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Exam 2 (Wed March 20th, 2019)
Chapters covered (6, 8, 9, 10, 12)
Chapter 6: Structures in Equilibrium
[Section 6.4]: Method of Sections
In the method of sections, a truss is divided into two parts
by
taking an imaginary “cut” (shown here as a-a) through the
truss
STEPS FOR ANALYSIS
1. Decide how you need to “cut” the truss. This is based on:
a) where you need to determine forces
b) where the total number of unknowns does not exceed
three (in general)
2. Decide which side of the cut truss will be easier to work
with (minimize the number of reactions you have to find)
3. If required, determine any necessary support reactions by
drawing the FBD of the entire truss and applying the E-of-E
4. Draw the FBD of the selected part of the cut truss. We
need
to indicate the unknown forces at the cut members.
5. Initially we may assume all the members are in tension,
as
we did when using the method of joints. Upon solving, if
the answer is positive, the member is in tension as per our
assumption. If the answer is negative, the member must be
in compression.
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6. Apply the scalar equations of equilibrium (E-of-E) to the
selected cut section of the truss to solve for the unknown
member forces. Please note, in most cases it is possible to
write one equation to solve for one unknown directly. So
look for it and take advantage of such a shortcut!
[Section 6.5]: Skipped
[Section 6.6]: Frames & Machines
STEPS FOR ANALYZING A FRAME OR MACHINE
1. Draw a FBD of the frame or machine and its members, as
necessary
a. Identify any two-force members
A two force member is a body that has forces (and only
forces, no moments) acting on it in only two locations.
In order to have a two force member in static
equilibrium, the net force at each location must be
equal, opposite, and collinear. This will result in all
two force members being in either tension or
compression
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b. forces on contacting surfaces (usually between a pin
and a member) are equal and opposite
c. for a joint with more than two members or an external
force, it is advisable to draw a FBD of the pin
d. Consider entire structure first and see if you can
quickly solve for an unknown
2. Develop a strategy to apply the equations of equilibrium
to
solve for the unknowns
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Chapter 8: Friction
[Section 8.1 – 8.2]: Characteristic of Dry Friction &
Problems
Dry Friction:
The maximum friction force is attained just before the block
begins to move (a situation that is called “impending
motion”).
The value of the force is found using Fs = s N, where s is
called
the coefficient of static friction
Once the block begins to move, the frictional force
typically
drops and is given by Fk = k N. The value of k (coefficient
of
kinetic friction) is less than s: k < s
Steps for solving equilibrium problems involving dry
friction:
1. Draw the necessary free body diagrams. Make sure that
you show the friction force in the correct direction (it
always opposes the motion or impending motion)
2. Determine the number of unknowns. Do not assume
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F = S N unless the impending motion condition is given
3. Apply the equations of equilibrium and appropriate
frictional equations to solve for the unknowns
IMPENDING TIPPING versus SLIPPING:
For a given W and h of the box, how can we determine if the
block will slide or tip first? In this case, we have four
unknowns
(F, N, x, and P) and only three E-of-E
Hence, we have to make an assumption to give us another
equation (the friction equation). Then we can solve for the
unknowns using the three E-of-E. Finally, we need to check
if
our assumption was correct
Assume: Slipping occurs
Known: F = s N
Solve: x, P, and N
Check: 0 x b/2
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[Section 8.3]: ANALYSIS OF A WEDGES
A wedge is a simple machine in which a small force P is used to
lift
a large weight W.
To determine the force required to push the wedge in or out, it
is
necessary to draw FBDs of the wedge and the object on top of
it.
It is easier to start with a FBD of the wedge since you know
the
direction of its motion
Note that:
1. the friction forces are always in the direction opposite
to
the motion, or impending motion, of the wedge
2. the friction forces are along the contacting surfaces
3. the normal forces are perpendicular to the contacting
surfaces
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Next, a FBD of the object on top of the wedge is drawn.
