129

Strain EnergyUNIT 8 STRAIN ENERGY

Structure 8.1 Introduction

Objectives

8.2 Potential Energy

8.3 Concept of Strain Energy 8.3.1 Strain Energy in Axial Loading 8.3.2 Strain Energy in Shear 8.3.3 Strain Energy in Bending 8.3.4 Strain Energy in Torsion

8.4 Suddenly Applied Loads

8.5 Impact Loads 8.5.1 Stresses Due to Axial Loads 8.5.2 Stresses Due to Bending Loads

8.6 Castigliano’s Theorem 8.6.1 Application of Castigliano’s Theorem to Angular Movements 8.6.2 Application to Truss Problems 8.6.3 Application to Beam Problems

8.7 Summary

8.8 Answers to SAQs

8.1 INTRODUCTION This unit introduces you to the concept of strain energy, how it is related to stress and strain and the method of estimating this energy in different modes of application of forces or loads on the body. It also explains to you how to obtain the stresses in the case of sudden and impact loads using the strain energy concept. Further, it deals with Castigliano’s theorem which can be employed to estimate the defections of deformable bodies under loads.

Objectives After studying this unit, you should be able to

• conceptualise the potential energy,

• estimate the strain energy under different types of loads,

• obtain the stresses in a body when subjected to suddenly applied or impact loads, and

• determine the deflections of beams through the application of Castigliano’s theorem.

8.2 POTENTIAL ENERGY

Energy is defined as capacity to do work. It may exist in many forms, e.g. mechanical (potential or kinetic), thermal, nuclear, chemical, etc. It is also well known that energy is conserved through transformation from one form to another. The potential energy of a body is the form of energy which is stored by virtue of the work which was previously been done on that body, e.g. in lifting it to some height above a datum. In this case, work is done on the body by displacing it against the force of gravity. This potential energy is

130

converted to an equivalent kinetic energy if the body is allowed to fall from the height by developing velocity of fall. Let us now consider the case of a deformable body subjected to certain external forces or what we call as loads. Because of the property of deformability of the body, each of the points of application of these forces may undergo certain displacement. Thus, the force moves through a certain distance, which is equivalent to doing work on the body, i.e. a certain energy imparted to the body. One may now ask a question, ‘where did all this energy go’? because it had to be conserved.

Forces and Stresses in Beams

This question, we shall answer in the next section.

8.3 CONCEPT OF STRAIN ENERGY

From some of the earlier units, we know that when a deformable body is subjected to external forces the body internally develops stresses and strains. For example, if we apply a compressive force on a spring, the spring deforms developing internally shear stresses and shear strains. At the same time we observe that the point of application of the compressive force moves through a distance of the spring deflection (amount of contraction). We also know that this compressed spring has a capacity to do work when released. This means that a certain amount of energy is stored in the spring in its strained state. Earlier we noted that work was done on the spring by the compressive force. Now, when we link the above two statements we realise that the work done on the spring is transformed into energy stored in it by virtue of the strain in the spring. This form of energy emanating out of the internal strain in the body is known as strain energy. Thus, strain energy is a particular form of potential energy which is stored within materials which have been subjected to strain as a result of work done on the material.

Also, we have seen that strain energy,

U = external work done.

Thus, for a gradually applied load the work done in straining the body is given by the shaded area under the load-extension graph of Figure 8.1.

12

U P= Δ

Load

, P

Extension, Δ Figure 8.1

The strain energy per unit volumes is after referred to as resilience. The resilience at the yield point or proof stress is termed as proof resilience. The strain energy stored upto the fracture of the material is generally known as toughness.

In the earlier units, it is seen that loads may be applied on a body in different modes, viz., as axial loads, as shear loads, as bending loads or as torsional loads.

In the following sections, we shall derive expressions for the strain energy stored under each of the above forms of loading.

8.3.1 Strain Energy in Axial Loading Consider the bar of length L and cross-sectional area A shown in Figure 8.2 to which an axial load P is applied. Under this load, the bar may extend or contract (extend if P is

131

Strain Energytension and contract if P is compression) by an amount Δ where the P – Δ relationship is

a shown in Figure 8.1.

L Δ

P

Cross-sectional Area A

P

Figure 8.2

The work done by the force P is equal to the strain energy stored in the bar.

Thus, 12

U P= Δ

But, we know that the deformation, Δ = strain × L

Stress P LLE A E

Δ = × = ×

∴ 21

2 2PL P LU PAE AE

= × =

If we rewrite the above expression in terms of the stress PA

⎛ ⎞⎜ ⎟⎝ ⎠

and the strain in the bar,

1 1 Stress × Strain × Volume2 2

P PU ALA AE

= × × × = ×

The strain energy per unit volume of the bar 1 Stress × Strain2

= ×

2 21

2 2 2E

Eσ ε

= × σ × ε = =

Hence, strain energy per unit volume of the material is as follows :

2

2(Stress) Young's modulus (Strain)2 Young's modulus 2

= = ××

8.3.2 Strain Energy in Shear Consider now an element of length ds of the bar subjected to a shear load Q at one end causing deformation through an angle γ (the shear strain) and a shear deflection δ, as shown in Figure 8.3.

