Virtual Displacement

Dr. Rakesh K Kapania Aerospace and Ocean Engineering Department Virginia Polytechnic Institute and State University, Blacksburg, VA

AOE 5024, Vehicle Structures

~v T~ (~v) ~v = Unit Normal

u =0 Deformed Structure With Virtual Displacement Superimposed

©Rakesh K. Kapania AOE 5024, Vehicle Structures 2 Virtual Displacement (contd...)

I Virtual displacement is an imaginary inﬁnitesimal displacement that is superimposed on the already displaced structure.

I Virtual displacement is consistent with the essential or kinematic (displacement) boundary conditions. They vanish where essential boundary conditions are speciﬁed.

I Applied loads do not change due to the action of these inﬁnitesimal displacements.

I Virtual displacements can be expressed as the δ operator.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 3 Virtual Work

I Work done on a structure by all the forces acting on the structure as the structure is given a virtual displacement.

I As a simple example, consider a simple linear spring with spring constant k subjected to a load Fs . The deﬂection x is given by x = Fs /k. Given a small virtual displacement δx, the work done by the external load, δWe , in moving through the virtual displacement δx is given as: δWe = Fs δx

©Rakesh K. Kapania AOE 5024, Vehicle Structures 4 Virtual Work (contd...)

I The virtual work done by the internal spring force, kx, is represented as δWi and can be expressed as:

δWi = kxδx

Since F = kx, the virtual work done by the external force is equal to the virtual work done by the internal force.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 5 Principle of Virtual Work

I At equilibrium: δWe = δWi This is the Principal of Virtual Work, PVW. This principle is equally valid for more general, complex structures; linear or nonlinear response; as well as Conservative or Non-Conservative forces.

I The principle of Virtual Work is used both to derive the governing equations for a structure and also for developing approximate methods.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 6 Principle of Virtual Work (contd...)

I The internal Virtual Work done can also be thought of as the change in the strain energy, δU, due to a virtual displacement. For example, for the case of spring, δU = Fs δx = δWi .

Note: Fs = kx for a linear spring

Hence δWe = δU is also a statement of PVW.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 7 Cantilever Beam Example

Here,

δw(0) = 0 δw 0(0) = 0

©Rakesh K. Kapania AOE 5024, Vehicle Structures 8 Cantilever Beam Example (contd...)

Consider a Cantilever beam, of length L and subjected to a distributed load q(x), a tip moment M in the anti-clockwise direction, and point loads F1 and F2 acting at distances x1 and x2 respectively. An example of valid transverse Virtual displacement δw(x) for this beam is: h πx i δw(x) = δA 1 − cos 2L

An Note that both δw(0) and δw 0(0) vanish, this is as it should be since at x = 0 both the transverse displacement w(x) and the slope dw/dx are speciﬁed to be zero in this case. It is interesting to see what happens at x = L.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 9 Cantilever Beam Example (contd...)

Since the end at x = L is free to both deﬂect and rotate, the assumed virtual displacement and rotation should not be allowed to vanish at the free end. If any of these vanish, then we are not satisfying the consistency requirements for virtual displacements.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 10 Virtual External Work

By distributed load,

Z L h πx i = δA q(x) 1 − cos dx 0 2L

h πx1 i By F , F , respectively are: F δA 1 − cos , and 1 2 1 2L h πx2 i F δA 1 − cos 2 2L And that due to moment M, d h πx i MδA 1 − cos | dx 2L x=L

©Rakesh K. Kapania AOE 5024, Vehicle Structures 11 Virtual External Work (contd...)

The virtual work done by external forces is then given as:

Z L h πx i δWe = δA q(x) 1 − cos dx 0 2L h πx1 i h πx2 i + F δA 1 − cos + F δA 1 − cos 1 2L 2 2L

d h πx i + M δA 1 − cos dx 2L x=L For a given set of loads, the external virtual work will become δA times some known quantity.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 12 Change in Strain Energy

The change in strain energy can be expressed as, " # 1 Z L d2w 2 δU = δ EI 2 dx 2 0 dx 1 Z L d2w d2w = EI .2 2 δ 2 dx 2 0 dx dx

Assuming the deﬂection proﬁle to be given as: h πx i w(x) = A 1 − cos 2L δU becomes: π 4 Z L πx δU = AδA EI cos2 dx 2L 0 2L ©Rakesh K. Kapania AOE 5024, Vehicle Structures 13 Change in Strain Energy (contd...)

The variation in the strain energy δU, after performing the integral, depends upon product of A, δA and some known quantity.

By equating δU = δWe , from the previous slide, we can obtain A; leading to a one-term approximate solution of the problem.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 14 Example

For x1 = L/3, x2 = 2L/3, F1 = F2 = F , and, M = 2FL

δWe becomes

Z L h πx i h π i = δA q(x) 1 − cos dx + F δA 1 − cos 0 2L 6

h π i d h πx i +F δA 1 − cos + 2FLδA 1 − cos 3 dx 2L x=L

" √ # Z L h πx i 3 3 δWe = δA q(x) 1 − cos dx + F δA π + − 0 2L 2 2

©Rakesh K. Kapania AOE 5024, Vehicle Structures 15 Example (contd...)

