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Chapter 5

Applications of Energy Methods

Widely used to obtain solutions to elasticity problems

Basis for the Finite Element Method

Energy is a scalar quantity

Can be used to obtain elastic deflections of statically indeterminate structures and to

determine redundant reactions

Two Principles will be presented:

Principle of Stationary Potential Energy

Castiglianos Theorem

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Consider a system with a finite number of DOF with the equilibrium configuration (x1, x2, , xn)

Impose a virtual (imagined) displacement such that (x1, x2, , xn) (x1+dx1, x2+dx2, , xn+dxn) where (dx1, dx2, , dxn) is the virtual displacement

The corresponding virtual work is

dW = Q1dx1+Q2dx2+ + Qidxi + + Qndxn (a)

where (Q1, Q2, , Qi, , Qn) are the components of the generalized

load

If dxi is a translation Qi is a force If dxi is a rotation Qi is a moment

5.1 Principle of Stationary Potential Energy

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For a deformable body, d W can be separated into

d W = d We + d Wi (b)

where

d We is the virtual work of the external forces

d Wi is the virtual work of the internal forces

The virtual work done by the external loads can be expressed as

d We = P1dx1+P2dx2+ + Pidxi + + Pndxn (c)

For an elastic system where the loading is path independent, the virtual work of the internal forces is equal to the negative of the

virtual change in the elastic strain energy, d U

d Wi = - d U (d)

where U=U(x1, x2, , xn) is the total strain energy in the system

For a conservative system, dW = 0 which is known as the Principal of Stationary Potential Energy

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For a conservative system,

(e)

(f)

For any system with finite DOF, if the components Qi vanish, then the system is in equilibrium a system is in equilibrium if

Eq. 5.1 is sometimes called Castiglianos First theorem

But the dxi can assume any value, so for the above equation to be true requires

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For a simple example of this theorem, consider a uniform bar loaded at its end by a load P

Let the bar be made of a nonlinear elastic material

The area below the curve represents the total strain energy U stored in the

bar,

By Eq. 5.1, P = U/ e

If the load-elongation data are plotted as s-e, then

and s = U0 /e (as given by Eq. 3.11)

See nice Example 5.1 in Boresi text Fig. 5.1

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5.2 Castiglianos Theorem on Deflections

The derivation of Castiglianos theorem on deflections is based on the concept of the complimentary energy of

the system

The complimentary energy C is equal to the strain energy U for a linear elastic material

In Castiglianos theorem, C is considered to be a function of the generalized forces (F1, F2, , Fn) acting on a system that is structurally stable (i.e. sufficiently

supported)

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(5.2)

Note: Castiglianos theorem is limited to small displacements

C also depends on the distributed loads, but these loads do not need to be considered explicitly in the

derivation

Castiglianos Theorem:

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The complimentary energy C for a structure composed of m members is:

Castiglianos theorem can be extended to compute the rotation of line elements in a system subjected to

couples (e.g., F1 and F2)

By Eq. 5.2

(a)

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Considering the magnitudes of F1 and F2 to be a function of S, then by the chain rule

(b)

If S = F1 = F2 =F, then F1/S = F2/S = 1,

(c)

Eqs. (a) and (c) imply

(a)

(d)

Because M=bF (5.3)

Eqs. 5.2 and 5.3 may be applied to structures with nonlinear elastic behavior

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5.3 Castiglianos Theorem on Deflections for Linear Load-Deflection Relations

If the structural response is assumed to be limited to linear elastic material behavior and to small

displacements then

the resulting load-deflection relation is linear the strain energy U equals the complimentary energy C

the Principle of Superposition applies

where U=U(F1 ,F2 , ,Fp, M1 , M2 , , Ms)

The strain energy U is:

(5.4)

(5.5)

(5.6)

Fig. 5.1

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Restricting the discussion to linear elastic, isotropic, homogeneous materials gives U0 to be

(5.7)

With load-stress formulas derived for the members of the structure, U0 may be expressed in terms of the

loads that act on the structure.

