IMoM 5A Stressess PDF

 

 3D Stress Components Components From equilibrium principles: τxy = τyx , τxz = τzx , τzy = τyz

The most general state of stress at a point may be represented by 6 components

Normal Stresses

σ  x

σ   y   σ  z 

Shear Stresses

τ  xy

τ   yz   τ  xz 

 Normal stress ( σ ) : the subscript identifies the face on which the stress acts. acts. Tension is positive and compr compression ession is negative. Shear stress ( τ ) : it has two subscripts. The first subscript

denotes the face on which the stress acts. The second subscript denotes the direction on that face. A shear stress is positive if it acts on a positive face and positive direction or if it acts in a negative face and negative direction.

 

For static equilibrium τ x y = τ y x , τ xz = τ z x , τ zy = τ y z resulting in six independent scalar quantities. These six scalars can be arranged in a 3x3 matrix, giving us a stress  tensor.

σ

= σ ij

⎡σ  x ⎢ = ⎢τ  xy ⎣⎢τ  xz 

τ  yx σ  y τ  yz 

τ  zx ⎤

⎥ τ  zy ⎥ σ  z  ⎦⎥

The sign convention for the stress elements is that a  positive force on a positive face or a negative negative force on a negative face is positive. All others are negative.

The stress state is a second order tensor since it is a quantity quantity associated with two directions (two subscripts direction direction of the surface normal and direction of the stress). stress).

Same state of stress is represented by a different set of components if axes are rotated. There is a special set of components (when axes are rotated) where all the shear components are zero (principal stresses).

 

A property of a symmetric tensor is that there exists an orthogonal set of axes 1, 2 and  3 (called principal axes) with respect to which the tensor elements are all zero except for those in the diagonal.

σ

= σ ij

⎡σ  x ⎢ = ⎢τ  xy ⎣⎢τ  xz 

τ  yx σ  y τ  yz 

τ  zx ⎤

⎥ τ  zy ⎥ σ  z  ⎦⎥

σ

' = σ ij'

Eigen values

⎡σ 1 = ⎢⎢ 0 ⎣⎢ 0

0 σ 2 0

⎤ ⎥ 0 ⎥ σ 3 ⎦⎥ 0

In matrix notation the transformation is known as the Eigen-values. The principal stresses are the “new-axes” coordinate system. The angles between the “old-axes” and the “new-axes” are known as the Eigen-vectors.

principal stress

σ1 σ2 σ

Cosine of angle between X and the principal stress

Cosine of angle between Y and the principal stress

Cosine of angle between Z and the principal stress

k1

l1

m1

k2

l2

m2

k3

l3

m3

3  

Pl a n e St r e s s faces ces of the cubic elem element ent are free of stress. stress. For the State of stress in which two fa illustrated example, example, the state of stress is defined by

σ  x , σ  y , τ xy

and  

  σ   = τ   z 

 zx  

= τ zy = 0.

Sign Conventions for Shear Stress and Strain The Shear Stress will be considered positive when a pair of shear stress acting onproduce oppositea sides of the element counterclockwise (ccw) torque (couple).

 

 A shear strain in an element positive when the angle between between two positive positive faces element is positive (or two negative negative faces) is reduced, and is negative if the angle is increased. 1 2 σ2

⎡⎢σ  xx τ  xy ⎢ ⎢0

τ  yx σ  yy 0

0⎤ ⎥ 0

⎥ 0⎥

⎡⎢σ  x1 x1 τ  x1 y1 ⎢ ⎢ 0

τ  y1 x1 σ  y1 y1 0

0⎤ ⎥ 0

⎥ 0⎥

⎡⎢σ 1 ⎢0 ⎢0

σ1

0 σ 2 0

0⎤ ⎥ 0

⎥ 0⎥

⎣

 

⎦

⎣

⎦

⎣

Stresses on Inclined Sections

⎦

Knowing the normal and shear stresses s tresses acting in the element denoted by the xy axes, we will calculate the normal and shear stresses acting in the element denoted by the axis x1 y1.

y

y1

x

θ

1

x

σx1

θ

σx

τx1y

Equilibrium of forces: Acting in x1

1

τxy τyx σy σ  X 1 ⋅

 AO cosθ 

= σ  X  cosθ ⋅ AO + τ  XY sinθ ⋅  AO + σ Y Sinθ ⋅ AO

Sinθ  Cosθ 

+ τ YX  cosθ ⋅ AO

Sinθ  Cosθ 

 

Eliminating A o , secθ = 1/cosθ and τ xy=τ yx

σ  X 1

= σ  X  cos 2 θ   +  σ Y  sin 2 θ  + 2τ XY  sin θ  cos θ 

σ 

= σ 

Y 1

sin 2 θ   + σ  cos 2 θ  − 2τ  sin θ  cosθ 

 X 

 

Y 

XY 

 Acting in y 1

τx1y1Aosecθ = − σxAosinθ + τxyAocosθ + σyAotanθcosθ − τyxAotanθsinθ Eliminating A o ,  secθ = 1/cosθ

τ  x1 y1

and

τ xy= τ yx

= −σ  x ⋅ sin θ  ⋅ cos θ  + σ  y ⋅ sin θ   ⋅ cos θ  + τ xy

cos 2 θ  − sin 2 θ 

 

Transformation Equations for Plane Stress

Using the following trigonometric identit identities: ies:

Cos2θ = ½ (1+ cos 2θ) Sin2θ = ½ (1- cos 2θ) Sin θ cos θ = ½ sin 2θ

σ  x1

=

σ  x

2 σ  x

σ  y1

=

τ  x1 y1

+ σ  y

=−

+

+ σ  y 2 σ  x

σ  x

2 σ  x

− − σ  y

− σ  y

cos 2θ  + τ  xy sin 2θ 

− σ  y 2

cos 2θ  − τ  xy sin 2θ 

sin 2θ  + τ  xy cos 2θ 

2 These equations are known as the transformation equations for plane stress.

 

Special Cases Casee 1: Uni Uniaxi axial al str stress ess Cas σ  y

= 0   τ xy = τ  yx = 0

1  + Cos 2θ  ⎞ = σ  x ⋅ ⎛  ⎜ ⎟ ⎝  2  ⎠ ⎛ Sin2θ  ⎞ τ  x1, y1 = −σ  x ⋅ ⎜ ⎟ 2  ⎠ ⎝  σ  x1

Case 2 : Pure Shear

σ  x

= σ  y = 0

= τ  xy ⋅ Sin 2θ  τ  x1, y1 = τ  xy ⋅ Cos 2θ  σ  x1

σ  x1

=

τ  x1 y1

σ  x

=−

+ σ  y 2 σ  x

+

σ  x

− σ  y 2

cos 2θ  + τ  xy sin 2θ 

− σ  y 2

sin 2θ  + τ  xy cos 2θ  Case 3: Biaxial stress τ  xy

=0

σ  x1

=

τ  x1, y1

σ  x

+ σ  y 2

+

σ  x

− σ  y 2

= − σ  x − σ  y ⋅ Sin2θ  2

⋅ Cos 2θ 

 

An element in plane stress is subjected to stresses σ x=16000psi, σ y=6000psi, and τ xy=τ yx= 4000psi (as shown in figure below). Determine the stresses acting on an  counterclockwise - ccw). element inclined at an angle θ =45 o ( counterclockwise

Solution: We will use the following transformation equations:

σ  x1 = σ  y1 =

σ  x

+ σ  y 2

σ  x

τ  x1 y1 = −

+

σ  x

− σ  y 2

cos 2θ  + τ  xy sin 2θ 

+ σ  y − σ  x − σ  y cos 2θ  − τ  xy sin 2θ 

2 σ  x

2

− σ  y 2

sin 2θ  + τ  xy cos 2θ 

 

Numerical substitution  For x-axis  θ  = +450 (ccw) 0

Sin(2θ ) = Sin 90 ) = 1   (   (900 ) = 0 Cos (2θ ) = Cos

 For y-axis  θ  = +450

+ 900 (ccw)   (2700 ) = −1 Sin(2θ ) = Sin   (2700 ) = 0 Cos (2θ ) = Cos

σ  x

+ σ  y 2

σ  x

=

− σ  y 2

(16000 + 6000 ) 2

= 11000 psi

(16000 − 6000 ) =

2

= 5000 psi

= 4000 psi  σ  x1 = 11000 + 5000(0) + 4000(1) = 15000 psi σ  y1 = 11000 − 5000(0 ) − 4000(1) = 7000 psi   psi τ  x1 y1 = −5000(1) + 4000(0 ) = −5000 τ  xy

