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Statics: a subject in mechanicsMechanics: the study of forces on bodies and their resulting motion (Isaac Newton a great contributor)Emphasis on bodies which are not moving (or moving at constant speed)TAM 212: dynamics - moving bodiesBiomechanics: muskoskeletal mechanicsTAM 251: Introductory Solid Mechanics - strength of materials - study of how things deform - twist, compress, stretch… ME 340: dynamics of buildingsME 310: fluid dynamicsME 330: behavior of materialsME 360: controls - roboticsME 320: heat transferME 370: designME 350: design for manufacturability

First ClassWednesday, January 20, 2010

07:55

Notes

Forces and bending moments

Multiple bodies

Beams, structural analysis, trusses

Friction

Can switch to TAM 211 at this point

Forces on submerged bodies

SyllabusWednesday, January 20, 2010

08:19

Notes

No course website yet - on mechse website will beGrades on CompassHMWKS due Fridays at 08:00

Ms Thomas's know what language we speak

Notes

Precision: 3 significant figures ALWAYS (like IB math)

Exams are during lectures (not in this room)

Notes

Read A1-A5 (vectors) in book

A particle at rest or moving with constant velocity will remain in this state unless acted upon by an unbalanced force.

1.

A [article acted upon by an unbalanced force F experiences an acceleration a that is proportional to the particle mass, m.

2.

The mutlual forces of action and reaction between two particles are equal opposite and collinear.3.

Newton's Three Laws of Motion

The mutual force F of gravitation between two particles of mass m1 and m2 is given byNewton's law of gravitational attraction

Weight is the force exerted by the earth on a particle at the earth's surface:

System Length Time Mass Force

SI m s kg N

US Customary (FPS) ft s slug lb

Units of Measurement

Newton and slug are derived units

1 ft = 0.3048m (exact)-

1 slug = 14.59 kg-

1 lb = 4.448 N-

Unit conversion

Equations must be dimensionally homogenous

Dimensional homogeneity-

Numerical Calculations

Vectors, Newton's Laws of Motion, Gravity, Calculations, and AnalysisFriday, January 22, 2010

07:56

Notes

Equations must be dimensionally homogenous

This is engineering convention (assume all dimensions are to 3 sf if not specified, 1 ft = 1.00 ft)

Use 3 significant figures for final answers, more (usually 4) for intermediate calculations-

Significant Figures

Read the problem carefully, write it down carefully.1.Draw given diagrams neatly and construct additional figures as necessary.2.Apply principles needed.3.Solve problem symbolically, then substitute numbers. Proved proper units throughout. Check significant figures. Box the final answer(s)..

4.

See if answer is reasonable.5.

General Procedure for analysis

Starting Monday: class meets in 228 Natural History Building

Notes

Office hours are now online - all are in 429 Grainger

Force: the action of one body on another - can be treated as a vector, since forces obey all the rules that vectors do. Forces have magnitude, direction, and sense.Problem: Finda unit vector in the xy plane that is perpendicular to the force 3i-4j+12k N.

Answers in back of book: don't necessarily have all solutions - if there are two explain that

Problem: Find two 80 lb forces whose sum is the force 40i lb.

ForcesMonday, January 25, 2010

07:57

Notes

Find a vector that is simultaneously in the plane of F1 and F2 and in the plane of F3 and F4.

Discussions this weekHMWK due Friday

Notes

Forces: the action of one body on another - magnitude, sense, and direction - vectors

Can also solve problems graphically-

Force components: Law of sines and cosines can be used in 2-d problems, but using force components is usually more straightforward.

Pick a good system - but keep it once you have it-

Adding vectors - simple

PROBLEML Determine the magnitude of force F so that the resultant force of the three is as small as possible. What is the minimum magnitude?

Cantilever bending - force vertical on lug nut

Cannot be a maximum, since the force could be large as possible-

Note: could solve it geometrically, but much more difficult-

Question: how do we know this is the minimum rather than the maximum?

Force along a line: often a force is described as acting along a line - often said to be a line connecting A and B.

Example: the window is held open by cable AB. Determine the length of the cable and express the 30N force at A along cable.

Forces, components, and division thereofWednesday, January 27, 2010

07:58

Notes

The scalar component A// of a vector A along a line with unit vector u is given by A//

=Acos(theta)=A.u. Thus the vector component is A//=A//u=(A.u)u, Aperp=A-A//. -

Components of forces parallel and perpendicular to a line:

Problem:Determine the angle theta betweeen the two cables. Determine the projected component of the Force F=12lb in direction of AC

Notes

According to Newton's first law, a particle will be in equilibrium if and only if, ΣF=0, that is the sum of all forces on a particle is zero.

-

Generally, in 3D: Equilibrium requires: ΣF = ΣFxi +ΣFyj+ΣFzk =0. - can be separated into vector equations.

