7-Strain Energy

CHAPTER 7STRAIN ENERGY

FACULTY OF MECHANICAL ENGINEERINGDIVISION OF ENGINEERING MECHANICS

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Strain EnergyStrain Energy

Materials for this chapter are taken from :

1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials” 5th Edition in SI

units

2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition

MEC411 – MECHANICS OF MATERIALS Ch 7 - 1



Work & EnergyWork & EnergyF

Consider a solid object acted upon by force, F, ata point, O, as shown in the figure.

F

Let the deformation at the point be infinitesimaland be represented by vector dr, as shown.F

dr The work done = F dr

For the general case: W = Fx dx

F

dx

y i.e., only the force in the direction of thedeformation does work.

dx

xz




Amount of Work DoneAmount of Work Done

Constant Force: If the Force is constant, the work is simply the product of theforce and the displacement, W = Fx

F

Displacementx

Linear Force: If the force is proportional to the displacement, the work is

1F

ooxFW21

=

Fo


Displacementxxo



Concept of Strain EnergyConcept of Strain Energy

The external work done on an elastic member incausing it to distort from its unstressed state iscausing it to distort from its unstressed state istransformed into strain energy which is a form ofpotential energypotential energy.

The strain energy in the form of elastic deformation isThe strain energy in the form of elastic deformation ismostly recoverable in the form of mechanical work.




Concept of Strain EnergyConcept of Strain Energy

Consider a simple spring system, subjected to a Force such that F isproportional to displacement x; F = kx.

Now determine the work done when F = Fo, from before:

1oo xFW

21

=

This energy (work) is stored in the spring and is released when the forceis returned to zero




Strain Energy DensityStrain Energy Density

y C id b f t i l t d by

a

Consider a cube of material acted upon by aforce, Fx, creating stress σx = Fx/a2 causing anelastic displacement, δ in the x direction, andt i δ/

a

strain εx = δ/a

δxFW21

=xa

y

x2

32 11 aeaeaU σσ ==y

Fx

22aeaeaU xxxx σσ

U 11 33xxxx eaae

VUu σσ

21/

21 33 ===

xa

δ

Where U is called the Strain Energy, and u is the Strain Energy Density.


δ



Example Linear Elastic MaterialsExample - Linear Elastic Materials

500

400CONTINUED

MPa

)

200

300

tress

(M

u =1/2(300)(0.0015) N.mm/mm3

100

200St

( )( )=0.225 N.mm/mm3

0 0100 0080 0060 0040 0020 0000

100

0.0100.0080.0060.0040.0020.000Strain




Example Elastic Perfectly PlasticExample - Elastic Perfectly Plastic

500

400CONTINUED

MPa

)

300

ress

(M u = 1/2(350)(0.0018)+350(0.0022)

100

200Str

=1.085 N.mm/mm3

0

100

0.0100.0080.0060.0040.0020.0000

Strain




St i E D t V i L diStrain Energy Due to Various Loading




Strain Energy Due to Axial LoadingStrain Energy Due to Axial Loading

axial ; ;F FLA AE

σ = Δ =ΔL

21

A AE

F LFA

212 2

F LU FAE

= Δ =

F = Axial Force (Newtons, N)A = Cross-Sectional Area Perpendicular to “F” (mm2)A = Cross-Sectional Area Perpendicular to F (mm )E = Young’s Modulus of Material, MPaL = Original Length of Bar, mm




Comparison Of Energy Stored In Straight And Stepped Bars

ΔaLΔbL/2 L/2

FA FAnA

AELFU

2

2

= +=nAELF

AELFU

22/

22/ 22

(a) (b)

AE2⎟⎠⎞

⎜⎝⎛ +

=nn

AELF

21

2

2

Note for n = 2; case (b) has U= which is 3/4 of case (a)AE

LF24

3 2




E l 6 1Example 6.1

The steel rods AB and BC are made of a steel for which the yield strength is σy =y

300 MPa and E = 200 GPa. Determine the maximum strain energy that can beacquired by the assembly without causing any permanent deformation when thelength a of rod AB is (a) 2 m, (b) 4 m.g ( ) ( )




Strain Energy due to TorsionStrain Energy due to Torsion

From definition:

T TLk and GJφ φ

= =

Therefore,

GφGkIφ

=

Strain energy is:

2 2

2 2 2T T T LU

k GJφ= = =




Example 6 2Example 6.2

Rod AC is made of aluminium (G=73 GPa) and is subjected to a torque T applied atd C K i th t ti BC f th d i h ll d h i id di t f 16end C. Knowing that portion BC of the rod is hollow and has an inside diameter of 16

mm, find the strain energy of the rod for a maximum shearing stress of 120 MPa.




Strain Energy due to Direct ShearStrain Energy due to Direct Shear

From definition:

/F F Aτ //

F F Ak and GL

τδ γ δ

= = =

Therefore,

GAkI

=I

Strain energy is:

2 2F F F L2 2 2F F F LU

k AGδ= = =

Alternatively allowing z to be variable strain energy is: Alternatively allowing z to be variable, strain energy is:

2

2 2 2 2F A A bxU dz dz

Gτδ δ τγ τ= = = =∫ ∫


2 2 2 2 zG∫ ∫



Strain Energy due to BendingStrain Energy due to Bending

From definition:

M2MdU d and ds Rdθ θ= =

Therefore,

2MdsdU

R=

Then,

2M E M dstherefore dU= =2

therefore dUI R EI= =

Strain energy is:

2

2M dxU

EI= ∫





Using E = 200 GPa, determine the strain energy due to bending for the steel beamd l di h i I 165 106 4and loading shown. given I = 165 x 106 mm4 .




