Determine the End support reactions and End support moments for the fixed beam shown in Fig. 1 belolw - span 5 m ; point load 150 kN acting at 3 m from LEFT end and at 2 m from RIGHT end.

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To determine the end support reactions and moments for a fixed beam with a point load, we can use static equilibrium equations.

### Given:
- Length of the beam, $$ L = 5 \, \text{m} $$
- Point load, $$ P = 150 \, \text{kN} $$
- Distance from the left end to the point load, $$ a = 3 \, \text{m} $$
- Distance from the right end to the point load, $$ b = 2 \, \text{m} $$

### Steps to Solve:

1. **Identify the Support Reactions:**
   For a fixed beam, there are reactions at both ends:
   - Vertical reaction at the left end, $$ R_A $$
   - Vertical reaction at the right end, $$ R_B $$
   - Moment at the left end, $$ M_A $$
   - Moment at the right end, $$ M_B $$

2. **Set Up Equilibrium Equations:**
   - **Sum of Vertical Forces:**
     $$
     R_A + R_B - P = 0
     $$
     $$
     R_A + R_B = 150 \, \text{kN} \quad \text{(1)}
     $$

- **Sum of Moments about Point A:**
     Taking moments about point A (counterclockwise moments are positive):
     $$
     M_A - P \cdot a + M_B = 0
     $$
     $$
     M_A - 150 \cdot 3 + M_B = 0
     $$
     $$
     M_A + M_B = 450 \, \text{kN} \cdot \text{m} \quad \text{(2)}
     $$

- **Sum of Moments about Point B:**
     Taking moments about point B:
     $$
     M_B - P \cdot b - M_A = 0
     $$
     $$
     M_B - 150 \cdot 2 - M_A = 0
     $$
     $$
     M_B - 300 - M_A = 0 \quad \text{(3)}
     $$

3. **Solve the Equations:**
   From equation (3):
   $$
   M_B = M_A + 300 \quad \text{(4)}
   $$

Substitute equation (4) into equation (2):
   $$
   M_A + (M_A + 300) = 450
   $$
   $$
   2M_A + 300 = 450
   $$
   $$
   2M_A = 150
   $$
   $$
   M_A = 75 \, \text{kN} \cdot \text{m}
   $$

Now substitute $$ M_A $$ back into equation (4):
   $$
   M_B = 75 + 300 = 375 \, \text{kN} \cdot \text{m}
   $$

Now substitute $$ M_A $$ into equation (1) to find $$ R_A $$ and $$ R_B $$:
   $$
   R_A + R_B = 150
   $$

We also need to find $$ R_A $$ and $$ R_B $$. We can use the moment about point A to find $$ R_B $$:
   $$
   R_B = \frac{P \cdot a}{L} = \frac{150 \cdot 3}{5} = 90 \, \text{kN}
   $$

Now substitute $$ R_B $$ back into equation (1):
   $$
   R_A + 90 = 150
   $$
   $$
   R_A = 150 - 90 = 60 \, \text{kN}
   $$

### Summary of Results:
- **Vertical Reaction at A, $$ R_A $$**: $$ 60 \, \text{kN} $$ (upward)
- **Vertical Reaction at B, $$ R_B $$**: $$ 90 \, \text{kN} $$ (upward)
- **Moment at A, $$ M_A $$**: $$ 75 \, \text{kN} \cdot \text{m} $$ (counterclockwise)
- **Moment at B, $$ M_B $$**: $$ 375 \, \text{kN} \cdot \text{m} $$ (counterclockwise)

These calculations provide the support reactions and moments for the fixed beam under the given loading conditions.
The vertical reaction at the left end is 60 kN upward, at the right end is 90 kN upward. The moment at the left end is 75 kN·m counterclockwise, and at the right end is 375 kN·m counterclockwise.

Determine the End support reactions and End support moments for the fixed beam shown in Fig. 1 belolw - span 5 m ; point load 150 kN acting at 3 m from LEFT end and at 2 m from RIGHT end.

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