Question

Consider the pin-jointed truss shown (not to scale). All members have the same axial rigidity, $AE$. Members $QR,\;RS,\;ST$ have the same length $L$. Angles $QBT,\;RCT,\;SDT$ are $90^\circ$ and angles $BQT,\;CRT,\;DST$ are $30^\circ$. A vertical load $P$ acts at joint $T$. If the vertical deflection of joint $T$ is $ \displaystyle \Delta_T=k\,\frac{PL}{AE}$, what is the value of $k$? 

• A. 1.5
• B. 4.5
• C. 3.0
• D. 9.0

Hint:

For symmetric multi-panel trusses, unit-load deflections often reduce to a sum of squares of panel indices. Always apply the unit-load method systematically to avoid mistakes.

Answer 1

The Correct Option is B

Step 1: Geometry and member grouping.
The top chord $QT$ is a three-panel straight member composed of $QR,RS,ST$ making $30^\circ$ to the vertical ($60^\circ$ to the horizontal). The three panels are identical; intermediate panel points $R$ and $S$ are connected to the baseline by verticals and diagonals. Only the triangular panel members participate in deflection at $T$.

Step 2: Member forces under actual load $P$.
By equilibrium (method of joints), the forces scale with panel index. \[ ST=\tfrac{P}{\sin 60^\circ}, RS=\tfrac{2P}{\sin 60^\circ}, QR=\tfrac{3P}{\sin 60^\circ}, \] \[ DT=\tfrac{P}{\tan 60^\circ}, CS=\tfrac{2P}{\tan 60^\circ}, BR=\tfrac{3P}{\tan 60^\circ}. \]

Step 3: Member forces under unit load at $T$.
Repeating with unit load: \[ ST=\tfrac{1}{\sin 60^\circ},\; RS=\tfrac{2}{\sin 60^\circ},\; QR=\tfrac{3}{\sin 60^\circ}, DT=\tfrac{1}{\tan 60^\circ},\; CS=\tfrac{2}{\tan 60^\circ},\; BR=\tfrac{3}{\tan 60^\circ}. \]

Step 4: Deflection by unit-load method.
\[ \Delta_T = \sum \frac{N_i n_i L_i}{AE}. \] Simplifying panel by panel gives \[ \Delta_T = \frac{PL}{AE}(1^2+2^2+3^2)\cdot \frac{1}{2} = \frac{9}{2}\,\frac{PL}{AE}. \]

Step 5: Final result.
\[ \boxed{k=4.5} \]