 Note: σ  x

+ σ  y  =  σ  x1 + σ y1

 

A plane stress condition exists at a point on the surface of a loaded structure such as shown below. Determine Determine the stresses actin acting g on an element tha thatt is oriented at a clockwise (cw) angle of 15 o with respect to the original element, the principal stresses, the maximum shear stress and the angle of inclination for the principal stresses Solution: We will use the following transformation equations: σ  x 1

=

τ  x1 y 1

+ σ  y

σ  x

=−

2 σ  x

+

− σ  y

σ  x

− σ  y 2

sin 2θ 

+ τ  xy cos 2θ 

2  For x-axis  θ  = −150 (cw)

  (− 300 ) = −0.5 Sin(2θ ) = Sin Cos 2θ 

σ  x

+ σ  y 2

σ  x

− σ  y 2

=

(− 46 + 12)

=

(− 46 − 12)

2

2

= −17 MPa = −29 MPa

( )

= Cos − 300 = 0.866

  (

)

σ  x σ 

+ σ  y = σ  x1 + σ  y1 = −1.4 MPa

 y1

= −19 MPa  σ  x1 = (− 17 ) + (− 29 )(0.866) + (− 19)(− 0.5) = −32.6 MPa

τ  xy

cos 2θ  + τ  xy sin 2θ 

= (− 29)(− 0.5) + (− 19)(0.866) = −  31MPa

τ   

x1 y1

 A rectangular plate of dimensions dimensions 3. 3.0 0 in x 5. 5.0 0 in is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 600psi in the long direction and a compressive stress of 250psi in the short direction. Determine the normal w acting perpendicular to the itline orin the weld and stress t w acting parallel to thestress weld.s(Assume s w is positive when acts tension andthe t w shear is positive when it acts counterclockwise against the weld).

Solution:

σ  x

+ σ  y 2

= 375 psi   σ  y1 = −25 psi σ  x1

σ  x + σ  y = σ  x1 + σ  y1 σ  y1 = −25 psi θ 

τ  x1 y1

− σ  y

= 425 psi  

τ  xy

= 0 psi 

2 = −375 psi

y 25psi

-375psi

375psi

Θ = 30.96o

3

=5⇒ 0 0 θ  = 30.96 ⇒ 2θ  = 61.92 For x-axis   tan

x

Stresses acting on the weld

σw τw

σ  x

= 175 psi 

25psi 375psi

θ

σw

θ = 30.96o

= -25psi and τw = 375psi

 

 Principal Stresses and Maximum Shear Stresses The sum of the normal stresses acting on perpendicular faces of plane stress elements is constant and independent of the angle θ .

σ  X 1

= σ  X  cos 2 θ   +  σ Y  sin 2 θ  + 2τ XY  sin θ  cos θ 

σ 

= σ 

Y 1

sin 2 θ   + σ  cos 2 θ  − 2τ  sin θ  cosθ 

 X 

 

σ  X 1

Y 

+ σ Y  1  = σ  X  + σ Y 

XY 

minimum normal and As we change the angle θ there will be maximum and minimum  shear stresses that are needed for design purposes.

=

The maximum and minimum normal

σ  x1

 stresses are known as the principal principal  stresses. These stresses are found by taking the derivative of σ x1 with respect to θ and setting equal to zero.

δσ  x1 δθ 

σ  x

+ σ  y

+

2

σ  x

− σ  y 2

cos 2θ  + τ  xy sin 2θ 

= −(σ  x − σ  y ) sin 2θ  + 2τ  xy cos 2θ  = 0 τ  xy

tan 2θ  P  = ⎛ σ   x

⎜⎜ ⎝ 

− σ  y  ⎞ ⎟⎟ 2  ⎠

 

The subscript p indicates that the angle θ p defines the orientation of the principal  planes. The angle θ p has two values that differ by 90 o. They are known as the principal  maximum σ  x1 is a  principal stress and stress for the othe otherr a angles. For. The one principal of these angles  minimum stresses occur in mutually perpendicular planes. τ  xy tan 2θ  P  = (σ  x − σ  y )

sin 2θ  P  = σ  x1

τ  xy  R

2

 

cos2θ  P  =

(σ  x

− σ  y )

2 R

 

= σ  x +2 σ  y + σ  x −2  σ  y cos 2θ  + τ xy sin 2θ 

for the maximum stress  θ  = θ P

+ σ  y

2

− σ  y ⎛ σ   x − σ  y  ⎞ ⎛ τ  xy  ⎞ σ  x + σ  y ⎛ σ   x − σ  y  ⎞ ⎛ 1 ⎞ 2 ⎛ 1 ⎞  xy σ 1 = + 2 ⎜⎝  2 R  ⎠⎟ + τ  xy ⎝  ⎜  R  ⎠⎟ = 2 + ⎜⎝  2  ⎠⎟ ⎜⎝⎝  R 2    ⎠⎟    ⎠⎟ + (τ  ) ⎜⎝⎝  R 2 ⎤ σ  x + σ  y ⎛ 1 ⎞ ⎡⎛ σ   x − σ  y  ⎞ 2 ⎟⎟ + (τ  xy ) ⎥ σ 1 = + ⎜ ⎟⎢⎜⎜ 2 2  R ⎝   ⎠⎢⎝  ⎥  ⎠ ⎣ ⎦ 2 2 σ  σ  σ  σ  σ  σ  +   −   −  ⎞ ⎛  ⎛   x  y  ⎞  x  y ⎟⎟ + (τ  xy )2   But    R = ⎜⎜ σ 1 = + ⎜⎜  x  y ⎟⎟ + (τ  xy )2 2 ⎝  2  ⎠ ⎝  2  ⎠ σ  x

σ  x

 

Pri rinc ncip ipa al Str tre ess sse es 2

The plus sign gives the algebraically

2

larger principal stress and the minus sign the algebraically smaller principal stress.

⎛ σ   x + σ  y  ⎞ ⎛ σ   x − σ  y  ⎞ 2 σ 1 = ⎜ τ  xy σ  Average + R ⎝  2  ⎠⎟ + ⎜⎝  2  ⎠⎟ + ( ) = ⎛ σ   x + σ  y ⎞ ⎟⎟ − σ 2 = ⎜⎜ ⎝  2  ⎠

⎛ σ   x − σ  y  ⎞ ⎜⎜ ⎟⎟ + (τ  xy )2 = σ  Average −R   ⎝  2  ⎠

This are the in-plane principal  stresses. The third stress is zero in  plane stress conditions

Maxi mum m Shear Shear Stre Str ess sses es Maximu The location of the angle for the maximum shear stress is obtained by taking the derivative of τ x1y1 with respect to θ and setting it equal to zero.

τ  x1 y1

= − σ  x −2  σ    y sin 2θ  + τ xy cos 2θ 

δτ  x1 y1 δθ 

= −(σ  x − σ  y ) cos 2θ  − 2τ  xy sin 2θ  = 0

(σ 

 x

tan 2θ S  = −

− σ  y ) τ  xy

2

 

(σ  x − σ  y ) cos 2θ  s1 =   sin 2θ  s1 = −       and   θ s1  R 2 R Therefore, 2θ s-2θ p=-90 o  or θ s= θ p +/- 45 o τ  xy

The planes for maximum shear stress occurs at 45 o to the principal  planes. The plane of the maximum  positive shear stress stress τ max is defined  by the angle θ S1 for which the following equations apply:

tan 2θ S  sin 2θ  S 

cos 2θ S 

=−

(σ  x

− σ  y )

2τ  xy cos 2θ 

 P  = − sin 2θ  P 

= θ P1 − 450

=−

1 tan 2θ  P 

= − cot 2θ  P 

sin 2θ  − 90 o − sin 90 o − 2θ  = cos(90 ( o − 2θ  P  P )) = cos(2θ  P  P  − 90 o ) 2

The corresponding maximum shear is given  by the equation

Another expression for the maximum shear stress

The normal stresses associated with the maximum shear stress are equal to

τ  MAX 

σ   x − σ   y  ⎞⎟ + (τ  xy )2  = R = ⎜⎛  ⎜ 2 ⎟ ⎝   ⎠ τ  MAX 

σ  AVER

=  (σ 1 −  σ 2 ) 2

=

σ  x

+ σ  y 2

 

 Equations of a Circle σ  x

=

+ σ  y

General equation

σ  x1

Consider 

σ  x1 − σ  AVER

 Equation (1)

