Chapter 14 Truss Analysis Using the Stiffness Method Structural Mechanics 2 Dept of Arch Eng, Ajou Univ Outline • • • • • • • • • Fundamentals of the stiffness method Member stiffness matrix Displacement and force transformation matrix Member global stiffness matrix Truss stiffness matrix Application of the stiffness method for truss analysis Nodal coordinates Trusses having thermal changes & fabrication errors Space-truss analysis 14-1 Fundamentals of the Stiffness Method • Subdivides the structure into a series of discrete finite elements represented by members and joints • The force-displacement relations – – – – – Equilibrium condition at each node Stiffness matrix Unknown displacement vector Given loading Unknown external and internal forces 14-2 Member Stiffness Matrix + = 14-2 Member Stiffness Matrix • For +ve displacement, dN in local x’ and y’ coordinates, the forces developed at the ends of the members are AE q'N dN L AE q'F dN L 14-2 Member Stiffness Matrix • For +ve displacement, dF at the far end, keeping the near end pinned, results in member forces AE dF q' ' N L AE dF q' 'F L 14-2 Member Stiffness Matrix • By superposition, the resultant forces caused by both displacements are AE AE qN dN dF L L AE AE qF dF dN L L 14-2 Member Stiffness Matrix • These load-displacement equations may be written in matrix form as q N AE 1 1 d N qF L 1 1 d F q k 'd AE 1 1 k ' : stiffness matrix L 1 1 14-3 Displacement and Force Transformation Matrices • Transformation of member forces q and displacement d defined in local coordinates to global coordinates • Global coordinates convention: +ve x to the right and +ve y upward 14-3 Displacement & Force Transformation Matrices • direction cosines xF x N x cos x L xF x N ( xF x N ) 2 ( y F y N ) 2 yF yN y cos y L yF yN ( xF x N ) 2 ( y F y N ) 2 14-3 Displacement & Force Transformation Matrices • Displacement Transformation Matrix d N DN x cos x DN y cos y 14-3 Displacement & Force Transformation Matrices • Displacement Transformation Matrix d F DFx cos x DFy cos y 14-3 Displacement & Force Transformation Matrices • Displacement Transformation Matrix Let x cos x ; y cos y d N DN x x DN y y ; d F DFx x DFy y In matrix form, DN x D 0 0 d N x y Ny D d F 0 0 x y Fx DFy d T D ※ [T] transforms the 4 global displacement into 2 local x’ displacement 14-3 Displacement & Force Transformation Matrices • Force Transformation Matrix QN x q N cos x q N x QFx qF cos x q F x QN y q N cos y q N y QFy qF cos y q F y 14-3 Displacement & Force Transformation Matrices • Force Transformation Matrix – In matrix form QN x x Q Ny y Q Fx 0 QFy 0 Q T q 0 0 q N x qF y T ※ [T]T transforms the 2 local forces q into 4 global force components Q 14-4 Member Global Stiffness Matrix Since q k d and d T D, q k T D Also since Q T q, T Force - Displacement is obtained as Q T k T D k D T where k T k T T 14-4 Member Global Stiffness Matrix • Performing the matrix operation yields 14-5 Truss stiffness matrix • Stiffness matrix [K] for entire truss can be obtained by assembling all member stiffness matrices [k] in global coordinates • The 4 code numbers to identify the 2 global degrees of freedom at each end of a member • Appropriate for analysis by computer programming Example 14.1 Determine the structure stiffness matrix. Use constant AE. (F) (Constrained) Two unknown displacement (N) (Constrained) Stiffness matrix: Member 1 Direction cosines: x 30 1; 3 y 00 0 3 (F) Example 14.1 Determine the structure stiffness matrix. Use constant AE. (F) (Constrained) Two unknown displacement (N) (Constrained) Stiffness matrix: Member 2 Direction cosines: x 30 0.6; 5 y 40 0.8 5 (F) Example 14.1 Determine the structure stiffness matrix. Use constant AE. • Algebraically added to form structure stiffness matrix 14-6 Application of the Stiffness Method for Truss Analysis • The relationship between global force components Q acting on a truss and its global displacements D Q K D • The structure stiffness equation 14-6 Application of the Stiffness Method for Truss Analysis • Expanding the structure stiffness equation Qk K11 Du K12 Dk Qu K 21 Du K 22 Dk • Often Dk = 0 since the supports are not displaced, then Qk K11 Du • Solving unknown displacement Du K11 Qk 1 • Then unknown reactions at supports Qu K 21 Du 14-6 Application of the Stiffness Method for Truss Analysis • The member forces can be determined using q k T D • Expanding this expression • Since with qN = -qF for equilibrium Example 14.3 Determine the force in each member. AE is constant. 