RE 30-Structural nalysis ecture Notes, Fall 009 nsgar Neuenhofer ssociate Professor, Department of rchitectural Engineering alifornia Polytechnic State University, San uis Obispo opyright 009 nsgar Neuenhofer (photo by author)
TE OF ONTENTS Review of Statics. Goal of structural analysis. Structural idealizations.3 Summary of properties of moment and shear force diagrams.4 Example. 3.5 Frames 6.6 Example. 9.7 Statical determinacy-instability-degree of indetermincay 0 Problems 4 Principal of Virtual Forces 8. Introductory remarks 8. Virtual work-principle of virtual forces 8.3 Procedure for analysis 9.4 Principle of virtual forces for trusses 9.5 Principle of virtual forces for beams 9.6 Principle of virtual forces for frames 9.7 Example. 0.8 Example..9 Integration tables 3.0 Summary 5. Example.3 7. Example.4 8.3 Example.5 0.4 Example.6 3.5 Shear deformation 35 Problems 38 3 The Force ethod 49 3. Introduction 49 3. Discussion 49 3.3 Example 3. 50 3.4 Example 3. 53 3.5 Example 3. (lternative) 55 3.6 Example 3.3 57 3.7 Example 3.4 60 3.8 Force method for arbitrary degree of statical indeterminacy 64 3.9 axwell's law 64 3.0 Summary of force method 65 3. Example 3.5 65 3. Example 3.6 70 3.3 Example 3.7 73 3.4 The force method for space frames 75 Problems 78
4 The Slope-Deflection ethod 88 4. General remarks 88 4. End moments for prismatic members 89 4.3 Example 4. 93 4.4 Example 4. 95 4.5 Example 4.3 97 4.6 Example 4.4 99 4.7 Summary of procedure 0 4.8 omparison between slope-deflection and force methods 03 4.9 Fixed-end moments 04 Problems 06 5 The oment-distribution ethod 3 5. Introduction 3 5. General description 3 5.3 Example 5. 4 5.4 Example 5. 6 5.5 Summary of steps 8 5.6 ore discussion and illustration 9 5.7 Example 5.3 0 5.8 Summary Problems 3 6 pproximate analysis of building frames under lateral load 6 6. Introduction 6 6. Discussion 6 6.3 The Portal ethod 8 6.4 The antilever ethod 33 Problems 43 7 Influence ines 45 7. Introduction 45 7. üller-reslau principle 48 7.3 Use of influence lines 49 7.4 Example 50 7.5 Properties of influence lines of statically determinate structures 5 7.6 Influence lines of statically indeterminate structures 5 Problems 56
Review of Statics. Goal of structural analysis The objective of a structural analysis is to determine the force (stress) and displacement (strain) demand of structures using a mechanical model. The analysis must be both as economical as possible and as accurate as necessary. Since the exact mechanical relations are extremely complicated, we rely on many approximations that are more or less accurate. With these approximations and simplifications we map the real structure onto our mechanical model. It is important that we are aware of these simplifications in order to judge whether or not certain models are appropriate. fter selecting an appropriate structural model, we analyze the model under the most unfavorable combination of loads. The results of structural analysis are the internal forces and deformations that form the basis for, say, reinforced concrete, steel or timber design. We have to always remember that the results of any structural analysis can never be better than the underlying model. crude model remains a crude model and yields crude results, no matter how many digits we include in our analysis, in short garbage in-garbage out. In this class we talk little about modeling but assume that we already have an appropriate mechanical representation of the real structure. Thus we are omitting a very important step of engineering work, if not the most important, a fact we should always be aware of.. Structural idealizations ll structures are three-dimensional. In structural analysis we usually work with one- or two-dimensional idealizations of the real structure. (a) 3-dimensional structural elements (rarely used in structural engineering) (b) -dimensional structural elements (plate, shells) y H requirement H H x y x (c) -dimensional structural elements Structural elements whose two dimensions (width and height) are small compared to their length are commonly called truss (axial force response) or beam (bending moment response) elements. In this class, we often use the term frame member or frame element to denote a combination of truss and beam elements. We represent a frame element by its axis, which is the connection of the centroids of the cross-sections. In this class, we will be dealing with one-dimensional structural elements (truss, beam and frame members) only. We use those to model plane (two-dimensional) structures and, to a smaller extent, space (three-dimensional) structures.
.3 Summary of Properties of oment and Shear Force Diagrams In the following, we summarize key properties of shear force and bending moment diagrams. In beam segments without distributed loading, the shear force is constant and the bending moment is linearly varying. In regions with a uniformly distributed load the shear force varies linearly and the bending moment is a quadratic parabola. In general, if the distributed load is of ordern, the functions for the shear force and bending moment are of ordern andn, respectively. t points where a concentrated force (a reaction force or an externally applied force) is applied the shear force is discontinuous. It jumps upward or downward according to the direction of the force. The moment function has a change in slope at that point but is continuous. n external moment causes a jump in the bending moment. It does not change the slope of the moment function, nor does it affect the shear force at that location. The shear force is the derivative of the bending moment. Hence the moment function is one degree higher than the shear force function. When the shear force is zero, the bending moment takes on its maximum. w P y w V = 0 V P y linear no change in slope quadratic max linear linear change in slope linear no change in slope Fig.. Properties of shear force and bending moment diagrams.
