MAE2103 - Engineering Mechanics I
Course Notes
Contents
Lecture 1: Introduction, units, linear algebra
0 Introduction
0.1
Overview . . . . . . . . . . . . . . . . . . . . . .
0.2
Units . . . . . . . . . . . . . . . . . . . . . . . .
1 Linear Algebra
1.1
Vectors . . . . . . . . . . . . . . . . . . . . . . .
1.2
Matrices . . . . . . . . . . . . . . . . . . . . . . .
Lecture 2: Vector products, matrix multiplication, linear systems
1.3
Dot product . . . . . . . . . . . . . . . . . . . . .
1.4
Cross product . . . . . . . . . . . . . . . . . . . .
1.5
Linear systems . . . . . . . . . . . . . . . . . . .
Lecture 3: Point equilibrium, 2D examples
2 Point Equilibrium
2.1
Free body diagrams . . . . . . . . . . . . . . . . .
2.2
Force bearing members . . . . . . . . . . . . . . .
2.3
Governing equations . . . . . . . . . . . . . . . .
2.4
Problem methodology . . . . . . . . . . . . . . .
2.5
2D examples . . . . . . . . . . . . . . . . . . . .
Lecture 4: 2D examples cont., 3D equilibrium, 3D examples
2.6
3D force-bearing members . . . . . . . . . . . . .
2.7
3D free body diagrams . . . . . . . . . . . . . . .
2.8
3D equilibrium equation . . . . . . . . . . . . . .
2.9
3D examples . . . . . . . . . . . . . . . . . . . .
Lecture 5: 3D examples cont., moment equilibrium
2.10 Static indeterminacy . . . . . . . . . . . . . . . .
3 Moments
3.1
Force-bearing members . . . . . . . . . . . . . . .
3.2
Moments . . . . . . . . . . . . . . . . . . . . . .
Lecture 6: Couple moments and distributed loads
3.3
Moment about a specific axis . . . . . . . . . . .
3.4
Couple moments . . . . . . . . . . . . . . . . . .
3.5
Reduction of forces and moments . . . . . . . . .
3.6
Distributed loads . . . . . . . . . . . . . . . . . .
Lecture 7: Rigid body equilibrium–2D
4 Rigid Body Equilibrium
4.1
Equilibrium equations . . . . . . . . . . . . . . . .
All content © 2016, Brandon Runnels
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. . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
1.1
1.1
1.1
1.2
1.2
1.7
2.1
2.1
2.5
2.8
3.1
3.1
3.1
3.2
3.3
3.4
3.4
4.1
4.4
4.4
4.5
4.5
5.1
5.2
5.3
5.3
5.4
6.1
6.1
6.2
6.4
6.5
7.1
7.2
7.2
T.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
4.2
2D Free-body diagrams . . . . . .
4.3
2D solution strategy . . . . . . .
4.4
2D examples . . . . . . . . . . .
Lecture 8: Rigid body equilibrium 2D & 3D
4.5
3D Free-body diagrams . . . . . .
4.6
3D solution strategy . . . . . . .
Lecture 9: 3D equilibrium examples, multi-body
4.7
3D examples . . . . . . . . . . .
4.8
Multi-body free-body diagrams . .
4.9
Multi-body equilibrium . . . . . .
4.10 Multi-body solution strategy . . .
Lecture 10: Equilibrium of multiple rigid bodies
4.11 Multi-body examples . . . . . . .
All content © 2016, Brandon Runnels
Course Notes - Table of Contents
solids.uccs.edu/teaching/mae2103
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equilibrium
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3
7.5
7.6
8.1
8.4
8.6
9.1
9.1
9.5
9.7
9.7
10.1
10.1
T.2
Lecture 1
0
Introduction, units, linear algebra
Introduction
Welcome to Engineering Mechanics I. This class is usually referred to as “Statics,” but we’ll be covering some extra
material that typically falls into the category of “Dynamics.” For the majority of this class, we will be looking at
mechanical systems that do not move, or are in “static equilibrium.”
0.1
Overview
The majority of the course (15 weeks) will be spent on the Statics portion of the class. The governing equations of
statics are:
X
X
F=0
M=0
(0.1)
where F are the force vectors and M are the moment vectors. In other words, “the sum of the forces and moments are
equal to zero.”
For dynamics, the governing equations are similar, except that we have time dependence. The governing equations
become:
X
F=
dP
dt
X
M=
dL
dt
(0.2)
where P is the momentum vector and L is the angular momentum vectors.
0.2
Units
Let’s review some of the basic units that we will use in this course:
Base units
Length
Time
Mass
Temperature
Derived units
Area,volume
Velocity, acceleration
Force
Pressure
Energy
Power
Metric
m
s
kg
◦
C, K
Metric
USCS
ft
s
slug
◦ ◦
F, R
USCS
m2 , m3
m/s
N = kgm
s2
Pa = mN2
J = Nm
W = Js
ft 2 , ft 3
ft/s, ft/s 2
lb = slugft
s2
psi = inlb2
ft lb
ft lb/s
Let’s make some notes about units. Specifically, note that units:
ˆ can be multiplied, divided, cancelled, even square-rooted
ˆ cannot be added or subtracted
Why do we care about units? Not only do they connect numbers to physically meaninful quantities, but they are a great
way to check your answer. If you find that you are adding different units together, or that the units of your answer are
wrong, you’ve probably made a mistake somewhere.
All content © 2016, Brandon Runnels
1.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
1
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
Linear Algebra
Linear algebra is a deep and elegant branch of mathematics that has a wide variety of applications. In mechanics, linear
algebra is a framework for describing quantities in a systematic way that makes analysis easy. In this section: introduce
only the linear algebra that will be useful in this course.
1.1
Vectors
Before defining vectors, we need to define scalar quantities:
Def 1.1. A scalar quantity is specified by its magnitude only
What are some examples of scalar quantites?
ˆ mass (m)
ˆ time (t)
ˆ volume (V )
ˆ speed (s)
What is a vector?
Def 1.2. A vector quantity is specified by both a magnitude and a direction. A vector quantity is expressed as v, v , v , ~v
∼
L
θ
What are some examples of vector quantities?
ˆ position (x)
ˆ velocity (v)
ˆ force (f)
Notice how I drew the angle with respect to a dotted line. Why draw the dotted line like that, instead of straight down,
or up? I implicitly chose a coordinate system.
1.1.1
Coordinate system
Def 1.3. A coordinate system is a convention for measuring locations in space.
Examples:
All content © 2016, Brandon Runnels
1.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
y
z
z0
L
y0
y0
θ
x0
y
x0
x
x
2D Cartesian coordinate system
Polar coordinate system
3D Cartesian system
Notes:
ˆ There are 3D polar coordinate systems too (spherical and cylindrical) but they are ugly and horrible and not too
useful right now. In general, we can get along pretty well by simply using trigonometry.
ˆ The number of dimensions must always match the number of coordinate variables.
ˆ When working in 3D, we use a right-handed system.
z
RH
z
x
LH
y
x
LH
x
z
y
x
RH
y
y
y
RH
z
x
z
It’s really important to get comfortable with this because it will be used a lot when we get to moments.
1.1.2
Component representation
There are two ways of representing vectors in component representation:
ˆ matrix form
ˆ unit vector form
Example 1.1
All content © 2016, Brandon Runnels
1.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
Write the following figure in component notation
4m
x
2m
2m
First: we need to define a coordinate system. Let the upward pointing axis be z. Then the vector is
" #
2m
x = 2m
4m
1.1.3
(1.1)
Converting to/from angular representation
Because of the way problems are specified, it is often necessary to convert to and from magnitude-angle notation to
coordinate notation (and back)
Example 1.2
Convert the vector to rectangular coordinates:
L=
8m
√
y = (8m) sin(60◦ ) = 4 3m
θ = 60◦
x = (8m) cos(60◦ ) = 4m
The x component is L cos θ and the y component is L sin θ so the resultant vector is
4m
√
x=
4 3m
(1.2)
We will have to do this for 3D vectors as well:
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1.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
Example 1.3
Find the component representation of the following vector:
z
L = 1m
φ = 60◦
y
x
θ = 45◦
We can find the z component directly:
z = (1m) sin(60◦ ) =
√
3/2m
(1.3)
It is more tricky to find the x and y magnitudes. First we have to find the magnitude of the “projected” vector,
which is (1m) cos(60◦ ) = 1/2m. Now, we can compute the x and y magnitudes:
1
x = 1/2m cos(45◦ ) = √ m
2 2
So the component representation is
1.1.4
1
y = 1/2m cos(45◦ ) = √ m
2 2

√ 
1/2√2m

x = 1/2
√ 2m
3/2m
(1.4)
(1.5)
Vector algebra
The nice thing about vectors (especially in the component notation) is that we can add, subtract, and scale them easily.
ˆ Vector addition:
h i
a
a = ax
y
h i
b
b = bx
y
(1.6)
then I can add them together as
h
i
a +b
c = a + b = ax + bx
y
y
(1.7)
Pictorally:
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1.5
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
a
b
a
b
a
c
=
a
+
b
b
ˆ Vector subtraction:
h
i
a −b
c = a − b = ax − bx
y
y
(1.8)
c=a−b
a
b
Note that in the picture, it is clear that a = b + c.
ˆ Scalar multiplication:
h
i
αa
αa = αax
y
(1.9)
Pictorally, mulitplying by a scalar simply changes the magnitude of the vector by the value of that scalar.
ˆ What about vector multiplication?
1.1.5
Unit vectors
Since we can add, subtract, and multiply vectors, we can define the following:
" #
" #
1
0
î = 0
ĵ = 1
0
0
" #
0
k̂ = 0
1
(1.10)
Then we can write the vector in the above example as
x = (2m)i + (2m)j + (2m)k
(1.11)
I prefer to avoid using unit vectors (so I probably won’t use them much in lecture) but there’s nothing wrong with them
and you should feel free to use them if you like.
All content © 2016, Brandon Runnels
1.6
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
1.2
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
Matrices
Matrices (and vectors) are useful tools for organizing the equations that we have to solve.
Def 1.4. A matrix is a set of numbers organized in a grid
(It’s a simple definition but it will do for now.)
Examples of matrices:
"
#
h
i
1 2 3
1 2
4 5 6
3 4
7 8 9
| {z }
| {z }
2x2
h
|
3x3
9
0
3
3
2
1
1
2
{z
i
(1.12)
}
2x4
Notes:
ˆ [# rows] x [# columns]
ˆ nx1: n-D vectors
ˆ nxn: square
ˆ Notation: A
ˆ Addition/subtraction/scalar multiplication works if same size
1.2.1
Matrix matrix multiplication
Matrices A and B can be multiplied if # colums of A
a
a12 ...
11
a21 a22 ...
..
 ..
..
.
.
.
ap1 ap2 ...
{z
|
pxq
The result is:
P
Pi a1i bi1
 i a2i bi1

..

P .
api bi1
| i
= # rows of B

a1q  b11 b12 ...
a2q  b21 b22 ...
..  
.
..
..
.
.  ..
.
apq
bq1 bq2 ...
}|
{z
qxr
P
Pi a1i bi2
i a2i bi2
..
P .
i api bi2
{z
...
...
..
.
...
pxr

b1r
b2r 
.. 
. 
bqr
}
(1.13)
P

Pi a1i bir
i a2i bir 

..

P .
a
b
i pi ir
}
(1.14)
a matrix with the number of rows of A and number of columns of B.
