Page 219 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
14:35
January 4, 2005
Brown.cls
Brown˙C05
PRINCIPAL STRESSES AND MOHR’S CIRCLE
U.S. Customary SI/Metric 201
= (7.65 + 6.85 + 1.69) kpsi = (73.35 + 65.58 + 16.44) MPa
= 16.19 kpsi = σ 1 = 155.37 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 =−13.2 ± 45 ◦ φ s = φ p ± 45 =−13.3 ± 45 ◦
= 31.8 ◦ or −58.2 ◦ = 31.7 ◦ or −58.3 ◦
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
(−58.2 ) will be chosen. (−58.3 ) 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.
0.89 8.67
0 0
76.8° 76.7°
16.19 155.37
–13.2° –13.3°
16.19 155.37

0 0
0.89 8.67

Step 9. Display the maximum and minimum Step 9. Display the maximum and minimum
shear stresses found in step 2, the average stress shear stresses found in step 2, the average stress
found in step 1 at the rotation angle (φ s ) chosen found in step 1 at the rotation angle (φ s ) chosen
in step 7 in a rotated element. in step 7 in a rotated element.

7.65 73.35
7.65 73.35
31.8° 31.7°
8.54 82.02
8.54 82.02
–58.2° –58.3°
7.65 73.35
7.65 73.35