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ME 270 Basic Mechanics I

Prof. Jones

Rm: ME 2008BPh: 494-5691

Email: [email protected]

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DISTRIBUTED LOADS

Learning Objectives

1). To determine the resultantof a given line loadand toevaluate thesupportreactionsacting on the body that

carries such a load.

2). To do an engineering estimateof the equivalent load and its

location.

Distributed Parallel Line Loads

L

oR w(x) dx = Resultant Force

L

rP co

M x w(x)dx = Resultant Moment about pt. P

Lc

o rP

L

o

x w(x) dx Mx

Rw(x) dx = Line of action for equivalent force

Note:

RF = area under the curve, q(x)

x = centroid of the area under the curve, q(x)

Determine direction ofR

F andO

RM by inspection

Use your intuition to check your answer.

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CENTROID AND CENTER OF MASS BY

INTEGRATION

Learning Objectives1). To determine the volume, mass, centroidand center of mass

using integral calculus.

2). To do an engineering estimateof the volume, mass, centroid

and center of mass of a body.

Definitions

Centroid: Geometriccenterof a line, area or volume.

Center of Mass: Gravitationalcenterof a line, area or volume.

The centroidand center of masscoincide when the densityis

uniformthroughout the part.

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Centroid by Integration

a). Line:

L dLz cL x x dL cL y y dL

b). Area:

A dAz cA x x dA cA y y dA

c). Volume:

V dVz cV x x dV

cV y y dV cV z z dV

where: x , y , z represent the centroid of the line, area orvolume.

c i(x ) ,

c i(y ) , c i

(z ) represent the centroid of the differential

element under consideration.

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Center of Mass by Integration

m dm dVzz

G c cm x x dm x ( dV)

c cm y y dm y ( dV)

c cm z z dm z ( dV)

Note:For a homogeneous body = constant, thus

m d V dV Vzz Tabulated values of the centroid and center of massof

several standard shapes can be found on the back inside

cover of the textbook.

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Arch Length

Planar Area

Body or Shell

of Revolution

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Arc Length

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Planar Area

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Body or Shell of Revolution

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CENTROID AND CENTER OF MASSBY COMPOSITE

PARTS

Learning Objectives

1). To evaluate the volume, mass, centroidand center of mass

of a composite body.

2). To do an engineering estimateof the volume, mass, centroid

and center mass of a composite body.

Definitons

Centroid: geometriccenter of a line, area or volume

Center of Mass: gravitationalcenter of a line, area or volume.

The centroidand center of masscoincide when the densityis

uniformthroughout the part.

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Centroid by Composite Parts

a). Line

L L ii 1

n

n

i c i

i 1L x L (x )

n

i c i

i 1L y L (y )

b). Area

A A ii 1

n

n

i c i

i 1

A x A (x )

n

i c i

i 1

A y A (y )

c). Volume

V Vii 1

n

n

i c i

i 1

V x V (x )

n

i c i

i 1

V y V (y )

n

i c i

i 1

V z V (z )

where,

x , y , z = centroid of line, area, or volume

c i(x ) , c i

(y ) , c i(z )

= centroid of individual parts.

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Center of Mass by Composite Parts

m m V1i 1

n

ii 1

n

i

n n

c i i c i i

i 1 i 1

m x (x ) m (x ) ( V )i

n n

c i i c i ii 1 i 1

m y (y ) m (y ) ( V )i

n n

c i i c i i

i 1 i 1

m z (z ) m (z ) ( V )i

where

x , y , z = center of mass of the composite body.

c i(x ) , c i

(y ) , c i(z )

= center of mass of individual parts.

Note:

Tabulated values of centroidand center of massof several standard

shapes can be found on the back inside cover of the textbook.

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94

FLUID STATICS

Learning Objectives

1). To evaluate the hydrostatic pressure loadingacting ona body that is immersed in a liquid.

2). To determine the resultantof a given line, pressure, or

body loadon a submerged body and to evaluate the

reactionacting on the body that carries such a load.

3). To do an engineering estimateof the equivalent loading.

Assumptions

1). The liquid is incompressible, (i.e., = constant).

2). The hydrostatic pressure loading always acts normalto

any submerged surface, regardless of orientation.

3). The hydrostatic (gage) pressure at a point in a liquid is

proportional to the depthhbelow the free surface.

Hydrostatic Pressure Distribution

p = hydrostatic (gage) pressure (pgage= pabspatm)

= density of the liquidg = acceleration of gravity

h = height below the free surface of the liquid

p =gh

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FRICTION

Learning Objectives

1). To understand the principles of Coulomb(dry)friction.2). To evaluate thefriction forcesrequired to hold a system

in static equilibrium.

