Design Of Hinged Base Data:H.E.B (300) P = 1.84 t H = 1.57 t a = 30 cm b = 30 cm Fcu = 250 kg/cm2 1. Base Dimension A = a +5 cm A = 30 +5 = 35 cm B1 B = b +5 cm B b B = 30 +5 = 35 cm a A1 = A +10 cm A A1 A1 = 35 +10 = 45 cm B1 = B +10 cm B1 = 35 +10 = 45 cm Area1 = 35*35 = 1225 cm2 Area2 = 45*45 = 2025 cm2 Area2 Area1 = 1.28 OK 2. Check Of concrete Capacity Pu C (0.67 * Fcu * Area1 ) Area2 Area1 Pu 0.6(0.67 * 0.25 *1225) *1.28 157.58t Pu 157.58 p 1.84 ok 3. Plate Thickness pU A1 2 * M * 2.5 * b * 2.5 M P Fy * Z P P ZP = M P b*tp 2 4 2 2.8 * 21 * t p 4 1.841225* 2.5 * 35 * 2.5 2 0.6 * 2.8 * t t p = 0.3 cm 4. Design Of weld 2 p Use Min. t p = 2 cm Ou Q 0.6 * N QU 1.84 0.6 *1.57 = 2.78 t Ruw,q Q Lw = 2.78 2 * 35 2 *1.1 0.7 * S * 0.4 * Fu = 0.3 cm Take Sweld Smin . 0.5cm 5. Design Of Bolts For bolts Grade (10.9) , M=24 RUV =1.84 / 2 = 0.92 t R shear 0.6(0.6Fub ) As * n 0.92 0.6(0.6 * 6) As *1 AS = 0.43 Use 2 M 20 R bearing br * d * t min * ( * Fu ) R bearing 0.7 * 2.0 * 2 * (1.2 * 4.4) = 14.7 t Rmin 14.7ton N Qu Rmin 1.84 0.12 14.7 Take N 2