9. TRUSS ANALYSIS... 1 9.1 PLANAR TRUSS... 1 9. SPACE TRUSS... 11 9.3 SUMMARY... 1 9.4 EXERCISES... 15 9. Trss analysis 9.1 Planar trss: The differential eqation for the eqilibrim of an elastic bar (above) showed that it has only axial forces and axial displacements de to applied forces and/or temperatre changes. A trss is a strctral system composed of straight two force members defined as having only two eqal and opposite end forces acting along the axis of the member and thereby casing only axial displacement of the member. These members are joined (in theory) only at their ends to span the space between concrrent spport points. Clearly, the two force members are the axial bars governed by the element matrices developed from the differential eqation of eqilibrim. A trss is jst mltiple axial bars joined together, bt they are not all aligned in the same direction and sally more than two bars are connected at a joint. The design of a trss sally reqires a series of analyses. Historically, the first is sally jst based on the assmed geometry (trss mesh), a constant assmed area for each member, and neglects the weight of the trss. Today, the FEA easily allows the inclsion of the member weights in the first analysis. Mltiple vertical and horizontal load cases are considered to determine the maximm tension force and maximm compression force in each member. Then, each member is assigned a new area (and weight) consistent with material stress limits in tension and compression, and each load case is re-rn. In the final iteration the compression members are checked for bckling and additional braces may be attached at the center of long compression members to redce their effective bckling length. An application script to carry ot atomated planar trss stdies is spplied with the name Planar_Trss.m. Modern trsses contine to have the centroidal axes of the trss members intersect at a single point. However, the members are often joined to plates that lie in the plane of the trss. Sch joints can actally transmit bending moments (from distribted weight or wind loads), so the final step in a design is re-check the trss by modeling it as a planar frame which incldes its bending moments of inertia. The qestion here is how can the bar matrices developed in the prior chapter be sed to analyze a trss strctre? The key to the process is to recall that the displacement of the bar is a vector qantity. Previosly, when two or more bars were joined together all of the displacement vectors at the joints were collinear. Consider a planar trss where all of the bars lie in the x-y plane. Some of the bars mst be inclined relative to the x-axis, as in Fig. 9.1-1. That means that the bar displacement vectors which previosly were all directed along the x-axis now have components in both the x- and y-directions. In other words, the bar matrices need to be revised to accont for being inclined in the x-y plane. 1
Figre 9.1-1 Direction cosines for an inclined trss member The position vector defining the length, L, between the two ends of the trss member now incldes horizontal and vertical components, x and y, that combine to form a right triangle that defines that total length: L = x + y and the direction cosines with respect to the x- and y-axes: cos θ x = x/l and cos θ y = y/l. Of corse, those length components depend on the differences in the x- and y-coordinates of the two end nodes. Let the axial displacement of the bar now be denoted as b. It is collinear with the length of the bar. Therefore, the displacement vector at either end of the member also now has two components, x and y, when viewed in the x-y plane containing the trss. From similar triangles, the two system (global) displacement components are x = b cos θ x = b x/l and y = b cos θ y = b y/l. Before extending the bar stiffness matrix to a trss stiffness matrix, consider what the above observations reqire as changes to a compter model of the strctre. Now, both the x- and y- coordinates of each node will be reqired. That means that the physical space dimension control integer has increased to two, n s =. Each trss node now has two generalized displacements, n