3. Internal Actions and Stress Resultants

Definition

Internal actions are unseen forces that act in the interior of a body. They can be:

• Normal (axial) force $N$
• Shear (transverse) force $T$
• Bending moment $M$

To find internal actions, we use sectioning: cutting a body into two parts across a cross-section.

Internal actions MUST vanish when summed up!

Cross-section-local coordinate space conventions

A positive cross-section has beam axis z going outward the body, while a negative cross-section has the beam z-axis going inward.

The Diagrams

The Goal

To determine the internal forces ($N,T,M$) at every single point along the beam’s length ($z$). This tells us where the structure is weakest (the “Critical Points”).

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A. The Setup (Method of Sections)

1. Define Coordinate System ($z$): Place an origin (usually at the left end, A). The variable $z$ travels along the beam axis. 2. Identify Discontinuities (Segments): You cannot write one single function for the whole beam if the loading changes. You must split the beam into segments wherever:

• A concentrated force or moment is applied (Point B).
• A constraint exists. (Points A, G)
• The geometry changes. (Point C)
• A distributed load starts or ends. (Points D, E)

B. Cutting the Beam

Step 1: Sectioning Pick a segment (e.g., Segment AC, where $0\le z\le 2a$). Imagine cutting the beam at an arbitrary distance $z$.

• Rule: Discard one half (usually the right). Keep the other (usually the left).
• Expose Forces: Replace the cut part with unknown internal forces: Normal ($N$), Shear ($T$), and Moment ($M$). Step 2: Equilibrium of the Cut Piece Write equilibrium equations for just that isolated piece to find functions for $N(z),T(z),M(z)$.
• $\sum F_z=0\to N$
• $\sum F_y=0\to T$
• $\sum M_{cut}=0\to M$ (Repeat for the next segment, e.g., CB)

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C. The Results (The Diagrams)

NOTE

The diagrams below are not related to the one above! Instead, they refer to this one:

Once you have the functions, you plot them. These plots reveal the physics of the beam.

1. Axial Load Diagram ($N$)

Shows tension (+) or compression (-).

• In this example, it is constant ($N=F$).

2. Shear Force Diagram ($T_y$)

Shows the force trying to slice the beam vertically.

The Jump Rule

At Point C (where force $F$ pushes down), the Shear Diagram JUMPS exactly by the amount $F$.

• Discontinuities in load = Jumps in Shear.

3. Bending Moment Diagram ($M_x$)

Notice that the above shear force diagram is the derivative of this bending moment diagram!

Shows the bending effort. This is the most important diagram for failure.

• Key Insight: The moment is Zero at free ends (A and B).
• The Peak: It peaks under the load (Point C). This is where the beam snaps ($M_{max}=-2Fa/3$).

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Properties of Diagrams

The Power of Derivatives

You do not need to cut the beam 10 times. The shape of the diagrams is mathematically predicted by the loads.

• Load ($q$) determines the Slope of Shear ($T$).
• Shear ($T$) determines the Slope of Moment ($M$).

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A. The Differential Rules

These rules apply to every infinitesimal slice of the beam ($dz$).

1. Slope of Shear = -Load $\frac{dT}{dz}=-q(z)$

• Translation: The value of the distributed load $q$ tells you how steep the Shear diagram is.
• Example: If $q=0$ (no load), the slope is 0 (Shear is horizontal).

2. Slope of Moment = Shear $\frac{dM}{dz}=T(z)$

• Translation: The value of the Shear $T$ tells you how steep the Moment diagram is.
• The “Order” Rule: The mathematical order increases by 1 each step.
  • Load: Constant $\to$ Shear: Linear $\to$ Moment: Parabolic.

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B. The “Graphical Method” (Sketching)

Use these relationships to spot errors or sketch rapidly.

Feature	Effect on Shear ($T$)	Effect on Moment ($M$)
No Load ($q=0$)	Constant (Horizontal Line)	Linear (Slope = $T$)
Uniform Load ($q=c$)	Linear (Sloped Line)	Parabolic (Curve)
Point Force ($F$)	JUMP (Vertical shift by $F$)	Kink (Change in slope)
Couple ($C$)	No change	JUMP (Vertical shift by $C$)

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C. Finding the Critical Point (Max Moment)

Critical Rule

Since $\frac{dM}{dz}=T$, the Maximum (or Minimum) Bending Moment occurs exactly where the Shear Force is ZERO.

$T(z)=0⟹M_{max}$

Strategy: Do not calculate moment everywhere. Find where $T$ crosses the axis, and calculate $M$ only at that specific point.