# Designing Lumber Beam-Columns

## Contents

## Question

Design a 3 m tall Pinned-pinned column subjected to a 70 kN vertical load and a 12kN lateral load at 2/3 of the height as displayed in the diagram. It can be considered that the loads are of standard duration and the column will be subjected to wet service conditions and is treated (unincised).

## Solution

### Applied Shear, Moment, & Axial Loads

We must find the maximum applied loads on the beam by drawing the axial, shear and moment diagrams as seen below.

We can see from these diagrams that

- $ {P_{f}}=70\hspace{5 mm}{kN} $ (Axial Load)

- $ {V_{f}}=8\hspace{5 mm}{kN} $ (Maximum Shear Load)

- $ {M_{f}}=8\hspace{5 mm}{kN/m} $ (Maximum Applied Moment)

For the member to be acceptable

- $ {P_{f}}< {P_{r}} $

- $ {V_{f}}< {V_{r}} $

- $ {M_{f}}< {M_{r}} $

Where $ {P_{r}} $,$ {V_{r}} $, & $ {{M_{r}} $ are the members resistances calculated using the wood design manual.^{[1]}

### Member Selection & Service Condition Factors

From the Selection tables in the **Wood Design Manual**. ^{[1]} we can choose an initial size of member to be analysed. From these tables it can be seen that the axial capacity of a 191x191 S-P-F select member is greater then the required axial capacity and therefore we can try this size of member.

A 191x191 S-P-F Select member can be classified as a *Post and Timber grade member from *Table 5.2.2.1 with the following specified strengths from *Table 5.3.1D in the Wood Design Manual ^{[1]}.*

- $ {f_{b}}=12.7\hspace{5 mm}MPa $ (Bending)

- $ {f_{v}}=1.2\hspace{5 mm}MPa $ (Shear)

- $ {f_{c}}=9.9\hspace{5 mm}MPa $ (Compression)

- $ {E}=8500\hspace{5 mm}MPa $ (Modulus Elasticity)

- $ {E_{05}}=6000\hspace{5 mm}MPa $ (Modulus Elasticity For Compression)

Also from our questions details and Tables 5.4.2 - Table 5.4.5 in the Wood Design Manual ^{[1]}
the following K Factors Can be used.

- $ {K_{Sb}}=1.0 $ (Bending Factor)

- $ {K_{Sv}}=1.0 $ (Shear Factor)

- $ {K_{Sc}}=0.91 $ (Compression Parallel to Grain Factor)

- $ {K_{SE}}=1.0 $ (Modulus Elasticity)

- $ {K_{T}}=1.0 $ (Treatment Factor)

- $ {K_{H}}=1.0 $ (System Factor)

- $ {K_{D}}=1.0 $ (Load Duration Factor)

### Axial Capacity Check

To Calculate the axial capacity of the selected 191x191 member the equation from Clause 5.5.6.2.3 in the Wood Design Manual ^{[1]} must be used. The equation is as follows:

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

where

$ \phi=0.8 $

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

- $ =9.9(1.0\times1.0\times0.91\times1.0) $
- $ =9.009/hspace{2mm}MPa $

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

- $ =191\times191 $
- $ =36481\hspace{5 mm}{mm^{2}} $

$ {K_{Zc}}=6.3(dL)^{-0.13} \hspace{5mm} \leq1.3 $

- $ =6.3(191\times3000)^{-0.13} $
- $ =1.124 $

Due to the fact that the member will be pinned-pinned the effective length Factor will be $ K=1.0 $ (Table A.5.5.6.1^{[1]}) and must be used to find the slenderness ratio of the member as follows (Clause 5.5.6.2.2 ^{[1]}). (Note: due to the fact it is a square member only one direction must be calculated)

$ {C_{c}}= \frac{Effective\hspace2mmLength}{Member\hspace{2mm}Width}\hspace{5mm}\leq50 $

- $ =\frac{1.0\times3000}{191} $

- $ =15.71\hspace{5mm}\leq50 $

Now that the slenderness ratio has been calculated it can be used in clause 5.5.6.2.4 ^{[1]} to find the Slenderness Factor $ {K_{c}} $ as follows.

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

- $ =[1.0+\frac{9.009\times1.124\times15.71^3}{35\times6000\times1.0\times1.0}]^{-1.0} $

- $ =0.842 $

With all of the components now found the axial capacity can be calculated.

$ {P_{r}}=0.8\times9.009\times36481\times1.124\times0.842 $

- $ =248.84\hspace{2mm}kN $

$ {P_{r}}\geq{P_{f}} $ Therefore the Axial Capacity is adequate.

