Internal Forces in Beams

Internal Forces in Beams

$Learning Objectives$

Explain how a section cut exposes internal normal force, shear force, and bending moment.
Apply a consistent internal-force sign convention to a cut beam segment.
Determine internal actions from equilibrium of either side of a section.
Construct and interpret shear-force and bending-moment diagrams.
Use the differential and area relationships connecting distributed load, shear, and moment.
Identify jumps, slopes, extrema, and boundary values without overgeneralizing the condition $V=0$ .

$Internal Normal Force$

The axial resultant

N

acting normal to a cut cross-section. Under the convention used here, positive

N

denotes tension.

$Internal Shear Force$

The transverse resultant

V

acting in the plane of a cut cross-section and required for equilibrium of the isolated beam segment.

$Internal Bending Moment$

The internal couple

M

required at a cut section to satisfy rotational equilibrium. Under the convention used here, positive

M

corresponds to sagging bending.

$Method of Sections for Beams$

To determine the internal actions at position $x$ :

compute the external support reactions first when required;
cut the beam at the target location;
isolate the simpler side of the cut;
replace the removed portion by the internal resultants $N$ , $V$ , and $M$ ;
apply planar equilibrium.

The internal actions on opposite faces of the cut are equal and opposite, consistent with Newton's third law.

$Sign Convention Used in This Lesson$

A sign convention must be declared because textbooks and software may use different conventions. Here:

positive $N$ is tensile;
positive $V$ acts downward on the cut face of a left segment and upward on the cut face of a right segment;
positive $M$ is sagging, acting counterclockwise on the cut face of a left segment and clockwise on a right segment.

With downward distributed load intensity $w(x)$ taken as positive, this convention gives $dV/dx=-w$ and $dM/dx=V$ .

$Do Not Mix Sign Conventions$

A shear or moment diagram is only correct relative to its declared convention. If another reference or software uses the opposite shear sign, convert consistently rather than mixing equations from two conventions.

$Shear and Moment Diagrams$

A shear-force diagram plots $V(x)$ and a bending-moment diagram plots $M(x)$ along the beam. These diagrams reveal:

where internal-force magnitudes are largest;
how point loads and distributed loads alter the response;
where bending changes from sagging to hogging;
which regions later require the greatest strength or reinforcement demand.

The diagrams describe section resultants, not stress directly. Stress also depends on cross-sectional geometry and material behavior.

$Interactive Exploration$

Switch between point and distributed loading and vary the load parameters. Compare the displayed loading, reactions, shear diagram, and bending-moment diagram as one coupled equilibrium model.

Tension-positive N · sagging-positive M

Sign convention: positive tension. On the left cut face, positive shear acts downward and positive moment counterclockwise; arrows reverse on the right face. Downward distributed load is positive, so

dV/dx=-w(x)

and

dM/dx=V(x)

Load w(x)0.00 kN/m

Shear V(x)-2.74 kN

Moment M(x)67.66 kN·m

10 m

20 kN

4.00 m

18 kN·m

4.5 m

Cursor shear

-2.74 kN

Cursor moment

67.66 kN·m

Left reaction

17.26 kN

Right reaction

32.44 kN

Maximum moment

69.03 kN·m at 4.00 m

Minimum moment

0.00 kN·m at 0.00 m

Zero-shear locations

None

Moment extrema are evaluated at continuous V=0 roots, member ends, load/support breakpoints, and both faces of concentrated-moment jumps; point-load cusps therefore remain eligible even when V changes sign discontinuously.

Piecewise shear equation

V(x)=17.258⟨x-0.00⟩⁰ + 32.442⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰ - 3.200⟨x-4.50⟩¹ - 0.400⟨x-4.50⟩² + 7.600⟨x-10.00⟩¹ + 0.400⟨x-10.00⟩²

Piecewise moment equation

M(x)=17.258⟨x-0.00⟩¹ + 32.442⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹ - 1.600⟨x-4.50⟩² - 0.133⟨x-4.50⟩³ + 3.800⟨x-10.00⟩² + 0.133⟨x-10.00⟩³ + 18.000⟨x-7.20⟩⁰

\frac{dV}{dx}=-w(x)

\frac{dM}{dx}=V(x)

Load-Shear Differential Relationship

Relates downward-positive distributed-load intensity to the slope of the shear diagram under this lesson's sign convention.

