The magnitude of reaction force at joint C of the hinge-beam shown in the figure is _______ kN (round off to 2 decimal places).

Correct Answer: 20

Step-by-Step Explanation:

Based on the provided beam diagram in the image, we have a hinge-beam with the following details:
- A fixed support at joint A (left end).
- A concentrated downward point load of 50 kN at a distance of 3 m from joint A.
- An internal hinge at joint B, located at a distance of 6 m from joint A (3 m to the right of the 50 kN load).
- A roller support at joint C (right end).
- A uniformly distributed load (UDL) of 10 kN/m acting downwards over the span BC, which has a length of 4 m.

An internal hinge cannot transfer any bending moment. Therefore, the bending moment at the internal hinge B must be zero. We can analyze the right part of the beam (span BC) independently by taking the summation of moments about the hinge B from the right side:

$\sum M_{B\text{(right)}}=0$

Let $R_C$ be the vertical reaction force acting upwards at the roller support C.

Taking moments about joint B:
- The upward reaction force $R_C$ at a distance of 4 m from B creates a counter-clockwise moment:
$\text{Counter-clockwise Moment}=R_C\times 4$

- The downward UDL of 10 kN/m acting over the 4 m span creates a clockwise moment. The equivalent point load of the UDL is:
$10\text{kN/m}\times 4\text{m}=40\text{kN}$
This equivalent load acts at the midpoint of span BC, which is at a distance of 2 m from joint B:
$\text{Clockwise Moment}=(10\times 4)\times 2$

Equating the clockwise and counter-clockwise moments:

$4R_C=10\times 4\times 2$

$4R_C=80$

$R_C=\frac{80}{4}=20\text{kN}$

Therefore, the magnitude of the reaction force at joint C is 20 kN.