Design for Horizontal Shear:

Horizontal shear, V =  P/40 = 1200/40 = 30 kN

∴ Force in lacing, F = + (V/N) cosec θ = 30/2 cosec 45^o

 = 21.213 kN

∴ Length of lacing, l = 230 √2 = 325 mm

t_min for lacing = 325/40 = 8.1 mm, adopt say 10 mm thickness.

Try 60 × 10 mm flat MS section for lacings.

r_min      =          t/√12=   10/√12 = 2.9 mm ∴ λ = 325/2.9 = 112

Corresponding value of σ_ac = 70.4 MPa (from Table 2)

∴          F_comp = 70.4 × 600 = 42,240 N > 21213 N   ∴          OK.

Allowable tensile stress in lacing, σ_at = 0.6 × f_y =0.6 × 250

= 150 MPa

∴          f_ten  = 150 × 600 = 90000 > 21213 N            ∴          OK.