Q

ds

δ

γ

Figure 8.3

132

Forces and Stresses in Beams Strain energy, dU, of the element = work done 1 1=

2 2Q Q dsδ = γ

But, shear strain, Shear Stress,=Shear Modulus,

QG GA

τγ =

∴ 21

2 2Q QdU Q ds

GA GA= × =

ds

∴ Total strain energy in the bar, resulting from shear is as follows :

2 2

0 2 2

L Q ds Q LUAG AG

= =∫

2

2Gπ

= × Volume of the bar

8.3.3 Strain Energy in Bending Let the element now be subjected to a constant bending moment M causing it to bend into an arc of radius R and subtending an angle dθ at the centre, as in Figure 8.4. The beam will also have moved through an angle dθ.

Strain energy of the element, dU = work done

12

= × moment × angle of rotation

dx

d

M

R

x

dθ

M

dθ

Figure 8.4

12

M d= × θ

1 (since, )2

dsM ds R dR

= × = θ

1 since,2

M M EM dsEI I

⎛ ⎞= × × =⎜ ⎟⎝ ⎠R

2

2M ds

EI=

Total strain energy in bending, 2

0 2

L M dsUEI

= ∫

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Strain Energy

For small curvatures this reduces to 2

0 2

L M dxUEI

= ∫

where dx is the undeformed element length.

If M is constant throughout the length L, then

2

2M LU

EI=

8.3.4 Strain Energy in Torsion The element ds is now considered subject to a torque T as shown in Figure 8.5 producing an angle of twist dθ.

ds

T

dθ

Figure 8.5

Strain energy = work done 12

T d= θ

But, from the theory of simple torsion, orT d TG d dsJ ds GJ

θ= × θ =

Where J is the polar moment of inertia and G is the shear modulus of the material.

Therefore, total strain energy resulting from torsion, 2

0 2

L T dsU GJ

= ∫

For constant T, we get 2

2T LU .

GJ=

Example 8.1

Determine the diameter of an aluninium shaft which is designed to store the same amount of strain energy per unit volume as a 50 mm diameter steel shaft of the same length. Both shafts are subjected to equal compressive axial loads. What will be the ratio of the stresses set up in the two shafts?

Take Esteel = Es = 201 GN/m2 and Ealuminium = Ea = 67 GN/m2.

Solution

Since the strain energy per unit volume of the two shafts is equal, we get

2 2 2

267 1or

2 2 201 3a s a a

a s ss

EE E E

σ σ σ= = =

σ=

∴ 3 or ss a

a

σσ = σ =

σ3

But we know stress, loadarea of cross-section

σ = ,

134

Forces and Stresses in Beams

Thus, 2 23

4 4s a

P PD D

⎡ ⎤⎧ ⎫⎡ ⎤⎛ ⎞ ⎢ ⎥⎪ ⎪⎢ ⎥⎜ ⎟ ⎢ ⎥⎪ ⎪⎢ ⎥⎜ ⎟ = ⎨ ⎬⎢ ⎥⎛ ⎞⎢ ⎥⎜ ⎟π π⎪ ⎪⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ ⎝ ⎠⎩ ⎭⎣ ⎦

i.e. 4

4 3a

s

DD

=

or ∴ 4 43 (50)aD = ×

1

4 4(3 (50) ) 65.8 mmaD = × =

The required diameter of the aluminium shaft is 65.8 mm.

Example 8.2

Two shafts are of the same material, length and weight. One is solid and of 100 mm diameter, the outer is hollow. If the hollow shaft is to be store 25% more energy than the solid shaft when transmitting torque, what must be its internal and external diameters? Assume the same maximum shear stress in both the shafts.

Solution

Let A be the solid shaft of diameter Da and B be the hollow shaft with external diameter Db and internal diameter db. Since both the shafts are to be the same material and of equal weight, their volumes should be equal.

Therefore, 2 2 2( )

4 4b b bD D dL Lπ π −

× = ×

or 2 2 2 2(100) mma b bD D d= − = 2

Since the maximum shear stress is the same for both the shafts, we have,

a bτ = τ

since2 2 2a a b b

a B

T D T D T r T DJ J J

× ×⎛ ⎞= τ = =⎜ ⎟⎝ ⎠J

Therefore, a b

b a

T D JT D J

= × a

b

But it is also given that the strain energy of B = 1.25 × strain energy of A.

Thus, 22 25 4or

2 4 2 5b a a

b B b

T L T L T JGJ GJ T J

⎛ ⎞= × = ×⎜ ⎟

⎝ ⎠a

b

From the relationship, this reduces to

2 4 4

4( )4 4

5 5b b b

a a a

D J DD J D

⎛ ⎞ −= × = ×⎜ ⎟

⎝ ⎠bd

Then, ∴ 4 4 2 2 2 2

24 2

( ) ( ) ( )4 45 5

b b b b b bb

a a

D d D d DDD D− +

= × = ×d−

or 2 4 4 2 2 24 ( ) because 105b b b b b aD D d D d D= × + − = = 4

135

Strain Energy

or ∴ 2 2 24 ( 15b b bD D D= + − 20 )

2 43 4 10 or 115.47 mmb bD D= × =

Therefore, ∴ 2 2 4(115.47) 10 or 57.74 mmb bd d= − =

Thus, the internal and external diameters of the hollow tube are 57.74 mm and 115.47 mm, respectively.

