Iks

Angle of shear plane 1 force relative to the grain direction in member 1 in the joint comprising members 1,2,1, 01.2 (in degrees)

Angle of shear plane 1 force relative to 021 = 45 - 012 02 1 = 11.31

the grain direction in member 2 in the joint comprising members 1,2,1, 021 (in degrees)

Design force in shear plane 2, Fv.sp.2

Fv .sp.2 = [(Fv.sp.1 • cos(01.2 • deg) - Fv.sp.2 • cos(45 • deg)) + (Fv.sp.1 • sin(01.2 • deg) - Fv.sp.2 • sin(45 • deg)) ]

Angle of shear plane 2 force relative to the horizontal plane in the joint comprising members 2,3,2

Horizontal component, Fv.sp.2 h

Fv.sp.2.h = Fv.sp.1 • cos (01.2 • deg) - Fv.d.2 • cos(45 • deg) Vertical component, Fv.sp.2.v

Fv.sp.2.v = Fv.sp.1 • sin (01.2 • deg) - Fv.d.2 • sin(45 • deg)

Angle of shear plane 2 force relative angle(Fv.sp.2 h, Fv.sp.2 v) to the horizontal plane (in degrees)

Angle of the shear plane 2 force relative to the grain of member

2 in the joint comprising members 2,3,2, 02.3.a

Angle of the shear plane 2 force relative to the grain of member

3 in the joint comprising members 2,3,2, 03.2.a

Angle of the shear plane 2 force relative to the grain of member 3 in the joint comprising members 3,2,3, 03,2,b

Angle of the shear plane 2 force relative to the grain of member 2 in the joint comprising members 3,2,3, 02.3.b

Embedment strength of the timber:

0 0

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