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Structural Displacements Displacements Structural Structural Displacements Displacements P Beam Displacement Truss Displacements 1 2 Several examples which demon- The deflections of civil engineer- strate the value of deflection ing structures under the action of analysis include (Tartaglione, usual design loads are known to 1991): be small in relation to both the 1. Wind forces on tall buildings overall dimensions and member hbkhave been known to pro duce lengths. “Why bother to compute excessive lateral deflections that deflections?” Basically, the have resulted in cracked windows design engineer must establish and walls, as well as discomfort to that the predicted design loads the occupants. will not result in large deflections that mayy, lead to structural failure, 2. Large floor deflections in a impede serviceability, or result in building are aesthetically an aesthetically displeasing and unattractive, do not inspire distorted structure. confidence, may crack brittle finishes or cause other damage, and can be unsafe. 3 4 1 3. Floor systems are often 6.Deflection computations serve to designed to support motor- establish the vibration and driven machines or sensitive dynamic characteristics of equipment that will run satisfac- structures that must withstand torily only if the support system moving loads, vibration, and undergoes limited deflections. shock environment -- inclusive of 4. Large deflections on a railway seismic design loads. or highway structural support Elastic Deformations ≡ structure system may impair ride quality, deflections disappear and the cause passenger discomfort, structure regains its original and be unsafe. shape when the actions causing 5.DfltiDeflection con tro l an d cam ber the deformations are removed. behavior of pre-stressed con- Permanent deformations of crete beams during various structures are referred to as stages of construction and load- inelastic or plastic deformations. ing are vital for a successful design. 5 6 This course will focus on linear Work-Energy Methods elastic deformations. Such deformations vary linearly with Work-energy methods for truss, applied loads and the principle of beam and frame structures are superposition is valid for such considered. Such methods are struct ures . Fu rthermore, since based on the principle of the deflections are expected to be conservation of energy, which small, deflections are measured states that the work done by a with respect to the original, unde- system of forces applied to a formed or reference geometry. structure (W) equals the strain energy stored (U) in the structure. This statement is based on slowly applied loads that do not produce kinetic energy, which can be written as W = U 7 8 2 A disadvantage of work-energy methods is that only one dis- placement component or rotation can be computed with each application. Work ≡ force (moment) times displacement (rotation) in the force (moment) direction Differential work of Fig. 1 can be expressed as Figure 1. Force versus Displacement Curves dW = P (dΔ) 9 10 For P = F (force), Δ equals Linear Elastic Structure displacement δ: δ W = 1 F δ 2 W = ∫ Fdδ (1) W=W = 1 M θ 0 2 For P = M (moment), Δ equals rotation θ: Complementary Work θ The area above the load- W = Mdθ (2) displacement diagrams of Fig.1 is ∫ known as complementary work, 0 W as shown in Fig 2. For a Eqs. (1, 2) indicate that work is linear-elastic system: simply the area under the force – 1 displacement (or moment – rota- W =W = P Δ 11 2 12 tion) diagrams shown in Fig. 1. 3 Fig. 2. Complementary Work Virtual Work complementary Virtual (virtual ≡ imaginary, not work Load real, or in essence but not in fact) work procedures can produce a W single displacement component at any desired location on the Displacement structure. To calculate the desired Direct use of work-energy displacement, a dummy or virtual calculations is only capable of load (normally of unit magnitude) calculating displacements at the is applied at the location and in location of an applied p oint force the direction of the desired and rotations at the point of displacement component. Forces application of a point couple; associated with this virtual force obviously a very restrictive are subscripted with a V. condition. Consequently, virtual work principles are developed in 13 14 the subsequent sections. Use of a virtual force in calcula- Alternatively, if virtual displace- ting virtual work is defined as the ments are applied then the virtual principle of virtual forces (which work is defined as the principle of will be the focus of this chapter): virtual displacements: Principle of Virt ual Forces Principle of Virtual If a deformable structure is in Displacements equilibrium under a virtual system If a deformable structure is in of forces, then the external work equilibrium and remains in done by the virtual forces going equilibrium while it is subject to a through the real displacements virtual distortion, the external equals the internal virtual work virtual work done by the external done by the virtual stress forces acting on the structure is resultants going through the real equal to the internal virtual work displacement differentials. done by the stress resultants. 15 16 4 The virtual work principles (forces Pv = virtual force or moment and displacements) are based on Δ = real displacement δ or conserving the change in energy rotation θ due to the applied virtual load or displacement, which can be P + Pv ex pressed mathematically as WV P WUVV= for the principle of virtual forces, W which is the focus of this chapter, where again the overbar signifies compltlementary energy. The rea l and virtual complementary exter- Δ nal work is shown schematically in Fig. 3. Fig. 3. Complementary Real and Virtual Works 17 18 Complementary Axial For a single axial force Strain Energy member in equilibrium For a single axial force subjected to a virtual force member subjected to a real FV, the virtual complemen- fFthlforce F, the complemen tary tary strain energy (virtual strain energy (internal work) complementary internal is work) is F U =δ UFVV= δ 2 FL ⇒=UF FL VVEA δ= EA Real and virtual complemen- tary strain energies for a FL2 ⇒=U single member are shown 2EA schematically in Fig. 4. 19 20 5 Fig. 4. Complementary Real and Virtual Strain Energies for UVi = Complementary Virtual a Single Truss Member Strain Energy for Truss Member i F + Fv δi = Real Member i Deformation UV F FLii δ=i ; for a mechanically U EAii loaded truss member δiiii=αL Δ T ; for a thermally δ loaded truss member For a truss structure: mm α=linear coefficient of UUVVivii==δ∑∑ F thermal expansion i1== i1 21 Δ=T change in temperature 22 Example Deflection δ=ΔifiL ; for a fabrication Calculation – Mechanically Loaded Truss Structure error ofΔ Lf in the truss member EA = constant Non-mechanical δi are positive if they produce a positive Rocker change in member length consistent with tension positive forces in truss members. Calculate the horizontal displace- ment at C. 23 24 6 Complementary Bending Example Deflection Strain Energy Calculation – Thermally Loaded Truss Structure EA = constant EA = constant α = constant M dU= dθ 2 M ddθ = x Calculate the horizontal displace- EI ment at G if the top chord mem- 1M bers are subjected to a temper- ⇒=UMdx 2EI∫ ature increase of Δ T = 100. L Equation of condition at C! 25 26 dUVV=θ M d M M = real bending equation due ⇒=UMdx to the external applied loading VV∫ EI L M = δ for a virtual moment M+MM + M V MV v equation due to a unit force dUV M for a point displacement δ calculation at the desired dU point in the assumed direc- tion of the displacement θ = M V for a virtual moment equation for a unit virtual dθ couple for a point rotation θ Fig. 6. Complementary Real and calculation at the desired Virtual Strain Energies for point in the assumed rotation a Differential Bending 27 direction 28 Segment 7 Beam Deflection Example For a multi-segment beam: Mi UMdxVVi= ∑ ∫ EIi i Li where i designates beam segment. Calculate the tip displacement andttifthd rotation for the can tilever beam. 29 30 Frame Deflection Example For a frame structure: m M j UMdxVVj= ∑ ∫ EIj j1= L j where m equals the number of frame members. Note, axial deformation has been ignored. Also, if a frame member is composed of multi- segments, then a summation Calculate the vertical displace- over the segments must also be ment and rotation at C. included. 31 32 8.
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