Question

Two plates \(20 \mathrm{~cm}\) wide and ' \(t\) ' thickness are joined by three bolts as shown in Figure. Each

bolt supports a shear load of \(1200 \mathrm{~kg}\) and the plates of the lap joint subjected to a tensile load

of P. An allowable tensile stress of the plate \(\sigma_{\text {allow }}=\underline{\text { A }} \times 100 \mathrm{~kg} / \mathrm{cm}^{2}\). Determine the required

plate thickness 't' by considering the critical failure conditions already discussed in the lesson.

Answer 1

Sol: Peale = 20cm load 1200kg Tallow = A x 100kg/cm2 = 5 x 100 = 500 kg 1cm² Student Number Note Value not given so Take Net Area of plate P Р Р— width of plate -300 of hole) Olum 20cm Allowable load taken by pa glot plate (Net area of Plate ) x Allowable Tensild stress) = 16 -3 do)xtx Tallow In Critical condition This Allowable load will be equal do max Shearing Stress of 3 Bolts Dia hole = 1cm 3 x max shearing stress Bolts Allowable Tensild Stress = 3x 1200 (5-3 dn) xt * Tau 20-3x1 x t x 500 = 3600 t = 4.24 mm

Answer 2

Sol: Peale = 20cm load 1200kg Tallow = A x 100kg/cm2 = 5 x 100 = 500 kg 1cm² Student Number Note Value not given so Take Net Area of plate P Р Р— width of plate -300 of hole) Olum 20cm Allowable load taken by pa glot plate (Net area of Plate ) x Allowable Tensild stress) = 16 -3 do)xtx Tallow In Critical condition This Allowable load will be equal do max Shearing Stress of 3 Bolts Dia hole = 1cm 3 x max shearing stress Bolts Allowable Tensild Stress = 3x 1200 (5-3 dn) xt * Tau 20-3x1 x t x 500 = 3600 t = 4.24 mm