1.1 A beam is 10 meters long. Support A is located 2 meters from the left end, and Support B is located 7.5 meters from the left end. Loads are applied as follows: 450 kg at the left end, 6 kN at 3.5 meters from the left end, 700 kg at 6 meters from the left end, and 5 kN at the right end. Calculate the reactions at Supports $$ A $$ and $$ B $$.

Question

UpStudy AI · Accepted Answer

To solve for the reactions at Supports $$ A $$ and $$ B $$, we will follow these steps:

1. **Identify the known values:**
   - Length of the beam, $$ L = 10 $$ m
   - Position of Support $$ A $$, $$ A = 2 $$ m from the left end
   - Position of Support $$ B $$, $$ B = 7.5 $$ m from the left end
   - Loads:
     - $$ W_1 = 450 $$ kg (converted to Newtons: $$ W_1 = 450 \times 9.81 $$ N)
     - $$ W_2 = 6 $$ kN (converted to Newtons: $$ W_2 = 6000 $$ N)
     - $$ W_3 = 700 $$ kg (converted to Newtons: $$ W_3 = 700 \times 9.81 $$ N)
     - $$ W_4 = 5 $$ kN (converted to Newtons: $$ W_4 = 5000 $$ N)

2. **Convert the weights to Newtons:**
   - $$ W_1 = 450 \times 9.81 = 4414.5 $$ N
   - $$ W_3 = 700 \times 9.81 = 6867 $$ N

3. **Set up the equilibrium equations:**
   - The sum of vertical forces must equal zero:
     $$
     R_A + R_B - W_1 - W_2 - W_3 - W_4 = 0
     $$
   - The sum of moments about Support $$ A $$ must equal zero:
     $$
     \sum M_A = 0
     $$
     Taking moments about point $$ A $$:
     $$
     -W_1 \cdot 0 - W_2 \cdot (3.5 - 2) - W_3 \cdot (6 - 2) - W_4 \cdot (10 - 2) + R_B \cdot (7.5 - 2) = 0
     $$

4. **Calculate the moments:**
   - Moment due to $$ W_2 $$:
     $$
     W_2 \cdot (3.5 - 2) = 6000 \cdot 1.5 = 9000 \, \text{N m}
     $$
   - Moment due to $$ W_3 $$:
     $$
     W_3 \cdot (6 - 2) = 6867 \cdot 4 = 27468 \, \text{N m}
     $$
   - Moment due to $$ W_4 $$:
     $$
     W_4 \cdot (10 - 2) = 5000 \cdot 8 = 40000 \, \text{N m}
     $$

5. **Substituting into the moment equation:**
   $$
   -9000 - 27468 - 40000 + R_B \cdot 5.5 = 0
   $$
   $$
   R_B \cdot 5.5 = 76468
   $$
   $$
   R_B = \frac{76468}{5.5}
   $$

6. **Calculate $$ R_B $$:**
   $$
   R_B = 13813.27 \, \text{N}
   $$

7. **Substituting $$ R_B $$ back into the force equilibrium equation:**
   $$
   R_A + 13813.27 - 4414.5 - 6000 - 6867 - 5000 = 0
   $$
   $$
   R_A + 13813.27 - 22881.5 = 0
   $$
   $$
   R_A = 22881.5 - 13813.27
   $$

8. **Calculate $$ R_A $$:**
   $$
   R_A = 9078.23 \, \text{N}
   $$

Now we have the reactions at Supports $$ A $$ and $$ B $$:
- $$ R_A \approx 9078.23 \, \text{N} $$
- $$ R_B \approx 13813.27 \, \text{N} $$

Thus, the reactions at Supports $$ A $$ and $$ B $$ are approximately $$ 9078.23 \, \text{N} $$ and $$ 13813.27 \, \text{N} $$, respectively.
The reactions at Supports A and B are approximately 9078.23 N and 13813.27 N, respectively.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Engineering Questions