Example 1.4
"
# h
i
0
1 3 2 2
3 = (1)(2) + (3)(1) + (2)(1)
0 5 9 1
(0)(2) + (5)(1) + (9)(1)
1
4
| {z }
|
{z
}
2x3
h
i
(1)(0) + (3)(3) + (2)(4)
7 17
=
14 51
(0)(0) + (3)(5) + (9)(4)
| {z }
(1.15)
2x2
3x2
Example 1.5
"
#" # "
# " #
(3)(0) + (1)(2) + (4)(5)
3 1 4 0
22
8 1 3 2 = (8)(0) + (1)(2) + (3)(5) = 17
1 4 0 5
8
(1)(0) + (4)(2) + (0)(5)
| {z } |{z}
| {z }
3x3
All content © 2016, Brandon Runnels
3x1
(1.16)
3x1
1.7
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 1
solids.uccs.edu/teaching/mae2103
Notice that the matrix “ate” the vector and turned it into another vector. This is actually a very nice way of looking
at matrices: we can think of them as machines that turn a vector into another vector.
All content © 2016, Brandon Runnels
1.8
Lecture 2
1.2.2
Vector products, matrix multiplication, linear systems
Determinants
You may have (hopefully) seen the determinant of a matrix before. It may seem like a rather odd quantity, but its
usefulness will become apparent as we use it in the course. In particular, we will use it a lot when computing cross
products and when solving linear systems. A rigorous definition of the determinant is beyond the scope of this course,
so instead we will introduce it by example.
Example 1.6
For a 2x2 matrix:
h
1
det 3
i 1
2
4 = 3
2
4 = (1)(4) − (2)(3) = 4 − 6 = −2
(1.17)
Example 1.7
For a 3x3 matrix: two
"
1
det 4
0
ways of computing
#
2 3
5 2 = (1)((5)(1) − (2)(3)) − (2)((4)(1) − (2)(0)) + (3)((4)(3) − (5)(0))
3 1
= (1)(−1) − (2)(4) + (3)(12) = 27
(1.18)
(1.19)
Alternatively, we can compute the diagonals
(1)(5)(1) + (2)(2)(0) + (3)(4)(3) − (1)(2)(3) − (3)(5)(0) − (2)(4)(1) = 27
(1.20)
Let’s make a few notes about the properties of the determinant. Again, these properties will come in handy later on.
ˆ If the vectors that make up two of the columns of the matrix are parallel, then the determinant will be zero.
ˆ If the third vector is equal to a combination of the first two, the determinant will be zero.
ˆ If A is a n × n matrix and α is a scalar, then det(αA) = αn det(A)
1.3
Dot product
Suppose we have two vectors:
" #
1
a= 2
3
How can we multiply them together?
" #" #
1 4
2 5
3 6
| {z }
bad
" #
4
b= 5
6
[1
|
" #
4
2 3] 5
6
{z
}
(1.22)
good!
This is called the dot (or inner) product, and it gives a scalar value:
" #
4
[1 2 3] 5 = (1)(4) + (2)(5) + (3)(6) = 32
6
All content © 2016, Brandon Runnels
(1.21)
(1.23)
2.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
This operation is called the dot product and is denoted by
1.3.1
a·b
(1.24)
a · b = (magnitude of a)(magnitude of b) cos(angle between a and b)
(1.25)
Angular representation
In polar form the dot product is computed simply as
Notes:
ˆ What if the angle between a and b is 90◦ ? Then a · b is zero.
ˆ What is a · b = 0? Then cos θ = 0. So a · b = 0 if and only if they are perpendicular (orthogonal) to each other.
1.3.2
Vector norm
We know how to convert from angular to component representation; how do we convert from component to polar?
3
2
√ 32 +
4
=
5
tan−1 (3/4) = 19.89◦
4
If we have a vector
" #
vx
v = vy
vz
then the magnitude is
q
vx2 + vy2 + vz2 =
√
v · v ≡ ||v||
(1.26)
(1.27)
This is called the vector norm. Some properties of the norm are:
ˆ ||v|| ≥ 0
ˆ ||v|| = 0 if and only if v = 0
ˆ ||αv|| = |α|||v||
ˆ ||u + v|| ≤ ||u|| + ||v|| (the triangle inequality)
With this terminology, we can write this very useful equation:
u · v = ||u||||v|| cos θ
All content © 2016, Brandon Runnels
(1.28)
2.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
1.3.3
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
Unit vectors
Whenver we have a nonzero vector, we can divide by its magnitude to get a unit vector. Given v as we had earlier, then
we define
" #
1 vx
vy
(1.29)
v̂ =
||v|| vz
Example 1.8
Compute the unit vector corresponding to the position vector
" #
1m
x = 2m
3m
The magnitude is
||x|| =
(Note that the units are in meters.)
Now we divide:
p
1m2 + 4m2 + 9m2 =
(1.30)
√
14m2 =
√
14m
(1.31)
 √ 
1/√14
x̂ = 2/√14
3/ 14
(1.32)
Note that the unit vector is (ironically) unitless. This should always be true of unit vectors.
Unit vectors are very nice because they allow us to decouple the magnitude of a vector from its direction.
Example 1.9
Consider a cable that is attached to a point on a wall as shown.
z
n
5m
y
f =tn
x
3m
4m
The cable is pulled so that it has a tension of t. Compute the force vector acting by the cable on the wall.
We will do this sort of thing a lot in this class! The thing to in this situation is recognize that we can write the
force vector in this way
f =tn
(1.33)
where t = ||f||, the magnitude of the force. We already know t – it was given to us. Now, we need to compute n.
How do we do this? We know that the force will be in the same direction of the cable. So, we can find the unit
vector that is parallel to the cable, and it will serve as a unit vector for the force too. The vector going along the
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2.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
cable is
"
# " # "
#
3m
0
4m
d = 4m − 0 = 3m
0
5m
−5m
(Remember to be careful about the signs!) Now, we just need to normalize d:
p
p
√
||d|| = (3m)2 + (4m)2 + (5m)2 = 25m2 + 25m2 = 5 2m
so our unit vector is
(1.35)

√ 
4/5√2
d
n=
=  3/5 √2 
||d||
−1/ 2
(1.36)

√ 
4t/5√2
f = t n = 3t/5√ 2
−t/ 2
(1.37)
and our force vector is
1.3.4
(1.34)
Projection
The dot product allows us to compute the projection of one vector onto another. It determines the “amount” of the
vector that is in the direction of the other. For example:
u
v
θ
|| c
||u
os
θ
The projection is ||u|| cos θ. We can write this in terms of the dot product:
||u|| cos θ =
||u||||v|| cos θ
u·v
=
= u · v̂
||v||
||v||
(1.38)
Example 1.10
Consider a train moving along a track, subjected to a constant force:
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2.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
h i
1m
xf = 1m
y
x
h i
0m
xi = 0m
h i
0N
f = 1N
What is the effective force that is acting in the direction of the motion of the car?
h i
1m
∆x = xf − xi = 1m
(1.39)
The projection is:
1
(1m)(0N) + (1m)(1N)
∆x · f
√
=√ N
=
||∆x||
2m
2
(1.40)
Note that the unit is Newtons, which is what we expect.
1.4
Cross product
We will also make extensive use of the vector cross product, especially when we start talking about moments. Here, we
will just define it and do an example.
1.4.1
Component representation
Suppose we have two vectors
" #
ux
u = uy
uz
" #
vx
v = vy
vz
(1.41)
The cross product between them is
"
uy vz − uz vy
u × v ≡ uz vx − ux vz
ux vy − uy vx
Fortunately, there’s a nice mnemonic
"
î
u × v = det ux
vx
#
(1.42)
for remembering how to compute the cross product using the î, ĵ, k̂ unit vectors:;
#
ĵ
k̂
(1.43)
uy uz = î (uy vz − uz vy ) + ĵ (uz vx − ux vz ) + k̂ (ux vy − uy vx )
vy vz
(1.44)
Pictorally:
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MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
u×v
v
u
v×u
Notes:
ˆ u × v is orthogonal to both u and v (you’ll show this in your homework)
ˆ How do we know which way it points? Right hand rule.
ˆ And v × u = −u × v
Example 1.11
Compute the cross product between
" #
1
a= 2
3
" #
2
b= 4
6
(1.45)
What do you think the cross product will be? It should be zero because b = 2a which means they point in the same
direction.
Let’s find out:
"
#
î ĵ k̂
a × b = 1 2 3 = î (12 − 12) + ĵ (6 − 6) + k̂ (4 − 4) = 0
(1.46)
2 4 6
exactly as we expected!
1.4.2
Scalar cross product
Unlike the dot product, the cross product does not generalize naturally to 2D (or any other dimension, for that matter).
When working with 2D vectors, we can write them in 3D:
" #
" #
h i
h i
ux
vx
ux
v
u = u → uy
v = v x → vy
(1.47)
y
y
0
0
Then their cross product is
"
î
u × v = ux
vx
ĵ
uy
vy
#
k̂
0 = k (ux vy − uy vx )
0
2D cross products have a k̂ component only, so just keep track of the magnitude:
h
i
u u
||u × v|| = vx vy
x
y
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(1.48)
(1.49)
2.6
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
1.4.3
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
Angular representation
This is where component representation really shines, because there’s just no easy way to compute the cross product in
angular representation. However, we do have the following important identity:
||u × v|| = ||u|| ||v|| sin θ
(1.50)
Notes:
ˆ When is the magnitude of the cross product maximized? When u and v are orthogonal (perpendicular) to each
other.
ˆ When is the magnitude of the cross product minimized? When u and v are parallel to each other. (compare to
dot product!)
Example 1.12
Given
a = [6
6
1]
b = [1
5
3]
(1.51)
So
"
î
a×b= 6
1
ĵ
6
5
# "
# "
#
18 − 5
13
k̂
1 = 1 − 18 = −17
30 − 6
24
3
(1.52)
Example 1.13
Compute the cross product between î and ĵ.
What do you think it will be?
" #
1
î = 0
0
" #
0
ĵ = 1
0
"
î
î × ĵ = 1
0
1.4.4
ĵ
0
1
(1.53)
#
k̂
0 = î (0) + ĵ (0) + k̂ (1) = k̂
0
(1.54)
Cross product with unit vectors
In the last example we showed that î × ĵ = k̂. We can also show that
î × î = 0̂
î × ĵ = k̂
î × k̂ = −ĵ
(1.55)
ĵ × î = −k̂
ĵ × ĵ = 0̂
ĵ × k̂ = î
(1.56)
k̂ × î = ĵ
k̂ × ĵ = −î
k̂ × k̂ = 0̂
(1.57)
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2.7
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
Let a = ax î + ay ĵ + az k̂ and b = bx î + by ĵ + bz k̂. Then we can compute the cross product by multiplying these directly:
a × b =(ax î + ay ĵ + az k̂) × (bx î + by ĵ + bz k̂)
ax bx (î × î) + ax by (î × ĵ) +ax bz (î × k̂) +
=
| {z }
| {z }
(1.58)
(1.59)
−ĵ
k̂
+ay bx (ĵ × î) +
ay by (ĵ ×
ĵ) + ay bz (ĵ × k̂) +
| {z }
| {z }
−k̂
(1.60)
î
+az bx (k̂ × î) +az by (k̂ × ĵ) +
ak bz (k̂ × k̂)
| {z }
| {z }
(1.61)
−î
ĵ
Collecting terms we get
a × b =(ay bz − az by ) î + (az bx − ax bz ) ĵ + (ax by − ay bx ) k̂
1.5
(1.62)
Linear systems
Something that we will have to do regularly is solve systems of linear equations. For example:
2x + 3y + z = 4
(1.63)
4x + 5y + z = 2
(1.64)
4y + 3z = 2
(1.65)
How do we solve this system?
