Question

Question 4 20 pts Determine the equivalent state of stress on an element at the point which represents (a) the principal stresses and (b) the maximum in-plane shear stress and the associated (c)average normal stress. Also, for each case, determine the corresponding "orientation" (angles) of the element with respect to the element shown and sketch the results on the element. 80 MPa 40 MPa Upload Choose a File

Answer 1

The Standard formalae have been used to solve this problem. Detailed solution is as shown in the attached images.

Eq- → Pg -2 2(-40) tan 20 p = 0 0 - 20p = 135° 1 Op = 67.5° ) or 200 = 315 lop = 157.5 principle planes are oriented 67.50 and 157.5° anticlockwise direction to the given plane . The in 4 : * Principle stresses: Put op, & Ope in eq (O), use the following formule ito = 0, VI = Calen) + Jersey + Pay = (20+ J1200)*+ [1034 - 40 I / 2x 40² (OP=1515 6.568 - 16.568 (66 = 61.5°) The principle stresses are 196.568 MPa & - 16.568 mpell J +16.568 MPa (compressive) negative sign that stress compressive say is N 96.568 MPa (tensile) (b),() Man. shear strus plone, For shear stress to be maximum, dho =0 → → prono) Coszo.cz) + Toy Ebin20)(2) = 0 tar 20 - (on-oy) 2 Tony

here, ton 20 = -280-0) - -u042 → 20 - us', 225" - 10 = 22.5° or 112.5° ) ". Shear stress is monimum in planes at angles 22.50 and 112.5° in anticlockwise direction, Putting these 'o' values in eq - , we get a mon = 100 1.22 = 1-(800) sin(us) - 4D cosus) = (4052 MPa (blans) Saug = 1.-22.5 est = oto ) + (60-8) 1 / 2 40 1 / 2 = 40 mia. Cans (c)) en the man i. The man associated shear average stress normal is nota mea, and stress is 40 MPa 112.5 ang = do mla ve mon = 4012 mea. (22.50 Tang=hompa