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 \(45^{\circ}\) 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.
(Figure not to scale)
After application of the boundary conditions, the stiffness matrix of the truss becomes:
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Write each member's 2x2 stiffness matrix using its own direction cosines, then add the two members' contributions at joint Q.
Step 1: Apply a unit horizontal displacement at Q (a physical, unit-load style method).
Give joint Q a unit displacement $u=1$ horizontally only ($v=0$), and work out how much each member stretches. For a member at angle $\theta$ to the horizontal, its elongation equals the component of the joint displacement along its own axis: $\delta = u\cos\theta + v\sin\theta$.
Step 2: Elongation of each member under this unit horizontal push.
Member PQ is at $\theta=45^\circ$, so $\delta_{PQ} = (1)\cos45^\circ = \frac{1}{\sqrt2}$.
Member QR is at $\theta=-45^\circ$ (sloping down from Q to R), so $\delta_{QR} = (1)\cos(-45^\circ) = \frac{1}{\sqrt2}$.
Both members stretch by the SAME amount for a unit horizontal push, since both make the same 45 degree angle with the horizontal, just mirrored.
Step 3: Force in each member, and its components back at Q.
Axial force in each member is $\frac{AE}{L}\delta = \frac{AE}{L}\cdot\frac{1}{\sqrt2}$.
The horizontal component of this force from each member is $\frac{AE}{L}\cdot\frac{1}{\sqrt2}\cdot\cos\theta = \frac{AE}{L}\cdot0.5$ for both members, so adding both gives the horizontal stiffness $K_{uu} = \frac{AE}{L}(0.5+0.5) = \frac{AE}{L}$.
The vertical component from PQ is $\frac{AE}{L}\cdot\frac{1}{\sqrt2}\cdot\sin45^\circ = 0.5\frac{AE}{L}$ (pulling up, since PQ slopes up), while from QR it is $\frac{AE}{L}\cdot\frac{1}{\sqrt2}\cdot\sin(-45^\circ) = -0.5\frac{AE}{L}$ (pulling down, since QR slopes down). These cancel, so $K_{vu}=0$.
Step 4: Repeat for a unit vertical displacement, using symmetry.
By the same logic, or by the mirror symmetry of the truss about the vertical line through Q, pushing Q vertically by a unit amount gives a vertical stiffness $K_{vv}=\frac{AE}{L}$ and a horizontal coupling term $K_{uv}=0$, exactly mirroring Step 3.
Step 5: Assemble the stiffness matrix.
Collecting $K_{uu}=\frac{AE}{L}$, $K_{uv}=K_{vu}=0$, $K_{vv}=\frac{AE}{L}$:
\[ K = \frac{AE}{L}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} \]
This confirms option (C) using a direct unit-displacement physical argument instead of assembling the direction-cosine matrices. The zero off-diagonal terms come straight from the two members' vertical pulls cancelling under a horizontal push, and vice versa, a direct result of the mirror symmetry of the 45-45 geometry.
Final Answer:
\[ K = \frac{AE}{L}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} \]