g =, and the nmber of nknowns has basically dobled at both the element and system levels. There are still only two nodes on the element, n n =, bt the nmber of independent degrees of freedom on the element has increased to for, n i = n g n n = = 4. Likewise, if the trss system has a total of n m nodes then the total nmber of displacement eqations for the trss, n d = n g n m, has dobled. Also, vector sbscripts for gathering and scattering element data doble becase their nmber is directly proportional to n g. Frthermore, when specifying an essential bondary condition (known displacement) at a node there has to be a way to distingish between horizontal or vertical components. For each trss element the connecting nodes are gathered and sed to get the nodal x- coordinates in order to compte the element length. The connections are sed again with the control integers to generate the system degree of freedom nmbers for the element. That is done with the get_element_index.m script given in Fig. 7.-1. That script is always sed for each element being assembled. Becase n g = the system eqation nmber for degree of freedom j at system node nmber N now is row (j) = n g (N 1) + j, 1 j n g. For a planar trss the degree of freedom j = 1 corresponds to a horizontal displacement and j = corresponds to the vertical displacement. The trss eqilibrim analysis reqires solving for all of the horizontal and vertical displacements. Since the node displacement components, x and y, are now the primary nknowns they need to be related to the bar displacement, b, by sing the orientation (direction cosines) of the bar element. From geometry, at any node the local bar axial displacement is related to the global x- and y-components by: b = x cos θ x + y cos θ y. This identity can be written in matrix form { b } L = [cos θ x cos θ y ] { x } or L = t(θ) G. (9.1-1) y G
Since the direction cosines are actally calclated from the length components of the trss it is practical to re-write the direction cosines and the member length components and the node displacement transformation as cos θ x C x = x L, cos θ y C y = sin θ x = y L { b } L = [C x C y ] { x } = 1 y G L [ x y] { x } or L = t(θ) G y G In Ex. 8.3-1 the strctral stiffness matrix of the two-noded bar was shown to be S e = Ee A e L e [ 1 1 1 1 ] That stiffness was obtained from the element s scalar contribtion to the integral eqilibrim form I S = T L S e L = { T 1 E e A e }b L e [ 1 1 1 1 ] { 1 }b In order to convert that original bar relation to a trss member stiffness relation the node transformation above mst be applied to both of the end nodes of the bar: x { 1 1 = [ cos θ x cos θ y y1 }b cos θ x cos θ ] { y } (9.1-) x y e L = [T(θ)] e G = [ t(θ) 1 = ( 4)(4 1) t(θ) ] G e In other words, the for trss components of displacement are redced to the two bar end displacements by mltiplying by a rectanglar geometry transformation matrix. The scalar integral contribtion to eqilibrim mst be the same in any coordinate system. So it is the same in the global system: I S = L T S e L = G T S G e G (9.1-3) 1 1 = (1 )( )( 1) = (1 4)(4 4)(4 1) Sbstitting the bar transformation: I S = ([T(θ)] G e ) T S e ([T(θ)] G e ) = G T S G e G gives the global trss stiffness as I S = G T [[T(θ)] T S e [T(θ)]] G = G T S G e G S G e = [T(θ)] T S e [T(θ)] (9.1-4) 4 4 = (4 )( )( 4) 3
It is probably easier to evalate the matrix S e, since we already have it, and T(θ), and have the compter do the mltiplications, bt the trss stiffness can be written in detail in terms of the length components as: S G e = EA L C x C x C x C y C x C x C x C y C x C y C y C y C x C y C y C y C x C x C x C y C x C x C x C y [ C x C y C y C y C x C y C y C y ] = EA [ L 3 x x x y x y y y x x x y x y y y x x x y x y y y ] (9.1-5) x x x y x y y y As a spot check, degenerate the trss into a single horizontal bar. Then, y = and x = L and the trss stiffness becomes L L 1 1 S e E A G = L 3 [ E A L L ] = [ ] L 1 1 which shows that a horizontal bar member has no vertical stiffness. Enforcing the EBCs that the two vertical displacements are zero redces the effective horizontal trss member stiffness back to that of the bar S e E A = [ 1 1 L 1 1 ] Bars and trss members can be sbjected to thermal loads. For the bar it was shown in section 8.4 that the thermal load is c α e = E Aα T { 1 1 } (9.1-6) This bar load is also transformed to a trss member load by sing the scalar work term in the governing integral form: I c = L T c e = G T c G e. Sbstitting the rectanglar member orientation transformation matrix gives: which converts any thermal load in a trss member to c G e = [T(θ)] T c e (9.1-7) 4 1 = (4 )( 1) c α e = EAα T { C x C y C x C y } = EAα T L x y { } x y When a trss has all members receive the same temperatre change and is free to expand it jst changes shape and no net member forces develop. However, if the trss is not free to expand then a temperatre change in even one member can prodce addition axial forces in some or all of the other members. 4
To form a kinematically stable trss the trss elements and/or the spport connections mst form a strctre made p of trianglar regions. A region with only for trss elements combined into the shape of a qadrilateral will be kinematically nstable and wold collapse nder its own weight. Mentioning the weight, for trsses it is assmed that the weight of a trss element is split eqally at its two ends as vertical forces: c G γ A L γ = { 1 w L } = { 1 } (9.1-8) 1 1 where γ denote the specific weight of the material, and w = γa is the weight per nit length. Now that each node has two degrees of freedom the packed integer code that flags the essential bondary condition at each node can have the following packed vales: a free joint, 1 only is given, 1 only v is given, 11 both and v are given (a pin joint), etc. Usally the given vales are zero, bt any small deflection vale can be imposed as reqired by the application. If any nit vale in the flag is replaced with a then that means that component is copled to another degree of freedom with a type mlti-point-constraint (MPC) The application software for planar trss analysis, with limited graphical spport, is spplied in the application library as Planar_Trss.m. As another illstration of its application it was applied to a seven bar trss shown in Fig. 9.1-6. All of the members had the same properties. The steel trss was pinned at the lower left, spported by a horizontal roller at the lower right. The top point carried a vertical load, the middle left point a horizontal load, and the weights were omitted. These data are sperimposed on the node displacement plot on the left of that figre. The right portion of the figre shows the compted axial forces in each bar member, with negative being compression. To review the inpts to the general finite element library, the inpt files are smmarized in Fig. 9.1-7. Example 9.1-1 Given: The two bar steel trss in Fig. 9.1- spports a downward vertical sign load of P = kn at their jnction (node 3). The two member connections to the vertical wall are pinned against displacement. Both elements have the same cross-sectional area, A =.1 m. Neglect the member weights. Find the deflection of the load, and the system reactions. Check the reaction forces sing statics. The trss data are: Node 1 3 Element Connections x (m) y (m) L (m) C x C y X(m) 3 1, 3 3 3 1 Y(m) 4 1,3 3-4 5 3/5-4/5 DOF 1, 3,4 5,6 Soltion: First, note that the two wall connections for the third side of a stable triangle. Steel has an elastic modls of E = e9 N m. The vertical sign load is at node 3 in the negative vertical direction. It corresponds to system DOF nmber of *(3-1) + = 6. The nll external horizontal point load there is at DOF nmber is *(3-1) +1 = 5. So the system external point load and reaction force vector is: c T P = [R x1 R y1 R x R y P] The nmerical vales of the data can be sbstitted to obtain the nmerical soltion, or an analytic soltion is available from a symbolic script by letting the inclined member have a length L so that the length of the horizontal member is 3L 5. Those reslts are rather messy and not 5
