Question

A rigid bar $AB$ of length 3 m is subjected to a uniformly distributed load of $100 \,\text{N/m}$. The bar is supported at $A$ (pin) and by rod $CD$ connected at $D$. The rod $CD$ has axial stiffness $40 \,\text{N/mm}$, and $C$ is pinned. Find the vertical deflection at point $D$ (in mm). 

Hint:

When dealing with inclined rods, always resolve the vertical component of the axial force for equilibrium, and project elongation to vertical deflection.

Answer 1

Step 1: Equivalent load on rigid bar.
UDL on $AB$: \[ w = 100 \,\text{N/m}, L = 3 \,\text{m} \] Total load: \[ W = wL = 100 \times 3 = 300 \,\text{N} \] Acts at midspan of bar, i.e., $1.5 \,\text{m}$ from $A$.

Step 2: Support conditions.
- At $A$: hinge support. - At $D$: rod $CD$ resists vertical displacement by axial force. Bar $AB$ is rigid, so deflection at $D$ must be consistent with rod elongation.

Step 3: Geometry of rod $CD$.
Coordinates: - $C(0,0)$, - $D(1,1)$ (since bar at height 1 m, at 1 m from $A$). Initial length of $CD$: \[ L_{CD} = \sqrt{1^2 + 1^2} = \sqrt{2} \,\text{m} = 1414 \,\text{mm} \]

Step 4: Stiffness of rod.
Axial stiffness given: \[ k = 40 \,\text{N/mm} \]

Step 5: Compatibility.
Let vertical deflection at $D = \delta$. This induces elongation in $CD$, so force in rod: \[ F = k \delta \]

Step 6: Equilibrium of bar.
Taking moment about $A$: \[ 300 \times 1.5 = F \times 1 \] \[ F = 450 \,\text{N} \]

Step 7: Deflection.
\[ \delta = \frac{F}{k} = \frac{450}{40} = 11.25 \,\text{mm} \] Wait — check geometry: Only vertical component resists load. Force in rod = $P$, vertical component = $P \frac{1}{\sqrt{2}}$. So equilibrium: \[ 300 \times 1.5 = \left(P \frac{1}{\sqrt{2}}\right) \times 1 \] \[ P = \frac{450\sqrt{2}}{1} = 636.4 \,\text{N} \] Rod extension: \[ \Delta L = \frac{P}{k} = \frac{636.4}{40} = 15.91 \,\text{mm} \] Vertical deflection of $D$: \[ \delta = \Delta L \cdot \sin 45^\circ = 15.91 \times \frac{1}{\sqrt{2}} \approx 11.25 \,\text{mm} \] Rounded to nearest integer: \[ \boxed{11 \,\text{mm}} \]