\frac{dV}{dx}=-w(x)

Variables

Symbol	Description	Unit
$V$	Internal shear force	kN
$w(x)$	Downward-positive distributed-load intensity	kN/m
$x$	Coordinate along the beam	m

Shear-Moment Differential Relationship

Relates shear to the slope of the bending-moment diagram.

\frac{dM}{dx}=V(x)

Variables

Symbol	Description	Unit
$M$	Internal bending moment	kN·m
$V(x)$	Internal shear force	kN
$x$	Coordinate along the beam	m

$Area Relationships$

Integrating the differential equations gives a direct geometric interpretation: the signed area under the load diagram changes shear, while the signed area under the shear diagram changes moment.

Load-Area Change in Shear

Relates the signed area under the distributed-load diagram to the change in internal shear.

$V(x_2)-V(x_1)=-\int_{x_1}^{x_2}w(x)\,dx$

Variables

Symbol	Description	Unit
$V(x_1)$	Shear at the start of the interval	kN
$V(x_2)$	Shear at the end of the interval	kN
$w(x)$	Downward-positive distributed-load intensity	kN/m

Shear-Area Change in Moment

Relates the signed area under the shear diagram to the change in bending moment.

$M(x_2)-M(x_1)=\int_{x_1}^{x_2}V(x)\,dx$

Variables

Symbol	Description	Unit
$M(x_1)$	Bending moment at the start of the interval	kN·m
$M(x_2)$	Bending moment at the end of the interval	kN·m
$V(x)$	Internal shear force over the interval	kN

$Point Loads and Applied Couples$

A concentrated transverse force produces a jump in the shear diagram equal to its signed magnitude. The bending-moment diagram remains continuous across an isolated point force when no concentrated couple acts there.

A concentrated applied couple produces a jump in the bending-moment diagram equal to the signed couple according to the adopted convention. It does not by itself create a jump in shear.

$Diagram Shape Rules$

Within an interval without concentrated discontinuities:

if $w=0$ , shear is constant and moment is linear;
if $w$ is constant, shear is linear and moment is quadratic;
if $w$ varies linearly, shear is quadratic and moment is cubic.

These shape checks are useful for detecting hand-calculation errors before numerical values are considered.

$Where Maximum Moment Can Occur$

At a smooth interior point where $M(x)$ is differentiable, a local extremum requires $dM/dx=V=0$ . However, the governing maximum absolute moment may also occur at a support or free-end boundary, immediately adjacent to an applied couple, or at another point where the moment function is not differentiable. Always check the full domain and relevant discontinuities.

$Boundary Conditions and Physical Checks$

Common idealized boundary values include:

an unloaded ideal pin or roller cannot transmit a reaction couple, so the beam bending moment at that end is zero unless an external end couple is applied;
a free end has zero internal shear and moment only if no force or couple is applied at that end;
a fixed support can develop a reaction force and reaction moment.

Use equilibrium and the actual end loading rather than memorized diagram shapes when determining boundary values.

$Constructing Shear and Moment Diagrams$

STEP-BY-STEP

Draw the beam and determine support reactions.
Mark every location where the loading expression changes.
Apply point-load jumps to the shear diagram with the declared sign convention.
Use $dV/dx=-w$ or signed load areas between discontinuities.
Apply concentrated-couple jumps to the moment diagram.
Use $dM/dx=V$ or signed shear areas to construct moment.
Check known boundary values, diagram closure, units, and global equilibrium.
Evaluate all candidate locations for the governing positive, negative, and absolute extrema.

$How to Use This Workflow$

Use the workflow to keep diagram jumps, continuous loading regions, and boundary checks in one consistent sequence. If the diagrams do not close or violate a known boundary value, revisit reactions and signs before interpreting extrema.

Shear-Force and Bending-Moment Diagram Workflow

Key Takeaways

A beam section cut exposes internal normal force, shear force, and bending moment.
Internal-action signs are convention-dependent and must remain consistent across equations and diagrams.
Under the convention used here, $dV/dx=-w$ and $dM/dx=V$ .
Point forces jump shear, while applied couples jump bending moment.
Diagram shape follows the order of the loading function: constant load produces linear shear and quadratic moment.
$V=0$ identifies smooth interior moment extrema, but boundaries and discontinuities must also be checked.
Shear and moment diagrams are section-resultant maps that later feed stress, strength, and serviceability calculations.