Lab6A Theory

8/9/2019 Lab6A Theory

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MAE 244 Stress transformation and circular beam under combined loading Lab6-A .

Stress transformation and circular beam under combined loading

Stress transformation

When the stress state for mechanical element can be analyzed in a single plane, the materials is

said to be subject to plan stress. The general state of plan stress at a point is represented by acombination of normal components, σ x , σ y, and shear-stress component, σ xy. Realize that if

these three stress components at a point are known for an element orientated in the x and y

direction, then the three stress components representing the same state of stress at the point onan element orientated in the x’ and y’ directions will be different. In this section, we will

demonstrate how to transform the stress components from one orientation of an element to a

different orientation.

Figure 1 plane stress transformation

Figure 2 Free-body diagram of the segment

Before the transformation equations are derived, we first establish a sign convention for stress

components. Figure 1 shows the positive sign convention. σ x is positive since it acts to the right

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on the right-hand vertical face (in other words, σ x points outward from the acting face). σ xy is

positive since it acts upward on the right-hand vertical face and it acts to the left on the bottomface. Given the state of the plan stress in Figure 1, the orientation of the inclined plan on which

the normal and shear stress component are to be defined using the angle, θ . We take a segment

from the element. The resulting free-body diagram of the segment is shown in Fig.2. Applying

the equation of force equilibrium to determine '' , y x σ σ , and '' y xσ , we obtain:

0cos)cos(

sin)cos(sin)sin(cos)sin(;0 ''

=∆−

∆−∆−∆−∆=∑θ θ σ

θ θ σ θ θ σ θ θ σ σ

A

A A A A F

x

xy y xy x x

0sin)cos(

cos)cos(cos)sin(sin)sin(;0 '''

=∆+

∆−∆−∆+∆=∑θ θ σ

θ θ σ θ θ σ θ θ σ σ

A

A A A A F

x

xy y xy y x y

Hence,

)2cos()2sin(2

)2sin(2

'' θ σ θ σ

θ σ

σ xy

y x

y x+−−=

If the normal stress ' y

σ can be obtained by substituting (θ =θ +90o) for θ , which yields,

Considering, 2/)2cos1(cos,2/)2cos1(sin,cossin22sin22

θ θ θ θ θ θ θ +=−== , we obtain:

(1)

(2)

(3)

In-plane principle stress: To determine the maximum and minimum normal stress, we will

differentiate Eq.(1) with respect to θ and set the equation to zero,

02cos2)2sin2(2

'

=+

−

−= θ σ θ σ σ

θ

σ

xy

y x x

d

d

Solving the equation, we get the orientation pθ θ = of the planes of maximum and minimum

normal stress:

2




y x

xy

y x

xy

pσ σ

σ

σ σ

σ θ

−=

−=

2

2/)(2tan (4)

The solution of this equation has two roots, 12

pθ and 2

2 p

θ . And they have a relationship of π θ θ +=

2122

p p . Therefore, 2/21

π θ θ += p p . The principle stresses corresponding to 1

2 p

θ and

22

pθ :

(5)

(6)

This particular set of stress values is called in-plane principle stress. And the corresponding

planes on which they act are called principle planes of stress on which no shear stress acts.

Mohr’s circle-Plane stress: In the above equation, the plane transformation of stresscomponents is particularly suited for graphical interpretation. From Eq.(1) and (3), the parameter

θ can be eliminated, and this results in,

xy y x

y x y x

x

222

2

)2

()2

( ''' σ

σ σ σ

σ σ σ +

−=+

+−

For a specific problem, σ x, σ y and σ xy are know constant, thus the above equation can berewritten,

[ ] 2220''' R

y xavg x=−+− σ σ σ (7)

Where

2

y x

avg

σ σ σ

+=

If we establish a coordinate axes with the normal stress σ positive to the right and '' y x

σ

positive downward so that the positive direction of θ is counterclockwise, the geometrical

representation of Eq.(7) is a circle (Fig.3), which has a center located at (σ avg , 0) and the radius

of R. This stress circle is called Mohr’s circle in honor of Otto Mohr, who first employed it to

study plane stress problem.

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In the Mohr’s circle, Point A has coordinate (σ x, σ xy), which represents the normal and shear

stress components on the element’s right-hand vertical face. At this point, the x’ axis coincides

with the x axis. Hence, this represents 0=θ . Point G has coordinate (σ y, σ xy), which representsthe normal and shear stress components acting on the top face of the element.

Figure 3 Mohr’s Circle

It is easy to find the principle stress from the Mohr’s circle. The principle stresses σ 1 and σ 2 are

determined from the two points of the intersection of the circle with the abscissa, where the shear stress is zero. Line OB represents σ 1,

Line OD represents σ 2,

The orientation of the plane of primary stress can be derived from the angle between CA and CB

is 2θ p, which is expressed by Equation (4).