.4 Example.: eam with internal hinges 50 kn 0 kn/m [m] 5.00.00 6.00.00 5.00 Fig.. eam with internal hinges. Problem: Draw the bending moment and shear force diagrams for the beam in Fig... Shown in Figure. is a beam with two internal hinges. The first important observation is that this structure is not a single rigid body. If we remove the beam from its supports, it ceases to be rigid. We can easily see, that we can rotate against because of the hinge. The consequence for calculating the support reactions is that we have to break the structure apart and look at free-body diagram of individual parts. Differently put, we cannot calculate the support forces by looking at the structure as a whole because there are more unknown forces than there are equilibrium equations. We have four unknown support reaction but only two equilibrium equations. 50 0 E F F F D FE FF Fig..3 Free-body diagram of whole structure. 50 0 F F F F F D F D FE FF Fig..4 Free-body diagram of individual parts. We cut the beam at the internal hinges resulting in the three free-body diagrams in Fig..4. In applying the equilibrium equations to the free-body diagrams it is important to find the right starting point. The center portiond of the beam contains only two unknown forces while the exterior portions contain three unknowns. Hence = 0 F = 0 6 3 =30 D 6 (.) F = 0 F = 3050 60 =80 y We have applied the 50-kN concentrated force to the center free-body diagram. pplying the force to the left-hand portion of the beam is an alternative that would affect the results for F but not the final results. We now apply forces F and F D to the exterior portions of the beam and solve for the support reactions F F and F, F, E F, 3
respectively. aution: Since the numerical results for F and F D are negative, we change the direction of the arrows representing these forces (see Fig..5). F F 80 30 FE F F Fig..5 Free-body diagrams of exterior parts after calculating F and F D. pplying the equilibrium equations to the free-body diagrams in Fig..5 yields = 0 F = F = 0 F =3 = 0 F = 4 F = 0 F = F E y F y We have now calculated all support reactions. efore drawing the shear force and bending moment diagrams, it is good practice to redraw the free-body diagrams and label all forces by their magnitude. (.) 50 0 3 80 80 30 30 4 Fig..6 Free-body diagrams with all forces known. 80 V 30 [N] 3 30 60 [Nm] 45 60 D E F Fig..7 Shear force and bending moment diagrams. Unlike the previous examples, in which we wrote explicit equations for the shear force and bending moment variation, we now use our experience gained in previous examples and try to draw the internal force moment diagrams directly from the free-body diagrams in Fig..6. 4
Shear force () t support the 3 force acts down. Hence the shear force jumps down by 3. () Nothing happens to the beam between supports and. Hence the shear force is constant. (3) t support a force acts up. Hence the shear force jumps from 3 to 3=80. (4) Nothing happens to the beam between support and hinge. Hence the shear force is constant. (5) t hinge the applied force 50 acts down. Hence the shear force jumps from 80 to 80-50=30. Note that the two 80 forces cancel when moving across hinge (see Fig..6). (6) lot happens between hinges and D. total force0 6 = 60 acts down. However, this force is uniformly distributed. This cause the shear force to go down gradually from 30 to 30-60=-30. Note that the two 30 forces cancel when moving across hinge D. (7) Nothing happens to the beam between hinge D and support E. Hence the shear force is constant. (8) t support E the 4 force acts up. Hence the shear force jumps from 30 to 304=. (9) Nothing happens to the beam between supports E and F. Hence the shear force is constant. (0) Finally, we check whether the shear force calculated between supports E and F is consistent with the support force at F. This is obviously the case. ending moment () t support the 3 force acts down. This force causes a linearly varying bending moment increasing from 0 at support to 3 5= 60 at support. Since the support force at acts downward (preventing uplift) it causes tension at the top and compression at the bottom of the beam. The bending moment is thus negative. The bending moment must not jump at support. On the other hand it must be zero sat the internal hinge. onsequently, the moment has to linearly decrease from 60 at support to zero at hinge. () The beam segmentd is a simply supported beam under a uniformly distributed 0-kN/m load spanning from hinge to hinge D. The moment variation is a quadratic parabola whose maximum value is 6 = w = 0 = 45 knm (.3) 8 8 This moment is positive since it produces tension at the bottom of the beam. (3) We now turn to the left-hand segment of the beam. t hinge D the 30 force acts down. This force causes a linearly varying bending moment increasing from 0 at hinge D to 30 = 60 at support E. Since the force at D acts downward, it causes tension at the top and compression at the bottom of the beam. The bending moment is thus negative. The bending moment must not jump at support E. On the other hand it must be zero at support F. onsequently, the moment has to linearly decrease from 60 at support E to zero at support F. s a quick check we argue along the same line starting from support F. t support F a -kn force acts down. This force causes a linearly varying bending moment increasing from 0 at support F to 5= 60 at support E. Since the support force at acts downward (preventing uplift) it causes tension at the top of the beam and the moment is negative. Whenever experience and problem complexity permit, we should try to distance ourselves from blind mathematics and adopt the approach followed in this example. Remarks: The bending moment diagram changes its slope at but does not change the slope at D. an you explain? The shear force jumps at but is continuous at D. an you explain? Fig..8 Sketch of deflected shape. 5
.5 Frames We now apply the concepts learned before to find the internal force diagrams of frame structures. w w (a) H (b) H x x y y Fig..9 Three-hinge frame. (a) system and loading; (b) free-body diagram of whole structure. Problem: Draw the shear force, axial force and bending moment diagrams for the frame in Fig..9(a). Solution: () s always, we first calculate the support reactions. Since there are four unknown reactions but only three equilibrium equations for the structure as a whole, we have to dismember the structure to find the reactions. We find the vertical reactions by looking at the structure as a whole (Fig..9b). Equilibrium for the whole structure also requires that the horizontal support forces and are equal and opposite. x = 0 = w F = 0 = w F = 0 = x y y y x x x (.4) w w x x y y H x w / / x w Fig..0 Three-hinge frame. ut at hinge to find horizontal reaction forces. () Next, we cut the structure at hinge and formulate equilibrium for one of the two resulting free bodies (Fig.0). Since we know the vertical reaction force from Eq. (.4), the two free-body diagrams in Fig..0 involve three unknown forces each. = 0 = w w x H x = w x = w (.5) 4 8H 8H due to y due to load Important: Since a hinge cannot transfer a moment, there is no internal moment in Fig..0 at point. lso note that by selecting as the moment reference point, we avoid referencing the forces and transferred though the hinge. 6 x y
