# Giao trinh bai tap chapter 4 hoạch định chiến lược tiếp thị giữa các tổ chức

M. Vable

Mechanics of Materials: Chapter 8

Stress Transformation
• Transforming stress components from one coordinate system to
another at a given point.
• Relating stresses on different planes that pass through a point.

σxx

T

T

P

P

τθx

σxx

τxθ

Cast Iron

Aluminum

Learning Objectives
• Learn the equations and procedures of relating stresses (on different
planes) in different coordinate system at a point.

• Develop the ability to visualize planes passing through a point on
which stresses are given or are being found, particularly the planes of
maximum normal stress and maximum shear stress.

August 2012

8-1

M. Vable

Mechanics of Materials: Chapter 8

Wedge Method
• The fixed reference coordinate system in which the entire problem is
described is called the global coordinate system.
• A coordinate system that can be fixed at any point on the body and has
an orientation that is defined with respect to the global coordinate system is called the local coordinate system.
Vertical plane

y
y
t

t

in

cl

In
pl

x

e

an

n

ed

n

Outward normal to
inclined plane

Horizontal plane

z
x

z
(a)

(b)

• Plane stress problem: We will consider only those inclined planes that
can be obtained by rotation about the z-axis.

August 2012

8-2

M. Vable

Mechanics of Materials: Chapter 8

C8.1
In the following problems one could say that the normal stress
on the incline AA is in tension, compression or can’t be determined by
inspection. Similarly we could say that the shear stress on the incline AA
is positive, negative or can’t be determined by inspection. Choose the
correct answers for normal and shear stress on the incline AA by inspection. Assume coordinate z is perpendicular to this page and towards you.
(a)

σ

y

(b)

y
τ
A

A
A
300

A

600

x

Class Problem 1
y
τ

(c)

(d)

A

300

August 2012

A

x

8-3

x

M. Vable

Mechanics of Materials: Chapter 8

General Procedure for Wedge Method
Step 1: A stress cube with the plane on which stresses are to be found, or are
given, is constructed.

Step 2: A wedge made from the following three planes is constructed:
• a vertical plane that has an outward normal in the x-direction,
• a horizontal plane that has an outward normal in the y-direction, and
• the specified inclined plane on which we either seek stresses or the
stresses are given.

Establish a local n-t-z coordinate system using the outward normal of the
inclined plane as the n-direction. All the known and the unknown stresses
are shown on the wedge. The diagram so constructed will be called a
stress wedge.
Step 3: Multiply the stress components by the area of the planes on which the
stress components are acting, to obtain forces acting on that plane. The
wedge with the forces drawn will be referred to as the force wedge.

Step 4: Balance forces in any two directions to determine the unknown
stresses.

Step 5: Check the answer intuitively.

August 2012

8-4

M. Vable

C8.2

Mechanics of Materials: Chapter 8

Determine the normal and shear stress on plane AA.

A

300

8 ksi
A

10 ksi

August 2012

8-5

M. Vable

Mechanics of Materials: Chapter 8

Stress Transformation By Method of
Equations
Stress Cube

Stress Wedge

σyy

y

τyx
B

t

τxy

σxx

Δy

σxx
C

τxy

θ
O
τyx

τxy

x

A

Δt
Δx

τyx

σyy

Force Wedge

σnn

θ

θ

σxx

τnt

n

σyy

Force Transformation
τnt (Δt Δz)

y

σnn(Δt Δz)

θ

t

θ

n
Fy

θ

co

Fx
θ

θ

Fy

τxy (Δt cosθ Δz)

θ

s
co

sin
Fx

σxx (Δt cosθ Δz)

Fy

sin

Fx

τyx (Δt sinθ Δz)

σyy (Δt sinθ Δz)
2

2

σ nn = σ xx cos θ + σ yy sin θ + 2τ xy sin θ cos θ
2

2

τ nt = – σ xx cos θ sin θ + σ yy sin θ cos θ + τ xy ( cos θ – sin θ )
Trigonometric identities

August 2012

8-6

x

M. Vable

Mechanics of Materials: Chapter 8

2

cos θ = ( 1 + cos 2θ ) ⁄ 2
2

2

cos θ – sin θ = cos 2θ

2

sin θ = ( 1 – cos 2θ ) ⁄ 2
cos θ sin θ = ( sin 2θ ) ⁄ 2

( σ xx + σ yy ) ( σ xx – σ yy )
σ nn = ---------------------------- + -------------------------- cos 2θ + τ xy sin 2θ
2
2
( σ xx – σ yy )
τ nt = – ---------------------------- sin 2θ + τ xy cos 2θ
2
Maximum normal stress
2τ xy
⎛ dσ nn

