Lecture 3
The Concept of Stress, Generalized Stresses and Equilibrium
Problem 3-1: Cauchy’s Stress Theorem
Cauchy’s stress theorem states that in a stress tensor field there is a traction vector t that linearly
depends on the outward unit normal n:
a. Express Cauchy’s stress theorem in index form.
b. Suppose the state of stress at a point in x, y, z coordinate system is given by the
matrix below. Determine the normal stress n and the shear stress on the surface
defined by .
Problem 3-1 Solution:
(a)
ti
n
ij i
(b)
A normal vector to a plane specified by
f x, y , z
ax by cz d
0
is given by
N
where
f
a
b
c
f denotes the gradient.
1
The unit vector of the surface
2 x y 3z 9
is
2
1
N
N
n
n
2
3
2
1
( 3) 2
2
1
1
14
3
n
The magnitude of normal stress
1
2
can be calculated by
n
t n
where t is the surface traction vector
t
n
20 10
10 30
10 0
10
0
50
2
1
1
14
3
1
14
1
t n
14
80
50
170
2
1
1
14
3
720
14
80
50
170
Then
n
n
The direction of the normal stress vector
n
51 Force / area
n
is n
n
n
2
51
1
14
3
2
Shear stress vector
can be simply calculated by subtract the normal stress vector
n
from the
traction vector t
t
1
14
80
50
170
2
360
1
7 14
3
1
14
160 / 7
10 / 7
110 / 7
3
Problem 3-2: Three invariants of a stress tensor
Suppose the state of stress at a point in a x, y, z coordinate system is given by
100 1 180
0 20 0
180
0
20
a. Calculate the three invariants of this stress tensor.
b. Determine the three principal stresses of this stress tensor.
Problem 3-2 Solution:
a) Invariants of stress tensor
Recall: these values do not change no matter the coordinate system selected
I1
x
y
100 20 20
z
I1
I2
60
2
x
y
y
z
z
x
2
yz
xy
2
zx
( 100 20) ( 100 20) 202 0 0 802
I2
I3
x
y
z
2
10000
2
xy yz zx
x yz
2
y zx
2
z xy
( 100 20 20) 2 0 0 20 ( 80) 2 0
I3
168000
b) Principal stresses
The eigenvalue problem for a stress tensor
4
100 0
0
20
80 0
80
0
20
is given by
det
Solve for
100
0
80
0
20
0
80
0
20
0
, we have three eigenvalues
1
140
2
60
3
20
The principal stresses are the three eigenvalues of the stress tensor
xp
140
yp
60
zp
20
5
Problem 3-4: Transformation Matrix
Suppose the state of stress at a point relative to a x, y, z coordinate system is given by:
15
10
10
5
Try to find a new coordinate system (x’, y’) that corresponds to the principal directions of the
stress tensor.
a. Find the principal stresses.
b. Determine the transformation matrix
c. Verify
.
.
Problem 3-4 Solution:
a)
Determine principal stresses
The eigenvalue problem for a stress tensor
15
10
10
5
10
10
5
is given by
det
Solve for
15
0
, we have two eigenvalues
1
19.14
2
9.14
The principal stresses are the two eigenvalues of the stress tensor
xp
19.14
yp
9.14
6
b)
Determine transformation matrix
Calculate the eigenvectors for the stress tensor
For
19.14
1
10
x1
5 19.14
x2
15 19.14
10
Set x1
0
1 , use the first equation
x2
0.414
Normalize eigenvector
1
1
12 ( 0.414) 2
0.92
0.38
1
For
9.14
1
10
x1
5 ( 9.14)
x2
15 ( 9.14)
10
Set x1
1
0.414
0
1 , use the first equation
x2
2.414
Normalize eigenvector
1
2
12 (2.414)2
1
Check
1
2
1
2.414
0.38
0.92
?
7
1
0.92 0.38 0.38 0.92 0
2
1
2
!
Transformation Matrix
L
0.92
L
b)
1
2
0.38
0.38 0.92
Verify
L
'
T
L
If we transform the given stress tensor, will we arrive at the principal stresses?
