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First Class
Wednesday, January 20, 2010
07:55
Statics: a subject in mechanics
Mechanics: 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 bodies
Biomechanics: muskoskeletal mechanics
TAM 251: Introductory Solid Mechanics - strength of materials - study of how things deform - twist,
compress, stretch…
ME 340: dynamics of buildings
ME 310: fluid dynamics
ME 330: behavior of materials
ME 360: controls - robotics
ME 320: heat transfer
ME 370: design
ME 350: design for manufacturability
Notes Page 1
Syllabus
Wednesday, January 20, 2010
08:19
Forces and bending moments
Multiple bodies
Beams, structural analysis, trusses
Friction
Can switch to TAM 211 at this point
Forces on submerged bodies
Notes Page 2
No course website yet - on mechse website will be
Grades on Compass
HMWKS due Fridays at 08:00
Ms Thomas's know what language we speak
Notes Page 3
Precision: 3 significant figures ALWAYS (like IB math)
Exams are during lectures (not in this room)
Notes Page 4
Vectors, Newton's Laws of Motion, Gravity, Calculations, and
Analysis
Friday, January 22, 2010
07:56 A1-A5 (vectors) in book
Read
Newton's Three Laws of Motion
1. A particle at rest or moving with constant velocity will remain in this state unless acted upon by an
unbalanced force.
2. A [article acted upon by an unbalanced force F experiences an acceleration a that is proportional
to the particle mass, m.
3. The mutlual forces of action and reaction between two particles are equal opposite and collinear.
Newton's law of gravitational attraction
The mutual force F of gravitation between two particles of mass m1 and m2 is given by
Weight is the force exerted by the earth on a particle at the earth's surface:
Units of Measurement
System
Length Time
Mass
Force
SI
m
s
kg
N
US Customary (FPS) ft
s
slug
lb
Newton and slug are derived units
Unit conversion
- 1 ft = 0.3048m (exact)
- 1 slug = 14.59 kg
- 1 lb = 4.448 N
Numerical Calculations
- Dimensional homogeneity
 Equations must be dimensionally homogenous
Notes Page 5
 Equations must be dimensionally homogenous
Significant Figures
- Use 3 significant figures for final answers, more (usually 4) for intermediate calculations
 This is engineering convention (assume all dimensions are to 3 sf if not specified, 1 ft = 1.00
ft)
General Procedure for analysis
1. Read the problem carefully, write it down carefully.
2. Draw given diagrams neatly and construct additional figures as necessary.
3. Apply principles needed.
4. Solve problem symbolically, then substitute numbers. Proved proper units throughout. Check
significant figures. Box the final answer(s)..
5. See if answer is reasonable.
Starting Monday: class meets in 228 Natural History Building
Notes Page 6
Forces
Monday, January 25, 2010
07:57
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.
Notes Page 7
Find a vector that is simultaneously in the plane of F1 and F2 and in the plane of F3 and F4.
Discussions this week
HMWK due Friday
Notes Page 8
Forces, components, and division thereof
Wednesday, January 27, 2010
07:58
Forces: the action of one body on another - magnitude, sense, and direction - vectors
Force components: Law of sines and cosines can be used in 2-d problems, but using force components is
usually more straightforward.
- Can also solve problems graphically
Adding vectors - simple
- Pick a good system - but keep it once you have it
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?
Question: how do we know this is the minimum rather than the maximum?
- Cannot be a maximum, since the force could be large as possible
- Note: could solve it geometrically, but much more difficult
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.
Notes Page 9
Cantilever bending - force vertical on lug nut
Components of forces parallel and perpendicular to a line:
- 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//.
Problem:
Determine the angle theta betweeen the two cables. Determine the projected component of the Force
F=12lb in direction of AC
Notes Page 10
Equilibrium, Free body diagrams, and Idealizations
Friday, January 29, 2010
07:59
Equilibrium:
- 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 +ΣFy j+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
Free Body diagram
- 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.
Idealization:
- 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.
Notes Page 11
Example Problems
Friday, January 29, 2010
08:03
In reality either rope could break, as there is a distribution of ropes
break at slightly different values - they follow a normal curve
Notes Page 12
Notes Page 13
Systems of Particles
Monday, February 01, 2010
07:59
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.
