# Designing Glulam Beam-Columns

## 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

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. Wood Design Manual 2010. Ottawa, ON: Canadian Wood Council (2010).