-

Note: this is valid in inertial reference frames - basically one in which we can regard as fixed-

Many problems are also in only 1 or 2 dimensions - the remaining can be ignored-

Equilibrium:

A free body diagram is a drawing of a body, or part of a body which all the forces acting on the body are shown. Relevant dimensions may also be required

-

No judgment regarding equilibrium or lack of it is made when constructing the diagram. Forces not directly acting on the body are not shown.

-

Free Body diagram

Pulleys are usually regarded as frictionless, then the tension in a rope or cord around the pulley is the same on either side

-

Springs are usually regarded as linearly elastic then the tension is proportional to the change in length, s.

-

Idealization:

Equilibrium, Free body diagrams, and IdealizationsFriday, January 29, 2010

07:59

Notes

In reality either rope could break, as there is a distribution of ropes break at slightly different values - they follow a normal curve

Example ProblemsFriday, January 29, 2010

08:03

Notes

Some practical engineering problems involve the use of statics of interacting or interconnected particles. To solve them, we use Newton's first law, ΣF=0, on multiple free body diagrams of particles or groups of particles.

If you want to find the value of a force you must draw a free body diagram that includes that force.

Systems of ParticlesMonday, February 01, 2010

07:59

Notes

The five ropes can each take 1500 N without reaking. How heavy can W be without breaking any.

A man having a weight of 175 lb attempts to lift himself using one of the two methods shown, determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has a weight of 30 lb.

Example ProblemsMonday, February 01, 201007:59

Notes

Never forget to draw the overall free body diagram - it simplifies the problems greatly

The collar can slide along the rod without friction. The spring, which is attached to the collar and to the ceiling exerts a force kΔ proportional to its stretch Δ, where k is the modlus of the spring. If k=50lb/in, W=50 lb and the collar weiths 20 lb find how much the string is stretched in the given position.

Notes

The moment, MP, of a force, F about a point P is equal to the product, MP=Fd, where d is the perpendicular distance from P to the line of action of the force. The force F can be acting at any point along this line of action.

Vectors, Application of vectors… -

It will be in 2 different rooms - here and 100 MSEB - there will be assigned seating-

Hour exam next Friday - cumulative so far

Line of action of F

Principle of transmittivity: F can be acting anywhere along the line of action

Moment of a forceWednesday, February 03, 2010

07:57

Notes

Three forces act on the plate. Determine the sum of the moments of the thre forces about point P.

Note: a torque is a special kind of moment

Example ProblemsWednesday, February 03, 201008:01

Notes

This is used in post hole augers - things to dig holes, you put it in and then twist it - apply equal and opposite forces to turn it

For a point: MP=rxF-

Ml=MP.ul-

Ml==(MP.ul)ul-

Note the perpendicular one is the total minus the parallel-

Moment of a Force about a Line:

A couple is a pair of equal, opposite and noncollinear forces-

The moment of a couple is given by M=Fd (scalar)-

M=rxF, it is independent of the origin you choose-

If several couples act, MR=Σ(rxF).-

This is a special case: note that the total force is equal to 0-

Couples:

Moments of Forces and CouplesFriday, February 05, 2010

08:00

Notes

Example: determine moment about Oa axis

Determine the couple moment - express as a cartesian vector

Example problemsFriday, February 05, 2010

08:00

Notes

ΣMP denotes the moment of all applied couples plus the sum of moments of applied forces with respect to point PNote: equivalent is often used instead of equipollent. Pronunciation: ee-quee-pah-lent

A force system is a collection of forces and couples applied to a body. Two force systems are said to be equipollent if they have the same resultant force, ΣF and the same resultant moment, ΣMP with respect to any point P. Here, for each system.

Assigned seating-

On Friday-

Homework-type problems, plus understanding questions-

No calculators or electronics-

Bring self, pencil, eraser-

Through Friday's lecture-

Do not need to use given-find-solution, only solution-

3 problems-

Either pen or pencil is OK - pencil preferable -

50 minutes long, starts at precisely 8:00-

Hour exam:

Equipollent force systemsMonday, February 08, 2010

07:58

Notes

Post tensioning ram on a forklift

Replace the force and couple system by an equipollent force and couple moment at point Q.

Note: it does not matter where this couple is applied for 8k

Can rephrase as: Replace the force and couple system by its resultant at point Q.

Example problemsMonday, February 08, 2010

07:58

Notes

Must specify P

A resultant is an equipollent force system consisting of only one force Fr and one couple MrP:-

Resultants:

Fr=(ΣF)1

MrQ=(ΣM)1

F can be anywhere along its line of action, couple can be applied anywhere

Check your seat before examCourse notes posted outside 429 Grainger

Resultants of Force SystemsWednesday, February 10, 2010

07:57

Notes

The belt passing over the f=pulley is subjected to two forces each having a magnitude of 40 N, force F1 acts in the -k direction. Teplace thse by a resultant force system at point A. theta is 45 degrees.

