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Check mode (j) based on 25% of Johansen load

(2 ■ My.Rk ■ fh.e.1.2.k ■ d) + Fax.Rk Fv.Rk.4 = 13.11 kN

2 ■ £2.1 ■ (1 + £2.1) +4 ■■(2 + ^ • My.Rk-£2.1

All modes are possible and the minimum characteristic strength of shear plane 1 is:

Fv.Rk.1.2 = min(Fv.Rk.1, Fv.Rk.2, Fv.Rk.3, Fv.Rk.4, Fv.Rk.3.3) Fv.Rk.1.2 = 6.12kN (mode (h))

Design force in shear plane 1 Design capacity of shear plane 1

KM. connection

Fv.R.d.sp.1 = 3.76 kN OK Design capacity of shear plane 1 exceeds the design force in shear plane 1; therefore OK.

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