The problem is to determine the maximum absolute bending moment in a beam subjected to a uniformly distributed load. The beam is 8 m long, with supports at A and D, and the distributed load, $P = 36 N/m$, is applied between B and C. The distance between A and B is 2 m, and the distance between C and D is 2 m. Therefore, the length of the beam subjected to the distributed load is $8 - 2 - 2 = 4 m$.

1. Problem Description

The problem is to determine the maximum absolute bending moment in a beam subjected to a uniformly distributed load. The beam is 8 m long, with supports at A and D, and the distributed load, $P = 36 N/m$, is applied between B and C. The distance between A and B is 2 m, and the distance between C and D is 2 m. Therefore, the length of the beam subjected to the distributed load is $8 - 2 - 2 = 4 m$.

2. Solution Steps

First, we need to determine the reactions at supports A and D. Let $R_A$ be the reaction at support A, and $R_D$ be the reaction at support D.

Sum of vertical forces:

$R_A + R_D - (36 N/m)(4 m) = 0$

$R_A + R_D = 144 N$ (Equation 1)

Sum of moments about point A:

$R_D (8 m) - (36 N/m)(4 m)(2 m + 2 m) = 0$

$8 R_D = (36)(4)(4)$

$8 R_D = 576$

$R_D = 576/8 = 72 N$

Substitute $R_D$ into Equation 1:

$R_A + 72 = 144$

$R_A = 144 - 72 = 72 N$

Now let's analyze the shear force and bending moment along the beam.

- From A to B (0 <= x <= 2):

$V(x) = R_A = 72 N$

$M(x) = R_A * x = 72x$

At x = 2 m (point B):

$V(2) = 72 N$

$M(2) = 72 * 2 = 144 Nm$

- From B to C (2 <= x <= 6):

Let $x'$ be the distance from B. Then $x' = x - 2$.

$V(x) = R_A - P x' = 72 - 36 (x - 2) = 72 - 36x + 72 = 144 - 36x$

$M(x) = R_A x - P x' (x'/2) = 72x - 36(x-2)((x-2)/2) = 72x - 18(x-2)^2 = 72x - 18(x^2 - 4x + 4) = 72x - 18x^2 + 72x - 72 = -18x^2 + 144x - 72$

To find the location of the maximum bending moment between B and C, we need to find where the shear force is zero.

$V(x) = 144 - 36x = 0$

$36x = 144$

$x = 144/36 = 4 m$

The maximum bending moment occurs at x = 4 m.

$M(4) = -18(4^2) + 144(4) - 72 = -18(16) + 576 - 72 = -288 + 576 - 72 = 216 Nm$

- From C to D (6 <= x <= 8):

Let $x''$ be the distance from D. Then $x'' = 8 - x$.

$V(x) = -R_D = -72 N$

$M(x) = R_D * x'' = 72(8-x)$

At x = 6 m (point C):

$V(6) = -72 N$

$M(6) = 72(8-6) = 72 * 2 = 144 Nm$

At x = 8 m (point D):

$V(8) = -72 N$

$M(8) = 72(8-8) = 0 Nm$

Comparing the bending moments:

At B: $M(2) = 144 Nm$

At C: $M(6) = 144 Nm$

At x=4: $M(4) = 216 Nm$

Maximum absolute moment is 216 Nm.

3. Final Answer

216 Nm