Note: use 4 decimal places to get principal stresses
L
T
L
0.9238
0.3827
0.3827
0.9238
15
10
0.9238
5
0.3827 0.9238
10
'
19.174
0
0
9.14
0.3827
19.174
0
0
9.14
8
Problem 3-5: Beam Equilibrium
Derive the equation of force and moment equilibrium of a beam using the equilibrium of an
infinitesimal beam element of the length dx.
Problem 3-5 Solution:
Consider an element of a beam of length dx subjected to
1. Distributed loads q
2. Shear forces V and V
3. Moments M and M
dV
dM
Equilibrium of forces in x and y direction
Fx
0
N
dN
Fy
0
V
qdx (V
N
0
dN
dx
dV )
0
0
dV
dx
q
( )
Moment equilibrium of the element at point O
M
0
M qdx(
dx
) (V
2
dV )dx M
dM
0
From equation (*),
dV
q
dx
dV
dV
dx
dx
qdx
10
Substitute the above equation into the moment equilibrium equation, we have
dM
q
(dx) 2 Vdx q(dx) 2
2
0
Ignore the second-order terms, we have
dM
dM
dx
Vdx
V
To sum up, the equations of force and moment equilibrium of a beam are
dN
dx
dV
dx
dM
dx
0
q
V
11
Problem 3-6: Moderately large deflections in beams
Explain which of the three equilibrium equations below is affected by the finite rotations in the
theory of the moderately large deflections of beam.
1. Moment equilibrium
2. Vertical force equilibrium
3. Axial force equilibrium
Problem 3-6 Solution:
Consider an element of a beam of length dx subjected to
1. Distributed loads q
2. Shear forces V and V
3. Moments M and M
dV
dM
and under moderately large deflection
Equilibrium of forces in t direction
Ft
0
V cos
qdx cos
dV
dx
(V
dV ) cos
0
q
(1)
Note: All terms involve cos
12
Moment equilibrium of the element at point O
M
0
M qdx(
dx
) (V
2
dV )dx M
dM
0
From equation (1),
dV
q
dx
dV
dV
dx
dx
qdx
Substitute the above equation into the moment equilibrium equation, we have
q
(dx) 2 Vdx q(dx) 2
2
dM
0
Ignore the second-order terms, we have
dM
Vdx
dM
dx
V
The moment equilibrium equation is not affected by moderately large deflection.
Force Equilibrium in x direction
Fx
0
( N dN) cos
dN
dx
N cos
0
0
(2)
The axial force equilibrium equation is not affected by moderately large deflection.
13
Define effective shear force V * as the sum of the cross-sectional shear V and the projection of
the axial force into the vertical direction
V* V
V
N sin
N
dw
dx
(3)
Force equilibrium in y direction
V * dV *
dV *
dx
V * qdx
q
0
q
dV
dx
0
Combining equation (3), we have
dV
dx
d
dw
N
dx
dx
dN dw
d 2w
N 2 q 0
dx dx
dx
use equation(2), we have the vertical force equilibrium equation
dV
dx
d 2w
N 2 q
dx
0
The vertical force equilibrium equation is affected by moderately large deflection.
14
Problem 3-7:
At full draw, and archer applies a pull of 150N to the bowstring of the bow shown in the figure.
Determine the bending moment at the midpoint of the bow.
Problem 3-6 Solution:
P is the pulling force at point A and T is the tension in the string. Force balance at point A
Fx
0
P 2T cos 70
0
15
T
P
2cos 70
150
2cos 70
219.3N
Free body diagram
where the tension T is projected in x and y direction as Tx and Ty.
Moment balance at point B
MB
0
Tx 0.7 Ty 0.35 M B
0
Finally, the bending moment at the midpoint of the bow is
MB
Tx 0.7 Ty 0.35
T cos 70
0.7 T sin 70
219.3 cos 70
124.6N m
0.35
0.7 219.3 sin 70
0.35
16
Problem 3-8:
A curved bar ABC is subjected to loads in the form of two equal and opposite force P, as shown
in the figure. The axis of the bar forms a semicircle of radius r.
