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How to Draw Cantilever Shear-Force and Bending-Moment Diagrams

By Clara Voss ·

A shear moment diagram for a cantilever beam becomes systematic when the coordinate direction and sign convention are fixed before any equations are written. Use this sequence:

  1. Define the coordinate and sign convention.
  2. Calculate the fixed-support reactions.
  3. Divide the beam wherever loading changes.
  4. Derive shear V(x) and bending moment M(x) in each region.
  5. Plot jumps, slopes, straight lines, and curves.
  6. Verify equilibrium, signed areas, continuity, boundary values, extrema, and units.

The examples below use one convention throughout so the equations and diagrams remain consistent.

What the diagrams show on a cantilever beam

A cantilever beam is fixed at one end and unsupported at the other. Here, the fixed support is on the left and the free end is on the right.

The shear-force diagram (SFD) plots internal transverse shear V against position. The bending-moment diagram (BMD) plots internal bending moment M against position. Together, they show how these internal actions vary from section to section along the beam. This article is limited to static transverse loading; it does not cover axial force, torsion, dynamics, stability, or large-deformation behavior. These definitions and limits are consistent with the Engineering Statics open textbook.

A fixed support restrains translation and rotation. Under general planar loading, it can therefore provide:

  • A horizontal reaction
  • A vertical reaction
  • A reaction moment

Only vertical loads and applied couples are used in the worked examples, so the horizontal reaction is zero. The vertical reaction balances the net vertical load, while the reaction moment balances the moments produced by the external loads.

Two moments must be distinguished at the support:

  • The support reaction moment is an external action exerted by the support on the beam.
  • The internal bending moment immediately inside the beam is the internal resultant exposed by cutting the beam just beside the support.

Their magnitudes balance as required by equilibrium, but their algebraic signs need not be written identically. The sign depends on the free body being considered and the declared convention.

Shear and moment diagrams help locate the largest internal-action magnitudes, changes in sign, discontinuities, and critical sections. They do not independently establish:

  • Bending or shear stress
  • Deflection or rotation
  • Local or lateral stability
  • Connection adequacy
  • Member capacity
  • Code compliance
  • Construction safety

Those questions require additional information, including section geometry, material properties, load combinations, boundary assumptions, and applicable design requirements.

Choose the coordinate system and sign convention before calculating

Different references can display opposite algebraic signs for the same cantilever without disagreeing about its physical behavior. The difference normally comes from the coordinate origin, load direction, cut-face convention, or plotting convention.

For every example in this article:

  • The fixed support is at x=0.
  • The free end is at x=L.
  • The coordinate x increases from left to right.
  • Distributed-load intensity w(x) is positive when acting downward.
  • Positive internal shear V acts upward on a left cut face and downward on a right cut face.
  • Positive internal bending moment M acts clockwise on a left cut face and counterclockwise on a right cut face.
  • Positive diagram ordinates are plotted above the baseline.
  • Negative diagram ordinates are plotted below the baseline.

Cut-face convention

The arrows below show the assumed positive internal actions. The two cut faces are equal and opposite because they act on adjoining pieces of the same beam.

Retained left segment                     Retained right segment

──────────────│ cut                 cut │──────────────
              ↓ +V                    +V ↑
              ↺ +M                    +M ↻
        right cut face            left cut face

On the retained left segment, the exposed face is a right cut face, so positive V points downward and positive M acts counterclockwise. On the retained free-end segment, the exposed face is a left cut face, so positive V points upward and positive M acts clockwise.

Under this convention,

dV ÷ dx = -w(x)

and

dM ÷ dx = V(x).

Therefore,

V(x₂)-V(x₁) = -\intx_₁x_²w(x)\,dx

and

M(x₂)-M(x₁) = \intx_₁x_²V(x)\,dx.

These load–shear–moment relationships are presented in the Engineering LibreTexts treatment of shear and moment diagrams.

Because downward w is defined as positive here, the change in shear equals the negative of the numerical area under the w-versus-x graph. If a signed load diagram instead plots downward loading below its baseline, its signed area can be used directly. Either approach works if the graphical and algebraic definitions agree.

The second area relationship is more direct: the signed area under the shear diagram equals the change in bending moment.