SAQ 1 (a) Calculate the strain energy stored in a bar of circular cross-section of

diameter and length of 0.2 m and 2 m, respectively, when it is

(i) subjected to a tensile load of 25 kN,

(ii) subjected to a torque of 25 kN m, and

(iii) subjected to a uniform bending moment of 25 kN m.

Take E = 208 GN/m2, G = 80 GN/m2 for the bar material.

(b) Compare the strain energies of two bars of the same material and length and carrying the same gradually applied compressive load if one is 25 mm diameter throughout and the other is turned to 20 mm diameter over half its length, the remainder being 25 mm diameter.

If both bars are subjected to pure torsion only, compare the torsional strain energies stored if the shear stress in both bars is limited to 75 N/mm2.

(c) A 50 mm square alloy-steel bar 1 m long is a part of a machine and must resist an axial load of 100 N m. What must be the proportional limit of the steel to safely resist the energy load elastically with a factor of safety of 4? E = 200 GPa.

8.4 SUDDENLY APPLIED LOADS

If a load P is applied gradually to a bar to produce an extension δ, the load-extension graph will be as shown in Figure 8.1. However, if a load P′ is applied suddenly to the bar (by suddenly we mean that the value of P′ is reached instantaneously but not gradually as is the case with P) so as to produce the same extension δ, the load extension graph now looks as in Figure 8.6.

P

P1

δ

Work Done, W = ½ P δ (Gradually Applied Load)

Work Done (Suddenly Applied Load) W ′ = P ′ δ

Extension

Figure 8.6

136

Since the bar is strained to the same extent in both the cases, the strain energy should be the same for the two cases. However, strain energy is equal to the work done by the load.

Forces and Stresses in Beams

∴ W W ′=

or 1 or2 2

PP P P′ ′× δ = δ =

This means that the suddenly applied load which is required to produce a certain value of instantaneous strain is half the equivalent value of static load (gradually applied load) required to produce the same strain. Vice-versa, one can say that the effect of a suddenly applied load is two times that achieved when the load is applied gradually. Thus, greater care must be exercised while designing structural elements that are susceptible to sudden loading conditions.

8.5 IMPACT LOADS

If a body that possesses kinetic energy by virtue of its motion strikes an elastic member and is brought to rest, it can be conservatively assumed that all the kinetic energy of the body is converted into strain energy in the member. Equally, if the body falls from a certain height, the potential energy that the body has lost or the kinetic energy gained by the body will be converted into strain energy in the member. Such a loading where there is a transfer of kinetic energy into strain energy is termed as impact loading.

8.5.1 Stresses Due to Axial Loads Consider the problem of a vertical rod of strength L and cross-sectional area A shown in Figure 8.7. A load W, concentric with the rod and free to slide vertically, is dropped from a height h on the end stop which may be rigid collar.

h

δ

L

Load, W

Rigid Collar

Figure 8.7

The rod is suspended from a rigid support. When the load is dropped it will produce a maximum instantaneous extension δ in the bar, at the point of impact. Then, the load would have done work equal to W (h + δ). This then must be equal to the strain energy stored in the bar. If σ is the instantaneous stress produced in the bar.

The strain energy, 2

2U AL

Eσ

= ×

Hence, 2

( )2

W h ALE

σ+ δ =

but 2

LE

σδ =

137

Strain Energy

Therefore, 2

2LW h AL

E Eσ σ⎛ ⎞+ = ×⎜ ⎟

⎝ ⎠

or 2 2 2 0W WEhA L

σ − × σ − =

Denoting 2 as stWA

σ , which would have been the stress in the bar if W were to have been

applied gradually, the above equation can be written as

2 22 0st stEhL

σ − σ σ − σ =

The solution of this equation results as follows :

2 21 1 1 1stst

Eh W EhAL A L

⎡ ⎤

W⎡ ⎤

σ = σ + + = + +⎢ ⎥ ⎢ ⎥× σ⎢ ⎥ ⎣ ⎦⎣ ⎦

The instantaneous deflection δ then can be obtained from

LE

σδ =

If the load is not dropped but suddenly applied from effectively zero height (h = 0 in the above equation), the stress developed would be

2 stσ = σ

It is a result which we have already predicted in the earlier section.

Thus, it can be seen that the stresses resulting from impact loads will be even higher and any design work where impact loading is possible should always employ a safety factor well in excess of two.

SAQ 2 An axial pull of 50 kN is suddenly applied to a steel rod 2 m long and 1000 mm2 in cross-section. Calculate the strain energy that can be absorbed, if E = 200 GPa.

8.5.2 Stresses Due to Bending Loads Consider a simply supported beam as shown in Figure 8.8 subjected to an impact load produced by a weight W falling from a height h. it will be assumed that the weight of the beam can be neglected in comparison with W and that the work done by W is completely transformed into strain energy of bending.

h

δ

W

Figure 8.8

138

Forces and Stresses in Beams

Then, the strain energy, U = Work done by W = W (h + δ)

where δ is the instantaneous maximum deflection under the load W.

