Lecture 5: 3-D Rotation Matrices.

Transcription

1 3.7 Transformation Matri and Stiffness Matri in Three- Dimensional Space. The displacement vector d is a real vector entit. It is independent of the frame used to define it. d = d i + d j + d k = dˆ iˆ+ dˆ ˆj+ dˆ kˆ ˆ ẑ o d We are interested in calculating what the global coordinate representation is based on elemental coordinates, and vice versa. ŷ

2 Solve for the global displacements one at a time: dˆ ˆ ˆ ( ) ( ˆ ) ˆ ( ˆ ) ˆ ( ˆ ) ( ˆ ) ( ˆ ) ˆ d i = d = i i d + j i d + k i d = i i j i k i d ˆ d dˆ d j d ( ˆ ) ˆ ( ˆ ) ˆ ( ˆ ) ˆ ( ˆ ) ( ˆ ) ( ˆ ) ˆ = = i j d + j j d + k j d = i j j j k j d dˆ dˆ d k = d ( ˆ ) ˆ = i k d + ( ˆj k) dˆ ( ˆ ) ˆ ( ˆ ) ( ˆ ) ( ˆ ) ˆ + k k d = i k j k k k d dˆ

3 Or, we could collect the three scalar equations into a matri form: ( ˆ ) ( ˆ ) ( ˆ ) ˆ d i i j i k i d d ( ˆ ) ( ˆ ) ( ˆ ) ˆ = i j j j k j d d ( ˆ ) ( ˆ ) ( ˆ ) ˆ i k j k k k d R 33 Looking at the columns of the rotation matri R it is apparent that it must be orthonormal.

4 The entries of the rotation matri are referred to as the direction cosines. ( iˆ i ) = iˆ i cosθ = cosθ ( ˆj i ) = ˆj i cosψ = cosψ ( ˆ ) ˆ k i = k i cosφ = cosφ ( iˆ j) = iˆ j cosθ = cosθ ( ˆj j) = ˆj j cosψ = cosψ ( ˆ ) ˆ k j = k j cosφ = cosφ ( iˆ k) = iˆ k cosθ = cosθ ( ˆj k) = ˆj k cosψ = cosψ ˆ ˆ k k = k k cosφ = cosφ ( ) Although it appears that there are nine values that make up the rotation matri onl three are independent. R r r r = r r = 1 r r = r r = 0 r r = r r = 0 r r = î ĵ ˆk

5 The course tetbook uses the variable θ to refer to the direction angles for the element ais. Note of the three direction cosines: cosθ, cos θ and cosθ, onl two are independent. To specif the orientation of the elemental frame, we would need one more direction cosine of either of the element or aes.

6 We need to go between element coordinates and global coordinates for an vector. ˆ cθ cθ cθ T ˆ = R = cψ cψ cψ ˆ cφ cφ cφ Appling the transformation above to the element node force and displacement vectors

7 f d1 f d 1 T T R 0 f1 AE R 0 d1 T = T 0 R f L R d f d f d T T f d ˆf ˆd

8 The revised element stiffness matri onl depends on the direction cosines of the ˆ ais. Tf = ktd ˆ T f = T kt ˆ d k Logan s notation. k C C C C AE C C = L C C SYM C C C C = = = cosθ cosθ cosθ

9 Logan notes how the 66 element stiffness matri for the spatial truss element can be broken down into a tiling of a 33 component. k AE λ λ = L λ λ C λ = C C

10 It onl remains to calculate the direction cosines. ˆ C = iˆ i = cosθ = C = iˆ j = cosθ = C = iˆ k = cosθ = 1 L 1 L 1 L Onl the node locations (in the nominal configuration) are required.

11 Sample problems: Eample 3.7 pg.# 89. P pg.# 13. Determine element lengths. Calculate the direction cosines of the ˆ ais for each element. E A = = 6 in psi

12 Element 1. (1) L = = in (1) (1) (1) C = = = 0.46 C (1) = = = C (1) = = = (1) L L L Element. () L = = 16.0 in () () 4 7 () C = = = C () = = = C () = = = () L 16.0 L 16.0 L 16.0 Element 3. (3) L = = 16.0 in (3) (3) (3) C = = = C (3) = = = C (3) = = = (3) L 16.0 L 16.0 L 16.0

13 Prior to assembl note the structure of the 3-D truss stiffness matri. k C C C C AE C C = L C C SYM C We need onl evaluate this 33 building block, λ. Eploit the homogeneous boundar conditions. d d d d d d d d d 1 = 1 = 1 = = = = 3 = 3 = 3 = 0.0 Combining steps 5 (assembl) and 6 (solution)

14 F1 d1 F 1 d 1 F 1 d1 F d F d F AE = d (1) F3 L d3 F 3 d3 F3 d3 F 4 d 4 F4 d4 F 4 d 4

15 The 33 block that must be assembled is the bottom right corner of each of the element stiffness matrices. For eample, consider the contribution of element 1. 0 (1) f C 1 C d1 (1) f C 1 C d 0 1 (1) f 1 AE C C 0 d 1 (1) = (1) f C 4 L d 4 (1) f 4 C d 4 (1) f SYM 4 C d4

16 Evaluating this 33 for all 3 elements: AE element AE element AE element

17 The reduced sstem becomes 0 F ( ) ( ) d4 0 AE F 4 0 ( ) 0 = d F ( ) ( ) d ( ) 4 = d 4 = d 4 0 d = = 4000 d d