Question

Dy = 40 ksí 30ksi SPERS2o2a C6340-01 | iTwº Due_03/31/2ozsTue, 03/24/2a2O (use the class sign convention, PA Prob#1(25 pts).U se compass and ruler. 19 For the state of Plane stress shown, find (a) the principal Planes, 60 R$i (b) the principal 5 tresses on the planes in (a), (c) the maximum Jhear stress and its associated normal stress. Lis the sign Convention for classe'.

Answer 1

Soph: sign convention : compression -ve' tension tue' -40KSC - - 30 ksi given : ox= 60 ksi y = -40ksi Trey = -30 ksi -Goksi Qops V 10ksi 60.31=opi ksi . Principal -48.31=002 ksi planes maxon stress shear plane principal planes: (a) Location of - tan 2 Opt =- 2 Txey >> Opi = 1 taāt lieu de los → opi = -15.48°] Op2= Opi+90° = -15.48 + 909 Op2 = 74.52° (6) magnitude of principal stresses : opilop2 = ( Ox ) Ź Cose-oy)2 + 4 Esey2 = ( 60+ (-40)) + < 100-(-40))2 + 4 (-30)2 X 2

spilop2 = 10 + 58.3095 opie 60-31 ksi? 10p2 = -48.31 ks 47 cross check : Sin Ox'= x costo + rysino + 2 trysino.coso = 60 cos? (74-52°) + (-40) sin? (74-52") +2(-30) sin (14-12") cos (74.52) lose' = -48-31 ksi so for Op2 = 74-52 = Op2 = -48,31 ksi Opi= -15.48° = Opi= 60.31 ksi eo S maxim shear stress (Imax): 60.31-(-48-31) Tmax = Ojos op2 = 6 2 Tmax = 54-31 ksi normal stresses on plane of max shear stress : r: Ox toy - 60+,(-40) To = 10 ksi 2
summary-

a) principal planes= -15.48° & 74.52°

b) principal stresses= 60.31 ksi & -48.31 ksi

c) max shear stress= 54.31 ksi

Normal stress on plane of max shear stress= 10 ksi