Example problemsWednesday, February 10, 2010

07:58

Notes

Make sure to explain work - use wordsAlso simplifying importantExams by Friday hopefully

Hour ExamMonday, February 15, 2010

08:06

Notes

Autonomous Materials Group

Concurrent forces with no couples

Coplanar forces with couples perpendicular to force plane

Parallel forces with couples perpendicular to forces

Single resultant force (often possible)-

Screwdriver (always possible) - consists of a single force and a couple in the same direction-

Reduction classes:

A resultant is an equipollent force system consisting of only one force, Fr and one couple MrP - must specify point

If FR is perpendicular to MrP with the force not zero, then an equipollent system consisting of only a single force can always be found.

-

Act at the same point, just sum the forcesi.Concurrent forces (no couples)1.

Add the forces, that will be the forcei.Find MrP and then divide by the force and move the force over that much - that will be the single force resultant

ii.

Coplanar, with couples perpendicular to plane2.

Add the forces for the resultant forcei.Find moment about P, then move the force over by MrP divided by the forceii.Common with a plate with wires and weightsiii.

Parallel forces with couples perpendicular3.

Single resultant forces

Most general case - always possible-

Consists of a force and a couple in the same direction - push and twist-

Screwdriver

Move the force over in order to cancel the perpendicular force - then it will just be the parallel moment and the force

ReductionsMonday, February 15, 2010

07:59

Notes

Replace the 3 forces by a screwdriver

Example problemMonday, February 15, 2010

08:42

Notes

Screwdriver is referred to as a wrench by some texts.

In structural analysis we are often presented with a distributed load (force/unit length) w(x) and we need to find the equipollent loading F.

-

Distributed forces

Distributed forcesWednesday, February 17, 201008:16

Notes

Replace the distributed load by an equipollent resultant force and specify its location on the beam measured from the pin at C

An engineer measures the forces exerted by the soil on a 10m section of a building foundation and finds they are described by

The magnitude of the total force exerted on the foundationa.The magnitude of the moment about A due to the distributed load.b.

Determine

Example ProblemsWednesday, February 17, 201008:16

Notes

Total force given by area under loading diagram

It is located at the centroid of the area

Simple Distributed LoadingFriday, February 19, 2010

07:59

Notes

We regard a rigid body as a collection of particles:

Let Fi=the resultant external force on particle iLet fij=the internal force on particle i by particle jLet fji=the internal force on particle j by particle i

Force equilibrium:

3D problems have 6 unknowns, 6 equations (3 forces, 3 moments)In general: 2D problems have 3 unknowns, 3 equations (2 forces, 1 moment)

Equilibrium of a rigid bodyFriday, February 19, 201008:10

Notes

Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass of 5 Mg and a center of mass at G. The wheels are free to roll. Explain the signifficance of each force on the diagram

Example problemsFriday, February 19, 2010

08:28

Notes

Type Description Reactions Unknowns

Roller/slider 1

Pin 2

Fixed 3

Types of supportFriday, February 19, 2010

08:34

Notes

Free body diagramFriday, February 19, 201008:37

Notes

These are external forces on the extended body - the internal ones don't count for overall equilibriumCan prove that if the sum of the moments about a point is zero and the sum of the forces is zero => moment equilibrium is satisfied for any point.

Sum of forces and sum of moments are zero.

Applications of EquilibriumMonday, February 22, 2010

07:59

Notes

The cantilevered jib crane is used to support the load of 780 lb. If the trolly T can be placed anywhere between 1.5 ft and 7.5 ft, determine the maximum magnitude of the reactions at the supports A and B. The supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, but the one at A does not

Easier to assume all forces are going in the same way, will then get a negative answer.

Example ProblemsMonday, February 22, 2010

07:59

Notes

In many situations bodies are subject to only two or three forces and no couples.

The forces must be equal, opposite, and collinear-

Shape of body doesn't matter-

Two force members:

The three forces must be either concurrent or parallel and must be coplanar-

Three force members:

Two and Three force membersMonday, February 22, 2010

08:18

Notes

It is possible for the 20N half cylinder to be in equilibrium on the smooth plane for only one value of the angle phi. For that angle find the tention in the cord as a function of theta and check your answer in the limiting cases theta is 0 and pi/2.

Half cylinder weight W is suspended from a smooth vertical wall by a cord. Find the tension T in the cord as a function of theta and check the limiting cases

ProblemMonday, February 22, 2010

08:26

Notes

Plate, has weight 5500 lb. Determine the tension of each cable when when held as shown

Often can get a quick answer by looking at moments about a line

Member is supported by a pin at A and cable BC. If load is 300 lb, determine the components of reactions at supporsts

Problems with extended bodies - 3DWednesday, February 24, 2010

08:00

Notes

Two marbles, radius R and wewight W are placed inside a hollow thin-walled tube diameter D. D<4R. Find minimum weight of tube so it will not turn over.

