3D problem: Fx Fy Fz. Forces act parallel to the members (2 5 ) / 29 (2 5 ) / 29

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1 problem: x y z 0 t each joint a a a a 5a j i W k y z x x y z Equations:S x =S y =S z =0 at each joint () Unknowns: Total of : Member forces,,, () Reactions : x, y, z, x, y, z, x, y, z (9) y z x W orces act parallel to the members j k a a a i 5a a W Member Unit Vector pointing away from Joint. j k i W n n n Wi 0 Member Member Member n ai ak a i k ( 5 ) / 9 ( 5 ) / 9 n ( i j 5 k ) 5 n ( i j 5 k ) 5

2 Joint i j k W n n n Wi 0 (i 5 k) / 9 ( i j5 k) / 5 ( i j5 k) / 5 Wi 0 S / 9 / 5 / 5 W 0 x S / 5 / 5 0 y S 5 / 9 5 / 5 5 / 5 0 z ( 5 /0) W ( 9 /5) W Joints,, a a a a 5a j i W k y z y z y z x n i j k 0 x y z n i j k 0 x y z / 9 0 x x y 0 y 5 / 9 0 z z n i j k 0 x y z n i j k 0 x x y z / 5 0 x x / 5 0 y y 5 / 5 0 z z x n i j k 0 x y z n i j k 0 x y z / 5 0 x x / 5 0 y y 5 / 5 0 z z 4

3 oes this procedure always work? P Equations (8): S x = S y =0 at each of the 4 joints Unknowns (8): 4 Member forces:,,,, 4 Reaction forces: x, y, x, y / 0 ( x) x / 0 ( y) y 0 ( x) x 0 ( y) y P / 0 ( x) / 0 ( y) 0 ( x) 0 ( y) { x0, fcd 0, fad0, facp, fbcp, yp, yp, xp } structure like this is Statically eterminate 5 P oes this procedure always work? Equations (8): S x = S y =0 at each of the 4 joints Unknowns (7): Member forces:,, 4 Reaction forces: x, y, x, y 0 ( x) x 0 ( y) y x 0 ( x) 0 ( y) y 0 ( x) 0 ( y) P 0 ( x) 0 ( y) No solution that satisfies all the equations. This is a mechanism 6

4 oes this procedure always work? P Equations (8): S x = S y =0 at each of the 4 joints Unknowns (9): 5 Member forces:,,,,, 4 Reaction forces: x, y, x y No joint, members cross / 0 ( x) x / 0 ( y) y / 0 ( x) x / 0 ( y) y P / 0 ( x) / 0 ( y) / 0 ( x) / 0 ( y) structure like this is Statically Indeterminate 7 Structures for which internal forces can be calculated using equilibrium equations for joints are statically determinate W structure structure M = 5 members M = 9 members R = reaction force components R = 6 reaction force components J = 4 joints J = 5 joints No. unknowns = M+R = 8 No. unknowns = M + R = 5 No. equations = J=8 No. equations = J = 5 8 4

5 How can you tell if a structure is statically determinate, indeterminate or a mechanism? Maxwell s conditions are a guide (not perfect but easy). Procedure count no. members M; count no. support force components R; count no. joints J. or problem: M+R>J (indeterminate) M+R=J (determinate) M+R<J (mechanism) problem: M+R>J (indeterminate) M+R=J (determinate) M+R<J (mechanism) 9 Examples where Maxwell works M=4 R= J=4 M+R=7 J=8 M+R<J Mechanism M=5 R= J=4 M+R=8 J=8 M+R=J eterminate M=8 R= J=5 M+R= J=0 M+R>J Indeterminate M=5 R=4 J=4 M+R=9 J=8 M+R>J Indeterminate 0 5

6 n example where Maxwell fails M=4 R=4 J=4 M+R=8 J=8 M+R=J???????? R T 0 ( x) x R T 0 ( y) y R T 0 ( x) x R T 0 ( y) y T T 0 ( x) 0 ( y) T 0 ( x) T 0 ( y) () () () (4) M=6, R=6, J=9 M+R<J Mechanism M=7, R=, J=7 M+R<J Mechanism M=6, R=5, J=7 M+R=J eterminate M=6, R=6, J=4 M+R=J eterminate? 6

7 Why do we care if a structure is determinate? Mechanism will move if subjected to forces. Sometimes OK. ad news if your house is a mechanism, though M=4 R= J=4 M+R=7 J=8 M+R<J Mechanism Why do we care? n indeterminate structure:. Is hard to assemble. Is hard to analyze. Weighs more than it has to 4. May have forces in members even without loading 5. Small motions at joints induce big forces 6. Thermal expansion of some members induce big forces M=8 R= J=5 M+R= J=0 M+R>J Indeterminate 4 7