= 0⎟ ⇒ tan 2θ p = ---------------------------⎜
( σ xx – σ yy )
d
θ

θ = θp

−τxy

-(σxx − σyy)/2
2θp

τxy

R
2θp
(σxx − σyy)/2

θ1 = θp
θ2 =90 + θp

R

σ xx – σ yy 2
( σ xx + σ yy )

2
τ 1, 2 = 0
σ 1, 2 = ---------------------------- ± -----------------------⎞ + τ xy

2
2
• Planes on which the shear stresses are zero are called the principal
planes.
• The normal direction to the principal planes is referred to as the principal direction or the principal axis.
• The angles the principal axis makes with the global coordinate system
are called the principal angles.
• Normal stress on a principal plane is called the principal stress.
• The greatest principal stress is called principal stress one.
• Only θ defining principal axis one is reported in describing the principal coordinate system in two-dimensional problems. Counterclockwise rotation from the x axis is defined as positive.
August 2012

8-7

M. Vable

Mechanics of Materials: Chapter 8

σ nn + σ tt = σ xx + σ yy = σ 1 + σ 2
• The sum of the normal stresses is invariant with the coordinate transformation.
⎧0
σ 3 = σ zz = ⎨
⎩ ν ( σ xx + σ yy ) = ν ( σ 1 + σ 2 )

Plane Stress
Plane Strain

In-Plane Maximum Shear Stress
Vertical plane

y
t

in

cl

In
pl

x

e

an

n

ed

n

Outward normal to
inclined plane

Horizontal plane
x
z

• The maximum shear stress on a plane that can be obtained by rotating
about the z axis is called the in-plane maximum shear stress.

( σ xx – σ yy )
τ nt = – ---------------------------- sin 2θ + τ xy cos 2θ
2
– ( σ xx – σ yy )⎞
⎛ dτ nt

σ1 – σ2
τ p = -----------------= 0⎟ ⇒ ⎜ tan 2θ s = -------------------------------⎟

2τ xy
2
⎝ d θ θ = θs

• maximum in-plane shear stress exists on two planes, each of which are
45o away from the principal planes.
Maximum Shear Stress
• The maximum shear stress at a point is the absolute maximum shear
stress that acts on any plane passing through the point.

August 2012

8-8

M. Vable

Mechanics of Materials: Chapter 8

Planes of Maximum Shear Stress
p2

p2

p1

p1

σ2 – σ3
τ 23 = – τ 32 = -----------------2
p3

p3
p2

p2

p1

p1

σ3 – σ1
τ 31 = – τ 13 = -----------------2

p3

p3

p2

p2

(In-plane)
p1

p1

σ1 – σ2
τ 21 = – τ 12 = -----------------2
p3

p3

σ1 – σ2 σ2 – σ3 σ3 – σ1 ⎞
, ------------------ , -----------------τ max = max ⎛ -----------------⎝

2
2
2
Plane Stress
⎧0
σ 3 = σ zz = ⎨
Plane Strain
⎩ ν ( σ xx + σ yy ) = ν ( σ 1 + σ 2 )
• The maximum shear stress value may be different in plane stress and
in plane strain.

August 2012

8-9

M. Vable

Mechanics of Materials: Chapter 8

C8.3
Determine the normal and shear stress on plane AA using the
method of equations (resolving problem C8.2).

A

300

8 ksi
A

10 ksi

August 2012

8-10

M. Vable

Mechanics of Materials: Chapter 8

Stress Transformation By Mohr’s Circle
( σ xx + σ yy ) ( σ xx – σ yy )
σ nn = ---------------------------- + -------------------------- cos 2θ + τ xy sin 2θ
2
2
( σ xx – σ yy )
τ nt = – ---------------------------- sin 2θ + τ xy cos 2θ
2

σ xx + σ yy 2
σ xx – σ yy 2
⎛ σ – ----------------------⎞

2
2
- + τ nt = -----------------------⎞ + τ xy
8.1
⎝ nn

2
2
• Each point on the Mohr’s circle represents a unique plane that passes
through the point at which the stresses are specified.
• The coordinates of the point on Mohr’s Circle are the normal and
shear stress on the plane represented by the point.
• On Mohr’s circle, plane are separated by twice the actual angle
between the planes.

