Question

ems 11.1 For the engineering components illustrated in Fig. 11.37 it is necessary to determine the normal and shear stresses at the points shown. Sketch sense of the stresses on each face. an element in each case showing the magnitude and O000o C 1.37 p 3 MN/m2 O00. p 1m t-15 mm 3 kN m 20 kN 1 m 20 kN 50 mm 100 mm 50 mm 20 kN 50 mm dia 2 m Y3 kN m o0000000

Answer 1

Intemal pressure = 3MN = 3 N/mm2 m2 b=3MN/m2 t=15mm A PD Hoop or circumferential stress D Im 2t 100 N/mm F 3x1000 10ONImm2 2X15 Longitu dnal stress = PD_ 4t 50 N/mm 3 x1000 50 Nmm2 50N/mm2 2*15 100NIMM2 20MN 2> 0 mm Point A lies on extreme fibre A No transverse shear stress atA 10omm due to transverse lood of 20KN 2m MA=20X103x j000 = 20X106 N-mm MA YA Bending stress at A INA X1000 20X10X50 = 24 N/mm2. 50 X 1003 24N/mm2 A 24 N/mm 12 Point B lies on neutral fibre so no bendine stress at B PLA71 Transverse shear stress at B = bINA = 20x103x (50x50) x 25 6N/mm2. 50X100 50 X 12 6N/mm2 At A 3KN-m 20 KN axial for ce will induce compressive 20KN force =20x103 =10.186 N/mm2 TT4X 502 50mm dia 3KN-m twisting moment will induce torsiona Shear stress 3KNM 6 16T = 16X 3x ID 122.23 N/mm2 Td3 TTx 503 10-186 N mm2 10-186 N/mm2 122-23 N/mm2 beooo