Notes Page 14
If you want to find the value of a force you must draw
a free body diagram that includes that force.
Example Problems
Monday, February 01, 2010
07:59
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.
Notes Page 15
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 Page 16
Moment of a force
Wednesday, February 03, 2010
07:57
Hour exam next Friday - cumulative so far
- Vectors, Application of vectors…
- It will be in 2 different rooms - here and 100 MSEB - there will be assigned seating
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.
Principle of transmittivity: F can be acting anywhere along the line of action
Notes Page 17
Line of action of F
Example Problems
Wednesday, February 03, 2010
08:01
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
Notes Page 18
Notes Page 19
Moments of Forces and Couples
Friday, February 05, 2010
08:00
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
Moment of a Force about a Line:
- For a point: MP=rxF
- Ml=MP.ul
- Ml==(MP.ul)ul
- Note the perpendicular one is the total minus the parallel
Couples:
- 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
Notes Page 20
Example problems
Friday, February 05, 2010
08:00
Example: determine moment about Oa axis
Determine the couple moment - express as a cartesian vector
Notes Page 21
Notes Page 22
Equipollent force systems
Monday, February 08, 2010
07:58
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.
ΣMP denotes the moment of all applied couples plus the sum of moments of applied forces with
respect to point P
Note: equivalent is often used instead of equipollent.
Pronunciation: ee-quee-pah-lent
Notes Page 23
Hour exam:
- 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
Example problems
Post tensioning ram on a forklift
Monday, February 08, 2010
07:58
Can rephrase as: Replace the force and couple system by its resultant at point Q.
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
Notes Page 24
Resultants of Force Systems
Check your seat before exam
Course notes posted outside 429 Grainger
Wednesday, February 10, 2010
07:57
Resultants:
- A resultant is an equipollent force system consisting of only one force Fr and one couple MrP:
 Must specify P
Fr=(ΣF) 1
MrQ=(ΣM)1
F can be anywhere along its line of action, couple can be
applied anywhere
Notes Page 25
Example problems
Wednesday, February 10, 2010
07:58
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.
Notes Page 26
Hour Exam
Monday, February 15, 2010
08:06
Make sure to explain work - use words
Also simplifying important
Exams by Friday hopefully
Notes Page 27
Reductions
Autonomous Materials Group
Monday, February 15, 2010
07:59
Reduction classes:
- Single resultant force (often possible)
 Concurrent forces with no couples
 Coplanar forces with couples perpendicular to force plane
 Parallel forces with couples perpendicular to forces
- Screwdriver (always possible) - consists of a single force and a couple in the same direction
A resultant is an equipollent force system consisting of only one force, Fr and one couple MrP - must
specify point
Single resultant forces
- 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.
1. Concurrent forces (no couples)
i. Act at the same point, just sum the forces
2. Coplanar, with couples perpendicular to plane
i. Add the forces, that will be the force
ii. Find MrP and then divide by the force and move the force over that much - that will be the
single force resultant
3. Parallel forces with couples perpendicular
i. Add the forces for the resultant force
ii. Find moment about P, then move the force over by MrP divided by the force
iii. Common with a plate with wires and weights
Screwdriver
- Most general case - always possible
- Consists of a force and a couple in the same direction - push and twist
Move the force over in order to cancel the perpendicular
force - then it will just be the parallel moment and the force
Notes Page 28
Example problem
Monday, February 15, 2010
08:42
Replace the 3 forces by a screwdriver
Notes Page 29
Distributed forces
Screwdriver is referred to as a wrench by some texts.
Wednesday, February 17, 2010
08:16
Distributed forces
- 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.
Notes Page 30
Example Problems
Wednesday, February 17, 2010
08:16
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
Determine
a. The magnitude of the total force exerted on the foundation
b. The magnitude of the moment about A due to the distributed load.
