7-59 Draw The Shear And Moment Diagrams For The Beam

Let's delve into the process of drawing shear and moment diagrams for a beam, focusing on the specific example designated as 7-59. This involves understanding the forces and moments acting on the beam, and then graphically representing how these quantities vary along its length. Shear and moment diagrams are essential tools in structural engineering, allowing engineers to visualize and analyze the internal stresses within a beam under load, crucial for ensuring structural integrity and preventing failure.

Understanding Shear and Moment

Before diving into the specifics of problem 7-59, it's crucial to define what shear force and bending moment represent.

• Shear Force (V): Shear force at any section of the beam is the algebraic sum of all the vertical forces acting to either the left or right of that section. It represents the tendency of one part of the beam to slide vertically relative to the adjacent part.

• Bending Moment (M): The bending moment at any section is the algebraic sum of the moments (about that section) of all the forces acting to either the left or right of that section. It represents the internal forces that resist bending of the beam.

Both shear force and bending moment are functions of the position along the beam, typically denoted as 'x'. The diagrams we create are graphs of V(x) and M(x).

General Steps for Drawing Shear and Moment Diagrams

Here's a breakdown of the typical procedure:

1. Determine the Support Reactions: This is the foundation. Before you can analyze the internal forces, you need to know the external reactions at the supports. This involves applying the equations of static equilibrium:

   • ΣF_x = 0 (Sum of horizontal forces equals zero)
   • ΣF_y = 0 (Sum of vertical forces equals zero)
   • ΣM = 0 (Sum of moments equals zero)

   Choose a convenient point about which to sum moments. The sign convention for moments is typically positive for counter-clockwise and negative for clockwise.

2. Establish Sections: Divide the beam into sections based on where the loading changes. This typically occurs at:

   • Concentrated loads (point loads)
   • Start or end of distributed loads (uniform or varying)
   • Hinge points (if any)

3. Determine Shear Force Equations: For each section, consider a cut at an arbitrary distance 'x' from the left support (or another convenient starting point). Sum the vertical forces to the left of the cut. This will give you an equation for V(x) for that section. Be careful with signs! Upward forces are typically considered positive.

4. Determine Bending Moment Equations: Similarly, for each section, sum the moments of all forces to the left of the cut about the cut location. This will give you an equation for M(x) for that section. Again, be consistent with your sign convention. A positive bending moment typically causes the beam to "smile" (concave up), while a negative bending moment causes it to "frown" (concave down).

5. Plot the Diagrams: Based on the equations you've derived, plot the shear force and bending moment diagrams. Key points to remember:

   • The slope of the shear diagram at any point is equal to the negative of the distributed load at that point (dV/dx = -w(x), where w(x) is the distributed load).
   • The slope of the moment diagram at any point is equal to the shear force at that point (dM/dx = V(x)).
   • Concentrated loads cause jumps in the shear diagram.
   • Concentrated moments cause jumps in the moment diagram.
   • The maximum bending moment usually occurs where the shear force is zero (or changes sign). This is a critical point to identify.

6. Check Your Work: Look for inconsistencies. Does the shear diagram close to zero at the end of the beam? Does the moment diagram start and end at zero (unless there's an applied moment at the ends)? These are good checks to ensure you haven't made any calculation errors.

Applying the Steps to Problem 7-59

Since I don't have the visual representation of problem 7-59, I'll create a hypothetical problem that incorporates common elements found in beam analysis. Let's assume the following:

Hypothetical Problem 7-59:

A simply supported beam, 6 meters long, is subjected to a uniformly distributed load of 2 kN/m over its entire length and a concentrated load of 5 kN at a point 2 meters from the left support. Draw the shear and moment diagrams.

Solution:

1. Determine the Support Reactions:

Let A be the left support and B be the right support. Let R_A and R_B be the vertical reaction forces at supports A and B, respectively.