Please
note that:
1. at the contacting surfaces between the wedge and the
object the forces are equal in magnitude and opposite in
direction to those on the wedge
2. all other forces acting on the object should be shown
To determine the unknowns, we must apply E of E, ( Fx = 0
and
Fy = 0) to the object and the wedge, as well as, the
impending
motion frictional equation, F = S N
Note: It is easier to apply E of E to the object first, since
you
probably have a know (i.e. weight, spring force, etc)
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[Section 8.4]: Skipped [Section 8.5]: Frictional Forces of Flat
Belts Consider a flat belt passing over a fixed curved surface with
the
total angle of contact equal to radians. If the belt slips or is
just about to slip, then T2 must be larger than T1 and the motion
resisting friction forces. Hence, T2 must be greater than T1
Here: T2 = T1 e
where is the coefficient of static friction between the belt and
the surface. Be sure to use radians when using this formula!!
[Sections 8.6 – 8.8]: Skipped
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Page Left Blank
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Chapter 9: Centroids & Center of Mass
[Section 9.1]: Centroid, composite areas, composite
volumes/lines, center of mass objects/composite area
The location of the center of gravity G with respect to the x,
y, z
axes is given by:
By replacing the W with a m (W=gm) in these equations, the
coordinates of the center of mass can be found:
Similarly, the coordinates of the centroid of volume, area,
or
length can be obtained by replacing W by V, A, or L,
respectively:
Centroid of Volumes:
Centroid of Areas:
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Centroid of Lines/Rods
CONCEPT OF CENTROID
The centroid, C, is a point which defines the geometric center
of
an object.
The centroid coincides with the center of mass or the center
of
gravity only if the material of the body is homogenous
(density
or specific weight is constant throughout the body).
If an object has an axis of symmetry, then the centroid of
object
lies on that axis. In some cases, the centroid is not located on
the
object
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STEPS TO DETERME THE CENTROID OF AN AREA
1. Choose an appropriate differential element dA at a
general
point (x,y). Hint: Generally, if y is easily expressed in
terms
of x (e.g., y = x2 + 1), use a vertical rectangular element
since the thickness of the element is dx. If the converse is
true, then use a horizontal rectangular element
2. Express dA in terms of the differentiating element dx or
dy
3. Determine coordinates ( �̃�, �̃�) of the differential
element’s
centroid in terms of the general point (x,y) i.e. �̃� = y/2
4. Express all the variables and integral limits in the
formula
using either x or y depending on whether the differential
element is in terms of dx or dy, respectively, and integrate
5. Use appropriate equations to determine centroid of the
body (𝑥,̅ �̅�)
Note: Similar steps are used for determining the Center of
gravity (CG) or center of mass (CM).
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Using vertical element: dA= y dx, but y = x3, so dA =
x3dx and the moment arms �̃� = x and �̃� = y/2
Using vertical element: dA = (y2 -y1) dx, but y2 = x and y1 =
x3, so dA = (x - x3) dx and the moment arms are: �̃� = x and
�̃� = y1 + (y2 -y1)/2 or in terms of x: �̃� = x3 + (x - x3)/2,
simplying �̃� = (x + x3)/2
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[Section 9.2]: Composite Bodies
CG / CM OF A COMPOSITE BODY: Consider a composite body
which consists of a series of particles(or bodies) as shown in
the
figure below. The net or the resultant weight is given as:
WR = W
Summing the moments about the y-axis, we get:
�̅�WR = �̃�1W1 + �̃�2W2 + ……….. + �̃�nWn
where �̃�1 represents x coordinate of W1, etc
By replacing the W with a M in these equations, the
coordinates
of the center of mass can be found
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STEPS FOR ANALYSIS COMPOSITE BODIES:
1. Divide the body into pieces that are known shapes.
Holes are considered as pieces with negative weight or size
2. Make a table with the first column for segment number,
the
second column for weight, mass, or area (depending on the
type of problem), the next set of columns for the moment
arms, and, finally, several columns for recording results of
simple intermediate calculations
3. Fix the coordinate axes, determine the coordinates of the
CG, CM or centroid of each piece, and then fill in the table
4. Use inside back cover of the book to find �̃� or �̃�
coordinates
of the shapes (triangle, rectangle, circle, semi-circle,
quarter
circle, etc)
5. Sum the columns to get �̅�, �̅�, and �̅�. For example:
�̅� = ( �̃�i Ai ) / ( Ai ) or �̅� = ( �̃�i Wi ) / ( Wi )
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Example of Center of Gravity – composite bodies
The densities of the materials are:
A = 150 lb / ft3 and B = 400 lb / ft3
Weight = W = (Volume in ft3)
Use CG equations to find:
�̅� = ( �̃�i Wi ) / ( Wi )
�̅� = ( �̃�i Wi ) / ( Wi )
�̅� = ( �̃�i Wi ) / ( Wi )
[Sections 9.3 – 9.5]: Skipped
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Chapter 10: Moments of Inertia
Moments of Inertia (MoI), Parallel Axis Theorem, Composite
Areas of simple objects, and Mass Moments of Inertia
[Section 10.1]: DEFINITION OF MOMENTS OF INERTIA (MoI) FOR
AREAS
The moments of inertia for the entire area are obtained by
integration:
Ix = A y2dA ; Iy = A x2dA
JO = A r2dA = A ( x2 + y2 ) dA = Ix + Iy
The MoI is also referred to as the second moment of an area
and
has units of length to the fourth power (m4 or in4)
For the differential area dA, shown in the figure below:
d Ix = y
2
dA
d Iy = x
2
dA
d JO
= r2
dA , where JO
is the polar moment of inertia about
the pole O or z axis.
