Internal Forces in Structural Members: Method of Sections and Applications

Internal Forces

Introduction to Internal Forces

Internal forces are the forces that act within a structural member when external loads are applied. Understanding these forces is essential for designing safe and efficient structures, such as beams and columns, which must resist various types of loads.

• Structural members like beams and columns are designed to support loads, often applied perpendicularly to their axes.

• Internal forces help determine the required thickness and material strength at different points in a structure.

• Common applications include bridge spans and billboard supports, where the shape and size of members are influenced by internal force distribution.

Types of Internal Forces

In two-dimensional structural analysis, three primary types of internal forces are considered:

• Normal (Axial) Force (N): Acts perpendicular to the section, either compressing or stretching the member.

• Shear Force (V): Acts tangentially or along the face of the section, causing sliding between adjacent parts.

• Bending Moment (M): Causes the member to bend, producing tension and compression within the cross-section.

These forces are typically shown on a Free Body Diagram (FBD) at the location of interest.

Method of Sections

The method of sections is a systematic approach for determining internal forces at a specific location within a structural member. It involves cutting the member at the point of interest and analyzing the resulting section.

• Take an imaginary cut at the location where internal forces are to be determined.

• Decide which resulting section (left or right) is easier to analyze.

• Draw a Free Body Diagram (FBD) of the chosen section, showing all external loads and internal forces (N, V, M) at the cut surface.

• Apply the Equations of Equilibrium (E-of-E) to solve for the unknown internal forces:

$\sum F_x = 0$ $\sum F_y = 0$ $\sum M = 0$

Steps for Determining Internal Forces

Follow these steps to analyze internal forces in a structural member:

1. Imaginary Cut: Select the location for the cut and decide which section to analyze.

2. Support Reactions: If necessary, determine support or joint reactions by drawing an FBD of the entire structure and solving for unknowns.

3. Section FBD: Draw an FBD of the chosen section, indicating N, V, and M at the cut surface.

4. Equilibrium Equations: Apply the E-of-E to solve for the internal forces.

Applications

Beams and Columns in Structures

Beams and columns are fundamental structural elements. Their design depends on the distribution of internal forces:

• Beams supporting bridge spans are often thicker at the supports than at the center, reflecting higher internal forces near supports.

• Columns supporting billboards or other loads are usually wider/thicker at the bottom, where internal forces are greatest due to the accumulation of load.

Worked Example: Internal Forces in a Beam

Example 1: Determining Internal Forces at Point C

Given: A beam loaded as shown. Find: Internal forces at point C. Plan: Use the method of sections.

1. Take an imaginary cut at C. Analyze the right section (from C to B) for simplicity.

2. Determine the support reaction $B_y$ using the FBD of the entire beam:

   • $\sum F_x = B_y = 0$

   • $\sum M_A = -B_y(9) + 18(3) = 0$

   • $B_y = 6$ kip

3. Draw the FBD of the right section, showing $N_C$, $V_C$, and $M_C$.

4. Apply equilibrium equations:

   • $\sum F_x = N_C = 0$

   • $\sum F_y = -V_C - 6 = 0 \implies V_C = -6$ kip

   • $\sum M_C = -6(4.5) - M_C = 0 \implies M_C = -27$ kip·ft

Result: At point C, $N_C = 0$, $V_C = -6$ kip, $M_C = -27$ kip·ft.

Example 2: Group Problem Solving

Given: A beam loaded as shown. Find: Internal forces at point C. Plan: Use the method of sections, analyzing the left section.

1. Take an imaginary cut at C. Analyze the left section (from A to C).

2. Determine support reactions $A_x$ and $A_y$ using the FBD of the entire frame:

   • $\sum F_x = A_x + 400 = 0 \implies A_x = -400$ N

   • $\sum M_B = -A_y(5) - 400(1.2) = 0 \implies A_y = -96$ N

3. Draw the FBD of the left section, showing $N_C$, $V_C$, and $M_C$.

4. Apply equilibrium equations:

   • $\sum F_x = N_C - 400 = 0 \implies N_C = 400$ N

   • $\sum F_y = V_C - 96 = 0 \implies V_C = 96$ N

   • $\sum M_C = 96(1.5) + M_C = 0 \implies M_C = -144$ N·m

Result: At point C, $N_C = 400$ N, $V_C = 96$ N, $M_C = -144$ N·m.

Concept and Attention Quizzes

Quiz Questions

• Given a column loaded with a vertical or horizontal force, identify sections where internal loads are the same, largest, or lowest.

• Determine the magnitude of internal loads (normal, shear, bending moment) at a specific point.

• Identify the force component acting tangent to a section (shear force).

Summary Table: Types of Internal Forces

Type of Force	Symbol	Direction	Effect
Normal (Axial) Force	N	Perpendicular to section	Compression or tension
Shear Force	V	Tangential to section	Sliding between parts
Bending Moment	M	About axis of section	Bending of member

Key Definitions

• Free Body Diagram (FBD): A graphical representation showing all forces and moments acting on a body or section.

• Equations of Equilibrium (E-of-E): Mathematical conditions for a body to be in static equilibrium: $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$.

• Method of Sections: Analytical technique for finding internal forces by cutting the member and analyzing the resulting section.

Additional info:

• These notes are foundational for engineering statics and mechanics of materials, not directly part of a standard Calculus curriculum, but they do use equilibrium equations and concepts that require algebraic manipulation and basic calculus for more advanced analysis (e.g., distributed loads, shear/moment diagrams).