x P V V P H P V w 8 H I II III w 8 H w 8 H x x w w w Fig.. Three cuts for internal forces. (3) With reaction forces known, we are now in the position to find the internal force variation by making cuts at arbitrary locations along the girder and the two columns. Figure. shows the resulting free-body diagrams. Equilibrium for the three sections requires Section I Fx = 0 V = w Fy = 0 P = w = 0 =w x 8H 8H Section II Fx = 0 P = w Fy = 0 V = w wx 8H x = 0 = w x 8 due tox due toy due tow (.6) Section III Fx = 0 V = w Fy = 0 P = w = 0 =w x 8H 8H Note that because of the 90 degree angle at the column-girder junction, the axial force in the column turns into a shear force in the girder, and the shear force in the column turns into an axial force in the girder. (4) efore drawing the internal force diagrams, it is important to emphasize that we should consider the internal force diagrams as a proper connection of values calculated at specific locations rather than plots of mathematical relations. olumns: The axial force in the columns is constant and equal to the vertical reaction forces. learly, the columns are in compression. The shear force in the columns is also constant and equal to the horizontal reaction forces. ccording to our sign convention, the shear force is negative for the left column and positive for the right column. The bending moment in the columns varies linearly from zero at the pin supports to a certain value at the column-girder junction. We calculate this value by substituting x= H into the moment expression for sections I or III, or x= 0 into the moment expression or Section II. Hence 7
= w H = w (.7) 8H 8 Girder: The axial force in the girder is constant and equal to the horizontal reaction forces. y inspection, we conclude that the girder is in compression. The shear force in the girder at the column junction is equal to the axial force in the columns. ecause of the uniformly distributed load acting on the girder, the shear force varies linearly and the bending moment varies quadratically and is zero at hinge. The properties just discussed lead directly to the internal force diagrams below. We include a sketch of the deflected shape of the frame for illustration. We observe that both columns and girder bend (the deflected shape is curved). lso note that the two portions of the girder rotate against each other at the hinge. We will discuss a method to calculate deflections in hapter. w 8 w 8 w w w w 8 H V 8 H w 8 H w P w Fig.. Internal force diagrams V,, P and deflected shape for three-hinge frame. 8
.6 Example. 60 kn m m m 60 kn m m m Fig..3 Sample frame structure. Problem: Find the three internal force diagrams for the frame structure shown. Solution: We obtain the support reactions from the 3 global equilibrium conditions plus the condition that the bending moment is zero at hinge (for parts or ). It is important to realize that we don t have to set up equations describing the variations of the internal forces as a function of a local coordinate x. This is what we did in introductory statics classes. ll we need to do is determine the bending moment at selected locations (e.g. at the column-girder connections and the points of application of external forces) and then properly connect those points. lways remember: We want to do structural engineering with as little math as possible and as much experience and intuition as possible. 80 80 70 0 0 0 0 60 V [knm] 40 [kn] 40 0 40 70 N [kn] sketch of deflected shape Fig..4 Internal force diagrams and sketch of deflected shape. 0 9
.7 Statical Determinacy-Instability-Degree of Indetermincay Trusses We need to be able to identify whether a structure is statically determinate, statically indeterminate, or unstable. Figures.5-.7 depict examples for each of those properties. The truss in Fig..5 is constructed by adding eight triangular panels. ll members are necessary for stability of the truss, i.e. if we remove one member from the structure it becomes unstable and collapses under applied loads. The truss is statically determinate. We will learn how to analyze this type of trusses in subsequent sections of this chapter. The trusses in Fig..6 are statically indeterminate. The structure in Fig..6(a) has one more support than necessary for stability. Since the indeterminacy is with respect to the reactions, the truss is termed statically indeterminate externally. The structure in Fig..6(b) has more members than necessary for stability. We can remove one of the top three members without sacrificing the stability of the structure. Since in this case the indeterminacy is with regard to the numbers of members, the truss is called statically indeterminate internally. nalysis of indeterminate structures is beyond the scope of this class. The trusses in Fig..7 are unstable. The structure in Fig..7(a) is externally unstable because it has no support that resists horizontal movement (we would have to turn one roller support into a pin support to make the truss statically determinate). The structure in Fig..7(b) is internally unstable because it has too few members. ore precisely, the square panel is unstable since it has no diagonal (we would have to add one diagonal to make the truss statically determinate). oth structures undergo excessive displacements without forces being applied to it. Figure.7 illustrates those displacement modes. lthough we will discuss formulas to calculate the degree of indeterminacy and instability in subsequent classes, the best method by far to assess those properties is to use intuition, inspection, insight and experience. Fig..5 Statically determinate truss. (a) (b) Fig..6 Statically indeterminate trusses. (a) (b) Fig..7 Unstable trusses with collapse modes. 0
There are formulas for determining the degree of statical indeterminacy. Unfortunately, the following equations are not foolproof since they constitute only a necessary but not a sufficient criterion for stability of the structure. Degree of statical indeterminacy for truss structures plane trusses n = s m nn s: Number of support reactions space trusses n = s m3 nn m: Number of truss members nn : Number of nodes (.8) n: Degree of statical indeterminacy Note: Usually, the best way of determining whether a structure is statically determinate, indeterminate or unstable is by inspection. n = 3 7 0 = 0 (statically determinate) n = 3 8 = (unstable) n = 3 6 9 = (statically indeterminate) Fig..8 Statically determinate, indeterminate and unstable truss structures.
eams and frames Degree of statical indeterminacy for plane frame structures formula similar to Eq. (.8) exists to calculate the degree of statical indeterminacy for beam and frame structures. n = s i 3m3 p s: Number of support reactions i: Number of internal forces at hinges m : Number of closed loops without hinge p: Number of parts n: Degree of statical indeterminacy (.9) n = 5 0 0 3 = (stat. indet. to nd degree) n = 5 4 0 3 3 = 0 (!! unstable!! UTION) n = 3 3 3 3 = (stat. indet. to th degree) 6 4 n = 3 8 03 7 = 0 (stat. det) n = 4 4 3 3 3 = (!! unstable!!, UTION) Fig..9 Statically determinate, indeterminate and unstable frame structures.