If We is an equivalent static load which, when applied gradually, produces the same deflection as the impact load W, the work done by We will be equal to the work done by W.

i.e. 1 ( )2 eW W hδ = + δ

Let δst be the deflection of the beam if W were applied gradually. Then, by the linear relationship of load to deflection, it can be written

oree

st s

WW W Wt

δ= =

δ δ δ×

Substituting this in the above work equation, we get

( )2 st

W W hδ× × δ = + δ

δ

On simplifying, we get

2 2 2st st hδ − δ δ − δ = 0

Solution of this quadratic equation gives

2 21st stst

hδ = δ ± δ +

δ

21 1stst

h⎛ ⎞δ = δ ± +⎜ ⎟⎜ ⎟δ⎝ ⎠

This then will be the instantaneous deflection under the impact load W.

Once δ is known one can obtain We and the maximum bending moment, Mmax produced in the beam by the application of We. Then, the maximum stress σmax occurring in the beam can be obtained from the bending equation as

maxmax

M CI

σ = ,

where I is the moment of inertia for the cross-section and C is the extreme fibre distance from the neutral axis.

SAQ 3 (a) A bar 3 m long and 5 cm diameter hangs vertically and has a collar securely

attached at the lower end. Find the maximum stress induced when,

(i) a weight 2.5 kN falls from a height of 12 cm on the collar, and

(ii) a weight of 25 kN falls from a height of 1 cm on the collar.

(b) An unknown weight falls through 10 mm onto a collar rigidly attached to the lower end of a vertical bar 3 m long and 600 mm2 in section. If the maximum instantaneous extension is known to be 2 mm, what is the corresponding stress and the value of the unknown weight.

Take E = 200 kN/mm2.

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Strain EnergyExample 8.3

What will be the instantaneous stress and elongation of a 25 mm diameter bar, 2.6 m long, suspended vertically, if a mass of 10 kg falls through a height of 300 mm onto a collar which is rigidly attached to the bottom end of the bar? Take E = 200 GPa.

Solution

We have 21 1W EA L

⎛ ⎞σ = + +⎜ ⎟⎜ ⎟

⎝ ⎠

hAW

9 2

210 9.81 4 2 200 10 0.3 (0.025)1 1

4 10 9.81 2.6(0.025)

⎛ ⎞× × × × × × π ×⎜ ⎟= + +⎜ ⎟× × ×π ⎝ ⎠

= 96.24 MPa

Now, 6

996.24 10 2.6 0.00125 m = 1.25 mm

200 10L

Eσ × ×

δ = = =×

Thus, the maximums tress and elongation are 96.24 MPa and 1.25 mm, respectively.

This elongation of 1.25 mm is very small compared to the height of fall namely 300 mm. Let us find out the stress if this elongation were to be neglected in the calculation. When it is neglected the work strain energy equation becomes.

2

2Wh AL

Eσ

= ×

or 9

22 2 10 9.81 0.3 200 10 4 96.04 MPa

(0.025) 2.6WhEAL

× × × × × ×σ = = =

π × ×

This shows that the error in the stress calculation is insignificant even if the elongation is neglected.

Example 8.4

When a concentrated force of 1 kN is applied at the midspan point of a simply supported beam a static deflection of 5 mm is produced. The same load produces a maximum stress of 158 MN/m2. Determine the magnitude of the instantaneous stress produced when a weight of 10 kg is allowed to fall through a height of 12 mm onto the beam at midspan. What will be the instantaneous deflection?

Solution

Let We be the equivalent gradually applied load which produces the same static deflection δmax as that of the impact load.

Then, max

1 kN 200 kN/m0.005 m

eW= =

δ

Now, equating the work done by the impact load to that done by the equivalent static load, we get

max max( )2

eWW hδ = + δ

i.e. 10 9.81 0.0122 200 1000 200

e e eW W W× ⎛ ⎞× = +⎜ ⎟⎝ ⎠

∴ 2 0.9162 0.47088 0e eW W− − =

140

Forces and Stresses in Beams Hence,

20.1962 (0.1962) 4 0.470882eW

+ + ×=

= 0.981 + 0.6932 = 0.791 kN

But max0.791 0.003955 m = 3.955 mm 4 mm

200 200eW

δ = = = ;

It is given that the load of 1 kN produces a maximum stress of 158 MN/m2. Therefore, a load of We should produce a maximum stress, σmax equal to (158 We) MN/m2.

∴ 2 2max 158 0.791 124.978 MN/m 125 MN/mσ = × = ;

Thus, the maximum stress and maximum deflection are 125 MN/m2 and 4 mm, respectively.

SAQ 4 (a) When a load of 20 kN is gradually applied at a certain point on a beam, it

produces a deflection of 13 mm and a maximum bending stress of 75 MN/m2. From what height can a load of 5 kN fall onto the beam at this point if the maximum bending stress is to be 150 MN/m2.

(b) A beam uniform cross-section with centroid at mid-depth and length 7 m, is simply supported at its ends and carries a point load of 5 kN at 3 m from one end. If the maximum bending stress is not to exceed 90 MN /m2 and the beam is 150 mm deep,

(i) working from the first principles, find the deflection under the load, and

(ii) what load dropped from a height of 75 mm onto the beam at 3 m from one end would produce a stress of 150 MN/m2 at the point of application of the load?

Take E = 200 GN/m2.

8.6 CASTIGLIANO’S THEOREM

In the previous sections we have seen that the strain energy stored in a elastic body is equal to the work done by the external forces on the body during the process of deformation. This suggests that there should be some way by which one can determine the displacements of the body given the strain energy stored in the body. A general theory for determining displacements of points in an elastic body was first stated by Castigliano. Castigliano stated that

“If the total strain energy of a body or a framework is expressed in terms of the external loads and is partially differentiate with respect to one of the loads, the result is the displacement of the point of application of the load in the direction of the load.”

