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# AS 1720.1-2010 Table 4.10(A)

## AS 1720.1-2010 Table 4.10(A)

(OP)
In Table 4.10(A), for "Two member", the effective timber thickness b_eff is given as 2 x t_1 for use to determine the characteristic capacity. I could not figure out the reason for this value being 2 x t_1 instead of t_1. Does anyone know the reason for the use of "2 x t_1"?

### RE: AS 1720.1-2010 Table 4.10(A)

In NZ code one of the equations for then using b_eff has a 0.5 factor, this scales it back to just the thickness as you infer as being the logical answer. I would assume Australian code has a similar equation for working out the bolt capacity?

### RE: AS 1720.1-2010 Table 4.10(A)

(OP)
The two timber members are connected by a bolt perpendicular to grain in the member with thickness t_1
b_eff = 2 x t_1
System capacity Q_skp = Q_kp, so no 0.5 factor in the System capacity Q_skp to scale back the thickness.

### RE: AS 1720.1-2010 Table 4.10(A)

I had a look at the Australian code, and it doesn't give the equations that I could see for Qkl and Qkp. Only tables are given, it's these equations in the NZ code that have the 0.5 factor. If there are equation let me know and I will compare.

I suspect overall the effect is due to some double curvature in the bolts bearing on the timber where the thickness of the second timber can govern or at least influence the behaviour.

### RE: AS 1720.1-2010 Table 4.10(A)

(OP)
It has something to do with the failure shape as you mentioned but I could not figure out the logic behind the values. Thanks Agent666.

### RE: AS 1720.1-2010 Table 4.10(A)

Take a look in clause C4.1 & C4.2 the equations are there for Qkl and Qkp. As noted previously there is a divide by 2 included in one of the governing equations to get you back to the thickness of the member which logically makes sense.

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