Degree of statical indeterminacy for space frame structures For three-dimensional structures the formula is n = s i 6m6 p s: Number of support reactions i: Number of internal forces at hinges m : Number of closed loops without hinge p: Number of members n: Degree of statical indeterminacy Example: (.0) Fig..0 Statically indeterminate three-dimensional frame. For the 3-D frame above we obtain n = 4 0 6 6 = 4 (stat. indeterminate to the 4th degree) (.) 3
Problems. D 0 kn/m E F G 6 m 5 m 3 m 3 m 4 m Solution: 53 88 80.5 0 40.8 40.8 35 35. N [knm] 9.3 [kn] 40.0 30.6.7 30.6 4.7 V [kn] 8.0 sketch of deflected shape. 0 kn/m D E F 3 m 3 m 50 kn 8 m 4 m (a) Show that the two structures above are statically determinate. (b) Find the bending moment, shear force and axial force diagrams of the two structures for the loading shown. Draw the bending moment diagram on the tension side of the member. 4
.3 (a) 6 m (b) 6 m 5 m 3 m 3 m 4 m 5 m 3 m 3 m 4 m 6 m 6 m (c) (d) 5 m 3 m 3 m 4 m 5 m 3 m 3 m 4 m (e) (f) ft ft 0 ft 0 ft (g) ft (h) ft 0 ft 0 ft Draw the bending moment diagrams of the above structures for the given loading. ll applied forces, moments and distributed loads have unit magnitude and intensity, respectively. 5
.4 (a) 0 kn/m 0 kn 0 kn.00 m 3.00 m.00 m.00 m.00 m.00 m 3.00 m.00 m 0 kn/m 0 kn 0 kn/m (b).00 m 3.00 m.00 m.00 m.00 m.00 m 3.00 m.00 m 5 knm 0 kn 0 kn 5 knm (c).00 m 3.00 m.00 m.00 m.00 m.00 m 3.00 m.00 m 0 kn/m (d).00 m 3.00 m.00 m.00 m.00 m.00 m 3.00 m.00 m 5 knm 5 knm 5 knm (e).00 m 3.00 m.00 m.00 m.00 m.00 m 3.00 m.00 m kn/m 0 knm kn/m (f) (g) 4.00 m.00 m.00 m 4.00 m Find the shear force and bending moment diagrams of the beams for the given loading. Sketch the deflected shape. 6
7 Solution: 5 0 5 5 0 0 0 0 5 5 0 0 5.667 0 0.667 0 8.33.67 30 0 5 5 0 5 5 0 0 0 0 V 0 0 0 V V V (a) (b) (c) (d) 5 6.5.5.5 6.5 5 6.5 6.5.5.5.5.5 5 5 3.5 3.5 V V (e) (f)
Principal of Virtual Forces If Newton s laws are the salt of the earth, energy principles are manna from heaven.. Introductory remarks If we apply external loads to a structure, these loads cause external displacements at their point of application. The product of external loads and corresponding displacements is the external energy or external workw e imposed on the structure. The applied loads also cause internal forces and corresponding deformations (strains). The product of internal forces and corresponding deformations, integrated over the entire structure is the strain energy stored in the structure, also called the internal energy or internal workw i. The energy principle or law of conservation of energy states that the external energy is equal to the internal energy stored in the structure. We assume that forces are applied slowly such that neither heat or kinetic energy is generated.. Virtual work-principle of virtual forces Virtual work is done by virtual (imagined) forces on actual (real) displacements or the work done by actual forces along virtual displacements. ccordingly, we distinguish between the principle of virtual forces and the principle of virtual displacements. In this class, we use the principle of virtual forces to calculate a specific displacement Δ (more precisely a single component of the displacement) or a rotationϕ at any point on the structure. When applying the principle of virtual force, we apply a virtual or dummy loadp at the point and in the direction of the desired displacement Δ or a dummy moment in the direction of a desired rotationϕ. With the virtual force acting, we apply the real loading to the structure. s the structure displaces under the real loading, the virtual force does external virtual W = P Δ (.) e work as it moves through the real displacement Δ of the structure. The dummy load also does internal virtual work W = () x κ()d x x N ()()d x ε x x V () x γ()d x x i (.) which is the work done by the internal virtual forces (virtual axial force N, virtual shear forcev and virtual bending moment along the strains (axial strain ε, shear strain γ and curvature κ ) caused by the real loading. Note that we denote virtual (dummy) quantities by the prime -symbol. In order for a system to be in equilibrium, the internal virtual workw i must be equal to the external virtual workw e. Hence (.3) P Δ = () x κ()d x x N ()()d x ε x x V () x γ()d x x It is important, that the domain of integration is the entire structure and x is a coordinate along a structural member. For convenience, if we let P =, we obtain (.4) Δ = ( x) κ( x)d x N ( x) ε( x)d x V ( x) γ( x)dx where, N andv are the internal virtual forces caused by a unit virtual force P =. Now we can use HOOKE s law to express the real strains in terms of the corresponding real internal forces Nx ( ) x ( ) Vx ( ) ε( x) = κ( x) = γ( x) = (.5) E EI G and obtain x () Nx () Vx () Δ = ( x) d x N ( x) d x k V ( x) dx EI E (.6) G κ ε γ The factork is a shape factor that accounts for the nonlinear distribution of shear stresses and shear strains (see RE, τ = VQ / It ) across the section. We will take a closer look at the shape factor in Section.5. 8
nalogously, we can use a unit virtual moment to calculate an unknown rotationϕ at any point on the structure. The virtual work equation becomes (analogous to Eq..3) ϕ = () x κ()d x x N ()()d x ε x x V () x γ()d x x (.7) where, N andv are the internal virtual forces caused by a virtual moment. We use the index to distinguish between the concentrated virtual moment that acts as a load on the left-hand side of the preceding equation and the internal virtual moment on the right-hand side. gain, we let = and obtain (.8) ϕ = ( x) κ( x)d x N ( x) ε( x)d x V ( x) γ( x)dx where, N andv are the internal virtual forces caused by a unit virtual moment =. We summarize: We use a virtual force to determine unknown displacements and a virtual moment to determine unknown rotations..3 Procedure for analysis () Determine the internal force diagrams of the structure for the actual loading () Place a unit load on the structure at the point and in the direction of the desired displacement (if we want to determine a rotation, we place a unit moment at that point, in direction of the desired rotation). Determine the internal force diagrams of the structure for the unit loading. (3) Evaluate the strain energy integrals (internal virtual work), which according to Eqs. (.6) and (.8) gives us the desired displacement or the desired rotation directly..4 Principle of virtual forces for trusses Truss members carry axial force only such that Eq. (.6) reduces to Nx ( ) Δ = N ( x) dx (.9) E Furthermore