Shear Forces and Bending Moments in Beams

Prerequisite knowledge: force equilibrium, SI dimensional analysis, algebra and first-year calculus.

Learning objectives: state a model and its validity range, calculate physically meaningful engineering quantities, reconcile equations with a free-body diagram or deformation compatibility, and identify at least two failure modes excluded by the idealization.

Unit context and engineering interpretation

Beam diagrams describe internal force resultants, not displacements. A beam can have a large bending moment yet small deflection if its flexural rigidity is high. Conversely, a low moment in a flexible long span can produce appreciable serviceability deformation. The first task in any beam problem is identifying correct supports and externally applied loads, including their units. A roller supplies a reaction force normal to its travel surface; an ideal pin supplies forces but not a moment; a built-in end may supply a moment as well. Real supports lie between these extremes.

A localized downward force produces a step in shear equal to its magnitude. A concentrated couple produces a jump in moment. Distributed forces determine diagram slopes, and the areas under w(x) and V(x) determine changes in shear and moment. These integral relationships create a powerful consistency check: the change in V over an interval must equal the negative total applied distributed load in that interval, and the change in moment must equal the integral of shear over that interval. A computational diagram should be subjected to these checks, not merely admired.

Live-load envelopes add a second layer of reasoning. A roadway beam might experience a vehicle in many possible locations. The largest midspan moment and the largest support shear may occur under different vehicle placements. Optimizing one load case does not automatically envelope all demands. For students, moving a point load in the simulator reveals this distinction. For professionals, code-prescribed load patterns, construction staging and system continuity must be incorporated in the structural model.

Lesson 5.1 — Beam loads, supports and sign conventions

Physical mechanism and governing idea

A beam supports transverse loading through shears and bending moments. External loads and reactions must balance before internal functions are calculated. A distributed load has units of force per length, so its resultant force is the integral of intensity along the beam. The diagrams depend on location and support restraint, not just the total applied force.

Governing relationship

dV/dx=-w(x); dM/dx=V(x) under a consistent sign convention

Reading the equation: Start from a properly oriented free body or a measured constitutive response, identify the force or displacement controlled externally, and decide which geometric quantity belongs in the denominator. Check units before substituting numbers, and compare the sign of the result with the physical action.

Worked interpretive example

A simply supported six-meter beam with uniform load 3 kN/m carries 18 kN total, producing 9 kN reactions at each end.

What the simplified formula misses

The same total force applied as a point load at an end gives very different internal diagrams. A reaction is not automatically half the total for unsymmetric loading.

Engineering reasoning exercise

1. Explain in your own words which observable physical quantity the formula in Lesson 5.1 is estimating. Specify where it would be measured.
2. Change exactly one of the underlying dimensions or loading parameters, predict the direction of change before calculating, and say which other variables are held constant.
3. Invent a connection, defect or boundary condition under which the approximation in this lesson becomes unreliable. Suggest the simplest more faithful model.
4. State a dimensional consistency check and one limiting case that should hold if the stated equations are applied correctly.

Lesson 5.2 — Concentrated loads and discontinuities

Physical mechanism and governing idea

A concentrated vertical force introduces a step in the shear-force diagram. The bending moment remains continuous across a simple point force in Euler-Bernoulli beam theory. A concentrated couple instead introduces a moment jump. These graphical features are strong debugging checks for a computer-generated diagram.

Governing relationship

V(x+) - V(x-) = -P for downward point load

Reading the equation: Start from a properly oriented free body or a measured constitutive response, identify the force or displacement controlled externally, and decide which geometric quantity belongs in the denominator. Check units before substituting numbers, and compare the sign of the result with the physical action.

Worked interpretive example

A 12 kN downward point load causes a 12 kN downward jump in V at its application point.

What the simplified formula misses

Drawing a jump in bending moment at an ordinary point force is a sign-convention or free-body mistake.

Engineering reasoning exercise

1. Explain in your own words which observable physical quantity the formula in Lesson 5.2 is estimating. Specify where it would be measured.
2. Change exactly one of the underlying dimensions or loading parameters, predict the direction of change before calculating, and say which other variables are held constant.
3. Invent a connection, defect or boundary condition under which the approximation in this lesson becomes unreliable. Suggest the simplest more faithful model.
4. State a dimensional consistency check and one limiting case that should hold if the stated equations are applied correctly.

