Page 238 - Mechanical Behavior of Materials

Section 6.2 Plane Stress 239

Solving these then gives
|σ 1 − σ 2 | σ 1 + σ 2
τ 3 = , σ τ3 = (6.12)
2 2
The absolute value is necessary for τ 3 , due to the two roots of Eq. 6.9.

Example 6.1
At a point of interest on the free surface of an engineering component, the stresses with
respect to a convenient coordinate system in the plane of the surface are σ x = 95, σ y = 25, and
τ xy = 20 MPa. Determine the principal normal and shear stresses and their coordinate system
rotations. Also determine the maximum normal stress and the maximum shear stress.
Solution Substitution of the given values into Eq. 6.6 gives the angle to the coordinate axes
for the principal normal stresses:

2τ xy 4
tan 2θ n = = , θ n = 14.9 ◦ (CCW) Ans.
σ x − σ y 7
Substitution into Eq. 6.7 gives the principal normal stresses:

2
σ x + σ y σ x − σ y 2
σ 1 ,σ 2 = ± + τ xy
2 2
σ 1 ,σ 2 = 60 ± 40.3 = 100.3, 19.7MPa Ans.

The corresponding planes and state of stress are shown in Fig. E6.1(b). Note that the direction
for the larger of the two principal normal stresses is chosen so that it is more nearly aligned with
the larger of the original σ x and σ y than with the smaller.
◦
Alternatively, a more rigorous procedure is to use θ = θ n = 14.9 in Eq. 6.4, which gives
◦
◦

σ = σ = σ 1 = 100.3MPa. Use of θ = θ n + 90 = 104.9 in Eq. 6.4 then gives the normal
x

stress in the other orthogonal direction, σ = σ = σ 2 = 19.7 MPa. The zero value of τ at θ = θ n
y
can also be verified by using Eq. 6.5.
For the equivalent representation where the maximum shear stress in the x-y plane occurs,
Eq. 6.8 gives
σ x − σ y 7
◦
tan 2θ s =− =− , θ s =−30.1 = 30.1 ◦ (CW) Ans.
2τ xy 4
The stresses for this rotation of the coordinate system may be obtained from σ 1 and σ 2 as
previously calculated and from Eq. 6.12:
|σ 1 − σ 2 | σ 1 + σ 2
τ 3 = =±40.3MPa, σ τ3 = = 60 MPa Ans.
2 2