Designing Glulam Beam-Columns

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Question

Design a 2.5 m tall column that is subjected to a 10 kN vertical load and a 15 kN lateral load at mid-height. The column is fixed at the bottom and free at the end (cantilever). service conditions are dry and the wood is treated. [1]

Solution

Load Combinations

from Clause 4.2.4 use: [1]

Case 1: 1.4D
Case 2: (1.25D or 0.9D) + 1.5L + 0.4W
Case 4: (1.25D or 0.9D) + 1.4W


Use the load combinations above to determine the governing factored Axial load, and solve for the accompanying Lateral load, Moment applied, and Shear.


Case 1 ($ K_D $ = 0.65)

$ {P_f}={1.4}\times{10} $


Case 2 ($ K_D $ = 1.15)

$ {P_f}={1.25}\times{10}+{0.4}\times{15}=18.5kN $
$ {P_L}={0.4}\times{15}=6kN $
$ {M_f}={6}\times{2.5/2}=7.5kNm $
$ {V_f}=6kN $


Case 4 ($ K_D $ = 1.15)

$ {P_f}={1.25}\times{10}+{1.4}\times{15}=34 kN $
$ {P_L}={1.4}\times{15}=21kN $
$ {M_f}={21}\times{2.5/2}=26kNm $
$ {V_f}=21kN $

Note:$ K_D $ factor is obtain using Clause 4.3.2.2[1]


Therefore, Case 4 will govern for all loads.

Selecting Section for Trial

In order to select our section, we first need to determine the EI (stiffness) required with respect to the allowable deflection.


$ \triangle_{Max}=\frac{Pb^2}{6EI}{3L-b} = \triangle_{Allowable}=\frac{L}{180} $

$ {EI_(req)}=1758\times{10^9}Nmm^2 $

Note: $ \triangle_{Max} $ was obtained using "Beam Diagrams and Formulae, Cantilevered Beam, #27"[1]


Therefore, Try Glulam D.Fir-L with a size of 175x228 with a grade of 18t-E. Section was chosen according to its EI value of $ 2210 x 10^9 Nmm^2 $.


The properties of this section is as follows:


$ {f_{b}}=24.3\hspace{5 mm}MPa $
$ {f_{c}}=30.2\hspace{5 mm}MPa $
$ {E}=8500\hspace{5 mm}MPa $
$ {E_{05}}=12006\hspace{5 mm}MPa $


$ {K_{SE}}=1.0 $
$ {K_{sb}}=1.0 $
$ {K_{SC}}=1.0 $
$ {K_{T}}=1.0 $ (unincised)
$ {K_{H}}=1.0 $
$ {K_{D}}=0.65 $ (For long term loads only)
$ {K_{D}}=1.15 $ (Structure has both wind and dead loads)


Note: All specified strengths and modification factors are found in Clause 6.3[1] and Clause 6.4[1] respectively.

Check Compression Resistance

From Clause 6.5.8.4.2[1] for compression resistance parallel to plain

$ {P_{r}}=\phi{F_{c}}{A}{K_{Zcg}}{K_{C}} $

Where;

$ \phi=0.8 $

$ {F_{C}}={f_{c}}({K_{D}}{K_{H}}{K_{Sc}}{K_{T}}) $

$ =30.2(KD\times1.0\times1.0\times1.0)={30.2K_D} $


$ {A}={B}\times{D} $

$ =175\times228=39900mm^2 $


The volume is

$ Z = 0.175 \times0.228\times2.5 = 0.1m^3 $


Therefore;

$ {K_{Zcg}}=0.68(Z)^{-0.13} \hspace{5mm} \leq1.0 $

$ =0.68(0.1)^{-0.13} $
$ =0.92\leq1.0 $ ok


From table A.5.5.6.1[1] and since the column is restrained against rotation at one end and free at the other $ K_e $ = 2.0. From Clause 6.5.8.2[1] the governing slenderness ratio is:

$ {C_{c}}= \frac{K_eL}{depth}\hspace{5mm}\leq50=\frac{2.0\times2500}{228}=21.9 $ ok

$ {C_{c}}= \frac{K_eL}{width}\hspace{5mm}\leq50=\frac{2.0\times2500}{175}=28.6 $ ok

Therefore, $ C_c $ with a value of 28.6 governs for slenderness.