How can we write this set of equations using matrices and vectors?
"
#" # " #
2 3 1 x
4
4 5 1 y = 2
0 4 3 z
2
| {z } |{z} |{z}
x
A
(1.66)
b
where A is called the coefficient matrix, and x is the “vector of unknowns.”
1.5.1
Cramer’s rule
Cramer’s rule is an easy way to solve linear systems of equations without too much work. Suppose we have this coefficient
matrix (note: this introduces some slightly new notation)
"
#
px qx rx
py qy ry ≡ [p q r]
(1.67)
pZ qz rz
so that we have the linear system
[p
q
r] x = b
(1.68)
Then the solution is given by
x1 =
det [b q r]
det [p q r]
x2 =
det [p b r]
det [p q r]
x3 =
det [p q b]
det [p q r]
(1.69)
(Notice that this can be generalized to n-dimensional systems. But in this class we’ll pretty much stick to 2 and 3.)
Example 1.14
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2.8
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 2
solids.uccs.edu/teaching/mae2103
Let’s solve the system we had before:
"
2
4
0
#" # " #
1 x
4
1 y = 2
3 z
2
3
5
4
(1.70)
To save some work:
det(A) = (2)(3)(5) + (3)(1)(0) + (1)(4)(4) − (2)(1)(4) − (3)(4)(3)(1)(5)(0) = 30 + 16 − 8 − 36 = 2
Now we only have to compute three more determinants:
"
#
4 3 1
det 2 5 1 = 60 + 6 + 8 − 16 − 18 − 10 = 30
2 4 3
"
#
2 4 1
det 4 2 1 = 12 + 0 + 8 − 4 − 48 − 0 = −32
0 2 3
"
#
2 3 4
det 4 5 2 = 20 + 0 + 64 − 16 − 24 − 0 = 44
0 4 2
So our solution is
(1.71)
(1.72)
(1.73)
(1.74)
" # "
#
15
x
y = −16
z
22
(1.75)
Notes:
ˆ What happens when the determinant of the coefficient matrix is zero? The solution vector is infinite
⇒ A solution exists for the system if and only if the determinant of the coefficient matrix is nonzero.
– Always a good idea to check the determinant first. It will be useful in solving the system, and it may help
you to find errors in your work.
ˆ What if the coefficient matrix is not square? You can only solve the system with square matrices.
⇒ Number of solutions must equal number of unknowns.
1.5.2
Proof of Cramer’s rule
One of the nice things about Cramer’s rule is that it’s very easy to prove. Suppose we have the 3D system
" #
x
det [p q r] y = p x + q y + r z = b
z
(1.76)
Let us consider the following determinant:
det [(p x + q y + r z) q
r]
which is equal to
det [b
q
r]
(1.77)
We know by the linearity of the determinant that we can write
det [(p x + q y + r z) q
r] = x det [p
q
r] + y det [q
q
r] + z det [r
q
r]
(1.78)
The latter two terms have a repeated column, so they must be zero. So we have
x det [p
q
x=
r] = det [b
det [b q r]
det [p q r]
q
r]
(1.79)
(1.80)
and so on for y and z.
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2.9
Lecture 3
2
Point equilibrium, 2D examples
Point Equilibrium
2.1
Free body diagrams
Def 2.1. A free body diagram (FBD) for an object is a simple diagram of an object’s geometry, constraints, forces,
and moments
2.1.1
For a point
Consider the following complex system:
fspring = tspring nspring
fcable = tcable ncable
θspring
θcable
θweight
fweight
For this problem, the FBD is the simplest figure possible that contains all the information we need to know to solve the
system.
Notes:
ˆ Drawing a FBD establishes a convention for your problem. (e.g. assume a member is in tension.)
ˆ More than one FBD may be necessary for a complex system
2.1.2
Diagram cutting
It is frequently useful to “cut” a diagram. It is easy to get confused about how to draw the forces. Consider the following
very simple system:
We wish to draw FBDs for each, so we “cut” the diagram in two. To account for the effect of the “other half” we must
add a force acting at the boundary.
Important point: the sum of the forces at a cut from both sides must equal zero.
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3.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
2.2
2.2.1
Course Notes - Lecture 3
solids.uccs.edu/teaching/mae2103
Force bearing members
Pivots
Pivots denoted like this:
Attributes:
ˆ constrained position in x,y – sustained force in x,y
ˆ free rotation
2.2.2
Rollers
Rollers denoted like this:
Attributes:
ˆ constrained position parallel to surface – sustained force parallel to surface
ˆ free position along surface – no force along surface
ˆ free rotation
2.2.3
Two force members (2FM)
Consider the following system:
two force member has two forces acting at each end
arrow pointing this way from beam’s perspective
but pointing the opposite direction from the pivot’s perspective
Notes:
ˆ 2FM cannot have forces acting midway, only at ends
ˆ forces are co-linear, equal, and opposite
ˆ forces are parallel with member (or else member would spin)
ˆ use convention of tension = positive. always make sure to be consistent!
⇒ if beam ends up being in compression, the tension will be negative.
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3.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
2.2.4
Course Notes - Lecture 3
solids.uccs.edu/teaching/mae2103
Cables
Cables have the following properties:
ˆ as 2FM (tension only)
ˆ weightless
ˆ constant tension
2.2.5
Pulleys
Pulleys look like this. (Draw a free body diagram)
freaction
fcable1
t
θ/2
θ/2
t
fcable2
Forces acting on pulleys:
ˆ cables (tension is constant)
ˆ reaction force from fixed mount
ˆ external forces
2.2.6
Springs
(Linear) springs are 2FMs with variable length
Li
k
Lf
ˆ Tension on a spring is given by
t = k∆L = k(Lf − Li )
(2.1)
ˆ Spring constant: units N/m
ˆ Tension or compression (but usually tension unless indicated by a guide)
2.3
2.3.1
Governing equations
Equilibrium
For point equilibrium:
X
f=0
(2.2)
(remember: this gives you 2 equations in 2D and 3 equations in 3D)
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3.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
2.3.2
Course Notes - Lecture 3
solids.uccs.edu/teaching/mae2103
Geometric
Equations that describe constrained geometry: e.g.
ˆ fixed cable length
ˆ roller on specified track
2.3.3
Constitutive
Equations that describe a member’s response to a loading: e.g.
ˆ spring stiffness equation
ˆ noncompressive forces on a 2-force member
ˆ maximum allowable tension
2.4
Problem methodology
1. Draw free body diagram
2. Identify unknown quantities
3. Write down equilibrium, geometric, and constitutive equations
4. Solve for unknowns
5. Test the solution (“sanity check”)
6. Substitute numerical values
2.5
2D examples
Example 2.1
Consider the following cable-pulley-spring system.
ˆ length of the cable is `
ˆ unstretched length of the spring is h
Find the equilibrium position of the pulley and the tension in the cable.
h
k
y
x
L
w
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3.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 3
solids.uccs.edu/teaching/mae2103
1. Free body diagram:
h i
0
fspring = tspring 1
fcable1 = tcable
√
1
x 2 +y 2
h
−x
y
i
fcable2 = tcable
√
1
(L−x)2 +y 2
h
L−x
y
i
h
i
0
w = −w
2. Unknown quantites: x, y , tcable
3. Equations:
(a) Equilibrium
fcable1 + fcable2 + fspring + w = 0
(2.3)
(b) Geometric
`1 + `2 =
p
x2 + y2 +
p
(L − x)2 + y 2 = `
(2.4)
(c) Constitutive
tspring = k((y + h) − h) = ky
(2.5)
4. Solve for unknowns:
x component of equilibrium:
tcable
t p cable (−x) + p
(L − x) + 0 + 0 = 0
2
2
(L − x)2 + y 2
x +y
x
L−x
p
=p
(L − x)2 + y 2
x2 + y2
(2.6)
(2.7)
guess: does x = L/2 work?
L/2
(L/2)
L
p
=p
→ yes! → x =
2
(L/2)2 + y 2
(L/2)2 + y 2
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(2.8)
3.5
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 3
solids.uccs.edu/teaching/mae2103
now, solve for y → use geometry equation
p
p
(L/2)2 + y 2 + (L/2)2 + y 2 = `
r
L2
+ y2 = `
2
4
L2
l2
+ y2 =
4
4
l 2 − `2
2
y =
4
√
2
2
` −L
y=
→ units?X
2
(2.9)
(2.10)
(2.11)
(2.12)
(2.13)
(2.14)
what do we want now? we can get tspring easily, so let’s do it!
√
k `2 − L2
tspring = ky =
2
what equations have we used? all but the y component of equilibrium.
what unknowns do we still need? just one: the tension of the cables.
√
X
tcable y
tcable y
k `2 − L2
Fy = p
+p
+
−w =0
2
L2 /4 + y 2
L2 /4 + y 2
{z
} |
{z
}
|
=`/2
(2.15)
(2.16)
`/2
√
2tcable y
k ` 2 − L2
+
=w
`/2
2
√
√
2tcable `2 − L k `2 − L2
+
=w
`
2
`
k
w
√
tcable =
−
→ units?X
2
`2 − L 2
5. Test solution
x=
ˆ intiuition? X
√
y=
L
2
(2.17)
(2.18)
(2.19)
(2.20)
`2 − L2
2
(2.21)
ˆ what if ` < L? is this physically possible?
tcable =
`
2
√
k
w
−
`2 − L 2
(2.22)
ˆ what if ` = L? does it make sense for the tension to be infinite?
ˆ when might tcable is negative? would our solution still be valid? does this make sense?
6. Now, (if provided), we plug in numbers.
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3.6
Lecture 4
2D examples cont., 3D equilibrium, 3D examples
Example 2.2
A mass of weight w was attached to the system as shown:
L
θ
θ
`
2
`
2
k
k
w
Before the weight was attached, both springs were pre-stretched by a distance y . Given values for k, L, y , and θ,
what is the value of w ?
1. Free body diagram:
The unit vectors for the two cables can be found easily:
− cos(θ)
f1 = t
sin(θ)
cos(θ)
f2 = t
sin(θ)
h
i
0
w = −w
2. Identify unknown quantity: w
3. Governing equations
(a) Equilibrium
f1 + f2 + w = 0
(2.23)
L
`
L
= cos θ =⇒ ` =
2
2
cos θ
(2.24)
(b) Geometric
(c) Constitutive: If not prestretched:
`
L
t = k∆x = k
−
2 2
(2.25)
But this was prestretched, so we have
`
L
t=k y+ −
2 2
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(2.26)
4.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 4
solids.uccs.edu/teaching/mae2103
4. Solve for unknowns: use the y component of the equilibrium equation
t sin(θ) + t sin(θ) − w = 0
L
w = k 2y + ` − L sin(θ) = k 2y +
− L sin(θ)
cos θ
(2.27)
(2.28)
5. Sanity checks:
(a) units? X
(b) what if θ = 0? then w = 0, which makes sense.
6. Now plug in values:
k = 30
So
N
m
θ = 30◦
L = 4m
w = (30N/m) 2(1m) +
y = 1m
4m
− 4m sin(30◦ ) = 39.28N
cos 30◦
(2.29)
(2.30)
Example 2.3
Consider the following system:
4
θ1
3
2
θ2
θ1
1
w
Compute the tension in each cable in terms of w , θ1 , θ2
If the cables break under a tension of 100N, find the maximum load allowable if θ1 = 30.00◦ , θ2 = 51.15◦
Before drawing the FBD, let’s compute the unit vectors:
h
i
h
i
cos θ
− cos θ
n1 = sin θ 1
n2 = ± sin θ 2
1
2
n3 = ±
h i
−1
0
h
i
cos θ
n4 = sin θ 1
1
(2.31)
Do they have magnitude of unity? Yes, sin/cos unit vectors in 2d always will.
