Question

A structure has the stresses on its most critical point (a) as shown. If the yield strength for the material is Sy=420 MPa. Determine the following 1. Draw Mohr's Circle for the stress element 2. Principal Stress element showing all stresses and angle 3. Maximum Shear Stress element showing all stresses and angle 4. Determine the factor of Safety of this structure against yielding. 5. If the structure fails, what would you do to make it Safe? 0x = 500 MPa Txy = 300 MPa Oy = 300 MPa

Answer 1

Solution, Tx=500mpa IH Tay= 30ompa I 300.041090D (300,0) 102, Tylyield Strength) = 420 MPa. o Mohr circle. 300 -500,0) Ty=300MPa T(MPa) o(MPa) (ه,) Principle stresses and angles. -100+400)? +(300)? -100 + 500 1400 MPa. 2 2 600 MPa. DI -100 - Vi400)4+ (300) no tan (2-09.) = 1300 400 > Op, = 18.43° clockwise OP2 got opi = 108.43°(clockwise from a-plane.

(3 manimum Shear stress (Iman) Radius of Mohr circle 4ooj +(300)? 2 SDO MPa. of plane angler at which monimum shear stress occures opi + 45º = 45+18.43 = 163.43 in clockwiser direction from X-Plane 4 Factor of safety against yeelding → Sy lul FOS = 420 600 a 0.7 - 3. To make structure safe, the safe, the metal strip is provided sourroundings to the material which will take some stress. Soba algoud