In this chapter, we seek to calculate the stress from bending which is caused by bending moments. As such, constructing the SF and BM diagrams is necessary to obtain the BM magnitude along the beam, so that we can calculate the resulting bending stress.
SF and BM diagrams were originally covered in Chapter 5.3 of Statics, so we will only give you a brief summary on how to construct SFBM diagrams here. You can refer to the original Chapter 5.3 for the full explanation.
Basically, there are two methods to construct SF and BM diagrams:
In this chapter, we seek to calculate the stress from bending which is caused by bending moments. As such, constructing the SF and BM diagrams is necessary to obtain the BM magnitude along the beam, so that we can calculate the resulting bending stress.
SF and BM diagrams were originally covered in Chapter 5.3 of Statics, so we will only give you a brief summary on how to construct SFBM diagrams here. You can refer to the original Chapter 5.3 for the full explanation.
Basically, there are two methods to construct SF and BM diagrams:
Basically this method works by directly constructing the SF diagram using the FBD, and BM diagram using both the SF diagram and FBD.
Two key relationships for this method are as follows:
The relationships basically say that the gradient of the SF diagram is equal to the –ve of the distributed load, while the gradient of the BM diagram is equal to the SF:
We also present to you a few examples of how different forces acting on a beam are represented in SF and BM diagrams:
If this all seem a little confusing to you, don't worry as we will be working through an example soon. But first, let's brush up our memory on sign conventions.
We touched on sign conventions for SFBM diagrams before in Chapter 5.2 of Statics. The positive directions for the normal force N, shear force V and bending moment M are shown below:
Let’s look at an example now to brush up your memory. If you need more questions to practice, there are more questions here.
Basically this method works by directly constructing the SF diagram using the FBD, and BM diagram using both the SF diagram and FBD.
Two key relationships for this method are as follows:
The relationships basically say that the gradient of the SF diagram is equal to the –ve of the distributed load, while the gradient of the BM diagram is equal to the SF:
We also present to you a few examples of how different forces acting on a beam are represented in SF and BM diagrams:
If this all seem a little confusing to you, don't worry as we will be working through an example soon. But first, let's brush up our memory on sign conventions.
We touched on sign conventions for SFBM diagrams before in Chapter 5.2 of Statics. The positive directions for the normal force N, shear force V and bending moment M are shown below:
Let’s look at an example now to brush up your memory. If you need more questions to practice, there are more questions here.