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[Section 10.3]: RADIUS OF GYRATION OF AN AREA:
For a given area A and its MoI, Ix , imagine that the entire
area is
located at distance kx from the x axis.
Then, Ix = 𝑘𝑥2A or kx = Ix / A). This kx is called the radius
of
gyration of the area about the x axis
Similarly:
kY = ( Iy / A ) and kO = ( JO / A )
The radius of gyration can be considered to be an indication of
the stiffness of a section based on the shape of the cross-section
when used as a compression member (for example a column).
The diagram indicates that this member will bend in the thinnest
plane.
The radius of gyration is used to compare how various structural
shapes will behave under compression along an axis. It is used to
predict buckling in a compression member or beam.
The smallest value of the radius of gyration is used for
structural calculations as this is the plane in which the member is
most likely to buckle.
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MoI FOR AN AREA BY INTEGRATION:
For simplicity, the area element used has a differential size
in
only one direction (dx or dy). This results in a single
integration
and is usually simpler than doing a double integration with
two
differentials, i.e., dx·dy
The step-by-step procedure is:
1. Choose the element dA: There are two choices: a vertical
strip or a horizontal strip. Some considerations about this
choice are:
a. A differential element parallel to the axis about which
the MoI is to be determined usually results in an easier
solution. For example, we typically choose a horizontal
strip for determining Ix and a vertical strip for
determining Iy
b. If y is easily expressed in terms of x (e.g., y = x2 +
1),
then choosing a vertical strip with a differential
element dx wide may be advantageous
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Example: Find: The MoI of the area about the x- and y-axes using
horizontal element of width dy
Iy = A x2dA
Here to calculate Iy it is easier to use vertical element with
width
dx, so dA = y dx where y = √2𝑥 so dA = √2𝑥 dx
Iy = 20 x2√2𝑥dx
[Section 10.2]: PARALLEL-AXIS THEOREM FOR AN AREA &
MOMENT
OF INERTIA FOR COMPOSITE AREAS
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This theorem relates the moment of
inertia (MoI) of an area about an axis
passing through the area’s centroid to
the MoI of the area about a
corresponding parallel axis. This
theorem has many practical
applications, especially when working
with composite areas
Consider an area with centroid C. The x' and y' axes pass
through C. The MoI about the x-axis, which is parallel to,
and
distance dy from the x' axis, is found by using the
Parallel-Axis
Theorem
IX = 𝐼x̅' + Ady2
IY = 𝐼y̅' + Adx 2
JO = 𝐽C̅ + Ad 2
[Section 10.4]: MOMENT OF INERTIA FOR A COMPOSITE AREA
A composite area is made by adding or subtracting a series
of
“simple” shaped areas like rectangles, triangles, and
circles.
For example, the area below can be made from a rectangle
minus
a triangle and circle
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The MoI about their centroidal axes of these “simpler”
shaped
areas are found in most engineering handbooks as well as the
inside back cover of the textbook.
