Page 111 - Shigley's Mechanical Engineering Design

bud29281_ch03_071-146.qxd 11/24/09 3:01PM Page 86 ntt 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles:

86 Mechanical Engineering Design
Answer Substituting φ p = 64.3 into Eq. (3–9) again yields τ = 0, indicating that −24.03 MPa
◦
is also a principal stress. Once the principal stresses are calculated they can be ordered
such that σ 1 ≥ σ 2 . Thus, σ 1 = 104.03 MPa and σ 2 =−24.03 MPa.
Since for σ 1 = 104.03 MPa, φ p =−25.7 , and since φ is defined positive ccw in the
◦
transformation equations, we rotate clockwise 25.7° for the surface containing σ 1 . We
see in Fig. 3–11c that this totally agrees with the semigraphical method.
To determine τ 1 and τ 2 , we first use Eq. (3–11) to calculate φ s :
1 −1 σ x − σ y 1 −1 80
◦
φ s = tan − = tan − = 19.3 , 109.3 ◦
2 2τ xy 2 2(−50)
◦
For φ s = 19.3 , Eqs. (3–8) and (3–9) yield
80 + 0 80 − 0
Answer σ = + cos[2(19.3)] + (−50) sin[2(19.3)] = 40.0MPa
2 2
80 − 0
τ =− sin[2(19.3)] + (−50) cos[2(19.3)] =−64.0MPa
2
Remember that Eqs. (3–8) and (3–9) are coordinate transformation equations. Imagine
that we are rotating the x, y axes 19.3° counterclockwise and y will now point up and
to the left. So a negative shear stress on the rotated x face will point down and to the
right as shown in Fig. 3–11d. Thus again, results agree with the semigraphical method.
◦
For φ s = 109.3 , Eqs. (3–8) and (3–9) give σ = 40.0 MPa and τ =+64.0 MPa.
Using the same logic for the coordinate transformation we find that results again agree
with Fig. 3–11d.

3–7 General Three-Dimensional Stress
As in the case of plane stress, a particular orientation of a stress element occurs in space
for which all shear-stress components are zero. When an element has this particular ori-
entation, the normals to the faces are mutually orthogonal and correspond to the prin-
cipal directions, and the normal stresses associated with these faces are the principal
stresses. Since there are three faces, there are three principal directions and three prin-
cipal stresses σ 1 ,σ 2 , and σ 3 . For plane stress, the stress-free surface contains the third
principal stress which is zero.
In our studies of plane stress we were able to specify any stress state σ x , σ y , and
τ xy and find the principal stresses and principal directions. But six components of
stress are required to specify a general state of stress in three dimensions, and the
problem of determining the principal stresses and directions is more difficult. In
design, three-dimensional transformations are rarely performed since most maxi-
mum stress states occur under plane stress conditions. One notable exception is con-
tact stress, which is not a case of plane stress, where the three principal stresses are
given in Sec. 3–19. In fact, all states of stress are truly three-dimensional, where
they might be described one- or two-dimensionally with respect to specific coordi-
nate axes. Here it is most important to understand the relationship among the three
principal stresses. The process in finding the three principal stresses from the six