CE 312 Lecture 16: Shear/Moment Diagrams I, Internal V/M (2026.09.23)
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Overview
Gregory Michaelson introduces internal shear and bending moment analysis as the basis for constructing diagrams used to design beams, then establishes the sign conventions and section-cut method needed to calculate internal forces. A worked beam section gives V = 15.8 kips and M = 309.6 kip-ft, while a differential-element derivation connects distributed load, shear, and moment to the graph shapes used in later lectures.
Key takeaways
- For a linear structure, multiplying every applied load by a factor multiplies every response by that factor; scaling only one load among several does not scale the full response uniformly.
- The chosen internal-force convention is downward positive shear and counterclockwise positive moment on a left-hand cut, with opposite directions on a right-hand cut.
- At the section 12 feet from A, the left segment's equilibrium gives V = 15.8 kips and M = 309.6 kip-ft.
- When a cut passes through a distributed load, calculate the resultant from only the retained portion; the illustrated partial load has a 13-kip resultant rather than 20 kips.
- For a beam element, dV = w dx and dM = V dx, linking load intensity to shear change and shear to moment change.
- Diagram shape can be inferred from these relationships: constant shear gives a linear moment segment, while linear shear gives a quadratic moment curve.
Chapters
- For a structure with a single 20-kip load, scaling the load scales every reaction and member force by the same factor.
- Michaelson notes that a virtual analysis can be obtained by dividing real-structure results by the applied load when only one load is present.
- Influence lines will apply the same linearity idea by moving a unit load across a structure and recording its response.
- Shear and moment diagrams plot the internal forces and moments that keep a beam in equilibrium.
- Design decisions such as beam depth, concrete rebar placement, and steel flange and web dimensions depend on maximum internal shear and moment.
- A concentrated force creates a jump in the shear diagram, while a concentrated moment creates a jump in the moment diagram.
- On a left-hand section cut, positive shear points downward and positive moment is counterclockwise.
- On a right-hand section cut, positive shear points upward and positive moment is clockwise.
- The sign convention makes diagram values consistent: negative calculated shear or moment is plotted below the reference axis.
- The reused simply supported beam has a known 35.8-kip reaction at A; the cut is 12 feet from A, leaving a 6-foot segment to the 20-kip point load.
- Using the left segment and summing vertical forces gives an internal shear of V = 15.8 kips.
- Summing moments at the cut gives M = 35.8 × 12 − 20 × 6 = 309.6 kip-ft.
- A right-side cut produces the same positive shear when its own sign convention is applied.
- Replace a 2-kip-per-foot load over 10 feet with its 20-kip resultant at the 5-foot centroid only when analyzing the entire loaded length.
- If a section cut retains only part of that load, recompute the resultant from the retained load area.
- For the illustrated partial segment, the correct resultant is 13 kips, not the original 20 kips.
- Michaelson isolates an infinitesimal beam segment of width dx, treating a variable distributed load w as approximately constant across that tiny interval.
- Vertical-force equilibrium gives dV = w dx, or equivalently the rate of shear change is governed by the load intensity.
- Moment equilibrium, with the negligible higher-order dx-squared term discarded, gives dM = V dx.
- These relationships explain why load, shear, and moment diagrams have linked slopes and shapes.
- The integral of a constant shear is linear moment, so constant shear corresponds to a straight segment in the moment diagram.
- Positive shear produces a positive moment-diagram slope; negative shear produces a negative slope.
- Integrating linear shear produces a quadratic, curved moment segment, and a shear sign change marks a change in the moment curve's direction.
- The next class will develop shear and moment diagram patterns for concentrated loads.
- A prerecorded Monday lecture will cover distributed loads, followed by triangular load distributions.
- A linearly varying triangular load produces a curved shear diagram and a higher-order curved moment diagram.
Transcript excerpt
0:07 Okay. Let's get started. I understand I understand a number of you just got done with the an exam. I'm I'm hoping everything's okay, sensing a little bit of surface tension. I'm sure it was a very high pressure environment. Just uh make sure you take it easy and go with the flow. I'm such a dork, I know. Just wait. I mean like I we're not done with this. Just wait till you know we take a moment and pause and Yeah, I know. Y'all y- Y'all are just getting started with all these really bad cheesy puns. Okay. All right, everybody. Um so, just a
0:55 heads-up, no, I do not have your exam graded. I do have a goal of getting it to you by Friday, mainly for two reasons. As I mentioned that I'm going to be going out of the country next week. So, I would like to kind of have your exam graded by then, but also midterm grades are due on Monday and I'd I'd like to get the exams graded before I post those. So, um I did want to show you something though on the exam. I mentioned it to a couple people uh outside in the hall as we were waiting to for the room to be available. And it's not as if anybody did anything wrong, but there was a thing I did on the exam that I kind of wanted to just illustrate and there's a reason for it that'll become clear later. But I kind of wanted to show you something. So, um
1:42 just as a thought experiment. Okay, so let's say that you have a truss. So, I've got the same truss drawn two different ways. And let's say the truss only has this single 20 kip load on it.
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Related notes
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Gregory Michaelson.