CE 312 Lecture 28: Exam 2 Review (2026.10.23)
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Overview
Gregory Michaelson reviews the CE 312 Exam 2 format and emphasizes support reactions, shear and moment diagrams, and virtual work for beam and frame deflections. A 24-ft simply supported beam example combines a 2 kip/ft distributed load, a 104 ft-kip applied moment, and virtual-work integrals to find a downward deflection of 0.029 inches at the midpoint.
Key takeaways
- For the 24-ft example beam, the 2 kip/ft load over 8 ft becomes a 16-kip resultant at 4 ft from A; moment and force equilibrium give reactions of 9 kips at A and 7 kips at E.
- A concentrated applied moment changes the moment diagram but not the shear diagram; at x = 16 ft, the 104 ft-kip moment creates a jump from -48 to 56 kip-ft.
- The real bending-moment functions are 9x - x^2 on 0 to 8 ft, 64 - 7x on 8 to 16 ft, and 168 - 7x on 16 to 24 ft.
- For midpoint deflection on a simply supported 24-ft beam, a unit virtual load at x = 12 ft produces virtual moments x/2 and (24 - x)/2 on the two half-spans.
- When real and virtual moment functions have different breakpoints, split the virtual-work integral at every breakpoint from either set; this example requires four intervals.
- With E = 4,000 ksi and I = 800 in^4, the worked virtual-work calculation gives a midpoint deflection of 0.029 inches downward.
Chapters
- Exam 2 follows the same format as the previous exam and is expected to fit the allotted time under the rule of three.
- Bring a straight edge for constructing shear and moment diagrams; prepare the formula sheet yourself rather than copying slides or having ChatGPT generate it.
- Support reactions remain necessary despite being covered on Exam 1, while shear and moment diagrams and virtual work are the main topics.
- Be able to find internal shear and moment at a point, derive their functions, and draw diagrams graphically.
- Virtual-work questions may require real and virtual structural analyses to calculate beam or frame displacement and rotation.
- Review collapsing distributed and triangular loads, the three support conditions, and how a cut at a hinge has zero internal moment.
- Use the diagram relationships: load gives the slope of shear, shear gives the slope of moment, and diagram areas give changes in shear or moment.
- The example asks for vertical deflection at midpoint C of a simply supported beam spanning 24 ft.
- A 2 kip/ft uniform load covers the first 8 ft, and a concentrated 104 ft-kip moment acts at x = 16 ft.
- The beam properties are E = 4,000 ksi and I = 800 in^4; the example is designed to combine reactions, diagrams, functions, and virtual work.
- Replace the 2 kip/ft load over 8 ft with a 16-kip resultant located 4 ft from support A.
- Taking moments about A gives 16 × 4 + 104 = Ey × 24, so the right reaction is Ey = 7 kips upward.
- Vertical equilibrium gives Ay + 7 = 16, so Ay = 9 kips upward; the horizontal reaction is zero.
- The shear starts at 9 kips, decreases linearly across the loaded 8-ft region, and reaches -7 kips; the applied moment at x = 16 ft does not change shear.
- The shear crosses zero at x = 4.5 ft and reaches -7 kips at x = 8 ft; its triangular areas are 20.25 and -12.25 kip-ft.
- Integrating shear gives moment values of 20.25 kip-ft at x = 4.5 ft, 8 kip-ft at x = 8 ft, and -48 kip-ft at x = 16 ft before the applied-moment jump.
- The 104 ft-kip applied moment creates an upward jump in the moment diagram from -48 to 56 kip-ft, after which the diagram returns to zero at the right support.
- For the loaded span, a section cut and moment equilibrium give M_AB = 9x - x^2 for 0 ≤ x ≤ 8 ft.
- The constant-shear regions have linear moment functions: M_BD = 64 - 7x for 8 ≤ x ≤ 16 ft, and M_DE = 168 - 7x for 16 ≤ x ≤ 24 ft.
- The functions capture the 104 ft-kip discontinuity at x = 16 ft; the second and third functions give -48 and 56 kip-ft there.
- Plotting the functions or checking points against the diagram verifies the equations and their ranges.
- To find displacement at midpoint C, apply a unit virtual load at x = 12 ft on a 24-ft simply supported beam.
- Symmetry gives virtual reactions of one-half at each support, producing a triangular virtual moment diagram with a peak of 6 ft.
- The virtual moment functions are M1 = x/2 for 0 ≤ x ≤ 12 ft and M2 = (24 - x)/2 for 12 ≤ x ≤ 24 ft.
- Because the real and virtual moment functions change at different locations, divide the beam into four shared intervals: 0 to 8, 8 to 12, 12 to 16, and 16 to 24 ft.
- Pair M1 with the real functions on the first two intervals and M2 with the real functions on the final two intervals.
- The method of virtual work sums the four integrals of the real and virtual moment products, with the stated conversion factor 1728/EI.
- The four integral contributions are 256, -416/3, -1984/3, and 1792/3, summing to 160/3.
- Using E = 4,000 ksi and I = 800 in^4 gives a midpoint deflection of 0.029 inches downward.
- The complete example demonstrates reactions, graphical diagrams with parabolic and linear regions, a concentrated-moment jump, section-based moment functions, and real-plus-virtual analysis.
Transcript excerpt
0:11 [clears throat] [laughter] >> All right, everybody. Hopefully the the turbulence of the last hour has calmed down a little bit. >> [clears throat] >> Look at it this way. Everybody in here for your last exam, it either went well or it was a minor loss, so. You haven't done minor losses yet? AH. I I THOUGHT I HAD A good pun. Alas, it will be a good pun. It'll be one later. Okay. Um So, I actually spent a good bit of time preparing an example for class day. I ended up redoing this example like three
0:58 times cuz my goal was in this one example to cover just about everything that you need to know on this exam and I wanted to make sure I came prepared. So, I spent that time instead of grading or finishing up the grading on submittal one, I focused my energy on that cuz I think that'll be Well, for one it's more critical path and two I I think it like net positive benefit to you right now. I mean, we don't need the project submittal graded until mass hands sort of knocked out, but you have an exam on Monday in here. Um homework 6.3 and 6.4 were due today. Hopefully those went well. Homework's caught up other than homework 6.2. That one's going to take a little while as I imagine you suspect. I still need to update attendance grades, I believe, but
1:47 other than that everything should be rocking and rolling. And with that, we're going to begin our exam reviews. Does anybody have any questions? Did do get the code? I Let me I left my my mouse in the um uh in my office, so hold on. Ah, there we go. Okay. So, let's start off with a really brief
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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.