How To Draw Shear And Bending Moment Diagrams

Understanding how structures respond to external forces is fundamental to civil and mechanical engineering. Shear and bending moment diagrams are vital visual tools used to analyze these internal forces and moments within a beam, which are critical for structural design and safety. Mastering the construction and interpretation of these diagrams allows engineers to predict stress distribution, identify critical points within a beam, and ensure structural integrity under various loading conditions.

Understanding Shear and Bending Moment

Shear force refers to the internal force acting perpendicular to the beam's longitudinal axis. This force results from the tendency of one part of the beam to slide or shear past the adjacent part. It is typically denoted by 'V'.

The bending moment, denoted by 'M', represents the internal moment acting about the beam's cross-section. This moment is a measure of the internal forces causing the beam to bend.

Both shear force and bending moment vary along the length of the beam, depending on the applied loads and support conditions. Shear and bending moment diagrams graphically represent these variations, showing the magnitude of V and M at every point along the beam. These diagrams are essential for:

• Determining maximum stress: The maximum shear force and bending moment dictate the maximum shear and bending stresses within the beam.
• Designing beam size: Engineers use these diagrams to determine the appropriate size and shape of the beam to withstand the applied loads without failure.
• Identifying critical locations: These diagrams pinpoint locations where the beam is most susceptible to failure, such as points of maximum moment or shear.
• Predicting deflection: The bending moment diagram is crucial for calculating the beam's deflection under load.

Prerequisites for Drawing Shear and Bending Moment Diagrams

Before diving into the steps, ensure you understand these fundamental concepts:

• Types of Supports: Know the reactions each support provides (roller, hinge, fixed).
• Types of Loads: Understand point loads, uniformly distributed loads (UDL), and varying loads.
• Equilibrium Equations: ΣFx = 0, ΣFy = 0, ΣM = 0 (sum of forces in x, y directions and sum of moments equal zero).
• Free Body Diagrams (FBD): Ability to draw FBDs representing the beam and applied loads.

Step-by-Step Guide to Drawing Shear and Bending Moment Diagrams

Here's a detailed guide, broken down into manageable steps, with illustrative examples:

1. Draw a Free Body Diagram (FBD) of the Beam:

• Represent the beam with a simple line.
• Show all applied external loads, including their magnitudes and locations.
• Replace supports with their equivalent reaction forces and moments. Remember:
  • Roller support: Vertical reaction force.
  • Hinge support: Vertical and horizontal reaction forces.
  • Fixed support: Vertical and horizontal reaction forces, and a moment reaction.

2. Calculate Support Reactions:

• Apply the equilibrium equations to the FBD to solve for the unknown support reactions.
• First, sum the moments about a support point to solve for one unknown reaction force. Choose a point that eliminates as many unknowns as possible.
• Then, sum the vertical forces to solve for the remaining vertical reaction force.
• Finally, sum the horizontal forces to solve for the horizontal reaction force (if any).
• Always double-check your results. Ensure that the sum of all vertical forces and the sum of all moments equal zero (or are very close, allowing for rounding errors).

Example 1: A simply supported beam of length 6m carries a point load of 10kN at 2m from the left support (A).

• FBD: Draw the beam with a length of 6m. Show the 10kN load at 2m from A. Represent the roller support at A with a vertical reaction RA, and the hinge support at the right end (B) with vertical reaction RB and horizontal reaction (which will be zero in this case as there are no horizontal forces).
• Reactions:
  • ΣMA = 0: -(10 kN * 2 m) + (RB * 6 m) = 0 => RB = 3.33 kN
  • ΣFy = 0: RA + RB - 10 kN = 0 => RA = 10 kN - 3.33 kN = 6.67 kN
  • ΣFx = 0: No horizontal forces, so the horizontal reaction at B is 0.

3. Determine Shear Force at Key Points:

• Start at the left end of the beam (x=0).
• Consider the forces acting upwards as positive and downwards as negative.
• Move incrementally along the beam, calculating the shear force (V) just before and just after each load or support.
• The shear force at a section is equal to the algebraic sum of all the vertical forces acting to the left of that section.
• Important: At the free end of a beam, the shear force is zero.

