Internal Forces in Structural Members - Theory & Concepts

Internal Forces in Structural Members

$Learning Objectives$

Determine normal force, shear force, and bending moment at a movable section cut.
Apply a consistent cut-face sign convention to left and right isolated segments.
Construct axial-force, shear-force, and bending-moment diagrams from concentrated, couple, and distributed loads.
Use piecewise equations and discontinuity rules to interpret structural diagrams.
Relate load, shear, and moment using differential and area relationships.
Analyze a determinate frame and a multi-span determinate beam idealization.

Section cut and internal N-V-M resultants

An imaginary cut exposes axial force, shear, and bending moment. Their piecewise variation must remain consistent with loads, jumps, slopes, and end conditions.

$Internal Resultants$

The normal force $N$ , shear force $V$ , and bending moment $M$ exposed at an imaginary cut through a planar structural member.

$Cut-Face Sign Convention$

Positive $N$ is tension. On the left cut face, positive $V$ acts downward and positive $M$ acts counterclockwise. On the right cut face, the arrows reverse. The scalar values obtained from either side must agree. Positive beam bending moment is sagging.

Section Equilibrium

The exposed internal resultants complete the equilibrium of either isolated segment.

\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0

Variables

Symbol	Description	Unit
$N$	Internal normal force	kN
$V$	Internal shear force	kN
$M$	Internal bending moment	$kN\cdot m$

$Guided Example: Cut a Simply Supported Beam$

STEP-BY-STEP

Solve the two support reactions from whole-beam equilibrium.
Place a cut at coordinate $x$ away from concentrated-force or couple discontinuities.
Draw the left segment using positive left-face directions for $N$ , $V$ , and $M$ .
Solve the three internal resultants from equilibrium.
Draw the right segment with reversed arrows and confirm the same scalar $N(x)$ , $V(x)$ , and $M(x)$ .
Move the cut across each load location and record the new piecewise equations.

$Simulation 1 Instructions: Movable Section Cut$

Move the cut along the beam, display $N$ , $V$ , and $M$ , and verify that the left-section and right-section calculations produce a near-zero consistency error.

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)-4.35 kN

Moment M(x)58.23 kN·m

10 m

20 kN

4.00 m

5.00 m

Left N

-0.00 kN

Left V

-4.35 kN

Left M

58.23 kN·m

Right N

0.00 kN

Right V

-4.36 kN

Right M

58.23 kN·m

Independent normalized equilibrium residual

8.16e-16

Cursor shear

-4.35 kN

Cursor moment

60.40 kN·m

Left reaction

15.65 kN

Right reaction

20.56 kN

Maximum moment

62.58 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)=15.645⟨x-0.00⟩⁰ + 20.555⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰ - 3.600⟨x-5.50⟩¹ + 3.600⟨x-10.00⟩¹

Piecewise moment equation

M(x)=15.645⟨x-0.00⟩¹ + 20.555⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹ - 1.800⟨x-5.50⟩² + 1.800⟨x-10.00⟩²

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

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

$Concept Check 1$

Why do the cut-face arrows reverse between the left and right isolated segments even though the reported scalar values of $N$ , $V$ , and $M$ agree?

$Axial-Force Diagram$

A plot of normal force $N(x)$ along a member, with positive values representing tension and negative values representing compression.

$Axial-Force Discontinuities$

A concentrated axial force creates a jump in the axial-force diagram. Between axial load application points, $N(x)$ is constant when no distributed axial load acts.

$Simulation 2 Instructions: Axial-Force Diagram$

Move the concentrated axial load, predict the constant regions and jump, then reveal the piecewise normal-force diagram.

Tension-positive N · axial equilibrium

Sign convention: positive normal force is tension. With no distributed axial load,

N(x)

remains constant between concentrated axial actions and changes only at an axial-force discontinuity.

