CIV 103.doc

TECHNICAL INSTITUTE FOR ENGINEERING AND PETROLEUM (THIEP)A subsidiary of Hajr for Technical Education and Training (Edexcel Center Number: 90729)

CIVIL ENGINEERING TECHNOLOGY

COURSE MANUAL

National/Diploma (NQF Level 3 BTEC)

STATICS(CIV 103)

Prepared byBamgbopa Ibrahim A. (MSc, PMP, LEED Green Associate)

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CHAPTER 1 - INTRODUCTION

Mechanics - the physical science which describes or predicts the conditions of rest or motion of bodies under the action of forces.A. Rigid bodies

1. statics2. dynamics

B. Deformable bodiesC. Fluid Mechanics

1. compressible - gas2. incompressible - liquids

In Statics we will assume the bodies to be perfectly rigid, no deformation.

This is never true in the real world, everything deforms a little when a load is applied.

These deformations are small and will not significantly affect the conditions of equilibrium or motion, so we will neglect the deformations.

Rigid body - a body is considered rigid when the relative movements between its parts are negli-gible.

Basic concepts: space, time, mass, force1. Space - the geometric region occupied by bodies whose positions are described by linear and

angular measurements relative to a coordinate system.

2) Time - the measure of the succession of events3) Mass - the measure oaf the inertia of a body, which is its resistance to a change of motion.

sometimes called "quantity of matter"4) Force - the action of one body on another

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y

z

x

Cartesian


Newton developed the fundamentals of mechanics. The concepts above, space, time, and mass are absolute, independent of each other in Newtonian Mechanics.

Newton's 3 Fundamental Laws

1st Law - A particle remains at rest or continues to move in a straight line with a constant speed if there is no unbalanced force acting on it (resultant force = 0).

2nd Law - The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force.

(1)

3rd Law - The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and act along the same line of action (Collinear).

F = mg weight

Or usingF = ma

At the surface of the Earth

a = g F = mg

W = mgg = 9.81 m/s² = 32.2 ft/ s²

System of Units

Base units are units of length, mass and time.

Length Mass TimeSI Units Meter (m) Kilogram (kg) Second (s)English Units Foot (ft) Slug (slug) Second (s)

Force: Newton (N)

1 N = (1 kg)(1 m/ s²) 1 Newton is the force required to give a mass of 1 kg an acceleration of 1 m/ s². Weight is a force.

The weight of 1 kg Mass is: W = mg W = (1 kg)(9.81 m/ s²) = 9.81 NCHAPTER 2 – FORCE VECTORS

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F = ma

W = mg


Scalars and VectorsVectors – A mathematical quantity possessing magnitude and direction.

Scalar – A mathematical quantity possessing magnitude only.

Name some vectors: forces, velocity, and displacementName some scalars: Area, volume, mass energy

Representation of vectorBold R – Word Processors Book uses this.Arrow – Long Hand, Word ProcessorsUnderline R – Long Hand, Typewriter, Word Processors

Magnitude of a VectorBook uses italics for all scalars

Types of Vectors1). Fixed (or bound) vectors – a vector for which a unique point of

application is specified and thus cannot be moved without modifying the conditions of the problem.

2). Free vector – a vector whose action is not confined to or associated with a unique line in space. (couple)

3). Sliding vector – a vector for which a unique line in space (line of action) must be maintained.

For 2 vectors to be equal they must have the same:1). Magnitude P P

2). Direction

They do not need to have the same point of application.

A negative vector of a given vector has same magnitude but opposite direction. P -P

P and –P are equal and opposite P + (-P) = 0

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Vector OperationsProduct of a scalar and a vector

P + P + P = 4P (the number 4 is a scalar)This is a vector in the same direction as P but 4 times as long.

(+n)P = vector same direction as P, n times as long(-n)P = vector opposite direction as P, n times as long

Vector AdditionThe sum of 2 vectors can be obtained by attaching the 2 vectors to the same point and constructing a parallelogram – Parallelogram law.

R Q

R = P + Q P R = resultant vector

Note: The magnitude of P + Q is not usually equal to .

Addition of vectors is commutative: P + Q = Q + P

Triangle Rule P

Q R R Q

P

Let’s add 3 vectors!

Parallelogram Law Q Q

Q P P P

S S S

R1 = Q + P R = R1 + S = Q + P + S

Triangle Rule Q Q P P S

R1 Q R

1

P R

S S R = R1 + S = Q + P + SR1 = Q + P

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Polygon Rule – Successive applications of triangle rule. Q Q P S P R S

Note: P +Q + S = (P + Q) + S = P + (Q + S) vector addition is associative

Vector Subtraction – the addition of the corresponding negative vector

P – Q = P + (-Q)

Resolution of vector into componentsA single vector can be represented by 2 or more vectors. These vectors are components of the original vector. Finding these is called resolving the vector into its components.

