ENG2002 - Electronic Structure

ENG2002: R.I. Hornsey and Mark A. Post Struct: 1

ENG2002 Chapter 6Forces and Structures


Overview• We would like to apply our increased

understanding of the microscopic properties of materials to larger scale structures

• This branch of engineering is called statics because nothing is supposed to move

• We will briefly review forces, moments and couples

• As well as equilibrium and the free body diagram• From there, we can consider more complex

issues, such as trusses and beams• This is a brief overview of a large field

and we can only touch a small part of the total subject


Where you’ve seen some of this before• Phys1010 covers

forces, vectors and motion equilibrium and elasticity

• See Halliday, Resnick and Walker Ch. 2 – 6 & 13• While we will give a brief reminder here, it is

assumed that you know this material already

V = (x, y)

0 (x,0)

(0,y)Vectors can be broken down into components, to simplify calculation

V = (x,y) = (x,0) + (0,y)


Forces• Forces are vectors

in order to be fully described, both the magnitude and direction of a force is required

• There are three types of vectors free vector – has a magnitude and direction, but

its line of action does not pass through a unique position in space (e.g. all points in a car have the same velocity)

sliding vector – has a unique magnitude, direction and its line of action passes through a unique point, but the point of application of the force can be anywhere along the line of application (e.g. horizontally pushing or pulling a car)

fixed vector – a vector with a specific point of application (e.g. a load on a beam)


Transmissability• In all the cases considered in this course, we will

assumed that bodies are rigid no forces are generated within the body

• Hence, it is assumed that forces are transmitted perfectly through the body “the external effect of a force on a body is the same for all

points of application of the force along its line of action”

• Two separate forces add, reaction forces cancel!• Note that this refers to the external effect

internally, the microscopic response of the body might be quite different (but has little effect on the appearance)

forces have the same effect reaction forces are equal

F FFTotal: 2F Total: 0

F


Types of force• A number of classifications of forces can be drawn• Contact (physical connection and pull/push)

versus non-contact (e.g. gravity)• Point forces versus distributed forces


The lines of actions of forces due to ropes attached to an eyelet are considered to pass through the same point

Such force systems are termed concurrent

Concurrent force systems can be replaced by a single resultant force by the vector sum of all forces

Now, imagine that all the forces are restricted to a single 2-D plane(such as this 2-D slide itself)

These forces are called coplanar

Finally, imagine two forces applied directly on the same straight line.

These forces are collinear

Types of force


O

A

A

F

MO O Fd

MO = F X d

Moment (or torque) is a vector quantity obtained by cross product.

If we only care about the magnitude, use

|MO| = Fd (= T)

O is the moment centred is the moment armA–A is the moment axis

Moment• Any force that is not through the centre of mass

of an object causes the object to want to rotate• The moment of the force F, about O (a.k.a. torque)

is a measure of tendency to rotate about axis A–A which is perpendicular to the plane containing F and O direction can be determined by the right-hand rule

d


The original Starship Enterprise NCC1701 has engines that clearly do not pass through the centre of mass.

The resulting moment should cause it to rotate!

This was fixed in the later NCC1701d model(also the impulse engines were placed in this line)

http://www.cs.umanitoba.ca/~djc/startrek/pics/BigEnPlanet.gifhttp://www.research.ibm.com/people/d/dfb/trek/ncc1701.jpg

A Space Moment


Principle of moments• States that the moment of the resultant of a

system of forces is equal to the vector sum of the moments of the individual forces about the same point or axis

a

b A

B

Rh

O

d

MR = Rd = R(h cos )

MA = Aa = A(h cos )

MB = Bb = B(h cos )

also Rcos = Acos + Bcos

substituting for the cosines from the top three expressions and multiplying through by h gives:

MR = MA + MB


Centre of Mass• For any object or collection of objects, there is a

location at which all the mass can be thought of as being concentrated so a force acting through this centre of mass will cause no

rotation because the inertial mass on all sides is balanced

• On earth for objects small compared with the earth’s curvature, the centre of gravity is essentially identical to the centre of mass

• A similar concept applies to geometrical shapes (lines, areas, volumes), but the term is centroid because a shape is a concept (has no physical existence) however, as long as mass is evenly distributed

(homogeneous) the calculation is the same!


