Examples Of Shear Force And Bending Moment Diagram

Shear force and bending moment diagrams are essential tools in structural engineering for analyzing the internal forces and moments within a beam or structural element subjected to external loads. These diagrams visually represent the distribution of shear force and bending moment along the length of the member, allowing engineers to determine critical sections, assess structural integrity, and design safe and efficient structures.

Understanding Shear Force and Bending Moment

Before diving into examples, let's briefly define shear force and bending moment:

• Shear Force (V): The algebraic sum of all transverse forces acting on either side of a section of the beam. It represents the internal force acting perpendicular to the beam's axis.
• Bending Moment (M): The algebraic sum of the moments of all forces acting on either side of a section of the beam. It represents the internal moment acting about the beam's axis, causing it to bend.

Key Concepts for Drawing Shear Force and Bending Moment Diagrams

To effectively draw and interpret shear force and bending moment diagrams, consider these key concepts:

• Sign Conventions: Consistent sign conventions are crucial. Commonly:
  • Shear Force: Positive if it causes a clockwise rotation of the beam element to the left of the section.
  • Bending Moment: Positive if it causes compression in the top fibers of the beam (sagging).
• Relationship between Load, Shear Force, and Bending Moment:
  • The slope of the shear force diagram at any point is equal to the negative of the distributed load at that point: dV/dx = -w(x)
  • The slope of the bending moment diagram at any point is equal to the shear force at that point: dM/dx = V(x)
  • The area under the shear force diagram between two points is equal to the change in bending moment between those points.
• Concentrated Loads and Moments:
  • A concentrated load causes a sudden vertical jump in the shear force diagram.
  • A concentrated moment causes a sudden jump in the bending moment diagram.
• Supports: The support conditions dictate the shear force and bending moment values at the supports.
  • Pin/Hinge Support: Zero bending moment.
  • Fixed Support: Non-zero shear force and bending moment.

General Steps for Drawing Shear Force and Bending Moment Diagrams

1. Determine Support Reactions: Calculate the vertical and horizontal reactions at all supports using static equilibrium equations (ΣF_x = 0, ΣF_y = 0, ΣM = 0).
2. Establish Sections: Divide the beam into sections based on the locations of concentrated loads, distributed loads, and supports. Each section will have its own shear force and bending moment equations.
3. Develop Shear Force Equations: For each section, determine the shear force V(x) as a function of the distance 'x' from the left end of the beam (or any convenient starting point). Consider all forces acting to the left of the section.
4. Develop Bending Moment Equations: For each section, determine the bending moment M(x) as a function of the distance 'x'. Consider all forces and moments acting to the left of the section.
5. Plot the Diagrams: Plot the shear force and bending moment diagrams based on the equations derived in steps 3 and 4. Pay attention to the sign conventions.
6. Identify Key Values: Determine the maximum and minimum values of shear force and bending moment, and their locations. These values are critical for structural design.

Examples of Shear Force and Bending Moment Diagrams

Let's explore several examples with varying load conditions:

Example 1: Simply Supported Beam with a Concentrated Load at Mid-Span

• Description: A simply supported beam of length L is subjected to a concentrated load P at its mid-span (L/2).

• Step 1: Determine Support Reactions

  • Due to symmetry, the vertical reactions at both supports are equal: R_A = R_B = P/2.

• Step 2: Establish Sections

  • Section 1: 0 ≤ x < L/2 (left of the load)
  • Section 2: L/2 < x ≤ L (right of the load)

• Step 3: Develop Shear Force Equations

  • Section 1: V(x) = R_A = P/2 (Constant positive shear force)
  • Section 2: V(x) = R_A - P = P/2 - P = -P/2 (Constant negative shear force)

• Step 4: Develop Bending Moment Equations

  • Section 1: M(x) = R_A * x = (P/2) * x (Linear increase)
  • Section 2: M(x) = R_A * x - P * (x - L/2) = (P/2) * x - Px + PL/2 = PL/2 - (P/2) * x (Linear decrease)

• Step 5: Plot the Diagrams

  • Shear Force Diagram: A constant positive value of P/2 from x = 0 to x = L/2, then a sudden drop to -P/2 and remains constant until x = L.
  • Bending Moment Diagram: A linear increase from 0 at x = 0 to PL/4 at x = L/2, then a linear decrease back to 0 at x = L.

