Shear Force and Bending Moment

Introduction

Beams are structural members designed primarily to carry loads acting transverse to their longitudinal axes. Such loads produce internal shear forces and bending moments, which govern beam strength, deformation, and serviceability. The ability to determine these internal actions is fundamental to the design of bridges, machine frames, shafts, floor systems, crane girders, and aircraft structures.

A beam analysis normally follows this sequence: classify the beam and loading, determine support reactions, cut the beam at selected sections, calculate internal shear force and bending moment, and represent their variation using diagrams. These diagrams provide an immediate picture of critical sections, maximum bending moments, points of contraflexure, and changes caused by concentrated or distributed loads.

This chapter uses static equilibrium and the sign conventions stated below. Unless otherwise noted, beams are assumed to be prismatic, loads are static, supports are idealized, and deformation is sufficiently small that the original geometry may be used for equilibrium calculations.

Learning Objectives

After studying this chapter, the reader should be able to:

• classify common beams, supports, and loading systems;
• draw a complete free-body diagram and calculate support reactions;
• define shear force and bending moment at any beam section;
• construct shear-force and bending-moment diagrams for common loading cases;
• derive and use the relationships among load intensity, shear force, and bending moment;
• identify maximum moments, zero-shear locations, and points of contraflexure;
• check beam diagrams using equilibrium, continuity, and dimensional reasoning;
• apply the method to a practical beam-design situation.

1. Beam Classification and Idealized Supports

A beam is classified according to its support arrangement and the way it is loaded. The most frequently used beam types are listed below.

An ideal pin support can resist two force components, usually horizontal and vertical, but cannot resist a couple. A roller support normally supplies one force reaction normal to its supporting surface. A fixed support can resist horizontal force, vertical force, and moment.

For a planar beam, the three independent equilibrium equations are

where ∑F_x and ∑F_y are sums of force components and ∑M_O is the sum of moments about any point O. A structure is statically determinate when all unknown reactions can be found from these equations alone.

Types of Loading

A concentrated force acts at one location and is measured in newtons or kilonewtons. A concentrated couple produces a pure moment and is measured in N·m or kN·m. A distributed load acts over a length and is measured in N/m or kN/m.

For a uniformly distributed load of intensity w acting over length L, the equivalent resultant is

and it acts through the centroid of the loaded region, at its midpoint for a uniform load. Here, w has units of force per length, L has units of length, and R has units of force. For a linearly varying triangular load with maximum intensity w₀ over length L, the resultant is

The resultant acts at the centroid of the triangle: L/3 from the larger-intensity end, or 2L/3 from the zero-intensity end.

A distributed load may also be replaced by a resultant for calculating support reactions. However, the original distribution must be retained when constructing shear-force and bending-moment diagrams.

2. Internal Shear Force and Bending Moment

Imagine cutting a beam at a section distance x from its left end. The portion on either side of the cut must remain in equilibrium. The internal actions exposed at the cut are an internal shear force V(x) and an internal bending moment M(x). Their values are equal and opposite on the two cut faces, according to Newton’s third law.

This section method is central to beam analysis. One may isolate either the left segment or the right segment, but the chosen sign convention must be used consistently.

A common convention for the left segment is:

• positive shear causes a clockwise tendency on the isolated left segment;
• positive bending moment produces sagging curvature, conventionally drawn as a smile;
• negative bending moment produces hogging curvature, conventionally drawn as a frown.

Different textbooks may use an alternative shear convention. The physical results are unchanged provided the convention is applied consistently.

For a section at x, the internal actions can be calculated from equilibrium of the portion to the left:

The exact signs depend on the adopted convention. A useful practical rule is to draw the assumed internal actions on the free-body diagram and let equilibrium determine their algebraic values.

3. Relationships Among Load, Shear, and Moment

Let q(x) denote the distributed load intensity, V(x) the internal shear force, and M(x) the internal bending moment. Consider a small beam element of length dx. Vertical equilibrium gives

Moment equilibrium gives

Thus, with the adopted sign convention,

These relationships are among the most important results in beam analysis. They imply the following:

• the slope of the shear diagram equals the negative load intensity;
• the slope of the moment diagram equals the shear force;
• the curvature of the moment diagram is governed by the load intensity;
• where q = 0, the shear diagram is horizontal and the moment diagram is linear;
• where q is constant, the shear diagram is linear and the moment diagram is parabolic;
• a concentrated force causes a jump in the shear diagram;
• a concentrated couple causes a jump in the moment diagram;
• a distributed load causes continuous variation of shear and moment, unless an idealized discontinuity is present.