1. Free body diagram:
We need two of them
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4.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 4
solids.uccs.edu/teaching/mae2103
h
i
cos θ
f4 = t4 sin θ 1
1
f3 = t3
h
−1
0
f2 = t2
i
h
− cos θ2
sin θ2
i
h
i
cos θ
f1 = t1 sin θ 1
1
h
i
0
w1 = −w
h
i
cos θ
f2 = t2 − sin θ2
2
2. Identify unknowns: t1 , t2 , t3 , t4
3. Write governing equations:
ˆ The equilibrium equations are:
h
i
h
i h
i
cos θ
− cos θ
0
F = f1 + f2 + w = t1 sin θ 1 + t2 sin θ 2 + −w = 0
1
2
h
i
h i
h
i
X
cos θ2
−1
cos θ
F = f2 + f3 + f4 = t2 − sin θ + t3 0 + t4 sin θ 1 = 0
2
1
X
(2.32)
(2.33)
Good news! We have four equations and four unknowns, so now we can solve the problem.
ˆ Constitutive: ti ≤ 100N
4. Solve for unknowns: start with the first equilibrium equation, write in matrix notation
h
ih i h i
cos θ1 − cos θ2 t1
0
t2 = w
sin θ1
sin θ2
(2.34)
Check the determinant:
h
cos θ
det sin θ 1
1
i
− cos θ2
sin θ2 = cos θ1 sin θ2 + cos θ2 sin θ1 = sin(θ1 + θ2 )
(2.35)
When would this be equal to zero? For instance, when θ1 = θ2 = 0, θ1 = θ2 = 90◦ . For now, we’ll assume
that those cases don’t happen and that the system is solvable.
h
i
w cos θ2
1
0 − cos θ2
det w
sin θ2 = sin(θ1 + θ2 )
sin(θ1 + θ2 )
i
h
w cos θ1
1
cos θ
0
det sin θ 1 w =
t2 =
1
sin(θ1 + θ2 )
sin(θ1 + θ2 )
t1 =
(2.36)
(2.37)
Great, we’ve solved for t1 and t2 . Now we need the other two, so let’s use the second equilibrium equation to
solve in terms of t2
h i
h
i
h
i
−1
cos θ
− cos θ
t3 0 + t4 sin θ 1 = t2 sin θ 2
(2.38)
1
2
h
ih i h
i
−1 cos θ1 t3
−t2 cos θ2
(2.39)
0 sin θ1 t4 = t2 sin θ2
Check the determinant:
det
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h
−1
0
i
cos θ1
sin θ1 = − sin θ1
(2.40)
4.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 4
solids.uccs.edu/teaching/mae2103
Does this make sense? Yes, because the problem would be poorly defined if the third and fourth cables are
parallel. Assume that this is not the case, and solve for t3 , t4
h
i t (− cos θ sin θ − cos θ sin θ )
t2 sin(θ1 + θ2 )
1
2
2
1
1
2
−t cos θ cos θ
det t 2sin θ 2 sin θ 1 =
=
2
2
1
− sin θ1
− sin θ1
sin θ1
h
i −t sin θ
1
t
sin
θ
2
2
2
2
−1 −t2 cos θ1
t4 =
det 0
t2 sin θ2 = − sin θ1 = sin θ1
− sin θ1
t3 =
(2.41)
(2.42)
Substituting for t2 gives the final result:
t1 =
w cos θ2
sin(θ1 + θ2 )
t2 =
w cos θ1
sin(θ1 + θ2 )
t3 =
w
tan θ1
t4 =
w sin θ2
sin(θ1 + θ2 ) tan θ1
(2.43)
t4 = 1.396w
(2.44)
5. Sanity check:
ˆ units? X
ˆ θ1 = θ2 = 0? makes sense–that problem would be poorly defined.
6. Plug in values: θ1 = 30.00◦ , θ2 = 53.15◦ : then
t1 = 0.6040w
t2 = 0.8723w
t3 = 1.732w
Apparently, for this case, cable 3 carries the majority of the weight, so it will fail first. Substituting t3 = 100N
and solving for w gives
100N
= 57.74N
(2.45)
w=
1.732
2.6
2.6.1
3D force-bearing members
2D members
All of the 3d force-bearing members carry over (2FM, pulleys, etc.)
2.6.2
3D pivot
The 3d pivot is the 3d analog of the 2d pivot:
ˆ Restricted motion in x, y , z, sustained force
ˆ Free rotation in x, y , z
2.7
3D free body diagrams
ˆ especially important to define coordinate system!
ˆ lots of boxes can be helpful
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4.4
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2.8
Course Notes - Lecture 4
solids.uccs.edu/teaching/mae2103
3D equilibrium equation
X
In 3D this gives us 3 equations:
X
fx = 0
2.9
X
f=0
(2.46)
X
fy = 0
fy = 0
(2.47)
3D examples
Example 2.4
A box of weight w is suspended using three cables as shown.
3m
2m
3m
2m
6m
w
Determine the tensions in the cables in terms of w . Which cable carries the most weight?
1. Free body diagram:
It is very important to define a coordinate system. Then the free body diagram is:
z
" #
1
f1 = t1 0
0
x
#
#
"
−6/7
−6
−2 = t2 −2/7
f2 =
3
" #
"3/7 #
−6/7
−6
3
3 = t3 3/7
f3 = √t49
2
2/7
" y#
0
w= 0
−w
"
√t2
49
2. Identify unknowns:
We want to solve for t1 , t2 , t3 (3 unknowns)
3. Write equations:
X
F = 0 (3 equations)
(2.48)
Great! We are all set, all we need to do is solve equilibrium.
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4.5
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 4
solids.uccs.edu/teaching/mae2103
4. Solve for unknowns
f1 + f2 + f3 + w = 0
" #
"
#
"
# " #
−6/7
−6/7
1
0
t1 0 + t2 −2/7 + t3 3/7 = 0
w
0
3/7
2/7
"
#" # " #
7 −6 −6 t1
0
t2 = 0
0 −2 3
t3
0 3
2
7w
As usual, compute the determinant to make sure we can solve the thing:
"
#
7 −6 −6
det 0 −2 3 = 7(−4 − 9) = −91
0 3
2
so it’s solvable. Now we can easily compute the coefficients:
"
#
0 −6 −6
1
210
30
1
w=
w
t1 = − det 0 −2 3 = − (7w (−18 − 12)) =
91
91
91
13
7w 3
2
"
#
7 0 −6
1
1
21
3 = − (−147w ) =
t2 = − det 0 0
w
91
91
13
0 7w 2
"
#
7 −6 −6
1
1
14
t3 = − det 0 −2 3 = − (−98w ) =
w
91
91
13
0 0 7w
(2.49)
(2.50)
(2.51)
(2.52)
(2.53)
(2.54)
(2.55)
5. Sanity check
ˆ units? X
ˆ negative value? negative weight (pushing up) X
(the greatest tension is in cable 1).
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4.6
Lecture 5
3D examples cont., moment equilibrium
Example 2.5
Consider the following cable-strut system:
L
L
W
H
y
w
Compute the tension (or compression) in the cable and the struts in terms of L, W , H, y , w .
1. Draw a free body diagram
"
fcable =
0
−(y + W )
−H
√ tcable 2 2
(W +y ) +H
#
z
x
y
"
fstrut2 =
"
fstrut1 = √
tstrut1
L2 +y 2 +H 2
L
−y
−H
#
"
0
w= 0
−w
√ 2tstrut22 2
L +y +H
−L
−y
−H
#
#
2. Identify unknowns: tstrut1 , tstrut2 , tcable
3. Write the equations:
X
f = fstrut1 + fstrut2 + fcable + w = 0
4. Solve. To make things easier, let’s define
p
`strut ≡ L2 + y 2 + H 2
Now, write equilibrium in matrix form:
"
L/`strut
−L/`strut
−y /`strut −y /`strut
−H/`strut −H/`strut
`cable ≡
0
−(y + W )/`cable
−H/`cable
p
(W + y )2 + H 2
#"
# " #
tstrut1
0
tstrut2 = 0
tcable
w
(2.56)
(2.57)
(2.58)
Check the determinant:
det =
LyH
L(y + W )H
L(y + W )H LyH
2LHW
− 2
− 2
=− 2
`2strut `cable
`strut `cable
`strut `cable `2strut `cable
`strut `cable
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(2.59)
5.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 5
solids.uccs.edu/teaching/mae2103
When would it be zero? If L = 0, H = 0, W = 0. We’ll assume that this is not the case.
"
#
0
−L/`strut
0
`2strut `cable
`2 `cable L(y + W )w
−y /`strut −(y + W )/`cable = − strut
tstrut1 = −
det 0
2LHW
2LHW
`strut `cable
−w −H/`strut
−H/`cable
p
(y + W ) L2 + y 2 + H 2
= −w
2HWp
(y + W ) L2 + y 2 + H 2
= tstrut1
tstrut2 = ... = −w
" 2HW
#
L/`strut
−L/`strut 0
`2strut `cable
Lyw + Lyw
`2 `cable
tcable = −
det −y /`strut −y /`strut 0 = − strut
−
2LHW
2LHW
`2strut
−H/`strut −H/`strut w
p
y (y + W )2 + H 2
=w
HW
(2.60)
(2.61)
(2.62)
(2.63)
(2.64)
5. Sanity check:
ˆ units?
tstrut1,2 = [force]
[length][length]
= force X
[length][length]
tcable = [force]
[length][length]
[length][length]
X
(2.65)
ˆ undefined value? yes, when H, W = 0 as we expected
ˆ sign? assuming values in the picture, tstrut1,2 < 0, tcable > 0 as expected.
ˆ what if y < 0? then all the tensions are compressive. Does this make sense?
ˆ zeros? what if y = 0? then tcable = 0. does this make sense?
ˆ what if y < −W ? then the cable is in compression but the struts are in tension. Does this make sense?
6. Substitute values. No values given.
2.10
Static indeterminacy
As we saw last time, we occasionally encounter cases where our static system is unsolvable, yet our physical intuition
tells us that the system is well-defined. For instance, consider the following
" # system:
0
z
f2 = t2 0
1
1
1
x
"
f1 =
t1
√
3
#
1
−1
1
w
1
1
y
" #
−1
t
f3 = √33 1
1
"
#
0
w= 0
−w
The forces are already given to us so we’ll dispense with the FBD for now, and simply write the equilibrium equations
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5.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 5
solids.uccs.edu/teaching/mae2103
for the node.
f1 + f2 + f3 + w = 0
√ " # " #
√
1/ √3 0 −1/√ 3 t1
0
−1/ 3 0 1/ 3  t2 = 0
√
√
t3
w
1/ 3 1 1/ 3
(2.66)

(2.67)
As usual, we end up with a linear system. Our first step to solving linear systems, as always, is to compute the determinant:
det = 0 + 0 +
1 1
− −0−0=0
3 3
(2.68)
The determinant is zero, which means that we will be unsuccessful in solving the system. Yet, looking at the picture, we
have no reason to suspect that it would not be in static equilibrium. In this case, there are a couple of reasons why we
cannot find a solution:
(1) There are multiple solutions to this problem. It could be that the load is carried entirely by the center cable, or
split evenly between the other two cables, or some combination of the above. In other words, there are an infinite
number of solutions, so we cannot use our equations of statics to find it. (When we get to mechanics of materials,
we will treat the cables as springs. This gives us the additional equations we need to solve the system.)