Calculating centre of mass• To derive the centre of mass, we make use of the

definition of centre of gravity the moment of the entire body (i.e. at COG) around an axis

is equal to the sum of all the moments of the elemental masses comprising the body

we assume that COG and centre of mass are the same, so we can divide through by g

About the y-axis:Moment of whole body is = mgxcom

Moment of all elements = g∫x dm

Hence xcom = (∫x dm)/m

Similarly:ycom = (∫y dm)/mzcom = (∫z dm)/m

G

mg

gdmxcom ycom

zcom

elemental mass at x, y, z

x

y

z


Couple• A couple consists of two equal, non-collinear

forces that are anti-parallel That is, facing in opposite directions but still parallel

• The net force in any direction is zero, so a couple gives rise to ‘pure’ rotation

• Couples are present whenever you have a moment without linear force resulting

F

F

dMO = Fd(NOT 2*Fd!)provided O is in thesame plane as the couple

O0 0


Equilibrium of a particle• We will consider only particles in equilibrium

vector sum of all external forces = resultant force = 0 a body is called a particle when its shape and size have no

influence on the problem

• Typically, the system to be analysed consists of a number of interacting bodies/particles

• To isolate the one of interest, we draw a free body diagram of just that object this diagram includes all the external forces on the body of

interest, whether directly applied to it or resulting from interaction with the other bodies (not shown on the FBD)

• Constructing the FBD requires two main steps …


Free body diagram• 1. Decide which body to isolate and sketch the

external boundary around that object• 2. Draw all the forces (including unknowns) with

vectors in their correct positions the position of forces is important if rotation can occur! fixed interfaces to objects exert an opposing force to

maintain the object's position, moments can cause spin forces at interfaces between objects are assumed only to act

normal to the surface (since we assume no friction) ropes, chains, etc are deemed to be totally flexible, and

transmit only axial forces in tension (positive) if forces are not completely balanced, angular or linear

acceleration occurs – sum the collinear forces and moments

• See the following examples …



Reactions at Supports and Connections

Ball, Roller, or Rocker Fixed support

Smooth surface


Reactions at Supports and Connections: Table 6.1 pp.239-241

Rough Surface Pin in a smooth guide


ExampleGiven W=100 Kg, determine the components of the

reactions at supports A and B. Neglect the masses of the members of the truss.


Summary• Forces are vectors, and are free, sliding, or fixed

depending on how they act on an object• Free bodies are considered to be rigid, and

transmit applied forces through their structure• Concurrent forces sum, coplanar forces share the

same plane, collinear forces share the same line• Moments are caused by non-collinear forces, and

cause rotational acceleration about a point If two non-collinear forces are balanced and do not cause

linear acceleration or force, they are called a couple

• The center of mass is an integrated “average” of where all the mass in an object is located


Structures• We now have some basic tools to use on the

analysis of simple structures• A structure can be thought of as a system of free

bodies held in equilibrium so a structure must be “dismantled” in order to examine the

internal forces

• First we will look at plane trusses e.g. some bridges, the frames that support a house roof

• Then we will consider concentrated loads on beams before moving on to distributed loading

• In all cases, the structures will be statically determinate i.e. sufficient variables are known for all unknowns to be found for plane trusses, this requires #members +3 <= 2*#joints


Warren truss

• For an informative look at different types of bridge and roof trusses, see http://pghbridges.com/basics.htm

http://www.aku.ac.ir/faculty1/aliniamm/Structural%20Slides/trusses/p/IMG0009.jpg


Pratt truss

http://www.bcsj.org/rr/bcsj/trackplan/bcsj_ver_17/MillCity/pix/pinpratt_01_m.jpghttp://www.aku.ac.ir/faculty1/aliniamm/Structural%20Slides/trusses/p/IMG0005.jpg


Other trusses

http://www.kamloopsjunction.com/howe_tsa.jpghttp://icampus1.mit.edu/2.973/images/Baltimore_bridge.jpg


The strength ofA wood truss!(why or why not wood?)