The integral forms are

Therefore, the change in shear equals the negative area under the load diagram, and the change in moment equals the area under the shear diagram. These are valid for statically determinate and indeterminate beams alike, provided the internal functions are correctly determined.

Boundary and Continuity Conditions

At a free end with no applied concentrated force or couple, both end shear and end moment are zero. At a simple support, the bending moment is zero unless an external couple is applied directly at the support. At a fixed end, the bending moment need not be zero.

A point at which the bending moment changes sign is called a point of contraflexure. Since the curvature changes from sagging to hogging or vice versa, such points are important in structural interpretation.

For a beam carrying only transverse loads, the maximum or minimum bending moment within a smooth interval generally occurs where

This condition identifies stationary values of M because dM/dx = V. It must be supplemented by examination of boundaries, load discontinuities, and sign changes.

4. General Procedure for Constructing Diagrams

A reliable procedure is as follows:

1. Draw the complete free-body diagram, including all external loads and reactions.
2. Replace distributed loads by equivalent resultants only when calculating reactions.
3. Use global equilibrium to determine unknown reactions.
4. Mark every location where loading changes: supports, point forces, couples, and ends of distributed loads.
5. Select a coordinate x and write expressions for V(x) and M(x) in each interval.
6. Evaluate key values at the ends of each interval.
7. Plot shear using the correct jumps and slopes.
8. Plot moment using the correct slopes, continuity, and curvature.
9. Check the diagrams using the load–shear–moment relationships and boundary conditions.

Piecewise equations are especially useful for computer implementation, design calculations, and locating extrema.

5. Worked Example 1: Simply Supported Beam with a Point Load

A simply supported beam AB is 6 m long. A downward point load of 12 kN acts 2 m from A. Determine the support reactions and construct the shear-force and bending-moment relationships.

Reactions

Let A be a pin and B a roller. There is no horizontal load, so the horizontal reaction is zero. Taking moments about A,

Therefore,

Vertical equilibrium gives

so

Shear Force

For 0 < x < 2 m, the left segment contains only R_A:

Immediately to the right of the 12 kN load, the shear becomes

At B, the upward reaction of 4 kN brings the shear back to zero.

Bending Moment

For 0 ≤ x ≤ 2 m,

For 2 ≤ x ≤ 6 m,

At the load point,

At B,

The maximum moment is 16 kN·m at the point load because the shear changes from positive to negative there. The shear diagram consists of horizontal segments with a 12 kN downward jump at x = 2 m. The moment diagram is triangular-linear, increasing to 16 kN·m and then decreasing to zero.

6. Worked Example 2: Uniformly Distributed Load

A simply supported beam of span 8 m carries a uniformly distributed load of 5 kN/m over its full length. Determine the reactions and the maximum bending moment.

The total load is

Because the load is symmetric,

At a section x from A, where 0 ≤ x ≤ 8 m, the load to the left is wx. Thus,

The bending moment is

or

The maximum occurs where V = 0:

so x = 4 m. Hence,

The shear diagram is a straight line from +20 kN at A to −20 kN at B. The moment diagram is a parabola, zero at both supports and maximum at midspan.

7. Worked Example 3: Cantilever with an End Load and Distributed Load

A 3 m cantilever carries a 6 kN downward point load at its free end and a uniformly distributed load of 2 kN/m over the full length. Determine the fixed-end reactions.

The distributed load has resultant

and acts 1.5 m from the fixed end. Vertical equilibrium gives the fixed-end vertical reaction

Taking moments about the fixed end,

The fixed-end reaction moment is 27 kN·m counterclockwise to balance the clockwise load moments. The horizontal reaction is zero.

The internal shear is largest in magnitude near the fixed end, while the bending moment is also largest there. This illustrates an important design principle: for a cantilever under downward loading, the fixed connection is usually the critical section.

8. Applied Case: Preliminary Design of a Floor Beam

Background

Consider an illustrative simply supported steel floor beam spanning 6 m between columns. The beam carries a characteristic uniformly distributed service load of 18 kN/m, including floor, finishes, partitions, and an allowance for occupancy.

Problem

The engineer must identify the reactions, maximum shear, and maximum bending moment for preliminary sizing. Detailed design would additionally require load combinations, lateral stability, deflection, vibration, connection design, fire requirements, and applicable building codes.