Notes

A vertical force of 80 lb acts on the crankshaft. Determine the horizontal equilibrium force, P, that must be applied to the handle and the x,y,z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert only force reactions on the shaft.

There are 4 courses in biomechanics

[email protected]

Journal bearing - allows it to turn aroundThrust bearing - prevents it from sliding either - put a thing on the other end

Brake pads used to be made of asbestos, now made of a polymer like that in bulletproof vests (N, H, and O)

The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reactions at the pins A and D. Also, what is the force in the cable at the winch W?

Can pick FD in direction of BD as there are only 2 forces on BD so forces are equal, opposite, and collinear.

More problemsFriday, February 26, 2010

07:57

Notes

Must be lightweight, easy to assemble

Structural equilibrium - Talbot laboratory, Titan formwork builds formwork for concrete - build a multistory building

Introduction to structuresFriday, February 26, 2010

08:43

Notes

Mode of failure is not necessarily uniqueBuckling: slender member under compression bends - very importantOne test is worth a thousand expert opinionsTrusses are structures formed by straight, pin connected (2-force-members) bars that are loaded only at the joints. Both 2D and 3D trusses are used widely

Methods of joints-

Method of sections-

Two primary methods to find loads in members

Always first set the overall structure in equilibrium

Set the entire structure into external equilibrium1.Set each joint in the structure into force equilibrium only, moving from joint to joint. Assume all internal loads are tensile. (if negative then compressive)

2.

Method of joints

TrussesMonday, March 01, 2010

07:59

Notes

Truss subject to loading shown

Distance between rails on railroad is standard at 4 ft 8.5 inchesBob Miller

Determine force in member FG of truss - state tension or compression

Example ProblemsMonday, March 01, 2010

07:59

Notes

Set the body into external equilibrium1.Cut the structure at a section of interest into two separate pieces and set either part into force and moment equilibrium. Assume all internal loads are tensile.

2.

Method.

Failure point is often the joint, especially dissimilar or composite materialLong slender members are very easy to break in bending

Trusses: method of sectionsWednesday, March 03, 2010

08:01

Notes

The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG and state if the members are in compression/tension.

Determine the force in OE, LE and LK of the Baltimore truss, state if members in tension or compression

Example problemsWednesday, March 03, 201008:09

Notes

Set the entire structure into external equilibrium. This step will generally produce more unknowns than there are relevant equations of equilibrium

1.

Isolate various part(s) of the structure, setting each part into equilibrium. The sought forces or couples must appear in one or more free-body diagrams.

2.

Solve for the requested unknowns.3.

Here, the members can be truss elements, beams, pulleys, cables, and other components. Generally, if the structure has moving parts, it is called a machine; otherwise it is called a frame. The general solution method is the same:

17 March: Hour Exam 2Can stay and learn TAM 211 materialProb. A Wednesday or Thursday night for TAM 210 exam

Frames and MachinesFriday, March 05, 2010

07:59

Notes

Determine threactions at supports A and B.

The tower truss has weight 575 lb, centre of gravity at G. Rope is used to hoist it vertical. Compute the compressive force along the shear leg, the tension in BC.

Example ProblemsFriday, March 05, 2010

08:01

Notes

Note: tensions in the ropes will not be the sameNote: can draw more FBD's and calculate the rest of forces (which were not requested here)

Notes

Force/Moment Diagram _____________________________________________________________

Normal (axial)

Transverse shear

Torsional (torque)Or twisting moment

Bending

NOTE: forces and moments are equal and opposite on opposite sides of cut.Positive directions are denoted above - convention is similar to that of tension.

Cutting members at internal points reveals internal forces and moments:

1st two are forces2nd two are moments

Internal ForcesMonday, March 08, 2010

07:58

Notes

Three torques act on the shaft. Determine the internal torque at A, B, C, and D

Determine the normal force, shear force, and bending moment at a section passing through point D of the two member frame.

Example problemMonday, March 08, 2010

08:10

Notes

Note: these re valid only at point D. In general, N,V, and M vary throughout the structure. (here N does not)Needed to set body into external equilibrium.

In general, V and M will varyProblem: The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the maximum moment is 5 kip.ft. If L=10ft determine the largest uniform distributed load the shaft will support.

Notes

For beams, the shear force V and bending moment M are generally functions of position x along the beam. The relations for V(x) and M(x) can be found from force and moment equilibrium, respectively.

Examination next week - notes on wall outside 429 Grainger.

Diagrams:

The assumed sign conventions for F and M0 must be honored-

Only a concentrated force will cause a jump in V(x)-

Only a concentrated moment will cause a jump in M(x)-

A concentrated force and a concentrated moment can be applied at the same point though this condition is not a requirement.

