Principle of Engineering ENG2301

Principle of Engineering ENG2301

Mechanics Section

Textbook: A Foundation Course in Statics and DynamicsAddison Wesley Longman 1997

Syllabus OverviewA StaticsB Dynamics

UnitsforceNewton (N)stressNewton per metre squared (N/m2 ) or Pascal, 1 Pa = 1 N/m2 (Pa)pressureNewton per metre squared (N/m2 ) or bar, 1 bar = 1x105 N/m2 (bar)moment, torque, couple Newton . Metre (Nm)

UnitsMost commonly used prefixesmicrox 10-6millix 10-3mkilox 103 kmegax 106 Mgiga x 109 G* Note Capitals and lower case letters are important

Scalars and VectorTwo kind of quantities:ScalarVectorScalar quantities have magnitude but no directional propertiescan be handled by ordinary algebra, e.g. c= a+b, c= 8 if a=3, b= 5e.g. time, mass, speed and energy etc. etc....

VectorAssociated with directions and magnitudee.g. Force, displacement, acceleration and velocityCan be represented by a straight line with arrowhead and the magnitude is shown by the length

Vector Addition and SubtractionBy Triangle or Parallelogram lawsAdditionV = V1 + V2 V is called the resultant vector

Vector Addition and SubtractionSubtractionV = V1 - V2 can be regarded as V = V1 + (- V2)- V2 is drawn in the opposite direction V is the resultant vector

Vector Addition and SubtractionAdding more than two vectorsV = V1 + V2 +V3 +V4

Resolution of VectorsAny vector can be resolved into componentsCommonly resolve into two components perpendicular to each otherV = Vx + VyVx = V cos Vy = V sin magnitude V = Vx2 + Vy2) = tan-1 (Vy /Vx )

Force and Newtons First LawFirst Law - If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest), or will move with constant speed in a straight line (if originally in motion).State of Equilibrium - Equilibrium exists when all the forces on a particle are in balance . The velocity of a particle does not change , if the particle is in Equilibrium .

Interpretation of First LawA body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant.For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero. A Force can be defined as 'that which tends to cause a particle to accelerate', assuming that the force is not in Equilibrium with other forces acting on the body.

ForceA force cannot be seen, only the effect of a force on a body may be seen.Force Units: S.I. Unit ,Newton, (N) or (kN)Force is a vector quantity. It has both magnitude and direction.

Force VectorsPolar and Rectangular Coordinates

Example 1Calculate the components in rectangular coordinates of the 600 N force.Solution

EMBED Equation.2

EMBED Equation.2

35(

600N

_892623455.unknown

_892623556.unknown

Example 2A force vector has the components 600 kN and 300 kN in the x and y directions respectively, calculate the components in polar coordinates.Solution

Resultant ForceParallelogram Method

Resultant ForceAlgebraic Method

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

EMBED Equation.2

_864824892.unknown

_864825569.unknown

_864825572.unknown

_864825574.unknown

_864825235.unknown

_864824688.unknown

_864824838.unknown

_864824685.unknown

Resultant ForceTriangle of Forces MethodOrder is not important

Example 3Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two forces shown in the diagram,a) Using the Parallelogram Methodb) Using the Triangle of Forces Methodc) Using the algebraic calculation methodSolution

Example 3 (Solution)Or -108.260 from +ve x axis

6kN

4kN

15(

30(

Equilibrium of Concurrent forcesEquilibrant E are equal and opposite to Resultant RE = -R

Conditions for EquilibriumCoplanar: all forces being in the same plane (e.g.only x-y plane, no forces in z direction)Concurrent: all forces acting at the same point (particle)

For three forces acting on a particle

Some Definitions Particle is a material body whose linear dimensions are small enough to be irrelevantRigid Body is a body that does not deform (change shape) as a result of the forces acting on it .

Polygon of ForcesEquilibrium under multiple forcesRigid body under concurrent forcesForces acting on particle

Resultant and EquilibrantResultant = - Equilibrant R = - F5

Example 4The diagram shows three forces acting on a particle . Find the equilibrant by drawing the polygon of forces.