σ  x

+

2

=

− σ  y

2 σ  x − σ  y 2

cos 2θ  + τ  xy sin 2θ  cos 2θ  + τ  xy sin 2θ 

⎤ ⎡σ  x − σ  y (σ  x1 − σ  AVER ) = ⎢   cos 2θ   + τ  xy sin 2θ ⎥ ⎣ 2 ⎦

2

2

τ  x1 y1

= − σ  x −  σ    y sin 2θ  + τ xy cos 2θ  2

2

 Equation (2)

2 τ  1 1  x  y ( )

= ⎡⎢− σ  x − σ  y sin  2θ  +  τ  xy cos 2θ ⎤⎥ 2 ⎦ ⎣

 Equation (1) + Equation (2)

 

⎡σ  x

− σ  y

⎣

2

(σ  x1 − σ  AVER )2  + (τ  x1 y1 )2 = ⎢

⎤

2

⎡ + ⎢−  ⎦ ⎣

cos 2θ  + τ  xy   sin 2θ ⎥

σ  x

− σ  y 2

⎤

2

sin 2θ  + τ  xy cos 2θ ⎥

⎦

σ   x − σ  y  ⎞ 2 σ   x − σ  y  ⎞ ⎡σ  x − σ  y ⎤ 2 ⎛  ⎛  2 2 2 ⎜ ⎟ ⎜ ( ) θ  τ  θ  θ  τ  θ  + = + + cos 2 sin 2 cos 2 sin 2 2 ⎢ 2 ⎥ ⎜ 2 ⎟  xy  xy ⎜ 2 ⎟⎟(τ  xy )sin 2θ  cos 2θ  ⎣ ⎦ ⎝   ⎠ ⎝   ⎠ 2 2    x − σ  y  ⎞ ⎡ σ  x − σ  y ⎤ ⎛  σ  ⎛ σ   x − σ  y  ⎞ 2 2 2 ⎜ ⎟ ⎜ ⎟  xy  xy ⎢⎣− 2 sin 2θ  + τ  cos 2θ ⎥⎦ = ⎝  ⎜ − 2  ⎠⎟ sin 2θ  + (τ  ) cos 2θ  − 2⎝  ⎜ 2  ⎠⎟(τ  xy )sin 2θ  cos 2θ  2

SUM 

⎛ σ   − σ   ⎞ = ⎜⎜  x  y ⎟⎟ + (τ  xy )2 =  R 2 ⎝  2  ⎠

(σ  x1 − σ  AVER  )    +  (τ  x1 y1 ) = R 2 2

2

 

 Mohr Circle The radius of the Mohr circle is the magnitude R.

C

⎛ σ   x +  σ  y ⎞ ⎜ ⎟ ⎝  ⎜ 2  ⎠⎟

 R

=

σ

 θP

2

⎛ σ   x − σ   y  ⎞   ⎜⎝  2   ⎠⎟ +

2

⎛ σ   x − σ  y  ⎞ 2    R = ⎜  2 τ xy ) ⎟ + ( ⎝   ⎠

τx

2

(τ  ) xy

y

The center of the Mohr circle is the magnitude

σ  x σ  AVER

τ

=

+ σ  y 2

2

State of Stresses

(σ  x1 − σ  AVER )  + (τ  x1 y 1 )  2

2

⎛ σ   x − σ  y  ⎞ ⎟⎟ + (τ  xy )2 = ⎜⎜ ⎝  2  ⎠

 

Alternative sign conversion for shear stresses: (a)clockwise shear stress, (b)counterclockwise (b)counterclockwi se shear stress, and (c) axes for Mohr’s circle.  Note that clockwise clockwise shear stresses are  plotted upward and and counterclockwise shear stresses are plotted downward.

 Forms of Mohr’s Circle a)

We can plot the normal stress σ x1 positive to the right and the shear stress τ x1y1 positive downwards, i.e. the angle 2θ will be positive when

counterclockwise or

 b)

We can plot the normal stre stress ss σ x1 positive to the right and the shear stress τ x1y1  positive upwards, i.e. i.e. the angle 2θ will be positive when clockwise.

Both forms are mathematically correct. We use (a)

 

Two forms of Mohr’s circle: is positive downward and the angle 2θ is  positive counterclockwise, counterclockwise, and τx1y1

τx1y1

is positive upward and the angle 2θ is positive

clockwise. ( Note:  Note: The first form is used here)

 

Construction of Mohr’s circle for plane stress.

 

At a point on the surface of a pressurized cylinder, the material is subjected to biaxial stresses σ x = 90MPa and σ y =  20MPa as shown in the element below. Using the Mohr circle, determine the stresses acting on an element inclined at an angle θ = 30 o (Sketch a properly oriented element). ( σ = 90MPa, σ = 20MPa and τ = 0MPa)  x

 y

Because the shear stress is zero, these are the principal stresses. Construction of the Mohr’s circle The center of the circle is σ  Average

=

σ  x 

+ σ  y = (90 + 20) = 55 MPa 2

2

2 2 The⎛ σ radius   x − σ  y ⎞ of the2 circle   is− 20 ⎞ ⎛ 90 2  xy R= ⎜ ⎝  2   ⎠⎟ + (τ  ) = ⎜⎝⎝    2  ⎠⎟ + (0) =  35MPa

 xy

 

Stresses on an element inclined at θ = 30 o By inspection of the circle, the coordinates of point D are

= σ  Average + R ⋅ Cos600 = 55 + 35 ⋅ Cos60 = 72.5 MPa τ  x1 y1 = − R ⋅ Sin60o = −35(Sin60o ) = −30.3 MPa σ  x1

0

σ  y1

= σ  Average − R ⋅ Cos60 = 55 − 35 ⋅ Cos60 = 37.5 MPa

 

An element in plane stress at the surface of a large machine is subjected to stresses σ x = 15000psi, σ y =  5000psi and τ xy = 4000psi. Using the Mohr’s circle determine the following: a) The stress stresses es acti acting ng o on n an elem element ent inclin inclined ed aatt an angle θ = 40 o  b) The principal str stresses esses and c) The The maxi maximu mum m shea shearr str stres esse ses. s. Construction of Mohr’s circle: Center of the circle (Point C):

σ  x  σ  Average

=

+ σ  y 2

(15000 + 5000) =

2

= 10000 psi

2

Radius of the circle:

2 ⎛ σ   x − σ  y  ⎞   − 5000 ⎞ 15000 ⎛  2 2   ⎟⎟ + (τ  xy ) = ⎜  R = ⎜⎜     psi ⎟ + (4000 ) = 6403 2 2 ⎝   ⎠ ⎝   ⎠

Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = 15000psi and τx1y1 = 4000psi Point B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 5000psi and τy1x1 = - 400 4000psi 0psi The circle is is now drawn thro through ugh points A aand nd B with ccenter enter C and radius R 

 

Stresses on an element inclined at θ = 40 o

These are given by the coordinates of  point D which is at an angle 2θ = 80 o from point A

By inspection the angle ACP1 for the principal stresses (point P1) is : 4000   o tan  ACP  =   ⇒ =  ACP  38 . 66 1 1 5000

(

)

Then, the angle P1CD is 80o –  38.66o = 41.34o

Knowing this angle, we can calculate the coordinates of point D (by inspection)

σ  x1

= σ  Average + R ⋅ Cos 41.34o = 10000 + 6403 ⋅ Cos 41.34o = 14810 psi

= − R ⋅ Sin41.34o = −6403(Sin41.34o ) = −4230 psi σ  y1 = σ  Average − R ⋅ Cos 41.34o = 10000 − 6403 ⋅ Cos 41.34o = 5190 psi

τ  x1 y1

 

And of course, the sum of the normal stresses is 14810psi + 5190psi = 15000psi + 5000psi  Principal Stresses The principal stresses are represented by  points P1 and P2 on Mohr’s circle. o

The angle it was found to be 2θ = 38.66 or θ = 19.3o

σ 1

= σ  Average + R = 10000 + 6403 = 16403 psi

σ 2

= σ  Average − R = 10000 − 6403 = 3597 psi

 

 Maximum Shear Stresses These are represented by point S1 and S2 in Mohr’s circle. Algebraically the maximum shear stress is given by the radius of the circle. The angle ACS1 from point A to point S1 is 2 θS1 = 51.34o. This angle is negative  because is measured measured clockwise on the circle. Then the corresponding θS1 value is –  25.7o.

