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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. A=300A = 300 mm2^2 and E=2.1×105E = 2.1\times10^5 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 [k]=AEL[c2cs−c2−cscss2−cs−s2−c2−csc2cs−cs−s2css2][k]=\dfrac{AE}{L}\begin{bmatrix}c^2&cs&-c^2&-cs\\cs&s^2&-cs&-s^2\\-c^2&-cs&c^2&cs\\-cs&-s^2&cs&s^2\end{bmatrix} with c=xj−xiLc=\dfrac{x_j-x_i}{L}, s=yj−yiLs=\dfrac{y_j-y_i}{L}.

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. AE=300×2.1×105=6.3×107AE=300\times2.1\times10^5=6.3\times10^7 N. Global dofs are (un,vn)(u_n, v_n) at each node nn.

Element 1 (nodes 1 to 3)

L=5002+4002=640.31 mm,c=500L=0.7809,s=400L=0.6247,AEL=98389.47 N/mmL=\sqrt{500^2+400^2}=640.31\ \text{mm},\quad c=\frac{500}{L}=0.7809,\quad s=\frac{400}{L}=0.6247,\quad \frac{AE}{L}=98389.47\ \text{N/mm}

Rows and columns correspond to (u1, v1, u3, v3)(u_1,\ v_1,\ u_3,\ v_3):

[k]1=[59993.5847994.86−59993.58−47994.8647994.8638395.89−47994.86−38395.89−59993.58−47994.8659993.5847994.86−47994.86−38395.8947994.8638395.89] N/mm[k]_1=\begin{bmatrix} 59993.58 & 47994.86 & -59993.58 & -47994.86 \\ 47994.86 & 38395.89 & -47994.86 & -38395.89 \\ -59993.58 & -47994.86 & 59993.58 & 47994.86 \\ -47994.86 & -38395.89 & 47994.86 & 38395.89 \end{bmatrix}\ \text{N/mm}

Element 2 (nodes 3 to 4)

The member is horizontal: L=500L=500 mm, c=−1c=-1 (from node 3 to node 4), s=0s=0, AE/L=126000AE/L=126000 N/mm. Only c2=1c^2=1 and cs=0cs=0 appear in the matrix, so the sign of cc does not matter. Rows and columns correspond to (u3, v3, u4, v4)(u_3,\ v_3,\ u_4,\ v_4):

[k]2=[1260000−12600000000−126000012600000000] N/mm[k]_2=\begin{bmatrix} 126000 & 0 & -126000 & 0 \\ 0 & 0 & 0 & 0 \\ -126000 & 0 & 126000 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\ \text{N/mm}

Checks

  • Both matrices are symmetric and every row sums to zero.
  • For element 1, c2+s2=0.6098+0.3902=1c^2+s^2=0.6098+0.3902=1.
  • A horizontal bar has no stiffness in the vv direction, so the vv rows of [k]2[k]_2 are zero.

Answer: [k]1[k]_1 and [k]2[k]_2 as above, with AE/L=98389.47AE/L=98389.47 N/mm and 126000126000 N/mm respectively.

✓ Answer checked by Claude Opus 5.5 (high effort)

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