very practical. As an edcational compromise, sbstitte the nmerical direction cosine vales and se the inclined as a reference length. The horizontal trss member stiffness and loads are S h = E A [ (3L 5) 1 1 1 1 ] = 15 E A [ 75 L 1 1 1 1 ], c h γ = w(3l 5) The inclined stiffness is 9 1 S i E A = 5 L [ 1 16 9 1 1 16 1 }, c h α = EAα T { 1 { 1 } 1 9 1 7 36 1 16 E A ] = 9 1 75 L [ 36 48 7 36 1 16 36 48 7 36 36 7 48 ] 36 36 48 Since the reactions are reqired the fll 6 6 eqilibrim system will be assembled even thogh only the partition at node 3 is needed to find the displacements: E A 75 L 7 36 36 48 7 36 [ 36 48 15 15 7 36 x 1 36 48 y1 15 x y 15 36 x 3 36 48] { y3 } R x1 3/5 R y1 R = x + wl 1 4/5 1 R y 3/5 + EAα T 1 + 3/5 { { 1 3/5} P} { 4/5 } Note that the rows and colmns associated with y are all zero. If there is no EBC applied to that degree of freedom the system stiffness will be singlar. In that case, the system wold be kinematically nstable since the horizontal bar cold rotate abot node 3. The pinned bondary condition prevents that rotation and renders the system non-singlar. Since no gravity (γ = ) or temperatre change ( T = ) was given the free displacements satisfy Mltiplying by the inverse of the stiffness gives E A 36 [15 75 L 36 48 ] { x3 y3 } = { P } { x3 y3 } = PL 8 EA { 36 15 } 6
a node deflection down and to the left. Mltiplying all of the displacements times the fixed partition of the stiffness gives the reactions as: R x1 3 R y1 P = R x 4 { 4 } 3 { R y} The reslts are sketched in Fig. 9.1-. By inspection, the sm of the horizontal forces is zero, as is the sm of the vertical forces, and the sm of the moments abot any of the three nodes. Figre 9.1- Two bar planar trss with a point load Figre 9.1-3 Two bar planar trss with self-weight Example 9.1- Given: For the trss given in Ex. 9.1- sbject to a point load determine the local axial displacements of each trss member and the axial force in each bar. Soltion: (9.1-) gives the two displacements of a bar in terms of the for displacements of the trss. For the inclined member (element 1) those displacements are { 1 = [ cos θ x }b cos θ y cos θ x x 1 y1 cos θ ] { y } = x y { 1 = PL }1 4 EA { 5 } 1 5 [3 4 3 4 ] PL { 8 EA 36 15 } 7
The mechanical strain in that element is ε = B e e = 1 [ 1 { L 4 EA 5 } = 5 and the force in 4 EA the element is F = EA ε = 5P 4 tension. Of corse, since it is the only member at the pin spport the pin reaction mst be providing the same force. The reaction magnitde at node 1 is R 1 = R x + R y = ( 3P 4) + P = 5P 4. Likewise, for the horizontal member (element ) the displacements are { 1 = [ 1 } 1 ] PL 8 EA { } = PL 36 8 EA { 36 } 15 1] PL Its strain is ε = 1 [ 1 1] PL { (3L 5) 8 EA 36 } = 3 P and the force is F = 3 P 4, compression. 4 EA By inspection that agrees with the reaction shown in Fig. 9.1-3. P Example 9.1-3 Given: The trss in Ex. 9.1- is sbjected only to a temperatre increase of T. Find the displacement of the free node and the reaction forces. Soltion: Starting with the system matrices only the thermal load vector has changed: c α T = EAα T [ 3 4 5 (3 + 5) ( 4 + )]/5 Retaining the free partition, the eqilibrim eqations are Ths, the free displacements are E A 36 [15 75 L 36 48 ] { x3 y3 } = EAα T 5 { x3 y3 } = L α T 5 The spport reaction portions of the system matrices are 7 EA 75 L [ 36 15 36 48 ] L α T 5 { 3 4 } = EAα T 5 8 { 3 4 } { 8 4 } 3 R x1 { 4 R y1 } = + 5 R x { R y} EAα T 5 3 { 4 } = 5 Therefore the external reactions are identically zero. Also the forces in each trss member are zero becase, for this connectivity set, each member is free to expand. Node 3 of the inclined member moves on an arc, centered at spport node 1, of radis (1 + α T )L and node 3 of the horizontal member moves on an arc, centered at spport node, of radis (1 + α T ) 3L 5. Where those two arcs intersect is where node 3 displaces to, withot casing any member forces. The free expansion leading to no member forces fond here wold not occr if any other restrained members were present in the trss. For example, if a third bar was added to the trss between node 3 and a new node 4 at the mid-point between nodes 1 and, then an increase in temperatre wold case non-zero reactions well as non-zero member forces. Sch thermal indced reaction forces are shown in Fig. 9.1-3, and they clearly satisfy Newton s eqations of eqilibrim by inspection.