August 2012

8-11

M. Vable

Mechanics of Materials: Chapter 8

Construction of Mohr’s Circle
Step 1. Show the stresses σxx, σyy, and τxy on a stress cube and label the vertical
plane as V and the horizontal plane as H.
Step 2. Write the coordinates of points V and H as

V (σxx , τxy )and H (σyy , τyx )
The rotation arrow next to the shear stresses corresponds to the rotation
of the cube caused by the set of shear stress on planes V and H.

y

σyy

τyx
τxy

H
σxx

V

τxy

V

τ

H
τyx

σxx

R
E

σyy

(C)
x

H
τyx

(CW)

σxx

C

D
R

τxy

σ
(T)

V

σyy
(CCW)

σ xx + σ yy σ xx – σ yy
-------------------------- -------------------------2
2

Step 3. Draw the horizontal axis with the tensile normal stress to the right and
the compressive normal stress to the left. Draw the vertical axis with
clockwise direction of shear stress up and counterclockwise direction of
rotation down.
Step 4. Locate points V and H and join the points by drawing a line. Label the
point at which the line VH intersects the horizontal axis as C.
Step 5. With C as center and CV or CH as radius draw the Mohr’s circle.

August 2012

8-12

M. Vable

Mechanics of Materials: Chapter 8

Principal Stresses & Maximum In-Plane Shear Stress from Mohr’s
Circle
τ

S1

(CW)

H
τ12

R
P3
σ3

(C)

P2

2θp C
E

D
2θp

τ21
V

σ1
(CCW)

(T)

τxy

R
σ2

σ

P1

σ xx + σ yy
-------------------------2

S2
σ xx – σ yy
-------------------------2

• The principal angle one θ1 is the angle between line CV and CP1.
Depending upon the Mohr circle θ1 may be equal to θp or equal to
(θp+ 90o).
Maximum Shear Stress
τ
(CW)

τ23
(C)

τ32

τ12
P3

σ
(T)

P1

P2

τ21

(CCW)

August 2012

τ13

8-13

τ31

M. Vable

Mechanics of Materials: Chapter 8

Stresses on an Inclined Plane
σyy

y

τ
τyx

H

σxx
V

A

V

β
H θA A

τxy
τyx

(CW)
τxy
σxx

A
H
2θA
2βA F D
C

(C)
E

x
V

σyy

σA

(CCW)

Sign of shear stress on incline:
Coordinates of point A: ( σ , τ )
A A
t

V A

n

τA

A V
n

τA

H

August 2012

H

8-14

t

τA
(T)
σ

M. Vable

Mechanics of Materials: Chapter 8

Principal Stress Element
τ

S1

(CW)

P2

p2

R
P2
(C)

C

E

y

τ12
D
2θ1

σ2

P1
H
V

V
σ1

P2

(T)
τ21

R

S1

σ

P1

p1

P1

S2

H

V

θ1

H

S2

x

(CCW)

y

σav

τ21
P1

σ1

τ21

y

σav

S1
P2
σ2

x

Cast Iron
P1

σxx

S1
P2 H
S2

August 2012

P1

θ1

x
P

σxx

S2
P2

σ2

θ1
P

σ1

σ xx
τ max = --------2
V P1
σ 1 = σ xx

S2 P1
P2

Aluminum
S2

8-15

M. Vable

Mechanics of Materials: Chapter 8

C8.4
In a thin body (plane stress) the stresses in the x-y plane are as
shown on each stress element. (a) Determine the normal and shear
stresses on plane A. (b) Determine the principal stresses at the point. (c)
Determine the maximum shear stress at the point. (d) Draw the principal
element.
20 ksi
30 ksi
A

42o

Fig. C8.4

August 2012

8-16

10 ksi

M. Vable

Mechanics of Materials: Chapter 8

Class Problem 2
Associate the stress cubes with the appropriate Mohr’s circle for stress.
40 ksi

40ksi
8 ksi

8 ksi

40 ksi

cube 1

cube 2

cube 3

8 ksi

circle B

circle A

circle C

circle D

Class Problem 3

Determine the two possible values of principal angle one (θ1) in each
question.
y

1

2
V

H
H
V

V
H

August 2012

24o

24o
x

V

8-17

H

M. Vable

Mechanics of Materials: Chapter 8

Class Problem 4
Explain the failure surfaces in cast iron and aluminum due to torsion by
drawing the principal stress element
T

T

τθx
τxθ
Cast Iron

Aluminum

August 2012

8-18

M. Vable

Mechanics of Materials: Chapter 8

C8.5
A broken 2 in x 6 in wooden bar was glued together as
shown. Determine the normal and shear stress in the glue.
F = 12 kips

F
6 in

60o

August 2012

8-19

M. Vable

Mechanics of Materials: Chapter 8

C8.6
If the applied force P = 1.8 kN, determine the principal
stresses and maximum shear stress at points A, B, and C which are on the
surface of the beam.
P
A
B
C

30 mm
15 mm
30 mm

0.4 m

0.4 m

August 2012

8-20

6 mm
6 mm

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