There is an infinite number of ways to resolve one vector.

etc.

2 cases of particular interest are:1). One of the 2 components is known. Easy (see above)2). The line of action of both components is know.

When would #2 happen? When you are given a coordinate system!

What are the x and y components of P if P = 1000 lbs, = 30o

Px = P cos 30o = 866 lbs Py = P sin 30o = 500 lbs

Note: Given Px and Py, what is P?;

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P2 = Px2 + Py

2=8662 + 5002 = 1000 lbs1). Given: The fixed structure shown below.

P = 500 NT = 200 N

Find: Combine P and T into a single force R

Law of cosines:

Law of sines:

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30o

Dir of T2?

Dir of T2?

Dir of T2?


2.) Given: A barge is pulled by 2 tugboats. The resultant of the forces exerted by thetugboats is a 5000 pound force directed along the center axis of the barge.

A 1

B

C

2

Find: a). tension in each rope if =45 degreesb). value of such that the tension in rope 2 is minimum.

5000a).

T2

b). 5000dir of

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3). Given: The vertical force F of 350 lbs acts downward at A on the two-membered frame.

Find: The magnitudes of the two components of F directed along AB and AC.

Addition of a System of Coplanar Forces

Vector NotationIn many problems it will be necessary to resolve a force into 2 components that are perpendicular to each other. y

xO

2 vectors, and that have the direction shown and magnitude 1 - unit vectors.

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75


The and provide direction!-suppose I want a vector 4 units long in the x-direction -a vector 5 units long in the negative y-direction

Is a unit vector? Yes

Suppose we have a force, with magnitude Fx , that lies on the x-axis. and one on the y-axis .

What is the resultant, ?

Remember: and are vectors and are magnitude of vectors, which are scalars

What does look like?y

What is the magnitude of ?

What is ?

x

Given what is and

What is ?

Coplanar force resultantsGiven 3 forces, and their resultant is

y

x

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So:

Sign convention! You have either + or - components.

From now on, drop magnitude (| |) sign for all scalars. All vectors will have arrows.

Vector Magnitude of P.

Once you have the components, the resultant vector can be sketched and found using:

1). Given: Replace the 6 kN and 4 kN forces by a single force, expressed in vector nota-tion.

4 kN y

6 kN

x

2). Given: Previous problem. Find using scalar notation.

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4kN y

6kN

x

Cartesian vectors

Right hand coordinate system z

x

y

3-dimensional vectors

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Note:

Direction cosines

is a unit vector

important!!

F => Magnitude => Direction

When in 3-D always use vectors!!

Find complete vectors!!-Don't try to find all the x components and then add them, then y's and z's.

Position vectorsDef: Position vector- a fixed vector which locates a point in space relative to another point.

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x

position vector

Most of the time, the position vector will not begin at the origin, but at a point other than the ori-gin.

y

x

z

Hint: Find x, y, z coordinates of all points first, then compute position vectors.

1). Given: The tower BC is anchored by the guy wire AB which is anchored to the tower at B and the ground at A.

B 40m

80m A

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30m

C

Find: a). The position vector b). The position vector

c). The unit vector

a). A (40,0,-30) B (0,80,0)

b).

c).

2). Given: The previous problem. Suppose the magnitude of the tension in the guy wire is 2500 lbs.

Find: a). the tension vector b). the angles x, y, and z of

a). lbs

b).

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Note: These are the direction cosines

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Force Couple Systems

Let us look at the following

A

O

We want for one reason or another, to have act at point 0. How do we do this?

We can move along its line of action, but we cannot just move it to 0 because if we do so we will modify the action of on the body.

Let do the following

d A

O

We haven't changed anything

Moment of a couple.

Magnitude O

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Thus any force acting on a rigid body may be moved to an arbitrary point 0, provided that a couple of moment equal to the moment of about 0 is added.

Since is a free vector it may be applied anywhere. For convenience the couple vector is usually attached at 0.

This is known as a force-couple system.

The opposite can also be done. Any force-couple system may be replaced by a single equivalent force. How is this done?

This is done by moving the force in the plane perpendicular to until its moment about 0 becomes equal to the moment of the couple to be eliminated.

1). Given:

Find: replace the couple and force by an equivalent, single force applied to the lever. Determine the distance from 0 to the point of application of this force.

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Force System Resultants

In Chap. 2 we showed that was the only condition for the equilibrium of a particle. Now, what about a rigid body?

F

F

, but not in equilibrium. The body rotates. So is a necessary condition, but not sufficient. There is a Torque or Moment on the body. This is what will be discussed in this chapter.

Moment of a force-scalar formulation

Definition: Moment - The measure of the tendency of the force to make a rigid body rotate about a point or fixed axis.