Figre 9.1-3 Three- and two-bar (right) reactions for niform temperatre increase Example 9.1-4 Given: The trss given previosly in Fig. 9.1-4 has an inclined roller spport that reqires a mltipoint constraint (MPC). Omitting the weight and the thermal loads, nmerically determine the displacements, the reactions, and the member axial loads. Check the reactions with Newton s Laws. Soltion: The property data in Fig. 7.7- were edited to set the specific weight and temperatre change to zero. The data sets were sbmitted to the planar trss script Planar_Trss.m which is inclded in the separate Application Library. The compted system displacements and reactions are: Compted Soltion: System Reactions Axial Force Node, displacements per node Node, DOF, Reaction Vale Elem, Vale 1.. 1 1-36 1..9713. 1-48 -1e5 3.86178.114796 3 1-64 3 6e4 3 48 Checking, the sm of the x-forces are F x = 1, 36, 64, =, as reqired. The sm of the y-forces are F y = 48, + 48, =, as reqired. Note that the reactions at node 3 from the MPC give a vector resltant of 8, N perpendiclar to the inclined spport srface. Checking the moments abot node one (with CCW positive) gives M z = 1, N 4 m + 48, N 3 m + 64, N 4 m = The external load and reactions are shown in Fig. 9.1-4. Note that the ratio of the displacement components at node 3 is.757 which is a.1% error in the MPC eqation. 9
Figre 9.1-4 Loads and reactions in a trss with a MPC Figre 9.1-5 Trss member axial force vales for Fig. 9.1-4 1
Figre 9.1-6 Displacement vectors (left) and member axial forces in a planar trss Figre 9.1-7 Data files for the seven bar planar trss model 9. Space Trss: Unlike beam and frame members, for any trss member only the vale of its cross-sectional area matters. The orientation of the area abot the local member axis does not matter. That means that the extension from planar trss analysis to space trss analysis is relatively simple. The z-coordinates of the nodes are reqired to define their locations and their differences on a member define a third direction cosine reqired for an expanded transformation 11
matrix to relate the local bar axial displacements to the six displacement components of the space trss member. Therefore (9.1-) expands to be: x 1 y1 { 1 = [ cos θ x cos θ y cos θ z z 1 ] }b cos θ x cos θ y cos θ z (9.-1) x y { z } The local bar stiffness matrix expands in the same way as (9.1-4) and (9.1-7): S G e = [T(θ)] T S e [T(θ)] 6 6 = (6 )( )( 6) c G e = [T(θ)] T c e 6 1 = (6 )( 1) Of corse, point forces at any node can be applied in any or all of the three directions. Likewise, any or all of the three displacement components at a node can have prescribed vales. That also means that the packed bondary condition flag assigned to each node now contains three digits: the left-most flags the x-displacement; the center one flags the y-displacement, etc. A packed flag vale of means that the joint is completely free, a vale of 1 means only the x- displacement is prescribed. A vale of 111 means all displacements at the node are prescribed. Clearly, any mixtre of prescribed displacements can be flagged by other combinations of the three digits. An application script Space_Trss.m was obtained from the planar trss script Planar_Trss.m by adding less than 1% of new calclations to tilize the z-coordinates to form the member transformation matrix an to extend helpfl plots from -D to 3-D. EXAMPLES 9.3 Smmary n_b n b Nmber of bondary segments n_d n d Nmber of system nknowns = n_g n_m n_e n e Nmber of elements n_g n g Nmber of generalized DOF per node n_i n i Nmber of nknowns per element = n_g n_n n_m n m Nmber of mesh nodes n_n n n Nmber of nodes per element n_p n p Dimension of parametric space n_q n q Nmber of total qadratre points n_r n r Nmber of rows in the B e matrix (and material matrix) n_s n s Dimension of physical space b = bondary segment nmber =sbset 1 e = element nmber = nion of sets
