Internal Forces in Structural Members - Theory & Concepts

Internal Forces in Structural Members - Theory & Concepts

Learning Objectives

Understand the definition and types of internal forces in structural members.
Apply the method of sections to determine internal forces at a specific point.
Construct shear and bending moment diagrams for beams under various loading conditions.
Utilize the relationships between load, shear, and moment to analyze beams graphically.
Apply singularity functions (Macaulay's Method) to express shear and moment as single equations across a beam's entire length.

This topic provides a comprehensive overview of internal forces in structural members, covering how to calculate normal, shear, and bending forces. It introduces the Method of Sections, differential relationships between load and shear, graphical construction of shear and moment diagrams, and the use of singularity functions.

Internal Forces

Forces that act within a material or structural member to hold it together when subjected to external loads. Engineers must analyze these forces to ensure the material will not break or deform excessively.

The Purpose of Analyzing Internal Forces

In previous topics, we focused on finding external support reactions and forces acting at joints. However, to design a structural member, an engineer must know the forces acting inside the material. This is the core study of internal forces.

Types of Internal Forces

Normal Force ( $N$ )

The force acting perpendicular (normal) to the cross-section of a member. It tends to stretch (Tension) or compress (Compression) the member.

Shear Force ( $V$ )

The force acting tangent (parallel) to the cross-section. It tends to slide one part of the member past the other.

Bending Moment ( $M$ )

The couple moment acting on the cross-section. It tends to bend the member.

Internal Force Components (2D)

When a rigid body is cut by an imaginary plane, the internal forces acting on that cross-section can be resolved into these three distinct components (for 2D coplanar systems): Normal Force ( $N$ ), Shear Force ( $V$ ), and Bending Moment ( $M$ ).

Standard Sign Convention for Beams

Consistency is critical when analyzing beams. The standard sign convention is:

Normal Force: Tension is positive (+), Compression is negative (-).
Shear Force: Positive shear causes a clockwise rotation of the isolated segment. (If you cut a beam, a downward force on the left face is positive; an upward force on the right face is positive).
Bending Moment: Positive moment causes the beam to bend concave upwards (like a smile $\smile$ ), putting the top fibers in compression and bottom fibers in tension. Negative moment causes it to bend concave downwards (like a frown $\frown$ ).

Sign Convention Traps

When analyzing beams, a common mistake is to ignore or inadvertently flip the sign convention during the method of sections. For example, if you analyze the right half of a cut beam, remember that the positive shear force points upwards, not downwards. Consistency is key to creating accurate shear and bending moment diagrams. Always draw assumed internal forces in the positive direction on your FBD.

Method of Sections for Internal Forces

Analyzing Forces at a Specific Point

To find the internal forces at a specific point within a member, we apply the method of sections. This technique exposes the internal forces by imagining a cut through the structure.

Method of Sections for Internal Forces

STEP-BY-STEP

Determine Support Reactions: Calculate all external support reactions by applying global equilibrium equations to the entire structure.
Make an Imaginary Cut: Section the member at the specific point where the internal forces need to be determined.
Isolate a Segment: Choose either the left or the right portion of the cut member to analyze. Usually, the segment with fewer external loads is easier to work with.
Draw Free-Body Diagram (FBD): Draw the FBD of the isolated segment, including all external applied loads, support reactions, and the unknown internal forces ( $N$ , $V$ , and $M$ ) at the cut cross-section. Assume they act in the positive directions according to the sign convention.
Apply Equilibrium Equations: Use $\sum F_x = 0$ , $\sum F_y = 0$ , and $\sum M = 0$ on the isolated segment to solve for the unknown internal normal force, shear force, and bending moment.

Shear and Moment Diagrams

Shear and Moment Diagrams

Graphs that plot the internal shear force ( $V$ ) and bending moment ( $M$ ) as functions of position ( $x$ ) along the length of a beam. They allow for the immediate identification of the maximum values needed for design.

The Need for Diagrams

Because internal forces vary along the length of a beam, engineers must plot these variations rather than merely finding the forces at a single point.

Relations among Load, Shear, and Moment

Differential Relationships

In the analysis of beams, the distributed load $w(x)$ , internal shear force $V(x)$ , and internal bending moment $M(x)$ are intrinsically related through differential calculus. These relationships simplify the construction of shear and moment diagrams and make drawing these diagrams much faster than cutting infinite sections.

Shear and Load Differential Relationship

The rate of change (slope) of the shear force with respect to position is equal to the negative of the distributed load intensity.

