Stress Transformations

M. Vable Mechanics of Materials: Chapter 8Pr

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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.

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.

σxx σxx

TT

τxθ

τθx

PP

Aluminum

Cast Iron

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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 sys-tem is called the local coordinate system.

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

Horizontal plane

Outward normal to inclined plane

Inclined plane

Ver

tical

pla

ne

x

x

n �

�

�

ntt

y

y

z

z

(b)(a)

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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 inspec-tion. Assume coordinate z is perpendicular to this page and towards you.

Class Problem 8.1

A

A

σ

300

τ

A

A

600

(b)(a)

x x

y y

A

A

τ

300

(c) (d)

x

y

A A

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General Procedure for Wedge MethodStep 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.

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C8.2 Determine the normal and shear stress on plane AA.

8 ksi

A

A 300

10 ksi

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Stress Transformation By Method of Equations

Trigonometric identities

x

yn

t

O A

B

Cθ

θσxxσxx

σyy

σyy

τxy

τxy

τyx

τyx

σxx

σnn

τnt

σyyτyx

τxy

Δt

Δx

Δy

θ

Stress Cube Stress Wedge

σnn

τnt

τxy (Δt cosθ Δz)

θ(Δt Δz)

(Δt Δz)

σyy (Δt sinθ Δz)τyx (Δt sinθ Δz)

σxx (Δt cosθ Δz)

Force Wedge

Fx

Fx sinθF x

cosθFy

F ysin

θ

Fy cosθ θ

θ

nty

x

Force Transformation

σnn σxx θ2cos σyy θ2sin 2τxy θ θcossin+ +=

τnt σxx θ θsincos– σyy θ θcossin τxy θ2cos θ2sin–( )+ +=

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Maximum normal stress

• 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 princi-pal 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 princi-

pal coordinate system in two-dimensional problems. Counterclock-wise rotation from the x axis is defined as positive.

θ2cos 1 2θcos+( ) 2⁄= θ2sin 1 2θcos–( ) 2⁄=

θ2cos θ2sin– 2θcos= θcos θsin 2θsin( ) 2⁄=

σnnσxx σyy+( )

2----------------------------

σxx σyy–( )

2-------------------------- 2θcos τxy 2θsin+ +=

τntσxx σyy–( )

2---------------------------- 2θsin– τxy 2θcos+=

θddσnn

θ θp=0=

⎝ ⎠⎜ ⎟⎛ ⎞

2θptan2τxy

σxx σyy–( )----------------------------=⇒

τxyR

R

(σxx − σyy)/2

-(σxx − σyy)/2 2θp2θp−τxy

θ1 = θp

θ2 =90 + θp

σ1 2,σxx σyy+( )

2----------------------------

σxx σyy–2

-----------------------⎝ ⎠⎛ ⎞

2τxy

2+±= τ1 2, 0=

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• The sum of the normal stresses is invariant with the coordinate trans-formation.

In-Plane Maximum Shear Stress

• 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.

• 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.

σnn σtt+ σxx σyy+ σ1 σ2+= =

σ3 σzz0ν σxx σyy+( ) ν σ1 σ2+( )=⎩

⎨⎧

= =Plane Stress

Plane Strain

Horizontal plane

Outward normal toinclined plane

Inclined planeV

ertic

al p

lane

x

x

n �

�

nt

y

z

τntσxx σyy–( )

2---------------------------- 2θsin– τxy 2θcos+=

θddτnt

θ θs=0=

⎝ ⎠⎜ ⎟⎛ ⎞

2θstanσxx σyy–( )–

2τxy-------------------------------=

⎝ ⎠⎜ ⎟⎛ ⎞

⇒ τpσ1 σ2–

2------------------=

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Planes of Maximum Shear Stress

• The maximum shear stress value may be different in plane stress and in plane strain.