(2) We note that even though the problem is drawn in 3D, it is effectively 2D – that is, it would be 2D if we rotated
our coordinate system by 45◦ . In that case, we would end up with only two equations for three unkonwns.
Systems like this are called statically indeterminant, and it is important to learn to recognize these systems. Note: a
helpful rule of thum for finding static determinacy (aside from writing out all the equations) is to conduct the thought
experiment of removing a member and seeing if the structure would collapse. If the structure always collapses upon the
removal of any member, it is generally statically determinant.
3
Moments
Up until now, we have only been able to look at systems where all of the force pass through a single point. This us useful
for a number of applications (e.g. trusses, as we’ll see in a few weeks) but it limits us to very simple systems. Here, we
will introduce another balance law that allows us to do equilibrium easily on complex bodies.
3.1
Force-bearing members
Before talking moments, let us introduce a number of objects that we will use when doing rigid body equilibrium.
3.1.1
Rigid bodies
Rigid bodies are bodies that can sustain arbitrary forces and arbitrary moments (we’ll explain those shortly.) We draw
rigid bodies in the following way:
We note that there are no restrictions on the location or direction of the forces, other than that they must all sum to
zero. We also note that rigid bodies are not allowed to deform: for instance, we could not have a rigid body that contains
a hinge. Rigid bodies are much more “sophisticated” than our previous members, 2FMs, which could only sustain two
colinear forces.
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5.3
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University of Colorado Colorado Springs
3.1.2
Course Notes - Lecture 5
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Rigid joints
Previously we looked at pivots, which sustain a 2D force and restrict 2D motion, but allow rotation. Now, we will consider
rigid joints that we will draw as:
ˆ Like with pivots, rigid joints restrict motion in 2D which mean that they sustain a force in 2D.
ˆ Unlike pivots, rigid joints restrict rotation as well. This means that they must sustain a “force” that opposes the
rotation: this is a moment.
3.1.3
Hinges
Hinges have similarities to both pivots and rigid joints. We will draw hinges as:
ˆ As with pivots and rigid joints, we see that hinges restrict motion in all directions. This means that it sustains a
3d force.
ˆ We see that there is a free rotation allowed about the hinge axis. That means that there is no “force” (or moment)
opposing the rotation in that direction.
ˆ However, we see that the hinge does not allow twisting or rotation about the other two axes. Therefore, it must
sustain a moment about the other two. Here, we start to see the need to express a “rotational force,” and this
example shows that it must be three-dimensional.
3.2
Moments
In the previous section we considered only systems of “concurrent forces;” that is systems, where all force vectors passed
through a common point. Because they all pass through a single point, there can be no rotation. Now, we will consider
more complex systems in which rotational degrees of freedom must be taken into account.
3.2.1
Motivation
Previously we talked about 2FMs and argued that the forces had to be equal and opposite, and colinear with the length
of the beam.
In the second case, the sum of the forces equals zero but the sum of the moments does not. Lack of force equilibrium
causes linear acceleration, lack of moment equilibrium causes angular acceleration.
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5.4
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University of Colorado Colorado Springs
3.2.2
Course Notes - Lecture 5
solids.uccs.edu/teaching/mae2103
Definition of a moment
Def 3.1. A moment is the conjugate force to rotational motion. The moment of a force about a point p is the cross
product of the distance vector (the vector pointing from p to the point of application of the force) with the force vector.
In vector notation:
M=r×f
(3.1)
where r is the vector pointing from the reference point to the point of application of the force, and is in units of distance.
A couple of things to note:
(1) We take the direction of r as convention: the arrow should always be at the same point as the force.
(2) The cross product is anticommunicative, that is a × b = −b × a. Therefore, it is very important that we stick with
the convention of M = r × f, not the other way around.
(3) We note that moment is a vector quantity and is only defined in 3D. (For 2D systems, the moment vector always
points in the z direction only.)
(4) units: n · M
NB: this is technically the same units as energy (Joules), but we do not refer to them as Joules, simply Newtonmeters. The reason for this is that
[moment] × [angle turned] = [energy]
(3.2)
but angles are in radians which are unitless. You can think of moments as being in units of Newton-meters/radian.
Example 3.1
Consider the following system:
f=
h
−1N
1N
d = (1m, 1m)
i
c = (1m, 2m)
b = (2m, 0m)
a = (0m, 0m)
Compute the moment of f about a, b, c.
(a) compute distance vector:
h
i h i
1m − 0m
1m
rad = 1m − 0m = 1m
(3.3)
compute cross product:
"
Ma = rad
î
= 1m
−1N
ĵ
1m
1N
# "
# "
#
0
0
k̂
0
0
=
0 =
1Nm + 1Nm
2Nm
0
(3.4)
Do we expect a positive value?
(b) compute distance vector
h i h i h
i
1m
2m
−1m
rbd = 1m − 0m = 1m
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(3.5)
5.5
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Course Notes - Lecture 5
solids.uccs.edu/teaching/mae2103
then compute cross product
"
Mb = rbd
î
= −1m
−1N
ĵ
1m
1N
# "
#
0
k̂
0
=0
0 =
−1Nm + 1Nm
0
(3.6)
h
i
−1m
0m
(3.7)
Why is the vector zero? do we expect this?
(c) compute distance vector
rcd =
then compute cross product
"
Mc = rcd
î
= −1m
−1N
ĵ
0
1N
# "
#
0m
k̂
0m
0 =
−1Nm
0
(3.8)
Note that the moment is negative. Why is this? Because of the right hand rule.
3.2.3
Polar form
Recall the identity:
||M|| = ||r × f|| = ||r||||f|| sin θ
(3.9)
Let’s check this with the above results:
Example 3.2
√
√
√
√
◦
2
2
2
2
(a) ||r
√ad || =√ 1m + 1m = 2m and ||f|| = 1N + 1N = 2N and sin θ = sin 90 = 1 so we have
( 2m)( 2N)(1) = 2Nm X
(b) What is the angle? Looks like 180◦ , but it is actually 0◦ . So sin θ = 0 and we get 0. X
√
√
√
(c) How about the last one? ||rcd || = 1m and sin θ = cos 45◦ = 1/ 2, so ||M|| = (1m)( 2N)(1/ 2) = 1Nm X
In the last example, notice that we lost the information about the negativity of the moment. When balancing moments
using the polar form, one must keep track manually of which moments are positive and which are negative. How to do
that? Right hand rule:
f
r
r
θ
positive moment
θ
f
negative moment
(Note: if you are consistent about defining your angles then you don’t have to worry, but this is usually a pain. It’s way
easier to just use vectors.)
3.2.4
Pictorally
The moment vector is orthogonal to both the distance vector and the force vector. But the distance vector changes
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5.6
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University of Colorado Colorado Springs
Course Notes - Lecture 5
solids.uccs.edu/teaching/mae2103
M=r×f
f
ˆ At the corner, M is in the “twisting direction”
ˆ At the base, M has twisting and bending components
ˆ The moment vector changes continuously throughout the body.
ˆ The moment of a force along the line of action is always zero.
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5.7
Lecture 6
3.2.5
Couple moments and distributed loads
Line of application
Let us suppose we are taking the moment of a force f about a point so that the distance vector is r.
f
r + αf
r
Sometimes the location of the force makes finding a specific distance vector tedious. Let us consider a different distance
vector that points from the same reference point to some arbitrary point along which f acts. We can represent this by
r̂ = r + α f
(3.10)
where α is some scalar multiplier (that, incidentally, should have units of distance/force). We can see that r̂ always
points to a point in the line along which f acts, called the line of application. What is the resulting moment if we use
this new distance vector?
0
:
M̂ = r̂ × f = (r + α f) × f = r × f + α
f×
f=M
(3.11)
Therefore, we see that we can take our distance vector to be any vector that points from the reference point to the line
of application of the force.
3.3
Moment about a specific axis
We can find the twisting moment about a specific direction d using projection:
Mtwist =
M·d
||d||
(3.12)
Example 3.3
(a) In the above example, compute the moment vector at the base of the pipe:x
z
c = (−5m, 5m, 1m)
b = (0, 0, 1m)
x
"
−1N
f = −1N
0
a = (0, 0, 0)
#
y
Compute moment about a: first compute
"
rac
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−5m
= 5m
1m
#
(3.13)
6.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 6
solids.uccs.edu/teaching/mae2103
so the moment is
"
Ma = rac
î
× f = −5m
−1N
# "
#
1Nm
k̂
1m = −1Nm
10Nm
0
ĵ
5m
−1N
(3.14)
(b) Compute the twisting moment along the pipe. Using the vector we got above, just take the projection. We
will use the vector rab :
" #
0
rab = 0
(3.15)
1m
so the projection is
Mtwist =
Ma · rab
(0m)(1Nm) + (0m)(1Nm) + (1m)(10Nm)
=
= 10Nm
||rab ||
1m
(3.16)
which, as we expect, is just the z component of the moment vector.
(c) Compute the bending moment at a. We can write the twisting moment in vector form by multiplying by the
unit vector:
"
#
0
rab
0
=
(3.17)
Mtwist = Mtwist n = Mtwist
||rab ||
10Nm
Let us assume that we can write
M = Mtwist + Mbend
(3.18)
So then we can find Mbend by subtraction:
"
Mbend = M − Mtwist
# "
# "
#
1Nm
0
1Nm
0
= −1Nm −
= −1Nm
10Nm
0
10Nm
(3.19)
[draw on picture]
This is the component of the moment that acts to bend the pipe at this point. (Notice how the vector is
orthogonal to the direction of the pipe.)
3.4
Couple moments
Consider the following system where two equal and opposite forces are applied at points seperated by the orthogonal
distance vector d.
−F
d
F
O
r0
What is the moment of these two forces about the origin? Proceed formally:
M = r0 × −F + (r0 + d) × F = −r0 × F + r0 × F + d × F = d × F
(3.20)
Notice that even though we were taking the moment about the origin, the distance from the origin (ri ) was completely
cancelled out. This is a result of the fact that the forces were equal and opposite. Whenever there are two equal and
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6.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 6
solids.uccs.edu/teaching/mae2103
opposite forces that are seperated by an offset vector, the result is a couple moment. There are some important things
to note about couple moments:
(1) If the moment is generated by forces with magnitude F , and the orthogonal offset vector is d with magnitude d,
then the magnitude of the couple moment is
||d × F|| = ||d||||F|| sin θ = ||d||||F|| = dF
(3.21)
(2) The direction of the couple moment must be orthogonal to both forces, and is therefore orthogonal to the plane
that they define. Use the right hand rule to remember the sign. Alternatively, if you are applying a couple moment
to a screwdriver, the moment points in the direction that the screw is moving.
(3) The location of a couple moment does not matter. As we saw above, because r0 cancelled out, the couple moment
has exactly the same effect about every point. (This is counterintuitive, but generally a pretty friendly fact. When
doing rigid body
Pairs of forces like this produce a couple moment.
Def 3.2. A couple moment is a pure moment that has no net force and can act at any point.
Pictorally:
M
−F
F
d
Example 3.4
In order to install a screw, a screwdriver exerts a couple moment with magnitude M as shown.
F
d
M
F
To stabilize the part, two forces are applied at each end as shown. What magnitude of force should be applied so
that the net moment is zero?