8 Why do we care W n determinate structure:. Won t fall down when it s loaded. Is easy to assemble. Is easy to analyze (yeah right) 4. Won t have forces in members even without loading 5. Small motions at joints won t induce forces 6. Thermal expansion of some members won t induce big forces UT it will become a mechanism if a member fails! 5 Stiffness ormulation of Truss analysis: forces and deflections W 4 W undamental Unknowns are Nodal (joint) isplacements 8

9 Elastic rods are linear springs E,, 0 K, 0 k E 0 d d T E E T o T 0 T k T Example P u y u x P E,, 0 E,, 0 45 o Method of Joints: x P y / cos 0 / Psin 0 P sin cos, P cos 9

10 eflection in orce in eflection in orce in : Matrix Equation: Stiffness Matrix u y E u uy ux E u u 0 y x 0 y E ux Pcos sin 0 u y P 0

11 Works for Statically indeterminate too. P u x u y P E,, 0 E,, 0 45 o In General. Express M member elongations i in terms of u (M eqns). Enforce the boundary constraints on the displacements (R eqns). Express M member forces ij in terms of i (M eqns) 4. Enforce static equilibrium at each node (J or J eqns) Total number of equations: M+R+J Unknowns Nodal displacements u: J Member elongations i: M Member forces ij : M Reaction forces R k: R Total number of unknowns: M+R+J

12 Member orces and elongations in terms of u: or a member connecting joints (j) and (k): (j) x j e (k) x k e ( x x ), n ( x x ) / jk k j jk k j jk Member Elongations (j) x j ength jk ength jk+ jk (k) x k (k) k k x +u efore isplacement fter isplacement (j) x u x u jk jk k k j j j j x +u k k j j k k j j ( x u x u ) ( x u x u ) k j k j k j k j x x ( x x ) ( u u ) u u ( u u ) ( x x ) / jk k j k j jk ( u u ) n jk k j jk

13 inally jk kj k j jk ( u u ) n x j u j n jk n jk x k u k u member rotation member elongation kj n ( x x ) /, x x jk k j jk jk k j xial orce in the member: (j) (j) (k) (k) jk jk jk jk E jk kj E k j jk ( ) jk u u n Vectorial force on joint j through the member jk: n jk jk jk

14 Static Equilibrium: The net force on every joint is zero. (j) (j) (k) (k) N k P j is the externally applied force (known) on joint j R k is the reaction force (unknown) on joint k E N jk j j P R 0 k j,,,... N jk jk k j jk j j n u u n P R 0 ( ) j,,,... N onstraints or each reaction force, there is a corresponding constraint: Using prescribed displacement information, the unknown reactions R j, (j=,, N) can be written in terms of the nodal displacements. (j) (j) u k 0 (k) (k) N jk E kj j k ( n u u kj k ) j n P N jk k E kj j kj R n u n j k R 0 4

15 The end results is a system of linear equations for the nodal displacements: [ K] u r The coefficient matrix [K] and right-hand side vector r are fully determined by: Truss geometry (x k ) Material properties (E) k pplied loads P k Prescribed isplacements nd If You Try Some Time u [ K] r You Might ind You Get What You Need: Element orces Element strains Reaction orces 5

16 EXMPE P P ll members have equal cross section and material E n Member Elongations () u () u n n () u n i j n i j n i ( () () u u ) n u u () () ( ) u u ( u () u u n u ) () () () () () ( u () u ) n u () () 6

17 Member orces () u () u E E u u () () E E u u u () () () E E () u () u P P Equilibrium R R R P 0, P 0 R / 0, R / 0 / 0, R / 0 E E u u () () E E u u u () () () E E () u 7

18 Equations for u () u () u u u () () () () () u u u P E P E u u u 0 () () () Matrix orm () 0 u P E () 0 u P () u 0 [ K] u r U := ( ) 4 ( P7 PP ) ( PP P ),, 56 8 ( PP ) 4 u U E 8

19 Member forces E E u u () () E E u u u () () () E E () u () 0 u E () u () 0 0 u := 4 P 4 P P P P,, P P What if the structure is a mechanism? You can follow the procedure and arrive at the system of equations: [ K] u r Since nonzero displacements can occur without inducing member forces, matrix K will be singular. The number of zero eigenvalues corresponds to the degree of indeterminacy: (number of missing members or reactions. Null vectors of K correspond to the motion allowed by the mechanism 9

20 EXMPE P P P P ll members have equal cross section and material E,, n i j n i n 4 i j P P P P Stiffness u P 0 0 u P K 0 u P 0 0 u P Eigenvalues: 5 0,,, 5 0

21 Null Vector u 4 u u u Example P P P P P M= R= J=4 M+R=6 J=8 M+R<J Mechanism

22 K has two zero eigenvalues; 0 0 0u P u P K u P u P u P Null vectors are: u u 0 u 0 0 u u u 0 u 0 u 0 u 0 u M= R= J=4 M+R=6 J=8 M+R<J Mechanism