Page 211 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
14:35
Brown.cls
Brown˙C05
PRINCIPAL STRESSES AND MOHR’S CIRCLE
U.S. Customary SI/Metric 193
σ xx + σ yy σ xx − σ yy σ xx + σ yy σ xx − σ yy
σ x x = + cos 2θ σ x x = + cos 2θ
2 2 2 2
+τ xy sin 2θ +τ xy sin 2θ
(4.8 + 9.6) kpsi (37.7 + 75.4) MPa
= =
2 2
(4.8 − 9.6) kpsi (37.7 − 75.4) MPa
◦
◦
+ cos 2(60 ) + cos 2(60 )
2 2
◦
◦
+(0 kpsi) sin 2(60 ) +(0MPa) sin 2(60 )
(14.4 kpsi) (113.1MPa)
= =
2 2
(−4.8 kpsi) (−37.7MPa)
◦
◦
+ cos (120 ) + cos (120 )
2 2
◦
◦
+(0 kpsi) sin (120 ) +(0MPa) sin (120 )
= (7.2 kpsi) = (56.55 MPa)
+(−2.4 kpsi)(−0.5) +(−18.85 MPa)(−0.5)
+(0 kpsi) +(0MPa)
= (7.2 + 1.2 + 0) kpsi = (56.55 + 9.43 + 0) MPa
= 8.4 kpsi = 66.0MPa
Step 5. Similarly, using Eq. (5.2), calculate the Step 5. Similarly, using Eq. (5.2), calculate the
rotated stress (σ y y ) from the unrotated stresses rotated stress (σ y y ) from the unrotated stresses
given in step 4. given in step 4.
σ xx + σ yy σ xx − σ yy σ xx + σ yy σ xx − σ yy
σ y y = − cos 2θ σ y y = − cos 2θ
2 2 2 2
− τ xy sin 2θ − τ xy sin 2θ
(4.8 + 9.6) kpsi (37.7 + 75.4) MPa
σ y y = σ y y =
2 2
(4.8 − 9.6) kpsi (37.7 − 75.4) MPa
◦
◦
− cos 2(60 ) − cos 2(60 )
2 2
◦
− (0 kpsi) sin 2(60 ) − (0MPa) sin 2(60 )
◦
(14.4 kpsi) (113.1MPa)
= =
2 2
(−4.8 kpsi) (−37.7MPa)
◦
◦
− cos (120 ) − cos (120 )
2 2
◦
− (0 kpsi) sin (120 ) − (0MPa) sin (120 )
◦
= (7.2 kpsi) = (56.55 MPa)
− (−2.4 kpsi)(−0.5) − (−18.85 MPa)(−0.5)
− (0 kpsi) − (0MPa)
= (7.2 − 1.2 − 0) kpsi = (56.55 − 9.43 − 0) MPa
= 6.0 kpsi = 47.1MPa