Normal force N(x)0.00 kN

10 m

20 kN

4.00 m

4.5 m

Cursor normal force

0.00 kN

Horizontal support reaction

-20.00 kN

Peak |N|

20.00 kN

Axial-force interpretation

N(x) is piecewise constant between concentrated axial actions. A concentrated axial force produces a jump in N; transverse shear and bending-moment diagnostics do not apply to this axial-only model.

N(x)=\text{constant between concentrated axial loads}

$Concept Check 2$

Why is the normal-force diagram constant between concentrated axial loads when no distributed axial load acts?

$Shear-Force Diagram$

A plot of internal shear $V(x)$ along a member. Point forces create jumps, while distributed loading controls its slope.

Load-Shear Differential Relationship

With downward distributed load taken as positive, shear decreases at the rate of load intensity.

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

Variables

Symbol	Description	Unit
$w(x)$	Downward distributed-load intensity	kN/m
$V(x)$	Internal shear-force function	kN

Change in Shear

The change in shear equals the negative signed area under the distributed-load diagram.

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

Variables

Symbol	Description	Unit
$x_1$	Start coordinate	m
$x_2$	End coordinate	m

$Simulation 3 Instructions: Shear-Force Diagram$

Predict jumps at reactions and point loads, then predict the slope under the distributed load before revealing the calculated shear diagram.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Shear V(x)-4.35 kN

10 m

20 kN

4.00 m

4.5 m

Cursor shear

-4.35 kN

Cursor moment

60.40 kN·m

Left reaction

15.65 kN

Right reaction

20.56 kN

Maximum moment

62.58 kN·m at 4.00 m

Minimum moment

0.00 kN·m at 0.00 m

Zero-shear locations

None

Piecewise shear equation

V(x)=15.645⟨x-0.00⟩⁰ + 20.555⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰ - 3.600⟨x-5.50⟩¹ + 3.600⟨x-10.00⟩¹

Piecewise moment equation

M(x)=15.645⟨x-0.00⟩¹ + 20.555⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹ - 1.800⟨x-5.50⟩² + 1.800⟨x-10.00⟩²

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

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

$Concept Check 3$

How does a constant downward load intensity change the shape of the shear diagram, and why?

$Bending-Moment Diagram$

A plot of internal bending moment $M(x)$ along a member. Its slope equals shear, and concentrated applied couples create direct jumps.

Shear-Moment Differential Relationship

The slope of the moment diagram equals the internal shear force.

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

Variables

Symbol	Description	Unit
$M(x)$	Internal bending-moment function	$kN\cdot m$
$V(x)$	Internal shear-force function	kN

Change in Moment

The change in moment equals the signed area under the shear diagram.

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

Variables

Symbol	Description	Unit
$M(x_1)$	Moment at the start coordinate	$kN\cdot m$
$M(x_2)$	Moment at the end coordinate	$kN\cdot m$

$Simulation 4 Instructions: Bending-Moment Diagram$

Move the load and cursor, locate zero shear, and compare it with the calculated maximum or minimum bending moment.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Moment M(x)60.40 kN·m

10 m

20 kN

4.00 m

4.5 m

Cursor shear

-4.35 kN

Cursor moment

60.40 kN·m

Left reaction

15.65 kN

Right reaction

20.56 kN

Maximum moment

62.58 kN·m at 4.00 m

Minimum moment

0.00 kN·m at 0.00 m

Zero-shear locations

None

Piecewise shear equation

V(x)=15.645⟨x-0.00⟩⁰ + 20.555⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰ - 3.600⟨x-5.50⟩¹ + 3.600⟨x-10.00⟩¹

Piecewise moment equation

M(x)=15.645⟨x-0.00⟩¹ + 20.555⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹ - 1.800⟨x-5.50⟩² + 1.800⟨x-10.00⟩²

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

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

$Concept Check 4$

Under what condition does a zero-shear location correspond to a local maximum or minimum of the moment diagram?

$Internal-Force Diagrams for Determinate Frames$

For a frame member, choose a local axis along the member. Resolve the section resultants into local normal and shear components, then calculate the local bending moment. Internal actions at a connecting pin or rigid joint appear as equal-and-opposite resultants on adjacent member free-body diagrams.