Similarly after substituting values for x1, x2, F1, F2, and, M , δU becomes: π 4 L δU = A.δA.EI 2L 2

By equation δWe = δU, δA cancels out and A becomes, " √ # L h πx i 3 3 R q(x) 1 − cos dx + F π + − 0 2L 2 2 A = π 4 L EI 2L 2

©Rakesh K. Kapania AOE 5024, Vehicle Structures 16 Example (contd...)

F x Assuming: q(x) = . , L L " √ # 3 3 8 − 4π + π2 π + − + 2 2 2π2 A = F π 4 L EI 2L 2 FL3 A = 1.3285 EI This is a one-term approximation to the deﬂection of the beam. As will be seen subsequently, the accuracy of the solution can be increased by adding more terms to the solution and applying the PVW as many times there are number of unknown coeﬃcients.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 17 Principle of Virtual Work, General Case

For general 3-D case, subjected to body forces Bi and surface forces (~ν) Ti , where ~ν represents the unit normal on the surface area S,

©Rakesh K. Kapania AOE 5024, Vehicle Structures 18 Principle of Virtual Work, General Case (contd...)

The external virtual work, due to virtual displacement δui , can be written as: ZZZ ZZ ZZ (~ν) (~ν) δWe = Bi δui dV + Ti δui dS + Ti δui dS V S1 S2

Since δui = 0 on S2 ZZZ ZZ (~ν) δWe = Bi δui dV + Ti δui dS V S1 The ﬁrst term represents the virtual work done by the body forces and requires an integral over the volume V ; the second term represents the virtual work done by surface traction and requires an area integral over the surface S.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 19 Principle of Virtual Work, General Case (contd...)

Recall: ai bi = a1b1 + a2b2 + a3b3.

If S = S1 ∪ S2 such that u is prescribed on S2, then δu vanishes on S2 and the second integral is then performed over only S1, the surface area over (~ν) which traction Ti is speciﬁed.

(~ν) Using Cauchy’s relation, Ti = τij νj , the external virtual work becomes: ZZZ ZZ δWe = Bi δui dV + τij νj δui dS V S

©Rakesh K. Kapania AOE 5024, Vehicle Structures 20 Principle of Virtual Work, General Case (contd...)

The surface integral in the above equation can be converted into a volume integral using Gauss Divergence Theorem.

ZZZ ZZZ δWe = Bi δui dV + (τij δui ),j dV V V ZZZ = [(τij,j + Bi ) δui + τij δui,j ] dV V

Recall: τij,j + Bi = 0 from equilibrium equations.

ZZZ δWe = τij δui,j dV V

©Rakesh K. Kapania AOE 5024, Vehicle Structures 21 Principle of Virtual Work, General Case (contd...)

One can write: 1 1 u = [u + u ] + [u − u ] i,j 2 i,j j,i 2 i,j j,i ui,j = eij + ωij This gives:

ZZZ δWe = τij δ (eij + ωij ) dV V ZZZ = τij δeij dV Recall : τij ωij = 0 V δWe = δU = δWi (Internal Virtual Work)

©Rakesh K. Kapania AOE 5024, Vehicle Structures 22 Principle of Virtual Work, General Case (contd...)

If a structure is in equilibrium and remains in equilibrium while it is subjected to a virtual distortion, the external virtual work, δWe , done by external forces acting on the structure is equal to the internal virtual work δWi done by the internal stresses.

Conversely; If δWe = δWi for an arbitrary virtual distortion then the body is in equilibrium.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 23 Principle of Virtual Work, General Case (contd...)

©Rakesh K. Kapania AOE 5024, Vehicle Structures 24 Principle of Stationary Potential Energy

The internal virtual work done can be written as; ZZZ δWi = τij δeij dV V ZZZ ∂u = δeij dV V ∂eij = δU

Here u is the strain energy density and U is the strain energy of the complete structure. δU represents the ﬁrst variation in the strain energy.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 25 Principle of Minimum Potential Energy (contd...)

Next, we assume that the forces (body or traction) are conservative forces such that these can be derived from a potential, V . To illustrate the concept of Potential of a load, consider a load Q acting on a structure and q is the corresponding displacement. A load is called a conservative load if it can be derived from a potential V such that:

∂V Q = − ∂q Obviously, V = −Qq.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 26 Principle of Stationary Potential Energy (contd...)

For a general three-dimensional case, we can write V as: ZZZ ZZ (~ν) V = − Bi ui d(Vol) − Ti ui dS Vol S1 When given a virtual displacement, δu, the variation in the potential is given as: ZZZ ZZ ∂ui (~ν) ∂ui δV = − Bi δuj d(Vol) − Ti δuj dS Vol ∂uj S1 ∂uj ZZZ ZZ (~ν) = − Bi δui d(Vol) − Ti δui dS Vol S1 = −δWe = −δU

©Rakesh K. Kapania AOE 5024, Vehicle Structures 27 Principle of Stationary Potential Energy (contd...)