since the axial force is constant in each member (except for the effect of self weight of the member which is typically small), we can drop the dependence on x and the above integral reduces to a summation N Δ = (.0) n i N i i = E i i i wheren is the number of members in the truss..5 Principle of virtual forces for beams eams carry shear forces and bending moments such that Eq. (.6) reduces to x ( ) Vx ( ) Δ = ( x) d x k V ( x) dx EI (.) G We will show in Section.5 of this chapter that the shear deformations are small for typical beams and frames whose section depth is small compared to the length. onsequently, in this class, we will always neglect shear deformation in beams and frames unless otherwise noted. With this approximation the virtual force expression for beams is simply x ( ) Δ = ( x) dx (.) EI.6 Principle of virtual forces for frames Ignoring shear deformations, the principle of virtual forces for frames (structure that carry both bending moments and axial forces) gives us x ( ) Nx ( ) Δ = ( x) d x N ( x) dx EI E (.3) 9
.7 Example. 0 kn 0 kn D E 0 kn 3 m 4 m 4 m Fig.. Example.. Problem: Find the vertical displacement at and the horizontal displacement at E of the truss above for the given loading. The axial stiffness is E= 0, 000 kn. Solution:. ember forces for real loading. We first find the member forces resulting for the given loading. 3.3 8.0 8.0 6.0 6.0 T T N 0.0 T T 0.0 [kn] Fig.. ember forces for real loading.. ember forces for virtual loading. Vertical displacement at To find the vertical displacement at, we apply in the vertical direction a unit virtual force at and calculate the resulting member forces resulting from this load. 0.60 0.667 0.60 D P = E 3 m 0.60 0.60 T T N 4 m 4 m 0.333 T T 0.333 Fig..3 ember forces for virtual loading (vertical displacement at ). 0
Horizontal displacement at E To find the horizontal displacement at E, we apply in the horizontal direction a unit virtual force at E and calculate the resulting member forces resulting from this load. P = D E 0.500 T 3 m 0.45 0.45 0.45 T T 0.45 4 m 4 m T 0.750 T 0.50 N Fig..4 ember forces for virtual loading (horizontal displacement at E ). 3. Summation Since the truss structure has seven members, we have to include seven terms in the virtual work summation (see Eqs..9 and.0) 7 N P Δ= N dx = N N (.4) i i i E E i = i i The calculations are best arranged in tabular form: # ember N [kn] N ( Δ ) N ( Δ ) E [m] N N ( Δ ) N N ( Δ ) 0.0 0.333 0.750 4.000 3.33 30.0 0.0 0.333 0.50 4.000 3.33 0.0 3 DE -3.3-0.667 0.500 4.000 35.47-6.6 4 D -8.0-0.60 0.45 3.606 39.0-9.3 5 D 6.0 0.60-0.45 3.606 3.0-9.77 6 E 6.0 0.60 0.45 3.606 3.0 9.77 7 E -8.0-0.60-0.45 3.606 39.0 9.3 E 66. 3.4 The vertical displacement at and the horizontal displacement at E are thus 66. knm Δ = 0.07 m=7 mm N N = = E 0, 000 kn 3.4 knm Δ = N N 0.003 m=.3 mm E = = E 0, 000 kn (.5) Note: Since the cross-sectional area and the modulus of elasticity E are constant throughout the structure, we can move the product of the two quantities (the axial stiffness E ) out of the integral. If either E or varies, we must include them in the summation. We can select the direction of the virtual force arbitrarily. The sign of the result automatically gives the correct direction of the displacement. Positive results obtained in this example for both displacements indicate that the displacements are in direction of the applied virtual force. When the real and virtual member forces have different signs, i.e. real member force is compression and virtual member force is tension or vice versa, the corresponding term in the summation is negative.
D E Fig..5 Deflected shape of truss. Figure.5 shows the deflected shape of the truss structure. Note that the horizontal displacement of joint E is much smaller than the vertical displacement at joint which is consistent with the analysis above..8 Example. w EI / / Δ =? Fig..6 Example.. Problem: Find the midspan displacement for a simply supported beam of span and flexural stiffness EI under a uniformly distributed loadw. Solution: To find a specific displacement of a beam we use Eq. (.).. ending moment diagram for real loading. We find the real moment diagram by simple statics. w w 8 Fig..7 oment diagram for real loading.. ending moment diagram for virtual loading. Since we want to find the vertical displacement at midspan, we apply at midspan a unit load in the vertical direction. We find the virtual moment diagram by simple statics. P = 4 Fig..8 oment diagram for virtual loading.
3. Integration. easuring a coordinate x from the left support, we can write for the two bending moment diagrams w wx /4 x ( ) = x 0 x ( x) = x= x 0 x (.6) / Integrating the product of the two moment functions in the preceding equation according to Eq. (.) gives (because of symmetry, we only need to consider half the beam) / w wx Δ = () x ()d x x x x dx EI = EI 0 0 / 3 4 4 4 w x x w 5w = dx EI = = 4 4 EI 96 56 384EI 0 (.7) This is the same result derived in RE 3 either by double integration or the moment area method. We observe, that in order to find displacements or rotations at specific points using the principle of virtual forces we need to evaluate the integral of products of simple functions (shapes), e.g. rectangles, triangles, trapezoids and parabolas. In the following section, we will learn how to evaluate these integrals without formally integrating..9 Integration Tables.9. Discussion We can write the integral that we need to evaluate when applying the principle of virtual work as n Δ = d * * x = α (summation over n segments of the beam) (.8) i i i i EI ( EI ) where i = * : typical value of virtual moment i * : typical value of virtual moment α : tabulated integration factor depending on the shape of and : ength of beam segment EI : Flexural stiffness of beam segment Hence we can always evaluate the virtual work integrals by simple table look-up..9. Example s an example, we derive the α -factor for two linear functions (triangles) where the triangles have opposite orientation. Recall that in the context of the principal of virtual forces one factor in the integral of Eq. (.9) is the real internal force, the other represents the virtual internal force. learly, from a mathematical viewpoint the above concept can be used whenever we have to integrate the product of two functions f ( x) and f () x over a given interval (Fig..9). Problem: Find the factor α such that f () x f ()d x x = α F F (.9) 0 3
