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GATE 2026 CE (CE1) – Question 38

Structural Analysis · Stiffness matrix method · 2 marks · Multiple choice

A plane truss consists of two linearly elastic, homogeneous, identical members, namely PQ and QR. Both members have length ($L$), cross-sectional area ($A$), and modulus of elasticity ($E$). The members are inclined at 45° as shown in the figure. The truss has hinge supports at P and R. The translational degrees-of-freedom ($u$ and $v$) are shown at joint Q.

After application of the boundary conditions, the stiffness matrix of the truss becomes:

A truss with a hinge at P on the left and a hinge at R on the right, joined by two members PQ and QR that meet at the apex Q. Both members make 45° with the horizontal. The degrees of freedom at Q are $u$ (horizontal) and $v$ (vertical).
  1. $\frac{AE}{L}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$
  2. $\frac{AE}{L}\begin{bmatrix} 1 & 0.5 \\ 0.5 & 1 \end{bmatrix}$
  3. $\frac{AE}{L}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
  4. $\frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$

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Correct answer: (C) $\frac{AE}{L}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Explanation

A member at angle $\theta$ adds $\frac{AE}{L}\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$ to the stiffness at Q. With $c = s = \frac{1}{\sqrt{2}}$, each member gives $\frac{AE}{L}\begin{bmatrix} 0.5 & \pm 0.5 \\ \pm 0.5 & 0.5 \end{bmatrix}$. The off-diagonal terms have opposite signs for the two members, because they lean in opposite directions, so they cancel. The sum is $\frac{AE}{L}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.