Question

Solid Mechanics 3 Assessment Task 1a - 2020 Milestone a Question 1. For each of the plane-stress conditions given below, construct a Mohr's circle of stress, find the principal stresses and the orientation of the principal axes relative to the xy axes and determine the stresses on an element, rotated in the x-y plane 60° counterclockwise from its original position: (a) dx = 200 MPa Oy - 300 MPa T .40 MPa (b) dx = 500 MPa Gy = -80 MPa Exy 400 MPa Question 2. For each of the plane-stress conditions given in Question 1, using the matrix transformation law, determine the state of stress at the same point for an element rotated in the x-y plane 60° counterclockwise from its original position. Question 3 The state of stress at a point of an elastic solid is given in the x-y-z coordinates by o. , 0 90 180 0] lol 0, 0 180 2400 MPa 0 0 0 0 0 120 Using the matrix transformation law, determine the state of stress at the same point for an element rotated about the z-axis (in the x-y plane) 30° anticlockwise from its original position Calculate the stress invariants and write the characteristic equation for the original state of stress Calculate the deviatoric invariants for the original state of stress Calculate the principal stresses and the absolute maximum shear stress at the point Calculate the stress invariants and the characteristic equation from the transformed state of stress

need solution for milestones A Q1

Answer 1

Solution- a) 5x = 200 MPa 6) by = 300 MPa c) Tay = -40rpa Contreof Mohe's Circle - [Cestne),o] - [so, c] -(-07 Radius of Mohr's Circle 1602200)*+ €40) - 64.03 MP- Scale - = 2.56 cm X anis 1m = 25 MPa Y axis 1 cm- 25 MPa Ymax. 1 10 120 19 14 15 (2017 x 1300, 40) Principal stresses T = 314 MPa T2 = 186 MPa Ymare 64 Mra

from the figure tan 20p= TP / 200 op: tan -25.67 X4 Plane Rotate 6o Counter clockwire direction the stresses are $ - 180-60x2) - (2 x 25-67) - 866° Tz' → 250 - 64 C5 (866) = 186.729 MPa Ty → 250 + 64 cos (66) = 313. 270 MPa Day' 64 Cos (8.66) = 63.270 MPa un 186MPA Tmax 314 MPa 313.210 MP x 186.729MP = 63-210 MP 0.60

6 Tu= 500MPa: Sy = - COMPa day = 400 MPa Contu ot Mohr Grebe · [fet 4,0] - [8 * , •7+ (2.,o] Reclusos Platha's Curcle - R- / (6:59) + tap? - Centre Tango Radius of Mohs's Grcle & = 494.06 Sale :- X axis 1cm= 100 MPa Y anis 1cm= 100MPa = (2.1,0) R=4.94 cm. mat 2 X 3 4 5 6 8 Principal stresses T = 104 MPa 12 = -284 MPa a = 494.06 MPa

from the figure tan 20p = = tant (405) - 20p Op = 27-02 2, 4 Plane Rotate 60° So. 0 = 180 - (Fox2) - (2x270_) 5.96 vzp= 210 - 494.06 Cos (596) = -281.381 MPa by = 210 + 494.06 Cos (5.96) = 901.389 MPa Tay = 494.06 Cos(5.96) = 491.359 MPa Principal stresses - eleunt strenes After sokation. -