Shear Force And Bending Moment Diagram Examples

Shear force and bending moment diagrams are essential tools in structural engineering for analyzing the internal forces and moments within a beam subjected to various loads. These diagrams provide a visual representation of how shear force and bending moment vary along the length of the beam, allowing engineers to determine critical locations and design structurally sound elements. Understanding how to construct and interpret these diagrams is crucial for ensuring the safety and stability of structures.

Understanding Shear Force and Bending Moment

Before delving into examples, it's essential to define shear force and bending moment.

• Shear Force (V): The shear force at a section of a beam is the algebraic sum of all the transverse forces acting on either side of the section. It represents the internal force that resists the tendency of one part of the beam to slide vertically past the other.
• Bending Moment (M): The bending moment at a section of a beam is the algebraic sum of the moments of all the forces acting on either side of the section, taken about that section. It represents the internal moment that resists the bending of the beam.

Sign Conventions

Consistent sign conventions are crucial for accurate diagram construction. The following are commonly used:

• Shear Force:
  • Positive: Upward forces to the left of the section or downward forces to the right of the section. This tends to cause a clockwise rotation.
  • Negative: Downward forces to the left of the section or upward forces to the right of the section. This tends to cause a counter-clockwise rotation.
• Bending Moment:
  • Positive: Causes compression in the top fibers of the beam and tension in the bottom fibers (sagging).
  • Negative: Causes tension in the top fibers of the beam and compression in the bottom fibers (hogging).

General Steps for Drawing Shear Force and Bending Moment Diagrams

1. Determine Support Reactions: Calculate the reactions at all supports using static equilibrium equations (sum of vertical forces = 0, sum of horizontal forces = 0, sum of moments = 0).
2. Establish Sections: Divide the beam into sections based on where loads are applied or where the beam's geometry changes.
3. Calculate Shear Force and Bending Moment Equations: For each section, determine the shear force V(x) and bending moment M(x) as functions of the distance x from the left end of the section.
4. Plot the Diagrams: Plot the shear force and bending moment values along the length of the beam, using the calculated equations.
5. Identify Critical Points: Note the locations of maximum shear force and bending moment, as these are critical for design.

Example 1: Simply Supported Beam with a Concentrated Load

Consider a simply supported beam of length L, with a concentrated load P applied at its center (L/2).

1. Support Reactions:
   • Due to symmetry, the reactions at both supports are equal: R_A = R_B = P/2.
2. Sections:
   • Section 1: 0 ≤ x ≤ L/2 (Left of the load)
   • Section 2: L/2 ≤ x ≤ L (Right of the load)
3. Shear Force and Bending Moment Equations:
   • Section 1 (0 ≤ x ≤ L/2):
     • V(x) = R_A = P/2 (Constant positive shear force)
     • M(x) = R_A * x = (P/2) * x (Linearly increasing bending moment)
   • Section 2 (L/2 ≤ x ≤ L):
     • V(x) = R_A - P = P/2 - P = -P/2 (Constant negative shear force)
     • M(x) = R_A * x - P * (x - L/2) = (P/2) * x - P * x + (P*L)/2 = (P*L)/2 - (P/2) * x (Linearly decreasing bending moment)
4. Plotting the Diagrams:
   • Shear Force Diagram: The shear force is constant at P/2 from x = 0 to x = L/2, then drops abruptly to -P/2 and remains constant until x = L.
   • Bending Moment Diagram: The bending moment increases linearly from 0 at x = 0 to (P*L)/4 at x = L/2. Then, it decreases linearly back to 0 at x = L.
5. Critical Points:
   • Maximum shear force: P/2 (magnitude)
   • Maximum bending moment: (P*L)/4 at the center of the beam.

Example 2: Simply Supported Beam with a Uniformly Distributed Load (UDL)

Consider a simply supported beam of length L, subjected to a uniformly distributed load w (force per unit length) over its entire length.

