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:**
     $$
     R_A + R_B - P = 0
     $$
     $$
     R_A + R_B - 150 = 0 \quad \text{(1)}
     $$

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

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

3. **Solve the Equations:**
   From equation (3):
   $$
   M_A = 150 \cdot 2 = 300 \, \text{kN m}
   $$

Substitute $$ M_A $$ into equation (2):
   $$
   300 - 150 \cdot 3 + M_B = 0
   $$
   $$
   300 - 450 + M_B = 0
   $$
   $$
   M_B = 150 \, \text{kN m}
   $$

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

We also need to find $$ R_A $$ and $$ R_B $$. We can use the moment about point B to find $$ R_A $$:
   $$
   R_A \cdot 5 - 150 \cdot 2 = 0
   $$
   $$
   R_A \cdot 5 = 300
   $$
   $$
   R_A = 60 \, \text{kN}
   $$

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

### Summary of Results:
- **Vertical Reaction at Left End, $$ R_A $$**: $$ 60 \, \text{kN} $$
- **Vertical Reaction at Right End, $$ R_B $$**: $$ 90 \, \text{kN} $$
- **Moment at Left End, $$ M_A $$**: $$ 300 \, \text{kN m} $$
- **Moment at Right End, $$ M_B $$**: $$ 150 \, \text{kN m} $$

These are the end support reactions and moments for the fixed beam under the given loading conditions.
The end support reactions and moments for the fixed beam are as follows: Vertical reaction at the left end, $$ R_A $$, is $$ 60 \, \text{kN} $$; at the right end, $$ R_B $$, is $$ 90 \, \text{kN} $$; moment at the left end, $$ M_A $$, is $$ 300 \, \text{kN m} $$; and at the right end, $$ M_B $$, is $$ 150 \, \text{kN m} $$.

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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