• ΣF_y = 0: R_A + R_B - (2 kN/m * 6 m) - 5 kN = 0 => R_A + R_B = 17 kN (Equation 1)

• ΣM_A = 0: R_B * 6 m - (2 kN/m * 6 m * 3 m) - (5 kN * 2 m) = 0 => 6R_B - 36 kN.m - 10 kN.m = 0 => 6R_B = 46 kN.m => R_B = 7.67 kN

Substituting R_B into Equation 1: R_A + 7.67 kN = 17 kN => R_A = 9.33 kN

2. Establish Sections:

We need to divide the beam into two sections:

• Section 1: 0 ≤ x < 2 m (from the left support to just before the concentrated load)
• Section 2: 2 m ≤ x ≤ 6 m (from the concentrated load to the right support)

3. Determine Shear Force Equations:

• Section 1 (0 ≤ x < 2 m):

  • V₁(x) = R_A - (2 kN/m * x) = 9.33 - 2x (kN)

• Section 2 (2 m ≤ x ≤ 6 m):

  • V₂(x) = R_A - (2 kN/m * x) - 5 kN = 9.33 - 2x - 5 = 4.33 - 2x (kN)

4. Determine Bending Moment Equations:

• Section 1 (0 ≤ x < 2 m):

  • M₁(x) = R_A * x - (2 kN/m * x) * (x/2) = 9.33x - x² (kN.m)

• Section 2 (2 m ≤ x ≤ 6 m):

  • M₂(x) = R_A * x - (2 kN/m * x) * (x/2) - 5 kN * (x - 2 m) = 9.33x - x² - 5x + 10 = 4.33x - x² + 10 (kN.m)

5. Plot the Diagrams:

Now, let's analyze the equations and plot the diagrams.

• Shear Diagram:

  • At x = 0: V₁(0) = 9.33 kN
  • At x = 2 m: V₁(2) = 9.33 - 2(2) = 5.33 kN
  • At x = 2 m (Section 2, just after the 5 kN load): V₂(2) = 4.33 - 2(2) = 0.33 kN (Note the jump of 5 kN downwards)
  • At x = 6 m: V₂(6) = 4.33 - 2(6) = -7.67 kN

  The shear diagram starts at 9.33 kN, decreases linearly to 5.33 kN at x=2m. It then drops suddenly by 5 kN to 0.33 kN and continues decreasing linearly to -7.67 kN at x=6m. This closing value confirms that the calculation of R_B is correct since the shear force at the end of the beam should equal the negative of the reaction force at B.

• Moment Diagram:

  • At x = 0: M₁(0) = 0 kN.m
  • At x = 2 m: M₁(2) = 9.33(2) - (2)² = 18.66 - 4 = 14.66 kN.m
  • At x = 6 m: M₂(6) = 4.33(6) - (6)² + 10 = 25.98 - 36 + 10 = -0.02 kN.m ≈ 0 kN.m (Rounding error)

  To find the maximum bending moment, we need to find where the shear force is zero in either section.

  • Section 1: V₁(x) = 9.33 - 2x = 0 => x = 4.665 m. However, this is outside the range of Section 1 (0 ≤ x < 2 m), so it's not a valid location for a maximum moment in Section 1.

  • Section 2: V₂(x) = 4.33 - 2x = 0 => x = 2.165 m. This is within the range of Section 2 (2 m ≤ x ≤ 6 m).

  Therefore, the maximum bending moment occurs at x = 2.165 m.

  • M₂(2.165) = 4.33(2.165) - (2.165)² + 10 = 9.374 - 4.687 + 10 = 14.687 kN.m (approximately)

  The moment diagram starts at 0 kN.m, increases to 14.66 kN.m at x=2m, reaches a maximum of approximately 14.69 kN.m at x=2.165m, and then decreases to 0 kN.m at x=6m.

6. Sketches of the Diagrams:

(Since I can't draw here, imagine the following):

• Shear Diagram: A line sloping downwards from 9.33 kN at x=0 to 5.33 kN at x=2. A vertical drop of 5 kN to 0.33 kN at x=2. A line sloping downwards from 0.33 kN at x=2 to -7.67 kN at x=6.

• Moment Diagram: A curve starting at 0 kN.m at x=0, increasing to 14.66 kN.m at x=2. A smooth curve reaching a peak of approximately 14.69 kN.m at x=2.165, and then decreasing back to 0 kN.m at x=6.

Important Considerations for Varying Load Types and Beam Configurations:

The example above is for a relatively simple case. Let's consider how things change for more complex scenarios:

• Varying Distributed Loads: If the distributed load is not uniform (e.g., a triangular or trapezoidal load), w(x) will be a function of x. This means the shear force equation will be more complex (typically quadratic), and the bending moment equation will be even higher order (typically cubic). You'll need to integrate w(x) to find the shear force and integrate the shear force to find the bending moment.