How to determine the stresses on arbitrary plane with Mohr’s Circle:

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Since the Mohr’s circle represents the stress-transformation equation graphically, it can be used

to determine the normal and shear stress components acting on any arbitrary plane (Figure 4).

The details are as follows:

1). Establish a coordinate system with the abscissa σ (normal stress) positive to the right and

the '' y x

σ (shear stress) positive downward.

2). Plot the center of the circle, C , which is located on the σ axis with the coordinate (σ avg , 0).3). Plot the reference point A(σ x, σ xy).4). Connect Point A to C and draw a circle with the radius of CA. Line CA is the reference line

(θ =0)

5). For any arbitrary normal and shear stress components acting on a specific plane defined by

the known angle θ , we can obtain the values of the normal and shear stress components

(σ x2, σ xy2). That is, we draw a line with the angle 2θ with respect to the reference lineCA, and to locate P .

6). The in-plane primary stress σ 1 and σ 2 are represented by Line OB and Line OD ,

respectively.

Figure 4 stresses on arbitrary plane with Mohr’s Circle

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Stress tensor

Stresses have both magnitude and direction. Figure 5 shows the resolution of the resultantstress T n into its normal and tangential components σn and τn. They can be further resolved into

stress components in the XYZ coordinate system. Therefore, at any point, the three-dimensional

stress state can be described using stress tensor [σ]. The first row corresponds to the stress statedescribed in the above figure.

Figure 5 The resolution of stress

Stress Transformation in Matrix Form

Equation (1-3) can also be written in the matrix form:

By using the notation m=cosθ and n=sinθ , the above two-dimensional stress transformationequations can also be written in the following general form:

(9)

Where the transformation matrix,

It will relate the vector of the stress components {σ'}, in the new (rotated) coordinate directions

to the vector of the stress components, {σ}, in the original reference coordinate system. The

inverse transformation is performed by using the “inverse” matrix of [T],

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(10)

Where the inversed transformation matrix

Strain Transformation in Matrix Form

The transformation equations for Cartesian strain components are identical with those for the

corresponding stress components. Just remember that for Engineering Shear Strain, it is twice

the stress term defined in the strain tensor, that is xy xyε γ 2= .

(11)

Where the transformation matrix

It will relate the vector of the strain components {ε' }, in the new (rotated) coordinate directions to

the vector of the strain components, {ε}, in the original reference coordinate system. The inverse

transformation can also be performed by using the “inverse” of the matrix [T],

(12)

Where the transformation matrix

All the properties associated with stress transformations can be extended directly to

strain transformation:

- Existence of Principal directions for normal strains, planes of Maximum shear strain- Use of Mohr’s circles

- Strain Invariant (independent to coordinate transformation)

Strain Measurement under Two-dimensional Stress State

Under two-dimensional stress state, σ x , σ y and σ xy are nonzero and the principal stress directions

are not known. Therefore, there are three unknowns that must be determined in order to specify

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the complete state of stress at the measured point. These unknowns can be the principal stress σ 1,

σ 2 and the principal angle θ or σ x , σ y and σ xy. The required strain can be obtained from a three-

element strain rosette mounted on the free surface of the specimen. Consider three gages alignedalong axes A, B and C, as shown in the following figure, the equations of strain transformation

give,

(13)

Or

(13)That is

(14)

For a given set of angles, β A , β B, and β C , the strain components (ε x , ε y and ε xy) can be obtained bysolving Equation (13) when normal strain ε A , ε B and εC are known.

(15)

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Then the principal strain ε1 , ε2 and the principal direction θ can be calculated as,

y x

xy

pε ε

ε θ

−=

22tan (16)

From the principal strain, principal stresses are obtained,

(17)

Two popular strain gage configurations

Case 1: Three-element rectangular rosette, β A =0º, β B =45º , and β C =90º :

Substitute the angles into Equation (13), we get

This yields,

9




(18)

For the principal angle θ p, two possible values exist. One is between ε 1 and the x axis while the

other is between ε 2 and the x axis. The range can be determined by the following equations,

C A B

o

p

o when ε ε ε θ +><< 2,900

C A B

o

p

o when ε ε ε θ +<<<− 2,090 (19)

1,0 ε ε ε ε θ =>= AC A

o

pand when

2,90 ε ε ε ε θ =<±= AC Ao

p and when

The corresponding principal stresses can be expressed in terms of the measured strains. The

expressions are,

(20)

Case 2: Delta rosette, β A=0º, β B=120º , and β C =240º :Measured normal strains are,

This yields,

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(21)

Where

C A B

o

p

o when ε ε ε θ +><< 2,900

C A B

o

p

o when ε ε ε θ +<<<− 2,090

1,0 ε ε ε ε θ =>= AC A

o

p and when

2,90 ε ε ε ε θ =<±= AC A

o

p and when

For principal stresses,

(22)

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Lab6A Theory