Under this article’s convention, ordinary downward loads on the cantilevers considered below produce:

  • Positive shear over the loaded part of the span
  • Negative bending moment, commonly called hogging

That does not mean cantilever moment is universally negative. Another reference may reverse the internal-action convention, take upward loading as positive, measure distance from the free end, or draw the moment diagram on the tension side.

The convention-independent results are generally:

  • Diagram shape
  • Continuity or discontinuity
  • Jump magnitude
  • Maximum absolute magnitude
  • Critical location
  • Whether a function is constant, linear, or curved

Converting formulas measured from the free end

Some references measure distance from the free end. Let that coordinate be \xi, increasing from the free end toward the support. Then

\xi = L-x.

For a free-end point load, the moment magnitude may be written

|M| = P\xi.

Substituting \xi=L-x gives

|M| = P(L-x).

The formulas describe the same result using different coordinate origins.

A repeatable workflow for drawing both diagrams

The following method applies to statically determinate cantilevers under transverse static loading.

1. Draw and label the beam

Show:

  • Fixed and free ends
  • Total length
  • Point-load magnitudes and locations
  • Distributed-load intensities and extents
  • Applied couples and directions
  • The x-axis
  • The adopted sign convention

Do not begin with a memorized formula before identifying the coordinate origin and load boundaries.

2. Draw the whole-beam free-body diagram

Replace the fixed support with its possible reactions and apply

\sum F_x = 0,\qquad \sum F_y = 0,\qquad \sum M = 0.

If no horizontal load acts, the horizontal reaction is zero. Use vertical-force equilibrium to determine the vertical reaction and moment equilibrium about the fixed support to determine the reaction moment. A fixed support provides the force and moment reactions required to prevent translation and rotation, as summarized by the Efficient Engineer’s diagram-construction guide.

3. Divide the beam into loading regions

Begin a new interval wherever:

  • A point load occurs
  • An applied couple occurs
  • A distributed load begins
  • A distributed load ends
  • A distributed-load expression changes

A concentrated action occupies no finite length, but the equations immediately to its left and right can differ.

4. Make a cut in each region

At a general location x, cut the beam and retain either side. Replace the removed material with internal shear V(x) and moment M(x).

For a cantilever, the free-end side of a cut often contains fewer loads and gives shorter equations. The free-end segment can be used even though x remains measured from the fixed support.

5. Derive V(x) and M(x)

Apply force and moment equilibrium to the retained segment. Write the valid interval beside every expression.

Do not continue an equation through a point load, applied couple, or distributed-load boundary without checking whether the free-body diagram has changed.

6. Evaluate important locations

Calculate ordinates at:

  • x=0, immediately inside the support
  • x=L^-, immediately inside the free end
  • Every point-load location
  • Every applied-couple location
  • The start and end of each distributed load
  • Any interior point where V=0

Evaluate both sides of a concentrated action when a jump can occur.

7. Plot jumps and connect the ordinates

Use the governing shape relationships:

  • No distributed load: constant shear
  • Constant shear: linear moment
  • Constant distributed load: linear shear
  • Linear shear: quadratic moment
  • Point force: jump in shear
  • Point couple: jump in moment

Do not connect known ordinates with a straight line unless the slope relationship shows that the function is linear.

8. Verify the result

Check:

  • Whole-beam equilibrium
  • Jump magnitudes
  • Diagram slopes
  • Signed areas
  • Continuity
  • Free-end conditions
  • Extrema
  • Units

These checks are independent ways to reveal an incorrect reaction, sign, interval, or plotted shape.

Worked case 1: Cantilever with a point load at the free end

Consider a cantilever of length L carrying a downward point load P at x=L:

fixed                                      free
|||||----------------------------------------●
x = 0                                      x = L
                                             ↓ P

Support reactions

Vertical equilibrium gives an upward reaction

R_y = P.

Moment equilibrium about the support gives a counterclockwise reaction-moment magnitude

M_R = PL.

The horizontal reaction is zero. These results agree with the standard free-end point-load cantilever formulas; for example, a 5 lb load on a 3 ft cantilever gives a 5 lb reaction and a 15 ft·lb reaction moment (Engineering LibreTexts).

Internal shear and moment

Make a cut at x, retain the free-end segment, and note that the distance from the cut to the load is L-x. For the internal field 0\le x<L,

V(x) = P

and

M(x) = -P(L-x).

