Page 216 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
Brown˙C05
Brown.cls
198
U.S. Customary 14:35 STRENGTH OF MACHINES SI/Metric
and so the minimum shear stress (τ min ) from and so the minimum shear stress (τ min ) from
Eq. (5.13) is Eq. (5.13) is
τ min =−τ max =−2.4 kpsi τ min =−τ max =−18.85 MPa
Step 5. Using Eq. (5.14), calculate the average Step 5. Using Eq. (5.14), calculate the average
normal stress (σ avg ) as normal stress (σ avg ) as
σ xx + σ yy (4.8 + 9.6) kpsi σ xx + σ yy (37.7 + 75.4) MPa
σ avg = = σ avg = =
2 2 2 2
14.4 kpsi 113.1MPa
= = 7.2 kpsi = = 56.55 MPa
2 2
Step 6. Display the maximum and minimum Step 6. Display the maximum and minimum
shear stresses found in step 4, the average stress shear stresses found in step 4, the average stress
found in step 5, and the rotation angle (φ s ) found in step 5, and the rotation angle (φ s )
chosen in step 3, on a rotated element. chosen in step 3, on a rotated element.

7.2 2.4 7.2 56.55 18.85 56.55

45° 45°

–45° –45°

7.2 2.4 7.2 56.55 18.85 56.55

Step 7. As a final check on the calculations, Step 7. As a final check on the calculations,
use Eq. (5.15) to find the maximum principal use Eq. (5.15) to find the maximum principal
stress (σ 1 ) as stress (σ 1 ) as
σ 1 = σ avg + τ max = (7.2 + 2.4) kpsi σ 1 = σ avg + τ max = (56.55 + 18.85) MPa
= 9.6 kpsi = 75.4MPa
anduseEq.(5.16)to findtheminimumprincipal anduseEq.(5.16)to findtheminimumprincipal
stress (σ 2 ) as stress (σ 2 ) as
σ 2 = σ avg − τ max = (7.2 − 2.4) kpsi σ 2 = σ avg − τ max = (56.55 − 18.85) MPa
= 4.8 kpsi = 37.7MPa

In Chap. 4 on combined loadings, a statement was made at the end of most of the examples
that the stress element shown would be the starting point for discussions in Chap. 5. Consider
one of these elements, one which does not have the unrotated shear stress (τ xy ) equal to
zero, nor a pressure acting on either side of the element, as was the case for Example 3.