https://file.aiquickdraw.com/mathbot/file/1726337844573rt6qbeol.jpg

determine the support reactions from this image.

To solve for the support reactions in this problem, we need to identify and analyze the given forces and supports. Here's a breakdown of the situation based on the provided image:

### Known Data:
- **Uniformly distributed load (UDL)** of 15 kN/m over a span of 4 meters starting from point A.
- **Concentrated load** of 20 kN acting at point D.
- The beam is supported at two points:
  - **Pin support at point A** (fixed support allowing rotation but no vertical/horizontal displacement).
  - **Roller support at point B** (allowing vertical displacement but no horizontal displacement).

### Geometry:
- **Total span of the beam** is 10 meters, divided as follows:
  - 4 meters from A to C (under the UDL).
  - 4 meters from C to B.
  - 2 meters from B to D.

### Step-by-Step Solution:
#### 1. Reactions at Supports
Let’s denote the following reactions:
- \( R_A \) = Vertical reaction at support A.
- \( R_B \) = Vertical reaction at support B.

Since point A is a pin support, it will also have a horizontal reaction \( H_A \), but because there are no horizontal forces acting on the beam, \( H_A \) will be zero.

#### 2. Equivalent Loads
- **UDL (15 kN/m over 4 m)**: This is equivalent to a single concentrated load of:
  \[
  W_{\text{UDL}} = 15 \, \text{kN/m} \times 4 \, \text{m} = 60 \, \text{kN}
  \]
  acting at the center of the UDL, which is 2 meters from point A.

#### 3. Moment Equation (Taking Moments about Point A)
To solve for \( R_B \), we use the equilibrium condition that the sum of moments about point A is zero:
\[
\sum M_A = 0
\]
Taking counterclockwise moments as positive:
\[
(60 \, \text{kN}) \times 2 \, \text{m} + (20 \, \text{kN}) \times 10 \, \text{m} - R_B \times 8 \, \text{m} = 0
\]
Solving for \( R_B \):
\[
120 + 200 - 8R_B = 0
\]
\[
R_B = \frac{320}{8} = 40 \, \text{kN}
\]

#### 4. Vertical Force Equilibrium
Now, using the vertical force equilibrium condition:
\[
\sum F_y = 0
\]
\[
R_A + R_B = 60 \, \text{kN} + 20 \, \text{kN}
\]
Substitute \( R_B = 40 \, \text{kN} \):
\[
R_A + 40 = 80
\]
\[
R_A = 80 - 40 = 40 \, \text{kN}
\]

#### Final Reactions:
- \( R_A = 40 \, \text{kN} \)
- \( R_B = 40 \, \text{kN} \)

These are the vertical reactions at supports A and B.

### Do you want more details or have any questions?

#### Follow-Up Questions:
1. What is the horizontal reaction at point A?
2. How would the solution change if the concentrated load were moved to the center of the beam?
3. What are the internal shear forces at points C and B?
4. How would you calculate the bending moment at point B?
5. What would happen if an additional support were added at point D?

#### Tip:
When dealing with uniformly distributed loads, converting them to point loads makes moment calculations easier!

Determine the support reactions from this image.

Learn how to calculate the support reactions for a beam subjected to a uniformly distributed load (15 kN/m) over 4 meters and a concentrated load of 20 kN. This detailed solution includes moment equations and vertical force equilibrium. Suitable for engineering students.

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