# Bearing Resistance

## Contents

## Bearing resistance of a sawn lumber beam-Example

All clauses, equations and variables were obtained from the *Wood Design Manual 2010*^{[1]}

### Example:

Find the maximum factored compressive load $ Q_f $ that a 3m long 191x292mm S-P-F No.1 beam can withstand if a 89x89mm S-P-F S.S stud is bearing on it when:

- The stud is at the center of the beam
- The stud is above the supports and the beam is resting on a 50x50mm bearing plate

Assume:

- Standard term loading
- Dry service conditions
- Untreated lumber

### Answer:

#### Part 1

Compressive resistance perpendicular to grain (bearing) of a sawn lumber member is given by *Clause 5.5.7.2* in the design manual.
The Bearing resistance equation is:
$ Q_r={\Phi}F_{cp}A_{b}K_{b}K_{Zcp} $^{[1]}

where:

- $ \Phi $= 0.8
^{[1]} - $ F_{cp}=f_{cp}(K_{D}K_{Scp}K_{T}) $
^{[1]}- $ f_{cp} $ is the specified strength in $ MPa $ when calculating for compression perpendicular to grain. From
*Table 5.3.1C*we see that $ f_{cp}=5.3MPa $^{[1]}for this member - $ K_D=1.0 $ since we have standard term loading and it is obtained from
*Table 4.3.2.2*^{[1]} - $ K_{Scp}=1.0 $since we have dry service conditions and it is obtained from
*Table 5.4.2*^{[1]} - $ K_T=1.0 $since the member is untreated and operating in dry service conditions. This is obtained from
*Table 5.4.3*^{[1]}

- $ f_{cp} $ is the specified strength in $ MPa $ when calculating for compression perpendicular to grain. From

Therefore,
$ F_{cp}=(5.3)(1)(1)(1)= 5.3 MPa $^{[1]}

- $ A_b $ is the bearing area in $ mm^2 $ through which the perpendicular load acts. In the case where the stud lies over the center of the beam,

$ A_b=(89)(191)=16999mm^2 $^{[1]}

- $ K_B $ is the length of bearing factor given in
*Clause & Table 5.5.7.6*. Since it is in an area of high bending stress(center of the beam),

$ K_B=1.0 $^{[1]}

- $ K_{Zcp} $ is the size bearing factor given in
*Clause & Table 5.5.7.5*. It relies on the ratio of the member's width to its depth. For our case,

$ \frac{w}{d}={\frac{191}{292}}{\approx}0.65 {\leq} 1 $

Therefore, from *Table 5.5.7.5*

$ K_{Zcp}=1.0 $^{[1]}

Since we now have all the variables, we can go back to our original equation and find $ Q_r $

```
$ Q_r=(0.8)(5.3)(16999)(1)(1)= 72.1 kN $
```

This is equal to the maximum $ Q_f $ that the member can sustain when the stud is at the center of the beam.

#### Part2

Compressive resistance of a sawn lumber member within a distance from the supports equal to the member depth is given by *Clause 5.5.7.3*. Here,

$ Q'_r=(2/3){\phi}F_{cp}A'_{b}K_{B}K_{Zcp} $^{[1]}

where:

- $ {\Phi}=0.8 $
^{[1]} - $ F_{cp}=f_{cp}(K_{D}K_{Scp}K_{T}) $. This is unchanged from before and has the same value.

Therefore,

$ F_{cp}=5.3MPa $^{[1]}

- $ A'_b=b(\frac{L_{b1}+L_{b2}}{2}) \leq 1.5b(L_{b1}) $ represents the average bearing area of unequal bearing areas on either side of the beam. The equation for that is given in
*Clause 5.5.7.4*.^{[1]}- $ b $ is the average bearing width given in $ mm $
^{[1]} - $ L_{b1} $ is the smaller bearing length given in $ mm $
^{[1]} - $ L_{b2} $ is the larger bearing length given in $ mm $
^{[1]}

- $ b $ is the average bearing width given in $ mm $

So,

- $ A'_b=b(\frac{L_{b1}+L_{b2}}{2})= ({\frac{50+191}{2})(\frac{50+89}{2})=8374.75 mm^2 \leq 1.5(120.5)(50)=9037.5mm^2 $. It satisfies the limitation O.K
- $ K_B=1.0 $ since the bearing is at the supports meaning it is less than 75mm from the members end. This can be seen from
*Clause 5.5.7.6(a)*^{[1]} - $ K_{Zcp}=1.0 $ since it unchanged from Part 1

Now that we have all the variables, we can calculate $ Q'_r $

```
$ Q'_r=(\frac{2}{3})(0.8)(5.3)(8374.75)(1)(1)= 23.7kN $
```

This is equal to the maximum $ Q_f $ that can be sustained when the stud is at the support.

A very similar procedure is used to calculate the bearing resistance of a Glulam member by using *Clause 6.5.9.2* for **Part 1** and *Clause 6.5.9.3* for **Part 2** while using the respective values relating to the particular Glulam member.^{[1]}

## 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}^{1.14}^{1.15}^{1.16}^{1.17}^{1.18}^{1.19}^{1.20}*Wood Design Manual*2010. Ottawa, ON, Canadian Wood Council (2010).