Notes Page 31
Simple Distributed Loading
Friday, February 19, 2010
07:59
Total force given by area under loading diagram
It is located at the centroid of the area
Notes Page 32
Equilibrium of a rigid body
Friday, February 19, 2010
08:10
We regard a rigid body as a collection of particles:
Force equilibrium:
Let Fi=the resultant external force on particle i
Let fij=the internal force on particle i by particle j
Let fji=the internal force on particle j by particle i
In general: 2D problems have 3 unknowns, 3 equations (2 forces, 1 moment)
3D problems have 6 unknowns, 6 equations (3 forces, 3 moments)
Notes Page 33
Example problems
Friday, February 19, 2010
08:28
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
Notes Page 34
Types of support
Friday, February 19, 2010
08:34
Type
Description Reactions Unknowns
Roller/slider
1
Pin
2
Fixed
3
Notes Page 35
Free body diagram
Friday, February 19, 2010
08:37
Notes Page 36
Applications of Equilibrium
Monday, February 22, 2010
07:59
Sum of forces and sum of moments are zero.
These are external forces on the extended body - the internal ones don't count for overall
equilibrium
Can 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.
Notes Page 37
Example Problems
Monday, February 22, 2010
07:59
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.
Notes Page 38
Two and Three force members
Monday, February 22, 2010
08:18
In many situations bodies are subject to only two or three forces and no couples.
Two force members:
- The forces must be equal, opposite, and collinear
- Shape of body doesn't matter
Three force members:
- The three forces must be either concurrent or parallel and must be coplanar
Notes Page 39
Problem
Monday, February 22, 2010
08:26
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
Notes Page 40
Problems with extended bodies - 3D
Wednesday, February 24, 2010
08:00
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
Notes Page 41
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 Page 42
Notes Page 43
More problems
Friday, February 26, 2010
07:57
There are 4 courses in biomechanics
[email protected]
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.
Journal bearing - allows it to turn around
Thrust 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.
Notes Page 44
Notes Page 45
Introduction to structures
Friday, February 26, 2010
08:43
Structural equilibrium - Talbot laboratory, Titan formwork builds formwork for concrete - build a
multistory building
Must be lightweight, easy to assemble
Notes Page 46
Trusses
Monday, March 01, 2010
07:59
Mode of failure is not necessarily unique
Buckling: slender member under compression bends - very important
One test is worth a thousand expert opinions
Trusses 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
Two primary methods to find loads in members
- Methods of joints
- Method of sections
Always first set the overall structure in equilibrium
Method of joints
1. Set the entire structure into external equilibrium
2. 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)
Notes Page 47
Example Problems
Monday, March 01, 2010
07:59
Truss subject to loading shown
Distance between rails on railroad is standard at 4 ft 8.5 inches
Bob Miller
Determine force in member FG of truss - state tension or compression
Notes Page 48
Notes Page 49
Trusses: method of sections
Failure point is often the joint, especially dissimilar or composite material
Long slender members are very easy to break in bending
Wednesday, March 03, 2010
08:01
Method.
1. Set the body into external equilibrium
2. 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.
Notes Page 50
Example problems
Wednesday, March 03, 2010
08:09
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
Notes Page 51
Notes Page 52
Frames and Machines
17 March: Hour Exam 2
Can stay and learn TAM 211 material
Prob. A Wednesday or Thursday night for TAM 210 exam
Friday, March 05, 2010
07:59
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:
1. Set the entire structure into external equilibrium. This step will generally produce more unknowns
than there are relevant equations of equilibrium
2. 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.
3. Solve for the requested unknowns.
Notes Page 53
Example Problems
Friday, March 05, 2010
08:01
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.
Notes Page 54
Note: tensions in the ropes will not be the same
Note: can draw more FBD's and calculate the rest of forces (which
were not requested here)
Notes Page 55
Internal Forces
1st two are forces
2nd two are moments
Monday, March 08, 2010
07:58
Cutting members at internal points reveals internal forces and moments:
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.
Notes Page 56
Example problem
Monday, March 08, 2010
08:10
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.
Notes Page 57
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 vary
Problem: 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 Page 58
Shear force and bending moment distributions
Examination next week - notes on wall outside 429 Grainger.
Wednesday, March 10, 2010
07:59
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.
Diagrams:
Notes:
- 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 Page 59
Example Problem II
Friday, March 12, 2010
08:30
The beam consists of three segments pin connected at B and E. Draw the shear force and bending
moment diagrams for the beam.
