1

Introduction to Virtual Work Virtual Work is a hypothetical concept that is used mainly for derivation and proof. However it will be seen that this concept can be used as a primary method of analysis. “Consider a structure having external loads Wi with equivalent (vector) displacements δi at the joints, and internal forces Pj with equivalent displacements ej in the members. If the loads Wi are in equilibrium with the forces Pj and the displacements δi are compatible with the displacements ej, then the principle of virtual work states that:

Σ Wi δi = Σ Pj ej

joints members If the conditions of equilibrium and compatibility are satisfied, neither of the sets of forces or displacements need actually exist, but may be only virtual. Combinations of Virtual forces with real displacements lead to indirect methods of analysis in which the final equations are transformed into a more easily managed form”1 “The key to the solution of statically indeterminate structures is the ability to determine structural deformations”2. The deflection of a structure may be a design criterion, those familiar with the design of slabs are usually aware of this. There are 2 basic approaches to the analysis of structural deformations, strain energy and virtual work. Strain energy is limited to the elastic range, hence the advantage of the Virtual work method, of being able to deal with conditions other then those within the elastic range. Virtual work is also considered to be the more powerful of the two methods. It must be stressed that it is important to qualitatively understand the applications of the concepts of virtual work to the problem at hand. Two conditions that must be satisfied in a structure are:

1. The applied forces and the internal forces (reactions) must be in equilibrium. 2. Simultaneously, the structural deformations must be compatible with the type

of supports The example below will illustrate these two conditions:

1 Solving problems in Structures, Croxton and Martin 2 Understanding structural analysis, 2nd edition, David Brohn

2

W

A B C

MA

VA VC

W

A B C

Load and reaction equilibrium

Deflected shape compatible with supports

3

Virtual States: The concept of a virtual state is useful in solving certain kinds of problems, particularly those of deformations. These virtual states need only satisfy one of the two conditions mentioned previously:

either equilibrium or compatibility. We will use the previous example to discuss these concepts. “The virtual state is an analytical device. Because only one state needs to be satisfied it is more easily evaluated. When we combine real and virtual states, we can easily develop theorems of great power and application in the solution of structural problems”3 Students are referred to the book, Understanding structural analysis, 2nd edition for the proof of two formulae which are discussed herein after:

3 Understanding structural analysis, 2nd edition, David Brohn

VIRTUAL FORCE In this virtual force state there is equilibrium of forces, since the load is supported by the moment and vertical reaction at A. However the deflection is no longer compatible with the rigid support at C

W

A B C

MA

VA

A B C VIRTUAL DISPLACEMENT A state of virtual displacement can be created by the introduction of a hinge at A and B, which is compatible with the support conditions. The introduction of these two hinges results in a mechanism, Hence this system satisfies compatibilty but not equilibrium

4

The following theorems are reproduced from Brohn’s Understanding structural analysis:

1. Theorem of VIRTUAL Displacements; In any structural system the external VIRTUAL work of the applied loads acting over the VIRTUAL displacement is equal to the internal VIRTUAL work of the VIRTUAL displacement acting against the internal forces

2. Theorem of VIRTUAL Forces; In any structural system the external VIRTUAL work of the external VIRTUAL forces acting over the real displacement is equal to the internal VIRTUAL work done against the internal VIRTUAL forces by the real displacement

General equation of VIRTUAL work: W=actual force in direction of VIRTUAL displacement * VIRTUAL displacement Refer to examples from Brohn, chapter 5 frame 22 and frame 60 Ultimate Mt of R 2Mu for sagging and Mu for hogging

W A B C

δ θ1 θ1

θ2

θ1=

θ2=

MU MU

2MU

Virtual system of displacements

Actual equilibrium system of internal forces

A B C

D

18

12 6

A B C

D

1

0.556 0.444

M

m

C

W

A B A B C D

3 1 5

18

5

Virtual work method for deflection (applied at joints); This method has been covered adequately in Structural Analysis 2 or 3 for determining the deflection in a structure at a particular point and in a particular direction. Remember the formula Δ =ΣnNL/AE (refer to STAN201 or STAN301 notes).

σ=N/A , ε = Δ/L and E= σ/ ε hence E=(N/A)/(Δ/L) and Δ=NL/AE

internal virtual work U= Δn, therefore U= dxAENnL

0∫

Example 1 (deflection of truss):

Δ N

L

3 2

5 4 6 1

Y

X Z

10 kN 10 kN 10 m 10 m 10 m

Calculate the vertical component of the deflection, at node 6, for the truss shown below for the given loading; assume 300W steel and 50 x 50 x 5 angles for all members

6

member N n L NnL

1-2 14.144 0.9428 14.144 188.61 2-3 10 0.666 10 66.60 3-4 14.144 0.4714 14.144 94.30 4-5 -10 -0.3333 10 33.33 5-6 -10 -0.3333 10 33.33 6-1 -10 -0.6666 10 66.66 2-6 -10 -0.6666 10 66.66 3-6 0 -0.4714 14.144 0.00 3-5 -10 0 10 0.00

