- 2070 Magh · 10 marks
Truss element stiffness matrices
Short answer
Use the standard plane-truss stiffness matrix. Element 1 (1-3) has AE/L=98389.47 N/mm, c=0.7809, s=0.6247. Element 2 (3-4) is horizontal with AE/L=126000 N/mm, so only the u-direction stiffness entries are nonzero; both matrices are symmetric and rows sum to zero.
Determine the stiffness matrices for the elements as shown in Fig. 1. mm and MPa.
[Figure: plane truss; node 1 at the bottom-left (supported), node 2 at 500 mm to the right of node 1, node 3 at 400 mm above node 2, node 4 at 400 mm above node 1 (supported); elements numbered 1 and 2 (diagonal from node 1 to node 3 and member 3-4 as drawn); a 50 kN horizontal load at node 2 and a 40 kN downward load at node 3]
The stiffness of a 2D truss element is with , .
Geometry and assumptions
Coordinates in mm: node 1 (0, 0), node 2 (500, 0), node 3 (500, 400), node 4 (0, 400). Element 1 is the diagonal 1-3; element 2 is the member 3-4, as drawn. N. Global dofs are at each node .
Element 1 (nodes 1 to 3)
Rows and columns correspond to :
Element 2 (nodes 3 to 4)
The member is horizontal: mm, (from node 3 to node 4), , N/mm. Only and appear in the matrix, so the sign of does not matter. Rows and columns correspond to :
Checks
- Both matrices are symmetric and every row sums to zero.
- For element 1, .
- A horizontal bar has no stiffness in the direction, so the rows of are zero.
Answer: and as above, with N/mm and N/mm respectively.
✓ Answer checked by Claude Opus 5.5 (high effort)
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