Thus,

|V(x)| = P, \qquad |M(x)| = P(L-x).

Immediately inside the fixed support,

V(0⁺) = P, \qquad M(0⁺) = -PL.

Immediately inside the loaded free end,

V(L⁻) = P, \qquad M(L⁻) = 0.

Moving from left to right across the downward point load,

V(L⁺)-V(L⁻) = -P.

The shear therefore drops from P inside the beam to the zero external baseline. The ordinary point force does not create a moment jump.

Shear-force and bending-moment diagrams

LOAD
      fixed                                      free
      |||||----------------------------------------●
                                                   ↓ P
      x=0                                         x=L


SFD: V(x)=+P for 0 ≤ x < L

 +P   ┌────────────────────────────────────────────┐
      │                                            │
  0 ──┴────────────────────────────────────────────┴──→ x
      0                                            L
                                                   ↓ jump P


BMD: M(x)=-P(L-x)

  0 ───────────────────────────────────────────────●──→ x
/
/
/
 -PL     ●──────────────────────────────────
         x=0                              x=L⁻

      straight line below baseline

The SFD is a constant rectangle of ordinate +P. The BMD is a straight line from -PL at the support to zero at the free end.

The moment diagram is linear because

dM ÷ dx = V = P.

A constant shear gives a constant moment slope.

Maximum values and area check

The maximum shear magnitude is

|V|_\max = P

throughout the span, while the maximum moment magnitude is

|M|_\max = PL

at the fixed support.

The signed shear area confirms the total moment change:

M(L⁻)-M(0⁺) = \int₀^L P\,dx = PL.

Because M(L^-)=0, the equation gives M(0^+)=-PL.

Worked case 2: Cantilever under a full-length uniform load

Now apply a constant downward distributed load w over the full length L:

        ↓  ↓  ↓  ↓  ↓  ↓  ↓  ↓  ↓
        w per unit length over full span
|||||----------------------------------------
x = 0                                      x = L

The units of w are force per length, such as kN/m or lb/ft.

Support reactions

The statically equivalent resultant is

W = wL,

acting at the centroid of the rectangular load distribution, L/2 from the support. Therefore,

R_y = wL

and

M_R = (wL)(L ÷ 2) = wL² ÷ 2.

Internal shear and moment

Cut the beam at x and retain the free-end segment. Its loaded length is L-x, so its resultant load is w(L-x). Equilibrium gives

V(x) = w(L-x), \qquad 0\le x\le L,

and

M(x) = -w(L-x)(L-x ÷ 2) = -w(L-x)² ÷ 2.

Consequently,

|V(x)| = w(L-x), \qquad |M(x)| = w(L-x)² ÷ 2.

The endpoint values are

V(0⁺) = wL, \qquad M(0⁺) = -wL² ÷ 2,

and, because the free end is unloaded,

V(L⁻) = 0, \qquad M(L⁻) = 0.

These full-span UDL formulas and the associated linear-shear and quadratic-moment behavior are included in standard cantilever formula collections such as Structural Basics.

Shear-force and bending-moment diagrams

LOAD
      ↓  ↓  ↓  ↓  ↓  ↓  ↓  ↓  ↓
      |||||----------------------------------------
      x=0                                         x=L


SFD: V(x)=w(L-x)

 +wL  ●
│\
│ \
│ \
│ \
  0 ──┴────●──────────────────────────────────────→ x
      0    L

      linear triangular diagram above baseline


BMD: M(x)=-w(L-x)²/2

  0 ──────────────────────────────────────────────●──→ x
      |                                        _/
      |                                    __/
      |                                ___/
      |                         ______/
 -wL²/2●_______________________/
       x=0                                  x=L

      parabolic diagram below baseline

Because

dV ÷ dx = -w,

the shear has a constant negative slope and is therefore linear. Its diagram is triangular.

Because

dM ÷ dx = V = w(L-x),

the moment is the integral of a linear function and is therefore quadratic. Its diagram is parabolic.

Maximum values

For a downward UDL over the full span,

|V|_\max = wL

and

|M|_\max = wL² ÷ 2,

both at the fixed support.

The area under the triangular shear diagram confirms the moment change:

M(L⁻)-M(0⁺) = \int₀^L w(L-x)\,dx = wL² ÷ 2.