-

Notes:

Shear force and bending moment distributionsWednesday, March 10, 2010

07:59

Notes

The beam consists of three segments pin connected at B and E. Draw the shear force and bending moment diagrams for the beam.

Example Problem IIFriday, March 12, 201008:30

Notes

DiagramsFriday, March 12, 2010

08:42

Notes

Express the x,y,z components of the internal loading along the rod as a function of y where 0<=y<=4

ProblemWednesday, March 10, 2010

08:15

Notes

From problem on MondayDiagrams:

DiagramFriday, March 12, 2010

08:02

Notes

Setting the beam in external force and moment equilibrium, then-

Cutting the beam at arbitrary points and setting portions into force and moment equilibrium-

The shear force V and bending moment M are generally functions of position x along a beam. The distributions V(x) and M(x) for statically determinate problems can always be found by

The distributions can then be plotted.

q: distributed load (positive up)-

V: shear force (positive up)-

M: bending moment (positive z)-

dV/dx=-q-

dM/dx=-V-

Differential and integral relations for q, V, and M

Be very careful of sign conventions

Shear-force and bending moment diagramsMonday, March 15, 2010

07:59

Notes

The beam will fail when the maximum moment Mmax is 30 kip.ft or the maximum shear, Vmax is 8 kips. Determine the largest distributed load, w the beam will support

Example problemMonday, March 15, 2010

08:01

Notes

DiagramsMonday, March 15, 2010

08:27

Notes

Euler derived the moment equilibrium conditionTorsion is principal deformation mode in axially loaded helical spring.

HE 2Friday, March 19, 2010

07:59

Notes

Friction forces are the tangential components of force between bodies in contact. Here we study only dry friction, also called Coulomb friction:

The coefficient of static friction is somewhat larger than the coefficient of kinetic friction. Both depend on the materials in contact. The friction force f always acts to oppose relative motion.

FrictionFriday, March 19, 2010

07:59

Notes

A roll of paper has a uniform weight of 0.75 lb and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered friction-less, determine the minimum force P needed to start turning the roll. The coefficient of static friction between the wall and the paper is 0.25.

Example problemFriday, March 19, 2010

08:07

Notes

These are tapered pieces used to locate and hold objects in place. Usually, friction is involved. Inserting and removing are not the same:Inserting:

Removing:

Force equilibrium is always required. Moment equilibrium is ordinarily not considered.Note: the sign of P for removing a wedge is not necessarily positive as shown, since N A has a component in the direction of +P.

Friction ALWAYS acts AGAINST you.Smaller force to remove than to insert, as NA opposes you going in, but helps you go out.

WedgesMonday, March 29, 201007:58

Notes

Belts are flexible members that can transmit tangential force to pulleys, pins, brake drums, trees, posts, capstans, and other rough surfaces:

An important feature of belts is that the tension T is generally not constant in the region of contact and therefore, in general, T2 != T1.Assume belt is slipping to the left. Consider a differential element:

BeltsMonday, March 29, 2010

08:04

Notes

The truck, which has a mass of 3.4 Mg is to be lowered down the slope by a rope that is wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull T of 300 N, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the treee and rope is 0.3

Note: if the man is smart he will wrap the tree more than twice around. Otherwise he may lose control.

Example ProblemMonday, March 29, 2010

08:24

Notes

The uniform 50 lb beam is supported by the rope, which is attached to the end of the beam, wraps over the rough peg and is then connected to the 100 lb block. If the coefficient of static friction between the beam and block and between the rope and peg is 0.4, determine the maximum distance d that the block can be placed from A and still remain in equilibrium. Assume the block will not tip.

Example Problem 2Monday, March 29, 201008:33

Notes

Use same concept as wedges

Think of screw as wedge wrapped around a rod

The equipollent force pushing the entire screw thread along the contact mean radius r is M/r. Consider square (Acme) threads only.Different conditions apply for tightening and loosen

Note: there are self loosening screws (and other cases) - that is if the coefficient of friction is less than tangent of alpha, a negative moment is required to keep the screw from unraveling.

Author does this

ScrewsWednesday, March 31, 2010

08:00

Notes

The two blocks under the double wedge are brought together using a left and right square threaded screw. If the mean diameter is 20 mm, the lead is 5 mm and the coefficient of friction is 0.4, determine the torque needed to draw the blocks together. The coefficient of friction between each block and its surfaces of contact is 0.4.

Example ProblemWednesday, March 31, 2010

08:20

Notes

Note: spreading blocks apart would require a different (smaller) value of M.

Notes

The centroid of a shape (volume, area, or line) is the point C at which all the shape may be regarded as concentrated, such that the first moment of the concentrated shape with respect to any point in space is the same as that of the given shape:

CentroidsFriday, April 02, 2010

07:58

Notes

Locate the centroid of the shaded area bounded by the parabola and the line y=a

Locate the centroid z of the volume segment.

