Bearing Resistance

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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:

  • Standard term loading
  • Dry service conditions
  • Untreated lumber

Beam under perpendicular compression

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]

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. 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).