Lesson 5.3 — Distributed loading and diagram shape

Physical mechanism and governing idea

The slope of shear equals negative load intensity, while the slope of bending moment equals shear. Uniform loading therefore creates a straight shear segment and a quadratic moment segment. A maximum or minimum smooth moment occurs where shear passes through zero, as long as no applied couple creates a jump.

Governing relationship

V(x)=R_A - integral_0^x w(s) ds; M(x)=integral_0^x V(s) ds

Reading the equation: Start from a properly oriented free body or a measured constitutive response, identify the force or displacement controlled externally, and decide which geometric quantity belongs in the denominator. Check units before substituting numbers, and compare the sign of the result with the physical action.

Worked interpretive example

For uniform load w on a simply supported beam, Mmax=wL²/8 at midspan.

What the simplified formula misses

The differential identities have convention-dependent signs. A diagram should be checked by returning to the original free-body equilibrium.

Engineering reasoning exercise

1. Explain in your own words which observable physical quantity the formula in Lesson 5.3 is estimating. Specify where it would be measured.
2. Change exactly one of the underlying dimensions or loading parameters, predict the direction of change before calculating, and say which other variables are held constant.
3. Invent a connection, defect or boundary condition under which the approximation in this lesson becomes unreliable. Suggest the simplest more faithful model.
4. State a dimensional consistency check and one limiting case that should hold if the stated equations are applied correctly.

Lesson 5.4 — Load envelopes and structural reasoning

Physical mechanism and governing idea

Real beams experience multiple load combinations and moving load positions. A single nominal diagram is only one scenario. Load envelopes capture maximum and minimum internal demands across specified combinations, while code factors and pattern loading remain separate decisions. A safe design begins with correct demand definitions and relevant limit states.

Governing relationship

M_envelope(x)=max over admissible load cases of |M_case(x)|

Reading the equation: Start from a properly oriented free body or a measured constitutive response, identify the force or displacement controlled externally, and decide which geometric quantity belongs in the denominator. Check units before substituting numbers, and compare the sign of the result with the physical action.

Worked interpretive example

Moving a concentrated load toward a support reduces its maximum simple-span moment even when the load magnitude is unchanged.

What the simplified formula misses

An envelope over only the cases you chose says nothing about omitted traffic positions, construction stages or redistribution due to cracking.

Engineering reasoning exercise

1. Explain in your own words which observable physical quantity the formula in Lesson 5.4 is estimating. Specify where it would be measured.
2. Change exactly one of the underlying dimensions or loading parameters, predict the direction of change before calculating, and say which other variables are held constant.
3. Invent a connection, defect or boundary condition under which the approximation in this lesson becomes unreliable. Suggest the simplest more faithful model.
4. State a dimensional consistency check and one limiting case that should hold if the stated equations are applied correctly.

Worked numerical casebook

The following 18 worked cases are intended for dimensional fluency and model interpretation, not as a substitute for unique field measurements. Cases change materials, loads and geometry and deliberately restate their validity conditions.

When reproducing a result, document a complete free body or section cut, record the input units, and calculate at least one quantity without looking at the worked result.

Case 5.01: Point-force and distributed-load reactions in a girder

A simply supported beam of length 4 m carries a 20 kN point load at x=2.00 m and a full-span uniform load 0 kN/m. Moment equilibrium about the left end gives RB=10.00 kN, and force equilibrium gives RA=10.00 kN. Just under the point load, the bending-moment expression yields M(a)=20.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.01: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.02: Point-force and distributed-load reactions in a roof tie

A simply supported beam of length 6 m carries a 16 kN point load at x=1.20 m and a full-span uniform load 1.5 kN/m. Moment equilibrium about the left end gives RB=7.70 kN, and force equilibrium gives RA=17.30 kN. Just under the point load, the bending-moment expression yields M(a)=19.68 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.02: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.03: Point-force and distributed-load reactions in a machine anchor

A simply supported beam of length 3 m carries a 20 kN point load at x=0.60 m and a full-span uniform load 4 kN/m. Moment equilibrium about the left end gives RB=10.00 kN, and force equilibrium gives RA=22.00 kN. Just under the point load, the bending-moment expression yields M(a)=12.48 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.03: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.04: Point-force and distributed-load reactions in a laboratory frame

