Statics

Statics is the study of forces on bodies at rest or moving at constant velocity (equilibrium). This is fundamental to structural analysis and machine design.

Key Principle: Equilibrium

For a body to be in equilibrium:

ΣF = 0   (sum of all forces equals zero)
ΣM = 0   (sum of all moments equals zero)

No acceleration means all forces and moments balance out.

Moments and Torque

What is a Moment?

A moment (or torque) is the rotational effect of a force. It depends on:

1. Magnitude of the force
2. Distance from the pivot point (moment arm)

M = F × d

Where:
- M = moment (N⋅m or lb⋅ft)
- F = force (N or lb)
- d = perpendicular distance from pivot to line of action (m or ft)

Sign Convention

Counterclockwise (⟲) → Positive moment
Clockwise (⟳) → Negative moment

Example: Opening a Door

         ← F = 50 N
        |
    [===|========]  ← door (width = 0.8 m)
        |
        ↓ hinge

Moment at hinge:

M = F × d
M = 50 N × 0.8 m = 40 N⋅m (counterclockwise)

Key Insight: Pushing farther from the hinge creates more torque. That's why doorknobs are placed far from hinges!

Moment of a Force About a Point

        ↑ F
        |
        |
    ----*--------  ← Point O
         d⊥

M_O = F × d⊥

Where d⊥ is the perpendicular distance from point O to the line of action of F.

Principle of Moments (Varignon's Theorem)

The moment of a resultant force equals the sum of moments of its components:

M_O = F_x × d_y + F_y × d_x

This is incredibly useful for complex force systems.

Types of Supports and Reactions

Understanding support reactions is crucial for analyzing structures.

Support Types

2D Equilibrium Equations

For planar (2D) problems, we have 3 equations:

ΣFx = 0   (horizontal forces)
ΣFy = 0   (vertical forces)
ΣM = 0    (moments about any point)

This means we can solve for up to 3 unknowns in a 2D static problem.

Beam Analysis

Beams are fundamental structural elements that resist bending.

Simply Supported Beam

   Point Load P
        ↓
    ====*====
    ↑       ↑
    R_A     R_B
    
    |<--a-->|<--b-->|
    |<------L------>|

Given: Beam length L, load P at distance 'a' from left
Find: Reactions R_A and R_B

Solution:

ΣFy = 0:  R_A + R_B - P = 0           ... (1)
ΣM_A = 0: R_B × L - P × a = 0         ... (2)

From equation (2):
R_B = (P × a) / L

From equation (1):
R_A = P - R_B = P - (P × a)/L = P(L - a)/L = (P × b)/L

Check: Sum of moments about B should also equal zero.

Cantilever Beam

        Load P
          ↓
    ┤=====*=====
    ↑M_A         (free end)
    ┤ R_A
    
    |<----L---->|

Given: Beam length L, load P at free end
Find: Reaction force R_A and moment M_A at fixed support

Solution:

ΣFy = 0:  R_A - P = 0
          R_A = P

ΣM_A = 0: M_A - P × L = 0
          M_A = P × L

Uniformly Distributed Load (UDL)

    w (N/m) - load per unit length
    ↓↓↓↓↓↓↓↓↓↓↓↓
    ============
    ↑         ↑
    R_A       R_B
    |<---L--->|

Total load: W = w × L (acts at center of beam)

For symmetric support:
R_A = R_B = W/2 = (w × L)/2

Example Problem: Beam with Multiple Loads

    100 N       200 N
      ↓           ↓
    ==*====*======
    ↑            ↑
    R_A          R_B
    
    |<-2m->|<-3m->|

Find: Reactions R_A and R_B

Solution:

ΣFy = 0:
R_A + R_B - 100 - 200 = 0
R_A + R_B = 300          ... (1)

ΣM_A = 0 (taking moments about point A):
R_B × 5 - 100 × 2 - 200 × 5 = 0
5 R_B = 200 + 1000 = 1200
R_B = 240 N

From equation (1):
R_A = 300 - 240 = 60 N

Verification (check ΣM_B = 0):

R_A × 5 - 100 × 3 - 200 × 0 = 0
60 × 5 - 300 = 0 ✓

Trusses

A truss is a structure made of slender members (bars) connected at joints (pins). Used in bridges, roofs, and towers.

Assumptions for Truss Analysis

1. All members are connected by frictionless pins
2. Loads applied only at joints
3. Members are two-force members (force along the member only)
4. Weight of members is negligible

Member Forces

• Tension (T): Member is being pulled (←→)
• Compression (C): Member is being pushed (→←)

Method of Joints

Analyze forces at each joint using equilibrium equations.