F f ( x) x f ( x) F Fig..9 Integrating the product of two linear functions. Writing the two triangular variations as functions of x gives () F f x = ( x) (.0) and ( ) F f x Integrating the product of the two functions yields = x (.) 3 F F F F F F x x f () x f ()d x x = x x dx = x x dx = 3 0 = F F 6 ( ) ( ) 0 0 0 (.) Hence, the α -factor for two triangles with different orientation is /6. Factors for the most common internal force distributions are listed on the next page. It is often necessary to split up the area under a moment diagram into several parts before applying the integration tables, e.g, subdividing a trapezoidal moment shape into two triangles. Note that it is unnecessary to find the two distances and when subdividing an internal force diagram that varies from a positive to a negative value (see Fig..0). a a b = b x x x a a = b b Fig..0 Subdividing trapezoid into two triangles. 4
b a a b Fig.. Integrating the product of trapezoidal functions. The integral of the product of the two moment functions is ( x) ( x)dx ab aa ba bb a ( a b) b( b a) = = 6 3 6 3 6 (.3) If and are identical trapezoids( a = a, b = b ), a case often encountered when analyzing a structure by the force method (hapter 3), the above formula reduces to ( x) ( x)dx = ( a b ab ) (.4) 3 Throughout this class, we will be using Eqs. (.3) and (.4) over and over again..0 Summary efore discussing more examples, we summarize the principal of virtual forces to calculate displacements at specific locations in a structure. P = P = P = bsolute displacement Relative displacement = = Relative rotation (hinge rotation) bsolute rotation Fig.. Principle of virtual force (basic cases). 5
Table.: -Tables ( α -values) i k i i i k k 3 4 k 6 4 k 4 3 S k 3 3 5 k S 3 5 7 48 k S 3 4 7 48 0 () x ()d x x = α i k i k Note : urved functions are quadratic parabolas with horizontal slope @ S 6
. Example.3 EI P ϕ / / Fig..3 Example.3. Problem: Find the rotation EIϕ at the two supports for the simply-supported beam under a concentrated force at midspan. Solution:. ending moment diagram for real loading. P EI / / Fig..4 ending moment diagram for real loading. P 4. ending moment diagram for virtual loading. Since we want to find the rotation at support, we apply a virtual moment = at that location. The direction of the unit virtual moment is arbitrary. The sign of the answer will indicate the correct sense of the rotation. positive results means the rotation is in the direction of the applied dummy moment, a negative result indicates the sense of rotation is opposite to the applied moment. Since we expect the rotation to be counterclockwise, let s apply the moment in the counterclockwise direction. The results should then be positive. Simple statics gives = / / 0.5 Fig..5 ending moment diagram for virtual loading. 3. Integration. To find the rotation ϕ, we have to integrate the product of the two moment functions in Fig..3 and.4 (same procedure as in Example.). Using Table., we find that the α -factor to integrate the product of two triangles shaped like those above is ¼. Hence P P EIϕ = ( x) ( x)dx = = (.5) 4 4 6 The end rotation of a simply supported beam under a concentrated force at mid-span is thus P ϕ = (.6) 6EI 7
. Example.4 w = kn/m EI 0 m 4 m Fig..6 Example.4. Problem: Use the principle of virtual forces to find the vertical displacement Δ and the rotationϕ at. Solution:. ending moment diagram for real loading. 0 =.5 8 8 4 = 8 [knm] Fig..7 ending moment diagram for real loading.. ending moment diagram for virtual loading. In order to find the vertical displacement at, we apply a virtual force at in the vertical direction (direction is arbitrary). We obtain the corresponding moment diagram by simple statics. 4 0 m 4 m P = Fig..8 ending moment diagram for virtual loading P =. In order to find the rotation at, we apply a virtual moment at (direction is arbitrary). We obtain the corresponding moment diagram by simple statics. 0 m 4 m = Fig..9 ending moment diagram for virtual loading =. 3. Integration Displacement Δ We compare the real moment diagram in Fig..6 to the virtual moment diagram in Fig..7. Since we do not have a single integration factor α for the entire beam, we have to integrate segments and separately. We consider the real moment diagram along segment as a superposition of a negative triangle of magnitude 8 and a positive parabola of magnitude.5. We consider the real moment diagram along segment as a superposition of a negative triangle of magnitude 8 and a positive parabola of magnitude. ll integration factors are α = /3(see Table.). 8
EI Δ = ( x) ( x) dx 4 8 0 4.5 0 4 8 4 4 4 = 3 triangle x triangle triangle x parabola triangle x triangle triangle x parabola due to P = = 30 500 3 ( 8 3) = 8kNm 3 The negative answer indicates that the vertical displacement of point is opposite to the direction of the applied virtual force. Point thus moves up. Rotationϕ We compare the real moment diagram in Fig..6 to the virtual moment diagram in Fig..8. gain, we consider the real moment diagram along segment a superposition of a negative triangle of magnitude 8 and a positive parabola of magnitude.5. We consider the real moment diagram along segment a superposition of a negative triangle of magnitude 8 and a positive parabola of magnitude. (.7) EIϕ = ( x) ( x) dx due to = = ( 8 0.5 0) 8 4 4 4.33 knm 3 = 3 triangle x tringle rectangle x tringle rectangle x parabola ϕ (.8) Δ Fig..0 Deflected shape with calculated displacement Δ and rotationϕ.. 9
.3 Example.5 F = 0 kn EI = 5000 knm 4 m m ϕ ϕ = ϕ ϕ ϕ Δ 5 m 5 m Fig.. Example.5: Sample structure and deflected shape (displacements vastly exaggerated). Problem: Find the vertical displacement Δ and the hinge rotation (angle change)ϕ at. Neglect axial deformation in all members. Solution: The structure is a frame structure. i.e. the member are subject to bending moment and axial force. To find a specific displacement of a frame by the principle of virtual forces we use Eq. (.3). Since we ignore axial deformation, we do not have to include the virtual work done by the axial forces such that the integral is that of Eq. (.).. ending moment diagram for real loading. fter finding the reaction forces, drawing the real moment diagram is straightforward. Note that in order to find the horizontal reaction forces, we have to dismember the structure and take moments about hinge. 0 6.667 6.667 4.67 4.67 5 5 Fig.. Reaction forces and moment diagram for real loading.. ending moment diagram(s) for virtual loading. To find the vertical displacement at, we apply a unit vertical force at (see Fig..). To find the rotation hinge, we apply a unit pair of moments at (see Fig..3). Note that the direction of both the applied force and moment is arbitrary. 30