Eqns. 5.4 and 5.5 can be used to find the displacements at the points of application of the

concentrated forces

or moments (5.4)

(5.5)

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1. Axial Loading

2. Bending of Beams

3. Torsion

Three types of loads are considered in this chapter

In practice, it is typically easier to find U for each type of load, then use linear superposition to combine their input

to the overall U for the structure.

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5.3.1 Strain Energy UN for Axial Loading

Consider an arbitrary section of a nonuniform rod

The total strain energy UN for the tension member is

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Rewriting UN for discussion purposes

Noting that dez = ezz dz

And recalling ezz = szz/E and szz = N/A, where A is the cross- sectional area at z

An example of such a discretization is

(5.9)

Must discretize the contributions to UN at abrupt changes in

Material properties

Load

Cross section

(5.8)

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Strain Energy for Axially Loaded Spring

Recalling for a spring: F = kx

(5.12)

Fig. 5.5

Because

= Uspring

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5.3.2 Strain Energies UM and US for Beams

Consider a beam of uniform (or slowly varying) cross section

Origin is at left end of the beam

Length of beam along the z axis

x and y axes are principal axes

Flexure formula is assumed to apply

x

xzz

I

yM=s

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Mx causes D*F* to rotate w.r.t. B*D* an amount df

For a linear elastic material, df varies linearly w.r.t Mx

The shaded area below the inclined straight line is equal to the strain energy dUM resulting from the bending of the segment dz

Plane sections are assumed to remain plane

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The strain energy UM for the beam caused by Mx becomes

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Eq. 5.13 is exact for pure bending

But is only approximate for shear loading

An exact expression for the strain energy due to the shear loading, Us, is difficult to obtain

use an approximate value

The shear force Vy shear stresses szy

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Recall shear stress is zero on the top and bottom and maximum at the N.A.

Define an average value of szy as t=Vy/A

Assume this average acts over the entire beam cross section

Assume the beam cross section to be rectangular and with thickness b for convenience

Due to the shear force, the relative displacement of faces D*F* and B*C* is dey

For linear elastic behavior, dey varies linearly w.r.t. Vy (see Fig 5.6e)

dUs is the strain energy for the beam segment dz

A correction (fudge) factor is introduced s.t. the expression for the shear strain energy dUs of the element is equal to kdUs

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Exact values for k are generally not available

For most problems, US is small in comparison to UM

A quick approximation for k is to use ratio of the shear stress at the N.A. to the average shear stress t=Vy/A

e.g., k for a rectangular cross section

(5.15)

This value is greater than the exact value (therefore is conservative as it will overpredict the situation)

Authors suggest using the known value when it is available

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To get an exact expression for Us the shear stress distribution must be known

The known shear stress distribution can be used in

(5.7)

Assuming the shear stress to be uniform across a width b

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5.3.3 Strain Energy UT for Torsion

Consider a circular rod in torsion

Sections BC and DF are a distance dz apart.

After the torsional loads are applied, these sections become B*C* and D*F*

Section D*F* is rotated db w.r.t B*C*

For linear elastic behavior, db varies linearly w.r.t. T

The shaded area is the strain energy dUT

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Realizing that b db = g dz

Recalling g = t/G and t = Tb/J

(5.16)

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5.4 Deflections of Statically Determinate Structures

Statically Determinate vs. Statically Indeterminate

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The strain energy for a structure is equal to the sum of the strain energies for each of its members

For the jth member: Uj = UNj + UMj + USj + UTj For this chapter, loadings will be assumed to be on or about

principal axes

Knowing Utotal,

the deflection qi of the structure at the location of a concentrated force Fi in the direction of Fi is

(5.17)

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the deflection qi of the structure at the location of a concentrated moment Mi in the direction of Mi is

(5.18)

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Procedure for the application of Castiglianos Theorem

1. Write an expression for each of the internal actions (axial

force, moment and torque) in each member of the structure in

terms of the applied external loads.

2. To determine the deflection qi (or rotation qi) of the structure at the location of a concentrated force Fi (or moment Mi ) and in

the directed sense of Fi (or Mi ), differentiate each of the

internal action expressions w.r.t. Fi (or Mi ).

