7-55 Draw The Shear And Moment Diagrams For The Beam

The determination of shear and moment diagrams for beams is a fundamental skill in structural engineering, providing a visual representation of the internal forces and moments acting along the beam's length under a given loading condition. Mastering this skill is crucial for ensuring structural integrity and safety, as these diagrams allow engineers to identify critical sections where maximum stresses occur, enabling them to design beams that can withstand applied loads without failure.

Understanding Shear and Moment Diagrams

Shear force at any section of a beam is the algebraic sum of all the vertical forces acting to the left or right of that section. It represents the internal resistance of the beam to transverse shear. The bending moment at any section is the algebraic sum of the moments of all forces acting to the left or right of that section about the centroid of the section. It represents the internal resistance of the beam to bending. Shear and moment diagrams plot these values along the length of the beam.

Significance in Structural Analysis

• Identifying Critical Sections: Shear and moment diagrams help pinpoint locations where shear force or bending moment is maximum, indicating where the beam is most susceptible to failure.
• Designing for Safety: By understanding the distribution of shear and moment, engineers can select appropriate beam sizes and materials to ensure the structure can safely handle the applied loads.
• Optimizing Material Use: Diagrams allow for the efficient allocation of material along the beam's length, potentially reducing costs and weight.

Example Problem: 7-55

Let's consider a specific example, similar to problem 7-55, to illustrate the process of drawing shear and moment diagrams. Assume we have a simply supported beam with a span of L = 8 meters, subjected to a uniformly distributed load (UDL) of w = 5 kN/m over its entire length, and a point load P = 10 kN at a distance of a = 3 meters from the left support.

Step-by-Step Solution to Draw Shear and Moment Diagrams

1. Calculate Support Reactions:

   • First, determine the reactions at the supports. For a simply supported beam, we have two vertical reactions, ( R_A ) at the left support and ( R_B ) at the right support.

   • Sum of vertical forces: ( R_A + R_B = w \cdot L + P )

   • Sum of moments about support A: ( R_B \cdot L = (w \cdot L) \cdot \frac{L}{2} + P \cdot a )

   • Solving these equations:

     • ( R_B = \frac{(5 \cdot 8) \cdot \frac{8}{2} + 10 \cdot 3}{8} = \frac{160 + 30}{8} = 23.75 , \text{kN} )
     • ( R_A = 5 \cdot 8 + 10 - 23.75 = 40 + 10 - 23.75 = 26.25 , \text{kN} )

2. Define Sections and Determine Shear Force Equations:

   • Divide the beam into sections based on where the loading changes. Here, we have two sections:
     • Section 1: 0 ≤ x < 3 m (from the left support to just before the point load)
     • Section 2: 3 m ≤ x ≤ 8 m (from the point load to the right support)
   • For Section 1 (0 ≤ x < 3 m):
     • Shear force ( V_1(x) = R_A - w \cdot x = 26.25 - 5x )
   • For Section 2 (3 m ≤ x ≤ 8 m):
     • Shear force ( V_2(x) = R_A - w \cdot x - P = 26.25 - 5x - 10 = 16.25 - 5x )

3. Define Moment Equations:

   • Calculate the bending moment equations for each section.
   • For Section 1 (0 ≤ x < 3 m):
     • Bending moment ( M_1(x) = R_A \cdot x - w \cdot x \cdot \frac{x}{2} = 26.25x - 2.5x^2 )
   • For Section 2 (3 m ≤ x ≤ 8 m):
     • Bending moment ( M_2(x) = R_A \cdot x - w \cdot x \cdot \frac{x}{2} - P \cdot (x - a) = 26.25x - 2.5x^2 - 10(x - 3) )

4. Calculate Shear Force and Bending Moment Values at Key Points:

   • Evaluate the shear force and bending moment equations at the boundaries of each section.
   • At x = 0 m (left support):
     • ( V_1(0) = 26.25 , \text{kN} )
     • ( M_1(0) = 0 , \text{kN·m} )
   • At x = 3 m (just before the point load):
     • ( V_1(3) = 26.25 - 5 \cdot 3 = 11.25 , \text{kN} )
     • ( M_1(3) = 26.25 \cdot 3 - 2.5 \cdot 3^2 = 78.75 - 22.5 = 56.25 , \text{kN·m} )
   • At x = 3 m (just after the point load):
     • ( V_2(3) = 16.25 - 5 \cdot 3 = 1.25 , \text{kN} )
     • ( M_2(3) = 26.25 \cdot 3 - 2.5 \cdot 3^2 - 10(3 - 3) = 56.25 , \text{kN·m} )
   • At x = 8 m (right support):
     • ( V_2(8) = 16.25 - 5 \cdot 8 = -23.75 , \text{kN} )
     • ( M_2(8) = 26.25 \cdot 8 - 2.5 \cdot 8^2 - 10(8 - 3) = 210 - 160 - 50 = 0 , \text{kN·m} )

