module 20: mohr’s circle transformationsejb9z/media/module20.pdfmodule 20: mohr’s circle...

Module Content:

Module Reading, Problems, and Demo:

MAE 2310 Str. of Materials © E. J. Berger, 2010 20- 1

Module 20: Mohr’s Circle TransformationsApril 16, 2010

1. The transformed stresses are called principal stresses and represent the “worst case” stresses in the structure.2. The principal (normal) stress and the maximum shear stress can be plotted on a special coordinate plane which graphically illustrates the state of stress.

Reading: Sec. 9.1-9.3Problems: 9-9 (plus principal stresses), 9-15Demo: stress visualizer (?)Technology: http://pages.shanti.virginia.edu/som2010

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! =

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "!(!xy!A sin ") cos "

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "!(!xy!A sin ") cos "

!(!y!A sin ") sin "

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "!(!xy!A sin ") cos "

!(!y!A sin ") sin " = 0

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "!(!xy!A sin ") cos "

!(!y!A sin ") sin " = 0

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, II• the mathematics follow an equilibrium calculation

• consider this example in the x’-direction:

• from here we use lots of trig identities:

2

!Fx

! = !x!!A!(!x!A cos ") cos "

!(!xy!A cos ") sin "!(!xy!A sin ") cos "

!(!y!A sin ") sin " = 0

sin2! =

1

2(1 ! cos 2!) cos2 ! =

1

2(1 + cos 2!)

sin 2! = 2 sin ! cos !

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transformation, III• the results:

3

!x! =!x + !y

2+

!x ! !y

2cos 2" + #xy sin 2"

!y! =!x + !y

2!

!x ! !y

2cos 2" ! #xy sin 2"

!x!y! = !

"x ! "y

2sin 2# + !xy cos 2#

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transform, IV• now what?

• we look for the angle which maximizes the transformed stresses...

4

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transform, IV• now what?

• we look for the angle which maximizes the transformed stresses...

4

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transform, IV• now what?

• we look for the angle which maximizes the transformed stresses...

4

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Transform, IV• now what?

• we look for the angle which maximizes the transformed stresses...

4

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

x

y

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

negative shear stress (Prob. 9-14)

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

negative shear stress (Prob. 9-14)

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Aside: Shear Stress Signs• when using the stress transformation equations, we must carefully consider the sign of the shear stress

• here is our convention: if the shear stress on the positive x-face points in the positive y-direction, then the shear stress is defined as positive

5

positive shear stress (Prob. 9-1)

x

y

negative shear stress (Prob. 9-14)

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

x

y

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

!x = 90 MPa

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

!x = 90 MPa

!y = 50 MPa

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

!x = 90 MPa

!y = 50 MPa

!xy = !35 MPa

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

!x = 90 MPa

!y = 50 MPa

!xy = !35 MPa

!x! =

90 + 50

2+

90 ! 50

2cos 60 + (!35) sin 60 = 49.7 MPa

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

An Example: Prob. 9-9• determine the stress on the plane of section A-B

6

30o

x

y

! = 30o! 2! = 60

o

!x = 90 MPa

!y = 50 MPa

!xy = !35 MPa

!x! =

90 + 50

2+

90 ! 50

2cos 60 + (!35) sin 60 = 49.7 MPa

!x!y! = !

90 ! 50

2sin 60 + (!35) cos 60 = !34.8 MPa

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

d!x!

d"= !

!x ! !y

2(2 sin 2") + 2#xy cos 2" = 0

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

d!x!

d"= !

!x ! !y

2(2 sin 2") + 2#xy cos 2" = 0

tan 2!p ="xy

(#x ! #y)/2

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

d!x!

d"= !

!x ! !y

2(2 sin 2") + 2#xy cos 2" = 0

tan 2!p ="xy

(#x ! #y)/2

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

d!x!

d"= !

!x ! !y

2(2 sin 2") + 2#xy cos 2" = 0

tan 2!p ="xy

(#x ! #y)/2

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximizing Stresses--θp

• how do we maximize the normal stress? differentiate wrt θ and set equal to zero

7

d!x!

d"= !

!x ! !y

2(2 sin 2") + 2#xy cos 2" = 0

tan 2!p ="xy

(#x ! #y)/2

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!

!

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!

!

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!

!

cos 2!p =("x ! "y)/2

!

"

!x!!y

2

#2

+ #2xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!

!

cos 2!p =("x ! "y)/2

!

"

!x!!y

2

#2

+ #2xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!

!

cos 2!p =("x ! "y)/2

!

"

!x!!y

2

#2

+ #2xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Principal (Normal) Stresses• now substitute θp into the stress transformation equations:

8

!1,2 =!x + !y

2±

!

"

!x ! !y

2

#2

+ "2xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximum Shear Stress• differentiate the shear stress equation wrt θ, set equal to zero

• solve for θs

• substitute into the shear stress equation

9

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximum Shear Stress• differentiate the shear stress equation wrt θ, set equal to zero

• solve for θs

• substitute into the shear stress equation

9

tan 2!s =!("x ! "y)/2

#xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximum Shear Stress• differentiate the shear stress equation wrt θ, set equal to zero

• solve for θs

• substitute into the shear stress equation

9

tan 2!s =!("x ! "y)/2

#xy

MAE 2310 Str. of Materials © E. J. Berger, 2010 20-

Theory: Maximum Shear Stress• differentiate the shear stress equation wrt θ, set equal to zero

• solve for θs

• substitute into the shear stress equation

9

!max =

!

"

"x ! "y

2

#2

+ !2xy

tan 2!s =!("x ! "y)/2

#xy