Example 1 (continued):

• V(x=0-): 0 kN (just before A)
• V(x=0+): +6.67 kN (just after A, due to reaction RA)
• V(x=2-): +6.67 kN (shear force remains constant until the 10kN load)
• V(x=2+): +6.67 kN - 10 kN = -3.33 kN (just after the 10kN load)
• V(x=6-): -3.33 kN (shear force remains constant until B)
• V(x=6+): -3.33 kN + 3.33 kN = 0 kN (just after B, the shear force returns to zero)

4. Draw the Shear Force Diagram:

• Draw a horizontal axis representing the length of the beam.
• Plot the shear force values calculated in the previous step.
• Connect the points with straight lines or curves, depending on the type of load.
  • Point load: Vertical jump in the shear force diagram.
  • Uniformly distributed load (UDL): Sloping straight line in the shear force diagram. The slope is equal to the magnitude of the UDL (w).
  • Linearly varying load: Curved line (parabola) in the shear force diagram.
• Shade the areas above and below the axis for clarity. This isn't strictly necessary, but it can help visualize positive and negative shear forces.

Example 1 (continued):

• Draw the horizontal axis.
• At x=0, plot +6.67 kN.
• Draw a horizontal line from x=0 to x=2.
• At x=2, drop vertically down to -3.33 kN.
• Draw a horizontal line from x=2 to x=6.
• At x=6, the diagram should return to zero.

5. Determine Bending Moment at Key Points:

• Start at the left end of the beam (x=0).
• Consider clockwise moments as positive and counter-clockwise moments as negative.
• The bending moment at a section is equal to the algebraic sum of the moments of all the forces acting to the left of that section, taken about that section.
• Important: The bending moment is always zero at a free end or a simple support (hinge or roller) unless there is an externally applied moment at that point.
• Crucial Relationship: The shear force (V) is the derivative of the bending moment (M) with respect to x: V = dM/dx. This means:
  • The bending moment is maximum or minimum where the shear force is zero or changes sign.
  • The area under the shear force diagram between two points represents the change in bending moment between those points. This is a very useful shortcut!

Example 1 (continued):

• M(x=0): 0 kN.m (simple support)
• M(x=2): (6.67 kN * 2 m) = 13.34 kN.m
• M(x=6): 0 kN.m (simple support)

6. Draw the Bending Moment Diagram:

• Draw a horizontal axis representing the length of the beam.
• Plot the bending moment values calculated in the previous step.
• Connect the points with straight lines or curves, depending on the type of load.
  • Point load: Sloping straight line in the bending moment diagram.
  • Uniformly distributed load (UDL): Curved line (parabola) in the bending moment diagram.
  • Linearly varying load: Cubic curve in the bending moment diagram.
• Shade the area for clarity.

Example 1 (continued):

• Draw the horizontal axis.
• At x=0, plot 0 kN.m.
• At x=2, plot 13.34 kN.m.
• At x=6, plot 0 kN.m.
• Connect the points with straight lines. The bending moment diagram is a triangle.

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

Consider a simply supported beam of length 4m carrying a uniformly distributed load (UDL) of 5 kN/m over its entire length.

1. FBD: Draw the beam with a length of 4m. Show the UDL of 5 kN/m acting downwards along the entire beam. Represent the supports with vertical reactions RA and RB.
2. Reactions:
   • Total load due to UDL = 5 kN/m * 4 m = 20 kN
   • Due to symmetry, RA = RB = 20 kN / 2 = 10 kN
3. Shear Force:
   • V(x=0+): +10 kN
   • V(x): 10 kN - (5 kN/m * x) (where x is the distance from the left support)
   • V(x=4): 10 kN - (5 kN/m * 4 m) = -10 kN
   • V(x=2): 10 kN - (5 kN/m * 2 m) = 0 kN (Shear force is zero at the midpoint)
4. Shear Force Diagram: A straight line sloping downwards from +10 kN at x=0 to -10 kN at x=4. It crosses the x-axis (V=0) at x=2.
5. Bending Moment:
   • M(x=0): 0 kN.m
   • M(x): (10 kN * x) - (5 kN/m * x * x/2) = 10x - 2.5x²
   • M(x=2): (10 kN * 2 m) - (2.5 kN/m * (2 m)²) = 20 - 10 = 10 kN.m (Maximum bending moment)
   • M(x=4): (10 kN * 4 m) - (2.5 kN/m * (4 m)²) = 40 - 40 = 0 kN.m
6. Bending Moment Diagram: A parabola, starting at 0 kN.m at x=0, reaching a maximum of 10 kN.m at x=2, and returning to 0 kN.m at x=4.

