{Zt}$( Size factor in tension ) [12]

$K_{Zt}=1.1$ for S.P.F ( 241\times241 ) tension parallel to grain, Table 5.4.5

• $A_{n}=0.85A_{g}$[13]

- Assuming connection can withstand the full tension load therefore $A_{n}=0.85A_{g}$

$A_{n}=0.85(241\times241)= 49368.9 \approx 49369 mm^{2}$

Calculating $T_{rn}$

-$T_{rn}=0.9(0.56828)(49369)(1.1)=277.7KN \approx 278 KN ($[14]

-According to Limit State Design:

$T_{rn} > T_{f}$ = 278 > 230

Therefore use S.P.F 241x241 since it is adequate for the conditions.

### 1.2 Designing for Glulam

Designing for glulam tension members differs from that of sawn lumber and timber since the specified strengths for gross and net section are different. Therefore a different criteria must be followed, whereas the tensile resistance of the gross section and net section are to be checked separately and for each to be greater than the maximum factored tensile force for the design to be adequate, the criteria may be illustrated as follows:

1. Factored tensile resistance of gross section $T_{rG}\geq$ factored tensile force $T_{f}$

2. Factored tensile resistance at net section $T_{rn}\geq$ factored tensile force $T_{f}$

It is very common that the strength at the net section is greater that that of the gross section, therefore the resistance of the gross section would frequently govern.

• The following example illustrates the calculations required for designing a tension member out of glulam 24fEX D.Fir-L to resist the SAME conditions as the ones given for the previous example for lumber at the ultimate limit state:

• Assuming the same conditions as the example given above but for glulam 24fEX D.Fir-L:
• Recall $T_{f}=230$KN and $K_{D}=0.9226$

- Refer to tables for glulam 20f-EX and 24f-EX D.Fir-L [15]

• TRY D.Fir-L 24f-EX (130x190) :

$T_{rN}=\phi F_{tN} A_{n}$[16]

or

$T_{rg}=\phi F_{tg} A_{g}$[17]

where:

• $A_{n}=0.85A_{g}$[19]

$A_{g}=130x190= 24700 mm^{2}$

$A_{n}=0.85(130x190)= 20995 mm^{2}$

• $F_{tn} = f_{tn} ( K_{D} K_{H} K_{st} K_{T} )$[20]
• $F_{tg} = f_{tg} ( K_{D} K_{H} K_{st} K_{T} )$[21]

$f_{tn}=20.4 and f_{tg}=15.3$[22]

• $f_{tn} and f_{tg}$=specified strength on tension parallel to grain for net area and gross area.
• for Douglas Fir-Larch 24f-EX, tension parallel to grain for net area $f_{tn}$ , gross area.$f_{tg}$.[23]
• $K_{D}=0.9225$[24]
• $K_{H}=1.1$ ( System factor: case 1 ) [25]
• $K_{st}=0.75$ ( Service condition factor: for Dry service condition, tension parallel to grain ) [26]
• $K_{T}=1.0$ ( Treatment factor : Treated but unincised )[27]

Therefore:

-$F_{tn} = 20.4\times( 0.9225\times1.0\times0.75\times1.0 )=14.114$ MPA

and

-$F_{tg} = 15.3\times ( 0.9225\times1.0\times0.75\times1.0 )=10.586$ MPA

$T_{rN}= 0.9\times(14.114)\times(20995)=266.69\approx 267$ KN

or

$T_{rg}= 0.9\times(10.586)\times(24700)=235.33\approx 235$ KN

- $T_{rg}$ would govern

-According to Limit State Design:

$T_{rg} > T_{f}$ = 235 > 230

Therefore use 24fEX D.Fir-L (130x190) since it is adequate for the conditions.

## References

1. Test Reference
2. "Wood Design Manual 2010" Table 4.3.2.2
3. "Wood Design Manual 2010"Clause 4.3.2.3"
4. "Wood Design Manual 2010" p.163-167
5. "Wood Design Manual 2010"Clause 5.5.9
6. "Wood Design Manual 2010"Clause 5.5.9
7. "Wood Design Manual 2010" Tables 5.3.1A, TO 5.3.1D, 5.3.2 and 5.3.3
8. "Wood Design Manual 2010"Clause 4.3.2.3"
9. "Wood Design Manual 2010" Clause 5.4.4 and Table 5.4.4
10. "Wood Design Manual 2010" Table 5.4.2
11. "Wood Design Manual 2010" Table 5.4.3
12. "Wood Design Manual 2010"Table 5.4.5
13. "Wood Design Manual 2010" Clause 4.3.8 reduction in cross sectional area
14. "Wood Design Manual 2010"Clause 5.5.9
15. "Wood Design Manual 2010"p.169
16. "Wood Design Manual 2010"Clause 6.5.11
17. "Wood Design Manual 2010"Clause 6.5.11
18. "Wood Design Manual 2010"Clause 6.5.11
19. "Wood Design Manual 2010" Clause 4.3.8 reduction in cross sectional area
20. "Wood Design Manual 2010"Clause 6.5.11
21. "Wood Design Manual 2010"Clause 6.5.11
22. "Wood Design Manual 2010" Table 6.3
23. "Wood Design Manual 2010" Table 6.3
24. "Wood Design Manual 2010"Clause 4.3.2.3"
25. "Wood Design Manual 2010" Clause 5.4.4 and Table 5.4.4
26. "Wood Design Manual 2010" Table 5.4.2
27. "Wood Design Manual 2010" Table 5.4.3