Since M(L^-)=0,

M(0⁺) = -wL² ÷ 2.

Point-load and UDL comparison

Loading case Shear magnitude Shear shape Moment magnitude Moment shape
Free-end point load P P Constant rectangle P(L-x) Linear
Full-span UDL w w(L-x) Linear triangle w(L-x)^2 ÷ 2 Parabolic

The progression follows directly from integration:

  • A point load creates an abrupt shear change.
  • No distributed load leaves shear constant.
  • Constant shear produces linear moment.
  • Constant distributed load produces linear shear.
  • Linear shear produces quadratic moment.

Worked case 3: An intermediate point load and piecewise equations

Consider a downward point load P at x=a, where

0<a<L.

fixed                    load                    free
|||||----------------------●------------------------
x = 0                    x = a                  x = L
                           ↓ P

This case demonstrates why one equation cannot automatically be used over an entire beam.

Support reactions

Whole-beam equilibrium gives

R_y = P

and a reaction-moment magnitude

M_R = Pa.

Region 1: Between the support and load

For

0\le x<a,

the retained free-end segment includes the point load. Its distance from the cut is a-x, so

V₁(x) = P

and

M₁(x) = -P(a-x).

At the support,

V₁(0⁺) = P, \qquad M₁(0⁺) = -Pa.

Immediately to the left of the load,

V₁(a⁻) = P, \qquad M₁(a⁻) = 0.

Region 2: Between the load and free end

For

a<x\le L,

retain the segment extending from the cut to the free end. No external force, distributed load, or couple acts on that segment. Equilibrium therefore requires

V₂(x) = 0

and

M₂(x) = 0.

This result applies only when the region is genuinely unloaded. An end force, end couple, distributed load, or additional point load would change the equations.

The complete piecewise functions are

V(x) = \begin{cases} P, & 0\le x<a,\ 0, & a<x\le L, \end{cases}

and

M(x) = \begin{cases} -P(a-x), & 0\le x\le a,\ 0, & a<x\le L. \end{cases}

The piecewise result—constant shear before the load and zero shear and moment in the unloaded free-end region—is also shown in an intermediate-load cantilever discussion.

Shear-force and bending-moment diagrams

LOAD
      fixed                    load                    free
      |||||----------------------●------------------------
      x=0                      x=a                    x=L
                                 ↓ P


SFD

 +P   ┌──────────────────────────┐
      │                          │
  0 ──┴──────────────────────────┴──────────────────────→ x
      0                          a                      L
                                 ↓ jump P


BMD

  0 ─────────────────────────────●──────────────────────→ x
/
/
/
 -Pa     ●────────────────
         x=0              x=a              zero to x=L

The shear diagram is constant at +P from the support to the load, jumps downward by P at x=a, and remains zero to the free end.

The moment diagram rises linearly from -Pa at the support to zero at the load, then remains at zero.

At the point load,

M(a⁻) = M(a⁺) = 0.

Moment is continuous, but its slope changes because

dM ÷ dx = V

and shear changes from P to zero.

Why extending the first equation gives a false result

If

M(x) = -P(a-x)

is incorrectly applied for x>a, then a-x becomes negative and the formula predicts a nonzero moment in the unloaded segment.

That is false because the free-body diagram used to derive the equation contained the point load. Once the cut moves to the right of x=a, the load is no longer on the retained free-end segment. The equilibrium problem has changed, so a new expression is required.

An equation belongs to a specific free-body diagram and interval, not merely to the beam as a whole.

Applied couples, multiple loads, and the load-to-shape map

Concentrated couples and multiple point loads use the same equilibrium method, but their diagram effects differ from those of concentrated forces.

Cantilever carrying only an end couple

Suppose a couple of magnitude M_0 acts at the free end, with no transverse force or distributed load.

Because there is no transverse loading,

V(x) = 0.

Since

dM ÷ dx = V = 0,

the internal moment is constant:

|M(x)| = M₀, \qquad 0\le x<L.

The algebraic sign depends on the applied couple’s direction. An end couple acting counterclockwise produces M(x)=+M_0 under the convention used here; reversing the applied couple reverses the sign. Zero shear and constant moment for this loading case are shown in the Structural Basics cantilever formula reference.