Determine the axial force N, shear force V, and bending moment M acting at a cross section
defined by the angle
Problem 3-8 Solution:
The free body diagram and force balance at angle :
We can determine axial force N, shear force V according to the force balance diagram
N
V
P sin
P cos
MB
0
Moment balance at point B
P r sin
M
0
We have the bending moment M acting at a cross section defined by the angle
M
P r sin
17
Problem 3-9:
The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth
surface about the z axis (which is vertical) with an angular acceleration . Each of the two arms
has weight w per unit length and supports a weight W 2.5wL at its end.
Derive formulas for the maximum shear force and maximum bending moment in the arms,
assuming b L / 9 and c L / 10
Problem 3-9 Solution:
Shear force density distribution and free body diagram:
Where weight at the tip of arm W is considered as lumped mass
18
Force balance at any point of radius r
b L
V
r
w *
W
r dr *
b L c
g
g
W
2g
2
b L
W
b L c
g
r2
Moment balance at any point of radius r
b L
M
r
w *
r
g
W
b L c
g
r * r dr *
W
g
1 *3
r
3
W
g
1
b L
3
b L
1 *2
rr
2
3
W
g
r
b L c r
b L c b L c r
1
r b L
2
r3
2
W
g
r2
b L c b L c r
By observation maximum shear force and bending moment occurs at r=b (closest point to the
center line)
Vmax
M max
W
g
1
b L
3
W
g
1
3
L
9
3
b L
W
2g
L
9
2
L
9
3
W
b L c
g
b2
2
2
L
L
9
Vmax
3.64
1
r b L
2
b3
3
L
W
2g
1
r
2
M max
2
L
9
W L
g 9
L
L
10
wL2
g
W
g
b2
2
L
3.72
L
9
b L c b L c b
2
W
g
L
9
L
L
10
L
L
10
wL3a
g
19
Problem 3-10:
A simple beam AB supports two connective loads P and 2P that are distance d apart (see figure).
The wheels may be placed at any distance x from the left support of the beam.
(a) Determine the distance x that will produce maximum shear force in the beam, and also
determine the maximum shear force Vmax
(b) Determine the distance that will produce the maximum bending moment, and also draw
the corresponding bending moment diagram. (Assume P=10kN, d=2.4m, and L=12m)
C
D
Problem 3-10 Solution:
a) Free body diagram:
D
C
RA RB are reaction forces.
Moment balance at point A:
MA
0
20
RB L
P x 2P x d
3Px 2 Pd
3Px
L
RB
2 Pd
L
Force balance in y direction
FY
RA
RA
0
3P RB
P 2
2d
L
3x
L
Shear force distribution
C
D
There are two possible maximum shear forces,
3Px
L
2 Pd
2d
and P 2
L
L
3x
, depending on
L
the position of the wheels x:
If x
0, the two possible maximum shear forces are Vmax
P max
If x
L d, the two possible maximum shear forces are Vmax
3
2d 2d
,
L
L
P max
3
In the range of 0 d L , by observation, the maximum shear force occurs when x
the left hand side of point B.
Vmax
P 3
d
L
10 3
2.4
12
28kN at x
L d Note : 0
x
d d
,
L L
L d , at
L d
21
b) From the shear force diagram, we know the bending moment diagram could be either i orii
The maximum bending moment occurs at either point C or D. Thus we need only calculate
moment at these two points to determine the maximum bending moment.
Knowing RA
2d
L
P 2
3Px
L
3x
and RB
L
MC
RA x
MD
RA x d
2 Pd
,
L
2d
3x 2
3
x
L
L
P
P d
P
3
2d
3dx
x
L
L
P
3
2d
d
L
3
3x 2
L
3
2d
d
L
5d
3x 2
x
L
L
Pd
Pd
Calculate maximum M C and M D
If x
0, M C
0, M D
If x
L d,
MC
2Pd 1
Pd 1
d
L
d
,MD
L
0
By observation, the maximum moment occurs at x=0, where
MD
2 Pd 1
d
L
2 10 2.4
M max
1
2.4
12
38.4N m
38.4 N m
22
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2.080J / 1.573J Structural Mechanics
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