Question

Q1 a) Figure Q1(a) shows the sheet-metal bracket. Analyze location of the center of gravity ( x, 7 )of the sheet-metal bracket if the material is homogeneous and has a constant thickness. (Support your answer with a free body diagram). Ignore the 6 angle. (10 marks) 60 mm 20 mm 60 mm 10 mm dia. holes 20 mm 20 mm 80 mm 20 mm 60 mm Figure Q1(a)

Answer 1

Let us manner. consider this problem in the following We dissociate the shut metal bracket in two parts. Part-1) The metal plate in the the zoy plane. 2 60mm loren Sinu this is symmetric cornering G Bomm plate the centre of gravity will lie on the axis of symmetry so G is given by (0,0,40) 40 s 20 mm 20mm Hi TIG ZIG The area of the plate will be useful further, so we calculate it. A = Area of acctangulare plate - 4x Arun of circle. A, = 80x120 - 4x ax(s)? So A = 9286 mmt 64 30 Part-2) The metal plate in the ni ning plane. ry The centre of gravity will again lie on the axis of symmetry so bez is given by (30,0,0) 226 1201 Yem The area of this plate is A, = 60x120 = 7200 mm².

Assuming constant thicknen valen to be tmm. The volume will be as follows. V = At = 9286 mm3 27 At= 72004 mm The centre of gravity will be given by. xa nov + a V₂ For composite of the two plates Vi + 201 Now = o.v, +30 V₂ 30 X 72oot = 13.102 mm Vtv 92 86 t + proot You a Jook + Year 0 mm. VTV ZG - ZIGV, +ZzG V₂ = 4o. V, + 0.42 - 40 x 9286 t -22.531 V + V Vit V2 9286t+ 720ot. so the unth of gravity is given by ( 13.102,0, 22. 531) mm.

Assuming constant thicknen valen to be tmm. The volume will be as follows. V = At = 9286 mm3 27 At= 72004 mm The centre of gravity will be given by. xa nov + a V₂ For composite of the two plates Vi + 201 Now = o.v, +30 V₂ 30 X 72oot = 13.102 mm Vtv 92 86 t + proot You a Jook + Year 0 mm. VTV ZG - ZIGV, +ZzG V₂ = 4o. V, + 0.42 - 40 x 9286 t -22.531 V + V Vit V2 9286t+ 720ot. so the unth of gravity is given by ( 13.102,0, 22. 531) mm.
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