STRAIN ENERGY FOR SIMPLE LOADS - PDF Introduction to Mechanics of Materials: Part I

APPENDIX

A.5 STRAIN ENERGY FOR SIMPLE LOADS

Fig. A.20

Consider a rod BC of length L and uniform cross-section area A, attached at B to a fixed support. The rod is subjected to a slowly increasing axial load F at C (Fig. A.20). The work done by the load F as it is slowly applied to the rod must result in the increase of some energy associated with the deformation of the rod. This energy is referred to as the strain energy of the rod. Which is defined by

0^xF d

Strain energy U= =

∫

x^(A.21)

Dividing the strain energy U by the volume V = A L of the rod (Fig. A.20) and using Eq. (A.21), we have

0 F d

A L

U x x

V =

∫

^(A.22)

Recalling that F/A represents the normal stress σ_x in the rod, and x/L represents the normal strain ε_x, we write

x x

0 d

U V

HV H

³

(A.23)

The strain energy per unit volume, U/V, is referred to as the strain-energy density and will be denoted by the letter u. We therefore have

x x

0 d

³

^HV H _(A.24)

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A.5.1 ELASTIC STRAIN ENERGY FOR NORMAL STRESSES

In a machine part with non-uniform stress distribution, the strain energy density u can be defined by considering the strain energy of a small element of the material with the volume

∆V. writing

lim _V 0 U

u ^{D →} V

= D

D or d d u U

= V . (A.25)

for the value of σ_x within the proportional limit, we may set σ_x = E ε_x in Eq. (A.24) and write

2 2x

x x

1Eε 1σ ε σ

2 2 ^x 2E

u . (A.26)

The value of strain energy U of the body subject to uniaxial normal stresses can by obtain by substituting Eq. (A.26) into Eq. (A.25), to get

ó dx

U =

∫

2E V^. ^(A.27)

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ELASTIC STRAIN ENERGY UNDER AXIAL LOADING

When a rod is acted on by centric axial loading, the normal stresses are σ_x = N/A from Sec. 2.2.

Substituting for σ_x into Eq. (A.27), we have

2 2 d 2EN

U V

∫

A or, setting dV A V= d , ²

2E d

L N

U V

∫

A ^(A.28)

If the rod hasa uniform cross-section and is acted on by a constant axial force F, we then have

2L 2E U N

= A (A.29)

4. Elastic strain energy in Bending

The normal stresses for pure bending (neglecting the effects of shear) is σ_x = My / I from Sec.

4. Substituting for σ_x into Eq. (A.27), we have

x 2 2

σ d d

2E 2E

U V M y V

³ ³

I _(A.30)

Setting dV = dA dx, where dA represents an element of cross-sectional area, we have

2 2

0 0

y d dx dx

2E 2E

L M L M

U A

I I

³ ³ ³

_(A.31)

Example A.08

Fig. A.21

Determine the strain energy of the prismatic cantilever beam in Fig. A.21, taking into account the effects of normal stressesonly.

Solution

The bending moment at a distance x from the free end is M= −F x. Substituting this expression into Eq. (A.31), we can write

2 2 2 3

0 0

F x F L

dx dx

2E 2E 6E

L M L

³

I I

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A.5.2 ELASTIC STRAIN ENERGY FOR SHEARING STRESSES

When a material is acted on by plane shearing stresses τ_xy the strain-energy density at a given point can be expressed as

xy d xy

W J

³

(A.32) where γ_xy is the shearing strain corresponding to τ_xy. For the value of τ_xy within the proportional limit, we have τ_xy = G γ_xy, and write

xy2

xy2 xy xy

1G 1

2 2 2G

u W

J W J .

(A.33) Substituting Eq. (A.33) into Eq. (A.25), we have

xy2 d U W2G V

³

(A.34) Elastic strain energy in Torsion

The shearing stresses for pure torsion areτ_xy = Tρ / J from Sec. 3. Substituting for τ_xy into Eq. (A.27), we have

2 2 2

d ñ d

2G 2 E

U V T V

G J

³

(A.35) Setting dV = dA dx, where dA represents an element of the cross-sectional area, we have

2 2

0 2 0

ρ d dx dx

2G 2G

L T L T

U A

J J

³ ³ ³

(A.36)

In the case of a shaft of uniform cross-sectionacted on by a constant torque T, we have

2L 2G U T

= J (A.37)

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Elastic strain energy in transversal loading

If the internal shear at section x is V, then the shear stress acting on the volume element, having a length of dx and an area of dA, is τ = V Q / I t from Sec. 4. Substituting for τ into Eq. (A.27), we have

2 2 2 2

2 2

d 1 d dx d dx

2G 2G 2G

V V A

V Q V Q

U V A A

I t I t

W § · ^§ ^·

¨ ¸

³ ³ ³ ³

^(A.38)

The integral in parentheses is evaluated over the beam’s cross-sectional area. To simplify this expression we define the form factor for shear

2 2 d

S A

f A Q A

I t

∫

^(A.39)

Substituting Eq. (A.39) into Eq. (A.38), we have

S V

U f

∫

A (A.40)

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Fig. A.22

The form factor defined by Eq. (A.39) is a dimensionless number that is unique for each specific cross-section area. For example, if the beam has a rectangular cross-section with a width b and height h, as in Fig. A.22, then

t= A b h,= _{1 b h}³

I =12

2 2

h y h b h

A y 2 b y y

2 2 2 4

Q y

 − 

     

= ′ ′= +   − =  − 

   

 

 

Substituting these terms into Eq. (A.39), we get

/ 2 2 2

2 2

/ 2 3

bh b h y b dy 6

4b 4 5

121 bh

h S

f ⁺

−

 

=  −  =

   

 

 

∫

(A.41)

Example A.09

Fig. A.23

Determine the strain energy in the cantilever beam due to shear if the beam has a rectangular cross-section and is subject to a load F, Fig. A.23. assume that EI and G are constant.

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Solution

From the free body diagram of the arbitrary section, we have V(x) = F.

Since the cross-section is rectangular, the form factor 6

S 5

f = from Eq. (A.41) and therefore Eq. (A.40) becomes

2 2

6 F 3 F L

5 2G 5 G

L shear

U =

∫

A = A

Using the results of Example A.08, with A = b h, 1 b h³

I =12 , the ratio of the shear to the bending strain energy is

2 3 2

3 F L

3 h E

5 G

F L 10 L G 6E

shear bending

U A

= =

Since G = E / 2(1+n) and n = 0.5, then E = 3G, so

2 2

3 h 3G 9 h 10 L G 10 L

shear bending

U = =

It can be seen that the result of this ratio will increasing as L decreases. However, even for short beams, where, say L = 5 h, the contribution due to shear strain energy is only 3.6% of the bending strain energy. For this reason, the shear strain energy stored in beams is usually neglected in engineering analysis.

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Dalam dokumen PDF Introduction to Mechanics of Materials: Part I (Halaman 165-172)