Structural Displacements

Beam Displacement1

Structural Displacements

P

T Di l2

Truss Displacements

The deflections of civil engineer-The deflections of civil engineering structures under the action of usual design loads are known to be small in relation to both the overall dimensions and member lengths “Wh bother to comp telengths. “Why bother to compute deflections?” Basically, the design engineer must establishdesign engineer must establish that the predicted design loads will not result in large deflections that may lead to structural failure, impede serviceability, or result in an aesthetically displeasing andan aesthetically displeasing and distorted structure.

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Several examples which demon-strate the value of deflectionstrate the value of deflection analysis include (Tartaglione, 1991):1. Wind forces on tall buildings have been known to produce excessive lateral deflections that have resulted in cracked windows and walls as well as discomfort toand walls, as well as discomfort to the occupants.

2 L fl d fl ti i2. Large floor deflections in a building are aesthetically unattractive do not inspireunattractive, do not inspire confidence, may crack brittle finishes or cause other damage,

4and can be unsafe.

3. Floor systems are often designed to support motor-driven machines or sensitive equipment that will run satisfac-torily only if the support systemtorily only if the support system undergoes limited deflections.

4 L d fl ti il4. Large deflections on a railway or highway structural support system may impair ride qualitysystem may impair ride quality, cause passenger discomfort, and be unsafe.

5.Deflection control and camber behavior of pre-stressed con-pcrete beams during various stages of construction and load-i it l f f l

5ing are vital for a successful design.

6.Deflection computations serve to establish the vibration andestablish the vibration and dynamic characteristics of structures that must withstand moving loads, vibration, and shock environment -- inclusive of seismic design loadsseismic design loads.

Elastic Deformations ≡ structure d fl ti di d thdeflections disappear and the structure regains its original shape when the actions causingshape when the actions causing the deformations are removed.

Permanent deformations ofPermanent deformations of structures are referred to as inelastic or plastic deformations.

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This course will focus on linear elastic deformations Suchelastic deformations. Such deformations vary linearly with applied loads and the principle of app ed oads a d t e p c p e osuperposition is valid for such structures. Furthermore, since the deflections are expected to be small, deflections are measured with respect to the original undewith respect to the original, unde-formed or reference geometry.

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Work-Energy MethodsWork-energy methods for truss, beam and frame structures are considered. Such methods are based on the principle of conservation of energy whichconservation of energy, which states that the work done by a system of forces applied to asystem of forces applied to a structure (W) equals the strain energy stored (U) in the structure.

This statement is based on slowly applied loads that do not produce pp pkinetic energy, which can be written as

8W = U

A di d f kA disadvantage of work-energy methods is that only one dis-placement component orplacement component or rotation can be computed with each application.pp

Work ≡ force (moment) times displacement (rotation) in the force (moment) directionforce (moment) direction

Differential work of Fig. 1 can be dexpressed as

dW = P (dΔ)

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Figure 1 Force versusFigure 1. Force versus Displacement Curves

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For P = F (force), Δ equals displacement δ:displacement δ:

W (1)δ

∫W = (1)

0

F d δ∫For P = M (moment), Δ equals rotation θ:

0

rotation θ:

W (2)M dθ

θ∫W = (2)

0

M dθ∫Eqs. (1, 2) indicate that work is simply the area under the force –di l t ( t t

11displacement (or moment – rota-tion) diagrams shown in Fig. 1.

Linear Elastic Structure

W = 12 F δ

W =212 M θ2

Complementary Workp y

The area above the load-displacement diagrams of Fig 1 isdisplacement diagrams of Fig.1 is known as complementary work,

as shown in Fig 2. For a Wlinear-elastic system:

W = =W 1 P Δ12

W = =W 2 P Δ

Fig. 2. Complementary WorkLo

ad complementary work

W

Direct use of work-energyDisplacement

Direct use of work energy calculations is only capable of calculating displacements at the location of an applied point force and rotations at the point of application of a point couple;application of a point couple; obviously a very restrictive condition. Consequently, virtual

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condition. Consequently, virtual work principles are developed in the subsequent sections.

Virtual Work

Virtual (virtual ≡ imaginary, not real, or in essence but not in fact) , )work procedures can produce a single displacement component at any desired location on the structure. To calculate the desired displacement a dummy or virtualdisplacement, a dummy or virtual load (normally of unit magnitude) is applied at the location and in ppthe direction of the desired displacement component. Forces

i d i h hi i l fassociated with this virtual force are subscripted with a V.

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Use of a virtual force in calcula-ting virtual work is defined as theting virtual work is defined as the principle of virtual forces (which will be the focus of this chapter):be t e ocus o t s c apte )

Principle of Virtual ForcesIf a deformable structure is in equilibrium under a virtual system of forces then the external workof forces, then the external work done by the virtual forces going through the real displacements g pequals the internal virtual work done by the virtual stress resultants going through the real displacement differentials.

