At a point on the free surface of a machine component the strain rosette shown in the figure below 60°60°For this PLANE STRESS condition:a) Determine the strain state (Ex,EyıYxy and E,) at this poinb) Determine the stress state (o,0y, Txy) at this pointc) Determine the principle angle, all the three principle strains, the shear angle, the maxin plane shear strain and the absolute max shear straind) Show the principle strains &p1,&p2 and the max in plane shear strain on an appropriateschematice) Determine all the three principle stresses, the max in plane shear stress, averagenormal stress and the absolute max shear stressf) Show the principle stresses op1Op2, the max in plane shear stress and the averagenormal stress on an appropriate schematicNOTE: In the case that an appropriate diagram(s) is (are) not properly drawn to guidelexplain your effort inwriting any comations there will be no credit nwanded for wour anwr.in Onestion 2 Ploalailioent Posette showWn in the figure below mecasures the following normal strain valuesEa = 710 µɛ Ep = 1005 µE E=75 µe%3D6060°aFor this PLANE STRESS condition:a) Determine the strain state (E,Ey,Yzy and e,) at this poi..b) Detemine the stress state (0,0y, Tyy) at this poin.c) Determine the principle angle, all the three principle strains the shear angle, the maxin plane shear strain and the absolute max shear strainShow the principle strains &,E and the max in plane shear strain on an appropriate

Solution “Since you have asked multiple question, we will solve the first question for you. If youwant any specific question to be solved then please specify the question number or postonly that question.given that εa= 740 μεεb= 1005 μεεc= 75 με We know that normal strain in any direction θ is given byεn=(1/2)(εx+εy)+(1/2)((εx-εy)cos 2θ+(γxy/2) sin 2θso normal strain at θ=00εa=(1/2)(εx+εy)+(1/2)((εx-εy)cos 2(0)+(γxy/2) sin 2(0)710 = (1/2)(εx+εy)+(1/2)((εx-εy)cos 2(0)+(γxy/2) sin 2(0)710= εx.........................(1) normal strain at θ=1300εb=(1/2)(εx+εy)+(1/2)(εx+εy) cos 2 (1300)+ (yxy/2) sin2(1300)1005=εx/2 + εy/2 + (εx/2 + εy/2)(-0.17)+ (yxy/2)(0.34)After solving710.35 = 0.415 εy+0.17 γxy...............................(2)normal strain at θ=2500εc=(1/2)(εx+εy)+(1/2)(εx-εy)cos2θ+(γxy/2) sin2θ75=(1/2)(εx+εy)+(1/2)(εx-εy)cos2(250)+(γxy/2) sin2(250)After solving-145.1= 0.69εy+0.32γ...................(3) Equating all the equation find the value εx= 710 εy= 1.62*104 γxy=-3.5*104