If we adopt a coordinate system such that the moment acts in the negative z direction, then the screwdriver moment
is
"
#
0
0
(3.22)
−M
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6.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 6
solids.uccs.edu/teaching/mae2103
The moment exerted by the forces is:
"
#
d/2
r1 = h/2
0
" #
0
F1 = −F
0
"
# "
#
0
î
ĵ
k̂
0
M1 = r1 × r1 = d/2 h/2 0 =
−dF /2
0
−F 0
"
#
−d/2
r2 = −h/2
0
" #
0
F2 = F
0
(3.23)
(3.24)
"
î
M2 = r2 × r2 = −d/2
0
ĵ
−h/2
F
# "
#
0
k̂
0
0 =
−dF /2
0
(3.25)
So for the moments to be equal:
M=−
dF
dF
M
−
= −dF =⇒ F = −
2
2
d
(3.26)
Note that we never quantified where the moment had to be applied.
3.5
Reduction of forces and moments
We can show that each set of forces and moments can be combined to form a single effective force+moment combination
acting at a particular point. For example, consider the following beam that is supporting a load w with two forces:
w
w /2
w
=
=
w
w /2
w
w
We can write the supporting forces as two separate forces or as a force+a couple at a single point.
It is often convenient to reduce a complex force system to a single force and a single moment. We will illustrate this
with an example:
Example 3.5
Consider the following rigid body with forces and moments applied as shown.
F1
b
θ1
M
H
a
L
L
F3
θ2
F2
Find the effective force+moment combination at points a and b.
(a) Let’s begin by writing out all of the forces and distance vectors:
"
#
"
#
cos θ1
cos θ2
f1 = F1 sin θ1
f2 = F2 − sin θ2
0
0
" #
" #
L
2L
r1 = H
r2 = 0
0
0
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"
0
f3 = −F3
0
" #
L
r3 = 0
0
#
(3.27)
(3.28)
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and the moment vector is
" #
0
M= 0
M
(3.29)
The total force is just the sum of all of the forces:
"
#
F1 cos θ1 + F2 cos θ2
fa = F1 sin θ1 − F2 sin θ2 − F3
0
And the total moment is the sum of all of the moments:
"
0
0
Ma = (r1 × f1 ) + (r2 × f2 ) + (r3 × f3 ) + M =
F1 (L sin θ1 − H cos θ1 ) − 2F2 L sin θ2 − F3 L + M
(3.30)
#
(3.31)
(b) What is the equivalent force vector acting at b? It is exactly the same: the total force vector is independent
of location.
fa = fb
(3.32)
Now, we will compute the effective moment. First, of course, we compute distance vectors, remembering that
the distance vector points to the force location:
#
" #
"
#
"
0
0
L
(3.33)
r1 = 0
r2 = −H
r3 = −H
0
0
0
What forces will contribute moments? Only f2 : the first force is acting at the same point, and the third force
is colinear, so the resultant moment is just
"
#
0
0
Mb = (r2 × f2 ) + M =
(3.34)
F2 (−L sin θ2 + H cos θ2 ) + M
3.6
Distributed loads
Most of the examples we have looked at so far have involved “point loads” – that is, loads that are concentrated at a
single point. They are convenient to work with mathematically, but in reality, loads are rarely that concentrated; they
always have a finite width. In this section, we are going to look at how to work with loads that are distributed and
change continuously in space.
3.6.1
Discrete case
We will begin by considering the discrete case and taking the limit. Consider the following beam loaded with boxes of
varying height {yi } and constant width ∆x
y2
y1
y0
∆x
x0
x1
y3
y4
x2
x3
x4
y7
y5
x5
y6
x6
x7
Let each box have a specific weight (that is, weight per unit volume or area, should have units of [force]/[volume] or
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[force]/[area]) of wi , so that box i has weight w yi ∆x. Each box exerts the following force vector:
"
#
0
fi = −wi yi ∆x
0
What is the effective (or total) force acting at the joint? To get this we just add all of them up:
"
# "
#
"
#
0
0
0
X
wi yi ∆x
f = −w1 y1 ∆x + w2 y2 ∆x + ... = −
0
0
0
i
(3.35)
(3.36)
What is the effective moment acting at the joint? For this, we need to compute a position vector for each box. The
position vector for each one is just:
" #
"
#
0
xi
0
ri = 0
so each box has a moment vector of
Mi = ri × fi =
(3.37)
0
wi yi xi ∆x
So, the total moment acting at the joint is:
"
# "
#
"
#
0
0
0
X
0
0
0
M=
+
+ ... =
w1 y1 x1 ∆x
w2 y2 x2 ∆x
wi yi xi ∆x
i
3.6.2
(3.38)
Location of effective load
In the above example, we computed the effective load and moment about the joint. Can we represent the effective load
using just a force at a location, as in the following picture?
L
F
=
We know that the total force should just be the same as the sum of all the other forces, that is What is the effective
force vector, feff ? We know that it must exert the same force as the distributed load, so we just have
"
#
0
X
wi yi ∆x
feff = −
(3.39)
0
i
What about the location of the force, L? We can solve for L using the fact that the effective force should generate the
same moment about the joint as the distributed load. If feff is located at a distance L from the joint, then it has distance
and moment vectors
" #
"
#
"
#
0
Leff
X
î
ĵ
k̂
0
reff = 0
Meff = Leff
(3.40)
0 = −Leff
P 0
0
wi yi ∆x
0 − i wi yi ∆x 0
i
We know that Meff = M so we have
"
− Leff
X
i
#
"
#
0
0
X
0
0
=−
wi yi ∆x
wi yi xi ∆x
i
(3.41)
Equating the z terms gives
Leff
X
wi yi ∆x =
i
X
wi yi xi ∆x
(3.42)
i
and solving for L gives
Leff
P
wi yi xi ∆x
= Pi
.
i wi yi ∆x
(3.43)
Do the units check out? They do: the top has units of [force][distance] and the bottom has units of [force].
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3.6.3
Course Notes - Lecture 6
solids.uccs.edu/teaching/mae2103
Continuous case
The previous example was an example of a load that was piecewise continuous, but we formulated it in a very specific
way so as to make the connection to continuous loads. Imagine that we refine the blocks so that they become narrower
and narrower, approaching a smooth curve on the top:
y (x)
L
We will take the limit as the width of these rectangles goes to zero. What is the total force exerted by this continuous
blob?
"
#
0
X
X
−wi yi ∆x
f = lim
or, looking at the y component,
fy = − lim
wi yi ∆x
(3.44)
∆x→0
∆x→0
0
i=0
i=0
This may look familiar from caclulus. In the limiting case, we have
xi → x
wi → w (x)
yi → y (x)
X
∆x → dx
i
Z
L
→
(3.45)
0
so in the limit, our total force can be evaluated as the integral
Z
fy = −
L
Z
w (x) y (x) dx
or, for all three components,
0
f=−
0
L
"
0
w (x) y (x) dx
0
#
(3.46)
where the integral sign can safely be taken into each term in the vector. Now, let us do the same thing for the moment:
taking the limit as ∆x → 0, we get
"
#
#
Z L"
0
0
0
Mz = − lim P 0
=−
(3.47)
∆x→0
0
w (x) y (x) x dx
i wi yi xi ∆x
What is the location of the effective force? Doing the exact same analysis gives us that
RL
w (x) y (x) x dx
L = R0 L
w (x) y (x) dx
0
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(3.48)
6.7
Lecture 7
3.6.4
Rigid body equilibrium–2D
Generalized case
We used the boxes as a specific case for a continuous load, but we could have loading scenarios that are more complex.
Let us consider a generalized distributed load f(x) where f (x) has units of force per length. Then we have
Z
feff = f(x)dx
(3.49)
(where f (x)dx is the differential force element) Let us take the moment about the origin; then what is our distance
vectors? The differential force element is always acting at x, so we just have r(x) = x. Then, we can write that
Z
Meff = x × f(x)dx
(3.50)
Example 3.6
As a sanity check, consider a load:
H
L
with y (x) = H and w (x) = w0 . Then:
Z
F =
L
L
Z
w (x) y (x) dx =
w0 H dx = w0 HL
0
(3.51)
0
as we expect, and
Z
Mz =
L
Z
w (x) y (x) x dx = wH
0
0
L
L
1
1
2
xdx = w0 Hx = w0 HL2
2
2
0
(3.52)
Now, plug in to find the location of the effective force:
L=
Mz
w0 HL2 /2
L
=
=
F
w0 HL
2
(3.53)
so the force is acting halfway, exactly as we expect.
Example 3.7
Consider the following load
w (x)
L
where w (x) = w0 cos(πx/2L).
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ˆ force:
Course Notes - Lecture 7
solids.uccs.edu/teaching/mae2103
L
Z
w0 cos(πx/2L)dx = w0
0
2L
sin(πx/2L)
π
L
=
0
2Lw0
2Lw0
sin(π/2) =
π
π
(3.54)
ˆ moment: use integration by parts
Z
L
Z
w0 cos(πx/2L) x dx =
0
0
L
w0 cos(πx/2L) |{z}
x dx
|
{z
}
du
v
L Z L
2Lw0
w0 2Lx
=
sin(πx/2L) −
sin(πx/2L)dx
π
π
0
0
L
w0 2L2
4L2 w0
cos(πx/2L)
+
=
2
π
π
0
=
(3.55)
w0 2L2
2w0 L2 (π − 2)
4L2 w0
=
−
π
π2
π2
(3.56)
(3.57)
(3.58)
ˆ effective location:
L=
4
π
2w0 L2 (π − 2)
L(π − 2)
×
=
π2
2Lw0
π
(3.59)
Rigid Body Equilibrium
This section is where we start to solve more practical and interesting problems. We have built up all the machinery that
we need to do static analysis on many types of objects.
4.1
Equilibrium equations
With point equilibrium, we were concerned only about equilibrium of forces. Now, (as we have been alluding to) we will
solve for moment equilibrium as well. The governing equations for the equilibrium of a single rigid body in 2D or 3D are
X
X
F=0
M=0
(4.1)
that is, the sum of the forces and moments must equal to zero. These are vector equations, so how many actual equations
is this?
2D [2 force equations (x,y)] + [1 moment equation (z)] = 3 equations
3D [3 force equations (x,y,z)] + [3 moment equations (x,y,z)] = 6 equations
There are a few things to note:
(1) Even for 2D problems, you should do everything in 3D. It’s more systematic and makes it easier to keep track of
moments.
(2) In general, you should be able to solve these equations easily, but occasionally you may need to use Cramer’s rule.
(3) In general, when solving by hand, you will only have to solve “reduced” problems in 3D with fewer equations and
unknowns.
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7.2
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4.1.1
Course Notes - Lecture 7
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Choosing a reference point
We have talked a lot about computing the moments of vectors “about a point.” Which point should you choose? It
turns out that if
X
X
f=0
M=0
(4.2)
where the moments computed about one point are zero, the moments computed about all points are zero. We’ll prove
this:
f1
f1
f2
f2
z
z
r2
r1
r1
r0 + r2
r2
r3
f3
r3
a
f3
r0 + r3
y
r
+
r1
x
r0
0
b
Suppose a body is subjected to a system of forces (and moments) and is in equilibrium, where all moments are computed
about the origin:
X
X
X
fi = 0
Mai =
ri × fi = 0
(4.3)
Then the sum of the moments computed about a different point (b) will be zero as well:
X
Mbi
0
X>
X
X
X
:0
fi = 0 X
=
(r0 + ri ) × fi =
r0 × fi + ri × fi = r0 ×
| {z }
| {z }
moment eq.