Mathematically speaking if the total strain energy, U, is a function of the loads Pi (i = 1 to n), then

mm

UP

∂δ =

∂

where δm is the displacement of the point of application of the load Pi in the direction of Pm.

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Strain EnergyLet us prove the theorem now. Consider the beam in Figure 8.9 with forces PA, PB, PB C,

etc. acting at points A, B, C, etc.

PB PA PC

C B A

(+ Δ PA)

δ A δ B δ C

Δ δ A Δ δ B

Δ δ C

Unloaded Beam Position Beam loaded with PA, PB, PC

Beam loaded with PA, PB, PC , etc. Plus Extra Load Δ PA

Figure 8.9

Let δA, δB, δB C, etc. be the deflections in the direction of loads at the points A, B, C, . . ., respectively.

Then, the strain energy of the system is equal to the work done.

1 1 1 . . .2 2 2A A B B C CU P P P= δ + δ + δ +

If one of the loads PA is now increased by an amount Δ PA, the changes in the deflection will be Δ δA, Δ δB, Δ δB C, etc. as shown in Figure 8.9. In increasing the load from PA to (PA + Δ PA), extra work is done and it is as follows (refer Figure 8.10 also).

Extra work done at 12A A AA P P⎛ ⎞= + Δ × Δδ⎜ ⎟

⎝ ⎠

Extra work done at B BB P= × Δ δ

Extra work done at C CC P= × Δ δ

Similarly, in the case of other load also. Then, the total extra work one,

1 . . .2A A B B B B C CP P P P= Δδ + Δ Δδ + Δδ + Δδ +

This should be equal to the increase in strain energy Δ U.

Neglecting the product of small quantities (Δ PA Δ δA) since they are very small as compared to the other terms.

. . .A A B B C CU P P PΔ = Δ δ + Δ δ + Δ δ +

PA

Load At A

PA + Δ PA

= (PA + ½ Δ PA) ΔδA Extra Work

Deflection at A δA δA + Δ δ A

PA

Load At B

= PB +Δ δB

Extra Work

δB δB + ΔδB

Deflection at B

Figure 8.10

142

Forces and Stresses in Beams

Let us now consider the case where we apply the loads (PA + Δ PA), PB, PB C, etc. gradually to the full value. As the final loads in the case are exactly the same as the earlier case and as all the loads are applied statically, the final deflections should also be the same.

Then, the strain energy in the case, U′ (neglecting product of small terms Δ PA Δ δA) would be as follows :

1 1 1( ) ( ) ( ) ( ) . . .2 2 2A A A A B B B C C CU P P P P′ = + Δ δ + Δδ + δ + Δδ + δ + Δδ +

1 1 1 1 12 2 2 2 2A A A A A A B B BP P P P P= δ + Δδ + Δ δ + δ + Δ Bδ

1 1 . . .2 2C C C CP P+ δ + Δδ +

1 1 1 1 1. . .2 2 2 2 2A A B B C C A A BP P P P P= δ + δ + δ + + Δδ + Δ Bδ

1 1. . .2 2C C AP P A+ Δδ + + Δ δ

1 1 1 1. . .2 2 2 2A A B B C C AU P P P P= + Δδ + Δδ + Δδ + + δA

However, since loads as well as deflections are the same in both the cases, the total strain energy should also be the same. Hence,

U U U′ = + Δ

or 1 1 1 1. . .2 2 2 2A A B B C C AU P P P PΔ = Δδ + Δδ + Δδ + + Δ δA

∴ 2 . . .A A B B C C AU P P P PΔ = Δ δ + Δ δ + Δ δ + + Δ δA

Earlier we have obtained

. . .A A B B C CU P P PΔ = Δ δ + Δ δ + Δ δ +

Now, from the last two expressions, we get

orA A AA

UU PP

ΔΔ = Δ δ = δ

Δ

In the limit, AA

UP

∂δ =

∂

Similarly, it can be shown that ,B CB C

U UP P

∂ ∂= δ =

∂ ∂δ , etc.

Thus, in general, MM

UP

∂= δ

∂.

As a simple example of the application of Castigliano’s theorem, let us consider a bar of cross-sectional area A and length L under axial tension P. From earlier section we have known that the strain energy U in this case is given as

2

2P LUAE

=

If we differentiate U with P, we get,

U PLP AE

∂=

∂

which we recognise as the extension of the bar under the load P, i.e. the displacement of the point of application of the load P.

143

Strain Energy8.6.1 Application of Castigliano’s Theorem to Angular Movements

The terms load and displacement in Castigliano’s theorem have a more general meaning than is indicated in the previous section. The load may be a couple and the displacement a rotation produced by the couple. Thus, as per Castigliano’s theorem “If the total strain energy, expressed in terms of the external moments, be partially differentiated with respect to one of the moments, the result is the angular deflection (in radians) of the point of application of that moment and in its direction.” This form of the statement will be useful in determining rotations in either bending or torsion problems.

