+1 vote
159 views
by
Engineering Mechanics all formula module wise

2 Answers

0 votes
by (98.9k points)
edited by

 1.1 System of Coplanar Forces:

Classification of Force Systems:

Concurrent Forces:

  • Resultant Force: \( \mathbf{R} = \sum \mathbf{F} \)
  • Moment About a Point: \( \mathbf{M} = \sum \mathbf{r} \times \mathbf{F} \)

Parallel Forces:

  • Resultant Force: \( R = \sum F_i \)
  • Location of Resultant: \( x = \frac{\sum F_i \cdot x_i}{\sum F_i} \)

Non-concurrent Non-parallel Forces:

  • Resultant Force: \( \mathbf{R} = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2} \)
  • Angle of Resultant: \( \tan \theta = \frac{\sum F_y}{\sum F_x} \)

Principle of Transmissibility:

The effect of a force on a rigid body is the same whether the force is applied at its original point or along its line of action.

Composition and Resolution of Forces:

\( \mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 + \ldots + \mathbf{F}_n \)

\( \mathbf{F}_x = \sum \mathbf{F}_i \cos \theta_i \)

\( \mathbf{F}_y = \sum \mathbf{F}_i \sin \theta_i \)

1.2 Resultant:

Resultant of Coplanar and Non-Coplanar Force Systems:

Concurrent Forces: \( \mathbf{R} = \sum \mathbf{F}_i \)

Parallel Forces: \( R = \sum F_i \)

Non-concurrent Non-parallel Forces: \( \mathbf{R} = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2} \)

Moment of Force about a Point:

\( \mathbf{M} = \mathbf{r} \times \mathbf{F} \)

Couples and Varignon's Theorem:

Couples: \( \mathbf{M} = \mathbf{F} \cdot d \)

Varignon's Theorem: \( \mathbf{R} = \sqrt{\mathbf{F}_1^2 + \mathbf{F}_2^2} \)

Distributed Forces in Plane:

For distributed loads, integrate to find resultant forces and moments.

Centroid:

First Moment of Area:

\( Q_x = \int_A x \, dA \)

\( Q_y = \int_A y \, dA \)

Centroid of Composite Plane Laminas:

\( \bar{x} = \frac{Q_x}{A} \)

\( \bar{y} = \frac{Q_y}{A} \)

 

 

 

 2.1 Equilibrium of System of Coplanar Forces:

Conditions of Equilibrium:

Concurrent Forces: \(\sum \mathbf{F} = 0\) and \(\sum \mathbf{M} = 0\) (Net force and net moment are zero)

Parallel Forces: \(\sum F = 0\) and \(\sum M = 0\) (Net force and net moment are zero)

Non-concurrent Non-parallel Forces: \(\sum \mathbf{F} = 0\) and \(\sum \mathbf{M} = 0\) (Net force and net moment are zero)

Couples: \(\sum \mathbf{M} = 0\) (Net moment is zero)

Equilibrium of Rigid Bodies:

Free Body Diagrams:

  • Identify and isolate the body
  • Show all external forces and couples
  • Apply conditions of equilibrium

2.2 Equilibrium of Beams:

Types of Beams:

Simple Beams: Supported at both ends

Compound Beams: Combinations of simple beams

Types of Supports and Reactions:

Supports:

  • Hinged Support (Pin)
  • Roller Support
  • Fixed Support

Reactions:

  • Vertical Reaction (\(R_V\))
  • Horizontal Reaction (\(R_H\))
  • Moment Reaction (\(M\))

Determination of Reactions:

For various types of loads on beams, use the equations:

  • Sum of Vertical Forces: \(\sum F_V = 0\)
  • Sum of Horizontal Forces: \(\sum F_H = 0\)
  • Sum of Moments: \(\sum M = 0\)

 

 

 

<!DOCTYPE html> Friction and Kinematics Formulas

03 Friction:

Static Friction:

Force of static friction (\(F_{\text{static}}\)):

\[ F_{\text{static}} \leq \mu_s \cdot N \]

Dynamic/Kinetic Friction:

Force of kinetic friction (\(F_{\text{kinetic}}\)):

\[ F_{\text{kinetic}} = \mu_k \cdot N \]

Coefficient of Friction:

Static friction coefficient (\(\mu_s\))

Kinetic friction coefficient (\(\mu_k\))

Angle of Friction:

Angle of friction (\(\theta\)):

\[ \tan \theta = \frac{{\text{Opposite side}}}{{\text{Adjacent side}}} = \frac{{\mu_s}}{{1}} \]

Laws of Friction:

  1. Friction is proportional to the normal force.
  2. Friction is independent of the apparent area of contact.
  3. Friction is independent of the sliding velocity.
  4. Friction depends on the nature of the surfaces in contact.