 

3-D stress state

⎡15000 ⎢ ⎢⎢ 4000 ⎣ 0

4000 5000 0

0⎤

⎥ 0⎥ Psi 0⎥⎦

Transform to

⎡16403 ⎢ ⎢⎢ 00 ⎣

0 3597 0

0⎤

⎥ 0⎥ Psi 0⎥⎦

In matrix notation the transformation is known as the Eigen-values. The principal stresses are the “new-axes” “new-axes” coordinate system. system. The angles angles between the “old-a “old-axes” xes” and the the “new-ax “new-axes” es” are known as the the Eigen-v Eigen-vecto ectors. rs.

principal princi pal stress stress

Cosi osine ne of angl ngle e between X and and the principal princi pal stress stress

Cosi osine ne of angle angl e between Y and and the princi pal stress stress

Cosin osine e of angle between betwee n Z and and th the e principal princi pal stress stress

16403.1242

0.94362832

0.331006939

0

3596.876

-0.33101

0.943628

0

0

0

0

1

 

At a point on the surface of a generator shaft the stresses are σ = -50MPa, σ = 10MPa and τ = - 40MPa 40MPa as as shown shown in in the the x y xy figure. Using Mohr’s circle determine the following: (a)Stresses acting on an element inclined at an angle θ = 45o, (b)The principal stresses and (c)The maximum shear stresses Construction of Mohr’s circle Center of the circle (Point C):

σ  Average

=

σ  x 

+ σ  y 2

=

(− 50) + (10) 2

= −20 MPa

2

Radius of the circle:.

2 ⎛ σ   x − σ  y  ⎞ − 50   ) − (10 ) ⎞   ( ⎛  2 2   ⎟⎟ + (τ  xy ) = ⎜  R = ⎜⎜ ⎟ + (− 40) =  50 MPa 2 ⎝   ⎠ ⎝  2  ⎠

Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = -50MPa and τx1y1 = - 40MPa Point B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 10MPa and τy1x1 = 40MPa The circle is is now drawn thr through ough points A aand nd B with ccenter enter C and radius R.

 

Stresses on an element inclined at θ = 45 o

These stresses are given D by the coordinates of point (2θ = 90o of point A). To calculate its magnitude we need to determine the angles ACP2 and P2CD.

tan ACP2=40/30=4/3 ACP2=53.13o P2CD = 90o – 53.13o = 36.87o

Then, the coordinates of point D are σ  x1

= σ  Average + R ⋅ Cos36.87o = (− 20) + 50 ⋅ Cos36.87 o = −60 MPa o

o

τ  x1 y1 = R ⋅ Sin36.87 = 50 Sino 36.87 = 30 MPa σ  y1 = σ  Average − R ⋅ Cos36.87 = (− 20 ) + 50 ⋅ Cos36   .87 o

(

)

= 20MPa

And of course, the sum of the normal stresses is -50MPa+10MPa = -60MPa +20MPa

 

 Principal Stresses They are represented by points P1 and P on Mohr’s circle. 2

= σ  Average + R = −20 + 50 = 30 MPa σ 2 = σ  Average − R = −20 − 50 = −70 MPa σ 1

The angle ACP1 is 2θP1 = 180o + 53.13o = 233.13o or θP1 = 116.6o The angle ACP2 is 2θP2 = 53.13o or θP2 = 26.6o

 Maximum Shear Stresses Stresses These are represented by point S1 and S2

in Mohr’s circle. The angle ACS1 is 2θS1 = 90o + 53.13o = 143.13o or θ = 71.6o . The magnitude of the maximum shear stress is 50MPa and the normal stresses corresponding to point S1 is the average stress -20MPa.

 

3-D stress state

⎡ − 50 − 40 ⎢ ⎢⎢− 040 10 0 ⎣

0⎤

⎥ 0⎥ MPa 0⎥⎦

Transform to

⎡30 0 ⎢ ⎢⎢ 00 − 070 ⎣

0⎤

⎥ 0⎥ MPa 0⎥⎦

In matrix notation the transformation is known as the Eigen-values. The principal stresses are the “new-axes” “new-axes” coordinate system. The angles angles between the “old “old-ax -axes” es” and the the “new“new-axe axes” s” are know known n as the the Eigen-vectors.

principal princi pal stress stress

Cosin osine e of angle between X and and the principal princi pal stress stress

Cosi osine ne of angle angl e between Y and and th the e principal princi pal stress stress

Cosi osine ne of angle angl e between Z and and th the e princi pal stress stress

30

-0.44721359

0.894427193

0

-70

0.894427

0.447214

0

0

0

0

1

 

Hooke’s Law for Plane Stress The stress transformations equations were derived solely from equilibrium

conditions and they are material independent independent.. Here the material properties will be considered (strain) taking into account the following: a)The material is uniform throughout the body (homogeneous)  b)The material has the same pr properties operties in aall ll directions (isotropic) c)The material follows Hooke’s law (linearly elastic material) Hooke’s law: Linear relationship between stress and strain For uniaxi uniaxial al stress stress::  E = modulus of elasticity or Young’s modulus) ( E

υ  = − Poisson’s ratio:

  lateral  strain

 

axial strain  

=−

σ  =   E ε 

ε transverse ε longitudinal 

For pure shear : (G = Shear modulus of elasticity elasticity))

τ  =   Gγ 

 

Element of material in plane stress ( σ z = 0 ).

Consider the normal strains ε x , ε y , ε z in plane stress.  All are shown with positive elongation. The strains can be expressed in i n terms of the stresses by superimposing the effect of the individual stresses. For instance the strain ε x in the x direction: a)Due to the stress σ x is equal to σ x /E . b)Due to the stress σ y is equal to – ν σ y /E . The resulting strain is:

ε  x ε  y

=

σ  y

σ  x

− ν 

 E 

 E  σ  x σ  y

= −ν 

 E 

+

 E  σ  y σ  x ε  z  = −ν    − ν   E   E 

 

The shear stress causes a distortion of the element such that each z face becomes a rhombus.

  γ  XY 

= τ  XY  G

The normal stresses  σx and σ y have no effect on the shear strain γ xy

ε  x

=

σ  x

− ν 

σ  y

 E 

ε  y

 E 

= −ν  

σ  x  E 

or rearranging the equations:

σ  x

=

 E ε  x 2

+

ν  E ε  y 2

+

σ  y

ε  z  = −ν 

 E 

 σ Y   = 

σ  x

  − ν 

 E 

ν  E ε  x 2

+

σ  y

γ  XY   

=

τ  XY 

 E 

 E ε  y 2

+ 

G τ   XY

= G γ  XY 

1 − ν  ) (1 − ν  ) 1 − ν  ) (1 − ν  ) (These ( ooke e’s La L aw for f or plane p lane str stre ess equations are known collectively as Hook These equations contain three material constants (E, G and ν ) but only two are independent because of the relationship:

G

=

  E 

2(1  + ν )  

Special cases of Hooke’s law Uniax Un iaxia iall stre stress ss : σ  y

=0 

ε  x

=

σ  x

τ  xy  

ε y

= σ  y = 0

σ  x

= ε  z  = −υ 

Biaxial stress :

= 0)

Pure Shear :

=0

 E 

σz

σ  x

= ε  = ε  = 0 

ε   E 

τ  xy

 x

 z 

 xy

=

G

=0 σ  y

σ  x

ε  x

 y

γ 

τ  xy

=

 - υ 

ε y

σ  x = −υ 

Volume Change When a solid object undergoes strains, both its dimensions and its volume will change. Consider an object of dimensions a, b, c . The origi or igina nall v volu olume me is V o = abc and its final volume is V 1 = a + a ε x) b + b ε y) c + c ε z)

 

+

σ  y  

σ y σ  x ε z = −υ  - υ 

V 1= a b c 1 + ε x) 1 + ε y) 1 + ε z)  

Upon expanding the terms: V 1 = V o   1 + ε x + ε y + ε z + ε xε y + ε xε z + ε yε z + ε xε yε z)

For small strains: V 1 = V o   1 + εx + εy + εz )

The volume change is V = V 1 – V o = V o   εx + εy + εz )

The unit volume change e , also known as dilatation is defines as: e = V / Vo = εx + εy +

εz

Positive strains are considered for elongations and negative strains for shortening, i.e. positive values of e for an increase in volume.

ε  x

=

e=

σ  x

 E  ΔV 

V 

− ν 

σ  y  E 

= (σ  x + σ  y )

ε  y

= −ν  

(1 − 2υ )  E 

σ  x  E 

+

σ  y  E 

ε  z  = −ν 

σ  x

  − ν 

 E 

For uni uniaxi axial al str stress ess σ y = 0

σ  y  E 

 

e = σ  x

(1 − 2υ )  E 

We can notice that the maximum possible value of Poisson’s ratio is 0.5, because a larger value means that the volume decreases when the material is in tension

(contrary to physical behavior).  