A simply supported beam of length 5 m carries a 12 kN point load at x=3.00 m and a full-span uniform load 1.5 kN/m. Moment equilibrium about the left end gives RB=10.95 kN, and force equilibrium gives RA=8.55 kN. Just under the point load, the bending-moment expression yields M(a)=18.90 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.04: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.05: Point-force and distributed-load reactions in a small footbridge

A simply supported beam of length 4 m carries a 8 kN point load at x=2.00 m and a full-span uniform load 4 kN/m. Moment equilibrium about the left end gives RB=12.00 kN, and force equilibrium gives RA=12.00 kN. Just under the point load, the bending-moment expression yields M(a)=16.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.05: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.06: Point-force and distributed-load reactions in a service gantry

A simply supported beam of length 8 m carries a 24 kN point load at x=2.00 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=18.00 kN, and force equilibrium gives RA=30.00 kN. Just under the point load, the bending-moment expression yields M(a)=54.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.06: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.07: Point-force and distributed-load reactions in a construction prop

A simply supported beam of length 3 m carries a 20 kN point load at x=1.80 m and a full-span uniform load 2 kN/m. Moment equilibrium about the left end gives RB=15.00 kN, and force equilibrium gives RA=11.00 kN. Just under the point load, the bending-moment expression yields M(a)=16.56 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.07: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.08: Point-force and distributed-load reactions in a testing rig

A simply supported beam of length 5 m carries a 16 kN point load at x=1.25 m and a full-span uniform load 2 kN/m. Moment equilibrium about the left end gives RB=9.00 kN, and force equilibrium gives RA=17.00 kN. Just under the point load, the bending-moment expression yields M(a)=19.69 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.08: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.09: Point-force and distributed-load reactions in a industrial platform

A simply supported beam of length 4 m carries a 24 kN point load at x=1.60 m and a full-span uniform load 4 kN/m. Moment equilibrium about the left end gives RB=17.60 kN, and force equilibrium gives RA=22.40 kN. Just under the point load, the bending-moment expression yields M(a)=30.72 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.09: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.10: Point-force and distributed-load reactions in a equipment mount

A simply supported beam of length 5 m carries a 20 kN point load at x=2.00 m and a full-span uniform load 1.5 kN/m. Moment equilibrium about the left end gives RB=11.75 kN, and force equilibrium gives RA=15.75 kN. Just under the point load, the bending-moment expression yields M(a)=28.50 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.10: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.11: Point-force and distributed-load reactions in a maintenance bracket

A simply supported beam of length 5 m carries a 8 kN point load at x=2.50 m and a full-span uniform load 0 kN/m. Moment equilibrium about the left end gives RB=4.00 kN, and force equilibrium gives RA=4.00 kN. Just under the point load, the bending-moment expression yields M(a)=10.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.11: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.12: Point-force and distributed-load reactions in a temporary support

A simply supported beam of length 4 m carries a 6 kN point load at x=2.80 m and a full-span uniform load 0 kN/m. Moment equilibrium about the left end gives RB=4.20 kN, and force equilibrium gives RA=1.80 kN. Just under the point load, the bending-moment expression yields M(a)=5.04 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.12: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.13: Point-force and distributed-load reactions in a pipe rack

A simply supported beam of length 6 m carries a 16 kN point load at x=3.60 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=18.60 kN, and force equilibrium gives RA=15.40 kN. Just under the point load, the bending-moment expression yields M(a)=36.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.13: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.14: Point-force and distributed-load reactions in a solar panel rail

A simply supported beam of length 5 m carries a 8 kN point load at x=2.50 m and a full-span uniform load 0 kN/m. Moment equilibrium about the left end gives RB=4.00 kN, and force equilibrium gives RA=4.00 kN. Just under the point load, the bending-moment expression yields M(a)=10.00 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.14: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.15: Point-force and distributed-load reactions in a transit shelter

A simply supported beam of length 5 m carries a 12 kN point load at x=3.00 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=14.70 kN, and force equilibrium gives RA=12.30 kN. Just under the point load, the bending-moment expression yields M(a)=23.40 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.15: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.16: Point-force and distributed-load reactions in a lightweight crane hook

A simply supported beam of length 6 m carries a 24 kN point load at x=3.60 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=23.40 kN, and force equilibrium gives RA=18.60 kN. Just under the point load, the bending-moment expression yields M(a)=47.52 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.16: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.17: Point-force and distributed-load reactions in a water-tank frame

A simply supported beam of length 8 m carries a 6 kN point load at x=1.60 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=13.20 kN, and force equilibrium gives RA=16.80 kN. Just under the point load, the bending-moment expression yields M(a)=23.04 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.17: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Case 5.18: Point-force and distributed-load reactions in a portal-frame member

A simply supported beam of length 6 m carries a 20 kN point load at x=1.20 m and a full-span uniform load 3 kN/m. Moment equilibrium about the left end gives RB=13.00 kN, and force equilibrium gives RA=25.00 kN. Just under the point load, the bending-moment expression yields M(a)=27.84 kN·m. To find the global maximum, also check where V changes sign and the span endpoints; the point-load coordinate is not automatically the maximum when w is present.

