STRAIN ENERGY METHODS

Introduction

Strain energy methods are used to determine the angular and linear

displacements in complicated structural elements, such as a structure having

members in different planes, straight or curved members. There is extensive

use of these methods in determining: (a) displacements and forces in members

of redundant structures, (b) displacement at any joint of a truss and (c)

deflection and slopes in beams and cantilevers having variable section along

length, etc.

We will discuss Castigliano’s theorem, Maxwell’s reciprocal theorem, strain

energy expressions due to axial loads, shear loads, bending moment, twisting

moment and the principle of virtual force for determination of deflection in

trusses.

Castigliano’s First Theorem

According to Castigliano’s first theorem, ‘if a body is subjected to forces

F1, F2, F3,…, Fn, and U is the strain energy stored in the body due to

these forces, then partial derivative of the strain energy U, with respect

to force Fi gives the displacement of the body in the direction of Fi, i.e.,

the displacement,

This theorem is extremely useful in determining displacements in

complicated structures subjected to external forces.

i

iF

U

Proof

Let us consider a beam AB of length L, initially straight and simply

supported at the ends. Say, a number of transverse loads W1, W2,

W3,…, Wn are gradually applied on the beam at different points a, b,

c,…, n and say deflections under the loads are y1, y2, y3,…, yn,

respectively, as shown in Figure 1 by aa′, bb′, cc′,…, nn′, respectively.

Figure 1 Beam subjected to various point loads

A B

L

Then, strain energy stored in the beam is:

That is, the area covered under the triangles shown dotted in Figure 2.

Figure 2 Graphs between loads and deflections

A B

Now let us only increase load W1 by δW1, and due to this increase in the load,

there is an increase in the deflection under the loads, W1, W2, W3,…, Wn, by

δy1, δy2, δy3,…, δyn, which is indicated by the rectangular portion shown

hatched in the diagram.

Additional strain energy,

however, δW1δy1 is a very small quantity; therefore, it is neglected.

A B

Therefore,

δU = W1δy1 + W2δy2 + W3δy3 + ⋯ + Wnδyn

Differentiating partially with respect to load W1,

(loads W2, W3, … Wn remain constant)

but because these loads remain unchanged

Taking and dividing throughout by ∂W1 δU = W1δy1 + W2δy2 + W3δy3 + ⋯ + Wnδyn

in the limit

Subtracting this from we get,

Similarly, it can be proved that

If a system of forces F1, F2,…, Fn, bending moments M1, M2, M3,…, Mn and

twisting moments, T1, T2, T3,…, Tn act simultaneously on a body, then

Castigliano’s first theorem can be extended to find angular rotation due

to bending moment, angular twist due to twisting moment also, that is,

angular rotation at location of Mi.

angular twist at the location of Ti.

Strain Energy Due to Axial Force

Trusses are made from bars of different lengths joined at ends. Each bar carries

some axial force, tensile or compressive, depending on the external loads acting

on the trusses. Figure 3 shows a bar of length L, area of cross-section A,

subjected to a tensile force P. If E is the Young’s modulus of the bar,

Figure 3 bar subjected to axial load

Change is length,

Strain energy,

axial extension, or displacement of bar along the force.

Example 10.1

A truss ABCD is shown in the figure. The cross section and material of each

member is the same. Load P is applied at joint C. Length of member CD is

a. Using Castigliano’s first theorem, determine the deflection in the truss at

joint C.

Length of members

DC = CB = BD = a

Length of AB =

Let us find forces in all the members of the truss. The following table lists

the values of forces in members.

C = Compressive

T = Tensile

Strain energy in members of the truss,

Deflection under the load,

From Page 10:

Strain Energy Due to Shear Stress

Consider a rectangular block of dimensions L × B × H as shown in Figure 4.

The block is fixed at the lower face and is subjected to a tangential force Q at

the top face. The block is distorted due to this shear force Q and is subjected to

shear strain f.

