The problem shows a continuous beam supported at points A, B, and C. The beam is subjected to a uniformly distributed load of 2 kN/m, a concentrated load of 10 kN, and the span lengths between the supports are given as 4 m, 2 m, and 4 m. The task is to presumably determine the reactions at the supports A, B, and C.

1. Problem Description

The problem shows a continuous beam supported at points A, B, and C. The beam is subjected to a uniformly distributed load of 2 kN/m, a concentrated load of 10 kN, and the span lengths between the supports are given as 4 m, 2 m, and 4 m. The task is to presumably determine the reactions at the supports A, B, and C.

2. Solution Steps

To solve for the reactions at supports A, B, and C, we can use the equations of equilibrium and the three-moment equation (also known as Clapeyron's theorem).

First, let's define the distances:

$L_{AB} = 4 + 2 = 6$ m

$L_{BC} = 4$ m

The uniformly distributed load (UDL) $w = 2$ kN/m acts over the entire length from A to C. The concentrated load $P = 10$ kN acts at the center span between A and B.

Applying the three-moment equation to spans AB and BC:

$M_A L_{AB} + 2M_B(L_{AB} + L_{BC}) + M_C L_{BC} = - \frac{wL_{AB}^3}{4} - \frac{wL_{BC}^3}{4} - \frac{P a b (L_{AB}+a)}{L_{AB}}$

where $a$ is the distance from support A to the concentrated load which is 4 m. And $b$ is distance from support B to the concentrated load which is 2 m.

Since the beam is simply supported at A and C, the moments at A and C are zero:

$M_A = M_C = 0$.

Therefore, the equation simplifies to:

$2M_B(L_{AB} + L_{BC}) = - \frac{wL_{AB}^3}{4} - \frac{wL_{BC}^3}{4} - \frac{P a b (L_{AB}+a)}{L_{AB}}$

Plugging in the values:

$2M_B(6 + 4) = - \frac{2(6)^3}{4} - \frac{2(4)^3}{4} - \frac{10 \cdot 4 \cdot 2 (6+4)}{6}$

$20M_B = - \frac{2(216)}{4} - \frac{2(64)}{4} - \frac{80(10)}{6}$

$20M_B = - 108 - 32 - \frac{800}{6}$

$20M_B = -140 - \frac{400}{3}$

$20M_B = - \frac{420 + 400}{3}$

$20M_B = - \frac{820}{3}$

$M_B = - \frac{820}{60} = - \frac{41}{3}$

$M_B = -13.67$ kN.m (approximately)

Now consider the free body diagrams of span AB and BC. Let $R_A, R_B, R_C$ be the vertical reactions at A, B, and C respectively. Let $R_{BA}$ be the reaction at B from AB and $R_{BC}$ be the reaction at B from BC. Then $R_B = R_{BA}+R_{BC}$.

Taking moments about B for span AB:

$R_A * L_{AB} - w*L_{AB}*L_{AB}/2 - P * 2 - M_B = 0$

$6R_A - 2 * 6 * 6/2 - 10 * 2 - (-13.67) = 0$

$6R_A - 36 - 20 + 13.67 = 0$

$6R_A = 42.33$

$R_A = 7.06$ kN (approximately)

$R_{BA} = w * L_{AB} + P - R_A$

$R_{BA} = 2 * 6 + 10 - 7.06$

$R_{BA} = 12 + 10 - 7.06$

$R_{BA} = 14.94$ kN (approximately)

Taking moments about B for span BC:

$-R_C*4 - w*4*4/2 - M_B = 0$

$-4R_C - 32 + 13.67 = 0$

$-4R_C = 18.33$

$R_C = -4.58$ kN

Because $R_C$ is negative, the arrow is pointing in the wrong direction. This means it's not reacting.

So $R_C = 0$.

Let's re-do the calculation:

$R_C * L_{BC} + w * L_{BC}^2/2 - M_B = 0$

$4R_C = 13.67 + 2 * 4^2/2$

$4R_C = 13.67 + 16$

$4R_C = 29.67$

$R_C = 7.42$ kN

$R_{BC} = 2 * 4 - R_C = 8 - 7.42 = 0.58$ kN

$R_B = R_{BA} + R_{BC}$

$R_B = 14.94 + 0.58$

$R_B = 15.52$ kN

$R_A + R_B + R_C = 10 + 2 * 10$

$R_A + R_B + R_C = 30$

$7.06 + 15.52 + R_C = 30$

$R_C = 30 - 22.58 = 7.42$ kN

3. Final Answer

$R_A = 7.06$ kN

$R_B = 15.52$ kN

$R_C = 7.42$ kN