The real one is at:Hwy. 71,Sioux Narrows,Ontario


Assumptions for truss calculations

truss members are connected by frictionless pins – no moment

loads only applied to ends of members

members are weightless

provides vertical support but no moment or horizontal force

provides vertical & horizontal support but no moment


Simple analysis

P

A

B

C

P

TAB TBC

TBCTAB

TAB TBC

TAC TAC

TAC TAC

Ax

Ay Cy

Note that there are no torques!


• It is important to identify which members are in tension and which in compression long, thin truss members are relatively weak in compression

and tend to buckle they need to be identified and strengthened if necessary

• The roller support not only allows for expansion of the structure it removes one unknown – the horizontal reaction – to make

the structure determinate, and most trusses use this concept

• Trusses are regarded as rigid if they maintain their shape when the supports are removed

rigid non-rigid


Method of joints• The diagram on the previous slide was an

example of the method of joints the structure is ‘exploded’ and the forces on each member

and joint are identified the equilibrium analysis of each joint gives enough

simultaneous equations to solve for unknowns it’s really the ‘brute force’ approach

• There are several conventions initially we assume all members are in tension – if it turns out

they are in compression, the sign (direction) will tell us we also draw the arrows as if members are in tension forces are labelled e.g. TAC regardless of whether they go

from A to C or C to A

• Choose a joint with few unknown forces to start


1kip = 1000 pounds per square inch (psi)


Zero-force members• Frequently the analysis can be simplified by

identifying members that carry no load two typical cases are found

• When only two members form a non-collinear joint and there is no external force or reaction at that joint, then both members must be zero-force

PA B

C

D

E

TCB

TCD

C

If either TCB or TCD ≠ 0, then C cannot be in equilibrium, since there is no restoring force towards the right.

Hence both BC and CD are zero-load members here.


• When three members form a truss joint for which two members are collinear and the third is at an angle to these, then this third member must be zero-force in the absence of an external force or reaction from a support

P

A BC

D

E

TBCTAB

TBD

B

Here, joint B has only one force in the vertical direction.

Hence, this force must be zero or B would move (provided there are no external loads/reactions)

Also TAB = TBC


• While zero-force members can be removed in this configuration, care should be taken any change in loading can lead to the member carrying a load the stability of the truss can be degraded by removing the

zero-force member (i.e. it's not “zero” force, just negligible...)

P

A BC

D

E

You may think that we can remove AD and BD to make a triangle …

This satisfies the statics requirements

However, this leaves a long CE member to carry a compressive load. This long member is highly susceptible to failure by buckling.


ExampleUse the method of joints to find the force in each member of the truss

of Fig.6-76a


Method of sections• When the truss exceeds a certain size, the

method of joints becomes unnecessarily tedious• If the joints are in equilibrium, so too is the truss

as a whole• Hence, we can also split it into two equilibrium

part-trusses provided we apply the appropriate forces to the cut ends and we can draw an individual free body diagram for the two

parts, including the external and reaction forces

• We can cut up to three members and still have enough equations to solve for the six unknowns the three reactions from the supports and the three forces in the members



• If no cut can be made through three or fewer members and which also passes through the member of interest then the method of joints must be used

• The method of sections is especially useful to find the force in a beam in the middle of a large truss when lots of joints must be analysed to get the same result and errors made in calculations for each joint will propagate,

possibly giving an unacceptably inaccurate result

Method of sections


ExampleAll members of the inverted Mansard truss of Fig.6-82a are made of

structural steel. Determine:

(a) The axial stress in member CH if the cross-sectional area of this member is 2.5 in2.

