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An explanation of how to compute principal stresses and their associated shear stresses using mohr's circle. It includes a staged procedure, additional properties, a numeric example, and a graphical solution using mohr's circle. The document also mentions the concept of principal stress elements and their relation to principal planes and principal shear planes.
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2
2
1 av^2 av
av av
xx
yy
yy
ASEN 3112 Lecture 6 – Slide 13
ASEN 3112 Lecture 6 – Slide 14
x
y
P σ xx^ =100 psi
τ xy = τ (^) yx =30 psi
σ (^) yy = 20 psi (a)
σ = normal stress
τ = shear stress
0 20 40 60 80 100
50 40 30 20 10 0 − − − − −
H
V
C
σ 1 = 110
σ 2 = 10
coordinates of blue points are H: (20,30), V:(100,-30), C:(60,0)
τ (^) min= −
τ (^) max= 50
Radius R = 50
2 θ 1 = 36.
2 θ 2 = 36.88 +180 = 216.
(b) Mohr's circle
(a) Point in plane stress
ASEN 3112 Lecture 6 – Slide 16
This topic be briefly covered in class if time allows,
using the following slides.
If not enough time, ask students to read Lecture notes
(Sec 7.3), with particular emphasis on the computation
of the overall maximum shear
ASEN 3112 Lecture 6 – Slide 17
The σ turn out to be the eigenvalues of the stress matrix.
They are the roots of a cubic polynomial (the so-called
characteristic polynomial)
The principal directions are given by the eigenvectors
of the stress matrix.
Both eigenvalues and eigenvectors can be numerically
computed by the Matlab function eig(.)
σ −σ
σ −σ
σ −σ
τ τ
τ
τ τ
τ
xx xy^ xz
yy
zz
yz
zx zy
C(σ) = det yx
= −σ + I (^) 1 σ − I (^) 2 σ + I = 0
ASEN 3112 Lecture 6 – Slide 19
(Yes, There Is More Than One)
ASEN 3112 Lecture 6 – Slide 20
σ 1 , σ 2 , σ =
where σ and σ are the inplane principal stresses obtained
as described earlier in Lecture 6.
Consider plane stress but now account for the third dimension.
One of the principal stresses, call it for the moment σ 3 , is zero:
The zero principal stress σ is aligned with the z axis (the
thickness direction) while σ 1 and σ 2 act in the x,y plane:
σ 1
σ 2
Thin plate σ^3
ASEN 3112 Lecture 6 – Slide 22
Let us now (re)order the principal stresses by algebraic value as
(A) Inplane principal stresses have opposite signs. Then the
zero stress is the intermediate one: σ 2 , and
To compute the overall maximum shear 2 cases are considered:
σ 1 − σ 3
2
σ
2
(B) Inplane principal stresses have the same sign. Then
If σ σ 0 and σ = 0, τ =
overall max
σ
2
If σ σ 0 and σ = 0, τ = −
overall 3 1 max
τ = τ =
overall inplane max max
σ 1 σ 2 σ 3
ASEN 3112 Lecture 6 – Slide 23