Example 6 4 bi l diExample 6.4 combine loadings

The 18 mm diameter steel rod BC is attached to the lever AB and to the fixed supportC Th if t l l AB i 9 id d 24 d U i E 200 GP GC. The uniform steel lever AB is 9 mm wide and 24 mm deep. Using E = 200 GPa, G =77 GPa, and the method of work and energy, determine the deflection of point A.




SummarySummary

Note : The constant K for the traverse shear option is shown in thei h b F S l i (K 1)


section on traverse shear above. For a Structural section (K = 1)



Supplementary Problems 6 1Supplementary Problems 6.1

1. Determine the strain energy inthe rod assembly. Portion AB issteel, BC is brass, and CD isaluminium. Est = 200 GPa, Ebrst br

=101 GPa, Eal = 73.1 GPa.

2. Determine the total axial andbending energy in the A-36 steelbeam. A = 2300 mm2, I = 9.5(106)( )mm4, E = 200 GPa.

3 I th bl h t T3. In the assembly shown torques TA

and TB are exerted on disks A andB respectively. Knowing that bothh ft lid d d fshafts are solid and made of

aluminium (G=73 GPa), determinethe total energy acquired by the


assembly.



Castigliano’s TheoremCastigliano’s Theorem

This method was discovered in 1879 by Alberto Castigliano todetermine the displacement and slope at a pt in a body.

It applies only to bodies that have constant temperature and materialwith linear-elastic behavior.

His second theorem states that displacement is equal to the first partialderivative of strain energy in body w.r.t. a force acting at the pt and indirection of the displacementdirection of the displacement.





Consider a body of arbitraryshape subjected to a series of nforces P1,P2, … Pn.

Since external work done byyforces is equal to internal strainenergy stored in body, byconservation of energy, U = Ui.conservation of energy, Ue Ui.

However, external work is afunction of external loadsfunction of external loadsUe = ∑ ∫ P dx.





So, internal work is also a function of the external loads.Thus

( )U U f P P P= =

Now, if any one of the external forces say Pj is increased by a differentialamount dP Internal work increases so strain energy becomes

( )1 2, ,...,i e nU U f P P P= =

amount dPj. Internal work increases, so strain energy becomes

( )2ii i i j

UU dU U dPP

δδ

+ = +

Further application of the loads cause dPj to move through displacement

( )i i i jjPδ

j

Δj, so strain energy becomes

( )3i j i j iU dU U dP+ = + Δ ( )3i j i j iU dU U dP+ + Δ





dUj = dPjΔi is the additional strain energy caused by dPj.

In summary Eqn (2) represents the strain energy in the body determined byIn summary, Eqn (2) represents the strain energy in the body determined byfirst applying the loads P1,P2, …,Pn, then dPj.

Eqn (3) represents the strain energy determined by first applying dP thenEqn (3) represents the strain energy determined by first applying dPj, thenthe loads P1,P2, …,Pn.

Si h l iSince theses two eqns are equal, we require

Uδ ii

j

UP

δδ

Δ =





Determine the horizontal displacement SolutionDetermine the horizontal displacementof joint C of steel truss shown. Thecross-sectional area of each member isalso indicated Take E t = 210(103)

External force P : Since horizontaldisplacement of C is to be determined a

Solution

also indicated. Take Est 210(10 )N/mm2.

displacement of C is to be determined, ahorizontal variable force P is appliedto joint C. Later this force will be setequal to the fixed value of 40 kNequal to the fixed value of 40 kN.





Internal forcesN: Using methodof joints, force Nin each memberis found. Resultsare shown intable:

N Lδ⎛ ⎞Castigliano’s Second

( )6556.7 10 N m0 0

hCN LNP AE

δδ

⎛ ⎞Δ = ⎜ ⎟⎝ ⎠

⋅= + +

∑Castigliano s SecondTheorem gives;

( )( )

2 3 2

3

0 0625 mm 210 10 N/mm

283.7 10 N m

= + +⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦

⋅+

( )2 3 21250 mm 210 10 N/mm

4.24 1.08 5.32 mmhC

+⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦

Δ = + =




Supplementary Problem 6 2Supplementary Problem 6.2

1. Members of the truss shown are made of1. Members of the truss shown are made ofsteel and have the cross-sectional areasshown. Using E = 200 GPa, determine thehorizontal deflection of joint C caused byhorizontal deflection of joint C caused bythe application of the 210 kN load. ACand AB have area of 1200 mm2 and BC hasan area of 1800 mm2.an area of 1800 mm

2. For the beam and loading shown,determine the slope at end A. Use E =p200 GPa.




Supplementary Problem 6 2Supplementary Problem 6.2

3 Two rods AB and BC of the same flexural3. Two rods AB and BC of the same flexuralrigidity EI are welded together at B. Forthe loading shown, determine (a) thedeflectio of oi t C (b) the lo e ofdeflection of point C, (b) the slope ofmember BC at point C.


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7-Strain Energy

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