1314

a Equation 4.6. b Equation 4.5, Figure 4.4b.

a Equation 4.6. b Equation 4.5, Figure 4.4b.

From the last column of the above table, it can be seen that fracture is not a problem for any of the trial sections.

Check for the limit state of block shear: Figure 4.4c shows a possible block shear failure mode. To avoid block shear failure the required strength of Pu = 104 kip should not exceed the design strength, ftPn, calculated using Equations 4.12a or 4.12b, whichever is applicable.

For the C8 x 11.5 section:

Anv = Agv - 5 + -J (0.220) = 2.72 in. Agt = (3)(0.220) = 0.66 in.2 Ant = Agt - 1^1 + 0 (0.220) = 0.41 in.2

Substituting the above into Equation 4.12b, since (FuAnt = 23.8 kip) is smaller than (0.6FuAnv = 94.7 kip), we obtain ftPn = 88.8 kip, which is less than Pu = 104 kip. The C8 x 11.5 section is therefore not adequate. A significant increase in block shear strength is not expected from the C9 x 13.4 section because its web thickness tw is just slightly over that of the C8 x 11.5 section. As a result, we shall check the adequacy of the C8 x 13.75 section instead. For the C8 x 13.75 section:

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