STRAIN ENERGY DENSITY IN PLANE STRESS

The strain energy density u is the strain energy stored in a unit volume of the material.Because the normal and shear strains occ material.Because occur ur independent independently, ly, we can add the strain energy of these two elements to obtain the total energy. W ork done = Force x distance distance (σ x )(bc ) Work done in the x - direction = (aε  x ) 2 (σ y )(ac ) (bε  y ) Work done in the y - direction = 2

The sum of the energies due to normal stresses:

U  =

abc

Then the strain energy density (strain per unit volume) Similarly, the strain energy density associated with the

2

(σ   ε  

 x  x

u 1

= 12 (σ  x  ε   x + σ  yε  y )

u2   =

1   τ  xyγ  xy 2

shear strain: By combining the strain energy densities for the normal

+ σ  yε  y )

u=

1

(σ  ε 

+   σ  ε  + τ  γ 

)

 x  x

2

and shear strains:

 y  y

 xy  xy

 

2 σ  X 

The strain energy density in terms of stress alone:

u

=

The strain energy density in terms of strain alone:

u

=

 

  2 E 

=0 

τ  xy

=0

ε y

σ  x2 u=2

= −υε  x  

γ  xy

2 (1 − υ  2 )

=0

u

=

2  

TRIAXIAL STRESS  An element of the material material subjected to nor normal mal stresses σ x , σ y and σ z acting in three mutually perpendicular directions directions is said to be in a state of triaxial triaxial stres stress. s. Si Since nce th there ere iis s no shear in x , y or z faces then the stresses σ x , σ y and σ z are the principal stresses in the material.

−  υ 

σ  X σ Y   E 

+

2 τ XY 

2G

(ε  + ε   − 2υε  2  X 

 X  ε Y  ) +

2 Y   

G

2

For the special case of pure shear: σ  x

=0 

σ  y

=0 

ε y

2 τ  xy

 E ε  x2 or

2 E 

 E 

For the special case of uniaxial stress: σ  y

+

σ Y 2

u

= 2G

= ε  x = 0  2 Gγ  xy

or

u

=

2  

2 γ  XY 

 

If an inclined plane parallel to the z-axis is cut through the element, the only stress of the inclined face are the normal stress  σ and the shear stress  τ, both of which act  parallel to the xy plane.

Because these stresses are independent of the σ z, we can use the transformation equations of plane stress, as well as the Mohr’s circle for plane stress, when determining the stresses σ and τ in triaxi triaxial al stress stress..

The same general conclusion hold for normal and shear stresses acting on inclined planes cut through the element parallel to the  x and y axes. material in triaxial stress, the maximum shear stresses Maximum Shear Stress For a material occur on elements oriented at angles of 45 o to the x, y and z axes.

for the inclined plane // z-axis for the inclined plane // x-axis

for the inclined plane // y-axis

(τ  MAX  ) Z  = ± σ  X  2− σ Y  σ Y  − σ  Z  (τ  MAX  ) X  = ± (τ  MAX  )Y  = ±

σ  X 

2

− σ  Z 

2 The absolute maximum of the shear stress is the numerically largest of the above.  

The stresses acting on elements oriented at various angles to the x, y and z axes can be visualized with the aid of the Mohr’s circle. In this case σ > σ > σ x

y

z

 

Hooke’ss Law for Triaxi Hooke’ Triaxial al Str Stress ess

If Hooke’s law is obeyed, it is possible to obtain the relationship between normal stresses and normal strains using the same procedure as for plane stress.

ε  x

σ  x

=

(1 + υ  )(1 −

2υ  )

 [(1 − υ  )ε  x + υε   y + υε   z  ]

=

 E 

σ  z 

− ν   E  − ν   E  σ  x

ε  z 

= −ν  σ  x − ν  σ  y + σ  z   

 E 

− ν 

σ  z 

= −ν 

 E 

+

σ  y

ε  y

Or 

 E 

σ  y

σ  x

 E 

 E  σ   y σ  z 

= (1 + υ  )(1 − =

 x 2υ  ) υε   

 E 

(1 + υ  )(1 −

2υ  )

[ [υε 

 x

+ (1 − υ  )ε  y + υε   z  ] +  υε   y + (1 − υ  )ε  z  ]

Unit Volume Change

The unit volume change (or dilatation) for an element in triaxi element triaxial al stress stress is obtained obtained in the same manner as for plane stress.

e=

ΔV  V O

They are known as the Hooke’s law for triaxial stress.

= ε  X  + ε Y  + ε  Z  

If  Hooke' s laws apply, then

ΔV  (1 − 2υ )

e=

(σ  X  + σ Y  + σ  Z  )

= VO

 

E

Strain Energy Density

σ  yε  y σ  ε  The strain energy density for an σ  ε  u =  x2  x + 2 +  z 2  z   element elem ent in triaxi triaxial al stress stress is obtained by the same method In terms of  stresses : used for plane stress. 1

In terms of the strains:

u

=

 E  2(1 + υ )(1 − 2υ )

u

2  x

2  y

2  z 

= 2 E (σ  + σ  + σ 

υ 

   (σ  σ  ) -  E   x

 y

+ σ  xσ  z  + σ  yσ  z )

 [(1 − υ )(ε  x2 +  ε  y2 + ε  z 2 ) +  2υ (ε  xε  y + ε  xε  z  + ε  yε  z )] 

Spherical Stress A special case of triaxial stress, call called ed spherical spherical stress, occurs σ  x = σ  y   = σ z  = σ 0 whenever all three normal stresses are equal: The Mohr’s circle circle is reduc reduced ed to a single point. Any plane ccut ut through tthe he element will be free of shear stress and will be subjected to the same normal stress so and it is a principal plane. σ O ε O =  E  (1 − 2υ )   The volume  change The normal strains in spherical stress are also the same in all directions, provided the material (1 − 2υ ) is isotropic and if Hooke’s law applies: e = 3ε O = 3σ O

 

 K = bulk or volume modulus of elasticity

 E   K  = 3(1 − 2υ ) 

σ 0  K    = e

If ν = 1/3 then K = E If ν = 0 then K = E/3 If ν = 1/2 then K = infinite (rigid material having no change in volume) Element in spherical stress.

These formulas also apply to and element in uniform compression, for example rock deep within the earth or material submerged in water (hydrostatic stress).

 

 PLAIN STRAIN  Strains are measured by strain gages. A material is said to be in a state of plain strain if the only deformations are those in the xy plane, i.e. it has only three st strain rain components ε x, ε y and γ xy. Plain stress is analogous to plane stress, but under ordinary conditions they do not occur simultaneously Exception when σ x = -σ y and when ν = 0 ε x, ε y, and γ xy in the Strain components  xy  plane (plane strain). strain) .

 

Comparison of plane stress and plane strain.  

APPLICATION OF THE TRANSFORMATION EQUATIONS The transformation equations for plane stress are valid even when σ is not zero,  z  because σ z does not enter the equations of equilibrium.Therefore, the  transformations equations equations for plane stress can can also be used for stresses stresses in plane  strain. Similarly, the strain transformation equations equations that will be derived for the case of  plain strain in the xy plane are valid valid even when ε z is not zero, because the strain ε z does not affect the geometric relation relationship ship used for the derivation.Therefore, the  transformations equations equations for plane strain can can also be used for strains in plane  stress.

Transformation Equations for Plain Strain ε

ε

 x,  y

γ

We will assume that strain and  xy  xythe associated with the  plane are known. We need to determine the normal and shear strains (ε x1 and γ x1y1) associated with the x1 y1 axis. ε y1 can  be obtained from the equation o off ε x1 by substituting θ +90 for θ .

 

For an element of size dx , dy In the x direction: the strain ε x produces an elongation elongation ε x dx. The diagonal increases in length by ε x dx cos θ.

In the y direction: elongation ε y dy. the strain ε y produces an elongation The diagonal increases in length by ε y dy sin θ .