Notes Page 60
Diagrams
Friday, March 12, 2010
08:42
Notes Page 61
Problem
Wednesday, March 10, 2010
08:15
Express the x,y,z components of the internal loading along the rod as a function of y where 0<=y<=4
Notes Page 62
Diagram
Friday, March 12, 2010
08:02
From problem on Monday
Diagrams:
Notes Page 63
Shear-force and bending moment diagrams
Monday, March 15, 2010
07:59
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
- 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 distributions can then be plotted.
Differential and integral relations for q, V, and M
- q: distributed load (positive up)
- V: shear force (positive up)
- M: bending moment (positive z)
- dV/dx=-q
- dM/dx=-V
Be very careful of sign conventions
Notes Page 64
Example problem
Monday, March 15, 2010
08:01
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
Notes Page 65
Diagrams
Monday, March 15, 2010
08:27
Notes Page 66
HE 2
Friday, March 19, 2010
07:59
Euler derived the moment equilibrium condition
Torsion is principal deformation mode in axially loaded helical spring.
Notes Page 67
Friction
Friday, March 19, 2010
07:59
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.
Notes Page 68
Example problem
Friday, March 19, 2010
08:07
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 frictionless, 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.
Notes Page 69
Notes Page 70
Wedges
Monday, March 29, 2010
07:58
These are tapered pieces used to locate and hold objects in place. Usually, friction is involved. Inserting
and removing are not the same:
Inserting:
Friction ALWAYS acts AGAINST you.
Smaller force to remove than to insert, as N A opposes you going
in, but helps you go out.
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.
Notes Page 71
Belts
Monday, March 29, 2010
08:04
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:
Notes Page 72
Example Problem
Monday, March 29, 2010
08:24
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.
Notes Page 73
Example Problem 2
Monday, March 29, 2010
08:33
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.
Notes Page 74
Screws
Wednesday, March 31, 2010
08:00
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
Author does this
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.
Notes Page 75
Example Problem
Wednesday, March 31, 2010
08:20
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.
Notes Page 76
Note: spreading blocks apart would require a different (smaller) value
of M.
Notes Page 77
Centroids
Friday, April 02, 2010
07:58
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:
Notes Page 78
Example Problems
Friday, April 02, 2010
08:05
Locate the centroid of the shaded area bounded by the parabola and the line y=a
Locate the centroid z of the volume segment.
Notes Page 79
HE 2:
y=0.9x+20 - adjusted
Avg: 75.7, median 76.3
For every dV with a positive value of y (or x) there is another dV with a negative
value of y (or x).
Notes Page 80
Another problem
Monday, April 05, 2010
08:00
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 l 1 for the smooth support at B in order to improve the accuracy of this experiment.
Best place for l 1 is probably l/2, since then (l/2-l 1)W1 vanishes. Then errors in W1
become irrelevant.
Versabar: weighing structures and lifting them
Notes Page 81
Composite bodies
Monday, April 05, 2010
08:12
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)
Notes Page 82
Example
Monday, April 05, 2010
08:16
Determine the location x bar of the centroid of the solid made from a cone cylinder and hemisphere.
Notes Page 83
Another problem
Monday, April 05, 2010
08:32
When material is subtracted the corresponding V i effectively changes sign.
Notes Page 84
Example: bridfeeder
5,7,9 rule for birdfeeders
5 feet off the ground, 7 feet from something
sideways, 9 feet above
Wednesday, April 07, 2010
08:08
The bird-feeder pole is made from a continuous length of uniform rod. Determine the location (x bar,ybar)
of the centroid C.
If have two bodies the centroid of the composite must lie along
the line connecting the two individual centroids
Notes Page 85
Moment of Inertia of an Area
Friday, April 09, 2010
07:56
There are also moments of inertia of mass (dynamics)
Motivation: beam bending:
Consider a beam in bending under the action of a bending moment M:
When doing testing, must approximate the actual field
conditions
Will bend (if pure moment is applied) into the arc of a circle
Tension at bottom, compression at top, nothing in the centre (centroid) this is called the neutral surface
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.