(4.4)
force eq.
There are a few things to note:
(1) Because you can choose any reference point, always try to choose one that makes computing moments easy.
(2) If you use all of your force equations, you can only use the moment equation once.
(3) Alternatively, you can use two moment equations, but then you can only use one of your force equations.
4.2
2D Free-body diagrams
Free-body diagrams are crucial for describing and quantifying the problem, especially when we have to take into account
the reactions of constraints.
4.2.1
Pivots
Reaction forces and moments
A pivot is a constraint that restricts motion in the x and y direction, but allows rotation in the z direction.
We draw pivots like this:
"
fpiv
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fpiv ,x
= fpiv ,y
0
#
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Course Notes - Lecture 7
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Because a pivot restricts motion in two directions, it has two force unknowns. It allows rotation, so it has
no moment unknowns. Therefore, we can replace a pivot with a force that has an x and y component.
Rollers
A roller is a constraint that restricts motion normal to the surface, but allows motion tangent to the
surface. It also allows rotation in the z direction. We draw rollers like this:
n
froll = froll n
Because a roller restricts motion on only one direction, it has only one force unknown. Therefore, we
replace a pivot with a force where the direction is known (the normal vector to the surface) but the
magnitude is not. For example, on a flat surface,
frol
Fixed joint
" #
0
= froll,y 1
0
A fixed joint is a constraint that restricts motion in two directions. It also restricts rotation. We might
draw a fixed joint like this:
"
fpiv
fpiv ,x
= fpiv ,y
0
"
#
Mpiv
0
0
=
Mpiv
#
Because motion is restricted in two directions, the force must have two unknowns. Because rotation is
restricted as well, the moment has one unknown.
Collar
A collar is a constraint that restricts motion in one direction, and also restricts rotation. We might draw
a collar like this:
Mcol
fcol = fcol ncol
What kind of reaction force would we use here? Remember, location is only constrained in one direction,
so we can only have one force unknown – it will end up being the same a roller. The difference is that
rotation is also constrained, so we will also include a couple moment.
There are a wide variety of types of constraints, and it will often be up to you to determine how to replace a given
constraint with an equivalent reaction force and moment. It may not always be obvious what kind of reaction is
necessary. For instance, consider this crowbar applied to these smooth surfaces.
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How do we replace this with reaction forces and moments? The keyword is “smooth” – we know that the surfaces can
only exert forces normal to them, so we treat them as sliders.
4.2.2
Static indeterminacy and insufficient constraints:
Consider the following beam constrained with two rollers and subjected to a loading force f. What are the reactions at
the rollers?
f
f
frola
" #
0
= frola 1
0
frolb
" #
0
= frolb 1
0
We only have two force unknowns, which means we have two unknowns and three equations, so we cannot solve our
system. In this case, it may seem pretty obvious that the system is unsolvable: with nothing holding the beam back, the
horizontal component of the force will push it away. In this case, we say that the system has insufficient constraints.
Consider the following beam subjected to a loading force f.
f
f
"
fpiva
fpiva,x
= fpiva,y
0
#
"
fpivb
fpivb,x
= fpivb,y
0
#
Can we solve the system for the reaction at the joints? No: we have four unknowns but only three equations. It does
not mean that the system is unphysical, it just means that there are an infinite number of equally physical solutions. In
this case, we say that the system is statically indeterminate.
Consider the following two cases:
How many unknowns do we have for these cases? In the first we have three force unknowns, in the second we have two
force unknowns and one moment unknown, so both are solvable. We’ll see these a lot: they are referred to as “simply
supported” (left) and “cantilever” (right).
4.3
2D solution strategy
Just like with point equilibrium, we have a recipe for solving 2D equilibrium problems.
1. Identify unknowns: they may be specified in the problem, or you may need to find the reaction forces/moments of
constraints in the problem.
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2. Draw a free body diagram: remember to replace all of the constraints with the appropriate reaction force/moment.
3. Write governing equations:
ˆ Force equilibrium
ˆ Moment equilibrium
ˆ Geometric
ˆ Constitutive
4. Solve: easiest way possible, Cramer’s rule if you must.
5. Sanity check: make sure the units come out right, and substitute a couple of “easy” values to see what happens
to the solution.
6. Substitute numbers
We’ll cement this solution strategy by doing a couple of examples in 2D.
4.4
2D examples
Example 4.1
Consider the following rigid body that is supported by a roller and a rigidly-attached collar sliding along a rod.
M
L
L
45◦
L
f
Determine the support reactions.
Proceed formally:
1. Identify unknowns: at the smooth support we have froll,y . The collar exerts a force along the normal direction
fcoll , and a moment Mcoll
2. Draw free-body diagram:
"
froll = froll
M
−1
0
0
#
L
L
Mcoll
fcoll =
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f√
coll
2
" #
1
1
0
L
#
0
f = −f
0
"
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University of Colorado Colorado Springs
Course Notes - Lecture 7
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3. Write equations: The equations of force equilibrium are:
√
# " #
"
0
fcoll / √2 − froll
= 0
fcoll / 2 − f
0
0
(4.5)
Let us select the collar as a point for computing the moments. Our distance vectors are:
" #
" #
2L
L
r= 0
rroll = L
0
0
(4.6)
so the moments are
"
0
0
M=r×f =
−Lf
#
"
Mroll = rroll × froll
0
0
=
Lfroll
Our equation of moment equilibrium is, then:
"
# " # "
# " #
0
0
0
0
0
0
M + Mroll +
+ 0 =
= 0
0
Mcoll
M
Mcoll + M + Lfroll − Lf
4. We can write the equilibrium equations in matrix form as:
√
"
#"
# "
#
0
froll
−1 1/√2 0
f
fcoll =
0 1/ 2 0
Lf − M
L
0
1 Mcoll
#
(4.7)
(4.8)
(4.9)
where we can solve using Cramer’s rule.
However, this is a simple system that can be solved easily simply by substitution: From y-force equilibrium:
√
fcoll = f 2
(4.10)
Substituting into x-force equilibrium gives:
fcoll
froll = √ = f
2
(4.11)
Mcoll = Lf − M − Lfroll = Lf − M − Lfroll = −M
(4.12)
Substituting into moment equilibrium gives:
So we finally have
fcoll
" #
f
= f
0
"
froll
−f
= 0
0
#
"
Mcoll
0
= 0
−M
#
(4.13)
5. Sanity check:
ˆ Units? X
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7.7
Lecture 8
Rigid body equilibrium 2D & 3D
Example 4.2
Find the reactions at the pivot and the roller supporting the following rigid body:
L
L/2
f
H
w = w0 xL
1. Unknowns: fpiv ,x , fpiv ,y , frol
2. Free body diagram:
first, let’s reduce the distributed load: find the total force
L
Z
Feff ,y =
0
find the total moment:
L
w0 x 2 w0 L
w0 x
dx =
=
L
2L 0
2
(4.14)
L
w0 x 2
w0 x 3 w 0 L2
dx =
=
L
3L 0
3
(4.15)
Meff
2
w0 L2 /3
= L
=
Feff
w0 L/2
3
(4.16)
L
Z
Meff ,z = −
0
and the effective location:
Leff = −
and use this in our FBD:
L
"
L/2
0
f = −f
0
#
H
"
2L/3
"
fpiv
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fpiv ,x
= fpiv ,y
0
#
Feff
#
0
= −w0 L/2
0
"
0
#
frol = frol,y
0
8.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 8
solids.uccs.edu/teaching/mae2103
3. Write equations:
force equilibrium:
"
fpivy
# " #
fpiv ,x
0
+ frol,y − f − w0 L/2 = 0
0
0
(4.17)
moment equilibrium: select the pivot as the point, so the distance vectors are
"
#
"
#
" #
L
L/2
2L/3
0
r= H
reff =
rrol = 0
0
0
0
and the moments are
"
#
0
0
M=
−fL/2
0
0
=
−w0 L2 /3
"
Meff
"
#
Mrol =
0
0
Lfrol,y
(4.18)
#
(4.19)
so the moment equilibrium is
# " #
0
0
0
= 0
0
−fL/2 − w0 L2 /3 + Lfrol,y
"
(4.20)
4. Solve: from x equilibrium we have that fpiv ,x = 0. What does this mean?
moment equilibrium gives:
frol,y =
1
1
f + w0 L
2
3
(4.21)
y force equilibrium gives:
1
1
1
f
w0 L
1
fpiv ,y = f + w0 L − frol,y = f + w0 L − f − w0 L = +
2
2
2
3
2
6
(4.22)
5. Sanity check: do the units check out? Yes. If I substitute f = w0 = 0 do I get zero for all of my reactions?
Yes.
We will do one more example of rigid body equilibrium in 2D before moving to 3D:
Example 4.3
Consider the following system where a load f with magnitude f is applied to a beam as shown:
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8.2
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Course Notes - Lecture 8
solids.uccs.edu/teaching/mae2103
L/2
f
L/2
θ1
θ2
w
Determine w as a function of L, f , θ1 , θ2
1. Identify unknowns: we are asked for w , but we will also need to find the reactions frol1 , frol2
2. Draw FBD:
"
frol2
−frol2
0
=
0
#
"
rrol2
"
0
#
"
0
f = −f
0
frol1 = frol1
0
L cos θ1
= L sin θ1
0
#
#
r = rrol2 /2
"
fcab
t cos θ2
= tn = t sin θ2
0
#
3. Write equations:
force equilibrium for the beam:
"
X
f = f + fcab + frol1 + frol2
0
= −f
0
#
"
# "
# "
# " #
0
t cos θ2
0
−frol2
0
+ t sin θ2 + frol1 +
= 0
0
0
0
0
(4.23)
force equilibrium for the box:
t=w
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(4.24)
8.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 8
solids.uccs.edu/teaching/mae2103
moment equilibrium: choose roller 1. compute r, rrol2 – all other r vectors are zero.
"
#
"
#
(L/2) cos θ1
L cos θ1
r = (L/2) sin θ1
rrol2 = L sin θ1
0
0
now compute the moments by taking the cross product:
#
"
# "
0
î
ĵ
k̂
0
M = r × f = (L/2) cos θ1 (L/2) sin θ1 0 =
− 21 Lf cos θ1
0
−f
0
"
# "
#
0
î
ĵ
k̂
0
Mrol2 = rrol2 × frol2 = L cos θ1 L sin θ1 0 =
Lfrol2 sin θ1
−frol2
0
0
(4.25)
(4.26)
(4.27)
so moment equilibrium gives us:
# "
# " #
0
0
0
0
0
=
+
= 0
1
0
Lfrol2 sin θ1
− 2 Lf cos θ1
"
X
M = M + Mrol2
(4.28)
4. Now solve:
moment-z equilibrium gives:
L
frol2 sin θ1 =
f cos θ1
1
f
L
=
f cos θ1 =⇒ frol2 =
2
2 sin θ1
2 tan θ1
(4.29)
force-x equilibrium gives:
t cos θ2 = frol2 =⇒ t =
f
frol2
=
cos θ2
2 tan θ1 cos θ2
(4.30)
which is our answer. Just for kicks, we can also get the reaction at the first roller from the force-y equilibrium
equation:
f
tan θ2
frol1 = f − t sin θ2 = f −
sin θ1 = f 1 −
(4.31)
2 tan θ1 cos θ2
2 tan θ1
4.5
3D Free-body diagrams
We will now turn our attention to rigid body equilibrium problems in 3D. Recall that
Pivots
Just like for the 2D case, a pivot is a constraint that restricts translation in all directions; in the 3D
case it restricts translation in x, y , z. On the other hand, it allows rotation about all axes. We draw
pivots like this:
z
x
y
" #
0
M= 0
" #
0
fx
f = fy
fz
A pivot restricts three degrees of freedom of the system, and as such it introduces three force unknowns. Therefore we replace the picture of a pivot with a force that has three unknown components.