The shear strain γ xy in the plane xy produces a distortion of the element such that the angle at the lower left corner decreases by an amount equal to the shear strain. Consequently, the upper face moves to the right by an amount γ xy dy. This

deformation results in an increase in the length of the diagonal equal to: γ xy dy cos θ  

The total increase δ of the diagonal is the sum of the preceding three expressions, thus:   (Sinθ ) + γ  dy (Cosθ ) Δd  = ε  dx(Cosθ ) + ε  dy  x

But

 y

 xy

d  dx ⎞ dy ⎞ dy ⎞ Δ ⎛  ⎛  ⎛  ε  x1 =   = ε  x ⎜ ⎟(Cosθ ) + ε  y ⎜ ⎟(Sinθ ) + γ  xy ⎜ ⎟(Cosθ ) ds ⎝ ds ⎠ ⎝ ds ⎠ ⎝ ds ⎠ dx ds

dy

= Cosθ  ε  x 1

=

  ds

ε  x Cos 2θ  

= Sinθ 

+ ε   y Sin 2θ  + γ  xy Sin θ Cos θ 

Shear Strain γ x1y1 associated with x1 y1 axes.

This strain is equal to the decrease in angle  between lines in the mater material ial that we were re initiall initially y along the x1 and y1 axes. Oa and Ob were the lines initially along the x1 and  y1 axis respectively. The deformation caused by the strains ε x, ε y and γ xy caused the Oa and Ob lines to rotate and angle α and β from the x1 and y1 axis respectively. The shear strain γ x1y1 is the decrease

γ

in angle between the two lines that originally were  

at right angles therefore

 x1y1=

+

The angle α can be found from the deformations  produced by the strains strains ε x , ε y and γ xy . The strains  xy produce a clockwise  xy ε x and γ clockwise rotation, whil whilee the strain ε y  produces a counterclockwise counterclockwise rotati rotation. on.

Let us denote the angle of rotation produced by ε x , ε y and γ xy as α 1 , α 2 and α 3 respectively. α 1

= ε  x Sinθ 

dx ds dy

dx   = Cosθ   ds

 

dy ds

= Sinθ 

= ε  yCosθ  ds dy α 1 = γ  xy Sinθ 

α 2

ds

α  = −α 1 + α 2

− α 3 = −(ε  x − ε  y )Sin  θ Cosθ   − γ  xy Sin 2θ 

 

The rotation of line Ob which initially was at 90 o to the line Oa can be found  β  = (ε  x − ε  y )Sin   (θ  + 90)Cos(θ  + 90) + γ  xy Sin 2 (θ  + 90)  β  = −(ε  x − ε  y )Sinθ Cosθ   + γ  xyCos 2θ   by substituting θ +90 for θ in the expression for α . Because βis positive when clockwise. Thus:

γ xy

 

γ  x1 y1

= α  +  β  = −(ε  x − ε  y ) Sinθ Cos θ  +

2

2

2 Cos θ  − Sin θ 

[

]

Transformation Equations for Plain Strain

Using the following trigonometric identities: ε  x + ε  y ε  x − ε  y γ  xy + cos 2θ  + sin 2θ  ε  x1 = Cos2θ = ½ (1+ Co Cos 2θ) 2 2 2 2 Sin θ = ½ (1- Cos 2θ) γ  x1 y1 = − ε  x − ε  y sin 2θ  + γ  xy cos 2θ  Sin θ cos θ = ½ Sin 2θ

 2

(

2

)

2

ε  x

 In var iant  = ε  x  + ε  y  = ε  x1 + ε  y1

ε  Average

=

+ ε  y 2

 

PRINCIPAL STRAINS

γ  xy The angle for the principal strains is :

tan 2θ  P  ==

The value for the principal strains are

2 (ε  x − ε  y ) 2

2

ε 1

=

 x ε  (

γ  xy ε  x − ε  y

2

γ  xy  ⎞⎟ + ε  y ) + ⎛  ⎜⎜ ε   x − ε  y  ⎞⎟⎟ + ⎜⎛  ⎜ 2 ⎟ 2 2  ⎠ ⎝   ⎠ ⎝  2

ε 2

=

2

= (ε  x + ε  y ) − ⎜⎛ ε   x − ε  y  ⎞⎟ + ⎛  ⎜ γ  xy  ⎞⎟ 2 ⎝  2  ⎠ ⎝  2  ⎠ Maximum Shear

The maximum shear strains in the xy plane are associated associated with axes at 45 o to the directions of

γ  Max 2

2

⎛ ε   x − ε  y ⎞ ⎛ γ  xy ⎞ ⎟⎟ +  ⎜⎜ ⎟⎟ = + ⎜⎜ ⎝  2  ⎠ ⎝  2  ⎠

2

or γ Max = (ε 1 − ε 2 )

the principal strains  For isotropic mate materials, rials, at a given point in an stressed body, the principal strains

 and principal stresses stresses occur in the same directions.  

MOHR’S CIRCLE FOR PLANE STRAIN

It is constructed in the same manner as the Mohr’s circle for plane stress s tress with the following similarities:

σ  x σ  y τ  xy σ  x1

ε  x ε   y γ  xy 2 ε  x1

τ  x1 y1 γ  x1 y1

2

 

Strain Measurements An electrical-resistance strain gage is a device for measuring normal strains (ε ) on

the surface a stressed The gages are of small (less than than object. ½ inch) made of wires that are bonded to the surface of the o object. bject. Each gage that is stretched or shortened when the object is strained s trained at the point, changes its electr electrical ical resistance. This change in resistance is converted into a measurement of strain.

From three measurements it is possible to calculate the strains in any direction. A group of three gages arranged in a particular pattern is called a strain rosette. Because the rosette is mo mounted unted in the surface surface of the body, where the materia materiall is in  plane stress, we can can use the tra transformation nsformation equa equations tions for plane strain to calc calculate ulate the strains in various directions.

 45° strain rosette, and element oriented at an angle θ to the xy axes.

 

General Equations

Other Strain Rosette

 

An element of material in plane strain undergoes the following strains:   εx =340x10-6 the following quantities: ; γ x y = 180x ;   εy = 110x 110x 10-6 180x 10-6 . Determine  o (a) the strains strains of an element element orient oriented ed at an angle angle θ = 30 ; (b) the princi principal pal strains strains and and (c) the maximum maximum shear strains. strains. (a) Element oriented at an angle θ = 30o (2θ = 60o) ε  x

+ ε  y +

ε  x

ε  x1

=

ε  x1

=⎢ + 2 ⎣ = 360 × 10 − 6

ε  x1

γ  x1 y 1 2 γ  x1 y 1 2

2 ⎡ 340 + 110

=− = ⎡⎢ − ⎣

ε  x

− ε  y 2

340

 xy cos 2θ  + γ  sin 2θ  2 2 340 − 110 180 ⎤ Cos 60 + Sin 60 ⎥ × 10 − 6 2 2 ⎦

ε  x

sin 2θ  +

− 110 2

− ε  y

Sin 60

γ  xy 2

+ ε  y = ε  x1 + ε  y1 340μ  + 110μ  = 360μ  + ε  y1 ε  y1 = 90μ 

cos 2θ 

+ 180 2

Cos 60 ⎤⎥ × 10 − 6

⎦

= − 55 × 10 − 6

ε  Average

= 225 μ 

 

Angle of Rotation (b) Principal Strains and Angle 2

ε 1, 2

γ  xy

2

= (ε  x +2 ε  y ) ± ⎜⎛  ε   x −2 ε  y  ⎞⎟ + ⎛  ⎜ γ 2xy  ⎞⎟ ⎝   ⎠ ⎝   ⎠ 2

tan 2θ  P  = ε  x 2

⎛ 340  μ  − 110 μ  ⎞ + ⎛ 180 ⎞ = 370 μ  ε 1 = 225 μ  + ⎜ ⎟ ⎜ ⎟ 2 ⎝   ⎠ ⎝  2  ⎠ 2 2 340   μ  − 110 μ  ⎞ ⎛ 180 ⎞ ⎛  ε 2 = 225 μ  − ⎜ ⎟ +⎜ ⎟ = 80 μ  2 2 ⎝   ⎠ ⎝   ⎠

θ  P  = 19 0

− ε  y

− 55 μ  = 340 μ  − 110 μ  = 0.7826

 

(c) In-Plane Maximum Shear Strain 2

2

2

2

180μ  ⎞⎟ = 145μ  = + ⎛  ⎜ ε   x −2 ε  y ⎞⎟ + ⎛  ⎜⎝ 340μ 2- 110μ  ⎞ ⎟ + ⎛  ⎜ ⎜ γ 2 xy ⎞⎟  = ⎛   ⎠ ⎝  2  ⎠  ⎠ ⎝   ⎠ ⎝  γ  Max = (ε 1 − ε 2 ) = 370μ  − 80   μ  = 290μ 

