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: Recall the stiffness matrix of a single 2D truss member.
A truss member can only carry axial force, so its stiffness matrix in global (\(u,v\)) coordinates at one end, with direction cosines \(l=\cos\theta\), \(m=\sin\theta\) (\(\theta\) measured from the \(+x\) axis to the member), is
\[ k = \frac{AE}{L}\begin{bmatrix} l^2 & lm \\ lm & m^2 \end{bmatrix} \]
This comes from resolving the axial stiffness \(AE/L\) of the member into its \(u\) and \(v\) components.
Step 2: Find the direction cosines of member PQ.
Member PQ runs from support P up to joint Q at \(45^{\circ}\) to the horizontal base PR. Measuring the angle from P towards Q (up and to the right), \(l=\cos45^{\circ}=\frac{1}{\sqrt2}\) and \(m=\sin45^{\circ}=\frac{1}{\sqrt2}\).
So \(l^2=0.5\), \(m^2=0.5\), \(lm=0.5\), giving the contribution of PQ at joint Q as
\[ k_{PQ} = \frac{AE}{L}\begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix} \]
Step 3: Find the direction cosines of member QR.
Member QR runs from Q down to support R, so measured from Q towards R, the member points right and DOWN at \(45^{\circ}\) below horizontal, the mirror image of PQ.
Here \(l=\cos(-45^{\circ})=\frac{1}{\sqrt2}\) (still positive) but \(m=\sin(-45^{\circ})=-\frac{1}{\sqrt2}\) (negative).
So \(l^2=0.5\), \(m^2=0.5\), but now \(lm=\frac{1}{\sqrt2}\times\left(-\frac{1}{\sqrt2}\right)=-0.5\), giving
\[ k_{QR} = \frac{AE}{L}\begin{bmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{bmatrix} \]
Step 4: Assemble the total stiffness at Q.
P and R are fixed (hinge supports remove all displacement there), so the only free degrees of freedom are \(u\) and \(v\) at Q. The stiffness matrix for these DOFs is simply the sum of the two members' contributions:
\[ K = k_{PQ}+k_{QR} = \frac{AE}{L}\left(\begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix}+\begin{bmatrix} 0.5 & -0.5 \\ -0.5 & 0.5 \end{bmatrix}\right) = \frac{AE}{L}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Step 5: Interpret the result and rule out other options.
This matches option (C). Pushing Q horizontally or vertically by a unit amount each needs a force \(AE/L\), with NO coupling between the two directions, because the symmetric 45-45 layout makes the cross terms from the two members cancel exactly.
Option (A) would result if the sign of \(lm\) were wrongly kept the same for both members (ignoring that QR slopes the opposite way to PQ), doubling the coupling instead of cancelling it. Option (B) incorrectly halves the diagonal terms while keeping a nonzero cross term. Option (D) would arise from subtracting the two members' matrices instead of adding them.