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Alternatively, if virtual displace-ments are applied then the virtualments are applied then the virtual work is defined as the principle of virtual displacements:tua d sp ace e ts

Principle of Virtual DisplacementsIf a deformable structure is in equilibrium and remains inequilibrium and remains in equilibrium while it is subject to a virtual distortion, the external ,virtual work done by the external forces acting on the structure is equal to the internal virtual work done by the stress resultants.

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The virtual work principles (forces and displacements) are based onand displacements) are based on conserving the change in energy due to the applied virtual load or due to t e app ed tua oad odisplacement, which can be expressed mathematically as

=V VW U

f h i i l f i l ffor the principle of virtual forces, which is the focus of this chapter, where again the overbar signifieswhere again the overbar signifies complementary energy. The real and virtual complementary exter-p ynal work is shown schematically in Fig. 3.

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Pv = virtual force or moment

Δ = real displacement δ or rotation θ

P + Pv

VWP

W

V

W

Δ

Fig. 3. Complementary Real d Vi t l W k

Δ

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and Virtual Works

Complementary Axial Strain EnergyFor a single axial force gmember subjected to a real force F, the complementary p ystrain energy (internal work) is

FU2

= δ

FLEA

δ =EA

2F LU⇒ =19

U2EA

⇒

For a single axial force member in equilibrium subjected to a virtual force FV, the virtual complemen-tary strain energy (virtual complementary internal work) is

V VU F= δFLU F⇒ V VU FEA

⇒ =

Real and virtual complemen-Real and virtual complementary strain energies for a single member are shown

20schematically in Fig. 4.

Fig. 4. Complementary Real and Virtual Strain Energies for ga Single Truss Member

F + F

F

F + FvUV

U

δFor a truss structure:

m mV Vi vi iU U F= = δ∑ ∑

21i 1 i 1= =∑ ∑

ViU Complementary Virtual=Strain Energy for Truss Member i

i Real Member i Deformationδ =

i iF L ; for a mechanicallyδ i ii

i i; for a mechanically

E Aloaded truss member

δ =

loaded truss member

i i i iL T ; for a thermallyδ = α Δi i i iL T ; for a thermallyloaded truss member

li i i

δ = α Δ

linear coefficient ofthermal expansion

α =

22T change in temperatureΔ =

i f iL ; for a fabricationδ = Δi f i

f

L ; for a fabrication

error of L in the truss

δ Δ

Δmember

Non-mechanical δi are positive if they produce a positiveif they produce a positive change in member length consistent with tension positive forces in truss members.

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Example Deflection Calculation MechanicallyCalculation – Mechanically Loaded Truss Structure

EA = constant

R kRocker

Calculate the horizontal displace-ment at C

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ment at C.

Example Deflection Calculation ThermallyCalculation – Thermally Loaded Truss Structure

EA = constantEA = constantEA constantα = constant

Calculate the horizontal displace-ment at G if the top chord mem-bers are subjected to a temper-ature increase of T = 100Δ

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ature increase of T = 100.Equation of condition at C!

Δ

Complementary Bending Strain Energy

MdU d2

= θ

Md dxEI

θ =

1 MU M dx2 EI

⇒ = ∫26

L2 EI

V VdU M d= θ

V VL

MU M dxEI

⇒ = ∫

M + Mv

L

MdUV

dU

Fig. 6. Complementary Real and Virtual Strain Energies for

dθ

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Virtual Strain Energies for a Differential Bending Segment

M l b di ti dM = real bending equation due to the external applied loading

VM = for a virtual moment equation due to a unit force

VMδ

equation due to a unit force for a point displacement calculation at the desired

δ

point in the assumed direc-tion of the displacement

= for a virtual moment equation for a unit virtual

VMθ

couple for a point rotation calculation at the desired point in the assumed rotation

θ

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point in the assumed rotation direction

F l i bFor a multi-segment beam:

iV Vi

ii

MU M dxEI

= ∑ ∫i

ii LE I

here i designates beamwhere i designates beam segment.

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Beam Deflection Example

Calculate the tip displacementCalculate the tip displacement and rotation for the cantilever beam.

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F f

m jM

For a frame structure:

j

m jV Vj

jj 1 L

MU M dx

EI== ∑ ∫

jj L

where m equals the number of frame members. Note, axial deformation has been i d Al if fignored. Also, if a frame member is composed of multi-segments then a summationsegments, then a summation over the segments must also be included.

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Frame Deflection Example

C l l h i l di lCalculate the vertical displace-ment and rotation at C.

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