Question

Question 3 68 MPa 56 MPa 12 MPa Figure 3. Stress components acting on an element. Figure 3 shows an element experiencing several stress components. Determine the following: 1. The stress components oxx, y, and Tyyacting on the element oriented at a counter clockwise angle 0 = 30° from the horizontal x axis 2. The principal stresses, the maximum shear stress and their associated angles Show all results on sketches of properly oriented elements. Note: Solutions MUST be obtained using Mohr's circle ONLY.

Answer 1

Detailed answer of above asked question is solved as below.

Given T 68 MPa. by = 12 MPa by -- 68 MPa Zxy = 56 MPa 12 Mpa 30 Put lem= lompa 56 MPa 6x = 12 = 12cm by = - 68 = -6.8cm Zxy = 56 5o = 5-6cm = 30 2 0 = 600 CCW. Mohor's circle > max. shear stress Emax & v'y' = 68MPA by 12Mpa. P. А 0 E S 20=60 Nommal Stress be H Ania major principal stress bne = minor principal stress → Steps 1) Draw origin o 4 toke suitable scole re 1cm= 10 mipat. plot s at &xx' = 12 MPa = 1.2cm from 'olf point pat by y'='68 MPa = -6.8cm from 'o' 2) From poind s4p draw perpendicular at exy = 56 MPa = 5-6 cm. in opposite direction 1.0 PQ = SR = 5.6cm. 3 join @4R and mark 'cl as centre of circle on horizontal x-axis where line R intersect to xraxis

4) Draw circle with radius (R=CQ from centre c 5) From point cto Diret tox-axis or radius represent max shear stress 6). From centre 'cl Draw line at 20=2x30 = 60'iech in 4th Quadrant, because of cow orientation of plane & Draw perpendicular to H de HE f). How, measurement from mohr's circle represent following values principal simesses -Major principal stress Ani= oAxscale = 9.65810 = 96.5 Mpa. -Minor principal stress = bn2= 6 Bx scale = 4:1810 = 41 MPa 2) Max. shear shress = Zmax= coxscale='6.75x10=67.5 Mipa 3) Normal Shress = 6n=oExscale = 0.4X10 4) Tangential stress = Ot= H Exscale s) Resultant stress = 68:0HX scale = 6x10 6) Angle of obliquity=0=LHOE = 840 = 17 MPa 5.9 x 10 = sompa = Gompa.