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DEVAPRAKASAM DEIVASAGAYAM Professor of Mechanical Engineering Room:11, LW, 2 nd Floor School of Mechanical and Building Sciences Email: [email protected] , [email protected] ME202: Engineering Mechanics (3:1:0:4) Devaprakasam D, Email: [email protected] , Ph: +91 9786553933

ME202 Engineering Mechanics L3

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Lectures on Engineering Mechanics

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  • DEVAPRAKASAM DEIVASAGAYAMProfessor of Mechanical Engineering

    Room:11, LW, 2nd FloorSchool of Mechanical and Building Sciences

    Email: [email protected], [email protected]

    ME202: Engineering Mechanics (3:1:0:4)

    Devaprakasam D, Email: [email protected], Ph: +91 9786553933

  • Express the 2D and 3D equilibrium equations for particle resulting from the application of Newtons 1st Law

    0

    0

    0

    Y

    X

    F

    F

    F

    3D

    0

    0

    0

    0

    Z

    Y

    X

    F

    F

    F

    F2D

    0

    0

    0

    ||

    F

    F

    F

    2 Independent Eqns 3 Independent EqnsDevaprakasam D, Email: [email protected], Ph: +91 9786553933

    2D and 3D Equilibrium Conditions

  • Employ the rules for drawing a free body diagram (FBD) and sketch the FBD for a particle.

    Identify the particle or body of interest Sketch the particle or body free of constraints Apply the external forces. Add dimensions, angle, slope, other details.

    Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Free Body Diagram (FBD)

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Free Body Diagram (FBD)

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Free Body Diagram (FBD)Coordinate SystemCoordinate systems are an important component in the creation of a free body diagram, and are used to designate a position of the body in space. A coordinate system needs to be established prior to drawing a free body diagram to aid tracking of the relative direction in which forces are acting and to make it easier to determine the components of forces if they are acting at angles.

    Drawing the BodyThe next step in creating a free body diagram is drawing a representation of the object being analyzed. A free body diagram is drawn as if it were isolated from its environment, creating a free body that displays only necessary information. This makes it clear what forces are acting on the body. The shape of the body is generally simplified, but represents the overall appearance and behavior of the object that is being analyzed. The body should be drawn at the origin of the coordinate system in order to distinguish between forces acting in positive and negative directions.

    Drawing the ForcesOnce an object is isolated from its environment, the forces that were acting on it need to be represented to provide accurate visualization of existing forces. The external forces and resulting reactions that act directly on the body are the only forces that should be drawn. These forces are represented by arrows positioned to indicate the point of application, the angle, and the magnitude.

    Solving the DiagramThe final step in creating a free-body diagram is to use Newtons Laws to solve for any unknown forces. Forces acting in the same plane are added together, keeping in mind whether they are directed in the positive or negative direction. The forces in each plane X, Y, and Z are summed, providing equations that contain both known and unknown values. For forces that act at an angle, the magnitude is divided into X, Y, and Z components based on the angle, and these components are included separately in the X, Y, and Z equations. Depending on whether the body is at constant velocity or accelerating, the total sum of the forces is equal to either zero or the product of mass and acceleration.

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    3D Force rectangular Components

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Examples

    Q1

    Q2

    Q3

    Q4

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    SolutionThe forces at A are,TAB, TAC, TAD and P

    P= P j. Express all the forces in unit vectors I, j, kAB= - 4.2 i- 5.6j ; |AB|= 7.00 mAC = 2.4 i+ -5.6j+ 4.2k; |AC|=7.4mAD=-5.6 j - 3.3 k; |AD|= 6.5m

    TAB = TAB AB = [-0.6i-0.8j] TABTAC = TAC AC =[0.3240i-756j+0.567k] TACTAD = TAD AD = [-0.861j-0.5076k] TAD

    Equilibrium condition

    0F TAB + TAC + TAD +Pj =0

    Substituting the values of TAB ,TAC , TADEquating the I, j, k components to zeroWe will get TAB =259 (known) , TAC =479.15N , TAB =535.66 N, P=1031N

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Home Work

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Moment

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Moment (M)

  • Devaprakasam D, Email: [email protected], Ph: +91 9786553933

    Moment (M)