Question

40 M 45 MP 50 MPA - For the given state of stress, Part A: determine analytically (using stress transformation equations): 1) the principal planes. 2) the principal stresses. 3) Sketch the stress element for the above condition 4) the orientation of the planes of maximum in-plane shearing stress, 5) the maximum in-plane shearing stress and the corresponding normal stress. 6) Sketch the stress element for the above condition Part B: Only use Mohr's circle to determine 1) the principal planes and the principal stresses. 2) the orientation of the planes of maximum in-plane shearing stress, 3) the maximum in-plane shearing stress and the corresponding normal stress. 4) Write down briefly the steps of drawing your Mohr's circle. (Show all relevant calculations) Points 50 Distribution Part A Calculation of all principal stresses and planes: 5 Sketch of the above element: 5 All required calculations for maximum shear stress criteria:5 Sketch of the above element: 5 Part B Drawing of the Mohr's Circle: 10 Determination of stresses and orientation: 15 Writing the steps of drawing: 5

Answer 1

Sign Convention- are positive outwards (2) C is popitive in anticlockwise direction. (1) xy How, given = - 60 mpa - – 40 mpa Tory 45 mpa

part a- stress transformation equesions, (1) + (8-5) 6320 + sy since (ox-by) sincot try C8320 て。 (1) The principle plomes, (at op, , OP2) tan 20p, Bay 2 Cry 2 xЧ tanzopi = -4.5 (-60 +40) => 20p, 77.47 38.73 51-260 and op = op t go = Her, o is measured from vertical plane in anti- clockwise direction. (2) Principle stresses -50-40) + (-) <3(-79-42) + us sin(-37-4 = -50 - 2-1695 - 43.928 - 96.097 mpul (at o - - 38.73) toyo) (-60340) (3 (10 (83 (102.52) + 45šin (102.5 = -50+ 2.1695 + 43.9239 -3.9005 mpal (at o = 31.26 51-26) Son + Top

(3) 38-73 96.097 mpa 3.9 mpa (4) (60+40)/2 To = tan osi Gry 20s, 12.5288" os = 6.2644° 099 96.2644° -(-60740) sin (12.5288) + 45 cos (12.5288) cos 2 2.1693 + 43.9284 max. in plane shear stress Tosi 46.0977 MPa = (-80440) cos (12.5288) + 45 sin (12.5288) (-60-40) Sosi + 2 = -50 + (-9.76187) + 9.76187 = -50 mpa Josu (In max shear stress plane both normal stresses are same)

46.09 mpa sompa Com sompa Port Bir 20 in mohr's For the sign convention we have used, circle is measured clockwise Centre of motio's Circle of (Tory ,o) - (atry, o) (*) - (-50,0) 2 04 Radius of mohr's Circle to + Cruz + 46.097 mPa. position of point A = position of point 3= (by, say) = (-40, -45) (-60,45) A Trax A appo (-50,0) -10 c (90) B in element is in Mohrs circle angle o

(1) for principle planes and principle stresse from Mohr's circle -45 9.5 tan 200 -10 38.73 20 p = - 47.47 2002 - 20pt 180 Op ₂ = 51.26° principle stresses, of -50- 46.097 96.097 MPa Parg tr 2 - sot 48.097 - 3.903 mpa. (2) АА tanzos, 510 AC 205, = 12.5288 os z 6.2644 Os2 - 96-26440 20s₂ = 205, + 180 roh 6 or -sor Targ - Tmax =tr 46.096 MPa corresponding normal stress (4) Stepsi (i) befine co-ordinate system and sign convention (ii) Find centre (C) of circk (ii) & ind radius (8) of circk (iv) Plot A and B to be considered as a base plane