Using these data and the Parallel-Axis Theorem, the MoI for
a
composite area can easily be calculated
STEPS FOR ANALYSIS:
1. Divide the given area into its simpler shaped parts
2. Locate the centroid of each part and indicate the
perpendicular distance from each centroid to the desired
reference axis
3. Determine the MoI of each “simpler” shaped part about the
desired reference axis using the parallel-axis theorem
(for example: Ix = 𝐼x̅' + Ady2)
4. The MoI of the entire area about the reference axis is
determined by performing an algebraic summation of the
individual MoIs obtained in Step 3. (Please note that MoI of
a hole is subtracted)
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NOTE: Information about the centroids of the simple shapes can
be
obtained from the inside back cover of the book
Example
Ix = 𝐼x̅' + Ady2
IXa
= (1/36) (300) (200)3
+ (½) (300) (200) (67)2
= 201.3 (10)6
mm 4
IXb
= (1/12) 300(200)3
+ 300(200)(100)2
= 800 (10)6
mm 4
(a) (b) (c)
IXc
= (1/4) (75)4
+ (75)2
(100)2
= 201.5 (10)6
mm 4
IX = I
Xa + I
Xb – I
Xc = 799.8 (10)
6
mm 4
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[Sections 10.5 – 10.7]: Skipped
[Section 10.8]: Mass Moment of Inertia (MMI)
Consider a rigid body and the arbitrary axis p shown in the
figure. The
MMI about the p axis is defined as I = m
r2
dm, where r, the “moment
arm,” is the perpendicular distance from the axis to the
arbitrary element
dm. The MMI is always a positive quantity and has a unit of kg
·m2
or
slug · ft2
Parallel-Axis Theorem
Just as with the MoI for an area, the parallel-axis theorem can
be
used to find the MMI about a parallel axis z that is a distance
d
from the z’ axis through the body’s center of mass G. The
formula is:
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Iz = I
G
+ (m) (d)
2
(where m is the mass of the body)
The radius of gyration is similarly defined as:
k = (I / m)
The MMI can be obtained by integration or by the method for
composite
bodies. The latter method is easier for many practical
shapes
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Chapter 12: Motion of a Point
[Sections 12.1 and 12.2]: Rectalinear Kinematics: Continuous
Motion of points - position, velocity, acceleration,
Position:
A particle travels along a straight-line path defined by the
coordinate axis s
The position of the particle at any instant, relative to the
origin,
O, is defined by the position vector r, or the scalar s. Scalar
s can
be positive or negative. Typical units for r and s are meters
(m)
or feet (ft)
The displacement of the particle is defined as its change in
position
Vector form: r = r’ - r
Scalar form: s = s’ - s
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The total distance traveled by the particle, sT, is a positive
scalar
that represents the total length of the path over which the
particle travels
Velocity:
Velocity is a measure of the rate of change in the position of
a
particle. It is a vector quantity (it has both magnitude and
direction). The magnitude of the velocity is called speed,
with
units of m/s or ft/s
The average velocity of a particle during a time interval t
is
vavg = r / t
The instantaneous velocity is the time-derivative of
position
v = dr / dt
Speed is the magnitude of velocity: v = ds / dt
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Average speed is the total distance traveled divided by
elapsed
time:
(vsp)avg = sT / t
Acceleration:
Acceleration is the rate of change in the velocity of a
particle. It
is a vector quantity. Typical units are m/s2 or ft/s2
The instantaneous acceleration is the time derivative of
velocity
Vector form: a = dv / dt
Scalar form: a = dv / dt = d2s / dt2
Acceleration can be positive (speed increasing) or negative
(speed decreasing)
As the text indicates, the derivative equations for velocity
and
acceleration can be manipulated to get :
a ds = v dv
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SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION
Differentiate position to get velocity and acceleration
Integrate acceleration to get velocity and position
Note that so and vo represent the initial position and velocity
of
the particle at t = 0
v = ds/dt ; a = dv/dt or a = v dv/ds
Velocity:
= t
o
v
v o
dt a dv = s
s
v
v o o
ds a dv v or
= t
o
s
s o
dt v ds
Position:
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CONSTANT ACCELERATION
The three kinematic equations can be integrated for the
special
case when acceleration is constant (a = ac) to obtain very
useful
equations. A common example of constant acceleration is
gravity;
i.e., a body freely falling toward earth. In this case, ac = g =
9.81
m/s2 = 32.2 ft/s2 downward. These equations are:
t a
v v
c
o + = yields =
t
o
c
v
v
dt a dv o
2 c
o
o
s
t (1/2) a
t v
s
s + + = yields =
t
o s
dt v ds o
) s
-
(s 2a
) (v
v
o c
2
o 2
+ = yields =
s
s
c
v
v o o
ds a dv v
