Internal Forces in Structural Members - Theory & Concepts

Internal Forces in Structural Members - Theory & Concepts

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 to ensure it won't break or deform excessively. This is the study of Internal Forces.

Types of Internal Forces

When a rigid body is cut by an imaginary plane, the internal forces acting on that cross-section can be resolved into three distinct components (for 2D coplanar systems).

Internal Force Components (2D)

Normal Force ( $N$ ): The force acting perpendicular (normal) to the cross-section. 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.

Important

Sign Convention:

Consistency is critical. The standard sign convention for beams 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 $\frown$ ).

Method of Sections for Internal Forces

To find the internal forces at a specific point, we use the method of sections.

Shear and Moment Diagrams

Because internal forces vary along the length of a beam, engineers use Shear and Moment Diagrams to plot the values of

V

and

M

as functions of position (

x

). This allows for the immediate identification of the maximum values needed for design.

Relationships between Load, Shear, and Moment

Calculus provides powerful relationships that make drawing these diagrams much faster than cutting infinite sections:

$ \frac = -w(x)

$ Note: The slope of the shear diagram at any point is equal to the negative of the distributed load intensity at that point.

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

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

$ \frac = V

$ Note: The slope of the moment diagram at any point is equal to the shear force at that point. (Maximum moment occurs where shear is zero!).

$ \Delta M = \int V , dx

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

Effect of Concentrated Forces and Moments

The rules governing Shear (

V

) and Moment (

M

) 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

Drawing shear and moment diagrams using integration equations can be slow. The graphical method, based on the relationships

dV/dx = -w(x)

and

dM/dx = V

, is much faster and relies on calculating areas.

Procedure

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)

For beams with multiple concentrated loads or varying distributed loads, writing separate shear and moment equations for every segment is tedious. Singularity functions (or Macaulay brackets) allow you to express the shear

V(x)

and moment

M(x)

across the entire length of the beam using a single equation.

Macaulay Brackets Definition

A singularity function is written as

\langle x - a \rangle^n

. Its value depends on the term inside the brackets:

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

(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 $n \ge 0$ .)

Important

Integration Rules: 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} \quad \text{for } n \ge 0

Application: When formulating the load function

w(x)

, concentrate forces

P

applied at

x=a

are represented by

-P\langle x-a \rangle^{-1}

, but practically, we write the shear function directly as:

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 + \dots

And integrating 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 + \dots

Interactive Shear and Moment Visualizer

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.

Shear & Moment Diagram Visualizer

Load Position ($x$)5 m

Load Magnitude ($P$)10 kN

Reactions:

Left Support ($R_1$): 5.00 kN

Right Support ($R_2$): 5.00 kN

Maximum Moment: 25 kN·m (at x = 5m)

Shear Diagram (V)

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Moment Diagram (M)

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Relations among Load, Shear, and Moment

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.

Load, Shear, and Moment Equations

Shear and Load: The rate of change of the shear force with respect to $x$ is equal to the negative of the distributed load intensity at that point: $\frac{dV}{dx} = -w(x)$ . Integrating this gives the change in shear between two points: $\Delta V = -\int w(x) \, dx$ (the negative area under the load diagram).
Moment and Shear: The rate of change of the bending moment with respect to $x$ is equal to the shear force at that point: $\frac{dM}{dx} = V(x)$ . Integrating this gives the change in moment between two points: $\Delta M = \int V(x) \, dx$ (the area under the shear diagram).

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 $dV/dx = -w(x)$ and $dM/dx = V$ .

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