A- Point of application of

Def: The Moment of about O is defined as:

The direction of is also defined as the direction which would bring in line with right hand rule.

Note: is a vector from O to any point on the line of action of .From the definition of a cross product

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From our diagram, r sin d, so

Units: N-m, ft-lbs, in-lbs

Where d represents the perpendicular from O to the line of action of . d is commonly known as the moment arm.

Moment of a force- vector formulation

Remember:

We also know:

Thus,

Again, use vectors in 3D!!

Resultant moment of a system of forces

1). Given

A

24" 100 lb

60o

Find: a). Moment of the 100 lb force about O.b). Magnitude of a horizontal force applied at A which create the same moment

about O. c). The smallest force applied at A which creates the same moment about O.d). Distance from O a 240 lb vertical force must act to create the same moment

about O.

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A

a). 24" 100

------------ O

P P

b). L 24

O

2 Ways1) Components

P=57.7 lbs2) Perpendicular line to force

P=57.7 lbs

A

c). 24 P

OWhy is P perpendicular to the lever?-P is smallest when d in M=Fd is a maximum. This occurs when P is perpendicular to the lever.

-If P is not perpendicular to the lever, we have components parallel and perpendicular to the lever, thus the parallel component is "robbing" the perpendicular component of some force which could contribute to creating the moment.

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d). A B

240

O

2). Given: The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. The tension in the cable is 1349N.

Find: Moment about the coordinate axes of the force exerted by the cable at C.

3-D use:

Note: could have been used also.

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Analysis of Structures

Now we are going to consider internal forces.

In the force analysis of structures it is necessary to dismember the structure and to analyze sepa-rate FBD of individual members in order to determine the forces internal to the structure.

This analysis calls for very careful observance of Newton's 3rd law, which states that each is ac-companied by an equal and opposite reaction.

3 categories of engineering structures:1). Trusses- support loads, stationary, 2 force members2). Frames- support loads, stationary, at least one multi-force members3). Machines- transmit and modify forces, at least one multi-force member

Def. of truss: a framework composed of members connected at their ends to form a rigid structure.Examples: Bridges, roof supports, derricksStructural members used: I-beams, channels, angles, barsFastened together at ends by: Welding , rivets, bolts

Weights of truss members are assumed to be applied to the joints, half the weight at each end. Many times weight is small compared to the forces the members support so it is neglected.

In the analysis, it is assumed that the members are pinned together.

What does this do? The forces at each end of a member reduce to a single force and no couple. Members become a 2 -force member. Equal, opposite, collinear (Newton's 3rd law)

tension Compression

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Triangles constitute a rigid truss. B C

Not rigid

A D

Simple Truss: add 2 new members, attach them to separate existing joints and connect them at a new joint.

B D

Total # of members, mm 2n - 3 n total # of joints

A C

Analysis of trusses by the method of joints

Let's look at the following plane truss C

A D B

F

Dismembering C

A D B

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F

Notice Newton's 3rd law between the pin and member equal and opposite

Since the entire truss is in equilibrium, each pin is in equilibrium.

Truss contains n Pins => 2n equations (x, y)Remember:

m 2n 3 => 2n m 3

Thus 2n or m 3 unknowns may be determined

m => all the forces in the members 3 => and

The entire truss is a rigid body in equilibrium thus we can write the following equation for the entire truss.

These contain no new information and so are not independent.But we can use these to determine reactions at the supports.

The arrangement of pins and members in a simple truss is such that it will always be possible to find a joint involving only 2 unknown forces.

Once these forces are determined their values are transferred to adjacent joints and this joint is analyzed. This is repeated until all the unknown forces are determined.

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Examples

1). Given: the following truss

Find: using the method of joints, find the force in each member, state compression or ten-sion.

Joint B

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These are forces on pins! Forces on members are:

AB in Tension lbs T BC in Compression

lbs C

Joint A

Check:

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2). Given: the following truss

5 kN 20 kN 5 kN

A B C

1.6m

D E F3m 3m

12 kN

Find: Using the method of joints, determine the force in each member, state compression, or ten-sion.

FBD 5 kN 20 kN 5 kN

A B C

y

D E F x

12 kN

Joint A

5 kN

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Joint D

5 kN

21 kN

Joint E

Truss and loading is symmetrical about the centerline.

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Method of Sections

Method of joints works well when you want to know the forces in all the members.

If we want to know a force in a particular member, the method of sections is more efficient.

Look at the following truss:

To solve for the reactions at E and G, we use the entire truss as a free-body.

By method of joints we use free-body diagrams of individual nodes.

A

We know that the entire truss is in equilibrium.

If this is the case, then each joint is also in equilibrium.

Well, if this is the case, then a portion of the truss smaller than the entire but larger than the joints must also be in equilibrium.

