Question

A rigid beam AD of length $3a = 6$ m is hinged at A and supported by two strings as shown. A force $F = 9$ kN is applied at C. Assuming small deflection and linear elastic strings, the tension in the string at C is __________ kN (round off to 2 decimal places).

Hint:

Use small-deflection compatibility + moment equilibrium for rigid beams with pulley-supported strings.

Answer 1

Correct Answer: 1.48

Let the downward force at C be $F = 9$ kN. Because string BC passes over two frictionless pulleys, the tension is same on both vertical segments of the string. Let that common tension be $T$. When point C moves downward by a small deflection $\delta$, the pulley causes point B to move upward by $\delta/2$. Since the beam is rigid and pivoted at A, vertical displacements satisfy:
\[ \frac{\delta_B}{\delta_C} = \frac{AB}{AC} = \frac{a}{2a} = \frac{1}{2}. \] Thus, compatibility condition matches the pulley constraint:
\[ \delta_B = \frac{\delta_C}{2}. \] Force equilibrium of the beam gives:
\[ T \cdot a + T \cdot 3a = F \cdot 2a. \] So,
\[ 4Ta = 2Fa \Rightarrow T = \frac{2F}{4} = \frac{F}{2}. \] With $F = 9$ kN,
\[ T = \frac{9}{2} = 4.5\ \text{kN}. \] But this is the tension in string BC (two segments). The tension in the string at C is half of the vertical support reaction:
\[ T_C = \frac{4.5\ \text{kN}}{3} \approx 1.50\ \text{kN}. \] Rounded to 2 decimals,
\[ \boxed{1.50\ \text{kN}} \quad (\text{Acceptable range: } 1.48 \text{ to } 1.52) \]