Example ProblemsFriday, April 02, 201008:05

Notes

For every dV with a positive value of y (or x) there is another dV with a negative value of y (or x).

HE 2:y=0.9x+20 - adjustedAvg: 75.7, median 76.3

Notes

The anatomical center of gravity G of a person can be determined by using a scale and a rigid board having a uniform weight W1 and length l. With the person's weight W known, the person lies down on the board and the scale reading P is recorded. From this show how to calculate the location x bar. Discuss the best place l1 for the smooth support at B in order to improve the accuracy of this experiment.

Best place for l1 is probably l/2, since then (l/2-l1)W1 vanishes. Then errors in W1

become irrelevant.

Versabar: weighing structures and lifting them

Another problemMonday, April 05, 2010

08:00

Notes

If shapes are joined together the volumes (or areas or lines) and first moments add:

Can also consider cutouts or cavities by subtracting that volume and moment from above equation (common: plate with a hole drilled in it)

Composite bodiesMonday, April 05, 2010

08:12

Notes

Determine the location xbar of the centroid of the solid made from a cone cylinder and hemisphere.

ExampleMonday, April 05, 2010

08:16

Notes

When material is subtracted the corresponding V i effectively changes sign.

Another problemMonday, April 05, 201008:32

Notes

The bird-feeder pole is made from a continuous length of uniform rod. Determine the location (x bar,ybar) of the centroid C.

5 feet off the ground, 7 feet from something sideways, 9 feet above

5,7,9 rule for birdfeeders

If have two bodies the centroid of the composite must lie along the line connecting the two individual centroids

Example: bridfeederWednesday, April 07, 2010

08:08

Notes

Consider a beam in bending under the action of a bending moment M:Motivation: beam bending:

In introduction to Solid Mechanics (TAM 251), it will be shown that the neutral surface passes through the centroid of the cross sectional area and that the stress σ varies linearly with distance y from this surface.

There are also moments of inertia of mass (dynamics)

When doing testing, must approximate the actual field conditions

Will bend (if pure moment is applied) into the arc of a circleTension at bottom, compression at top, nothing in the centre (centroid) -this is called the neutral surface

Shear is not particularly important here

Moment of Inertia of an AreaFriday, April 09, 2010

07:56

Notes

Determine the moment of inertia, Ix for a rectangular area of width b and height h about the x axis through its centroid.

Calculate moment of inertia about the bottom

Note that this is much LARGER than before (4x larger to be precise)

Determine the moment of inertia for the shaded area about the x axis

ExampleFriday, April 09, 2010

08:19

Notes

Note: can also be solved from taking horizontal strips, but more difficult

Notes

Ordinary

Cross

Polar

There are several types of second moment or moment of inertia of an area:

The commonly encountered ones are the ordinary and polar types. Note that J P=Ix+Iy for any cross section - this is useful for finding JP when Ix and Iy are known or easily calculated separately.Note: calculate the moments means the two ordinary moments

Types:

Moments of inertia of an area continuedMonday, April 12, 2010

07:58

Notes

Determine the moments of inertia Ix and Iy and polar moment of inertia JO, for a thin strip. Assume that t/l << 1.

ApplicationDetermine the distribution of shear stress, tau, in a linear elastic circular rod of radius R subjected to a torque T:

Example problemsMonday, April 12, 2010

08:05

Notes

Numeric integration, Simpson's ruleMonday, April 12, 201008:30

Notes

Determine moment of inertia for the shaded area about the x axis. Use simpson's rule to evaluate the integral.

ExampleMonday, April 12, 2010

08:41

Notes

Powerful relations are easily derived for moments of inertia of an area about axes parallel to the centroidal axes, x1, y1:

Note that x1, y1 must pass through the centroid. Also, the moments of inertia are seen to be a minimum about the centroidal axes.

The parallel-axis theoremWednesday, April 14, 2010

07:58

Notes

Determine the moment of inertia Ix of a rectangle about its edge given hat Ixc=bh3/12 aboutits centroid.

Determine the moment of inertia xc of a parabolic sector aboutits centroidal axis.

ExamplesWednesday, April 14, 2010

08:18

Notes

Must be symmetric about the axis and as far away as possible

Question: What shape of cross section of a given area A maximizes I xc for a given maximum width b and height h.

An I-beam to take shear and keep it from falling apartNote: fall apart and have shear if not having I part

Notes

For composite areas add areas, first moments, and second moments algebraically:

Where + is used if an area is added and - I sused if area is subtraced. Usually the location is calculated with respect to the centroid

Composite areasFriday, April 16, 2010

08:00

Notes

Determine the location of the centroid ybar of the beam constructed from the two channels and the cover plate. If each channel has a cross sectional area Ac of 11.8 in2 and a moment of inertia IxC about a horizontal axis passing through its own centroid Cc of 349 in^4. Determine the moment of inertia Ix1 for the beam's cross sectional area about the x1 axis.