p1

p2

p3 p3

p2

p1

τ23 τ32–σ2 σ3–

2------------------= =

rotation about principal axis 1

p2

p3

p1

p2

p3

p1

τ31 τ13–σ3 σ1–

2------------------= =

rotation about principal axis 2

p2

p3

p1

p2

p3

p1

τ21 τ12–σ1 σ2–

2------------------= =

rotation about principal axis 3(In-plane)

τmax maxσ1 σ2–

2------------------

σ2 σ3–2

------------------σ3 σ1–

2------------------, ,⎝ ⎠

⎛ ⎞=

σ3 σzz0ν σxx σyy+( ) ν σ1 σ2+( )=⎩

⎨⎧

= =Plane Stress

Plane Strain

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

8 ksi

A

A 300

10 ksi

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Stress Transformation By Mohr’s Circle

8.1

• 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.

σnnσxx σyy+( )

2----------------------------

σxx σyy–( )

2-------------------------- 2θcos τxy 2θsin+ +=

τntσxx σyy–( )

2---------------------------- 2θsin– τxy 2θcos+=

σnnσxx σyy+

2------------------------–⎝ ⎠

⎛ ⎞ 2τnt

2+σxx σyy–

2-----------------------⎝ ⎠

⎛ ⎞2

τxy2+=

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Construction of Mohr’s CircleStep 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

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.

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.

V (σxx , τxy )and H (σyy , τyx )

σ (T)(C)

τ

R

R

CD

E

V

H

τxy

τyx

σyy

σxx

σxx σyy+

2--------------------------

σxx σyy–

2--------------------------

(CW)

(CCW)

x

y

σxxσxx

σyy

σyy

τxy

τxy

τyx

τyx

H

H

V V

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

• 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

σ (T)(C)

τ

R

R

C DE

V

H

P1P2

S1

S2

σ2

σ1

τ12

τ21

σ3 2θp

2θpP3

(CW)

(CCW) σxx σyy+

2--------------------------

σxx σyy–

2--------------------------

τxy

σ(T)

τ12τ13

τ31

τ(CW)

(CCW)

(C)P1

τ21

P3τ23τ32

P2

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Stresses on an Inclined Plane

Sign of shear stress on incline:Coordinates of point A:

σ (T)(C)

CD

E

V

H

2θA

A

σA

τA

F

τ(CW)

(CCW)

2βA

x

y

σxxσxx

σyy

σyy

τxy

τxy

τyx

τyx

θAβA

A

H

H

V V

σA τA,( )

AV

H

A V

H

τA

τA

n t

nt

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Principal Stress Element

σ (T)(C)

τ

R

R

C DE

V

H

P1P2

S1

S2

σ2

σ1

τ12

τ21

2θ1

(CW)

(CCW)

VVH

Hθ1

p2P1

P1

P2

P2

S1

S2

x

yp1

P2P1

P1

P2S1

S2

σ2

σ1

σ2

σ1

σav

σav

τ21τ21

θ1 x

y

θ1 x

y

σ1 σxx=P1

τmaxσxx

2---------=S1

S2

S2

P2

P1P2

H V

σxx σxx

P P

P1

S2

Cast Iron

Aluminum

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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 principal stresses at the point. (b) Determine the maximum shear stress at the point. (c) Draw the principal element. (d) Determine the normal and shear stresses on plane A.

Fig. C8.4

42o

A10 ksi

20 ksi

30 ksi

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Class Problem 8.2

Associate the stress cubes with the appropriate Mohr’s circle for stress.

Class Problem 8.3

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

40 ksi

8 ksi

cube 1

40ksi

8 ksi

40 ksi

8 ksi

cube 2 cube 3

circle B circle C circle Dcircle A

V

H

24o

V

H

24o

1 2

x

y

H

H

V V

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Class Problem 8.4

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

TT

τxθ

τθx

AluminumCast Iron

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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.

6 inF

60o

F = 12 kips

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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.

6 mm6 mm

30 mm

30 mm

0.4 m 0.4 m

A

BC

15 mm

P

August 2014 8-20

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Stress Transformations