To determine the reactions at the supports of the beam, let's analyze the loading step-by-step.

Given:

1. Triangular Distributed Load:

   • Maximum intensity = 1400 N/m (at the left end)
   • Spans over 2 m.

2. Point Load:

   • 2000 N, located at 2.5 m from the left end (halfway between the right support and the triangular load).

3. Supports:

   • Pin support at the left (A), capable of vertical and horizontal reactions $R_A^V$ and $R_A^H$.
   • Roller support at the right (B), providing only a vertical reaction $R_B^V$.

Steps to Solve:

1. Calculate the Resultant of the Triangular Load:

For a triangular load, the total force $W_t$ is given by:

$W_t = \frac{1}{2} \times \text{Base} \times \text{Height}$

$W_t = \frac{1}{2} \times 2 \, \text{m} \times 1400 \, \text{N/m} = 1400 \, \text{N}$

This resultant force acts at one-third of the base from the larger end (i.e., 2/3 of 2 m from A). Thus:

$\text{Distance from A to the centroid} = \frac{2}{3} \times 2 = 1.33 \, \text{m}$

2. Set up Equilibrium Equations:

• Sum of Vertical Forces: R_A^V + R_B^V = 1400 \, \text{N} + 2000 \, \text{N} = 3400 \, \text{N} \tag{1}

• Sum of Moments about A (taking counterclockwise moments as positive): $\sum M_A = 1400 \, \text{N} \times 1.33 \, \text{m} + 2000 \, \text{N} \times 2.5 \, \text{m} - R_B^V \times 3 \, \text{m} = 0$

  $(1862 + 5000) - 3 R_B^V = 0$

  $6862 = 3 R_B^V$

  $R_B^V = \frac{6862}{3} = 2287.33 \, \text{N}$

3. Calculate $R_A^V$:

Using Equation (1):

$R_A^V + 2287.33 = 3400$

$R_A^V = 3400 - 2287.33 = 1112.67 \, \text{N}$

4. Horizontal Force:

There are no horizontal forces in this problem, so:

$R_A^H = 0$

Final Reactions:

• $R_A^V = 1112.67 \, \text{N}$
• $R_B^V = 2287.33 \, \text{N}$
• $R_A^H = 0$

Do you have any questions or need further explanation?

5 Related Questions:

1. How would the reactions change if the triangular load was reversed (maximum at the right)?
2. How do you compute the moment of a uniformly distributed load?
3. What if the beam was simply supported with no roller—how would that affect reactions?
4. Can we solve this using shear force and bending moment diagrams?
5. What happens to reactions if the length of the beam increases?

Tip: For triangular loads, remember that the resultant force acts at 1/3 of the base from the larger end.