Figure 4 Rectangular block subjected to shear force

Work done on the block = = Strain energy stored

× ds LBH (multiplying and dividing by LBH)

where , shear stress developed

shear strain, in reality

angle f is very very small

within the elastic limit.

dsQ21

therefore,

because shear strain,

where G is the shear modulus.

Shear strain energy per unit volume,

The distribution of shear stress across the section of the beam is

generally non-uniform; therefore, the shear strain energy must be

integrated over this whole section of a body. Yet shear strain energy due

to shear deformation (particularly in beams) is very small and many

times it can be ignored.

GU s 2

2

Example 10.2

A beam of rectangular cross-section of breadth b, depth d and length l is simply

supported at its ends. It carries a concentrated load W at its centre. Determine

the shear strain energy in the beam and find the deflection due to shear.

G = Modulus of rigidity for the beam

The figure shows a beam AB of length l, simply supported at the ends and

carrying a concentrated load W at its centre C. Shear force between A and C is

+W/2 and between C and B, shear force is –W/2;

Consider a section X-X at a distance of x from the end A.

Shear force,

Now, consider a small length dx. Let us determine the shear strain energy

for the portion AC. Figure (b) shows the section of the beam. Consider a

layer of thickness dy at a distance of y from the neutral axis.

Shear stress at the layer = bI

yAFx

2W

Fx

where

Fx = Shear force at the section = W/2

A ӯ = First moment of the area above the layer under

consideration about neural axis

I

=

Moment of inertia =

b = Breadth

Shear stress,

= bI

yAFx

byd

A

2

2

2

yd

yy

Volume of the layer = bdxdy

Shear strain energy in the layer

Total shear strain energy for the portion AC

Since the beam is symmetrically loaded, shear force in the portion CB is the

same, that is, –W/2. Shear strain energy for the beam,

Deflection at the centre due to shear,

GbdlW

UU ss 203

22

GbdWl

W

U ss 10

3

Strain Energy Due to Bending

Consider a bar of length L, initially straight, subjected to a gradually increasing

bending moment, M. As the bending moment increases, curvature in the bar

increases or the angular rotation f goes on increasing. Say, at a particular instant

(Figure 5):

Figure 5 Bar subjected to bending moment

Bending moment = M

Angular rotation = ø

Radius of curvature = R

Work done on the bar =

however, ø = (arc length)

Since ø is very small and

the stress in the bar remains within

the elastic limit.

Work done = Energy stored in the bar

From the flexure formula,

Therefore, strain energy,

From “Fixed Beams in Bending”, page 12

R

yEEy

IM

x

dx

d

1

dx

d

ddx

yy

yydx

dy

dx

du

1

ydddydu

difference between the two fiber lengths is du

EILM

U2

2

If we consider a beam subjected to transverse

loads W1, W2,…, Wn, where the radius of curvature

goes on changing from one section to the other,

then the strain energy due to bending,

where Mx is the bending moment at any section X-X and dx is the small

length along the axis of the beam.

Say, δi is the deflection under the load Wi. Then,

Example 10.3

A circular section cantilever of length l, free at one end is fixed at the other end,

with diameter d for half of its length and diameter 2d for the rest of its length. It

carries a concentrated load W at the free end. If E is the Young’s modulus of the

material, determine the deflection and the slope at the free end.

The figure shows a cantilever ABC, fixed at end C and free at end A, with

diameter d for half of its length AB and diameter 2d for next half of its length

BC. Since we have to find out the slope at the free end A, let us apply a fictitious

moment M = 0 at the free end.

Portion AB

Bending moment,

Mx = M + Wx, where x = 0 to

taking the origin at A. x

Mx

Portion BC

Strain energy,

where

Deflection at A,

x

Mx

however, M = 0

Substituting the values of I1 and I2,

Slope at end A,

however, M = 0

Substituting the values I1 and I2,

Example 10.4

A circular ring of mean radius R and second moment of area of its cross-

section I, with a slit at one section is shown in the figure. Points A and B are

subjected to forces P as shown.