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For instance:

This section must also be in equilibrium.

So, we can cut the truss into sections and solve for the unknowns.

What is the only limitation of this method?We can only cut such that there are 3 unknowns because:

If I wanted to find FBD what would I do?FBD is only unknown!!

What about FCE? FCE only unknown!!!

What about FBE? FBE only unknown!!

Examples1). Given: The truss shown below

Find: Determine the force in members EF and GI.

FBD

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Member EF

Member GI

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2). Given :

Find: Determine the force in members FK and JO

FBD

3). Given: the truss shown below

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Find: Force in members DF,EF, and EG.

FBD - whole structure => Ax = 0 Ay = 0

Member DF (enlarged view)

Member EF Member EG

Center of Gravity and Centroid

Center of gravity- the point at which the weight of the body can be concentrated.

Composite plates and wires

In most cases a flat plate can be divided up into common shapes. We can use this fact to find the center of gravity and/or the centroid.

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Equating moments

Solve for C.G.

Following the previous development

Solve for centroid

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Examples 1). Calculate the x and y coordinate of the centroid of the shaded area.

Part # Area (mm2)1 /4)(240)2 = 45,239 0 0 0 02 -0.5(180)(90) = -8,100 -90/3 = -30 0 243,000 0

Centroid:

2). Find the y coordinate, measured up from the bottom, of the area shown below.

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Solution:

Part # Area (mm2)1 10(160) = 1,600 5 5(1600) = 8,0002 2[10(20)] = 400 15 15(400) = 6,0003 2[10(50)] = 1,000 35 35(100) = 35,0004 2[10(120)] = 2,400 70 70(2,400) = 168,000

Centroid:

3) Given: the machine element shown below 25 kg/m 2.5(10 ) kg/mm 2.5(10 ) g/mm

Find: C.G.

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semicircle

Part # mass (g)I 0.754 6 78.98 -40 4.524 59.55 -30.16II -0.543 6 62 -40 -3.258 -33.666 21.72III 1.2 6 37 -40 7.2 44.4 -48IV 2.4 50 6 -40 120.0 44.4 -96V -0.543 55 6 -40 -29.865 -3.258 21.72

To-tals

3.268 98.601 81.426 -130.72

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Moments of Inertia of Areas

Consider the beam shown below

Internal forces are distributed forces whose magnitudes vary linearly with the distance y from an axis passing through the centroid of the section.

Total resultant:

1st moment, Qx

The moment:

2nd moment, Qxx

Moment of inertia

Moment of inertia of an area by integration

dA = (a - x)dy dA = ydx

Rectangular moments of inertia

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Polar moment of inertia

Radius of gyration of an area

Has moment of inertia

Concentrate this area in a thin strip

If a strip is to have same moment of inertia,the strip should be placed as shown.

By def:

Thus:

or

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Radius of gyration - the distance, from an axis that passes through the CG, at which all the area or mass of an object could be concentrated without changing its moment of inertia.

Parallel- Axis Theorem

Consider:

Moment of inertia with respect to AA'

MOI, wrt 1ST moment total areacentroidal axis wrt BB' = 0

centroid, C, islocated on BB'

Parallel axis theorem

The moment of inertia of an area wrt to any given axis AA' is equal to the MOI of the area with respect to the centroidal axis plus . Product of area, A, and the square of the distance d be-tween the 2 axes.

The parallel axis theorem can be applied only if one of the two parallel axes passes through the centroid of the area.

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Moments of inertia by composite areas

A composite area, A, made of several components areas

The integral representing the MOI of A may be subdivided into integrals computed over

Thus, the MOI of A with respect to a given axis can be obtained by summing the MOI of with respect to the same axis.

MOI of common shapes on inside of back book cover.

2 things:1). Before adding the MOI of the component areas, the parallel axis theorem should be used to transfer each MOI to the desired axis.2). The radius of gyration of a composite area is NOT equal to the sum of the radii of gyration of the component areas.

Amenable to spreadsheets.

Example1). Determine and of the area shown with respect to centroidal axis.

Solution:

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Part # A (mm4) (mm4) (mm4) (mm4)1 4,800 40 192,000 (60)(803)/12 =

2.56(106) 4800(76-40)2 =

6.22(106)(603)(80)/12 =

1.44(106)0

2 7,200 100 720,000 180(403)/12 =960(103)

7200(100-76)2 =4.15(106)

(1803)(40)/12 =19.44(106)

0

Totals 12,000 912,000 3.52(106) 10.37(106) 20.88(106) 0

First we find the centroid to be:

Now using the centroid we find the Ad2 terms for each part. Then we sum the MOI and Ad2 terms for each part to find the MOI about the centroidal axis for the entire section:

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CIV 103.doc