8.6.2 Application to Truss Problems In this section, we shall apply Castigliano’s theorem to determine the deflections in a truss. Let a truss with m members be subjected to a system of loads P1, P2, . . . , etc. The forces in the members of the truss can be obtained by standard methods. Let Fi be the force in the ith member due to the above system of forces. Let Li and Ai be respectively the length and area of cross-section of the member. Then the strain energy stored in this member will be

2

2i i

i

F LA E

where E is the Young’s modulus of the material. Then, the total strain energy stored in the truss can be written as follows :

2

1 2

mi i

i i

F LUA E=

= ∑

Now, as per Castigliano’s theorem the deflection δn under the load Pn in its direction will be

2

1 12

m mi i i i i

ni in n i i n

F L F LUP P A E A E P= =

⎛ ⎞ ⎛ ∂∂ ∂δ = = ∑ = ∑ ⎜ ⎟ ⎜∂ ∂ ∂⎝ ⎠ ⎝

F ⎞⎟⎠

Let us observe this equation closely.

i

n

FP

∂∂

is the rate at which Fi varies with Pn with all other loads remaining constant or in

other words it is the force in the member when a unit load is applied instead of Pn, while all other loads are zero. The above concept can still be applied to find the deflection in any direction at any joint of the truss even when no load is acting in that direction or at that point. In this case, we introduce a fictitious force Q at the pint in the direction in which we need the deflection. Then we determine the member forces for the load system which now includes Q also.

Then, 2

1 2

mi i

i i

F LUA E=

= ∑ (Fi now includes the effect of Q also)

Therefore, 1

mi i i

Qi i

F L FUQ A E=

∂∂δ = = ∑

∂ ∂Q

Q

However, in actual problem there is no Q or its value is zero. So we put Q = 0 in the above expression and obtain δQ. The concept of unit load can be brought in even here. Now when we include Q in the load system Fi can be written as

i iP iF F F= + (principle of superposition)

144

where FiP and FiQ are the member forces only due to the Ps and Q, respectively. Forces and Stresses in Beams

Then, 2( )

2iP iQ i

i i

F F LU

A E+

= ∑

and 1

( )( )

m iP iQ iQ i

i i

F F LP iQF F

A E Q=

+ ∂δ = ∑ × +

∂

1

( )m iP iQ i iQiP

i i

F F L FFA E Q Q=

+ ∂⎛ ⎞∂= ∑ × +⎜ ⎟∂ ∂⎝ ⎠

1

( )m iP iQ iQQ i

i i

F F FL

A E Q=

+ ∂δ = ∑ × ×

∂

(since FiP does not involve Q and, hence, 0iPFQ

∂=

∂).

When we put Q = 0 in the above expression it reduces to

1

m iQiP iQ

i i

FF LA E Q=

∂⎛ ⎞⎛ ⎞δ = ∑ ⎜ ⎟⎜ ⎟ ∂⎝ ⎠ ⎝ ⎠

iQFQ

∂

∂ is nothing but the member force due to a unit load in the direction of Q at the point

of application of Q while all other loads are zero.

Example 8.5

In the truss shown in Figure 8.11, the cross-sectional area of its top and bottom chords is equal to 2.4 × 103 mm2, while the cross-sectional area of its diagonal and vertical members is equal to 1.6 × 103 mm2. The modulus of elasticity of the material form which the members are made is E = 200 GPa. Compute the vertical and horizontal components of the deflection at joint 7.

Solution

Vertical Deflection at Joint 7

Since there is a 40 kN acting vertically at joint 7, this is a case of straight forward application of Castigliano’s theorem.

As a first step, the forces in the members of the truss (F) are determined under the action of the two 40 KN loads that are acting on the truss. These

forces are shown in column 2 of Table 8.1. Then, the respective FLAE

are

calculated. These are presented in column 3.

Figure 8.11

3 @ 2.4 m = 7.2 m

40 k N

1

2

3 5 7

4 6

2

10

5

1

4

9

6

78

3

3.2 k N

40 k N

145

Strain EnergyTable 8.1

Member

F* (kN)

A (102

mm2)

L (mm)

FLAE

FP

∂

∂ . ∂

∂

FL FAE P

1 − 60 2.4 2400 − 0.30 − 1.5 0.450

2 − 60 2.4 2400 − 0.30 − 1.5 0.450

3 − 30 2.4 2400 − 0.15 − 0.75 0.1125

4 − 100 1.6 4000 − 1.25 − 1.25 1.5625

5 40 1.6 3200 0.40 0 0.0000 6 50 1.6 4000 0.625 1.25 0.78125 7 − 40 1.6 3200 − 0.40 − 1.00 0.400

8 50 1.6 4000 0.625 1.25 0.78125 9 120 2.4 2400 0.60 2.25 1.350

10 30 2.4 2400 0.15 0.75 0.1125

7(vertical)

FL FAE P

∂δ = ∑

∂⎛ ⎞ ⎛⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

+ 6.000**

mm

* Sign convention : − ve compression and + ve tension.