Determine the relative displacement between points A and B. Only the strain

energy due to bending is to be taken into account.

Consider an element of length ds = Rdθ at

angle θ from the vertical axis.

Bending moment of the force P on the element.

Mx = P(R – Rcos θ) = PR(1 – cos θ) ds

Strain energy,

Relative displacement between points A and B,

Strain Energy Due to Twisting Moment

Figure 6 shows a shaft of diameter d and length l subjected to a gradually

increasing twisting moment (torque). As the twisting moment increases, the

angular twist also increases gradually.

At a particular stage, the torque applied is T and angular twist in the shaft is θ.

Figure 6 Angular twist

Work done on the shaft = U, strain energy stored in the shaft

From torsion formula,

Strain energy due to

twisting moment,

where J = Polar moment of inertia of the shaft

If the torque or the section vary along the length of the shaft,

Angular twist due to the twisting moment,

Example 10.5

A circular bar of diameter d is bent at right angle. It is fixed at one end and a

load W is applied at the other end as shown in the figure. Determine the

deflection under the load W, if E = Young’s modulus and G = Shear modulus of

the material.

Let us calculate the strain energy:

Portion BC

Portion AB

U2 =Strain energy due to bending

Wb

Straight portion AB is also subjected to a twisting moment, T = Wb

U3 = Strain energy due to twisting moment

where

Total strain energy,

Deflection under the load,

Wb

Maxwell’s Reciprocal Theorem

This theorem states that ‘work done by the forces of the first state on the

corresponding displacements of the second state is equal to the work done by

the forces of the second state on the corresponding displacements of the first

state’. In symbols

P1δII

1 + P2δII

2 = P3δI3 + P4δ

I4

Figure 7 (a) shows an elastic body supported and subjected to forces P1 and P2

at points 1 and 2, respectively, due to which displacements at four points are

δI1, δ

I2, δ

I3 and δI

4, respectively. Figure 7 (b) shows the same elastic body

subjected to forces P3 and P4 at points 3 and 4, respectively, due to which the

displacements at four points are δII1, δ

II2, δ

II3 and δII

4, respectively.

Figure 7

Let us say that forces P1 and P2 are applied

first, then strain energy

Now, in the second step, forces P3 and P4

are applied due to which displacements are

δII1, δ

II2, δ

II3 and δII

1, respectively, the

strain energy

Total strain energy,

In the second manner of loading, we assume

that P3 and P4 forces are applied first on the

body and then the forces P1 and P2 are

applied.

Total strain energy,

Total energy stored in the body does not

depend on the order in which the forces are

applied; therefore, from both total strain

energy equations,

P1δII

1 + P2δII

2 = P3δI3 + P4δ

I4

Maxwell’s reciprocal theorem can be proved for

any number of forces.

Example 10.6

A beam of length l is simply supported at the ends. Loads W1 and then W2 are

applied in two states on points 1 and 2 as shown.

1.When load W1 is applied at point 1, in the first

state equation for deflection,

deflection at position (2) δ12, at

or,

EI

M

dx

yd

2

2

l l l

l l

l

2. When load W2 is applied at point 2 in the second state, the equation for

deflection,

(can be derived)

Deflection at position (1) δII1 at

Now,

From these expressions it is obvious that W1δII

1 = W2δI2,

which goes to prove Maxwell’s reciprocal theorem.

l l l

l l

l

Principle of Virtual Forces Applied to Trusses

The principle of virtual forces can be used to determine the displacement

of any point in a truss subjected to external loads. To illustrate this

principle, vertical displacement of joint C caused by the real load P at

joint B will be determined, in the following example. The load P at B can

cause only the axial forces in members; it is only necessary to consider

the internal virtual work due to axial load. To obtain this virtual work, we

will assume that each member has constant cross-sectional area A, and

the virtual load n and real load N are constant throughout. To obtain the

internal virtual work so as to get displacement of joint C, we apply a unit

load to joint C. Due to this unit load at C, axial loads are determined in

each member of the truss, as in member AB, axial force due to unit load,

n = 0.