549.49 deflection =NnL/AE

A= 4.80E-04 M^2 E= 2.00E+08

deflection= 5.72 mm

7

member N n L NnL

1-2 14.144 0.9428 14.144 188.61 2-3 10 0.666 10 66.60 3-4 14.144 0.4714 14.144 94.30 4-5 -10 -0.3333 10 33.33 5-6 -10 -0.3333 10 33.33 6-1 -10 -0.6666 10 66.66 2-6 -10 -0.6666 10 66.66 3-6 0 -0.4714 14.144 0.00 3-5 -10 0 10 0.00

549.49 deflection =NnL/AE

A= 4.80E-04 M^2 E= 2.00E+08

deflection= 5.72 mm

10

14.14 14.14

5 6 4

-9.999 -.0004 -9.999

-10 -10 -10

2 3

1

Y

X Z

0.6666

0.4714 0.9428

5 6 4

-0 -0.4714 -0.6666

-0.3333 -0.6666 -0.3333

2 3

1

Y

X Z

8

Virtual work method for deflection in rigidly jointed structures, where the loads are applied at points other then the joints. Here bending moment and other stress resultants will also be present. Due to the variable nature of the resultants it becomes necessary to integrate along the length of the member. The integrands are derived by partial differentiation of the expressions for strain energy (eg axial force: P2dx / 2EA)

Axial Force: nNdx / AE Bending Moment: mMdx / EI Shear Force: sSdx / GA Torque: tTdx / GJ (Lower case italics indicate stress resultants due to the unit / virtual load)

Total deflection is given by:

δi = Σ∫ pPdx / AE + Σ∫ mMdx / EI + Σ∫ sSdx / GA + Σ∫ tTdx / GJ The derivation of most of these formulae can be found in Brohn chapter 5, frames 33 to 52. For rotation a unit moment is used instead of a unit load

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Example 1 (eg 6.2 from SPIS) Show that the midspan deflection of a simply supported beam with a uniformly distributed load is 5wL4/384EI

δ = 20∫L/2 mMdx / EI Where for x <= L/2, M= wL/2*x – w*x2/2 m= ½*x= x/2

δ= w∫(Lx2-x3)dx/2EI = 5wL4/384EI Using table of integration (T6.2, P12, SPIS): ∫mMdx = αLac = 5/12*L*(wL2/8)*(1*L/4) = 5wL3/384EI Example 2 (eg 6.8 from SPIS) δH= [∫ mHMdz/2EI]AB + [∫ mHMdz/2EI]BC δH= [αLac/2EI]AB + [αLac/2EI] BC δH= * + 0 * δH= [0.5*h*(-1h)*(-2wh2)+0]/2EI= wh4/2EI δV= * + * δV = [1*h*(-2h)*(-2wh2)]/2EI + [1/4*2h*(-2h)*(-2wh2)]/EI δV = 4wh4/2EI+2wh4/EI= 4wh4/EI assume w=5 kN/m, h=3m, E=200x103N/mm2 and I=328x106mm4 Calculated deflection=4wh4/EI=4*5*3000^4/200E3/328E6= 24.7mm at node 10 vert

w 1

x

M m

x

w

h 2I I

A

B C

2h

-2wh2

M

-1*h

-2h

mH mV

10

Tutorial 1 a) Determine the slope at C, for the frame shown in example above.

θC= (1*h*(-1)*(-2wh2))/2EI+(1/3*2h*(-1)*(-2wh2))/EI θC= wh3+4wh3/3EI θC= 7wh3/3EI radians assume w=5 kN/m, h=3m, E=200x103N/mm2 and I=328x106mm4 Ma=90kNm θC= 7*5*3000^3/3/200E3/328E6=0.004802 =================== NODAL POINT COORDINATES ================ Node no. X-coord Y-coord Z-coord Node no. X-coord Y-coord Z-coord m m m m m m 1 0.000 0.000 0.000 2 0.000 1.000 0.000 3 0.000 2.000 0.000 4 0.000 3.000 0.000 5 1.000 3.000 0.000 6 2.000 3.000 0.000 7 3.000 3.000 0.000 8 4.000 3.000 0.000 9 5.000 3.000 0.000 10 6.000 3.000 0.000 ===============SECTION PROPERTIES =================== Section : A Section designation: 305x305x137 H1 A Ay Ax Ixx Iyy J Material m^2 m^2 m^2 m^4 m^4 m^4 17.40E-3 0.000 0.000 328E-6 106E-6 2.51E-6 Steel:300W Section : B Section designation: 305x305x240 H1 A Ay Ax Ixx Iyy J Material m^2 m^2 m^2 m^4 m^4 m^4 30.50E-3 0.000 0.000 642E-6 202E-6 13.0E-6 Steel:300W =============== NODAL POINT DISPLACEMENTS at SLS ============= Node Lcase X-disp. Y-disp. Z-disp. X-rot. Y-rot. Z-rot. mm mm mm rad. rad. rad. 10 1 3.06 -24.25 0.00 0.0000 0.0000 -0.0047

w

h 2I I

A

B C

2h

-2wh2

M

-1

Mdue to rotation

1

11

Maximum Deflections for Load Case 1:X :3.06 mm at node 4Y :-24.25 mm at node 10

2

1

8 96 74

3

105

Y

XZ

54 7

2

1

6

3

2.5

90

1040 22.5

62.598 10

90

90

90

Y

XZ