P =.667.667 0.47 0.47 0.5 0.5 Fig..3 Reaction forces and moment diagram for virtual loading P =. =.000 0.667 0.667 0.667 0.667 0 0 Fig..4 Reaction forces and moment diagram for virtual loading =. 3. Integration. Using the integration factors for triangles with equal and opposite orientation, we get EI Δ = dx = 6.667.667 4 6.667.667 5 = 73.80 Δ 3 3 73.8 Δ = = 0.035 m 5000 EIϕ = ϕ dx = 6.667 0.667 4 6.667 ( 0.667.0) 5 =99.44 3 6 99.44 ϕ = =0.00 rad=.4 5000 Remarks Watch for the proper sign in the integral expression. When the real and virtual moment diagrams have opposite signs, the corresponding contribution in the integral (summation) is negative. Since we neglect axial and shear force deformation, it is not necessary to determine the real and virtual axial force and shear force diagrams. In order to avoid mistakes, we should select a certain order in the summation and stick with it in all our work, e.g. α -factor x real moment x virtual moment x member length. The positive answer for Δ indicates that the displacement is in direction of the applied virtual force (down). The negative answer forϕ indicates that the change in angle is in the direction opposite to the applied virtual moment (compare the deflected shape in Fig..0 to the applied virtual moment in Fig..3). (.9) 3
.4 Example.6 I 50 kn/m I D 00 kn I EI E = 0, 000 knm = 50, 000 kn 4 m E F m 6 m m Fig..5 Example.3. Problem: Determine the vertical displacement at locations,,, Dof the frame structure for the given loading. Neglect axial force deformation except in member F. Solution:. ending moment diagram for actual loading. The structure is statically determinate such that we obtain the bending moments by statics. Since we consider axial deformation only in memberf, we do not have to draw a complete axial force diagram but use the bending moment diagram to show the axial force N in memberf. 00.0 5.0 [knm] N = 83.3 kn F 33.3 Fig..6 ending moment diagram for real loading.. ending moment diagram for virtual loading. We use the principal of virtual forces to calculate the deflection at locations,,, D. Hence we need to analyze four virtual load cases (unit force at,,, D) resulting in the four virtual moment diagrams shown in the figure below. ike the real moment diagram, they are obtained by elementary statics. gain, we do not need the complete axial force diagrams (only the virtual axial force in memberf ), since we ignore axial deformation except in memberf. 3
.000 P = N = 0.333 m 6 m m 4 m.667 for Δ P = N = 0 4 m for Δ m 6 m m.000 P = N = 4 m for Δ m 6 m m.000 P = N =.333 m 6 m m 4 m 0.667 for ΔD Fig..7 Virtual bending moment diagrams. 3. Perform integration For each of the four virtual bending moment diagrams we have to calculate the strain energy integral Δ = x ( ) ( x)dx N( xn ) ( x)dx EI E The results are (.30) 33
Δ = ( 00 5) 6.00 33.33.667 4.47 0000 6 3 0.333 83.33 4 50000 =. mm Δ = 33.33 4.47 0000 3 = 34.8 mm Δ = 83.33 4 50000 =.7 mm (.3) Δ = ( 00 5) 6.00 00. 00 33.33 0.667 4.47 D 0000 3 3 3.333 83.33 4 = 35. 3mm 50000 4. Summary Fig..8 Sketch of deflected shape. 34
.5 Shear deformation.5. Shape factork If we include flexural, axial and shear virtual work, the expression for a specific displacement in a frame structure is N V P Δ= dx N dx k V dx EI E G where NVare,, the internal bending moment, axial force and shear force, respectively, and the prime -symbol denotes the corresponding virtual forces. Selecting P = gives (.3) Δ = Δ Δ Δ = dx N N dx k V V dx flexural axial shear EI E G The shape factork accounts for the nonlinear shear stress and shear strain distriubution across the section. In what follows, we derive the shape factor for a rectangular section. (.33) Derivation ofk for rectangular cross-section The internal virtual work done by virtual shear stresses τ along real shear strains γ is ( τγd ) W dx i = (.34) Since we don t want to evaluate the volume integral in the previous equation in our displacement calculations, we calculate the shape factor based on the requirement τ V τ d dx = k V dx G (.35) G or τ V τ d = kv (.36) G G alculatingk based on the preceding equation allows us to perform the integration along the member, as if the shear stresses were uniformly distributed across the section. From the shear formula (see RE, 3) V Q τ = I t we calculate the maximum shear stress, that occurs at the centroid of the cross section, as (.37) τ max =.5 V (.38) d τ V τmax =.5 y τ() y = τ max 4 d b Fig..9 Shear stress distribution across rectangular section. Knowing that the shear stress is parabolically distributed across the section (see figure above), we can write for the shear stress variation across the section 35
τ() y = τ 4 =.5 4 d d max y V y d d y Using HOOKE s law gives τ() () y.5 V 4 y γ y = = d d y G G d (.39) (.40) Replacing V byv in Eq. (.39) we obtain for the virtual shear stresses V τ () y =.5 4 d y (.4) Substituting the expressions for the real shear strains and the virtual shear stress in Eq. (.40) and (.4), respectively into the integral in Eq. (.36) yields d / d / 4 V V 9 y V V 9 y y τ () y γ()d y = 4 d = b 8 6 dy G 4 G 4 4 d d d d / d / V V 9 8 6 V V V V = b d = = k 4 5 5 G G G (.4) The shape factor for a rectangular section is thusk = 6/5..5. Significance of shear deformation In order to shed some light on the significance of shear deformation we consider the cantilever structure in Fig..9. The structure has uniform rectangular cross sectiond xb and is subject to a force P at the top. Interest is in the relative contribution of shear and flexural displacements to the total displacement for varying aspect ratio / d. We obtain the total displacement at the top by combining flexural and shear deformations 3 V P P (.43) Δ = Δ Δ = dx k V dx =. total flex shear EI G 3EI G P V P P d Fig..30 antilever structure subject to shear and bending. Using the relation between the modulus of elasticity and the shear modulus for concrete E E E G = = = ( υ) ( 0.5).5 concrete/masonry (.44) we obtain P 3 3.5 Δ total = Δ flex Δ shear =. P = P 3 P (.45) 3EI E 3 EI E 36
With I I bd bd d 3 = = = (.46) the ratio of flexural deformation to shear deformation becomes Δ Δ flex shear = = = 3 3 9 3.333 I d d (.47) The relative contribution of shear and flexural deformation to the total deformation is then.333 4 Δ flex d d Δ 3 = = = = Δtotal.333 3 4.333 3 4 d d d d shear Δtotal (.48) Δ Δ flex total Δ, Δ shear total Δ Δ flex total 0.8 d 0.6 0.4 d d 0. Δ Δ shear total 0 0 3 4 5 d Fig..3 Relative contribution of shear and flexural deformations to total deformation. The figure above plots the relative contribution of shear and flexural displacement to the total displacement. For an aspect ratio of / d = for example, the shear and flexural displacements are about 43% and 57%, respectively. 4 d = d = d = d = 0.5 Fig..3 Increasing shear deformation with decreasing aspect ratio / d. 37
