Page 250 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
Brown˙C05
Brown.cls
232
U.S. Customary 14:35 STRENGTH OF MACHINES SI/Metric
Example 5. For the biaxial stresses (σ) and Example 5. For the biaxial stresses (σ) and
(2σ) given in Example 3 that are the principal (2σ) given in Example 3 that are the principal
stresses (σ 1 ) and (σ 2 ), find the maximum shear stresses (σ 1 ) and (σ 2 ), find the maximum shear
stress (τ max ) if the element is actually a triaxial stress (τ max ) if the element is actually a triaxial
stress element using Eq. (5.20), where stress element using Eq. (5.20), where
2σ = 16 kpsi = σ 1 2σ = 112 MPa = σ 1
σ = 8 kpsi = σ 2 σ = 56 MPa = σ 2
p i =−4 kpsi = σ 3 p i =− 28 MPa = σ 3
solution. solution.
Step 1. Using Eq. (5.20), the maximum shear Step 1. Using Eq. (5.20), the maximum shear
stress becomes stress becomes
σ 1 − σ 3 [16 − (−4)] kpsi σ 1 − σ 3 [112 − (−28)]MPa
τ max = = τ max = =
2 2 2 2
20 kpsi 140 MPa
= = 10 kpsi = = 70 MPa
2 2
Step 2. Compare this value for the maximum Step 2. Compare this value for the maximum
shear stress with that found in Example 8, where shear stress with that found in Example 8, where
σ 8 kpsi σ 56 MPa
τ max = = = 4 kpsi τ max = = = 28 MPa
2 2 2 2
Notice that the maximum shear stress found using Eq. (5.20) is two and a half times
the maximum shear stress found in Example 8 where the stress element was treated as a
biaxial stress element. This is obviously a nontrial difference and cannot be ignored in a
design.
Without providing the proof, the maximum shear stress found using Eq. (5.20) acts on a
plane rotated 45 about the maximum principal axis (σ 1 ). This means the material tears at
◦
◦
a45 angle in the cross section rather than straight across the cross section. This is seen in
actual failures and is a telltale sign that the vessel failed due to excessive pressure.
Three Dimensional Stress. While it is difficult to imagine a combination of loadings
that would produce a set of stresses, three normal (σ xx ,σ yy ,σ zz ) and three shear stresses
(τ xy ,τ xz ,τ yz ) acting on the six sides of an element, the three associated principal stresses
(σ 1 ,σ 2 ,σ 3 ) could be found by solving the following cubic equation. (A trial-and-error
approach works well.)
3 2 2 2 2
σ − (σ xx + σ yy + σ zz )σ + (σ xx σ yy + σ xx σ zz + σ yy σ zz − τ xy − τ xz − τ )σ
yz
2
− (σ xx σ yy σ zz + 2τ xy τ xz τ yz − σ xx τ 2 − σ yy τ 2 − σ zz τ ) = 0
yz xz xy
If these three principal stresses are ordered such that (σ 1 >σ 2 >σ 3 ), then the maximum
shear stress is given by Eq. (5.20), repeated here.
σ 1 − σ 3
τ max = (5.20)
2
Again, it is very unusual to be faced with having to analyze an element with this level
of complexity, but it is comforting to know that the methods, both analytical and graphical,
presented in this section, could be used if necessary.