• Step 6: Identify Key Values

  • Maximum Shear Force: P/2
  • Maximum Bending Moment: PL/4 at x = L/2 (at the location of the concentrated load).

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

• Description: A cantilever beam of length L is subjected to a uniformly distributed load w (force per unit length) along its entire length.

• Step 1: Determine Support Reactions

  • Vertical Reaction at the fixed end (A): R_A = wL (total load)
  • Moment Reaction at the fixed end (A): M_A = (wL) * (L/2) = wL²/2 (total load times the distance to the centroid of the load)

• Step 2: Establish Sections

  • One section is sufficient: 0 ≤ x ≤ L, where x is measured from the free end.

• Step 3: Develop Shear Force Equations

  • V(x) = -w * x (Linearly increasing negative shear force) (The negative sign is because the load is acting downwards, and we are considering the section from the free end)

• Step 4: Develop Bending Moment Equations

  • M(x) = - (w * x) * (x/2) = -wx²/2 (Quadratically increasing negative bending moment) (Again, negative because it causes tension on the top fibers)

• Step 5: Plot the Diagrams

  • Shear Force Diagram: A linear increase from 0 at x = 0 to -wL at x = L (the fixed end).
  • Bending Moment Diagram: A quadratic increase from 0 at x = 0 to -wL²/2 at x = L (the fixed end).

• Step 6: Identify Key Values

  • Maximum Shear Force: wL (at the fixed end)
  • Maximum Bending Moment: wL²/2 (at the fixed end)

Example 3: Overhanging Beam with a Concentrated Load

• Description: An overhanging beam is supported at points B and C, and has a length AB that overhangs support B. A concentrated load P is applied at the free end A. The length of AB is 'a' and the length of BC is 'L'.

• Step 1: Determine Support Reactions

  • Sum moments about point C: R_B * L - P * (L + a) = 0 => R_B = P * (L + a) / L
  • Sum of vertical forces: R_B + R_C - P = 0 => R_C = P - R_B = P - P * (L + a) / L = -Pa/L

• Step 2: Establish Sections

  • Section 1: 0 ≤ x < a (from A to B)
  • Section 2: 0 ≤ x < L (from B to C, using a new 'x' starting at B)

• Step 3: Develop Shear Force Equations

  • Section 1: V(x) = -P (Constant negative shear force)
  • Section 2: V(x) = -P + R_B = -P + P(L+a)/L = Pa/L (Constant positive shear force)

• Step 4: Develop Bending Moment Equations

  • Section 1: M(x) = -P * x (Linearly increasing negative bending moment)
  • Section 2: M(x) = -P * (a + x) + R_B * x = -Pa - Px + P(L+a)x/L = -Pa + (Pa/L)x

• Step 5: Plot the Diagrams

  • Shear Force Diagram: Constant -P from A to B, then jumps to Pa/L and remains constant from B to C.
  • Bending Moment Diagram: Linearly increasing negative moment from 0 at A to -Pa at B. Then a linearly increasing (less negative or positive) moment from -Pa at B to 0 at C.

• Step 6: Identify Key Values

  • Maximum Shear Force: P (in the overhanging portion)
  • Maximum Bending Moment: Pa (at support B)

Example 4: Simply Supported Beam with a Linearly Varying Load

• Description: A simply supported beam of length L is subjected to a linearly varying load, with the load intensity increasing from 0 at support A to w at support B.

• Step 1: Determine Support Reactions

  • Total Load: W = (1/2) * L * w = wL/2
  • The centroid of the triangular load is located at 2L/3 from A.
  • Sum moments about A: R_B * L - (wL/2) * (2L/3) = 0 => R_B = wL/3
  • Sum of vertical forces: R_A + R_B - wL/2 = 0 => R_A = wL/2 - wL/3 = wL/6

• Step 2: Establish Sections

  • One section is sufficient: 0 ≤ x ≤ L

• Step 3: Develop Shear Force Equations

  • The load intensity at any point x is: w(x) = w * (x/L)
  • The total load up to point x is: W(x) = (1/2) * x * w(x) = (1/2) * x * (wx/L) = wx²/(2L)
  • V(x) = R_A - W(x) = wL/6 - wx²/(2L)

• Step 4: Develop Bending Moment Equations

  • M(x) = R_A * x - W(x) * (x/3) = (wL/6) * x - (wx²/(2L)) * (x/3) = wLx/6 - wx³/(6L)

• Step 5: Plot the Diagrams

  • Shear Force Diagram: A parabolic curve decreasing from wL/6 at x=0 to -wL/3 at x=L.
  • Bending Moment Diagram: A cubic curve, starting at 0 at x=0, reaching a maximum value, and then decreasing back to 0 at x=L.