• Cantilever Beams: A cantilever beam is fixed at one end and free at the other. The fixed end provides both vertical and moment reactions. When calculating shear and moment equations, it's often easiest to start from the free end and work towards the fixed end. The shear and moment are zero at the free end.

• Overhanging Beams: These beams extend beyond one or both of the supports. The analysis is the same as for simply supported beams, but you need to carefully consider the sections beyond the supports when determining shear and moment equations.

• Internal Hinges: An internal hinge (or pin) introduces a point where the bending moment must be zero. This means you'll need to analyze the beam in segments separated by the hinge. The shear force may or may not be continuous at the hinge, depending on the loading.

• Applied Moments: A concentrated moment applied to the beam causes a sudden jump in the bending moment diagram at the point of application. The shear diagram is not affected by applied moments.

• Inclined Loads: If you have inclined loads, resolve them into horizontal and vertical components. The horizontal components will affect the axial force in the beam (which is typically ignored in simple beam bending analysis, but may be important in some cases), and the vertical components will contribute to the shear force and bending moment.

Common Mistakes and How to Avoid Them

• Incorrect Support Reactions: This is the most common source of error. Double-check your equilibrium equations and ensure you've considered all forces and moments. Use free body diagrams to visualize the forces clearly.

• Sign Conventions: Be consistent with your sign conventions for shear and moment. Changing conventions mid-problem will lead to incorrect results.

• Incorrectly Calculating Area Under Distributed Load: Remember that the force due to a distributed load is the area under the load curve. For uniform loads, this is simply the load intensity multiplied by the length. For varying loads, you'll need to use appropriate geometric formulas (e.g., area of a triangle for a triangular load) or integration.

• Forgetting Concentrated Loads/Moments: Make sure you include all concentrated loads and moments in your shear and moment equations. These cause jumps in the diagrams and must be accounted for.

• Not Checking Boundary Conditions: The shear and moment diagrams should satisfy the boundary conditions of the beam. For example, the bending moment should be zero at a simply supported end. If they don't, you've made a mistake.

• Algebra Errors: Simple algebraic errors can easily creep into the equations. Carefully check your work at each step.

Advanced Techniques and Software

While the manual calculation of shear and moment diagrams is fundamental, engineers often use software for more complex structures. Here are some advanced techniques and software used in practice:

• Finite Element Analysis (FEA): FEA software divides the structure into small elements and uses numerical methods to solve for the stresses and strains. This is essential for complex geometries, material properties, and loading conditions. Examples include ANSYS, Abaqus, and SolidWorks Simulation.

• Structural Analysis Software: Software packages specifically designed for structural analysis often have built-in tools for generating shear and moment diagrams. Examples include SAP2000, ETABS, and RISA.

• Influence Lines: Influence lines are diagrams that show the variation of a specific response (e.g., shear force, bending moment, reaction force) at a particular point in the structure as a unit load moves across the structure. They are useful for analyzing structures subjected to moving loads.

Practical Applications

Shear and moment diagrams are not just theoretical exercises. They are crucial for:

• Determining Maximum Stresses: The maximum bending moment directly relates to the maximum bending stress in the beam. This is a critical factor in determining if the beam will fail.

• Selecting Appropriate Beam Size: By knowing the maximum bending moment and shear force, engineers can select a beam size and material that can safely withstand the applied loads.

• Designing Reinforcement: In reinforced concrete beams, the bending moment diagram is used to determine the amount and placement of reinforcing steel needed to resist the tensile stresses.

• Analyzing Deflections: The bending moment diagram is used in conjunction with the beam's flexural rigidity (EI) to calculate the deflection of the beam. Excessive deflection can cause serviceability problems.

• Understanding Failure Modes: The shear and moment diagrams help engineers understand how a beam is likely to fail under load.

Conclusion

Understanding and accurately drawing shear and moment diagrams is a fundamental skill for any structural engineer or anyone involved in the design and analysis of structures. While the process can seem tedious at first, with practice, it becomes a powerful tool for visualizing and analyzing the internal forces within a beam. By carefully following the steps outlined above, paying attention to sign conventions, and checking your work, you can confidently create shear and moment diagrams and use them to ensure the safety and integrity of your designs. Remember that this explanation uses a hypothetical 7-59 problem. Always refer to the specific details of your actual 7-59 problem for an accurate solution. Good luck!