Two structural members are connected by a hinge and they are externally loaded as shown in figure A and B are roller support and C is pin support. Neglect mass of members, the magnitude of vertical reaction force at C in kN is

The correct answer is 5.

Step-by-Step Explanation:

1. System Identification and Support Reactions:
Based on the provided diagram, the system consists of two structural members connected by an internal hinge at point D:
- Member AD (on the left)
- Member DBC (on the right)
The supports and loading details shown in the image are as follows:
- A roller support at A, which provides a vertical reaction force, denoted as $R_A$.
- A roller support at B, which provides a vertical reaction force, denoted as $R_B$.
- A pin support at C, which provides vertical and horizontal reaction forces. Since there are no horizontal external loads, the horizontal reaction is zero, leaving only the vertical reaction force, denoted as $R_C$.
- A downward vertical point load of 10 kN is applied at a distance of 0.5 m to the right of support A.
- A downward vertical point load of 25 kN is applied at a distance of 0.2 m to the right of support B (or 0.6 m to the left of support C).
- The hinge D is located at a distance of 1.0 m from support A (0.5 m + 0.5 m) and 0.2 m to the left of support B.

2. Analysis of Member AD:
Since D is an internal hinge, the bending moment at D is zero. Let us isolate member AD and take the sum of moments about the hinge D:

$\sum M_D=0$

Taking clockwise moments as positive:

$R_A\times 1.0-10\times 0.5=0$

Solving for $R_A$:

$R_A=5\text{kN}$

Using vertical equilibrium for member AD:

$\sum F_y=0\Rightarrow R_A-10+V_D=0$

where $V_D$ is the vertical shear force at the hinge D acting on member AD.

$5-10+V_D=0\Rightarrow V_D=5\text{kN}$ (acting upwards on member AD)

3. Analysis of Member DBC:
By Newton's third law, the force exerted by member AD on member DBC at hinge D is equal and opposite. Thus, a downward vertical force of $V_D=5\text{kN}$ acts at point D of member DBC.
Now, let us consider the equilibrium of member DBC. Taking the sum of moments about support B:

$\sum M_B=0$

Taking clockwise moments as positive:

$-V_D\times 0.2+25\times 0.2-R_C\times (0.2+0.6)=0$

Substitute $V_D=5\text{kN}$ into the equation:

$-5\times 0.2+25\times 0.2-R_C\times 0.8=0$

Simplify the terms:

$-1.0+5.0-0.8R_C=0$

$4.0=0.8R_C$

$R_C=\frac{4.0}{0.8}=5\text{kN}$

Thus, the magnitude of the vertical reaction force at C is 5 kN.