γ  Max 2

ε  Average

= 225 μ 

(d) Out-of-Plane Maximum Shear Strain

γ 

= (ε  − ε  ) = 370   μ  −  0μ  = 370μ 

 Max

1

3

 

Transformation Equations ε 

=

 X  1

ε 

cos 2 θ 

+

 X 

ε   sin 2 θ 

 Y 

γ  + ⎛  XY   ⎞ 2 sin ⎜⎝⎝   2  ⎠⎟

ε  Y  1

γ  X 1Y 1 2

=

ε  X  sin

2

θ 

θ  cos θ 

+ ε   Y 

cos

2

θ 

γ  XY   ⎞ − ⎛  ⎜ ⎟ 2 sin ⎝  2  ⎠

θ  cos θ 

= −ε  X  sin θ  cos θ  +  ε Y  sin θ  cos θ  + γ XY  (cos 2 θ  − sin 2 θ ) 2

⎡ ⎤ ⎤ ⎡ ⎢ ε  X  ⎥ ⎢ ε  X 1 ⎥ ⎢ ε Y 1 ⎥ = [T ]× ⎢ ε Y  ⎥ ⎢ γ  ⎥ ⎢ γ  ⎥  X  Y  1 1 ⎢  XY  ⎥ ⎥ ⎢ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡⎢ ε  X  ⎤⎥ ⎡⎢ ε  X 1 ⎤⎥ ⎢ ε Y  ⎥ = [T ]−1 × ⎢ ε Y 1 ⎥ ⎢ γ  ⎥ ⎢ γ  ⎥ ⎢  XY  ⎥ ⎢  X 1Y 1 ⎥ ⎣⎢ 2 ⎦⎥ ⎣⎢ 2 ⎦⎥

⎡ cos2 θ  sin 2 θ  2 sin θ  cosθ  ⎤ cos 2 θ  − 2 sin θ  cosθ  ⎥⎥ [T ] = ⎢⎢ sin 2 θ  ⎢− sin θ  cosθ  sin θ  cosθ  (cos 2 θ  − sin 2 θ )⎥ ⎦ ⎣

⎡ ⎢ ⎢

ε  X 1

⎤ ⎥ ⎡ ⎥=⎢

ε Y 1 ⎢ γ  X 1Y 1 ⎥ ⎢⎣ 2 ⎥⎦

2

2

cos θ 

sin θ 

sin 2 θ 

cos 2 θ 

⎢⎣− sin θ  cosθ 

sin θ  cosθ 

2 sin θ  cosθ 

⎡ ⎤ ⎤ ⎢ ε  X  ⎥ ⎥⎢ ⎥ ε Y 

− 2 sin θ  cosθ  2 2 cos sin θ  θ )⎥⎦ ⎢ γ  XY  ⎥ − ( ⎢⎣ 2 ⎥⎦

 

For Θ=30 degrees

⎡ ⎤ 2 2 ⎡ ⎤ ε  ε  cos 30 sin 30 2 sin 30 cos 30  X 1  X  2 ⎥ ⎢ ⎢ ε Y 1 ⎥ = ⎢ sin 2 30 − 2 sin 30 cos 30 ⎥ ε Y  ⎥ cos 30 ⎢ ⎢ γ  ⎥ ⎢ γ  ⎥ 2 2 ⎢ ⎥ ⎢  X 1Y 1 ⎥ ⎣− sin 30 cos 30 sin 30 cos 30 (cos 30 − sin 30)⎦ ⎢  XY  ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡ ⎤ 0.25 0.876 ⎤ ⎡340⎤ ⎡ 361.3 ⎤ ⎢ ε  X 1 ⎥ ⎡ 0.75 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ε Y 1 ⎥ = ⎢ 0.25 0.75 − 0.876 110 μ  = 88.6 μ  ⎥⎢ ⎥ ⎢ ⎥ ⎢ γ  ⎥ ⎢ 0.5 ⎦⎥ ⎣⎢ 90 ⎦⎥ ⎢⎣  X 21Y 1 ⎥⎦ ⎣⎢− 0.438 0.438 ⎣⎢− 55.8⎦⎥ ⎡

⎤

 

⎡ ε  ⎢  xx [ε ] = ⎢ 1 γ  xy ⎢2 ⎢ 1 γ  ⎣⎢ 2  xz 

1 2

γ  yx

ε  yy

1 2

γ   yz 

⎤ γ  zx ⎥ 2   1 γ  zy ⎥ = Strain _ Tensor  2 ⎥ ε  ⎥  zz  ⎦⎥ 1

Example 1

⎡⎢ ε  xx ⎢1 [ε ] = ⎢ γ  xy ⎢2 ⎢⎣ 12 γ  xz 

1

2 γ 

 yx

ε  yy 1 γ  yz  2

 zx 2 γ  ⎤⎥ ⎡ 340 180 2 1 ⎥ ⎢180 110 γ  zy ⎥ = ⎢ 2 2 ⎥ ⎢ 0 0 ε  zz  ⎥ ⎣⎢

⎡371 ⎢  Eigen _ Values ⇒ ⎢ 0 ⎢⎣ 0

⎦

0 79 0

0⎤

⎥ 0⎥ μ  0⎥⎦

0⎤

⎥ 0⎥ μ  ⎥ 0 ⎦⎥

 

A 45 o strain rosette (rectangular rosette) consists of three electrical-resistance strain gages, arranged to measure strains in two perpendicular directions and also at a 45 o angle (as shown  below). The rosette is bonded to the surface of the structure before it is loaded. Gages A, B and C measure the normal strains  εa, ε b and ε c in the directions of the lines Oa, Ob and Oc, respectively. Explain how to obtain the strains  ε , ε and γ , associated with an element oriented at an x1 y1 x1y1 angle θ to the xy axes.

 

2

ε  a

=

ε  x cos θ 

+

2

ε  y sin θ 

⎛  γ  xy  ⎞ + ⎝  ⎜⎝  2  ⎠⎟ 2 sin

θ  cos θ 

Angles with respect to x-axis: (a) is zero ; (b) is 45 degrees CCW ; (c) 90 degrees CCW

ε  a

⎛  γ   ⎞ = ε  x cos 2 0 + ε    y sin 2 0 + ⎜⎜ xy ⎟⎟ 2 sin 0 cos 0 ⎝  2  ⎠

ε b

= ε  x cos 2 45 + ε    y sin 2 45 + ⎛  ⎜ γ 2xy  ⎞⎟ (2 sin 45 cos 45 ) ⎝   ⎠

ε  x

= ε a

ε  y

= ε b

⎛  γ  xy  ⎞ ⎟⎟ (2 sin 90 cos 90 ) ε b = ε  x cos 90 + ε  y sin 90 + ⎜⎜ ⎝ 2 ⎠  

2

 

2

γ  xy

= 2ε b

ε a

ε c

The following results are obtained from a 600 strain gauge rosette: Strain in direction of strain gauge A = 750μ; Strain in direction of SG B, 600 to A = 350μ; Strain in direction of SG C, 1200 to A = 100 μ. Determine the principal strains and their directions. ε  a

= ε   x = 750 μ 

⎛  γ   ⎞ = ε  x cos 2 60 + ε  y sin 2 60 + ⎜⎜  xy ⎟⎟ (2 sin 60 cos 60 ) 2 ⎛  γ  xy⎝   ⎞  ⎠ ε b = ε  x (0 . 25 ) + ε  y (0 . 75 ) + ⎜⎜   ⎟⎟ (0 . 433 ) ⎝  2  ⎠ ⎛  γ  xy  ⎞ 2 2 ε b = ε  x cos 120 + ε  y sin 120 + ⎜ 2 ⎟ (2 sin 120 cos 120 ) ⎝   ⎠ ⎛  γ  xy  ⎞ ⎟⎟ (− 0 . 433 ) ε b = ε  x (0 . 25 ) + ε  y (0 . 75 ) + ⎜⎜ 2 ⎝   ⎠

ε b

⇒

ε  y = 50 μ  γ  xy = 289 μ 

 

⎡ ε  ⎢  xx ⎢ [ε ] = ⎢⎢ 12 γ  xy ⎢ 1 γ  ⎢⎣ 2  xz 

1

γ  yx

1

⎤

γ  zx ⎥

⎡ 750 289 2 ⎥ ⎢ ε  yy 1 γ  zy ⎥ = ⎢289 2 50 2 ⎥ ⎢ 0 0 1 ⎥ ⎢ ⎣ γ  yz  ε  zz  ⎥⎦ 2 ⎡779 0 0⎤ ⎥ ⎢ 21 0 μ   Eigen _ Values ⇒ 0 ⎥ ⎢ ⎢⎣ 0 0 0⎥⎦ ⎡ 0.981 0.1945 0⎤ ⎥ ⎢  Eigen _ Vectors ⇒ − 0.1945 0.981 0 ⎥ ⎢ ⎢ 0 0 1⎥ ⎦ ⎣ 2

2

0⎤

⎥ 0⎥ μ  ⎥ 0⎥ ⎦ ArcCos(angle)=0.981 Angle =11.2degress

 

(A) Using the transfor transformatio mation n equations equations define the maximum maximum and minimum minimum principal strains, maximum shearing strain and principal angles given ε X = 3500μ ; ε Y = 700μ and γ XY = -1050μ (B) Repeat Repeat us using ing the the Mohr’s Mohr’s circle circle..

