Page 220 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
14:35
Brown.cls
Brown˙C05
STRENGTH OF MACHINES
202
As already mentioned, it is very difficult in actual practice to have a combination of
loadings that produce nonzero values of all three stresses (σ xx ),(σ yy ), and (τ xy ) on a plane
stress element. However, to provide another example using the equations presented in this
section, consider the following rather contrived set of unrotated stresses.
U.S. Customary SI/Metric
Example 5. For the normal and shear stresses Example 5. For the normal and shear stresses
given below, find the principal stresses (σ 1 ) and given below, find the principal stresses (σ 1 ) and
(σ 2 ), maximum and minimum shear stresses (σ 2 ), maximum and minimum shear stresses
(τ max ) and (τ min ), and the special angles (φ p ) (τ max ) and (τ min ), and the special angles (φ p )
and (φ s ), and display these values in appropri- and (φ s ), and display these values on appropri-
ate rotated plane stress elements, where ate rotated plane stress elements, where
σ xx = 10 kpsi σ xx = 75 MPa
σ yy =−3 kpsi σ yy =−25 MPa
τ xy =−4 kpsi τ xy =−30 MPa
displayed in the following element: displayed in the following element:
3 25
4 30

10 10 75 75

4 30

3 25
solution solution
Step 1. Calculate the average normal stress Step 1. Calculate the average normal stress
(σ avg ) from Eq. (5.14) as (σ avg ) from Eq. (5.14) as
σ xx + σ yy [10 + (−3)] kpsi σ xx + σ yy [75 + (−25)]MPa
σ avg = = σ avg = =
2 2 2 2
= 3.5 kpsi = 25 MPa
Step 2. Calculate the maximum shear stress Step 2. Calculate the maximum shear stress
(τ max ) from Eq. (5.12) as (τ max ) from Eq. (5.12) as

2 2

σ xx − σ yy 2 σ xx − σ yy 2
τ max = + τ xy τ max = + τ xy
2 2

2 2
10 − (−3) 75 − (−25)

2
2
= + (−4 ) kpsi = + (−30 ) MPa
2 2

2
2
2
2
= (6.5) + (−4 ) kpsi = (50) + (−30 ) MPa

= (42.25) + (16 ) kpsi = (2,500) + (900 ) MPa

= (58.25) kpsi = (3,400) MPa
= 7.6 kpsi 7.5 kpsi = 58.3MPa 58 MPa