6. Bending Moment and Shear Force
Bending moment and shear force analysis plays a critical role in solid mechanics by enabling the evaluation of the internal forces and stresses within a structural element subjected to external loads. These forces can arise from various external loads applied to the structure, including forces, moments, and distributed loads, and their distribution and magnitude must be understood to design and analyze various structures like beams, frames, and trusses. The calculation of bending moment and shear force allows engineers and designers to ensure that a structure can withstand the loads it is subjected to and prevent failure due to excessive stresses.
In mechatronics engineering, bending moment and shear force analysis are especially important for the design and analysis of mechanical components such as frames, beams, and joints. Structural components that support moving parts in mechatronic systems like robotic arms and CNC machines must be designed to withstand both external loads and internal forces and stresses caused by the motion and weight of the components. By analyzing the bending moment and shear force in these components, engineers can determine the ideal dimensions, materials, and shapes to ensure that they meet the necessary strength and durability requirements.
Additionally, mechatronic systems often involve the integration of various components like sensors, actuators, and control systems, which can affect the mechanical behavior of the system. Incorporating the principles of solid mechanics into the design and analysis of mechatronic systems enables engineers to ensure that the various components of the system work together in a coordinated and efficient manner, leading to a safe and reliable operation of the system. Overall, the study of bending moment and shear force in solid mechanics is critical in designing and analyzing the mechanical components of mechatronic systems, ensuring their optimal performance, and ultimately leading to better-designed products.
6.1. Types of Beam Supports and Configurations
6.1.1. Types of Beam Supports
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Pinned support: A pinned support, also known as a hinge, allows rotation but has no vertical or horizontal displacement. It can be thought of as a pivot point that allows the beam to rotate freely. The reaction force at a pinned support will have both vertical and horizontal components but no moment.
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Roller support: A roller support allows rotation and horizontal displacement but not vertical displacement. It can be thought of as a wheel or a roller that allows the beam to move horizontally but prevents it from moving vertically. The reaction force at a roller support will have only a horizontal component.
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Fixed support: A fixed support, also known as rigid support, does not allow rotation or any displacement. It can be thought of as a wall or a floor that prevents the beam from moving or rotating. The reaction force at a fixed support will have both vertical and horizontal components as well as a moment.
6.1.2. Types of Beam Configurations
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Cantilever beam: A cantilever beam is fixed at one end and free at the other. It is supported at only one end and can move and deform freely at the other end.
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Simply supported beam: A simply supported beam is supported at two points, typically at each end of the beam. The support at each end is usually a pinned or a roller support.
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Overhanging beam: An overhanging beam is a beam that extends beyond its supports on one or both sides. The overhanging part of the beam is not supported by any other means, and its load is transferred to the supports.
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Rigidly fixed beam or built-in-beam: A rigidly fixed beam, also known as a built-in-beam, is a beam that is fixed at both ends and cannot rotate or displace. It is typically used in structures that require high stiffness and stability.
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Continuous beam: A continuous beam is a beam that is supported by three or more supports, with at least one of them being a fixed support. The supports are usually evenly spaced, and the beam is typically continuous over the supports.
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The main similarity between types of supports and configurations is that they both affect the behavior of a beam under load. The choice of beam support and configuration can significantly impact the strength, stiffness, and deformation of the beam. The main difference between types of supports and configurations is in their function and location. Beam supports are typically located at the endpoints or along the length of the beam, while beam configurations relate to the arrangement of the beam and the way it is loaded. Moreover, while there are only a few types of beam supports, there are many types of beam configurations that can be used depending on the specific application. Depending on the type of support, the reaction to the applied force will be different. If a certain degree of freedom is restrained, there will be a reaction force at that point. For example, for pinned support, rotations are allowed, so there is no reaction moment, but there will be vertical and horizontal reaction forces since displacement in these directions is prevented. Understanding the various types of supports is important when designing and analyzing structures, as it helps to ensure that the beam will be able to support the external loads it is subjected to without failing due to excessive stresses. |
6.2. Components of Internal Forces
When a beam is loaded, internal forces develop to maintain equilibrium. There are two components of internal forces:
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Vertical forces (shear): A shear force is a force that acts perpendicular to the longitudinal axis of the beam, resulting in deformation of the shape of the beam along its length. Shear force causes the beam to bend or twist and is measured in Newtons or pounds.
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Horizontal/normal forces (moment): A moment is a force that acts perpendicular to the plane of the cross-section of the beam, causing it to bend. Moment is measured in Newton-meters or foot-pounds.
6.3. Ways to Load a Beam
A beam can be loaded in several ways, including:
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Concentrated forces: A concentrated force is a force that acts on a small area of the beam’s cross-section. Examples include point loads or external forces applied at a specific location on the beam.
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Distributed forces: A distributed force is a force that acts over a larger area of the beam’s cross-section. Examples include uniform loads that act on the entire length of the beam, such as uniformly distributed loads, linearly varying loads, or partially distributed loads that act on only a portion of the beam, and so on.