$Simulation 5 Instructions: Determinate Frame Diagrams$

Use the combined axial and transverse loading idealization to reveal $N$ , $V$ , and $M$ diagrams together and inspect how each load contributes to a different internal resultant.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Beam BC normal NkN

Beam BC shear VkN

Beam BC moment MkN·m

Column AB normal NkN

Column AB shear VkN

Column AB moment MkN·m

Aₓ

-8.00 kN

Aᵧ

8.00 kN

Cᵧ

12.00 kN

Force residual

0.00e+0

Moment residual

0.00e+0

10 m

5.0 m

20 kN

8 kN

4.00 m

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

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

$Concept Check 5$

Why can a frame member carry axial force, shear, and bending moment simultaneously even when an ideal truss member carries only axial force?

$Piecewise Structural Function$

A load, shear, or moment function represented by different equations over intervals separated by supports, load starts and ends, point forces, or applied couples.

$Discontinuity and Shape Rules$

A concentrated vertical force creates a shear jump but no direct moment jump.
A concentrated applied couple creates a moment jump but no shear jump.
A uniform distributed load makes $V(x)$ linear and $M(x)$ quadratic.
A linearly varying distributed load makes $V(x)$ quadratic and $M(x)$ cubic.
At a simple support or internal hinge, the ideal bending moment is zero unless an applied couple acts at that point.

$Simulation 6 Instructions: Synchronized Cursor$

Move one cursor across $w(x)$ , $V(x)$ , and $M(x)$ . Compare the current values, the shear slope under distributed loading, and the moment slope indicated by shear.

Tension-positive N · sagging-positive M

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

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)

$Concept Check 6$

At a point where $w(x)=0$ but $V(x)$ is positive, what are the local slopes of the shear and moment diagrams?

$Simulation 7 Instructions: Point Load$

Move a point load along a simply supported beam. Predict reaction changes, the downward shear jump at the load, and the two linear moment segments before revealing them.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Load w(x)0.00 kN/m

Shear V(x)-8.00 kN

Moment M(x)44.00 kN·m

10 m

20 kN

4.00 m

4.5 m

Cursor shear

-8.00 kN

Cursor moment

44.00 kN·m

Left reaction

12.00 kN

Right reaction

8.00 kN

Maximum moment

48.00 kN·m at 4.00 m

Minimum moment

0.00 kN·m at 0.00 m

Zero-shear locations

None

Piecewise shear equation

V(x)=12.000⟨x-0.00⟩⁰ + 8.000⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰

Piecewise moment equation

M(x)=12.000⟨x-0.00⟩¹ + 8.000⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹

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

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

$Concept Check 7$

Why is the moment diagram continuous beneath a point load even though the shear diagram jumps there?

Applied-Couple Jump Rule

Under the stated sign convention, a positive clockwise applied couple creates an upward jump of equal magnitude in the internal moment diagram and no shear jump.

\Delta M=M_0,\qquad \Delta V=0

Variables

Symbol	Description	Unit
$M_0$	Applied clockwise couple	$kN\cdot m$
$\Delta M$	Moment-diagram jump	$kN\cdot m$
$\Delta V$	Shear-diagram jump	kN

$Simulation 8 Instructions: Applied Couple$

Change the couple magnitude and sign. Verify that the shear remains continuous while the moment diagram jumps by the applied couple.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Load w(x)0.00 kN/m

Shear V(x)-1.80 kN

Moment M(x)9.90 kN·m

10 m

20 kN

4.00 m

18 kN·m

4.5 m

Cursor shear

-1.80 kN

Cursor moment

9.90 kN·m

Left reaction

-1.80 kN

Right reaction

1.80 kN

Maximum moment

10.80 kN·m at 4.00 m

Minimum moment

-7.20 kN·m at 4.00 m

Zero-shear locations

None

Piecewise shear equation

V(x)=-1.800⟨x-0.00⟩⁰ + 1.800⟨x-10.00⟩⁰

Piecewise moment equation

M(x)=-1.800⟨x-0.00⟩¹ + 1.800⟨x-10.00⟩¹ + 18.000⟨x-4.00⟩⁰

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

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

$Concept Check 8$

Why does a pure applied couple alter moment equilibrium without adding any net vertical force to the beam?