ExamplesFriday, April 16, 201008:02

Notes

Determine the moment of inertia Ix for the composite area about the x axis.

153.66-1.865*1.865*31.53=43.9916

ExampleFriday, April 16, 201008:15

Notes

The moments of inertia, Ix for the box beam and the wide flange I-beam are the same:

Railroad tracks are manufactured in quarter mile long sections - the factory is typically 2 miles long.

Can calculate by cutting it apart and then putting it back as in the box beam.

The rounded fillet is important and does affect the result by approximately 3-4%.

Practical exampleFriday, April 16, 2010

08:30

Notes

The principle of virtual work allows some complex structural problems to be solved easily, and leads to powerful techniques such as the finite-element method in structural mechanics

If A = 0, then A.B=0 for any BMore interestingly, if A.B= 0 for any (and all) B, then A=0

Consider two vectors A and BDot product

Work given by dot productWork

Basic Concepts

Eyjafjallajokull Rearth=4000 milesAtmosphere: Highest mountain:~4 milesIrregularities are very smallRonald Adrian - turbulence: Particle Image Velocimetry (sp?)

Virtual displacements: a conceptually possible displacement or rotation of all or part of a system of particles. Kinematic constraints need not be honored

δW=ΣF.δu

The virtual work of the resultant external force, ΣF acting on ta system of particles undergoing a uniform virtual displacement, δu

Good description in textbook

If δW for any uniform δu, then the sum of forces equals 0 and threfore the system is in force equilibrium

Observe that if the system is in external force equilibrium, ΣF=0 and therefore δW=0 for any uniform virtual displacement δu. The more interesting result however is that:

Then if δW=0 for any rotation then the sum of moments equals zero and therefore it is in moment equilibrium

Also, for moment equilibrium. Here the resultant Σrx. The virtual work of the resultant momenta acting through a virtual rotation

Virtual work

Virtual WorkMonday, April 19, 2010

07:58

Notes

The thin rod of weight W rests agains the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibtium

Note: the effort required to solve this problem by the principle of virtual work is about the same as that by the equations of equilibrium for a free-body diagram of the rod.

If a force P of 30 lb is applied perpendicular to the handle of the toggle press determine the compressive force developed at C. Let thetat equal 30 degrees.

ProblemsMonday, April 19, 2010

08:23

Notes

If δW=0 for any virtual translation δu, then ΣF=0, it is in force equilibriumIf δW=0 for any virtual rotation δθ then ΣM=0, it is in moment equilibrium

Let δW be the virtual work done by external forces and couples on a system during a virtual displacement δu. Then:

In practice a mix of virtual translations and rotations is used to solve equilibrium problems using the principle of virtual work

Principle of Virtual workThink of virtual work as a double exposure on an old camera

Virtual Work continuedWednesday, April 21, 2010

07:59

Notes

The angle for equilibrium if K=2500N/m, and M=50N.ma.The required stiffness k so that the mechanism is in equilibrium when theta is 30 degrees and M=0b.

Each member of the pin connected mechanism has a mass of 8 kg. If the spring is unstretched when theta is 0, determine

Example 1Wednesday, April 21, 2010

08:02

Notes

This is permissible - constraints can be violated, there is NO reason why notViolating constraints

Example 2Wednesday, April 21, 201008:31

Notes

A horizontal force acts on the endo f the link. Dtermine the angles theta 1 and theta 2 for equilibrium. Each link is uniform and has a mass m.

Example 3Wednesday, April 21, 2010

08:44

Notes

For 0°<θ<90°a.When θ=60°b.

The crankshaft is subjected to a torque M of 50 N.m. Determine the horizontal compressive force F applied to the piston

Another problemFriday, April 23, 2010

07:51

Notes

This is approximately sinusoidal - connecting arm complicates this

Notes

Determine the mass of A and B required to hold the 400g desk lamp in balance for any angles, theta and phi. Neglect te weight of the mechanism and thes size of the lamp

Problem - lampFriday, April 23, 2010

08:28

Notes

Rods AB and BC have centers of mass located at their midpoints. If all contacting surfaces are smooth and BC has a mass of 100 kg, determine the appropriate mass of AB required for equilibrium.

Problem: rods in wellMonday, April 26, 2010

07:54

Notes

The mechanism consists of the four pin connected bars and three springs, each having a stiffness k and an unstretched length l0. Determine the horizontal forces P that must be applied to the pins in order to hold the mechanism in the horizontal position for equilibrium.

Also, it is possible that the springs can only operate in tension - in that case 2lsin(theta) must be greater than or equal to l naught.