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

Variables

Symbol	Description	Unit
$V$	Internal shear force	-
$x$	Position along the beam	-
$w(x)$	Distributed load intensity	-

Change in Shear

The change in shear between two points equals the negative area under the load diagram between those points.

\Delta V = -\int w(x) \, dx

Variables

Symbol	Description	Unit
$\Delta V$	Change in shear force	-
$w(x)$	Distributed load intensity	-
$dx$	Differential length element	-

Moment and Shear Differential Relationship

The rate of change (slope) of the bending moment with respect to position is equal to the shear force. Maximum moment occurs where shear is zero.

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

Variables

Symbol	Description	Unit
$M$	Internal bending moment	-
$V(x)$	Internal shear force function	-
$x$	Position along the beam	-

Change in Moment

The change in moment between two points equals the area under the shear diagram between those points.

\Delta M = \int V(x) \, dx

Variables

Symbol	Description	Unit
$\Delta M$	Change in bending moment	-
$V(x)$	Internal shear force function	-
$dx$	Differential length element	-

Effect of Concentrated Forces and Moments

The rules governing shear and moment diagrams are distinct at points where concentrated loads or moments are applied:

Concentrated Force ( $P$ ): Causes an abrupt jump (discontinuity) in the shear diagram. The change in shear $\Delta V$ is equal to the magnitude of the force. (An upward force causes an upward jump). It changes the slope of the moment diagram but does not cause a jump in the moment itself.
Concentrated Couple Moment ( $M_0$ ): Has no effect on the shear diagram. It causes an abrupt jump in the moment diagram. By standard beam sign convention, an applied clockwise external moment causes an upward (positive) jump in the internal moment diagram.

Graphical Method (Area Method) for Shear and Moment Diagrams

The Graphical Area Method

Drawing shear and moment diagrams using integration equations can be slow. The graphical method, based on the relationships $\frac{dV}{dx} = -w(x)$ and $\frac{dM}{dx} = V(x)$ , is much faster and relies on calculating areas.

Graphical Method for Diagrams

STEP-BY-STEP

Find External Reactions: Draw the Free-Body Diagram of the entire beam and determine all support reactions.
Set Up Axes: Draw a vertical reference line at the left end of the beam. Draw a horizontal axis below the beam for the Shear ( $V$ ) diagram, and another horizontal axis below that for the Moment ( $M$ ) diagram.
Draw Shear Diagram ( $V$ ): Start at $V=0$ at the left end. Whenever you encounter a concentrated upward load or reaction, jump up by that amount. Whenever there is a downward point load, jump down. If there is a distributed load $w(x)$ between two points, calculate the area under $w(x)$ . Subtract this area from the shear value at the start of the section to find the shear value at the end of the section ( $\Delta V = -\int w \, dx$ ). If $w(x)$ is uniform (constant), the shear diagram is a straight line with a constant slope of $-w$ . If $w(x)$ is triangular, the shear diagram is a parabola.
Find Zero-Shear Points: Identify the exact $x$ -coordinates where the shear diagram crosses zero. These points are critical because they correspond to the maximum (or minimum) bending moments. Use similar triangles or the equation $\Delta V = w \cdot x$ to find these distances.
Draw Moment Diagram ( $M$ ): Start at $M=0$ at a pinned or roller end (unless there is an applied couple moment). If it's a fixed end, start at the calculated reaction moment. Calculate the area of each shape (rectangle, triangle, spandrel) in the shear diagram. The change in moment between two points is equal to the area of the shear diagram between them ( $\Delta M = \text{Area}_V$ ). If the shear is a constant horizontal line (degree 0), the moment is an inclined straight line (degree 1). If the shear is an inclined straight line (degree 1), the moment is a parabola (degree 2). If an external concentrated clockwise couple moment is applied to the beam, the moment diagram jumps UP by that amount. A counter-clockwise moment causes a jump DOWN.

Singularity Functions (Macaulay's Method)

Singularity Functions (Macaulay Brackets)

Mathematical brackets written as $\langle x - a \rangle^n$ that allow engineers to express the shear $V(x)$ and moment $M(x)$ across the entire length of a beam using a single equation, instead of writing separate equations for every segment.

Evaluating Macaulay Brackets

The value of a singularity function depends on the term inside the brackets:

If $x < a$ , then $\langle x - a \rangle^n = 0$
If $x \ge a$ , then $\langle x - a \rangle^n = (x - a)^n$

(Note: By convention, for $n < 0$ , the function strictly represents concentrated effects like point moments or forces, but for standard shear/moment equations, we primarily use $n \ge 0$ .)

Integration Rule for Singularity Functions

Singularity functions integrate similarly to standard polynomials, but the brackets remain intact.

