We are given a frame structure with a uniformly distributed load of 4 kN/m on the vertical member AB, a point load of 12 kN at point D, and two pinned supports at A and F. We need to find the support reactions $H_A$, $V_A$, $H_F$, and $V_F$. The height of the vertical members is 8 m, and the horizontal member has a total length of 6 m + 3 m + 3 m = 12 m.

1. Problem Description

We are given a frame structure with a uniformly distributed load of 4 kN/m on the vertical member AB, a point load of 12 kN at point D, and two pinned supports at A and F. We need to find the support reactions $H_A$, $V_A$, $H_F$, and $V_F$. The height of the vertical members is 8 m, and the horizontal member has a total length of 6 m + 3 m + 3 m = 12 m.

2. Solution Steps

First, let's calculate the equivalent point load of the distributed load.

$W = 4 \text{ kN/m} \times 8 \text{ m} = 32 \text{ kN}$

This force acts at the midpoint of AB, which is 4 m from A.

Now, we apply the equilibrium equations to the entire structure:

$\sum F_x = 0$

$H_A - 32 + H_F = 0$

$H_F = 32 - H_A$

$\sum F_y = 0$

$V_A - 12 + V_F = 0$

$V_A + V_F = 12$

$\sum M_A = 0$ (Taking moments about point A)

$-(32 \text{ kN} \times 4 \text{ m}) - (12 \text{ kN} \times 9 \text{ m}) + V_F \times 12 \text{ m} = 0$

$-128 - 108 + 12 V_F = 0$

$12 V_F = 236$

$V_F = \frac{236}{12} = \frac{59}{3} \approx 19.67 \text{ kN}$

Now, we can find $V_A$:

$V_A = 12 - V_F = 12 - \frac{59}{3} = \frac{36-59}{3} = -\frac{23}{3} \approx -7.67 \text{ kN}$

The negative sign indicates that the direction of $V_A$ is actually downwards.

To find $H_A$ and $H_F$, we can sum moments about F:

$\sum M_F = 0$ (Taking moments about point F)

$V_A \times 12 + H_A \times 8 - 32 \times 8 - 12 \times 3 = 0$

$-\frac{23}{3} \times 12 + H_A \times 8 - 256 - 36 = 0$

$-92 + 8H_A - 292 = 0$

$8 H_A = 384$

$H_A = \frac{384}{8} = 48 \text{ kN}$

Now, we can find $H_F$:

$H_F = 32 - H_A = 32 - 48 = -16 \text{ kN}$

The negative sign indicates that the direction of $H_F$ is actually to the left.

3. Final Answer

$H_A = 48 \text{ kN}$ (to the right)

$V_A = -7.67 \text{ kN}$ or $7.67 \text{ kN}$ (downwards)

$H_F = -16 \text{ kN}$ or $16 \text{ kN}$ (to the left)

$V_F = 19.67 \text{ kN}$ (upwards)