Question

Q2. The state of stress on the surface of part of an engineering component is shown in Fig. 22. 94 MPa 51 MPa 25° 63 MPa Fig. Q2 - The State of Stress on the Surface of an Engineering Component (a) Using graph paper construct a Mohr's Stress Circle for the element and hence determine the magnitudes of the principal stresses and orientations of the principal planes. Clearly label these features on your diagram and sketch the state of stress for an element lying on the principal plane. [14 marks) (b) What is the maximum shear stress in the plane of the element and is this the maximum shear stress in the material? If not, what is the maximum shear stress? (5 marks] contd./

Answer 1

qumpa Sol Giren data, B Simpa. -A a - 63MPa 63mpa 63mpa.t Ty = 94 mpa. Tocy impa " SIMPa. qumpa et sign Convension, Ave. -Ve. + C= 63 +94 ,5) = (705.0) 2 2 Her to each other. *. A,B are planes, which are A = (0,~) = (63, -sı) ; B = (0, 1) = (94,51). diameter of Mohr Circle. Distance ble A and B is Radius R= Txury 9 t Isy 63-94 +(51) 53.3 Mpa 2

( Tang, Taman) = (78.5, 58.3) .. (94,51) P ( 131:28, 7.42) 78-5 25.2 OS pl.-1318 53-3 781510 ( 63, 51) 5 = 1318MP; T2 = 25.2 mpa i tan 20p, 20p, 73.61 78-63 In component op, = 36.8° from A 2001 20p, t. 180 20% 253.61 >> Op. 126.8. frond! 5 F 131.8 tipa od 25.2 mpa 13650 25.2 -T2 9 = 131.8iupa

b ) Tmax (in-plane) - R- c 0-0 = 53-3 Mpa 2 8₂-0 Habs) in Material - Max lord {15 or 1663 | --La 131.8 1370 Tabs = 65.9 apa from 'B' bro in 2 R "Normal Stress will be CP's R. Cos go = 53.3.Coggi - 52-78 = pp' = R singo 53.3 sing 7.42 from origin (0,r) (78.5 to $2.78 7.42) ) = (131:28, 7:46) 0.5 x 200 stress Given; 100 mpa Max Normal as Not Sk@ satisfied (131-28 > 100) not safe