6 y ① [2] 2 ② 1 5 Since D3=D4=D5=D6=0 and Q1=0, Q2=-2kN, 0 3 0 4 4 Dk [1] ③ 0 5 x 3 0 6 Structure stiffness equation: 0 1 Qk 2 2 Example 14.3 - solution • Separating and solving unknown displacements 0 0.405 0.096 D1 0 AE 2 0.096 0.128 D2 0 4.505 19.003 D1 ; D2 AE AE • Substituting for unknown reactions Example 14.3 - solution • Then reactions at supports are Q3 1.5kN ; Q4 0kN ; Q5 1.5kN ; Q6 2.0kN • The force in each member For member 1, x 1, y 0, L 3m q1 1.5kN For member 2, x 0.6, y 0.8, L 5m q2 2.5kN 14-7 Nodal Coordinates • A truss supported by a roller placed on an incline • The condition of zero displacement at node 1 is defined only along the y” axis, while the displacement along the x” axis will have displacement components along both global coordinates axes x and y 14-7 Nodal Coordinates • Consider truss member 1 having a global coordinate system x, y at the near node and a nodal coordinate system x”, y” at the far node 14-7 Nodal Coordinates d N D N x cos x D N y cos y d F D F x " cos x " D F y " cos y " DN x d N x y 0 0 DN y d F 0 0 x y DFx " D Fy " 14-7 Nodal Coordinates QN x q N cos x QN y q N cos y QFx " q F cos x" QFy " qF cos y" QN x x QN y y QFx " 0 Q 0 Fy " 0 0 q N x " q F y" 14-7 Nodal Coordinates • Member stiffness matrix T k T k 'T Example 14.6 Determine the support reactions Example 14.6 Determine the support reactions Member 1 x 1, y 0, x" 0.707, y " 0.707 Example 14.6 Determine the support reactions Member 2 x 0, y 1, x" 0.707, y " 0.707 Example 14.6 - solution • Member 3 x 0.8, y 0.6 Example 14.6 - solution • Assembling the matrices to determine the structure stiffness matrix 157.5 127.3 352.5 D1 ; D2 ; D3 AE AE AE Q4 31.8kN ; Q5 7.5kN ; Q6 22.5kN 14-8 Trusses having thermal changes and fabrication errors • Superposition method – Calculate fixed end forces to prevent movement of the nodes by temperature or fabrication – Calculate the displacement due to equal but opposite forces placed on the truss at the nodes – Determine the actual forces in the members and the reactions by superposing the results 14-8 Trusses having thermal changes and fabrication errors • For statically indeterminate truss, increment in length of a member due to T is L = TL • A decrease in length due to a compressive force q0 is L' = q0L/AE • Equating the two gives q0 = AE T 14-8 Trusses having thermal changes and fabrication errors • This force will hold the nodes of the member fixed (q N ) 0 AET (q F ) 0 AET 14-8 Trusses having thermal changes and fabrication errors • If a truss member is made too long by an amount L before it is fitted into a truss, the force q0 needed to keep the member at its design length L is q0 = AEL /L AEL (q N ) 0 L AEL (q F ) 0 L 14-8 Trusses having thermal changes and fabrication errors • If the member is too short, then L becomes negative and these forces will reverse • In global coordinates, these forces are 14-8 Trusses having thermal changes and fabrication errors • The initial force-displacement relationship due to temperature changes and fabrication errors Q K D Q0 • Where Qo is the initial fixed-end forces caused by temperature changes and fabrication errors 14-8 Trusses having thermal changes and fabrication errors • Carrying out the multiplication on the right hand side, Qk K11 Du K12 Dk Qk 0 Qu K 21 Du K 22 Dk Qk 0 • Using the superposition procedure, the unknown displacements are determined from the first equation by subtracting K12Dk and (Qk)0 from both sides and then solving for Du 14-8 Trusses having thermal changes and fabrication errors • The member forces are determined by superposition q k T D q0 • The force at the far end of the member Example 14.7 Determine the force in member 1 and 2 if member 2 was made 0.01 m too short before it was fitted into place. Use AE = 8103kN. Since L = -0.01m, x 0.8 y 0.6 3m Forces due to shortage in global coordinates 4m Example 14.7 Determine the force in member 1 and 2 if member 2 was made 0.01 m too short before it was fitted into place. Use AE = 8103kN. Structure stiffness matrix: Example 14.7 Determine the force in member 1 and 2 if member 2 was made 0.01 m too short before it was fitted into place. Use AE = 8103kN. Partitioning the matrices to obtain the equation for the unknown displacement D1 0.003704m D2 0.002084m Example 14.7 Determine the force in member 1 and 2 if member 2 was made 0.01 m too short before it was fitted into place. Use AE = 8103kN. • Member 1 x 0, y 1, L 3m, AE 8 10 kN 3 q1 5.56kN • Member 2 x 0.8, y 0.6, L 5m, AE 8 10 kN 3 q2 9.26kN 14-9 Space-truss analysis • To account for the 3-D aspects of the problem, additional elements must be included in the transformation matrix [T] 14-9 Space-truss analysis • The direction cosines x cos x xF x N L xF x N ( xF x N ) 2 ( y F y N ) 2 ( z F z N ) 2 yF yN y cos y L yF yN ( xF x N ) 2 ( y F y N ) 2 ( z F z N ) 2 z cos z zF zN L zF zN ( xF x N ) 2 ( y F y N ) 2 ( z F z N ) 2 14-9 Space-truss analysis • The transformation matrix in 3D • Member stiffness matrix 14-9 Space-truss analysis • Member stiffness matrix in global coordinates