Analysis

The load is symmetric, so each support carries

The shear function is

The maximum positive shear magnitude at a support is 54 kN. Setting V = 0 gives x = 3 m. The moment function is

At midspan,

Approach and Outcome

The preliminary design must provide adequate resistance to the 54 kN shear and 81 kN·m bending moment, subject to the relevant design code and safety factors. A selected section must also satisfy serviceability requirements. If the beam is too flexible, occupants may experience noticeable vibration or floor deflection even when strength is adequate.

Learning Points

The case demonstrates that support reactions are not merely bookkeeping quantities: they define maximum end shears. The zero-shear location identifies the maximum sagging moment for this symmetric loading. The loading assumptions must be documented because changes in occupancy, partitions, or beam spacing directly alter the diagrams and design actions.

9. Practical Interpretation and Design Relevance

Shear force is particularly important near supports, load introduction points, web openings, and connections. Bending moment is often critical near midspan of simply supported beams and near fixed supports or internal supports of continuous beams. A beam may be safe in bending but unsafe in shear, or vice versa.

The shear-force diagram supports the design of webs, shear connections, bolts, welds, and bearing regions. The bending-moment diagram supports selection of beam depth and section modulus. In elastic bending design, the normal bending stress is commonly related to moment by

where σ is normal stress, M is bending moment, y is distance from the neutral axis, and I is the second moment of area. This equation belongs to bending-stress analysis and requires elastic behavior, small deformation, and a beam material obeying the relevant assumptions. It should not be used as a substitute for determining M from equilibrium.

Deflection analysis requires additional properties, particularly Young’s modulus E and the second moment of area I. Shear-force and bending-moment diagrams are therefore an essential intermediate stage between external loading and deformation or stress design.

10. Common Errors and Quality Checks

Typical errors include placing a distributed-load resultant at the wrong location, omitting a reaction, using inconsistent signs, confusing force units with moment units, and drawing a curved moment diagram under a point load. The following checks are effective:

• The algebraic sum of all external vertical forces must be zero.
• The algebraic sum of external moments about any point must be zero.
• The net change in shear must correspond to the applied loads.
• The area under the shear diagram between two sections must equal the change in moment.
• A simple support with no applied couple normally has zero moment.
• A free end without an applied force or couple has zero shear and moment.
• Moment units must be force × length, such as kN·m.
• A point force creates a shear jump but does not create a moment jump.
• A concentrated couple creates a moment jump but does not create a shear jump.

For hand calculations, dimensional checks and limiting cases are valuable. For example, if a beam load is increased while geometry is unchanged, reactions, shear, and bending moment should increase proportionally in a linear static model.

11. Limitations and Contemporary Practice

The elementary beam model idealizes supports, loads, geometry, and material behavior. Real supports have flexibility, loads may be distributed over finite contact areas, and beams may experience torsion, axial force, local buckling, residual stress, or dynamic loading. Continuous and fixed beams generally require compatibility equations, the force method, slope-deflection, moment distribution, or matrix structural analysis in addition to equilibrium.

Engineering software can generate shear and moment diagrams rapidly, but software output does not replace engineering judgment. The analyst must verify load paths, units, boundary conditions, mesh or element idealization, and sign conventions. Digital twins and structural-monitoring systems increasingly compare measured strains or accelerations with analytical predictions, but sensor data also require calibration and uncertainty assessment. AI-assisted design can help identify patterns or optimize sections, yet the responsible engineer remains accountable for assumptions, code compliance, safety factors, and the final decision.

Chapter Summary

Shear force and bending moment describe the internal actions produced when a beam carries transverse loading. Analysis begins by classifying the beam and supports, representing loads correctly, and applying global equilibrium to determine reactions. A section cut then reveals the internal shear and bending moment.

The fundamental differential relationships are dV/dx = −q and dM/dx = V. Consequently, load controls the slope of the shear diagram, shear controls the slope of the moment diagram, and the area under one diagram gives the change in the next. Point forces produce jumps in shear, concentrated couples produce jumps in moment, and distributed loads produce smooth variation.

Maximum bending moments commonly occur where shear is zero, but boundaries and discontinuities must also be checked. Reliable analysis combines equations, diagrams, boundary conditions, dimensional checks, and physical interpretation. These results form the basis for later study of bending stress, beam deflection, structural safety, and mechanical design.

Key Terms

Beam: A structural member primarily carrying transverse loads.

Support reaction: Force or moment exerted by a support on a beam.

Pin support: Support capable of resisting two planar force components.

Roller support: Support providing one reaction, normally perpendicular to its surface.