Shear is not particularly important here
Notes Page 86
Example
Friday, April 09, 2010
08:19
Determine the moment of inertia, I x 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
Notes Page 87
Note: can also be solved from taking horizontal strips, but more difficult
Notes Page 88
Moments of inertia of an area continued
Monday, April 12, 2010
07:58
Types:
There are several types of second moment or moment of inertia of an area:
Ordinary
Cross
Polar
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 J P when Ix and Iy are known or easily calculated separately.
Note: calculate the moments means the two ordinary moments
Notes Page 89
Example problems
Monday, April 12, 2010
08:05
Determine the moments of inertia Ix and Iy and polar moment of inertia JO, for a thin strip. Assume that
t/l << 1.
Application
Determine the distribution of shear stress, tau, in a linear elastic circular rod of radius R subjected to a
torque T:
Notes Page 90
Numeric integration, Simpson's rule
Monday, April 12, 2010
08:30
Notes Page 91
Example
Monday, April 12, 2010
08:41
Determine moment of inertia for the shaded area about the x axis. Use simpson's rule to evaluate the
integral.
Notes Page 92
The parallel-axis theorem
Wednesday, April 14, 2010
07:58
Powerful relations are easily derived for moments of inertia of an area about axes parallel to the
centroidal axes, x 1, y1:
Note that x 1, y1 must pass through the centroid. Also, the moments of inertia are seen to
be a minimum about the centroidal axes.
Notes Page 93
Examples
Wednesday, April 14, 2010
08:18
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.
Notes Page 94
Question: What shape of cross section of a given area A maximizes I xc for a given maximum width b and
height h.
Must be symmetric about the axis and as far away as possible
An I-beam to take shear and keep it from
falling apart
Note: fall apart and have shear if not
having I part
Notes Page 95
Composite areas
Friday, April 16, 2010
08:00
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
Notes Page 96
Examples
Friday, April 16, 2010
08:02
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 A c 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 I x1 for
the beam's cross sectional area about the x 1 axis.
Notes Page 97
Example
Friday, April 16, 2010
08:15
Determine the moment of inertia I x for the composite area about the x axis.
153.66-1.865*1.865*31.53=43.9916
Notes Page 98
Practical example
Friday, April 16, 2010
08:30
The moments of inertia, Ix for the box beam and the wide flange I-beam are the same:
Can calculate by cutting it apart and then putting it
back as in the box beam.
Railroad tracks are manufactured in quarter mile long sections - the factory is typically 2 miles long.
The rounded fillet is important and does affect the result by
approximately 3-4%.
Notes Page 99
Virtual Work
Eyjafjallajokull
Monday, April 19, 2010
07:58
Rearth=4000 miles
Atmosphere:
Highest mountain:~4 miles
Irregularities are very small
Ronald Adrian - turbulence: Particle Image Velocimetry (sp?)
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
Basic Concepts
Dot product
Consider two vectors A and B
If A = 0, then A.B=0 for any B
More interestingly, if A.B = 0 for any (and all) B, then A=0
Work
Work given by dot product
Virtual displacements: a conceptually possible displacement or rotation of all or part of a system of
particles. Kinematic constraints need not be honored
Virtual work
The virtual work of the resultant external force, ΣF acting on ta system of particles undergoing a
uniform virtual displacement, δu
δW=ΣF.δu
Good description in textbook
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:
If δW for any uniform δu, then the sum of forces equals 0 and threfore the system is in force
equilibrium
Also, for moment equilibrium. Here the resultant Σrx. The virtual work of the resultant momenta
acting through a virtual rotation
Then if δW=0 for any rotation then the sum of moments equals zero and therefore it is in
moment equilibrium
Notes Page 100
Problems
Monday, April 19, 2010
08:23
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.