1. Support Reactions:
   • The total load on the beam is w*L. Due to symmetry, the reactions at both supports are equal: R_A = R_B = (w*L)/2.
2. Section:
   • We only need one section from 0 ≤ x ≤ L due to the continuous nature of the load.
3. Shear Force and Bending Moment Equations:
   • V(x) = R_A - w*x = (w*L)/2 - w*x (Linearly decreasing shear force)
   • M(x) = R_A * x - (w*x) * (x/2) = (w*L/2) * x - (w*x^2)/2 (Parabolically increasing then decreasing bending moment)
4. Plotting the Diagrams:
   • Shear Force Diagram: The shear force decreases linearly from (w*L)/2 at x = 0 to -(w*L)/2 at x = L. The shear force is zero at x = L/2.
   • Bending Moment Diagram: The bending moment increases parabolically from 0 at x = 0 to a maximum value at x = L/2, then decreases parabolically back to 0 at x = L.
5. Critical Points:
   • Maximum shear force: (w*L)/2 (magnitude)
   • Maximum bending moment: Occurs where V(x) = 0, which is at x = L/2. M_max = (w*L^2)/8.

Example 3: Cantilever Beam with a Concentrated Load at the Free End

Consider a cantilever beam of length L, fixed at one end and with a concentrated load P applied at the free end.

1. Support Reactions:
   • Vertical reaction at the fixed end: R_A = P (upward)
   • Moment reaction at the fixed end: M_A = P*L (counter-clockwise, negative)
2. Section:
   • We only need one section from 0 ≤ x ≤ L (where x is measured from the free end).
3. Shear Force and Bending Moment Equations:
   • V(x) = -P (Constant negative shear force)
   • M(x) = -P*x (Linearly increasing negative bending moment)
4. Plotting the Diagrams:
   • Shear Force Diagram: The shear force is constant at -P along the entire length of the beam.
   • Bending Moment Diagram: The bending moment increases linearly from 0 at the free end (x = 0) to -P*L at the fixed end (x = L).
5. Critical Points:
   • Maximum shear force: P (magnitude)
   • Maximum bending moment: P*L at the fixed end.

Example 4: Overhanging Beam with a UDL and a Concentrated Load

Consider an overhanging beam with a length of L1 + L2. The beam is supported at two points, A and B, where the distance between A and B is L1. The overhang portion has a length of L2. The beam is subjected to a uniformly distributed load w over the entire length (L1 + L2) and a concentrated load P at the free end of the overhang.

1. Support Reactions:
   • First, calculate the total load due to the UDL: w(L1 + L2)*.
   • Take moments about support A to find the reaction at support B (R_B):
     • R_B * L1 - w*(L1 + L2) * (L1 + L2)/2 - P * (L1 + L2) = 0
     • Solve for R_B.
   • Calculate the reaction at support A (R_A) using the vertical equilibrium equation:
     • R_A + R_B - w*(L1 + L2) - P = 0
     • Solve for R_A.
2. Sections:
   • Section 1: 0 ≤ x ≤ L1 (Between support A and support B)
   • Section 2: L1 ≤ x ≤ L1 + L2 (Overhang portion)
3. Shear Force and Bending Moment Equations:
   • Section 1 (0 ≤ x ≤ L1):
     • V(x) = R_A - w*x
     • M(x) = R_A * x - (w*x^2)/2
   • Section 2 (L1 ≤ x ≤ L1 + L2): Let x' be the distance from support B (so x' = x - L1).
     • V(x') = R_B - w*(x' + L1) - P OR, in terms of x: V(x) = R_B - w*x - P
     • M(x') = R_B * x' - w*(x' + L1) * (x'/2 + L1/2) - P * (x') OR, in terms of x: M(x) = R_B * (x - L1) - w*x*(x/2) - P*(x - L1). This M(x) is taken relative to the left-most end of the beam. A simpler form, taken from the right-most end (distance x'' from the right end) is: M(x'') = -w*(x''^2)/2 - P*x'' where x'' = (L1 + L2) - x.
4. Plotting the Diagrams:
   • Plot the shear force and bending moment diagrams based on the equations derived for each section. Pay attention to the sign conventions.
5. Critical Points:
   • Identify the points of maximum positive and negative bending moments, and maximum shear forces. These points are crucial for design.