Fixed support: Support that restrains translation and rotation.

Concentrated force: Idealized force acting at one point.

Distributed load: Load intensity spread over a length.

Resultant load: Single force equivalent to a distributed loading for equilibrium purposes.

Shear force: Internal transverse force at a beam section.

Bending moment: Internal moment resisting beam curvature.

Shear-force diagram: Graph showing shear force as a function of position.

Bending-moment diagram: Graph showing bending moment as a function of position.

Sagging moment: Positive bending moment under the stated convention.

Hogging moment: Negative bending moment under the stated convention.

Point of contraflexure: Location where bending moment changes sign.

Statically determinate beam: Beam whose reactions are found from equilibrium alone.

Zero-shear section: Section where V = 0, often associated with an extreme moment.

Prismatic beam: Beam having constant cross-sectional geometry along its length.

Section method: Procedure using a cut beam segment to determine internal actions.

Review Questions

Short Answer Questions

1. What is the difference between a concentrated force and a distributed load?
2. What reactions can an ideal fixed support provide in a planar beam problem?
3. State the differential relationships among q, V, and M.
4. What change occurs in a shear-force diagram at a downward point load?
5. Why is the bending moment normally zero at a simple support?

Long Answer Questions

1. Explain the complete procedure for constructing shear-force and bending-moment diagrams for a statically determinate beam.
2. Derive the relationships dV/dx = −q and dM/dx = V using equilibrium of a differential beam element.
3. Discuss how concentrated forces, concentrated couples, and distributed loads affect shear and moment diagrams.
4. Analyze a simply supported beam carrying a full-span uniformly distributed load and obtain expressions for V(x) and M(x).
5. Explain the engineering significance of maximum shear, maximum bending moment, and points of contraflexure.

Analytical / Application Questions

1. A 10 m simply supported beam carries a 20 kN point load 4 m from the left support. Determine both reactions and the maximum bending moment.
2. A 5 m cantilever carries a uniformly distributed load of 3 kN/m over its entire length. Determine the fixed-end shear and moment reactions, and describe the shapes of its shear and moment diagrams.
3. A simply supported beam carries a nonuniform triangular load. Explain how its resultant magnitude and line of action would be found, and describe how the loading affects the curvature of the bending-moment diagram.

Multiple Choice Questions with Answers

Question 1

Which support normally provides one vertical reaction and no reaction moment?

A. Fixed support

B. Roller support

C. Built-in support

D. Pure couple support

Correct Answer: B. Roller support

Question 2

A uniformly distributed load of 4 kN/m acts over 6 m. What is its equivalent resultant load?

A. 10 kN

B. 12 kN

C. 24 kN

D. 48 kN

Correct Answer: C. 24 kN

Question 3

With the sign convention used in the chapter, the relationship between distributed load and shear is:

A. dV/dx = q

B. dV/dx = −q

C. dV/dx = M

D. dV/dx = −M

Correct Answer: B. dV/dx = −q

Question 4

What does a concentrated downward force produce in a shear-force diagram?

A. A vertical jump downward

B. A vertical jump upward

C. A parabolic segment

D. No change

Correct Answer: A. A vertical jump downward

Question 5

For a simply supported beam without an applied couple at the support, the bending moment at the support is generally:

A. Maximum

B. Equal to the shear force

C. Zero

D. Infinite

Correct Answer: C. Zero

Question 6

A simply supported beam is 8 m long and carries a full-span load of 5 kN/m. What is the reaction at each support?

A. 10 kN

B. 20 kN

C. 30 kN

D. 40 kN

Correct Answer: B. 20 kN

Question 7

For the beam in the preceding question, where does the maximum bending moment occur?

A. At the left support

B. At 2 m from the left support

C. At midspan

D. At the right support

Correct Answer: C. At midspan

Question 8

The area under a shear-force diagram between two sections equals the:

A. Distributed load intensity

B. Change in bending moment

C. Support reaction only

D. Second moment of area

Correct Answer: B. Change in bending moment

Question 9

A concentrated applied couple on a beam causes which idealized diagram change?

A. A jump in shear only

B. A jump in moment only

C. A parabolic shear variation

D. No change in either diagram

Correct Answer: B. A jump in moment only

Question 10

For a 3 m cantilever carrying a 6 kN end load, the magnitude of the fixed-end moment due to that load is:

A. 2 kN·m

B. 9 kN·m

C. 18 kN·m

D. 36 kN·m

Correct Answer: C. 18 kN·m