# Bearing Resistance

## 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:

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

Assume:

• Dry service conditions
• Untreated lumber

#### 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]

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]

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