7-70 Draw The Shear And Moment Diagrams For The Beam

Let's tackle the challenge of drawing shear and moment diagrams for beams, a fundamental skill in structural analysis. Mastering this technique allows engineers to predict a beam's internal forces and bending moments under various loading conditions, crucial for ensuring structural integrity and preventing failure.

Understanding Shear and Moment Diagrams

Shear force diagrams (SFD) visually represent the internal shear force along the length of a beam. Shear force, denoted as V, is the algebraic sum of the vertical forces acting on either side of a section of the beam.

Bending moment diagrams (BMD), on the other hand, illustrate the internal bending moment along the beam's length. Bending moment, denoted as M, represents the algebraic sum of the moments of all forces acting on either side of a section of the beam.

These diagrams provide vital information for:

• Determining maximum shear force and bending moment: These values are critical for selecting the appropriate beam size and material to withstand the applied loads.
• Identifying critical locations: Points of maximum shear or moment are potential failure locations.
• Understanding beam behavior: The shape of the diagrams reveals how the beam responds to the applied loads.

Essential Steps for Drawing Shear and Moment Diagrams

Let's break down the process into manageable steps, illustrated with examples.

1. Determine Support Reactions:

The first step is to calculate the reactions at the beam's supports. This ensures the beam is in static equilibrium (sum of forces and moments equals zero). Common support types include:

• Pin Support: Provides resistance in both vertical and horizontal directions (two reaction forces).
• Roller Support: Provides resistance only in the vertical direction (one reaction force).
• Fixed Support: Provides resistance in both vertical and horizontal directions, as well as a moment reaction (three reaction forces).

Example:

Consider a simply supported beam (pinned at A, roller at B) with a length of 10 meters and a concentrated load of 20 kN applied at the midpoint.

• Sum of Vertical Forces = 0: Ra + Rb - 20 kN = 0 (where Ra and Rb are the vertical reactions at A and B, respectively)
• Sum of Moments about A = 0: (Rb * 10 m) - (20 kN * 5 m) = 0 => Rb = 10 kN
• Substituting Rb in the first equation: Ra + 10 kN - 20 kN = 0 => Ra = 10 kN

Therefore, both supports have a vertical reaction of 10 kN.

2. Establish Section Cuts:

Next, divide the beam into sections based on changes in loading. Each section requires its own shear and moment equations. A section cut is an imaginary cut made perpendicular to the beam's axis, exposing the internal shear force (V) and bending moment (M) at that location.

Important Considerations:

• A new section is required whenever there's a change in applied load (concentrated load, distributed load, etc.) or a change in the beam's geometry (e.g., a hinge).
• The location of the section cut is denoted by 'x', representing the distance from the left end of the beam to the cut.
• Maintain consistency in your sign convention for shear and moment. A common convention is:
  • Shear: Positive if it causes clockwise rotation of the beam element.
  • Moment: Positive if it causes compression in the top fibers of the beam (sagging).

3. Derive Shear Force Equations (V(x)):

For each section, sum the vertical forces acting to the left of the section cut. Remember to include the support reactions and any applied loads within that section. Solve for V in terms of 'x'.

Example (Continuing from Step 1):

• Section 1: 0 < x < 5 m
  • Sum of Vertical Forces to the Left = 0: 10 kN - V = 0 => V(x) = 10 kN
• Section 2: 5 m < x < 10 m
  • Sum of Vertical Forces to the Left = 0: 10 kN - 20 kN - V = 0 => V(x) = -10 kN

4. Derive Bending Moment Equations (M(x)):

For each section, sum the moments about the section cut, considering all forces acting to the left of the cut. Solve for M in terms of 'x'. Remember that moments are force times distance.

Example (Continuing from Step 3):

• Section 1: 0 < x < 5 m
  • Sum of Moments about the Cut = 0: (10 kN * x) - M = 0 => M(x) = 10x kN.m
• Section 2: 5 m < x < 10 m
  • Sum of Moments about the Cut = 0: (10 kN * x) - (20 kN * (x - 5 m)) - M = 0 => M(x) = -10x + 100 kN.m

5. Plot the Shear Force and Bending Moment Diagrams:

Using the equations derived in steps 3 and 4, plot the shear force and bending moment values along the length of the beam.

• Shear Force Diagram (SFD): Plot V(x) versus x. The y-axis represents the shear force value, and the x-axis represents the position along the beam.
• Bending Moment Diagram (BMD): Plot M(x) versus x. The y-axis represents the bending moment value, and the x-axis represents the position along the beam.

Important Considerations:

• Concentrated loads cause jumps in the shear force diagram.
• Concentrated moments cause jumps in the bending moment diagram.
• The area under the shear force diagram between two points is equal to the change in bending moment between those points.
• The slope of the bending moment diagram at any point is equal to the shear force at that point.
• Points where the shear force diagram crosses the zero axis are potential locations of maximum bending moment.

6. Identify Key Values:

Determine the maximum positive and negative shear forces, and the maximum positive and negative bending moments. These values are essential for structural design.

Example (Continuing from Step 4 and 5):

• Maximum Shear Force: 10 kN (positive) and -10 kN (negative)
• Maximum Bending Moment: Occurs at x = 5 m (midpoint): M(5) = 10 * 5 = 50 kN.m

Example: Uniformly Distributed Load

Let's consider a simply supported beam of length L with a uniformly distributed load (UDL) of 'w' kN/m along its entire length.