Concept of Cone of Friction:

The cone of friction represents all possible directions of the frictional force.

Equilibrium of Bodies on Inclined Plane:

Forces along the inclined plane:

\[ F_{\text{parallel}} = W \cdot \sin \theta \] \[ F_{\text{perpendicular}} = W \cdot \cos \theta \]

Equations for equilibrium:

\[ \sum F_x = 0 \] \[ \sum F_y = 0 \] \[ \sum M = 0 \]

Application to problems involving wedges and ladders.

04 Kinematics of Particle:

Motion of Particle with Variable Acceleration:

Acceleration (\(a\)) as a function of time:

\[ a = \frac{dv}{dt} \]

General Curvilinear Motion:

Components of acceleration:

\[ a_t = \frac{dv}{dt} \] \[ a_n = \frac{v^2}{r} \]

Tangential & Normal Component of Acceleration:

Tangential component (\(a_t\)) and normal component (\(a_n\)) of acceleration.

Motion Curves:

Acceleration-time (\(a-t\), velocity-time (\(v-t\), and displacement-time (\(s-t\) curves.

Application of Concepts of Projectile Motion:

Related numerical formulas.

0 votes
by (98.9k points)
<!DOCTYPE html> Mechanics Formulas

Module 1: Statics

1.1 System of Coplanar Forces:

Classification of Force Systems:

  • Concurrent Forces: \(\mathbf{R} = \sum \mathbf{F}\)
  • Parallel Forces: \(R = \sum F_i\)
  • Non-concurrent Non-parallel Forces: \(\mathbf{R} = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}\)

Principle of Transmissibility:

The effect of a force on a rigid body is the same whether the force is applied at its original point or along its line of action.

Composition and Resolution of Forces:

\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 + \ldots + \mathbf{F}_n\)

\(\mathbf{F}_x = \sum \mathbf{F}_i \cos \theta_i\)

\(\mathbf{F}_y = \sum \mathbf{F}_i \sin \theta_i\)

1.2 Resultant:

Resultant of Coplanar and Non-Coplanar Force Systems:

Concurrent Forces: \(\mathbf{R} = \sum \mathbf{F}_i\)

Parallel Forces: \(R = \sum F_i\)

Non-concurrent Non-parallel Forces: \(\mathbf{R} = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}\)

Moment of Force about a Point:

\(\mathbf{M} = \mathbf{r} \times \mathbf{F}\)

Couples and Varignon's Theorem:

Couples: \(\mathbf{M} = \mathbf{F} \cdot d\)

Varignon's Theorem: \(\mathbf{R} = \sqrt{\mathbf{F}_1^2 + \mathbf{F}_2^2}\)

Distributed Forces in Plane:

For distributed loads, integrate to find resultant forces and moments.

Centroid:

First Moment of Area:

\(Q_x = \int_A x \, dA\)

\(Q_y = \int_A y \, dA\)

Centroid of Composite Plane Laminas:

\(\bar{x} = \frac{Q_x}{A}\)

\(\bar{y} = \frac{Q_y}{A}\)

Module 2: Equilibrium

2.1 Equilibrium of System of Coplanar Forces:

Conditions of Equilibrium:

Concurrent Forces: \(\sum \mathbf{F} = 0\) and \(\sum \mathbf{M} = 0\)

Parallel Forces: \(\sum F = 0\) and \(\sum M = 0\)

Non-concurrent Non-parallel Forces: \(\sum \mathbf{F} = 0\) and \(\sum \mathbf{M} = 0\)

Couples: \(\sum \mathbf{M} = 0\)

Equilibrium of Rigid Bodies:

Free Body Diagrams:

  • Identify and isolate the body
  • Show all external forces and couples
  • Apply conditions of equilibrium

2.2 Equilibrium of Beams:

Types of Beams:

Simple Beams: Supported at both ends

Compound Beams: Combinations of simple beams

Types of Supports and Reactions:

Supports:

  • Hinged Support (Pin)
  • Roller Support
  • Fixed Support

Reactions:

  • Vertical Reaction (\(R_V\))
  • Horizontal Reaction (\(R_H\))
  • Moment Reaction (\(M\))

Determination of Reactions:

For various types of loads on beams, use the equations:

  • Sum of Vertical Forces: \(\sum F_V = 0\)
  • S

Related questions

+1 vote
1 answer 66 views
+1 vote
1 answer 42 views
+1 vote
1 answer 40 views
+1 vote
1 answer 37 views

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

500 users

...