Example 10.7

Due to the real

load P at B, axial

forces are

determined with

the help of a force

polygon for the

truss as shown.

Say, for member AB, axial force due to real load, N = –P (compressive load)

Internal virtual work for a member

Virtual work for the entire truss is, therefore,

1 = External virtual load acting on the truss joint, in the direction of

displacement ∆

∆ = Joint displacement caused by the real loads on the truss

n = Internal virtual force in a truss member caused by external virtual

unit load

N = Internal axial force in a truss member due to real loads

L = Length of a member

A = Cross-sectional area of a member

E = Young’s modulus of elasticity of a member

Here,

Forces in members due to real load and unit load

Virtual work equation for the truss shown is

, if AE is constant for the truss for all members

P = 10 kN

l = 2 m = 2,000 mm

A = 500 mm2

E = 200 kN/mm2

∆ = Vertical displacement at joint C

For truss let us take

Example 10.8

A cantilever of length l, fixed at one end and propped at the other end, carries a

concentrated load W at its centre and a uniformly distributed load w per unit

length from the centre up to the fixed end. If EI is the flexural rigidity of the

cantilever determine the reaction at the prop.

The figure shows a cantilever ABC of length l, fixed at the end C and propped at

end A, carrying loads as given in the problem. Let us first determine the strain

energy due to bending. Say the reaction at the prop = R.

Portion AB (Origin at A)

Portion BC (Taking origin at B)

Bending moment,

Total strain energy = U1 + U2

at the propped end (no deflection)

therefore,

or,

Reaction at the prop,

Example 10.9

A bar is bent in the shape as shown, with radius of the bend R and length of

the straight portion l. Determine the horizontal deflection due to the force P

applied at the end A, if EI is the flexural rigidity of the bar. Consider only

the strain energy due to bending.

Portion AC Consider an element of length = Rdθ = ds

The bending moment on ds = PR sin θ

Strain energy

ds

Portion CB (origin at C),

Bending moment, Mx = Px

Strain energy,

Total strain energy,

Horizontal deflection at A,

Example 10.10

The figure shows a steel rod bent into the form of three quarters of a circle

of radius r. End A of the rod is fixed, while end B is constrained to move

vertically. If a load W is applied at the end B, determine the vertical

deflection at the end B. EI is the flexural rigidity of the rod.

The figure shows a rod bent into the form of

three quarters of a circle. Since the end B is

constrained to move only vertically, a horizontal

reaction R will be offered by the constraint.

Consider an element of length ds = rdθ at an

angle θ to the vertical.

Bending moment,

Mθ = W × r sinθ – R × (r – rcosθ),

where R is the reaction at B.

ds

Strain energy,

Since there is no horizontal deflection,

or

Arranging,

Reaction,

Vertical deflection,

putting the value of reaction, R

Example 10.11

The frame, as shown in the figure, is

supported by a fixed end at G and is free

at end A. A vertical load 12 kN acts at

point C. Determine the horizontal

deflection at A using Castigliano’s

theorem, E = 200 kN/mm2 and

I = 2,000 cm4.

Consider only the bending energy.

Apply imaginary load, P ≅ 0 at A

At the fixed end G, upward reaction, R1 = 12 kN

Horizontal reaction, R2 = P

Fixing moment, MG = 12 × 40 = 480 kN cm

Forces and moments as 3 portions are shown

Moment, My = −Py (taking A as origin)

Member BC

Member AB,

Member CD

Member DG

y varies from 0 to 80 cm

Using Castigliano’s theorem, δH at end A

however, P = 0

= Horizontal deflection at point A

The calculation of strain energy for any member, that is, AB, BC, CD or

DG, the origin for calculation of bending moment can be taken at any

end.