Page 218 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
Brown˙C05
Brown.cls
200
U.S. Customary 14:35 STRENGTH OF MACHINES SI/Metric
and use Eq. (5.16) to calculate the minimum and use Eq. (5.16) to calculate the minimum
principal stress (σ 2 ) as principal stress (σ 2 ) as
σ 2 = σ avg − τ max = (7.65 − 8.54) kpsi σ 2 = σ avg − τ max = (73.35 − 82.02) MPa
=− 0.89 kpsi =− 8.67 MPa
Step 4. Before going further, check that the Step 4. Before going further, check that the
values for the principal stresses (σ 1 ) and (σ 2 ) values for the principal stresses (σ 1 ) and (σ 2 )
satisfy Eq. (5.17). satisfy Eq. (5.17).
σ 1 + σ 2 = σ xx + σ yy σ 1 + σ 2 = σ xx + σ yy
(16.19 − 0.89 ) kpsi = (15.3 + 0) kpsi (155.37 − 8.67) MPa = (146.7 + 0) MPa
15.3 kpsi ≡ 15.3 kpsi 146.7MPa ≡ 146.7MPa
and they do. and they do.
Step 5. Using Eq. (5.9), calculate the rotation Step 5. Using Eq. (5.9), calculate the rotation
angle (φ p ) for maximum and minimum princi- angle (φ p ) for maximum and minimum princi-
pal stresses as pal stresses as
2τ xy 2 (−3.8 kpsi) 2τ xy 2 (−36.7MPa)
tan 2φ p = = tan 2φ p = =
σ xx − σ yy (15.3 − 0) kpsi σ xx − σ yy (146.7 − 0) MPa
−7.6 kpsi −73.4MPa
tan 2φ p = =−0.497 tan 2φ p = =−0.500
15.3 kpsi 146.7MPa
2 φ p =−26.4 ◦ 2 φ p =−26.6 ◦
φ p =−13.2 ◦ φ p =−13.3 ◦
Step 6. Without the benefit of the graphical Step 6. Without the benefit of the graphical
picture of Mohr’s circle, the only way to tell picture of Mohr’s circle, the only way to tell
which principal stress this value of the rotation which principal stress this value of the rotation
angle (φ p ) is associated with, is to substitute angle (φ p ) is associated with is to substitute
this angle in Eq. (5.1) and see which stress is this angle in Eq. (5.1) and see which stress is
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
(15.3 + 0) kpsi (146.7 + 0) MPa
= =
2 2
(15.3 − 0) kpsi (146.7 − 0) MPa
◦
◦
+ cos 2(−13.2 ) + cos 2(−13.3 )
2 2
◦
◦
+(−3.8 kpsi) sin 2(−13.2 ) +(−36.7MPa) sin 2(−13.3 )
= (7.65 kpsi) = (73.35 MPa)
◦
◦
+(7.65 kpsi) cos (−26.4 ) +(73.35 MPa) cos (−26.6 )
◦
◦
+(−3.8 kpsi) sin (−26.4 ) +(−36.7MPa) sin (−26.6 )
= (7.65 kpsi) = (73.35 MPa)
+(7.65 kpsi)(0.896) +(73.35 MPa)(0.894)
+(−3.8 kpsi)(−0.445) +(−36.7MPa)(−0.448)