5. Find the Point of Zero Shear (and Maximum Moment):

   • In Section 1:
     • Set ( V_1(x) = 0 \Rightarrow 26.25 - 5x = 0 \Rightarrow x = \frac{26.25}{5} = 5.25 , \text{m} )
     • However, this point is outside the range of Section 1 (0 ≤ x < 3 m), so we look in Section 2.
   • In Section 2:
     • Set ( V_2(x) = 0 \Rightarrow 16.25 - 5x = 0 \Rightarrow x = \frac{16.25}{5} = 3.25 , \text{m} )
     • This point is within the range of Section 2 (3 m ≤ x ≤ 8 m).
   • Calculate the maximum bending moment at x = 3.25 m:
     • ( M_2(3.25) = 26.25 \cdot 3.25 - 2.5 \cdot (3.25)^2 - 10(3.25 - 3) )
     • ( M_2(3.25) = 85.3125 - 26.40625 - 2.5 = 56.40625 , \text{kN·m} )

6. Draw the Shear Force Diagram:

   • Plot the shear force values along the length of the beam.
     • At x = 0 m, ( V = 26.25 , \text{kN} )
     • At x = 3 m (just before the point load), ( V = 11.25 , \text{kN} )
     • At x = 3 m (just after the point load), ( V = 1.25 , \text{kN} )
     • At x = 8 m, ( V = -23.75 , \text{kN} )
   • Connect the points with straight lines, noting the jump at the location of the point load.

7. Draw the Bending Moment Diagram:

   • Plot the bending moment values along the length of the beam.
     • At x = 0 m, ( M = 0 , \text{kN·m} )
     • At x = 3 m, ( M = 56.25 , \text{kN·m} )
     • At x = 3.25 m, ( M = 56.40625 , \text{kN·m} ) (maximum moment)
     • At x = 8 m, ( M = 0 , \text{kN·m} )
   • Connect the points with curves, ensuring the curve is smooth and reflects the parabolic nature of the moment distribution due to the UDL.

Additional Considerations

• Concentrated Moments: If the beam is subjected to a concentrated moment, the bending moment diagram will have a sudden jump at that location.
• Hinges: At a hinge, the bending moment is always zero. This condition helps in determining support reactions and drawing accurate moment diagrams.
• Cantilever Beams: For cantilever beams, the process is similar, but the boundary conditions are different. At the fixed end, both shear force and bending moment are non-zero, while at the free end, they are typically zero unless there is a load applied directly at the free end.

Practical Implications

Drawing shear and moment diagrams is not just an academic exercise; it has profound practical implications in structural design and analysis. Understanding the distribution of internal forces and moments allows engineers to:

• Select Appropriate Materials: Choose materials with sufficient strength to withstand the maximum shear and bending stresses.
• Optimize Beam Dimensions: Determine the optimal cross-sectional dimensions of the beam to minimize material usage while ensuring structural integrity.
• Design Reinforcements: In reinforced concrete beams, the diagrams guide the placement and amount of reinforcing steel needed to resist tensile stresses caused by bending.
• Assess Structural Safety: Evaluate the safety and stability of existing structures under various loading conditions.

Common Mistakes and How to Avoid Them

• Incorrect Support Reactions: A common mistake is calculating incorrect support reactions. Always double-check your calculations and ensure that the sum of vertical forces and moments equals zero.
• Sign Conventions: Be consistent with your sign conventions for shear force and bending moment. Inconsistent sign conventions can lead to errors in the diagrams.
• Missing Concentrated Loads or Moments: Ensure that all loads and moments are accounted for when writing the shear and moment equations.
• Incorrectly Drawing the Diagrams: Pay attention to the shape of the diagrams. Shear diagrams should be linear for UDLs and constant for point loads, while moment diagrams should be parabolic for UDLs and linear for point loads.
• Forgetting Boundary Conditions: Remember that the bending moment at a simple support or hinge is always zero, and the shear force and bending moment at the free end of a cantilever beam are typically zero.

Advanced Techniques

• Using Software: Modern structural analysis software can automatically generate shear and moment diagrams for complex loading conditions and beam geometries.
• Influence Lines: Influence lines are used to determine the maximum shear force and bending moment at a specific location in a beam due to a moving load.
• Moment Distribution Method: This method is used to analyze indeterminate beams and frames by distributing moments until equilibrium is achieved.
• Finite Element Analysis (FEA): FEA is a numerical technique used to analyze complex structures by dividing them into small elements and solving for the stresses and strains in each element.

Conclusion

Shear and moment diagrams are essential tools for structural engineers, providing a visual representation of the internal forces and moments acting within a beam. By understanding how to draw and interpret these diagrams, engineers can design safe, efficient, and reliable structures. The example problem illustrates the step-by-step process of drawing these diagrams for a simply supported beam with a combination of uniformly distributed load and point load. Continued practice and familiarity with different loading conditions will enhance your ability to analyze and design structural elements effectively. Remember to pay attention to details, double-check calculations, and use software tools to verify your results when dealing with complex structures.