How many unknown displacement coordinates does the truss have, and what are they?

Two: horizontal u and vertical v of joint A

Three: u, v of A and rotation of A

What contribution does each bar make to the structure stiffness matrix?

k[c^2 cs; cs s^2] with k = EA/L and c, s the direction cosines

4EI/L on the diagonal and 2EI/L off it

k[1 -1; -1 1] with no direction cosines

How are the displacements of A obtained once [K] is built?

Take the average of the three bar elongations

What are the displacements of A?

u = 1.570 mm right, v = 0.395 mm up

u = 0.395 mm right, v = 1.570 mm up

u = 1.570 mm left, v = 0.395 mm down

2074 Chaitra · 8 marks

Truss: stiffness method, displacement at node A

Short answer

Stiffness matrix assembly for three bars gives the displacements of joint A: horizontal 1.570 mm to the right and vertical 0.395 mm upward.

Analyze the truss shown in figure below by the stiffness matrix method and find the vertical and horizontal displacement at node A. Given $A = 800\ \text{mm}^2$ , $E = 200\ \text{kN/mm}^2$ . [Figure: three bars AB, AC and AD meeting at joint A, 1 m above the supports B, C and D on a horizontal line; horizontal spacing 1.0 m, 0.5 m and 0.5 m as marked; 200 kN horizontal load at A.]

Assumed geometry (the figure is only described): joint A is 1.0 m above the support line. Measured from the vertical through A, support B is 1.0 m to the left, C is 0.5 m to the right and D is 1.0 m to the right (the spacings 1.0, 0.5 and 0.5 m run from B to the foot of A, to C, to D). So A is at (1.0, 1.0), with B(0, 0), C(1.5, 0) and D(2.0, 0). The 200 kN load acts horizontally to the right at A. $E=200$ kN/mm² $=2\times10^8$ kN/m², $A=800$ mm² $=8\times10^{-4}$ m², so $EA=1.6\times10^5$ kN.

Degrees of freedom

Joint A is free to move: $u$ (horizontal) and $v$ (vertical). The supports are fixed. Coordinates: 1 = $u$ , 2 = $v$ .

Member data (direction cosines taken from A towards the support)

Bar	L (m)	c	s	k = EA/L (kN/m)	$kc^2$	$kcs$	$ks^2$
AB	1.4142	-0.7071	-0.7071	113137.1	56568.5	56568.5	56568.5
AC	1.1180	0.4472	-0.8944	143108.4	28621.7	-57243.3	114486.7
AD	1.4142	0.7071	-0.7071	113137.1	56568.5	-56568.5	56568.5

Structure stiffness matrix (sum over the bars of $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$ )

[K]=\begin{bmatrix}141758.8&-57243.3\\-57243.3&227623.8\end{bmatrix}\ \text{kN/m}

Displacements of A

\begin{bmatrix}141758.8&-57243.3\\-57243.3&227623.8\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}200\\0\end{Bmatrix} \;\Rightarrow\; u=1.5703\ \text{mm},\quad v=0.3949\ \text{mm}

Member forces

Elongation $\delta=-(uc+vs)$ ; force $F=k\delta$ (tension +):

Bar	δ (mm)	Force (kN)
AB	1.3896	157.22
AC	-0.3491	-49.95
AD	-0.8311	-94.03

Equilibrium check at A: $\sum F_x=200+\sum F c=0.000$ and $\sum F_y=\sum F s=0.000$ ✓.

Answer: horizontal displacement of A $u=1.570$ mm (to the right), vertical displacement $v=0.395$ mm (up).

✓ Answer checked by Claude Opus 5.5 (high effort)

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