A simply supported beam of 8m length is subjected to three concentrated loads of 10kN, 15kN, and 20kN at distances 2m, 5m, and 7m from the left support W, respectively. Calculate the reactions at the

To solve question number 2, we need to determine the reactions at the supports, and then draw the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD).

The beam is simply supported at W (pin support) and C (roller support).
The loads are:

• 10 kN at 2 m from W
• 15 kN at 5 m from W (2 m + 3 m)
• 20 kN at 7 m from W (2 m + 3 m + 2 m) The total length of the beam is 8 m (2 m + 3 m + 2 m + 1 m).

Step 1: Determine Reactions at Supports

Let $R_W$ be the vertical reaction at W and $R_C$ be the vertical reaction at C.

Apply the equilibrium equations:

1. Sum of vertical forces equals zero ($\sum F_y = 0$):

   $$R_W + R_C - 10\, \text{kN} - 15\, \text{kN} - 20\, \text{kN} = 0$$ $$R_W + R_C = 45\, \text{kN} \quad (1)$$

2. Sum of moments about W equals zero ($\sum M_W = 0$, taking clockwise moments as positive):

   $$(10\, \text{kN} \times 2\, \text{m}) + (15\, \text{kN} \times 5\, \text{m}) + (20\, \text{kN} \times 7\, \text{m}) - (R_C \times 8\, \text{m}) = 0$$ $$20\, \text{kN}\cdot\text{m} + 75\, \text{kN}\cdot\text{m} + 140\, \text{kN}\cdot\text{m} - 8R_C = 0$$ $$235\, \text{kN}\cdot\text{m} - 8R_C = 0$$ $$8R_C = 235\, \text{kN}\cdot\text{m}$$ $$R_C = \frac{235}{8}\, \text{kN}$$ $$R_C = 29.375\, \text{kN}$$

Substitute $R_C$ into equation (1):

$$R_W + 29.375\, \text{kN} = 45\, \text{kN}$$ $$R_W = 45\, \text{kN} - 29.375\, \text{kN}$$ $$R_W = 15.625\, \text{kN}$$

The reactions at the supports are:

• $R_W = \boxed{15.625\, \text{kN}}$
• $R_C = \boxed{29.