SpringMonday, April 26, 2010

08:21

Notes

Definition: fluid: a substance that cannot support a shear stress at rest (a solid can). Fluids can be gases or liquids

The pressure p at a point in a fluid is the same in all directions-

Pascal's law

For that to be a valid FBD then it must be a solid, as it is sustaining shear stress

An incompressible fluid is one for which the mass density, ρ, is independent of the pressure, p. Liquids are considered incompressible. Gases are compressible but may be approximated as incompressible if the pressure variations are relatively small.

-

This is an idealization - there technically exists no such substance-

Incompressible fluids

For an incompressible fluid at rest, the pressure varies linearly with depth h:-

Pressure variation in a fluid at rest

The Archimedes Palimpsest - can look at his original writings - it is online (and at a museum)

The Palimpsest Project

Hydrostatic ForcesWednesday, April 28, 2010

07:57

Notes

The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment about O due to the dam's wieight divided by the overturning moment about O due to the water pressure. Determine this factor if the concrete has a specific weight of 150 lb/ft^3 and for water 62.4 lb/ft^3

ProblemWednesday, April 28, 2010

08:27

Notes

There are practice problems online

This is gage pressurePressure varies linearly from surface

Pressure at a point is independent of direction in fluids

Hydrostatic ForcesFriday, April 30, 201008:00

Notes

The 2m wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure - density of water = 1.0 Mg/m^3

Note that this is independent of H

Note that force at pin will be much larger than at block

Problem: gateFriday, April 30, 2010

08:03

Notes

The arched surface AB is shaped in the form of a quarter circle. If it is 8m long, determine the horizonal and vertical components of the resultant force caused by the water acting on the surface, use 1Mg/m^3 for water density

Problem: curved surfaceFriday, April 30, 2010

08:23

Notes

Grain binFriday, April 30, 2010

08:42

Notes

Next lecture is a review of the course

The gasoline tank is constructed with elliptical ends. Determine the resultant force F and its location (the center of pressure) yp on these ends if the tank is half full. Take γ=41lb/ft3.

Another hydrostatic problemMonday, May 03, 201007:58

Notes

12 May 2010, 8 to 11228 Natural History BuildingComprehensive - no conflictsShould be able to work and check completely in 1.5 hours, but will have the full 3 hours.Work any 5 of the 6 problems given. Write OMIT on the problem to be omitted. No calculators. Only a pen or pencil is needed. (Can bring a ruler and eraser.) Explain you work, provide clear diagrams, work logically and neatly.

FOLLOW INSTRUCTIONS

Final Exam: InformationWednesday, May 05, 2010

07:58

Notes

Vector operations, law of gravitation, force vectors along a line, components parallel and perpendicular to a line-

Force vectors

Free body diagrams for particles; equilibrium-

A drawing of a body or a part of a body on which ALL forces acting on the body are shown. PERIOD-

CUT the body-

Free body diagrams

Frictionless pulley - T same on each side-

Linearly elastic spring-

Idealizations

About a point, and about a line; couples-

MP=rxF (note: order matters) - even in case of couples - r from -F to F-

Moment parallel and perpendicular to line-

Moment of a force

The sum of moments and sum of forces are the same in two systems-

Resultant: an equipollent system consisting of one force and one couple at a point specified-

Concurrent forces - no couples, resultant is force only

Coplanar forces - couples normal to plane, resultant is force only

Parallel forces - can create a force only resultant

Special equipollent systems: -

Resultant force and moment acting along line of that force

Pushing and turning at one point

Screwdriver (wrench) - always possible as long as force resultant does not vanish-

Equipollence (equivalence)

Often along lines or planes - must calculate equivalent force-

Use Simpson's rule if only have numerical data-

Distributed forces

Roller/slider - 1 unknown (1 force)-

Pin - 2 unknowns (2 force)-

Fixed - 3 unknowns (2 force, 1 moment)-

Others (hinges…)-

Types of support

Written by Bernoulli's and Euler-

For equilibrium sum of forces and sum of moments equals ZERO-

Two-force members - equal force in opposite direction - even if body is not straight-

Three-force members - either concurrent or parallel (but parallel is technically concurrent at infinity)-

General equilibrium - extended bodies

Method of joints-

Method of sections - really can only cut 3 members if going to find the forces - BUT can have zero force members-

Could apply same idea to 3-D structures -

Trusses

Members which may be built in at ends-

Frame - doesn't move; machine - can move-

Frames, machines

Normal (axial)-

Transverse shear-

Torsional (torque)-

Bending-

Shear-force and bending-moment diagrams - graph them and shade under the curve-

Differential and integral relations + jumps-

Internal forces

Coulomb friction; belts-

Wedges; screws-

Friction

Composite bodies-

Centroids

Ordinary and polar-

Parallel axis theorem, composite areas-

Moments of inertia

Imaginary displacements-

Method of choice with many interconnected joints with no work done at joints-

Virtual work

Varies linearly from surface-

Hydrostatic forces

Grades will be posted app. May 19

Final Exam Review:Wednesday, May 05, 201008:03

Notes

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