1. Support Reactions:

• Total Load = w * L
• Due to symmetry, Ra = Rb = (w * L) / 2

2. Section Cut:

• Cut the beam at a distance 'x' from the left support (0 < x < L).

3. Shear Force Equation (V(x)):

• Sum of Vertical Forces to the Left = 0: (wL/2) - w*x - V = 0 => V(x) = (wL/2) - wx

4. Bending Moment Equation (M(x)):

• Sum of Moments about the Cut = 0: (wL/2) * x - (wx)(x/2) - M = 0 => M(x) = (wLx/2) - (wx^2/2)

5. Plotting the Diagrams:

• SFD: The shear force diagram will be a straight line with a negative slope. V(0) = wL/2 and V(L) = -wL/2.
• BMD: The bending moment diagram will be a parabola. M(0) = 0, M(L) = 0. The maximum bending moment occurs at the center (x = L/2).

6. Key Values:

• Maximum Shear Force: wL/2 (positive and negative)
• Maximum Bending Moment: M(L/2) = (wL(L/2)/2) - (w(L/2)^2/2) = wL^2 / 8

Example: Overhanging Beam

Consider an overhanging beam with a pin support at A, a roller support at C, and a concentrated load P at the free end B. Let the length AC be 'L' and the overhang CB be 'a'.

1. Support Reactions:

• Sum of Vertical Forces = 0: Ra + Rc - P = 0
• Sum of Moments about A = 0: (Rc * L) - (P * (L+a)) = 0 => Rc = P(L+a)/L
• Substituting Rc: Ra + P(L+a)/L - P = 0 => Ra = P - P(L+a)/L = -Pa/L

Note that Ra is negative, indicating it acts downwards.

2. Section Cuts:

• Section 1: 0 < x < L (between A and C)
• Section 2: L < x < L+a (between C and B)

3. Shear Force Equations (V(x)):

• Section 1: V(x) = -Ra = Pa/L
• Section 2: V(x) = -Ra - Rc = -Pa/L - P(L+a)/L = -P

4. Bending Moment Equations (M(x)):

• Section 1: M(x) = -Ra * x = (-Pa/L) * x
• Section 2: M(x) = -Ra * x + Rc * (x-L) = (-Pa/L) * x + (P(L+a)/L) * (x-L) = P(L+a-x)

5. Plotting the Diagrams:

• SFD: The shear force is constant between A and C and then changes abruptly at C and remains constant between C and B.
• BMD: The bending moment is linear in both sections.

6. Key Values:

• The maximum bending moment will occur at the support C or at the free end B depending on the values of 'L' and 'a'.

Common Loading Scenarios and Their Impact on Diagrams

Understanding how different loading types affect the shape of SFD and BMD is crucial:

• Concentrated Load:
  • SFD: Sudden vertical jump at the point of application.
  • BMD: Abrupt change in slope at the point of application.
• Uniformly Distributed Load (UDL):
  • SFD: Linearly varying shear force.
  • BMD: Parabolically varying bending moment.
• Concentrated Moment:
  • SFD: No effect.
  • BMD: Sudden vertical jump at the point of application.
• Linearly Varying Load:
  • SFD: Parabolically varying shear force.
  • BMD: Cubically varying bending moment.

Tips and Tricks for Accuracy

• Start with the reactions: Accurately calculating support reactions is fundamental. Double-check your calculations.
• Maintain a consistent sign convention: Choose a sign convention and stick to it throughout the problem.
• Check for equilibrium: At any section, the sum of forces and moments to the left should equal the sum of forces and moments to the right.
• Relate SFD and BMD: Remember that the slope of the BMD is equal to the shear force. This relationship can help you identify errors.
• Use software: Software packages like AutoCAD, SolidWorks, or dedicated structural analysis programs can assist in drawing accurate diagrams, especially for complex loading scenarios. However, it's essential to understand the underlying principles before relying solely on software.

Advanced Considerations

• Influence Lines: Influence lines are graphs that show the variation of a specific reaction, shear force, or bending moment at a specific point on a beam as a unit load moves across the beam's span. They are useful for analyzing structures subjected to moving loads.
• Statically Indeterminate Beams: These beams have more supports than required for static equilibrium. Solving for reactions and drawing shear and moment diagrams requires considering the beam's deflection characteristics using methods like the moment distribution method, the slope-deflection method, or the finite element method.
• Dynamic Loading: When beams are subjected to dynamic loads (e.g., impact loads or vibrating loads), the analysis becomes more complex and requires considering the beam's mass and damping properties.

Common Mistakes to Avoid

• Incorrect support reactions: This will propagate errors throughout the entire solution.
• Inconsistent sign convention: Leading to incorrect shear and moment equations.
• Forgetting to include loads: Omitting loads within a section will result in inaccurate diagrams.
• Incorrectly calculating moments: Ensure you're using the correct distances when calculating moments.
• Not checking for equilibrium: Failing to verify equilibrium at sections can lead to undetected errors.

Conclusion

Drawing shear and moment diagrams is a cornerstone skill for structural engineers. By mastering the steps outlined above, understanding the impact of different loading scenarios, and avoiding common mistakes, you can confidently analyze beams and ensure the safety and stability of structures. While software can be a valuable tool, a solid understanding of the fundamental principles is essential for accurate analysis and informed decision-making. Remember to practice regularly with various beam configurations and loading conditions to solidify your understanding. Don't hesitate to consult textbooks, online resources, and experienced engineers for further guidance and clarification. With dedication and practice, you'll become proficient in creating and interpreting shear and moment diagrams, a skill that will serve you well throughout your engineering career.