\int \langle x - a \rangle^n \, dx = \frac{\langle x - a \rangle^{n+1}}{n+1}

Variables

Symbol	Description	Unit
$x$	Position along the beam	-
$a$	Position of the discontinuity (e.g., load location)	-
$n$	Power of the singularity function ( $n \ge 0$ )	-

Shear Function using Macaulay Brackets

When formulating the shear function, concentrated forces and distributed loads are represented as terms with singularity functions.

V(x) = R_1 - P_1 \langle x-a_1 \rangle^0 - P_2 \langle x-a_2 \rangle^0 - w_0 \langle x-a_3 \rangle^1

Variables

Symbol	Description	Unit
$V(x)$	Internal shear force function	-
$R_1$	Reaction force at the support	-
$P_1, P_2$	Concentrated point forces	-
$w_0$	Uniform distributed load intensity	-
$a_1, a_2, a_3$	Locations of the applied loads	-

Moment Function using Macaulay Brackets

Integrating the shear function yields the single moment equation for the entire beam.

M(x) = R_1 x - P_1 \langle x-a_1 \rangle^1 - P_2 \langle x-a_2 \rangle^1 - \frac{w_0}{2} \langle x-a_3 \rangle^2

Variables

Symbol	Description	Unit
$M(x)$	Internal bending moment function	-
$R_1$	Reaction force at the support	-
$P_1, P_2$	Concentrated point forces	-
$w_0$	Uniform distributed load intensity	-
$a_1, a_2, a_3$	Locations of the applied loads	-

Interactive Shear and Moment Visualizer

Interactive Shear and Moment Visualization

Explore how the position and magnitude of a single point load on a simply supported beam affect the internal shear and bending moment across its entire length. Note how the moment is maximum exactly where the shear diagram crosses zero.

Interactive Physics Simulation

Shear Force & Bending Moment Diagram Simulator

Apply a point load P at any position along a simply supported beam. Observe the resulting reaction forces, step shear diagram V, and parabolic bending moment diagram M.

Point Load Position (x)5.0 m

Load Magnitude (P)15 kN

Shear & Moment Calculus Relationships

Shear is the spatial derivative of Bending Moment:

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

Change in Moment is the area under Shear curve:

M(x) = \int V(x) \, dx

Notice that the maximum bending moment occurs exactly at the point where the shear diagram crosses zero ($x = 5\text{ m}$).

Shear Diagram V (kN)

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Bending Moment Diagram M (kN·m)

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Reaction R1 (Left)

7.50 kN

Reaction R2 (Right)

7.50 kN

Max Bending Moment

37.50 kN·m

Key Takeaways

Internal Forces must be determined to design structural members against failure.
The three 2D components are Normal Force ( $N$ ), Shear Force ( $V$ ), and Bending Moment ( $M$ ).
A strict Sign Convention (Tension positive, clockwise shear positive, "smile" bending moment positive) is required for consistency.
The Method of Sections involves finding support reactions, cutting the member, exposing internal forces on the cut face, and applying equilibrium equations.
Shear and Moment Diagrams map these forces over the entire member. Calculus dictates that $\frac{dV}{dx} = -w(x)$ and $\frac{dM}{dx} = V(x)$ .

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Internal Forces in Structural Members - Theory & Concepts

Learning Objectives

Internal Forces

The Purpose of Analyzing Internal Forces

Types of Internal Forces

Normal Force (NNN)

Shear Force (VVV)

Bending Moment (MMM)

Internal Force Components (2D)

Standard Sign Convention for Beams

Sign Convention Traps

Method of Sections for Internal Forces

Analyzing Forces at a Specific Point

Method of Sections for Internal Forces

Shear and Moment Diagrams

Shear and Moment Diagrams

The Need for Diagrams

Relations among Load, Shear, and Moment

Differential Relationships

Shear and Load Differential Relationship

Change in Shear

Moment and Shear Differential Relationship

Change in Moment

Effect of Concentrated Forces and Moments

Graphical Method (Area Method) for Shear and Moment Diagrams

The Graphical Area Method

Graphical Method for Diagrams

Singularity Functions (Macaulay's Method)

Singularity Functions (Macaulay Brackets)

Evaluating Macaulay Brackets

Integration Rule for Singularity Functions

Shear Function using Macaulay Brackets

Moment Function using Macaulay Brackets

Interactive Shear and Moment Visualizer

Interactive Shear and Moment Visualization

Shear Force & Bending Moment Diagram Simulator

Normal Force ( $N$ )

Shear Force ( $V$ )

Bending Moment ( $M$ )