Extension task 5.18: Replace one assumed boundary condition with a compliant or imperfect connection. Which input would you need to measure, and which claim above would cease to be justified?

Engineering case study: Temporary construction loading on a floor beam

During renovation, material pallets are placed close to one end of a simply supported beam. A uniform load approximation may miss concentrated reaction and moment peaks. The engineer should represent the actual pallet footprint or applied point-force idealization, solve support reactions and plot piecewise shear and bending moments. Repositioning the load changes support demands and bending behavior even when total weight remains unchanged.

Deliverable: Develop two loading cases: one centered and one near a support. Compare reactions, maximum moment location and the shear jump. State why this demonstration cannot replace a building-code load combination and restraint investigation.

Case-study discussion

Start by differentiating data from assumptions: measured loads and geometry should be separated from guessed restraints and material properties. Build one simple calculation whose result can be checked independently. List at least three mechanisms by which the idealization could fail, and prioritize them by whether they would change the engineering conclusion. Translate the numerical demand into a practical question about safety, serviceability, testing or maintenance. Keep a clear audit trail so a reviewer can follow the unit conversions, load path and model limitations.

Calculation verification

• Differentiate your bending function to recover the shear function away from concentrated loads.
• Verify RA + RB equals total applied downward load and that the end moments vanish for simple supports.
• Check positivity, zero-load behavior, symmetry, and asymptotic scaling whenever applicable.
• Do not use a graph or slider response as the only evidence that an equation is correct.

Structured unit assessment

Answer the following without access to the provided worked solutions. Each question needs a diagram, assumptions and a clear interpretation.

1. Use a load or material value from Case 5.02, change one geometric dimension by 20%, and calculate the resulting response with explicit units.
2. Explain how the result of Lesson 5.1 would change if one apparently rigid support had finite stiffness.
3. Choose a nonuniform or damaged member for which the idealization in Lesson 5.2 becomes inaccurate; describe a better analytical or experimental representation.
4. Derive a useful limiting case from the relationship in Lesson 5.3 and compare with a limiting setting in any associated simulation.
5. A reviewer reports a one-thousandfold discrepancy. Show one plausible N–kN, m–mm or N·m–N·mm conversion error that could create it for a calculation in this unit.
6. For the case study in Unit 5, distinguish two possible failure modes and explain why demonstrating adequate performance against one does not guarantee the other.
7. Create an experimental plan to validate one of the numerical predictions in Unit 5; name the sensor, its calibration requirement and an expected source of uncertainty.
8. State two ways that manufacturing tolerance, support flexibility or temperature could alter a result computed from one equation in Unit 5.

Associated interactive laboratories

Laboratory 04: Beam Shear–Moment Designer. Equilibrium gives reactions; a virtual cut through the beam yields shear and bending moment. A concentrated downward load produces a shear step, whereas uniform loading makes V linear and M quadratic.

Investigation: Set the distributed load to zero and move the point load from midspan to one quarter-span. Predict reaction changes and locate the largest positive bending moment.

Open the [working HTML model](../labs/lab-04-beamvm.html), record inputs and CSV data, and compare numerical outputs with an independent calculation. Identify the limiting assumptions before discussing how a real physical experiment would differ.

Unit research project

Prepare a technical notebook on Shear forces and bending moments in beams for a civil engineering structure of your choice. Include a drawing with geometric dimensions, a carefully numbered input table and a free-body diagram. Develop two load or restraint scenarios, compute the most relevant stress or deformation variables, cross-check them with an independent limiting case, and identify all omissions that would prevent formal design approval. Where an associated laboratory exists, export a CSV series and include an annotated plot. Write a short conclusion that distinguishes what your simplified model genuinely predicts from what would need a professional code-based check.

Project marking rubric