This gives:

δ(U + V ) = 0

The sum Π = U + V is called the Total Potential Energy. At equilibrium δΠ = 0, provided the forces are conservative, the displacements satisfy the essential boundary conditions, and the variation of the displacement that causes variations in the total potential energy vanish where ever essential boundary conditions are speciﬁed.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 28 Dummy Displacement Method

Principle of Virtual Work (PVW) can be used to calculate unknown forces if the deformation are known. Let a structure is in equilibrium under a set of forces Qi (could be a moment also). Apply a virtual known distortion on the structure such that δqi 6= 0, but all other q’s are zero. From PVW: ZZZ Qi δqi = τij δeij dV V

Often δqi = 1. This is the so-called Dummy Displacement Method

©Rakesh K. Kapania AOE 5024, Vehicle Structures 29 Example

Consider a set of N bars meeting at a point and ﬁxed at the other end.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 30 Example (contd...)

Let the nth bar has length Ln and makes an angle αn with the x-axis. The end where all the bars meet is subjected to a force P making an angle β with the x axis and we neglect body forces.

Let u and v represents the deﬂections in the x and y directions respectively, at A.

The nth bar thus, makes an angle αn and is displaced by u and v.

©Rakesh K. Kapania AOE 5024, Vehicle Structures 31 Example (contd...)

The axial strain, en, and stress σn, assuming u and v are very small, in the nth bar, can be written as:

u v en = cos αn + sin αn Ln Ln u v σn = En cos αn + En sin αn Ln Ln

Here, we have assumed each axial bar can be made of diﬀerent material and En is the Young’s modulus for the material of that bar.

Give the end a virtual displacement δu in the x direction only, i.e. δv = 0. The external virtual work is given by:

δWe = (P cos β)δu

Virtual strain in the nth bar due to the virtual displacement δu is given as:

δu δen = cos αn Ln

The internal virtual work for the total system can be obtained by summing the virtual work done by the internal stress σn in the n th bar as:

n=N u v δu δWi = ∑ En cos αn + En sin αn AnLn cos αn n=1 Ln Ln Ln

Here An is the area of cross-section of the nth bar.

From the PVW, δWe = δWi

n=N ! n=N ! EnAn 2 EnAn P cos β = ∑ cos αn u + ∑ sin 2αn v (1) n=1 Ln n=1 2Ln

Note the above equation represents one equation in two unknowns u and v. We need one more equation.

To get the second equation, apply a virtual displacement δv in the y-direction only. The external virtual work now is δWe = P sin βδv.

The PVW will yield:

n=N ! n=N ! EnAn EnAn 2 P sin β = ∑ sin 2αn u + ∑ sin αn v (2) n=1 2Ln n=1 Ln

Application of Principle of Minimum Potential Energy

V = − P cos β u − P sin β v

ZZZ Z eij U = τij deij dVol Vol 0

N Z en = ∑ En en den An Ln n=1 0 N 2 en = ∑ En An Ln n=1 2

u cosαn v sinαn en = + Ln Ln

N En An 2 U = ∑ [u cosαn + v sinαn] n=1 2Ln

Π = U + V N En An 2 = −P u cosβ − P v sinβ + ∑ [u cosαn + v sinαn] n=1 2Ln

δΠ = 0

∂Π ∂Π = δu + δv ∂u ∂v

δu and δv are independent of each other

∂Π = 0 ∂u ∂Π = 0 ∂v

N ∂Π En An = 0 ⇒ −P cosβ + ∑ 2 [u cosαn + v sinαn] cosαn = 0 ∂u n=1 2Ln

N ! N ! En An 2 En An ∑ cos αn u + ∑ sinαn cosαn v = Pcosβ n=1 Ln n=1 Ln

N ! N ! ∂Π En An En An 2 = 0 ⇒ ∑ sinαn cosαn u + ∑ sin αn v = Psinβ ∂v n=1 Ln n=1 Ln

Consider an arbitrary structure subjected to a set of generalized forces, Q1, Q2, ... Qi ... Qn. Note by generalized forces we mean a linear combination of physical forces and moments.

Let q1, q2, ... qi ... qn be the set of corresponding co-ordinates or displacements. ©Rakesh K. Kapania AOE 5024, Vehicle Structures 40 Castigliano’s Theorem (contd...)

Castiglianos’s two theorems state:

∂U Qi = First Theorem ∂qi ∂U∗ qi = Second Theorem ∂Qi

Here U and U∗ represent, respectively, Strain and Complementry Strain Energy. For linear elastic structure, U = U∗.

We can also get the two desired equations for the truss bar problem by applying Castigliano’s ﬁrst theorem. Recall:

N En An 2 U = ∑ [u cos αn + v sin αn] n=1 2Ln The force component in the x and y direction as the generalized forces Q1 = P cos β and Q2 = P sin β. Also, q1 = u, and q2 = v.

N ∂U En An P cos β = = ∑ [u cos αn + v sin αn] cos αn ∂u n=1 Ln N ∂U En An P sin β = = ∑ [u cos αn + v sin αn] sin αn ∂v n=1 Ln All the three methods yield the same set of two equations. ©Rakesh K. Kapania AOE 5024, Vehicle Structures 42