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:

The correct stiffness matrix of the truss after the application of the boundary conditions is:

$$\left[K\right]=\frac{AE}{L}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

Step-by-Step Derivation and Analysis:

1. Identify the Boundary Conditions and Active Degrees of Freedom:
As shown in the truss diagram, joints P and R are fixed hinge supports. Since they are constrained in both the horizontal and vertical directions, their displacement degrees of freedom are zero:
$u_P=v_P=0$
and
$u_R=v_R=0$
Consequently, the only free coordinates (active degrees of freedom) in the entire truss system are the horizontal displacement $u$ and the vertical displacement $v$ at joint Q. Thus, the reduced stiffness matrix of the structure corresponds to the joint Q and has a size of 2 × 2.

2. General Element Stiffness Contribution:
For a plane truss member of length $L$, cross-sectional area $A$, and modulus of elasticity $E$, oriented at an angle $\theta$ relative to the positive horizontal axis, the stiffness matrix contribution to the displacements at one of its nodes is given by:

$${\left[k\right]}_{\text{node}}=\frac{AE}{L}\left[\begin{array}{cc}{\cos }^2\theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & {\sin }^2\theta \end{array}\right]$$

3. Compute Stiffness Contribution of Member PQ:
Member PQ has length $L$ and is inclined at an angle of $\theta =45°$ relative to the horizontal line connecting P and R.
Using:
$\cos 45°=\frac{1}{\sqrt{2}}$ and $\sin 45°=\frac{1}{\sqrt{2}}$
We get:
${\cos }^{2}45°=0.5$
${\sin }^{2}45°=0.5$
$\sin 45°\cos 45°=0.5$
Substituting these values, the stiffness contribution of PQ at joint Q is:

$${\left[k\right]}_{PQ}=\frac{AE}{L}\left[\begin{array}{cc}0.5& 0.5\\ 0.5& 0.5\end{array}\right]$$

4. Compute Stiffness Contribution of Member QR:
Member QR is inclined at an angle of $45°$ relative to the horizontal PR at support R. If we orient the member vector pointing from support R to joint Q, its direction angle is:
$\theta =180°-45°=135°$
Using:
$\cos 135°=-\frac{1}{\sqrt{2}}$ and $\sin 135°=\frac{1}{\sqrt{2}}$
We obtain:
${\cos }^{2}135°=0.5$
${\sin }^{2}135°=0.5$
$\sin 135°\cos 135°=-0.5$
Substituting these values, the stiffness contribution of QR at joint Q is:

$${\left[k\right]}_{QR}=\frac{AE}{L}\left[\begin{array}{cc}0.5& -0.5\\ -0.5& 0.5\end{array}\right]$$

5. Structural Stiffness Matrix Assembly:
The structural stiffness matrix $\left[K\right]$ for the active degrees of freedom $u$ and $v$ is assembled by summing the contributions from both member PQ and member QR:

$$\left[K\right]={\left[k\right]}_{PQ}+{\left[k\right]}_{QR}$$

$$\left[K\right]=\frac{AE}{L}\left[\begin{array}{cc}0.5+0.5& 0.5-0.5\\ 0.5-0.5& 0.5+0.5\end{array}\right]=\frac{AE}{L}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$