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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

• 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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

• 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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

sθ

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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

( 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

rotation about principal axis 1

p1

p1

σ2 – σ3

τ 23 = – τ 32 = -----------------2

p3

p3

p2

p2

rotation about principal axis 2

p1

p1

σ3 – σ1

τ 31 = – τ 13 = -----------------2

p3

p3

p2

p2

rotation about principal axis 3

(In-plane)

p1

p1

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

0.4 m

August 2012

8-20

6 mm

6 mm

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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

• 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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

• 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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

sθ

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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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.

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

( 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

rotation about principal axis 1

p1

p1

σ2 – σ3

τ 23 = – τ 32 = -----------------2

p3

p3

p2

p2

rotation about principal axis 2

p1

p1

σ3 – σ1

τ 31 = – τ 13 = -----------------2

p3

p3

p2

p2

rotation about principal axis 3

(In-plane)

p1

p1

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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)

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

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

Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm

0.4 m

August 2012

8-20

6 mm

6 mm

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