30(

84 kN

122 kN

45 kN

20(

50(

Newtons Third LawThe forces of action and reaction between bodies in contact have the same magnitude, but opposite in direction.

BANG!

Solid Surface

Hammer

Action and Reaction

Free Body Boundary

Free Body diagram of Surface

Free Body Diagram of Hammer

Force on Hammer caused by Surface

Force on Surface caused by Hammer

Hammer

Solid Surface

Free Body DiagramFree Body Diagram - used to describe the system of forces acting on a body when considered in isolationRmgRRR

Free Body Diagram

W

Boundary

Free Body

Wa Action

Wr Reaction

W

August 30, 1999/~WRO1074

System of Particles or BodiesTwo or more bodies or particles connected together are referred to as a system of bodies or particles.

External Force External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other bodies and the reactions due to supports.

Transmissibility of Force

Load and ReactionLoads are forces that are applied to bodies or systems of bodies.Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body.

Tensile and Compressive ForcesPush on the body which is called a compressive force

Pull on a body which is called a tensile force

Tensile Force

Action

Reaction

Compressive Force

Reaction

Action

Procedure for drawing a free body diagramStep 1: Imagine the particle to be isolated or cut free from its surroundings. Draw or sketch its outlined shape.

Step 2: Indicate on this sketch all the forces that act on the particle. These forces can be applied surface forces, reaction forces and/or force of attraction.

Procedure for drawing a free body diagram

Procedure for drawing a free body diagramStep 3: The forces that known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that unknown.

Example 5

Ring

Ceiling Support

Weight = 10 N

Free Body boundary

Reaction Force = 10 N

Gravity Load =10 N

Action Load = 10 N

Ceiling Support

Ring

Weight = 10 N

Example 6

tow rope

tow rope

Ring.

Barge

Tug No.1

Tug No.2

60(

25(

Example 6 (Solution)resultant R of the two forces in tow ropes No.1 & No. 2 from the components in the x and y directions:

tow rope

tow rope

Ring.

Barge

Tug No.1

Tug No.2

60(

25(

Example 6 (Solution)Equilibrant E = - R

E = 32.43 kN

42.1(

42.1(

R = 32.43 kN

Example 6 (Solution)Resultant R is the sum of the actions of the tow ropes on the bargeEquilibrant E is the reaction of the barge to the ropesE = - R

tow rope

tow rope

Ring.

Barge

Tug No.1

Tug No.2

60(

25(

Moment and CoupleMoment of ForceMoment M of the force F about the point O is defined as: M = F d where d is the perpendicular distance from O to FMoment is directional

Moment and CoupleMoment = Force x Perpendicular Distance

B

A

F

(

M = F.d

= F.r.cos (

r

d

Resultant of A System of ForcesAn arbitrary body subjected to a number of forces F1, F2 & F3.Resultant R = F1 + F2 + F3Components Rx = F1x + F2x + F3x Ry = F1y + F2y + F3y

Resultant of A System of ForcesResultant moment Mo = Sum of MomentsMo = F1 d1 + F2 d2 + F3 d3Mo = R d

CoupleFor a CoupleR =F = 0But Mo 0Mo = F(d+l) - Fl = FdMoment of couple is the same about every point in its plane

Example 7Calculate the total (resultant) moment on the body.

15 N

A

50 mm

100 mm

300 mm

30 N

170 mm

30 N

15 N

Example 7 (Solution)Taking moments about the corner A

Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm(resultant force =0, moment is the same about any point).

Equilibrium of MomentsThe sum of all the moments is zero when the body is in moment equilibrium. or

If the body is in equilibrium the sum of the moments of all the forces on acting on a rigid body is the same for all points on the body. It does not matter at which point on a rigid body you choose for taking moments about

Example 8Calculate the resultant moment and the equilibrant moment.

