Example 1

Example 1:

Beam with Uniformly Distributed Load (UDL)

Problem Statement

A continuous beam $AB - BC$ has equal spans $L = 6$ m each. Span AB is subjected to a uniformly distributed load w=10$w=10$ kN/m.

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  Span BC has no external load. Supports A and C are fixed, while B is continuous.

📌 Objective: Find the end moments $M_{AB}, M_{BA}, M_{BC}, M_{CB}$.

Step 1: Identify the Beam and Boundary Conditions

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  Fixed at A and C: $\theta_A = 0$, $\theta_C = 0$.

  
  


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  B is continuous: $M_{BA} = M_{BC}$ (Moment equilibrium at B).

  
  

Step 2: Calculate Fixed-End Moments (FEMs)

For a UDL on a fixed-fixed beam, the fixed-end moments are:

$$F_{AB} = -\frac{wL^2}{12}, \quad F_{BA} = \frac{wL^2}{12}$$
$$F_{BC} = 0, \quad F_{CB} = 0$$

Substituting values:

$$F_{AB} = -\frac{(10)(6^2)}{12} = -30 \text{ kNm}, \quad F_{BA} = +30 \text{ kNm}$$
$$F_{BC} = 0, \quad F_{CB} = 0$$

Step 3: Write Slope Deflection Equations

Using the slope deflection formula:

$$M_{AB} = \frac{2EI}{L} (2\theta_A + \theta_B) + F_{AB}$$
$$M_{BA} = \frac{2EI}{L} (\theta_A + 2\theta_B) + F_{BA}$$
$$M_{BC} = \frac{2EI}{L} (2\theta_B + \theta_C) + F_{BC}$$
$$M_{CB} = \frac{2EI}{L} (\theta_B + 2\theta_C) + F_{CB}$$

Substituting $L = 6$, $E = EI$:

$$M_{AB} = \frac{2EI}{6} (2(0) + \theta_B) - 30$$
$$M_{BA} = \frac{2EI}{6} (0 + 2\theta_B) + 30$$
$$M_{BC} = \frac{2EI}{6} (2\theta_B + 0) + 0$$
$$M_{CB} = \frac{2EI}{6} (\theta_B + 0) + 0$$

Simplifying:

$$M_{AB} = \frac{EI}{3} \theta_B - 30$$
$$M_{BA} = \frac{2EI}{3} \theta_B + 30$$
$$M_{BC} = \frac{2EI}{3} \theta_B$$
$$M_{CB} = \frac{EI}{3} \theta_B$$

Step 4: Apply Boundary Conditions

Since B is continuous, moment equilibrium at B:

$$M_{BA} + M_{BC} = 0$$

Substituting values:

$$\frac{2EI}{3} \theta_B + 30 + \frac{2EI}{3} \theta_B = 0$$
$$\frac{4EI}{3} \theta_B = -30$$
$$\theta_B = -\frac{30 \times 3}{4EI} = -\frac{90}{4EI} = -\frac{45}{2EI}$$

Step 5: Calculate Final Moments

Substituting $\theta_B = -\frac{45}{2EI}$:

$$M_{AB} = \frac{EI}{3} \times \left(-\frac{45}{2EI}\right) - 30 = -\frac{45}{6} - 30 = -7.5 - 30 = -37.5 \text{ kNm}$$
$$M_{BA} = \frac{2EI}{3} \times \left(-\frac{45}{2EI}\right) + 30 = -\frac{90}{6} + 30 = -15 + 30 = 15 \text{ kNm}$$
$$M_{BC} = \frac{2EI}{3} \times \left(-\frac{45}{2EI}\right) = -15 \text{ kNm}$$
$$M_{CB} = \frac{EI}{3} \times \left(-\frac{45}{2EI}\right) = -7.5 \text{ kNm}$$

✅ Final Moments:

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  $M_{AB} = -37.5$ kNm

  
  


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  $M_{BA} = +15$ kNm

  
  


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  $M_{BC} = -15$ kNm

  
  


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  $M_{CB} = -7.5$ kNm