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Concentrated moments: A concentrated moment is a moment that is applied at a specific location on the beam, resulting in the deformation of the beam’s shape at that location.
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In this resource, we may not explicitly consider the weight of the beam itself in the calculations of bending moment and shear force. However, it is important to note that the weight of the beam can have a significant effect on the internal forces and stresses within the beam, particularly in cases where the beam is relatively long or heavy. In some cases, the weight of the beam may be neglected if it is relatively small compared to the other external loads acting on the beam. In general, it is good engineering practice to consider all of the external loads acting on a structure, including the weight of the beam itself, in order to ensure that the structure is designed to withstand all of the loads it is subjected to. |
6.4. Examples of Beam Configurations
6.4.1. Cantilever and a Point Load
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Example 1 |
Figure 8 shows a cantilever beam of span \(2.5 m\) carrying point loads. (i) Sketch the shear force and bending moment diagrams. |
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Figure 8. Cantilever point load.
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Solution:
Figure 9. Cantilever point load.
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6.4.2. Cantilever and a Linearly Varying Load
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Example 2 |
Figure 10 shows a cantilever beam AB of \(3 m\) span is subjected to a gradually varying load from \(W_1 = 5.5 kN / m\) to \(W_2 = 2.5 kN / m\). (i) Sketch the shear force and bending moment diagrams. |
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Figure 10. Gradually varying load.
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Solution:
Figure 11. Gradually varying load.
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6.4.3. Simply Supported Beam and a Point Load
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Example 3 |
A simply supported beam AD of span \(4 m\) is carrying two point loads as shown in Figure 12. \(W_1 = 2.5kN\), \(W_2 = 5kN\), \(x_2=x_3=1.5 m\) (i) Sketch the shear force and bending moment diagrams. |
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Figure 12. Simply supported beam.
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Solution:
Figure 13. Simply supported beam.
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6.4.4. Simply Supported Beam and a Uniformly Distributed Load
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Example 4 |
In Figure Figure 14, simply supported beam \(7 m\) long is carrying a uniformly distributed load of \(5.5 kN / m\) over a length of \(3.5 m\) from the right end. Draw the shear force and bending moment diagrams for the beam and also calculate the maximum bending moment on the section. |
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Figure 14. Cantilever point load.
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Solution:
Figure 15. Cantilever point load.
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Example 5 |
A simply supported beam \(5 m\) long is loaded with a uniformly distributed load of \(11 kN / m\) over a length of \(2.5 m\) as shown in Figure 16. Draw shear force and bending moment diagrams for the beam indicating the value of maximum bending moment. |
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Figure 16. Simply supported uniformly distributed load.
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Solution:
Figure 17. Simply supported uniformly distributed load.
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Example 6 |
A simply supported beam of \(4.5 m\) span is carrying loads as shown in Figure 18. Draw shear force and bending moment diagrams for the beam. |
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Figure 18. Simply supported uniformly distributed load.
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Solution:
Figure 19. Simply supported uniformly distributed load.
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6.4.5. Pinned and Roller Supports
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Example 7 |
Figure 20 shows a beam with a roller support. Use graphical method to sketch shear force and bending moment diagrams. Show all the calculations involved. |
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Figure 20. Roller support.
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Solution:
Figure 21. Roller support.
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6.4.6. Overhanging beams
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Example 8 |
An overhanging beam ABC is loaded as shown in Figure 22. Draw the shear force and bending moment diagrams and find the point of contraflexure, if any. |
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Figure 22. Overhang uniformly distributed point load.
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Solution:
Figure 23. Overhang uniformly distributed point load.
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Example 9 |
A beam ABCD, \(5 m\) long is overhanging by \(1.5 m\) and carries loads as shown in Figure 24. \(W_1 = 2.5 kN/m\), \(W_2 = 4.5 kN\), \(x_1=x_2 = 1.5 m\), \(x_3=2m\). (i) Find the reaction forces and sketch the shear force and bending moment diagrams. |
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Figure 24. Overhang uniformly distributed point load.
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Solution:
Figure 25. Overhang uniformly distributed point load.
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Example 10 |
Figure 26 shows an overhang beam subjected to point and uniformly distributed loads. \(W_1=1000N/m\), \(W_2=7000N\), \(W_3=4000N\), \(W_4=1500N/m\), \(x_1=x_3=4m\), \(x_2=x_4=3m\) (i) Find the reaction forces and sketch the shear force and bending moment diagrams. Indicate the numerical values at all important sections. |
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Figure 26. Overhang uniformly distributed point load.
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Solution:
Figure 27. Overhang uniformly distributed point load.
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Comments (7)
Ae we doing chapters 6, 7, 8 & 9?
No. That's for Solid II.
Kindly confirm for us the formula for volumetric strain
Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?
Solid Mechanics II starts from chapter 6.
Could you confirm if in example 1 the shear force diagram is correct since from my shear force calculations the forces are positive meaning the diagram ought to be upwards
Yes, in example 1, the shear force diagram is correct. There are two point loads acting downwards, which tend to shear the bar in the negative direction.