Question

Q4. (a) Plot a Mohr circle for the plane stress element which has been rotated 40 degrees in the anti-clockwise direction as shown in Figure Q4(a). (8 marks) (b) By using Mohr Circle, determine and sketch; all the stresses (Ox, O, and tyy) of the same element in the orientation as shown in Figure Q4(b). (4 marks) (i) the principal stress and its orientation. (4 marks) (i) the maximum shear stress, the corresponding normal stress, and its orientation. (4 marks) (20 marks) 60 MPa 30 MPa 190 Oy 85 MPa Exy FIGURE 04(a) FIGURE Q4(b)

Answer 1

QUESTION 4 (a)

In the above given question the element is rotated by angle $\Theta$ in anticlockwise direction so here we need to draw Mohr's circle for the given element

The problem is solved in the following images:-

Qu. 4(a) Tn Bomia oy - 60 rupa Try = 8empa la Sorra > Bompo 18 SMPC Centre - = 30-60 = -15 MPa 2 2 Radius: Cry on't a Tag? : /6-60-30f +4x83 2 - 96.18 MPa AY (30,85) Here cam=20mPa. pe مرد کا (89,0) O 10 28 20pi 660,85)

The question is solved by drawing Mohr's circle and the value of $\sigma$ _p1 = 81 Mpa and value of $\sigma$ _p2 = -111 Mpa

now QUESTION 4(b) (i)

As element is now rotated by 40^{$^{\circ}$} to clockwise direction the value of rotation will be 80^{$^{\circ}$} in mohr's circle and so value of  $\sigma$_x = 81 Mpa and value of $\sigma$ _y = -111Mpa and value _xy = 0

7

4 (b) (ii)

it is also the value of principle stress and its orientation is $\theta$ _p= 100^{$^{\circ}$}  

4 (b) (iii)

The value of maximum shear stress
7
_max= ($\sigma$_{p max} - $\sigma$ _{p min} )/2

= [81-(-111)]/2 = 96 Mpa

By mohr's circle the radius of mohr's circle will give maximum shear stress value.