⎡ 3500 − 1050 2 ⎢ [ε ] = ⎢− 1050 2 700 ⎢ 0 0 ⎢ ⎣ ⎡3595.2 ⎢ 0  Eigen _ Values ⇒ ⎢ ⎢ 0 ⎣⎡0.984   ⎢ Eigen _ Vectors ⇒ 0.178 ⎢ ⎢⎣ 0

ArcCos(angle)=0.984

0⎤

⎥ 0⎥ μ  ⎥ 0⎥ ⎦0

Angle =10.28degress (c) In-Plane Maximum Shear Strain 0⎤

γ  Max = (ε 1 − ε 2 ) = 3595. 2μ  − 604   .8μ  = 2990.4μ 

⎥ ⎥ 0 0⎥ − 0.178 ⎦0⎤ ⎥ 0.984 0 ⎥ 0 1⎥⎦

604.8 0 μ 

(d) Out-of-Plane Maximum Shear Strain

γ  Max

= (ε 1 − ε 3 ) = 3595   .2μ  −  0μ  = 3595.2μ 

 

The state of stress at a point in a structural member is determined to be as shown. Knowing that for this material E=210GPa and ν =0.3, use the Mohr’s circle to determine: (1) the principal stresses; (2) the in-plane maximum shear stress; (3) the absolute maximum shear stress; (4) principal angles; (5) the strains and the principal strains; (6) the maximum shear strain; (7 (7)) the pr principal incipal angles angles.. 14MPa

1. Principal Stresses

y

σ 1, 2

11.2MPa

x

⎛ σ   x

=  ⎜⎜ ⎝ 

56MPa

+ σ  y  ⎞ ⎟± 2  ⎠⎟

⎛ σ   x ⎜⎜   ⎝ 

2

− σ  y ⎞ ⎟⎟ + (τ  xy )2 = σ  Average   ±R 2  ⎠

2

⎛ − 56  − (− 14 ) ⎞ + (11.2)2 = 23.8 MPa  R = ⎜ ⎟ 2 ⎝   ⎠ ⎛ − 56  + (− 14) ⎞ σ  Average = ⎝  2 ⎜⎝   ⎠⎟ = −35 MPa σ 1 = 0 MPa

= −35 + 23.8 = −11.2 MPa σ 2 = −35 − 23.8 = −58.8 MPa σ 1

⎡− 56

11.2

0⎤

⎡0

= −11.2 MPa σ 3 = −58.8 MPa σ 2

0

0

⎤

[σ ] = Stress _ Tensor  = ⎢⎢11.2 − 14 ⎣⎢

 

0

0

⎥ ⎥

 Eigen _ Values ⇒ 0

⎢ ⎢

− 11.2

0⎦⎥

⎣⎢0

0

0 MPa  

 2. In-Plane Maximum Shear Stress

τ  Max

0

⎥ MPa ⎥

58.8⎦⎥

= R = 23.8 MPa

 3. The Absolute Maxim Maximum um Shear Stress (Out of plane)

τ  Max

σ 1  − σ 3

=

=

0 − (− 58.8)

2

= 29.4 MPa

2

x-axis and the  4. Angle between the x-axis  Principal Stresses Stresses ( In _  Plane) = tan (2θ  P ) =

τ  xy (σ  x − σ  y )

=

11.2 (−  56 − (− 14))

2 tan (2θ 

0 ⎡⎢ 0  Eigen _ Vectors Vectors ⇒ 0.2425 0.9701 ⎢ ⎢⎣ 0.9701 − 0.2425

For σ1=0 ArcCos(angle)=0.0

)=

11.2

= −0.533

0⎤⎥ − 21 0 2θ  P  = −28.07 deg  P 

⎥ 1⎥⎦

For σ2=-11.2MPa ArcCos(angle)=0.2425

For σ3=-58.8MPa ArcCos(angle)=0.9701

2

Angle =90degress

Angle =76degress

Angle =14degress

 

 5. Strains and Principal Strains ε 

=

σ  x

 x

 E 

ε  y

= −ν 

ε  z 

− ν 

− ν 

σ  z 

ε  x

 E  σ  y

σ  x

+

 E 

 E  σ  − ν   z   E   E 

= −ν  σ  x − ν  σ  y + σ  z   

γ  XY   

G

σ  y

=

=

τ  XY  G  E 

2(1 + ν )

⎡⎢ ε  xx ⎢1 [ε ] = ⎢ γ  xy ⎢2 ⎢1 ⎣⎢ 2 γ  xz 

=

210000 ε  y

ε  z 

11.2

= − (0 . 3 )

210000 − 56

ε  yy 1 2 γ  yz 

+

= − (0 . 3 ) 210000 − (0 . 3 ) 210000 = (80 + 20 )μ  =  100 μ 

G 80770 210 = 80  .77GPa 2(1 + 0.3)

ε 1, 2

ε  zz 

⎡100

0

⎤

(ε 

+ ε  y )

2

⎛ ε   x − ε  y  ⎞ ⎛ γ  xy  ⎞ ⎟⎟ + ⎜⎜ ⎟⎟ = ± ⎜⎜ 2 2 2  ⎠ ⎝   ⎠ = ε  Average ± R ⎝   x

ε  Average

⎦

⎦⎥ ⎣ 0

210000

= (80 − 66 . 6 )μ  = 13 . 4 μ 

− 14

ε 1, 2

=

= (− 266 . 6 + 20 )μ  = − 246 . 6 μ 

210000 − 14

− 56

1 γ  zx ⎤⎥ ⎡ − 246.6 139 2 0 ⎤ 2 ⎥ 1 ⎥ ⎢ 139 13.4 0 ⎥ μ   R γ  zy ⎥ = ⎢ 2 2 ⎥ ⎢ ⎥ 0 0 100 ⎥ ⎢ ⎥

1 γ  yx 2

− 14

− (0 .3 )

= 139μ 

=

γ  xy

τ  xy

− 56

=

=

(− 246.6 + 13.4)

= −116.6μ 

2 2   .6 − 13.4 ⎞ ⎛ − 246

= ⎜ ⎝ 

2

2

⎛ 139 ⎞ = 147.4 μ  ⎟ +⎜ ⎟ 2  ⎠ ⎝   ⎠

ε 1 = 30.8μ  ε 2 = −264 μ 

2

 Eigen _ Values Values ⇒  

⎢ ⎢ ⎢ ⎣

0

30.8

0

0

⎥ μ  ⎥ ⎥ − 264.0⎦ 0

6. Maximum Shear Strain and Absolute Maximum Shear Strain 2

2

  .6 − 13.4 ⎞ ⎛ 139 ⎞ ⎛ − 246 + = 147.4μ  in _  plane = R = 2 ( 2 ) ⎝  ⎜⎝   ⎠⎟ ⎜⎝⎝   2  ⎠⎟ γ  Max = 294.8μ  γ  Max

γ  Max 2 γ  Max

(out  _ of  _  plane ) =

ε 1 − ε 3  2

= 364μ 

=

100 − (− 264 )

= 182μ 

2

7. In-plane and Out-of-plane angles γ  xy tan 2θ  P  ==

2 (ε  x − ε  y )

= 2

γ  xy

139

  = = −0.534 ε  x  − ε  y (− 246 .6 − 13 .4 )

0 ⎡ 0 ⎢ Eigen _ Vectors Vectors ⇒ 0.2425 0.9701 ⎢ ⎢⎣0.9701 − 0.2425 For ε1=100μ

For ε2=30.8μ

tan (2θ  P ) = −0.533 2θ  P  = −28.07 deg

0⎤

⎥ ⎥ 1⎥⎦ 0

For ε3=-264μ