Example 5: Cantilever Beam with a Linearly Varying Load

Consider a cantilever beam of length L fixed at one end, subjected to a linearly varying load, with intensity w = kx where k is a constant and x is the distance from the free end. The load is zero at the free end and maximum (kL) at the fixed end.

1. Support Reactions:
   • Total load: The total load is the area under the linearly varying load distribution, which is (1/2) * L * (kL) = (kL^2)/2.
   • Vertical reaction at the fixed end: R_A = (kL^2)/2 (upward)
   • Moment reaction at the fixed end: M_A. To find this, consider the centroid of the triangular load distribution, which is located at 2L/3 from the free end. Therefore, M_A = (kL^2)/2 * (2L/3) = (kL^3)/3 (counter-clockwise, negative).
2. Section:
   • We only need one section from 0 ≤ x ≤ L (where x is measured from the free end).
3. Shear Force and Bending Moment Equations:
   • The load intensity at a distance x from the free end is w(x) = kx. The total load acting up to x is the area of the triangle: (1/2) * x * (kx) = (kx^2)/2.
   • V(x) = - (kx^2)/2 (Negative shear force)
   • M(x) = - (kx^2)/2 * (x/3) = - (kx^3)/6 (Negative bending moment). The x/3 term represents the distance from the section to the centroid of the triangular load distribution up to that point.
4. Plotting the Diagrams:
   • Shear Force Diagram: The shear force varies quadratically from 0 at the free end to -(kL^2)/2 at the fixed end.
   • Bending Moment Diagram: The bending moment varies cubically from 0 at the free end to -(kL^3)/6 at the fixed end.
5. Critical Points:
   • Maximum shear force: (kL^2)/2 (magnitude) at the fixed end.
   • Maximum bending moment: (kL^3)/6 at the fixed end.

Tips for Drawing Accurate Diagrams

• Start with Support Reactions: Always calculate support reactions accurately before proceeding. An error in the reactions will propagate through the entire diagram.
• Check for Discontinuities: Concentrated loads and moments cause discontinuities in the shear force and bending moment diagrams, respectively.
• Relate Shear and Moment: The slope of the bending moment diagram at any point is equal to the shear force at that point. This relationship can help verify the accuracy of the diagrams. Also, the area under the shear force diagram between two points is equal to the change in bending moment between those two points.
• Use Software: Structural analysis software can be used to verify hand calculations and generate accurate shear force and bending moment diagrams for complex loading conditions.

Common Mistakes to Avoid

• Incorrect Sign Conventions: Using inconsistent sign conventions will lead to incorrect diagrams.
• Forgetting Distributed Loads: Properly account for the contribution of distributed loads when calculating shear force and bending moment equations. Remember to treat the distributed load as acting at its centroid when calculating moments.
• Ignoring Support Conditions: Accurately represent the support conditions (e.g., fixed, pinned, roller) and their corresponding reactions.
• Incorrectly Calculating Areas: When using the area under the shear force diagram to find the change in bending moment, ensure the areas are calculated correctly, especially for complex shapes.

Applications of Shear Force and Bending Moment Diagrams

• Structural Design: These diagrams are used to determine the maximum shear force and bending moment in a beam, which are essential for selecting appropriate beam sizes and materials.
• Stress Analysis: The bending moment is directly related to the bending stress in the beam. The maximum bending moment corresponds to the maximum bending stress, which is a critical parameter for structural integrity.
• Deflection Analysis: The bending moment diagram is used in conjunction with the material properties and beam geometry to calculate the deflection of the beam under load.
• Failure Prediction: By analyzing the shear force and bending moment diagrams, engineers can identify potential failure points in the beam and take measures to prevent failure.

Understanding and accurately constructing shear force and bending moment diagrams are fundamental skills for any structural engineer. These diagrams provide a clear and concise representation of the internal forces and moments within a beam, enabling engineers to design safe and efficient structures. Through careful calculation, consistent application of sign conventions, and practice with various examples, one can master the art of shear and moment diagram creation and application.