### Flexural Capacity Check

Similarly to Axial Capacity, to Find the Flexural Capacity an equation from Clause 5.5.4.1 ^{[1]} can be used.

$ {M_{r}}=\phi{F_{b}}{S}{K_{Zb}}{K_{L}} $

Where

$ \phi=0.9 $

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

- $ =12.7(1.0\times1.0\times1.0\times1.0 $
- $ =12.7\hspace{2mm}MPa $

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

- $ =\frac{191\times191}{6} $

- $ =1161311.83\hspace{2mm}mm^3 $

from Table 5.4.5 and Clause 5.5.4.2.1 ^{[1]}

$ {K_{Zb}}=1.3 $ & $ {K_{L}}=1.0 $

With all of the components now found the Flexural capacity can be calculated.

$ {M_{r}}= 0.9\times12.7\times1161311.83\times1.3\times1.0 $

- $ =17.26\hspace{2mm} {kN}\cdot{m} $

$ {M_{r}}\geq{M_{f}} $ Therefore the Flexural Capacity is adequate.

### Axial & Bending load Interaction Check

Due to the fact the member is under axial and bending loads an interaction equation which takes into account the effect of the applied moment on the axial capacity must be considered. This equation is found in clause 5.5.10^{[1]} and must be satisfied to deem the design acceptable.

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

where $ {P_{E}} $ is the Euler Buckling load or the maximum load the member can resist before buckling and is calculated as follows.

$ P_{E}=\frac{\pi{E_{05}}{K_{SE}}{K_{T}}{I}}{L_{E}^2} $

- $ =\frac{\pi\times6000\times1.0\times1.0\times\frac{191^4}{12}}{(1.0\times3000)^2} $

- $ =232.28\hspace{2mm}kN $

With the Euler Buckling Load now found the interaction equation can be checked.

$ (\frac{70}{249})^2+(\frac{8}{17.26})(\frac{1}{1-\frac{70}{232.28}}) = 0.742 $

$ 0.742\leq1.0 $ Therefore the interaction is acceptable and can be resisted.

### Shear Capacity Check

To calculate the shear capacity an the equation from clause 5.5.5.1 ^{[1]} will be used.

$ V_{r}=\phi{F_{V}}\frac{2}{3}A_{n}K_{ZV} $

Where,

$ \phi=0.9 $

$ F_{V}=f_{v}(K_{D}K_{H}K_{Sv}K_{T}) $

- $ =1.2(1.0\times1.0\times1.0\times1.0) $
- $ =1.2\hspace{2mm} MPa $

$ A=36481\hspace{2mm}m^2 $

and from Table 5.4.5 ^{[1]},

$ K_{Zv}=1.3 $

With all of the components now found the Shear capacity can be calculated.

$ V_{r}=0.9\times\1.2\times\frac{2}{3}\times36481\times1.3 $

- $ =34.15\hspace{2mm}kN $

$ {V_{r}}\geq{V_{f}} $ Therefore the Shear Capacity is adequate.

### Serviceability Check

As for serviceability the deflection must be checked against the allowable deflection. The maximum deflection for the simply supported beam can be calculated using the formula below, found in the simple beam diagrams in references chapter 11 of the wood design manual ^{[1]}.

Allowable Deflection for serviceability is

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

- $ =\frac{3000}{180} $

- $ =16.66\hspace{2mm}mm $

and the Maximum Deflection is

$ \triangle_{Max}=\frac{Pa}{3EI}\frac{(l^2-a^2)^3}{(3l^2-a^2)^2} $

- $ =\frac{12000\times1000}{3\times8500\times(\frac{191\times191^3}{12})}\times\frac{(3000^2-1000^2)^3}{(3\times3000^2-1000^2)^2} $

- $ =11.315\hspace{2mm}mm $

$ {\triangle_{Allowable}}\geq{\triangle_{Max}} $ Therefore the deflection is within the acceptable limits.

### Conclusion

Since,

$ {P_{f}}< {P_{r}} $ , $ {V_{f}}< {V_{r}} $ , $ {M_{f}}< {M_{r}} $ , $ {\triangle_{Allowable}}\geq{\triangle_{Max}} $ ,

and the axial-moment interaction equation was satisfied, it can be said that the 191x191 S-P-F Select member is an effective design to resist the applied loads on this beam column.

## References

- ↑
^{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}*Wood Design Manual 2010*. Ottawa, ON: Canadian Wood Council (2010).