3). To determine the properties of a system or its loads for

which a system will be in a condition of impending

motion.

Introduction

1). Contact surfaces are rough and dry.

2). Friction opposesmotion (or impending motion).

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Dry Friction

1). Coulombs Law

a). Friction forces act in a direction oppositeto that in

which the surfaces move, or would tend to move,

relative to each other.

b). Consider a block, of weight W, on a rough

horizontal surface subjected to a horizontal force P.

F

PStatic Dynamic

f = fMAX= SN

f = kN

Impending Slipping

1

1

W

Nf

P

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i). For no relative motion, (i.e., for static equilibrium),

ii). For impending motion(i.e., at the instant before slipping)

The maximum value of F for static equilibrium is

iii). After motionbegins

sF P N

s sF P N=F

k kF = F N

s s

F N

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2. Systems with Friction

a). Given external loads, evaluate the friction force (and

determine if the body is in static equilibrium)

Note: to be in static equilibrium:

sF N

i). Procedure:

1). Assume static equilibrium

2). Solve for F from equilibrium equations3). Check assumption

If F > sN, then assumption incorrect and F = kN

Ifs

F N , then assumption correct and calculated F

is correct.

b). Given condition of impending motion

Note: F must be indicated in correct direction

s sF F N

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FRICTION: SLIPPING VS. TIPPING

Learning Objectives

1). To evaluate two types of impending motion,slipping vs.tipping, to determine which will occur first.

2). To do an engineering estimateof whether a system will slip

or tip.

Two Types of Impending Motion

1). Slipping2). Tipping

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Notre Dame Scissors-Lift Tipping Incident

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Notre Dame Report Spreads Responsibility in

Student Death

ByLYNN ZINSER

Published: April 18, 2011

High wind, a lack of current weather information and a hydraulic lift susceptible to tipping eachcontributed to the death of a student video trainer at Notre Dame in October, the university saidMonday after a six-month internal investigation into the accident.

A report about the October death of a Notre Dame student videographer concluded severalfactors, but not one person, was responsible.Notre Dames reportsaid no one person wasresponsible for the death of the student, 20-year-old Declan Sullivan,who was filming footballpractice from a liftextended to its full height of 40 feet in high wind on Oct. 27 when it toppled.

Several flaws were exposed that need to be acknowledged and addressed, the report stated.

Responsibility for these issues is shared by many individuals.

The president of Notre Dame, the Rev. John I. Jenkins, said at a news conference: In the grief

and distress that follows a tragic accident, it is common to seek the individual or individualsresponsible and assign blame. We have reached the conclusion that no one acted in disregard forsafety.

The report said that the lift on which Sullivan was perched was more susceptible to tipping inhigh wind than other models used by Notre Dame and that its use that day was based on anunderestimation of the wind speed. The university had no way to measure on-field windreadings, so the investigation disclosed that staff members were using weather reports fromearlier in the afternoon, before 2 p.m. Central. Practice did not begin until 3:45 p.m. Wind froman earlier report was much lower than the 53-miles-per-hour gust that knocked over the lift.

Among the recommendations offered in the investigation report is making sure real-time weatherinformation is available to the football staff and establishing wind limits for lift use. Notre Dame,however, has not used the lifts since Sullivans death, instead installing remote-controlledcameras mounted on 50-foot poles to film practice. The investigation was led by the Notre Dameexecutive vice president John Affleck-Graves and was reviewed by Peter Likins, an engineer anda former president of theUniversity of Arizona.Notre Dame said Likins was not paid for thereview.

Likinss opinions were released with the report. As reflected in this report, there were a number

of issues that led to the loss of a bright and energetic young man, including the implementation

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of the football programs wind-safety procedure without continuous access to real-time weatherinformation at critical periods of time, Likins wrote. What is clear, however, is that there were

a series of factors in the aggregate that led to this tragedy. Though a needless loss of life cries outfor one to shoulder blame, the facts here do not support any single individual finding of fault.

The report acknowledged that Coach Brian Kelly was advised by three people about the safety ofholding practice outside that day: Clad Klunder, the director of football operations; Jim Russ, thehead athletic trainer; and Tim Collins, the director of football video and film. The investigationfound that none advised Kelly that practice should have been held indoors because of theconditions.

Any time there is a death associated with your working on a daily basis, it profoundly affectsyou personally, Kelly said at the news conference. Youre never quite the same. But I think

were all collectively focused on making sure nothing like this happens again.

Collins, the report said, forgot to schedule the annual inspection for the lift in August 2010, but it

also said that oversight was not a factor in the accident.

