Question

The state of stress at a point on a body is given by the following stress components: 0 = 15 MPa, Oy = -22 MPa and Try = 9 MPa Matlab input: sx = 15; sy = -22; txy = 9; 1) Determine the principal stresses 01 and 02. 1 = MPa 02= MPa 2) Sketch the principal stress element, defined by the rotation @pl. y Enter the rotation @pi (-360º < Opl < 360°): Opl = Add stress components: C Delete button: x Note: the angle Aplis defined by the angle from the c-axis to the normal vector of the plane with maximum normal stress (vector pı). The principal stress g1 is exerted at the plane defined by Pi. 3) Determine the absolute value of the maximum shear stress Tmaz. Tmax = MPa 4) Sketch the maximum shear stress element, defined by the rotation 0g. Enter the rotation 0(-90° <0, < 90°): 0,= Add stress components: Delete button: x

Answer 1

Gineni x= 15 MPa y = -22 MPa Txy 9 MPa Principal stressed 89,2= ² a toj + scogoodt + 4 Exy² لا - = = ( 15-22 + 1615 +22)2 +4xg? 9,= 17.073 MPa [ 15-22 Jis +22² + 4xg² I o 8 = -24.073 mpa Location is given by, 2 Principal plane Tan20p - 2 lxy og o x = 2 X 9 15 +22 12.97 po 5=24.073 MPa D47.0.73mpa 5€ 17.072 mpa 12.97 12.97 0=15mpa X Oy=+22 mpa

② Absoulute Tmax = larger q *5*112/19/13 1/17,073 = Larges -24,073 17-073 + 24.073 f 2 2 Absolute Maximum Shear stress a 20.573 mpal

is shear stress Location of maximiem 14 ginen hen by tan 20 -1 (15 +22) 2 Exy 2x9 ofroy 0 = - 32.03 S 9 s*=htra 2 = 17.073 -24.073 2 32.09 2 Iman =-3.5 MPa