$Simulation 9 Instructions: Distributed Load$

Change the starting and ending load intensities to transition from uniform to triangular loading. Predict the shear slope and moment curvature before revealing the diagrams.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Load w(x)13.25 kN/m

Shear V(x)0.19 kN

Moment M(x)157.78 kN·m

10 m

20 kN/m

5.0 kN/m

4.5 m

Cursor shear

0.19 kN

Cursor moment

157.78 kN·m

Left reaction

75.00 kN

Right reaction

50.00 kN

Maximum moment

157.78 kN·m at 4.51 m

Minimum moment

0.00 kN·m at 0.00 m

Zero-shear locations

4.51

Piecewise shear equation

V(x)=75.000⟨x-0.00⟩⁰ + 50.000⟨x-10.00⟩⁰ - 20.000⟨x-0.00⟩¹ + 0.750⟨x-0.00⟩² + 5.000⟨x-10.00⟩¹ - 0.750⟨x-10.00⟩²

Piecewise moment equation

M(x)=75.000⟨x-0.00⟩¹ + 50.000⟨x-10.00⟩¹ - 10.000⟨x-0.00⟩² + 0.250⟨x-0.00⟩³ + 2.500⟨x-10.00⟩² - 0.250⟨x-10.00⟩³

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

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

$Concept Check 9$

How does changing a uniform load into a triangular load change the polynomial degree of the shear and moment diagrams?

$Gerber Beam Idealization$

A multi-span determinate beam formed by inserting internal hinges so that the structure can be separated into statically determinate segments while transmitting shear and axial force but no hinge moment.

$Simulation 10 Instructions: Multi-Span Determinate Beam$

Treat the two displayed spans as members connected by an internal hinge. Predict the zero hinge moments and solve each determinate span before revealing the moment diagrams.

Tension-positive N · sagging-positive M

dV/dx=-w(x)

and

dM/dx=V(x)

Solve the suspended span first. Its hinge reaction acts upward on that span and downward with equal magnitude on the supported overhang. Both ideal hinge faces have zero bending moment.

Supported overhang moment0.00 kN·m

Suspended-span moment0.00 kN·m

Hinge force on left

-7.500 kN

Hinge force on right

7.500 kN

Action-reaction residual

0.00e+0

Largest hinge moment

0.00e+0 kN·m

20 kN

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

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

$Concept Check 10$

What internal action is released by an ideal hinge, and which force components can still pass through it?

$Common Internal-Force and Diagram Mistakes$

Mixing left-face and right-face signs without reversing the arrows.
Treating a point load as a distributed-load area or drawing a sloped shear segment where no distributed load exists.
Drawing a moment jump under a point force instead of under an applied couple.
Forgetting reaction jumps at supports.
Identifying a maximum moment solely from a plotted sample without checking zero shear and discontinuities.
Using one global equation without activating and deactivating piecewise load terms correctly.
Assigning nonzero moment to an ideal internal hinge.

Internal-Force and Diagram Workflow

Derive axial-force, shear-force, and bending-moment functions region by region, with explicit sign conventions, discontinuity rules, and equilibrium checks.

Key Takeaways

A section cut exposes $N$ , $V$ , and $M$ , and either isolated side must give consistent scalar results.
Point forces create shear jumps; applied couples create moment jumps without shear jumps.
Distributed load controls shear slope, and shear controls moment slope.
Zero shear is a candidate location for a moment extremum when no moment discontinuity occurs there.
Piecewise equations and direct residual checks make structural diagrams auditable.
Determinate frames and hinged multi-span systems are analyzed member by member using rigid-body equilibrium.

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