• Step 6: Identify Key Values

  • To find the location of maximum bending moment, set dM/dx = 0:
    • dM/dx = wL/6 - wx²/(2L) = 0 => x² = L²/3 => x = L/√3 ≈ 0.577L
  • Maximum Bending Moment: Substitute x = L/√3 into the M(x) equation:
    • M_max = (wL/6) * (L/√3) - (w(L/√3)³)/(6L) = wL²/(9√3)

Example 5: Beam with a Combination of Loads

• Description: A simply supported beam of length L has a concentrated load P at L/4 from the left support and a UDL of w over the remaining 3L/4 length.

• Step 1: Determine Support Reactions

  • Total load from UDL: (3L/4) * w = 3wL/4
  • Sum moments about A: R_B * L - P * (L/4) - (3wL/4) * (L/4 + 3L/8) = 0
  • R_B * L - PL/4 - (3wL/4) * (5L/8) = 0
  • R_B = P/4 + 15wL/32
  • Sum of vertical forces: R_A + R_B - P - 3wL/4 = 0
  • R_A = P + 3wL/4 - R_B = P + 3wL/4 - P/4 - 15wL/32 = 3P/4 + 9wL/32

• Step 2: Establish Sections

  • Section 1: 0 ≤ x < L/4
  • Section 2: L/4 < x ≤ L

• Step 3: Develop Shear Force Equations

  • Section 1: V(x) = R_A = 3P/4 + 9wL/32
  • Section 2: V(x) = R_A - P - w(x - L/4) = 3P/4 + 9wL/32 - P - wx + wL/4 = -P/4 + 17wL/32 - wx

• Step 4: Develop Bending Moment Equations

  • Section 1: M(x) = R_A * x = (3P/4 + 9wL/32) * x
  • Section 2: M(x) = R_A * x - P(x - L/4) - w(x - L/4)² / 2 = (3P/4 + 9wL/32)x - Px + PL/4 - w(x² - Lx/2 + L²/16)/2 = (-P/4 + 9wL/32)x + PL/4 - wx²/2 + wLx/4 - wL²/32 = (-P/4 + 17wL/32)x + PL/4 - wx²/2 - wL²/32

• Step 5: Plot the Diagrams

  • Shear Force Diagram: The shear force diagram will have a constant positive value in the first section, then a jump down due to the concentrated load, followed by a linearly decreasing shear force due to the UDL.
  • Bending Moment Diagram: The bending moment diagram will have a linear increase in the first section, and a curved shape (quadratic) in the second section.

• Step 6: Identify Key Values

  • The location of maximum bending moment will occur where the shear force is zero in the second section. Solve V(x) = 0 for x in Section 2's shear equation.
    • -P/4 + 17wL/32 - wx = 0 => x = (-P/4 + 17wL/32) / w = -P/(4w) + 17L/32
  • Substitute the calculated x into the bending moment equation for Section 2 to find the maximum bending moment.

Practical Applications

Shear force and bending moment diagrams are crucial in:

• Structural Design: Determining the maximum shear forces and bending moments to select appropriate beam sizes and materials.
• Stress Analysis: Identifying critical sections where stresses are highest.
• Deflection Calculations: Used in conjunction with material properties to predict beam deflections under load.
• Failure Analysis: Understanding the potential failure modes of beams under different loading scenarios.
• Bridge Design: Analyzing the effects of moving loads on bridge structures.
• Machine Design: Designing machine elements such as shafts and axles.

Software Tools

While manual calculation is important for understanding the fundamentals, numerous software tools are available to automate the process of generating shear force and bending moment diagrams:

• AutoCAD Structural Detailing
• SAP2000
• ETABS
• RISA
• SkyCiv

These tools allow for complex loading scenarios and provide detailed analysis reports.

Conclusion

Understanding shear force and bending moment diagrams is paramount for any structural engineer or anyone involved in structural design and analysis. These diagrams provide a visual representation of the internal forces and moments within a structural member, enabling engineers to ensure structural integrity, safety, and efficiency. The examples presented here illustrate various load conditions and provide a foundation for analyzing more complex structural systems. While software tools can automate the process, a solid understanding of the fundamental principles is essential for accurate interpretation and informed decision-making.