§13.6 Complex Stresses 333 AB 28 No M B=28 02 BC Fig.13.9.Mohr's stress circle. (4)Join AB and BC. (5)The point P where this line cuts the a axis is then the centre of Mohr's circle,and the line is the diameter;therefore the circle can now be drawn. Every point on the circumference of the circle then represents a state of stress on some plane through C. Proof Consider any point on the circumference of the circle,such that PO makes an angle 20 with BC,and drop a perpendicular from O to meet the o axis at N. Coordinates of O: ON OP+PN=(ax+a)+R cos(20-B) =(x+)+R cos 20 cos B+R sin 20 sin B But R cosB=(ax-)and R sinB=txy ON =(x+a)+(ax-)cos 20+txy sin 20 On inspection this is seen to be eqn.(13.8)for the direct stress on the plane inclined at to BC in Fig.13.5. Similarly, QN R sin (20-B) =R sin 20 cos B-R cos 20 sin B =(ox-a)sin 20-txy cos 20 Again,on inspection this is seen to be egn.(13.9)for the shear stress te on the plane inclined at 0 to BC.$1 3.6 Complex Stresses 333 I I c v 01 1 I I Fig. 13.9. Mohr’s stress circle. Join AB and z. The point P where this line cuts the a axis is then the centre of Mohr’s line is the diameter; therefore the circle can now be drawn. circle, and the Every point on the circumference of the circle then represents a state of stress on some plane through C. Proof Consider any point Q on the circumference of the circle, such that PQ makes an angle 28 with E, and drop a perpendicular from Q to meet the a axis at N. Coordinates of Q: ON = OP+PN = ~(0,+a,)+R~0~(28-fl) = $(ax + a,) + R cos 28 cos p + R sin 28 sin p But R cos p = +(a, - a,) and R sin p = T,~ .. On inspection this is seen to be eqn. (13.8) for the direct stress BC in Fig. 13.5. ON = +(ax + a,) ++(a, - a,)cos 28 + T,, sin 28 on the plane inclined at 8 to Similarly, QN = R sin (28 - p) = Rsin28cos~-Rcos28sin~ = +(ax - a,) sin 28 - T,~ cos 28 Again, on inspection this is seen to be eqn. (13.9) for the shear stress TO on the plane inclined at t) to BC