Representative case: counterclockwise end couple

LOAD
      |||||----------------------------------------↺ M₀
      x=0                                         x=L


SFD

  0 ─────────────────────────────────────────────────→ x
      V(x)=0 throughout


BMD

 +M₀ ┌─────────────────────────────────────────────┐
     │                                             │
  0 ─┴─────────────────────────────────────────────┴──→ x
     0                                             L
                                                   jump M₀

The moment diagram is a constant rectangle. At the applied couple, it jumps by magnitude M_0 between the internal value and the zero external baseline. Shear does not jump because a couple has no net transverse force.

Point force versus point couple

The distinction is fundamental:

  • A point force causes a jump in shear.
  • An ordinary point force does not itself create a moment jump.
  • A point couple causes a jump in moment.
  • A point couple does not directly create a shear jump.

The jump’s algebraic direction depends on the action direction and sign convention, while its magnitude equals the corresponding concentrated action. These separate force and couple jump rules are set out in Purdue’s mechanics-of-materials lecture notes.

Compact load-to-shape map

Loading condition Effect on shear V Effect on moment M
No distributed load V is constant M is linear if V\ne0
Constant distributed load V is linear M is quadratic
Point force V jumps M remains continuous, but its slope changes
Point couple No direct shear jump M jumps
Zero shear over an interval V=0 M is constant
Constant nonzero shear V=constant M is a straight line

The table identifies behavior between and at loading changes. Signs and ordinates must still be calculated.

Multiple point loads

For downward point loads P_i at distances a_i from the fixed support, equilibrium gives

R_y = \sum_i P_i

and

M_R = \sum_i P_i a_i.

These are magnitude equations for loads acting in the same direction. If some forces act upward or applied moments oppose one another, use signed quantities.

Even when the support reactions are found by one summation, internal equations must still be written separately between load locations. Moving a cut across a point load changes which actions are present on the retained segment.

Superposition

Superposition does not eliminate the need to:

  • Preserve signs
  • Respect each load’s position
  • Treat discontinuities correctly
  • Use one coordinate definition
  • Check the combined result against equilibrium

Triangular-load formulas are not tabulated here because the support moment depends on which end of the load carries the peak intensity. A formula detached from its load orientation can be misleading.

How to check a cantilever shear and moment diagram

A completed diagram should satisfy several independent checks. If one fails, return to the free-body diagrams and region equations rather than adjusting the sketch by eye.

Check global equilibrium

The support force must balance the net transverse load:

\sum F_y = 0.

The support reaction moment must balance the external moments about the fixed support:

\sum M_support = 0.

For the standard downward cases:

  • Free-end point load: R_y=P, M_R=PL
  • Full-span UDL: R_y=wL, M_R=wL^2/2
  • Intermediate point load: R_y=P, M_R=Pa

Omitting the fixed-support reaction moment is a major cantilever-analysis error.

Check jumps

At a point force, compare the shear immediately on both sides. The jump magnitude must equal the point-force magnitude.

At an applied couple, compare the bending moment immediately on both sides. The jump magnitude must equal the couple magnitude.

Do not introduce a shear jump merely because a couple is present, or a moment jump merely because an ordinary point force is present.

Check slopes

Use

dV ÷ dx = -w(x)

and

dM ÷ dx = V(x).

These equations give immediate visual tests:

  • No distributed load means horizontal shear.
  • Constant downward w means shear slopes downward from left to right.
  • Positive shear means moment rises as x increases.
  • Negative shear means moment falls as x increases.
  • Zero shear means moment has zero slope.

A full-span UDL cannot produce a linear moment diagram under these assumptions because its shear varies linearly.

Check signed areas

Between x_1 and x_2,

\Delta V = -\intx_₁x_²w(x)\,dx

and

\Delta M = \intx_₁x_²V(x)\,dx.

For the end point-load case, the rectangular shear area is PL, matching the moment change from -PL to zero.

For the UDL case, the triangular shear area is

1 ÷ 2(L)(wL) = wL² ÷ 2,

matching the change from -wL^2/2 to zero.

Check continuity

At an ordinary point force:

  • Shear may be discontinuous.
  • Moment remains continuous.
  • The moment slope changes because shear changes.