** Positive sign indicates that the deflection is in the direction of the load. i.e. downwards.

Next, all the external loads are replaced by a single unit load acting vertically down at joint 7 and the member forces are obtained. The result

will be equal to FP

∂∂

. These values are presented in column 6 of the

Table 8.1. Then, the rest of the calculations are completed to obtain

.FL FAE P

∂∑

∂ which according to Castigliano’s theorem will be equal to the

vertical deflection at joint 7. We can see this deflection is 6 mm downwards. Horizontal Deflection at Joint 7

In this case, there is no horizontal load at joint 7. So to apply Castigliano’s theorem, we have to first introduce a fictitious horizontal load Q at joint 7, then find the members forces and differentiate these forces with respect to

Q. Finally, .FL FAE Q

∂∑

∂ is determined. Then, Q unit is made equal to zero in

this expression to get δ7 (horizontal). However, here we use the unit load concept and solve this problem. But you are advised to check the result by the above procedure. Here also, the first step is to determine the member forces (F) under the external load system. Thus, column 2 of the table remains the same. Now the external load system is replaced by a single horizontal unit load (because we require horizontal deflection). Let us assume that this unit load acts left to right. Now, the member forces are determined again under the action of

this unit load alone. The result will be equivalent to 1Q

FQ =

⎛ ⎞∂⎜ ⎟∂⎝ ⎠

at 7. These values are shown in column 6 of Table 8.2.

Now the horizontal deflection at joint 7 is given by .FL FAE Q

⎛ ⎞∂⎜ ⎟∂⎝ ⎠

. The results

are presented in Table 8.2. From this table, it can be seen that horizontal deflection of joint 7 is 0.75 mm towards the left.

Table 8.2

146

Forces and Stresses in Beams

Member

F* (kN)

A (102

mm2)

L (mm)

FLAE

∂

∂

FQ

.FL FAE Q

∂

∂

1 − 60 2.4 2400 − 0.30 1 − 0.30

2 − 60 2.4 2400 − 0.30 1 − 0.30

3 − 30 2.4 2400 − 0.15 1 − 0.15

4 − 100 1.6 4000 − 1.25 0 0.0

5 40 1.6 3200 0.40 0 0.0

6 50 1.6 4000 0.625 0 0.0

7 − 40 1.6 3200 − 0.40 0 0.0

8 50 1.6 4000 0.625 0 0.0

9 120 2.4 2400 0.60 0 0.0

10 30 2.4 2400 0.15 0 0.0

7(horizontal)

FL FAE Q

∂δ = ∑

∂

⎛⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎞

− 0.75* mm

* Negative sign indicates that the deflection is opposite to the direction of unit load.

SAQ 5 Determine the vertical and horizontal deflection of point A in the pin-joined frame shown in Figure 8.12. All members have EA = 20 MN.

Given is AB = AC = BC = 1 m and DB = 0.5 m.

A

B

C

50 kN

Figure 8.12

8.6.3 Application to Beam Problems In beams, the strain energy stored is mostly due to bending with shear contributing less and this contribution is generally neglected. Thus, the strain energy U can be written as

2 2

0 02 2

L LM ds M dxUEI E

= =∫ ∫ I (for very small curvatures)

The bending moment will be a function of the loads. Then, by Castigliano’s theorem the deflection under one of the loads, P, in its direction will be

2

0 0

.2

L LU M dx M M dxP P EI EI P

∂ ∂ ∂δ = = =

∂ ∂ ∂∫ ∫

147

Strain EnergyIf we have to find the deflection at a point where there is no load, a fictitious load Q has

to be introduced at that point in the direction in which the deflection is needed and U is

computed. Then, UQ

∂∂

is computed. If in this expression Q is made equal to zero, the

result will be the deflection at the pint of application of Q.

2

0 2

L M dxUEI

= ∫

2

0 0 00 0

.2

L L

Q Q Q

U M dx M M dxQ Q EI EI Q= = =

⎡ ⎤ ⎡⎡ ⎤∂ ∂ ∂ ⎤= =⎢ ⎥ ⎢ ⎥⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫ ∫δ =

From the similarities with the truss problems, it can easily be recognised that MQ

∂∂

will

be the moment at a section due to a unit load at the point of application of Q with no other loads present on the beam. We shall now apply these concepts to some practical problems.

Example 8.6

Determine the vertical deflection at the tip of the cantilever shown in Figure 8.13.

L

E, I

P

Figure 8.13

Solution

Measuring the distance x from the free end, the bending moment at any section distant x from the tip is given as

M Px= −

Then, 2 2 2

0 02 2 6

L L 2 3M ds P x P LU dxEI EI

= = =∫ ∫ EI

Hence, the vertical deflection below

3

3U PLPP E

∂= =

∂ I

The positive sign shows that the deflection is in the direction of P, i.e. downwards.

It may be seen that UP

∂∂

can be written as follows :

0

( )L Px x dx

EI−⎛ ⎞ −⎜ ⎟

⎝ ⎠∫

where andPx M MxEI EI P

∂− = − =

∂

148

Example 8.7 Forces and Stresses in Beams

Determine the deflection (vertical) at a distance ‘a’ from the tip in Example 8.6.

Solution

Since in this case there is no load at the point where the deflection is needed, we introduce a fictitious load Q as shown in Figure 8.14.