Example 3: Overhanging Beam

Consider an overhanging beam ABC, where AB = 4m and BC = 2m. A UDL of 2 kN/m acts over the entire length AC. There are roller supports at A and B.

1. FBD: Draw the beam. Show the UDL acting along the entire 6m length. Represent the roller supports at A and B with vertical reactions RA and RB.
2. Reactions:
   • Total load = 2 kN/m * 6 m = 12 kN
   • ΣMA = 0: -(2 kN/m * 6 m * 3 m) + (RB * 4 m) = 0 => RB = 18/4 = 9 kN
   • ΣFy = 0: RA + 9 kN - 12 kN = 0 => RA = 3 kN
3. Shear Force:
   • V(x=0+): +3 kN
   • V(x): 3 - 2x (for 0 ≤ x ≤ 4)
   • V(x=4-): 3 - (2*4) = -5 kN
   • V(x=4+): 3 - (2*4) + 9 = 4 kN
   • V(x): 4 - 2(x-4) (for 4 ≤ x ≤ 6, measuring x from A) or V(x) = 2(6-x) (measuring x from C)
   • V(x=6): 0 kN
   • Shear force is zero at x = 1.5m and x=6m (end C).
4. Shear Force Diagram: A straight line from +3 kN at A to -5 kN just before B. A jump of +9 kN at B, and then a straight line down to 0 kN at C.
5. Bending Moment:
   • M(x=0): 0 kN.m
   • M(x): 3x - x² (for 0 ≤ x ≤ 4)
   • M(x=4): 3(4) - 4² = -4 kN.m
   • M(x): 2(6-x)*(6-x)/2 = (6-x)² (for 4 ≤ x ≤ 6, measuring x from C)
   • M(x=6): 0 kN.m
   • Maximum bending moment occurs where shear force is zero (x=1.5m): M(1.5) = 3(1.5) - (1.5)² = 2.25 kN.m
6. Bending Moment Diagram: A parabola from 0 kN.m at A to -4 kN.m at B, and then a parabola back to 0 kN.m at C. The diagram will have a local maximum at x=1.5m.

Tips and Tricks for Accuracy

• Sign Conventions: Be consistent with your sign conventions for shear force and bending moment.
• Units: Always include units in your calculations and diagrams.
• Accuracy: Calculate reactions carefully and double-check your work.
• Relationship between V and M: Use the relationship V = dM/dx to check for errors. If the shear force is constant, the bending moment should be linear.
• Software: Use structural analysis software (like SAP2000, or online calculators) to verify your hand calculations and diagrams, especially for complex structures.
• Practice: The more you practice, the more comfortable you will become with drawing shear and bending moment diagrams.

Common Mistakes to Avoid

• Incorrect Support Reactions: This is the most common source of error. Double-check your equilibrium equations and FBD.
• Sign Errors: Be vigilant about using the correct sign conventions.
• Forgetting Distributed Loads: Remember to convert distributed loads into equivalent point loads when calculating reactions and moments.
• Incorrectly Applying the Relationship Between V and M: Misunderstanding that the area under the shear diagram represents the change in bending moment, not the bending moment itself.
• Ignoring Internal Hinges: Internal hinges (if present) introduce a point where the bending moment is zero.

Advanced Concepts

• Influence Lines: Influence lines are used to determine the maximum shear force and bending moment at a specific point in a beam due to a moving load.
• Moment Distribution Method: This method is used to analyze indeterminate beams and frames.
• Finite Element Analysis (FEA): FEA software can be used to analyze complex structures with various loading conditions.

Conclusion

Shear and bending moment diagrams are indispensable tools for structural engineers. While the initial steps may seem intricate, consistent practice and a firm grasp of the fundamental principles will lead to proficiency. By understanding these diagrams, engineers can design safe, efficient, and reliable structures that meet the demands of modern construction and engineering. The ability to accurately construct and interpret these diagrams is a cornerstone of structural analysis and design, ensuring the stability and integrity of buildings, bridges, and other essential infrastructure. Remember to always double-check your calculations, utilize available software for verification, and continually refine your understanding through practice and exploration of advanced concepts.