The report discussed Sullivans reluctance to film on the lift in the wind that day, expressed intwo messages he posted on his Twitter feed that afternoon, but said the investigators could notconclude that Sullivan was genuinely concerned for his safety. One of Sullivans messages saidthe winds were terrifying and the other included the phrase I guess Ive lived long enough.

The IndianaOccupational Safety and Health Administrations investigation reported thatSullivan had also expressed his worries to an assistant video coordinator.

In March,that investigation led OSHA to levy $77,500 in finesagainst Notre Dame for safety

violations related to Sullivans death. The university has requested a formal hearing with OSHAover the findings.

A version of this article appeared in print on April 19, 2011, on page B14 of the New York edition with the headline: Notre Dame Report Spreads

Responsibility in Student Death.

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Free Body Diagrams

Impending Tipping Impending Slipping

Assumptions

Impending Tipping Impending Slipping

N & F act at tippingedge. F = Fs= SN

F SN q b

Results

Impending Tipping Impending Slipping

Find PTand F Find Ps and q

If F Fs

, then tipping If q < b, then slipping

assumption is correct (P = PT) assumption is correct (P = Ps)

If F > Fs, then slipping If q > b, then tipping

occurs first. then tipping occurs first.

Comments

1). Generally easiest to assume tipping first, unless youre

reasonably sure the system will slip.

b

PT PSW W

N

qN

f f

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Impending Tipping Impending Slipping

Impending Tipping Impending Slipping

N and f @ Tipping Edge

f fMAX

f = fMAX

q is unknown

fA= fAmax

fB= fBmax

NA 0

NA= fA = 0

fBfBmax

PT PS

PT PS

WW

W W

q

f

N

f

N

fB

NB NA NB

FA fB

b

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WEDGES

Learning Objectives

1). To determine theforcerequired to insert and/or remove a wedge.2). To determine whether a wedge isself-locking.

3). To determine the minimum coefficient of frictionnecessary

for a wedge to be self-locking.

4). To determine the minimum force necessary to hold a non-

self-locking wedge in place.

Definitions

Wedge: a simple machine designed to affect a small change in

the position of a system. Wedges often experience large

normal and friction forces.

Self-Locking Wedge: a wedge in which the friction forces large

enough to prevent it from being squeezedout.

Remarks

1). Friction forces always opposesthe direction of impending

motion.

2). Evaluating the condition of impending motion outis the

only way to determine if a wedge is self-locking.

3). F = Fs= sN for impending motion.

4). The weight of the wedge is often neglected because of the

large normal and frictional forces acting on the wedge.

5). Wedges typically have half angles ( ) of about 6 , so that

between the wood and wedge need be only about 0.1 to be self-lockin

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TRUSSES

Learning Objectives

1). To identifyzero-force membersin a structure.2). To recognizeplanarandspace(i.e., three-dimensional)

truss structures.

3). To understand the assumptionsmade in modeling trusses.

4). To understand whystructures are often designed as trusses.

Definitions

Zero-Force Members: structural members that support Noloading but aid in the stability of the truss.

Two-Force Members: structural members that are: a) subject to

no applied or reaction moments, and b) are loaded only at

2pin joints along the member.

Multi-Force Members: structural members that have a) appliedor reaction moments, or b) are loaded at more than two

points along the member.

Truss: a rigid framework of straight, lightweight 2-force

members that are joined together at their ends.

Frame: a rigid framework of straight and/or curved members

intended to be a stationary structure for supporting a load.

Machine: an assembly of rigid members designed to do

mechanical work by transmitting a given set of input

loading forces into another set of output forces (Dynamics).

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Simple and Compound Trusses

Simple Truss: a truss whose number of members is given by m

= 2j 3, where m = no. of members and j = no. of joints. (For

simple space trusses the relationship is given by m = 3j

6).

Compound Truss: a truss formed from two or more simple

trusses.

Newtons Third Law

For each action there is an equaland oppositereaction (i.e.,

F FA ABody 1 Body 2 ).

Assumptions for Modeling

1). All members are straight.

2). All connections are modeled as pin joints.

3). The centerlines of all members must be concurrentat the

joint.

4). External loads act only at thejoints.

5). Weight of members is negligiblecompared with external

loads.

Advantage of Truss Structures

Truss structures can span longdistances without intermediate

supports (e.g., bridge and roof trusses) and can carry heavyloads with lightweight members.

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Applications

See text book.

Two Methods of SolutionsMethod of Joints

Method of Sections

Static Indeterminacy/Partial Constraint

A truss is internally indeterminateif:

m > 2j

3 (for planar trusses) where m = no. of members

m > 3j 6 (for space trusses) where j = no. of joints

A truss is improperly constrainedif:

m < 2j 3 (for planar trusses) where m = no. of members

m < 3j 6 (for space trusses) where j = no. of joints

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METHOD OF JOINTS

Learning Objectives

1). To employ the methods of jointsto evaluate the axial force

carried by each member in a truss.

