3. Fixed and Continuous Beams, Theory of Structures, CE 4003 (Same as CC 4003)

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Subject: Theory of Structures (CE 4003 Same as CC 4003)

Branch: Civil Engineering 🏗️
Semester: 4th Semester 📚

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3. Fixed and Continuous Beams

A fixed beam is a type of beam that is fixed at both ends, meaning it can't rotate at the supports. A continuous beam is a beam that spans over more than two supports (usually three or more), and it experiences moments and reactions at all supports.

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3.1 Concept of Fixity and Effect of Fixity 🔩

• Fixity means the beam is held tightly at its supports and cannot rotate. In this case, both shear force and bending moment are resisted by the supports.
• Effect of Fixity:
  • Fixing the ends of a beam increases its stiffness, and the beam will resist bending more effectively.
  • The bending moment at the fixed ends does not become zero (as opposed to a simply supported beam, where the moment is zero at the supports).
  • The beam will experience internal moments at the fixed ends, and these are known as Fixed End Moments (FEM).

Example: Imagine a beam fixed at both ends. If a point load is applied in the middle, the supports resist bending, and there will be moments at both ends (unlike a simply supported beam where there are no moments at the ends).

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3.2 Advantages and Disadvantages of Fixed Beam Over Simply Supported Beam ⚖️

Advantages:

1. Higher Load-Carrying Capacity: Fixed beams can carry more load because they are more resistant to bending.
2. Reduced Deflection: Fixed beams deflect less compared to simply supported beams for the same load.
3. No Moment at Supports: In a fixed beam, there is no moment at the supports because the fixed ends resist rotation.

Disadvantages:

1. More Complex Design: Fixed beams are harder to design and analyze compared to simply supported beams.
2. Increased Material Usage: They require more material due to their increased stiffness and strength requirements.
3. Difficult to Repair: Fixed beams are difficult to replace or repair since the supports are rigid.

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3.3 Principle of Superposition 🔀

The Principle of Superposition states that the response (deflection, shear force, and bending moment) of a structure subjected to multiple loads is the sum of the responses caused by each load acting independently.

Example:

If a fixed beam has two loads, one in the middle and one at the end, the deflection at a point on the beam is the sum of the deflections caused by each load applied separately.

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3.4 Fixed End Moments from First Principle for Beam Subjected to

When a fixed beam is subjected to loads, there are moments at the ends of the beam due to the fixity.

3.4.1 Point Load ⚡

• A point load $P$ at the center of a fixed beam causes fixed end moments at both supports.
• The fixed end moments for a point load $P$ at the center are:
  • At left support: $M_A = - \frac{P \cdot L}{4}$
  • At right support: $M_B = \frac{P \cdot L}{4}$
    

Where:

• $L$ = length of the beam
• $P$ = point load

Diagram:

  |           |       Load P
  |___________|__________|
  |     A     |     B    |
Fixed End Moment -M/4   +M/4

3.4.2 UDL over Entire Span 🔲

• A Uniformly Distributed Load (UDL) $w$ applied over the entire length of a fixed beam creates fixed end moments as well.
• The fixed end moments for a UDL are:
  • At left support: $M_A = - \frac{w \cdot L^2}{12}$
  • At right support: $M_B = \frac{w \cdot L^2}{12}$

Where:

• $w$ = load per unit length
• $L$ = length of the beam

Diagram:

  |           |      w (uniform load)
  |___________|__________|
  |     A     |     B    |
Fixed End Moment -wL^2/12  +wL^2/12

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3.5 Application of Standard Formulae for a Fixed Beam in Finding

3.5.1 End Moments 🛠️

Using standard formulae, we can easily calculate the end moments for different load types. For a fixed beam subjected to a point load or UDL, the moments at the ends are already mentioned in the previous sections.

3.5.2 End Reactions 🏋️‍♀️

End reactions for a fixed beam can be calculated using equilibrium equations. For example:

• For a fixed beam under a central point load, the reactions at both ends are equal.

  • $R_A = R_B = \frac{P}{2}$

• For a UDL, the reactions can be calculated by dividing the total load by 2, as it is uniformly distributed.

  • $R_A = R_B = \frac{wL}{2}$

3.5.3 Drawing S.F. and B.M. Diagrams 📐

For Shear Force (S.F.D.) and Bending Moment (B.M.D.) diagrams:

• SFD: The shear force varies linearly under a UDL and is constant under a point load.
• BMD: The bending moment is maximum at the center of the beam for both a point load and UDL.

Example: For a fixed beam with a point load in the center, the S.F.D. is constant on either side of the load, and the B.M.D. has a parabolic shape with a peak in the middle.

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3.6 Definition, Effect of Continuity, Nature of Moments Induced Due to Continuity, Concept of Deflected Shape, Practical Examples 🌍

Definition of Continuity:

• Continuity refers to a beam that is supported by more than two supports. For example, a continuous beam has at least three supports, and the moments and reactions at these supports are interconnected.

Effect of Continuity:

• Continuous beams resist more bending and deflect less than simply supported beams.
• The bending moments are spread over several supports, which reduces the maximum bending at any single support.

Nature of Moments Induced Due to Continuity:

• In a continuous beam, moments are induced due to the support conditions at each joint or connection.
• The fixed-end moments at each support depend on the interaction of all the spans in the continuous beam.

Concept of Deflected Shape:

• The deflected shape is the curve that represents how a beam deforms under load. For fixed and continuous beams, the deflection is minimized because of the multiple supports and fixity.
• The deflected shape will be flatter in a continuous beam than a simply supported beam under the same load.

Practical Examples:

• Fixed Beam Example: A bridge with fixed supports at both ends to carry a large load.
• Continuous Beam Example: A multi-span highway bridge where the beam spans over more than two supports.

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Revision Points 📝

1. Fixed Beams have fixed supports at both ends, causing fixed end moments at the supports.
2. Continuity in beams means they are supported by more than two points, leading to more efficient load distribution and reduced bending moments.
3. The Principle of Superposition helps in determining the overall response when multiple loads act on a structure.
4. End Moments and Reactions for a fixed beam can be calculated using specific formulas for point loads and UDLs.
5. Shear Force and Bending Moment Diagrams give insights into how the beam will behave under loads and help in designing the structure.
6. Continuity improves the strength and reduces the deflection of beams, and is used in multi-span bridges and other long-span structures.

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Question and Answer Examples ❓

1. Q: What is the effect of fixity on a beam?

   • A: Fixity at both ends of a beam causes fixed end moments, reduces deflection, and increases the beam's load-carrying capacity.

2. Q: How do you calculate end moments for a fixed beam subjected to a point load at the center?

   • A: The fixed end moments are $M_A = - \frac{P \cdot L}{4}$ and $M_B = \frac{P \cdot L}{4}$, where $P$ is the load and $L$ is the beam length.

3. Q: What happens to the bending moment in a continuous beam?

   • A: The bending moment is distributed across multiple supports, reducing the peak moments compared to a fixed beam with the same load.

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These notes should help you get a clear understanding of fixed and continuous beams, their effects, and how to analyze them.

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