Two structural members are connected by an internal hinge and are loaded externally as shown in the figure. A and B are roller supports and C is a pin support. Neglecting the mass of the members, find the magnitude of the vertical reaction force at C, in kN.
Show Hint
Split the structure at the hinge first, since it carries no moment, then work out the reactions piece by piece.
Step 1: Split the structure at the hinge.
The hinge cannot carry a moment, so treat the beam as two separate pieces: a left piece from \( A \) to the hinge, and a right piece from the hinge to \( C \), joined only by a shared vertical force at the hinge.
The left piece carries the \( 10 \) kN load at \( 0.5 \) m from \( A \), with the hinge sitting \( 1.0 \) m from \( A \), so the load is \( 0.5 \) m from the hinge.
Step 2: Solve the left piece for the reaction at A.
Taking moments about the hinge for the left piece alone (the hinge carries no moment, so this check is valid): \( R_A \times 1.0 - 10 \times 0.5 = 0 \).
This gives \( R_A = 5 \) kN.
Step 3: Find the force transferred through the hinge.
Vertical balance of the left piece: \( R_A + F_{hinge} - 10 = 0 \), so \( F_{hinge} = 10 - 5 = 5 \) kN acting upward on the left piece from the right piece.
By Newton's third law, the left piece pushes down on the right piece at the hinge with the same \( 5 \) kN.
Step 4: Solve the right piece by taking moments about B.
The right piece carries the \( 5 \) kN hinge force (\( 0.2 \) m to the left of \( B \)), the \( 25 \) kN load (\( 0.2 \) m to the right of \( B \)), and reactions \( R_B \) and \( R_C \), with \( C \) sitting \( 0.8 \) m to the right of \( B \).
Taking moments about \( B \): \( 5 \times 0.2 - 25 \times 0.2 + R_C \times 0.8 = 0 \).
Final Answer:
The vertical reaction at the pin support C balances the loads carried through the hinge and across the right span.
\[ \boxed{R_C = 5 \text{ kN}} \]