Problems. 0 k k/ft (a) 6 ft 6 ft ft (b) 0 k 0 k k/ft 4 ft 4 ft 4 ft 6 ft 6 ft (c) (d) 5kft 0 k ft 5kft (e) ft ft (f) 0 k ft k/ft 0 k ft 5kft k/ft 0 k ft (g) ft ft (h) Use the principle of virtual forces to calculate the midspan deflection EI Δ and the two end rotations EI ϕ and EI ϕ of the simple beams above for the given loading. Solution: a b c d e f g h 3 EI Δ [k-ft ] 360 70 63.3 35 45 45 90 35 EIϕ 90 7 60 40.5 0 30 7 [k-ft ] [k-ft ] EIϕ 90 7 60 3.5 0 0 0 Note: Positive values for Δ : displacement is down Positive values for ϕ : rotation is clockwise Positive values for ϕ : rotation is counter-clockwise 38
. 4kN/m 0 kn (a) EI =0, 000 knm 5kN 3 m 3 m 6 m Find the horizontal displacements at and for the frame shown. Neglect axial deformation in all members. Solution: Δ = 0.00 m = 0.8 " Δ = 0.08 m =.".3 0 kn EI = 0, 000 knm E= 5, 000 kn 5kN 3 m 3 m 6 m Find the horizontal displacements at and for the frame shown. Neglect axial deformation in the bending members, but consider axial deformation in the truss members. Solution: Δ = 0.04 m = 0.94 " Δ = 0.046 m =.8" 39
.4 00 k 00 k 0 k/ft 5 ft alculate the horizontal and vertical displacement at. The structure has uniform 3 ft x 3 ft square cross section. Use E= 576, 000 k/ft. Solution: Δ = 0.0336 ft=0.4 in Δ = 0.050 ft=0.3 in h.5 6 ft v 0 k 3 4 0 k 6 ft E=0, 000 k 8 ft For the simple truss shown, find the horizontal and vertical displacement of point. Solution: Δ = 0.034 ft = 0.4" Δ = 0.008 ft = 0.097 " H, V,.6 50 kn 0 kn/m 0 kn/m [m] D E F 3.00.00.00.00.00 3.00 G Find EI ϕc, whereϕ c is the angle change at hinge due to the given loading. The beam has constant flexural stiffnes EI. Solution: 40
EIϕ c =.6 knm.7 kn/m F G D EI =5, 000 knm E 4 m 4 m 3 m 3 m 3 m 3 m Find the horizontal displacement and rotation of point E of the above frame structure ( EI = 5, 000 knm ). Neglect axial deformation. Solution: Δ = = E 3 0.007 m ϕ 0.889 0 rad(cw) E.8 w b G = = 0.65E d The simply supported beam above has a rectangular cross sectiond xb. For / d = 0,5,and, find the relative contribution of bending and shear to the total displacement. Solution: / d 0 5 Δ [%] 98.5 94. 7.7 39.5 F Δ [%].5 5.8 7.3 60.5 S.9 F EI F r Find the horizontal displacement at for the semi-circular arch. onsider the three forces separately and ignore axial deformation. Solution: 3 3 3 r πr πr due to F : Δ = F due to F : Δ = F due to F : Δ = F 3 3 EI 4EI EI F 3 4
.0 0 kn/m 0 kn 0 kn D EI = 0, 000 knm E = 0, 000 kn 4.00 alculate the vertical displacement at locations and,, D of the given frame structure. Neglect axial force deformation in the girder D. Solution: Δ = 6.3 mm Δ = 6. mm Δ = 6.4 mm Δ = 8.4 mm. [m].00.00.00 5.00.00.00.00 D 0.50 m 3 4 5 6 7 8 9 0 3 4 5 6 7 8 0 @.00 m 9 0 3.00 m 0 9.0 T 9.0 T 8.7 T 8.7 T 33 T 33 T 6 T 6 T 78 T 78 T.5 0.0 34.9 T 0 50..50 40.4 0.0 35.0 T 50.3 5.55 33.4 0.0 5.0 T 3 55 7.50 4.5 0.0 8. T 57 83 8.57 8.4 0.0 3.8 T 87.0 T 86 0 alculate the vertical displacement of joint 8 of the truss structure above ( E= 00 N ). ll applied forces have magnitude 0 kn and the resulting member forces are provided by the figure above (forces in kn). The joints of the bottom chord lie on a quadratic parabola with zero slope at node. Solution: Δ =9.7 mm 8 4
. F F F F E F F The 3-D truss structure above (in RE circles known as the -node model) has square roof and wall panels of dimension, i.e. all angels are either 45 or 90 degrees. Find the horizontal and vertical displacements of points and (in terms of FEand, ). Hint: First analyze the roof plane as a plane truss with roller supports representing the wall diagonals and find the reaction forces at the roller supports..3 0 kn/m 4 m 4 m Δ ϕ Using the principle of virtual forces, find the: (a) vertical displacement at ; (b) hinge rotation at ; (c) rotation at. The flexural stiffness is EI = 0, 000 knm Solution: (a) Δ = 0.075 m (b) ϕ = 0.047 (c) ϕ = 0.03 ccw ϕ.4 (a) w w (b) Using the principle of virtual forces, find the vertical displacement of the tip of the cantilever for the two loading conditions. The beam has constant flexural stiffness EI. Solution: w w (a) Δ = (b) Δ = 0 EI 30 EI 4 4 43
.5 0. N/m 0.50 m.00 m E = 30, 000 N/m 8.00 m b = m Using the principle of virtual forces, find the vertical displacement of point of the cantilever beam above. The beam has uniform modulus of elasticity E and widthb. The bottom face of the beam is a quadratic parabola with zero slope at. Subdivide the beam into 0 segments and use the SIPSON integration rule to evaluate the virtual force integral numerically. Solution: Δ = 0.00 m.6 0 kn/m D 3.00 m 3.00 m 3.00 m Using the principle of virtual forces, find the midspan vertical displacement for spans andd above. The beams have uniform flexural stiffness EI. Solution: Δ = 36. 35.0 D EI Δ = EI.7 P, Δ E, rigid α tension rod stabilizes a rigid post as shown. (a) For given values Eα,,,, derive a parameterk such that P = k Δ. (b) Find the angle α that maximizesk. Solution: E = (a) k cos αsin α (b) α 35 44
.8 Two brackets support shelves of length as shown in the picture. () Where must the brackets be placed such that middle and end deflections are equal? () Where must the brackets be placed such that the stresses in the shelves at the location of the bracket and in the middle are equal? onsider the loading of the shelves uniform..9 k k k 4k k 4k k k k k k k The -node truss structure shown above is loaded by a uniformly distributed load acting on the roof. Using the concept of tributary area we have determined the corresponding k, k or 4 k point loads, acting on the truss joints. olumns are removed and replaced by trusses as shown on the next two pages. For the ten designs, use the principle of virtual forces to calculate the vertical displacement of point for the given loading. ll members of the truss have axial stiffness E. Note that for clarity the following structural elements are not included in the diagrams: (a) diagonals in the wall planes (b) diagonals in the roof plane (except for designs "Skewed" and "Ooo-a-a") where the roof diagonals also act as the chords of the trusses (c) the out-of-plane bracing of the truss joints below the roof plane. 45