Bondary, element, and system nknowns: δ b b δ e e δ Boolean extraction arrays: δ b β b δ, δ e β e δ Geometry: Γ b Ω e Bondary segment Ω = e Ω e Soltion domain Ω e Element domain Γ = b Γ b Domain bondary Interpolation: Local gradient: (x) = H(r) δ e = (x) T = δ et H(r) T (r) r = ( H(r) r) δ e Physical gradient: (x) x = ( (r) r)( r x) = ( (r) r) ( x r) 1 (x) x = [( G(r) r) x e ] 1 ( H(r) r) δ e B(r) δ e Local Line Element Stiffness Matrix: S e k = dh(r) T dh(r) L E(x) A(x) dx e A symmetry plane has a zero normal displacement. Thermal loads for linear qadratic bar: dx dx c α e = E e A e α e T e { 1 1 } Eqilibrim eqations for a linear bar element: E e A e L e [ 1 1 1 1 ] { 1 } = { P 1 } + 1 1 [ P 6Le 1 ] {q 1 q } + E e A e α e T e { 1 1 } Planar trss element: n g =, n n =, n i = 4, stiffness, thermal and gravity loads cos θ x C x = x L, cos θ y C y = sin θ x = y L, cos θ z C z = z L Transforming from local bar to global trss matrices S e G = [T(θ)] T S e [T(θ)], c e G = [T(θ)] T c e, [T(θ)] = [ t(θ) t(θ) ] Planar trss bar withot previos axial load: S e = Ee A e 1 1 [ Le 1 1 ], ce = E e A e α e T e { 1 1 }, t(θ) = [cos θ x (cos θ y = sin θ x )] { 1 = [ cos θ x }b cos θ y x 1 y1 cos θ x cos θ ] { y } x y 13
Global planar stiffness matrix, thermal and gravity load vectors: S e G = [T(θ)] T Ee A e 1 1 EeAe Le [ ] [T(θ)] = 1 1 (4 )( )( 4) = (4 4) L e [ C x C x C x C y C x C x C x C y C x C y C y C y C x C y C y C y, C x C x C x C y C x C x C x C y C x C y C y C y C x C y C y C y ] C x c e α = [T(θ)] T E e A e α e T e { 1 1 } = Ee A e α e T e C y, c C e γ = x { C y} (4 )( 1) = (4 1) Space trss bar withot previos axial load: γ A L { 1 } 1 S e = Ee A e 1 1 Le [ 1 1 ], ce = E e A e α e T e { 1 1 }, t(θ) = [cos θ x cos θ y cos θ z ] x 1 y1 { 1 = [ cos θ x cos θ y cos θ z z 1 ] }b cos θ x cos θ y cos θ z x y { z } Planar trss bar with previos tension load N; axial stiffness and initial (geometric) stiffness: 1 S e = Ee A e L e [ 1 1 ], S 1 e i = Ne L e [ 1 1 1 ] 1 t(θ) = [ C x C y C y C x ] 14
9.4 Exercises Index application library, 9 bar, bckling, 1 condction, 13 direction cosines, displacement, 1 displacement transformation, 3 displacement vector, 1 displacements, 7 eqilibrim,, 6 essential bondary condition, Exercises, 15 get_element_index.m, gradient, 13 gravity load, 13 inclined roller, 9 interpolation, 13 kinematically nstable, 6 L3_C, 13 L4_C, 13 member weight, 1 MPC, 5, 9 msh_bc_xyz.txt, 11 msh_ebc.txt, 11 msh_load_pt.txt, 11 msh_properties.txt, 11 msh_typ_nodes.txt, 11 Newton s Laws, 9 packed flag, 1 packed integer code, 5 pin spport, 5, 6, 8 planar trss, 1, 9, 13 Planar_Trss.m, 1, 9 point load, 5 reaction, 5 roller spport, 5, 9 seven bar trss, 5 space trss, 11 Space_Trss.m, 1 specific weight, 5 steel, 5 stiffness matrix,, 3 smmary, 1 temperatre, 4 thermal load, 4, 8, 13 transformation matrix, 4 trss element, 13 trss element stiffness, 3 trss member, trss member force, 1 two bar trss, 5 two force member, 1 nion, 1 15