Q P

a

L

Figure 8.14

Then, and 0 for 0MM Px x aQ

∂= − = ≤

∂≤

and ( ) and ( ) forMM Px Q x a x a a x LQ

∂= − − − = − − ≤ ≤

∂

∴ 0 0 0

[ ( ) ] [ ( )](0)L a LM M Px Q x a Px x adx dx dx

EI Q EI EI∂ − − −⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠∫ ∫ ∫

− −

2

0

( ) ( )L Px x a Q x a dxEI

⎛ ⎞− + −= ⎜ ⎟⎜ ⎟∫

⎝ ⎠

0 00

( ).L L

aQ

M M Px x adx dxEI Q EI

=

⎡ ⎤∂ −δ = =⎢ ⎥

∂⎢ ⎥⎣ ⎦∫ ∫

2 2 3 3

0

( )2 2 3 6

L L

a

P x x P L ax a dxEI EI

⎡ ⎤⎛ ⎞ ⎡⎢ ⎥= − − = + −⎜ ⎟ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣⎣ ⎦∫

2

2L a ⎤

⎦

Example 8.8

Determine the slope of the cantilever given in Example 8.6 at its free end.

Solution

To determine rotation in a beam we should have a couple acting at that point. Since at the free end, there is no couple, we introduce a fictitious couple M′ as shown in Figure 8.15.

P

AM′

Figure 8.15

L

θA

149

Strain Energy

Now and 1MM Px MM

∂′= − − = −′∂

∴ 0 00 0

( ) 1.L L

AM M

M M Px M dxEI M EI

′ ′= =

⎡ ⎤ ⎡ ⎤′∂ + ×θ = =⎢ ⎥ ⎢ ⎥

′∂⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫

2

2APL

EIθ = (Positive sign indicates rotation in the sense of M′, i.e. clockwise).

Thus, we see that the tip slope of a cantilever loaded with a concentrated vertical

force P at its free end is 2

2PL

EI.

SAQ 6 (a) Using Castigliano’s theorem, determine the deflection at the centre of the

beam loaded as shown in Figure 8.16. The EI is constant.

Figure 8.16

(b) A steel rod, of flexural rigidity EI, forms a cantilever ABC lying in a vertical plane as shown in Figure 8.17. A horizontal load P acts at C.

Calculate

(i) the horizontal deflection of C,

(ii) the vertical deflection of C, and

(iii) the slope at B.

Consider the strain energy resulting from bending only.

L/2 L/2

W Per Unit Length

P

b

B

C

a

Figure 8.17

150

Forces and Stresses in Beams 8.7 SUMMARY

In this unit, we have

(a) introduced the concept of potential energy,

(b) introduced the concept of strain energy,

(c) shown the relationship between flexural work and strain energy,

(d) derived the expressions for strain energy in different modes of loading,

(e) seen the differences between gradually applied loads, suddenly applied loads and impact loads,

(f) derived the expressions for instantaneous stresses and deflection for axial and bending impact loads,

(g) stated and proved Castigliano’s theorem, and

(h) applied Castigliano’s theorems to determine the deflections in trusses and beams.

8.8 ANSWERS TO SAQs

SAQ 1

(a) 0.096 N m, 49.7 N m, 38.2 N m

(b) 0.78, 0.22

(c) 235 MPa

SAQ 2

50 kN m

SAQ 3

(a) 142.83 N/mm2, 143.6 N/mm2

(b) 133.3 N/mm2, 6666.7 N

SAQ 4

(a) 78 mm

(b) 24 mm, 1.45 kN

SAQ 5

9.2 mm, 1.45 kN

SAQ 6

(a) 45

786WL

EI

(b) 2 2

( 3 ), ,3 2Pa Pab Paba b

EI EI+

EI

151

Strain EnergyFURTHER READING

Prasad, Jainti, Strength of Materials, CBS Publishers and Distributors.

Timoshenko, Stephan, Strength of Materials – Part I and II, CBS Publishing and Distributors.

Schaum’s Outline Series, (1989), Strength of Materials, Second Edition, McGraw-Hill Book Company.

Popov, E. P., (1993), Mechanics of Materials, Second Edition, Prentice-Hall of India Private Limited.

Ryder, G. H., (1993), Strength of Materials, Educational Low Priced Books Scheme.

Case, John, Chilver, L., Ross, Carl, T. F., (1993), Strength of Materials and Structures, Third Edition, Educational Low Priced Books Scheme.

Forces and Stresses in Beams FORCES AND STRESSES IN BEAMS

Computation of appropriate kinds of stresses in a structural member is a pre-requisite for designing the structural member. These stresses are developed in the member due to various forces acting on it. In this block, an attempt has been made to familiarise you with various types of forces and stresses developed in the structural members.

This block consists of five units.

Unit 4 deals with the most important parameters used in designing, namely shear force and bending moment. Unit starts with the classification of supports, beams and loadings. It covers all types of possible combinations of loads and beams for shear force and bending moment computation through a large number of examples.

Unit 5 covers the details of bending stresses developed across the cross-section of a beam. It also discusses the theory of simple or pure bending and concept of moment of resistance. Unit concludes with some practical applications of these concepts.

Unit 6 deals with stress distribution in the beams. In the beginning, unit covers the concept of section modulus and gives the bending stress distribution diagrams for all important geometric shapes of beam sections. It also covers the flitched beam and shear stress distribution pattern of different cross-sections of the beam.

Unit 7 discusses the effect of combination of direct and bending stresses on the members and gives the distribution pattern of these stresses across the cross-section. It also evolves the condition for no tension in a section.

Unit 8 presents some concepts of strain energy useful in axial loading, shear, bending and torsion. It also covers the Castigliaon’s theorem applied to angular movement, truss and beam problems.

152