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Exercise 17 Stress-Strain Relations: Hooke's Law 1

Exercise 17

Stress-Strain Relations: Hooke's Law

17-1 Introduction

In Exercise 16, we've learned that, if we apply uniaxial normal forces along X-direction, the only nonzero stress

components is X

and the nonzero strain components are

X=

X

Eand

Y=

Z=

X=

X

E

where E is the Young's modulus (16-3) and is the Poisson's ratio (16-4) of the material. For structural steel,E = 200

GPa and = 0.3. Similarly, if we apply uniaxial normal forces along Y-direction, the only nonzero stress components is

Y

and the nonzero strain components are

Y=

Y

Eand

Z=

X=

Y=

Y

E

Similarly, if we apply uniaxial normal forces along Z-direction, the only nonzero stress components is Z

and the

nonzero strain components are

Z=

Z

Eand

X=

Y=

Z=

Z

E

Now, consider that we apply multiple normal forces along all three direction. The three normal stresses (

X

,

Y,

Z) are all nonzero and the three normal strains are

X=

X

E

Y

E

Z

E

Y=

Y

E

Z

E

X

E (1)

Z

=

Z

E

X

E

Y

E

Explanation of the above equations is easy. For example, the first equation states that the normal strain X

is a

combination of three contributions: X

, Y

, and Z

. Without loss of generality, we assume that all three stress are

tensile, then the contribution ofX

is to elongate a strain ofX

E , the contribution of Y

is to shrink an strain of

Y

E , and, similarly, the contribution of Z

is to shrink an strain ofZ

E .

Shearing (twisting) doesn't involve Poisson's effects. The relations between the shear stresses and the shearstrains are simply

XY

=

XY

G,

YZ=

YZ

G,

ZX=

ZX

G (2)

where Gis the shear modulus (Section 16-5). Eqs. (1-2) are called the Hooke's lawfor an isotropic material. In this section, we will verify the Hooke's law usingresults of simulation conducted in Exercise 14 (the C-Bar).

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2 Copyright by Huei-Huang Lee

17-2 Start Up

[1] LaunchWorkbench.

[2] Open theproject "CBar,"

saved in Exercise14.

[3] Double-click to start up

.

17-3 Strain Components

Remember that we were investigating the point as

shown [1]. At the end of 14-5, we've written down

the six strain components. They are tabulated in the

table below [2].

[1] The origin of ishere. We are studyingthe strains and stresses

at this point.

StrainComponent

Strain Value(Dimensionless)

X 13.40510

6

Y 100.1510

6

Z 49.77310

6

XY

262.02106

YZ 0

ZX

0

[2] The straincomponents.

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Exercise 17 Stress-Strain Relations: Hooke's Law 3

17-4 Stress Components

[1] Highlight and

insert a .

[2] Select any one ofthe paths. Here weselect .

[3] Right-click andselect .

Repeat this step fourmore times.

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4 Copyright by Huei-Huang Lee

[4] Highlight and select

for.

[5] Highlight and select

for.

[6] Highlight and select for

.[7] Highlight , select for, and select

for.

[8] Highlight , select

for, and select

for.

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Exercise 17 Stress-Strain Relations: Hooke's Law 5

[11] Solve.

[9] Select six stress resultsobjects and right-click-select

.

[10] Results objectsfor six stresscomponents.

[12] Highlight .

X= 9.4217 MPa .

[13] Highlight .

Y= 22.767 MPa .

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6 Copyright by Huei-Huang Lee

[14] Highlight .

Z= 0.29805 MPa ,

which is a very small,

yet theoretically a

nonzero value.

[15] Highlight .

XY

= 20.156 MPa .

[16] Highlight .

YZ= 0 . Although it is displayed

as a nonzero value due to the

numerical nature, it is

theoretically a zero value.

[17] Highlight .

XZ

= 0 . Although it is

displayed as a nonzero value

due to the numerical nature, it

is theoretically a zero value.

StressComponent

Stress Value(MPa)

X 9.4217

Y22.767

Z-0.29805

XY

-20.156

YZ

0

ZX 0

[18] The stresscomponents aretabulated here.

X

Y

9.4217 MPa9.4217 MPa

22.767 MPa

22.767 MPa

20.156 MPa

20.156 MPa

20.156 MPa

20.156 MPa

[19] The

stressesapplied on the

material.

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Exercise 17 Stress-Strain Relations: Hooke's Law 7

17-5 Verify Hooke's Law

Using the stress components in 17-4[18], we may calculate the strain components according to Hooke's law (Eqs.

17-1(1-2)). The normal strains are

X=

X

E

Y

E

Z

E=

9.4217

200,000 0.3

22.767

200,000 0.3

0.29805

200,000=13.40510

6

Y=

Y

E

Z

E

X

E=

22.767

200,000 0.3

0.29805

200,000 0.3

9.4217

200, 000=100.1510

6

Z=

Z

E

X

E

Y

E=

0.29805

200,000 0.3

9.4217

200,000 0.3

22.767

200,000= 49.77310

6

and the shear strains are

X=

X

G=

20.156

76923= 262.0210

6

Y=

Y

G=

0

76923= 0

Z=

Z

G=

0

76923= 0

The above calculations are consistent with the strain components tabulated in 17-3[2]. Note that the shear modulusG

is calculated by ANSYS using

G=E

2(1+)=

200,000

2(1+ 0.3)= 76,923 MPa

Wrap UpClose , save the project, and exit Workbench.