Page 222 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
Brown˙C05
Brown.cls
204
U.S. Customary 14:35 STRENGTH OF MACHINES SI/Metric
determined. Substituting gives determined. Substituting gives
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
[10 + (−3)] kpsi [75 + (−25)]MPa
= =
2 2
[10 − (−3)] kpsi [75 − (−25)]MPa
◦
◦
+ cos 2(−15.8 ) + cos 2(−15.5 )
2 2
◦
◦
+(−4 kpsi) sin 2(−15.8 ) +(−30 MPa) sin 2(−15.5 )
◦
= (3.5 kpsi) + (6.5 kpsi) cos (−31.6 ) = (25 MPa) + (50 MPa) cos (−31.0 )
◦
◦
◦
+(−4 kpsi) sin (−31.6 ) +(−30 MPa) sin (−31.0 )
= (3.5 kpsi) + (6.5 kpsi)(0.852) = (25 MPa) + (50 MPa)(0.857)
+(−4 kpsi)(−0.524) +(−30 MPa)(−0.515)
= (3.5 + 5.5 + 2) kpsi = (25 + 43 + 15) MPa
= 11 kpsi = σ 1 = 83 MPa
So the rotation angle found in step 5 is for the So the rotation angle found in step 5 is for the
maximum principal stress (σ 1 ). maximum principal stress (σ 1 ).
Step 7. Using Eq. (5.11), the rotation angle Step 7. Using Eq. (5.11), the rotation angle
(φ s ) for the maximum shear stress becomes (φ s ) for the maximum shear stress becomes
◦
◦
◦
◦
φ s = φ p ± 45 =−15.8 ± 45 ◦ φ s = φ p ± 45 =−15.5 ± 45 ◦
= 29.2 ◦ or −60.8 ◦ = 29.5 ◦ or −60.5 ◦
where for reasons that will be presented in the where for reasons that will be presented in the
discussion on Mohr’s circle, the negative value discussion on Mohr’s circle, the negative value
(−60.8 ) will be chosen. (−60.5 ) will be chosen.
◦
◦
Step 8. Display the principal stresses (σ 1 ) and Step 8. Display the principal stresses (σ 1 ) and
(σ 2 ) found in step 3 at the rotation angle (φ p ) (σ 2 ) found in step 3 at the rotation angle (φ p )
found in step 5,and verified in step 6, in a rotated found in step 5, and verified in step 6, in a rotated
element. element.
4 33
0 0
74.2° 74.5°
11 83
–15.8° –15.5°
11 83
0 0

4 33