3. Substitute the expressions for internal actions obtained in

Step 1 and the derivatives obtained in Step 2 into Eq. 5.17 (or

Eq. 5.18) and perform the integration. The result is a

relationship between the deflection qi (or rotation qi) and the externally applied loads.

4. Substitute the magnitudes of the external loads into the result

obtained in Step 3 to obtain a numerical value for the

displacement qi (or rotation qi).

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5.4.1 Curved Beams Treated as Straight Beams

The magnitude of UM for curved beams is derived in Chapter 9

The error in using Eq. 5.13 to determine UM is negligible if the radius of curvature of the beam is more than twice its

depth.

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5.4.2 Dummy Load Method and Dummy Unit Load Method

So far, Castiglianos Theorem has been used to find deflections and rotations at the locations of concentrated

forces and moments

However, the displacement at a point other than where a concentrated load is applied may be desired

Modified procedure that requires use of a fictitious load

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1. Apply a fictitious force Fi (or moment Mi) at the point of interest for finding

the displacement.

2. Write an expression for each of the internal actions (axial force, moment

and torque) in each member of the structure in terms of the applied

external loadsincluding the fictitious force (or moment).

3. To determine the deflection qi of the structure at the location of a fictitious

force Fi (or Mi) and in the directed sense of Fi (or Mi), differentiate each of

the internal action expressions w.r.t. Fi (or Mi).

4. Substitute the expressions for internal actions obtained in

Step 2 and the derivatives obtained in Step 3 into Eq. 5.17 (or Eq. 5.18)

and perform the integration. The result is a relationship between the

deflection qi (or rotation qi) and the externally applied loads, including the fictitious force Fi (or moment Mi) .

5. Because the fictitious force (or moment) does not act on the structure, set

its value to zero in the relation obtained in Step 4. Then substitute the

magnitudes of the external loads into the result obtained in Step 3 to obtain

a numerical value for the displacement qi (or rotation qi).

Procedure for the application of Castiglianos Theorem with Dummy Load

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(5.20b)

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5.5 Statically Indeterminate Structures

A statically determinate structure may be made statically indeterminate by the addition of a member, internal action

and/or support.

Likewise, a statically indeterminate structure may be made determinate if certain redundant members, internal actions

or supports are removed.

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Likewise, for every redundant internal moment or redundant reaction (M1, M2 , )

(5.21)

It can be shown that for every redundant internal force or redundant reaction (F1, F2 , )

(5.22)

To show that Eqs. 5.21 and 5.22 apply to internal forces, consider the truss with the redundant internal for NBE

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Make a section cut through some point H of member BE

Apply equal and opposite reactions NBE and NBE

Because the component of the deflection of point h along member BE is not zero, it is not obvious that

(a)

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However, NBE = NBE = NBE

Likewise it can be shown for internal member forces N, V and M

(5.23)

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In the application of Eqs. 5.21 and 5.22 to the system with redundant supports of redundant members, it is assumed that the

unloaded system is stress-free.

However, in certain applications, these conditions do not hold.

Consider

Initially, the right end of the beam may be lifted off the support, or the end may exert a force on the beam due to either settling or

thermal expansion (or contraction).

As a result, the end of the beam (in the absence of the redundant support) may be raised a distance q1 above the location of the

support before the beam is loaded

or it may be a distance q2 below the support location

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If the displacement magnitudes q1or q2 are known, then the reaction R for the loaded beam may be computed by the relations

(5.24)

Where the minus sign indicates that the displacement q1and the force R have opposite senses.

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5.5.1 Deflections of Statically Indeterminate Structures

A structure is not altered if a redundant member or redundant support is removed and replaced by external forces and moments

that are identical to the forces and moments exerted by the deleted

parts.

By setting the deflections at the redundant supports to known values, expressions can be determined to find the associated

reactions.

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Example 5.15 Uniformly Loaded Propped cantilever Beam

Fig. E5.15

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Fig. P5.66

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