Info • In the first line of fasteners:

• In the second line of fasteners:

rmax n

• The maximum value will be:

(Note that for the example used the first line is the line furthest from the centroid and the second line is the line closest to the centroid. For connections with greater numbers of lines of fasteners, the process must be extended to cover all of the lines in the zone.)

In equation (12.18) and (12.19), n is as defined in equation (12.16) and the number of fasteners in the shear plane is nn in the first line and nt2 in the second line.

The shear force in member 1 will generate splitting forces in the member within the boundary area of the connection, however, the splitting equation in EC5, equation (8.4), does not apply to this loading condition. Assuming tension splitting will not be critical, the strength check will relate to a confirmation that the shear strength of the member within the boundary area is not exceeded and for member 1 this will be

3 KM, connection where, for a member with a rectangular cross-section, b (mm) thick and h (mm) deep, kmod is the modification factor for load duration and service classes as given in Table 2.4, Km is the partial coefficient for material properties, given in Table 2.6, and fv k is the characteristic shear strength of the member and values for the shear strength of timber and wood-based structural products are given in Chapter 1, hef is the effective depth, allowing for fasteners holes.

When considering member 2, for the loading configuration shown, the maximum shear force perpendicular to the grain within the boundary area of the connection will occur in the shear zone below the centre of rotation as shown in Figure 12.8a.

As for member 1, consider the connection to be subjected to the moment and the lateral forces separately. Under the action of the design moment Md, the lateral force in fastener i in the zone area will be as shown in Figure 12.8a. The radius of the fastener from the centroid of the connection is ri and is at an angle pi to the vertical. The horizontal design force per shear plane in the fastener, FM,hi,d, shown in Figure 12.8b, will be

FM,h,i,d = Fi cos pi = Fm,d,max —— ' I— = Fm,d,max ■T— (12.22)

rmax ri rmax where \yi | is the absolute value of the y coordinate of fastener i.

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