(b) The deformation of member BH if the cross-sectional area of this member is 2.5 in2.


Example (Cont’d)


Summary• We have now followed a path from atomic

bonding, through material properties and basic mechanical relations to an understanding of complex engineering structures

• In the next step, we look in more detail at the beams that comprise the trusses in particular, we are interested in how forces are transmitted

though beams and what effect the beam shape has on the strength


Beams


Overview• In this chapter, we consider the stresses and

moments present in loaded beams shear stress and bending moment diagrams


Beams with Concentrated Loads• A beam is defined as a slender structural member• For trusses we assumed that what happened in

the members was unimportant to the system although the nature of the members clearly affects the

strength of a truss here we look at what happens when beams are loaded

• We will look at the internal forces – axial and shear – and moments in the loaded beam

• Three types of beam are statically determinate …

F


simply supported beam

cantilever beam

combination beam


Cantilever

Free body diagram of the cantilever


If we cut the beam at an arbitrary point, we can determine the system of forces and moments required to maintain equilibrium

This system must include axial, shear and moment


Sign conventions


Shear Force & Bending Moment Diagrams

• In order to determine whether the beam can support the loads required, we need to determine the distribution of stress in the beam i.e. P, V, and M as a function of x

• Let’s take the following example: we first cut the beam at an arbitrary

location between the left end andthe force 0<x<(2/3)L

and then we cut the beam in anarbitrary location on the otherside of the force (2/3)L<x<L


• For 0 < x < (2/3)LP = 0V = (1/3)FM = (1/3)*F*(x)

• For (2/3)L < x < L P = 0 V = (-2/3)F M = (2/3)*F*(L - x)


Shear and Moment – Point Force

P = 0

(beam is horizontal)

V = Fleft side reaction

= F/3

M = F * dfrom left side

=(1/3)*F*(x)

P = 0

(beam is horizontal)

V = -Fright side reaction

= -2F/3

M = F * dfrom right side

= (2/3)*F*(L - x)


Distributed Loads on Beams


Example

Determine the resultant of this system of distributed loads and locate its line of action w.r.t the left support of the beam.


Example Write equations for the shear force V and the bending moment M for

any section of the beam:

In the intervals AB, BC and CD



General Equations

• For a generalised distributed load of w(x) N/m:

€

dPdx

=0

dVdx

=−w

dMdx

=V

(no axial force)

(shear increases with x)

(moment = shear x distance)


ExampleA beam is loaded and supported as shown below:

(a) Write equations for the shear force and the bending moment for any section of the beam in the interval AB.

(b) Repeat (a) for interval BC

(c) Draw complete shear force and bending moment diagrams for the beam.


Shear andMoment –

Distributed Force

Source: Riley, Sturges and Morris Text

PAB

= 0

Between A-B,

Distance is x

VAB

= -500

MAB

= -500*x

PBC

= 0

Between B-C,

Distance is (x-4)

VBC

= -500-200*(x-4)

MBC

= -500*x

-200*(x-4)*(x-4)/2( distributed force centroid

is at (x-4)/2 )


Summary of Shear and Moment• For a beam in static equilibrium, every part of the

beam must have balanced shear and moment• For point forces on a beam:

shear is constant between forces so that V = F (0th-order) moment changes linearly with distance (force x distance)

from pin joints or free beam ends, as M = F*d (1st-order)

• For distributed forces on a beam: shear changes linearly with distance along the part of the

beam with a distributed force because V = Fd*d (1st-order)

moment increases as a curve along the part of the beam with a distributed force because M = (F

d*d)*(d/2) (2nd-order)

It can be assumed that the point of distributed force action is halfway along the part of the beam with distributed force

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ENG2002 - Electronic Structure