At a concentrated couple:

  • Shear does not jump directly.
  • Moment may be discontinuous.

Where a distributed load begins or ends, the shear slope normally changes without an instantaneous shear jump, provided no point force acts at the same location.

Check free-end conditions carefully

At an unloaded free end, the internal limiting values are

V(L⁻) = 0

and

M(L⁻) = 0.

These are not universal free-end values:

  • An applied end force produces nonzero internal shear at L^-.
  • An applied end couple produces nonzero internal moment at L^-.
  • Both may be present simultaneously.

The correct boundary value follows from equilibrium of a short segment immediately beside the end.

Check possible extrema

Because

dM ÷ dx = V,

an interior moment extremum can occur where

V = 0.

Also check:

  • Fixed and free boundaries
  • Both sides of discontinuities
  • Ends of distributed-load regions
  • Locations where an equation changes

The fixed support carries the maximum moment magnitude for the standard downward point-load and full-span UDL cases. It need not govern every system with opposing loads, partial loading, or applied couples.

Check dimensions

Every equation must preserve units:

Quantity Dimensions Example units
Distributed load w Force/length kN/m, lb/ft
Point load or shear V Force kN, lb
Bending moment M Force × length kN·m, ft·lb

Thus,

w(L-x)

has force units, while

w(L-x)² ÷ 2

has force-length units. A dimensional mismatch reveals an incorrect expression even when the plotted shape appears plausible.

Common errors to avoid

  1. Omitting the fixed-support reaction moment
  2. Changing coordinate origins without transforming the formula
  3. Combining signs from different conventions
  4. Applying one equation through a load boundary
  5. Confusing a point force with a point couple
  6. Drawing a linear moment diagram under a full-span UDL
  7. Assuming free-end shear and moment are always zero
  8. Failing to evaluate both sides of a concentrated action
  9. Using force units for moment or load intensity
  10. Treating the diagrams as proof that a member is safe

Shear and moment diagrams identify internal actions. Further analysis is required to determine stresses, deformation, stability, capacity, connection demands, code compliance, and safe construction requirements.

Cantilever shear and moment diagram FAQs

Should x be measured from the fixed end or the free end of a cantilever?

Either origin is valid if it is clearly defined and used consistently.

This article uses x=0 at the support and x=L at the free end. If another solution measures \xi from the free end, convert with

\xi = L-x.

Thus,

|M| = P\xi

and

|M| = P(L-x)

describe the same free-end point-load moment magnitude.

Why is bending moment zero at an unloaded free end?

Isolate a very short segment immediately beside the free end. If no end couple or other loading acts on that segment, there is no external moment for the internal bending moment to balance. Therefore,

M(L⁻) = 0.

Similarly, if no end force acts,

V(L⁻) = 0.

An applied end force or couple changes the corresponding boundary value.

Does a point load create a jump in the bending-moment diagram?

No. An ordinary transverse point force creates a jump in shear, while bending moment remains continuous. Because

dM ÷ dx = V,

the jump in shear produces an abrupt change in the moment diagram’s slope.

A concentrated applied couple is the action that creates a moment jump.

Where is the maximum bending moment in a cantilever beam?

For the standard downward cases considered here, the maximum bending-moment magnitude occurs at the fixed support:

  • Free-end point load: PL
  • Full-span UDL: wL^2/2
  • Intermediate point load at x=a: Pa

This is not universal for every combined loading system. In general, check boundaries, both sides of discontinuities, and interior points where V=0.

What are the shear and moment diagrams for a cantilever carrying only an end moment?

For an end couple of magnitude M_0 with no transverse load,

V(x) = 0

throughout the span, while

|M(x)| = M₀

is constant. The SFD lies on the zero baseline. The BMD is a constant rectangle whose sign depends on the applied couple’s direction. At the end couple, moment jumps by magnitude M_0; shear does not jump.

The dependable method is always the same: declare x and the sign convention, calculate the support reactions, divide the beam at every loading change, derive V(x) and M(x) region by region, plot the required jumps and curves, and verify equilibrium, slopes, areas, endpoint conditions, continuity, and units. A free-end point load gives constant shear and linear moment; a full-span UDL gives linear shear and parabolic moment; an intermediate point load requires piecewise diagrams; and an applied couple jumps moment rather than shear.