Notes Page 101
Notes Page 102
Virtual Work continued
Wednesday, April 21, 2010
07:59
Principle of Virtual work
Let δW be the virtual work done by external forces and couples on a system during a virtual
displacement δu. Then:
If δW=0 for any virtual translation δu, then ΣF=0, it is in force equilibrium
If δW=0 for any virtual rotation δθ then ΣM=0, it is in moment equilibrium
In practice a mix of virtual translations and rotations is used to solve equilibrium problems using
the principle of virtual work
Notes Page 103
Think of virtual work as a double exposure on an old
camera
Example 1
Wednesday, April 21, 2010
08:02
Each member of the pin connected mechanism has a mass of 8 kg. If the spring is unstretched when
theta is 0, determine
a. The angle for equilibrium if K=2500N/m, and M=50N.m
b. The required stiffness k so that the mechanism is in equilibrium when theta is 30 degrees and M=0
Notes Page 104
Notes Page 105
Example 2
Wednesday, April 21, 2010
08:31
Violating constraints
This is permissible - constraints can be violated, there is NO reason why not
Notes Page 106
Example 3
Wednesday, April 21, 2010
08:44
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.
Notes Page 107
Another problem
Friday, April 23, 2010
07:51
The crankshaft is subjected to a torque M of 50 N.m. Determine the horizontal compressive force F
applied to the piston
a. For 0°<θ<90°
b. When θ=60°
Notes Page 108
This is approximately sinusoidal - connecting arm complicates this
Notes Page 109
Problem - lamp
Friday, April 23, 2010
08:28
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
Notes Page 110
Problem: rods in well
Monday, April 26, 2010
07:54
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.
Notes Page 111
Spring
Monday, April 26, 2010
08:21
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.
Notes Page 112
Hydrostatic Forces
Wednesday, April 28, 2010
07:57
Definition: fluid: a substance that cannot support a shear stress at rest (a solid can). Fluids can be gases
or liquids
Pascal's law
- The pressure p at a point in a fluid is the same in all directions
For that to be a valid FBD then it must be a solid, as it is sustaining
shear stress
Incompressible fluids
- 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
Pressure variation in a fluid at rest
- For an incompressible fluid at rest, the pressure varies linearly with depth h:
Notes Page 113
The Palimpsest Project
The Archimedes Palimpsest - can look at his original
writings - it is online (and at a museum)
Problem
Wednesday, April 28, 2010
08:27
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
Notes Page 114
Hydrostatic Forces
There are practice problems online
Friday, April 30, 2010
08:00
Pressure varies linearly from surface
This is gage pressure
Pressure at a point is independent of direction in fluids
Notes Page 115
Problem: gate
Friday, April 30, 2010
08:03
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
Notes Page 116
Problem: curved surface
Friday, April 30, 2010
08:23
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
Notes Page 117
Grain bin
Friday, April 30, 2010
08:42
Notes Page 118
Another hydrostatic problem
Next lecture is a review of the course
Monday, May 03, 2010
07:58
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.
Notes Page 119
Notes Page 120
Final Exam: Information
Wednesday, May 05, 2010
07:58
12 May 2010, 8 to 11
228 Natural History Building
Comprehensive - no conflicts
Should 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.
Notes Page 121
FOLLOW INSTRUCTIONS
Final Exam Review:
Wednesday, May 05, 2010
08:03
Force vectors
- Vector operations, law of gravitation, force vectors along a line, components parallel and perpendicular to a line
Free body diagrams
- 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
Idealizations
- Frictionless pulley - T same on each side
- Linearly elastic spring
Moment of a force
- 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
Equipollence (equivalence)
- 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
- Special equipollent systems:
 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
- Screwdriver (wrench) - always possible as long as force resultant does not vanish
 Resultant force and moment acting along line of that force
 Pushing and turning at one point
Distributed forces
- Often along lines or planes - must calculate equivalent force
- Use Simpson's rule if only have numerical data
Types of support
- Roller/slider - 1 unknown (1 force)
- Pin - 2 unknowns (2 force)
- Fixed - 3 unknowns (2 force, 1 moment)
- Others (hinges…)
General equilibrium - extended bodies
- 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)
Trusses
- 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
Frames, machines
- Members which may be built in at ends
- Frame - doesn't move; machine - can move
Internal forces
- 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
Friction
- Coulomb friction; belts
- Wedges; screws
Centroids
- Composite bodies
Moments of inertia
- Ordinary and polar
- Parallel axis theorem, composite areas
Virtual work
- Imaginary displacements
- Method of choice with many interconnected joints with no work done at joints
Hydrostatic forces
- Varies linearly from surface
Notes Page 122
Grades will be posted app. May 19