From Clause 6.5.8.5[1] the slenderness factor is calculated as follows

$ {K_{c}}=[1.0+\frac{F_{c}{K_{Zcg}}{C_{c}^3}}{35{E_{05}}{K_{SE}}{K_{T}}}] $


$ K_D=0.65 $

$ =[1.0+\frac{{30.2\times0.65}\times0.92\times28.6^3}{35\times12006\times1.0\times1.0}]=2.0 $


$ K_D=1.15 $

$ =[1.0+\frac{{30.2\times1.15}\times0.92\times28.6^3}{35\times12006\times1.0\times1.0}]=2.78 $



Calculating compression resistance at both $ K_D $ factors:


$ K_D=0.65 $

$ {P_{r}}=0.8\times30.2\times0.65\times39900\times0.92\times2.0\times10^{-3}=1153kN $

A compression resistance value of 1153 kN is much greater than our factored compression value of 34 kN therefore, our section is safe in axial.


$ K_D=1.15 $

$ {P_{r}}=0.8\times30.2\times1.15\times39900\times0.92\times2.78\times10^{-3}=2835kN $

The section under the duration factor of 1.15 is also safe with a compression resistance of 2835kN

Check Flexural Capacity

From Clause 6.5.6.5[1]

$ {M_{r_1}}=\phi{F_{b}}{S}{K_{x}}{K_{Zbg}} $
$ {M_{r_2}}=\phi{F_{b}}{S}{K_{x}}{K_{L}} $


Where;


$ {F_{b}}={f_{b}}({K_{D}}{K_{H}}{K_{Sb}}{K_{T}}) $

$ =24.3(1.15\times1.0\times1.0\times1.0=27.9 MPa $


$ {S}=\frac{{b}\times{d}^{2}}{6} $

$ =\frac{175\times228^2}{6}=15.16\times10^5mm^3 $


from clause 6.5.6.5.2[1] we have a straight member, therefore;

$ {K_x}=1.0 $


$ K_{Zbg}=1.03\times{(BL)}^{-0.18}\hspace{5mm}\leq1.0 $

$ =1.03\times{(0.175\times2.5)}^{-0.18}=1.20 $

The value we got exceeds the limit, therefore;

$ K_{Zbg}=1.0 $


Need now to calculate the lateral stability factor $ K_L $ from Clause 6.5.6.4.4[1] but first need to find $ C_B $ factor from Clause 6.5.6.4.3[1]

Also from table 6.5.6.4.3[1]

$ L_e = 1.92\times {L_u} $

$ =1.92\times2500 = 4800 $

$ C_B=\sqrt{\frac{{L_e}d}{b^2} $

$ =\sqrt{\frac{{4800}\times228}{175^2}}=6.0 $

therefore, as should in Clause 6.5.6.4.4 (a)[1], when $ C_B $ does not exceed 10 then $ K_L=1.0 $


Calculating Moment Resistance

$ {M_{r_1}}={M_{r_2}}=0.9\times27.9\times{(15.16\times10^5)}\times1.0\times1.0\times10^{-6}=38kNm $

The Moment Resistance with a value of 38kNm is greater than the factored moment which has a value of 26kNm, therefore, the section is safe against moment.


Check Interaction Between Bending and Compression

From Clause 6.5.12[1]

$ (\frac{P_{f}}{P_{r}})^2+(\frac{M_{f}}{M_{r}})(\frac{1}{1-\frac{P_{f}}{P_{E}}})\leq1.0 $

Where;

$ =\frac{\pi^2\times12006\times1.0\times1.0\times\frac{175\times228^3}{12}}{(2.0\times2500)^2}=819.26kN $



$ (\frac{33.5}{2835})^2+(\frac{26.25}{38})(\frac{1}{1-\frac{33.5}{819.26}}) = 0.72\leq1.0 $


Serviceability Limits

$ \triangle_{Allowable}=\frac{L}{180} $

$ =\frac{2500}{180}=13.89mm $


$ \triangle_{Max}=\frac{Pb^2}{6EI}{3L-b} $


Where;

$ {I}=\frac{{175}\times{228}^{2}}{12}=172.847\times10^6mm^4 $

$ \triangle_{Max}=\frac{{(15\times10^3)}175^2}{6\times13800\times{(172.847\times10^6)}}{((3\times2500)-175)}=0.23mm $

The Max deflection of the member (0.23mm) does not exceed the allowable deflection (13.89mm), therefore,

Use D.Fir-L 175x228 18t-E.

Reference

  1. 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 Wood Design Manual 2010. Ottawa, ON: Canadian Wood Council (2010).