B

A

5 N

15 N

10 N

1.25 m

1.0 m

0.5 m

2.5 m

3.0 m

Example 8 (Solution)Take moment about ATake moment about B

B

A

5 N

15 N

10 N

1.25 m

1.0 m

0.5 m

2.5 m

3.0 m

Example 8 (Solution)Note that the body is not in vertical and horizontal equilibrium.There is no unique value for the resultant moment. The value depends on where the resultant force acts, ie., depends on the perpendicular distance between the resultant force and the point for taking moment.Therefore, the moments about A and B are different.

Example 9Cantilever beamFind the reaction force and moment at the built in end

Point Load of weight W

Built in End

or Fixed End

Wall

Beam

d

W

Example 9 (Solution)Taking moment about A

C

A

B

Reaction V

Free Body Diagram of Beam

Moment Reaction MA

W

d

General Equations of Equilibrium of a Plane (Two Dimensional) Rigid Body(Non-concurrent forces)For complete equilibrium, all 3 equations must be satisfied

Types of Beam SupportsSimply supported beam

Types of Beam Supports

TIME \@ "MMMM d, yyyy" June 6, 1996/ FILENAME \* MERGEFORMAT Document4

M

Rx

Ry

Ry

Rx

Ry

a moment

Two components of force and

perpendicular forces

Both parallel and

to surface only.

Force perpendicular

Fix Support

Pin Joint Fixed to the ground

Pin Joint supported by a roller

Types of Supports and ConnectionsSimply supported beam

Types of Loading on Beams


W / unit length

Uniformly distributed loads

Another way of representing

W / unit length

W / unit length

W

W

b)

a)

Diagrammatic representation

Distrubuted load (uniform)

Diagrammatic representation

Point load



supported beam with a

concentrated load W

loads and supports by forces

Free body diagram replacing

Diagrammatic view of simply

b

a

B

C

A

W

W

C

B

A

B

R

A

R



distributed load w

supported beam with uniformly

(a+b)/2

F= w(b-a)

w N/m

loads and supports by forces

Free body diagram replacing

Diagrammatic view of simply

b

a

B

A

Example 10Find the reactions at the supports for the beam shown in the diagram.


7 m

3.5m

3.5m

5 kN/m

20 kN

Example 10(Solution)


3.5m

3.5m

7 m

B

A

R

R

35 kN

20 kN

Example 11Express F in terms of m, a and b.

W

F

a

b

Example 11(Solution)Ratio a/b is called Mechanical Advantage

R

a

b

F

W

Example 12Find reaction forces at supports A and B.

Example 12 (Solution)Consider the sum of vertical forces and horizontal forces are zero, and since RAx = 0 as point B can only take up vertical force.RAx = 200 x sin30o = 100 N

RAy + RBy = 350 + 200 cos30o = 350 + 173.2 N = 523.2 N

Example 12 (Solution)Taking moment about A,RBy x 0.3 = 350 x 0.15 + 200 cos30o x 0.4RBy x 0.3 = 121.8 N RBy = 405.9 NTherefore the reaction at point B is 405.9N upward.

RAy = 523.2 - 405.9 = 117.3 N

Example 12 (Solution)The reaction at point A is 117.3 N upward and 100 N to the left.Resultant at A is:RA = (117.32 + 1002) = 154.1 N

angle q = 49.55o100117.3q

Example 13Find reaction forces at supports A and B.

B

A

55 kN

750 mm

600 mm

Example 13 (Solution)

HB

VA

HA

55 kN

750 mm

600 mm

Example 14

Example 14 (Solution)

Example 14 (Solution)Fy = 0:Ry - 75 = 0Ry = 75 NFx = 0:T - Rx = 0T = Rx MO = 0:75 x 250 - T x 100 x cos 200 = 0

ThereforeRx = T = 200 NR = (Rx2 + Ry2) = 214 N = tan-1 (Ry/Rx) = 20o 33

Example 14 (Solution)Reaction R is that exerted by the frame on the bellcrank, which is equal and opposite to that on the chassis.

Example 14 (Graphical Solution)Having determined the line of reaction R, a scaled force polygon can be drawn.By measurement, T = 200 N, R = 214 N

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