2). To identifyzero-force membersin a truss.

3). To do an engineering estimateof the load distribution in a

truss.

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Procedure

1). Draw a FBD of the entire trussshowing the reaction forces

at the supports and the externalloads. Write the

equilibrium equations and solve for as many unknowns aspossible.

2). Identify anyzero-force membersand any members that

carry the same load as other members or external loads.

3). Draw a FBD of each jointin the truss. Be sure to abide by

Newtons Third Law (reactions between interacting

members are equal and opposite).

4). Make aplanfor solving the member loads. Start with the

joint with the least number of unknowns (this frequently

occurs at the supports). In solving the equilibrium

equations, avoid joints that have more than two unknowns

acting on it. Remember that since the forces at each joint

are concurrent (i.e., they intersect at the joint), only two

equilibriumequations can be utilized( F 0 and F 0x y , no moment equation exists).

5). When through solving, go back and state whether eachmember is in tensionor compression. (That is, if a

negative value is found for a member. Then you assumed

the wrong direction).

HINT: When drawing the FBDs of the joints, assume all

members are initially in tension(i.e., show all member

forces acting away from the joint). Then,

if load ispositive member is in tension.

if load is negative member is in compression.

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METHOD OF SECTIONS

Learning Objectives

1). To employ the method of sectionsto evaluate the axial forcecarried by selected members in a truss.

2). To do an engineering estimateof the load in select members

of a truss.

Procedure

1). Draw a FBD of the entire trussshowing the reaction forces

at the supports and the externalloads. Write theequilibrium equations and solve for as many unknowns as

possible.

2). Locate the force members to be evaluated. Identify whether

any of these forces can be determined by observation (e.g.,

zero-force members).

3). Identify section to be used and draw a FBD of the section

including any support reactions, external loads and internal

forces of sectioned members. Remember, the cutting plane

must cut through the members of interest. Also the cutting

plane need not be straight, it may be curved.

4). Write the equilibrium equations for one of the two sections.

The equations for either half of the section will yield the

same member forces.

5). Three equilibrium equations are available, so up to three

unknowns can be solved with a single section.6). At times more than one section may be necessary.

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FRAMES AND MACHINES

Learning Objectives

1). To evaluate the unknown reactionsat thesupportsand theinteraction forcesat the connection pointsof a rigid frame in

equilibrium by solving the equations of staticequilibriumof the

overall structureand each individual member.

2). To do an engineering estimateof these quantities.

Definitions

Two-Force Member: a structural member that is loaded only at

two pin joints along the member.Multi-Force Member: a structural member that is loaded at

more than two points along the member.

Truss: a rigid framework of straight, lightweight two-force

membersthat are joined together at their ends.

Frame: an assembly of rigid members (of which at least one is a

multiforce member) intended to be a stationary

structure for supporting a load.

Machine: an assembly of rigid members designed to do

mechanical work by transmitting a given set of input

loading forces into another set of output forces.

Newtons Third Law

Newtons Third Law: For each action there is an actionand

oppositereactionF F

A ABody 1 Body 2e j

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Frames

In frames, we are often interested not only in the reaction forces

at the supports but also in the interaction forces between

members and the loads carried by any two-force members.

Procedure:

1). Inspect structure for two-force members.

2). Draw FBDs of the entire structureand of each member. Be

sure the interaction forces between members are equal in

magnitude, opposite in direction and collinear (i.e., satisfy

Newtons Third Law).3). Count the number of unknowns and equations available for

each FBD. Successively write and solve the equilibrium

equating corresponding to the FBDs of interest.

Note:

1). For a structure composed of N members, will be N + 1

sets of equilibrium equations and FBDs. Only N sets ofequations are independent.

2). If all external reactions on a frame can be determined, then

the internal forces between members may be determined

from either member.

3). If there are more unknowns than available equations

Statistically Indeterminate. This is not always true.

Sometimes by disassembling the frame, the forces can bedetermined using the equilibrium equations.

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INTERNAL FORCE/COUPLE ANALYSIS

Learning Objectives

1) To calculate the internal forcesand momentsat a given point within an

object.

2) To do an engineering estimateof these quantities.

Newtons First Law

Given no net force, a body at rest will remain at rest and a body moving at a constant

velocity will continue to do so along a straight path R = F = 0, M = 0

Newtons Third Law

For each action, there is an equal and opposite reaction.

F 0 o

M 0

xF = 0 xM = 0

yF = 0 yM = 0

zF = 0 zM = 0