Example: Simple Truss

        2000 N
          ↓
        B *
       /|\
      / | \
     /  |  \
  A *   |   * C
    ↑   |   ↑
   R_A  |  R_C
    
  |<-2m>|<-2m>|
  Height = 2m

Step 1: Find Reactions

By symmetry: R_A = R_C = 1000 N

Step 2: Analyze Joint A

At joint A:
- R_A = 1000 N (upward)
- F_AB (along member AB, unknown)
- F_AC (along member AC, horizontal, unknown)

Angles: θ = 45° (for member AB)

ΣFx = 0:  F_AC - F_AB cos(45°) = 0
ΣFy = 0:  R_A + F_AB sin(45°) = 0

From ΣFy:
1000 + F_AB × 0.707 = 0
F_AB = -1414 N (negative = compression)

From ΣFx:
F_AC = -1414 × 0.707 = -1000 N (tension)

Continue this process for each joint to find all member forces.

Method of Sections

Cut through the truss and analyze a section as a free body. Useful for finding forces in specific members without analyzing the entire truss.

      Cut line
    - - - - - - 
         ↓ P
       B *
      /|  \
     / |   \
  A *  |    * C
    ↑  |    ↑

Analyze left or right section, apply equilibrium equations (ΣFx, ΣFy, ΣM = 0).

Friction

Friction is the force that resists sliding between surfaces.

Static vs Kinetic Friction

Usually: μ_s > μ_k (harder to start moving than keep moving)

Friction Force

F_friction ≤ μ_s × N

Where:
- F = friction force
- μ_s = coefficient of static friction
- N = normal force

Example: Block on Incline

       [====] m = 10 kg
      /
     /  θ = 30°
    /________
    
μ_s = 0.5

Will the block slide?

Forces parallel to plane:
- Component of weight: W∥ = mg sin(θ) = 10 × 9.81 × sin(30°) = 49.05 N
- Maximum friction: F_max = μ_s × N = μ_s × mg cos(θ)
  F_max = 0.5 × 10 × 9.81 × cos(30°) = 42.5 N

Since W∥ > F_max (49.05 > 42.5), the block will slide.

Centroids and Centers of Gravity

The centroid is the geometric center of a shape. The center of gravity is where the total weight acts.

Simple Shapes

Composite Shapes

For complex shapes, divide into simple parts:

x̄ = (Σ A_i × x_i) / (Σ A_i)
ȳ = (Σ A_i × y_i) / (Σ A_i)

Where:
- A_i = area of part i
- x_i, y_i = centroid of part i

Example: L-Shaped Beam

    [====2====]  ← Rectangle 1: 4m × 1m
    [         ]
    [ 1       ]  ← Rectangle 2: 1m × 3m
    [         ]
    [=========]

Find centroid from bottom-left corner:

Rectangle 1: A_1 = 4 × 1 = 4 m², centroid at (2, 3.5)
Rectangle 2: A_2 = 1 × 3 = 3 m², centroid at (0.5, 1.5)

x̄ = (4 × 2 + 3 × 0.5) / (4 + 3) = (8 + 1.5) / 7 = 1.36 m
ȳ = (4 × 3.5 + 3 × 1.5) / 7 = (14 + 4.5) / 7 = 2.64 m

Practice Problems

Problem 1: Beam Reactions

A 6 m beam is supported at both ends. A 500 N load is applied 2 m from the left end. Find the reactions.

<details> <summary>Solution</summary>

ΣM_A = 0:  R_B × 6 - 500 × 2 = 0
           R_B = 1000/6 = 166.7 N

ΣFy = 0:   R_A + 166.7 - 500 = 0
           R_A = 333.3 N

</details>

Problem 2: Moment Calculation

A 100 N force is applied perpendicular to a wrench 0.3 m from the bolt. What is the moment on the bolt?

<details> <summary>Solution</summary>

M = F × d = 100 N × 0.3 m = 30 N⋅m

</details>

Problem 3: Friction

A 50 kg box sits on a floor (μ_s = 0.4). What horizontal force is needed to start it moving?

<details> <summary>Solution</summary>

N = mg = 50 × 9.81 = 490.5 N
F_max = μ_s × N = 0.4 × 490.5 = 196.2 N

Need at least 196.2 N to overcome static friction.

</details>

Key Takeaways

✓ Equilibrium: ΣF = 0 and ΣM = 0
✓ Moments: M = F × d (force times perpendicular distance)
✓ Support reactions: Different supports provide different constraints
✓ Beams: Sum forces and moments to find reactions
✓ Trusses: Method of joints or sections for member forces
✓ Friction: F ≤ μ × N, static friction prevents motion

Next Steps

Ready to add motion to the analysis:

• Chapter 03: Dynamics: forces and motion
• Practice more beam and truss problems
• Learn about shear force and bending moment diagrams (advanced statics)

Pro Tip: Always start with a clear FBD and label all forces before writing equations!