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8.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Journal bearing
Course Notes - Lecture 8
solids.uccs.edu/teaching/mae2103
A journal bearing is similar to the collar in 2D. A smooth journal bearing restricts motion in two
directions: transverse to the hole and normal to the surface. It allows translation through the bearing.
Therefore it has two force degrees of freedom. A bearing allows axial rotation, but does not allow
rotation in any other direction. We will (attempt to) draw a journal bearing like this:
z
x
"
y
Mx
M= 0
Mz
#
" #
fx
f= 0
fz
The bearing must include two force unknowns and two moment unknowns, so we have four unknowns
altogether. (Sometimes we’ll add the constraint that the bearing is not allowed to sustain any
moments at all – we sometimes say that this means it is “well-designed.” In that case, we assume
that the moments are zero, and so we only have two unknowns.)
Square journal
bearing
The square journal bearing is similar to the regular journal bearing, except that rotation is not allowed
in the axial direction either. We might draw a square journal bearing like this:
" #
z
Mx
M = My
y
x
Mz
" #
fx
f= 0
fz
No rotation is allowed at all, so the moment has three unknowns. The force vector is the same as
for the journal bearing, resulting in five total unknowns.
Hinge
A hinge is similar to a journal bearing, except that it constrains motion in all directions, allowing
rotation in exactly one direction. We might draw a hinge like this:
" #
z
Mx
M= 0
y
x
Mz
" #
fx
f = fy
fz
A hing introduces three force unknowns and two moment unknowns.
Fixed joint
A fixed joint is a joint that does not allow for any rotations or any moments. It’s kind of tricky to
draw a “general” fixed joint but an example of one might look like this:
" #
Mx
z
M = My
Mz
y
x
" #
fx
f = fy
fz
A fixed contact introduces three force and three moment unknowns, and causes the problem to be
fully constrained. (If you ever encounter a problem with more than one constraint, and one of them
is a fixed joint, you know right off that it will be statically indeterminant.)
There are multiple other types of 3D constraints, and it is good to get some practice figuring out how to turn a constraint
into a set of reaction forces and moments.
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8.5
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
4.6
Course Notes - Lecture 8
solids.uccs.edu/teaching/mae2103
3D solution strategy
The 3D solution strategy is identical to that for 2D; especially if you use the 3D vector formulation in 2D, you’ll see
that the procedure is identical. The only difference is that both force and moment balance is significant for all three
dimensions, so you end up with a total of six equilibrium equations.
All content © 2016, Brandon Runnels
8.6
Lecture 9
4.7
3D equilibrium examples, multi-body equilibrium
3D examples
Example 4.4
Consider the following frame that is mounted using a pivot, a journal bearing, and a roller and is subjected to a
force f with magnitude f .
`
f
L
W
W
W
Determine the reactions at the supports.
I’m going to switch the order of (1) and (2) here and draw the FBD first:
1. Free body diagram:
"
#
fpivx
fpiv = fpivy
fpivz
"
0
f= 0
−f
`
#
"
"
W
z
x
Mjou =
y
fjou
W
"
#
fjoux
= 0
fjouz
Mjoux
0
Mjouz
#
frol =
0
0
#
frolz
L
W
2. Identify unknowns: fpivx , fpivy , fpivz , fjoux , fjouy , frolz , Mjoux , Mjouz = 8 unknowns. Uh oh, we have too many
unknowns, which means that this structure is statically indeterminate. This makes sense: imagine that the
bearing was “pre-stressed”, causing an additional moment in the system.
To solve this, we need to make some assumptions about the problem. We will make the intuitive assumption
that the moment exerted on the journal bearning is zero. If this assumption is safe, then we will be able
to solve the problem. If not, we’ll have to go back and check our assumptions.
Assume: Mjoux = Mjouy = 0. (always state assumptions clearly in your work!)
3. Write equations:
force equilibrium:
"
fpiv + fjou + frol + f =
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fpivz
fpivx + fjoux
fpivy
+ fjouz + frolz − f
#
" #
0
= 0
0
(4.32)
9.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
moment equilibrium: as usual, we must choose a reference point. Choosing the origin,
" #
"
#
"
#
0
−`
−L
r= W
rjou = 2W
rrol = 3W
0
0
0
(4.33)
so the moments are
#
# "
−Wf
î
ĵ
k̂
M = r × f = −` W
0 = −`f
0
0
0 −f
# "
#
"
2Wfjouz
î
ĵ
k̂
0
Mjou = rjou × fjou = 0
2W
0 =
−2Wfjoux
fjoux
0
fjouz
"
# "
#
3Wfrolz
î
ĵ
k̂
Mrol = rrol × frol = −L 3W
0 = Lfrolz
0
0
0
frolz
"
sum of the moments:
"
# "
# "
# "
# " #
−Wf + 2Wfjouz + 3Wfrolz
2Wfjouz
−Wf
3Wfrolz
0
X
0
−`f + Lfrolz
M = −`f +
+ Lfrolz =
= 0
0
0
−2Wfjoux
0
−2Wfjoux
(4.34)
(4.35)
(4.36)
(4.37)
4. Solve:
Y-force: fpivy = 0
Z-moment: fjoux = 0
Y-moment: −`f + Lfrolz = 0 =⇒ frolz = `f /L
f + 2
fjouz + 3
frolz = 0 =⇒
X-moment: −
W
W
W
fjouz =
f
3`f
1
(f − 3frolz ) = −
2
2
2L
(4.38)
0
*=
fjoux
0 =⇒ fpivx = 0
X-force: fpivx + Z-force: fpivz + fjouz + frolz − f = 0 =⇒
fpivz = f − fjouz − frolz
1
= f − f (1 − 3`/L) − `f /L
2
f`
f
= +
2 2L
(4.39)
(4.40)
(4.41)
(4.42)
Example 4.5
Determine the tensions in the cables and the reaction of the strut for the following system.
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9.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
H
L
W
L
W
w
1. Determine unknowns: fpivx , fpivy , fpivz , tcab1 , tcab2 , tcab3
2. Draw free body diagram:
"
fpiv
"
fcab1 =
0
0
fpivx
= fpivy
fpivz
#
"
fcab3 =
#
#
0
0
fcab3
tcab1
fcab2 = tcab2 ncab2
W
L
L
W
"
0
w= 0
−w
#
where the unit vector ncab2 is
ncab2 = √
and ` =
√
1
4W 2 + H 2
"
0
−2W
H
#
0
= −2Wtcab2 /`
Htcab2 /`
"
=⇒ fcab2
#
(4.43)
4W 2 + H 2 .
3. Write giverning equations:
force equilibrium:
"
X
f = fpiv + w + fcab1 + fcab2 + fcab3 =
fpivz
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fpivx
fpivy − 2Wtcab2 /`
− w + tcab1 + Htcab2 /` + fcab3
#
(4.44)
9.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
moment about the pivot: compute distance vectors
" #
" #
"
#
2L
2L
2L
rw = W
rcab1 = 0
rcab2 = 2W
0
0
0
"
rcab3
L
= 2W
0
#
(4.45)
so the moments are:
"
î
Mw = rw × w = 2L
0
ĵ
W
0
"
Mcab1 = rcab1 × fcab1
Mcab2 = rcab2 × fcab2
Mcab3 = rcab3 × fcab3
k̂
0
−w
#
"
−wW
= 2Lw
0
# "
#
(4.46)
#
0
î ĵ
k̂
= 2L 0
= −2Ltcab1
0
0
0 0 tcab1
"
# "
#
2WHtcab2 /`
î
ĵ
k̂
= 2L
= −2LHtcab2 /`
2W
0
−4WLtcab2 /`
0 −2Wtcab2 /` Htcab2 /`
"
# "
#
2Wtcab3
î
ĵ
k̂
= −Ltcab3
= L 2W
0
0
0
0
tcab3
(4.47)
(4.48)
(4.49)
and the moment equation is:
# " #
−wW + 2WHtcab2 /` + 2Wtcab3
0
M = 2Lw − 2Ltcab1 − 2LHtcab2 /` − Ltcab3 = 0
0
−4WLtcab2 /`
"
X
we notice right away that tcab2 = 0 which greatly simplifies the analysis. Rewriting the above:
"
# " #
−wW + 2Wtcab3
0
X
M = 2Lw − 2Ltcab1 − Ltcab3 = 0
0
0
(4.50)
(4.51)
4. Solve:
moment-x:
− wW + 2Wtcab3 = 0 =⇒ 2tcab3 =
w
2
(4.52)
moment-y:
2L
w − 2L
tcab1 − L
tcab3 = 0 =⇒ tcab1 =
3w
4
(4.53)
force-x:
fpivx = 0
(4.54)
fpivy − 2Wtcab2 /` = 0 =⇒ fpivy = 0
(4.55)
force-y:
force-z:
fpivz − w + tcab1 + tcab3 = 0 =⇒ fpivz = w − tcab1 − tcab3 = w −
w
3w
w
−
=−
2
4
4
(4.56)
5. Sanity check: units, signs, etc. make sense.
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9.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
4.8
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
Multi-body free-body diagrams
Up until now we have treated single rigid bodies, but now we will consider bodies that can be treated as two (or more)
rigid bodies attached with some kind of connection mechanism. When we draw our free body diagrams, we must always
draw at least as many diagrams as we have rigid bodies. In general, we will always draw one free body diagram for every
rigid body in the problem. How do we handle the connections between them? To properly do this, we need to account
for connecting forces. Some examples follow:
Pinned Joints Consider the following two members that are connected using a pin. Consider just the member on the
right: from “its perspective” the pinned joint acts exactly like a pivot, and similarly for the member on
the left. Just like with pivot constraints, we replace this with a force. The difference is that the force
exerted by the left member on the right member must be equal and opposite to the force exerted by
the right member on the left. So, when we draw our FBD, we split it up in the following way:
"
#
fpinx
fpin = fpiny
0
=
+
"
−fpin
−fpinx
= −fpiny
0
#
Note that each member fixes the location of the other, implying that a force is sustained by the pinned
joint. On the other hand, they are free to rotate, so neither beam can exert a moment on the other.
What about when there are more than one object joined with one pin, as in the following example:
fpina
a
c
fpinb
b
fpinc
Here it may be helpful to think of the pin as an additional rigid body, where the forces acting on it are
−fpina , −fpinb , −fpinc . We know by force equilibrium that we must then have
fpina + fpinb + fpinc = 0
(4.57)
Note that this adds an additional equation. Note also that if fpinc were absent, we would recover
fpina = −fpinb , exactly as we expect.
Slider
The slider is analagous to the familiar roller joint: two members are connected with a pin, allowing
motion of one member along the slot.. The free body diagram will be similar to that for the roller,
noting that the forces again must be equal and opposite.
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9.5
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
fsli = fsli nsli
=
+
−fsli
We note that the fixed relative position normal to the slot requires a sustained force normal to the slot.
Free relative position parallel to the slot implies that there is no force parallel to the slot. Free relative
rotation implies that there is no sustained moment. So a slider introduces one force unknown and no
moment unknowns.
Frictionless
contact
Consider two members that have smooth contact with each other and a wide contact area. Keywords:
“smooth” implies that there is free tangential motion, and “wide contact area” implies that rotation is
restricted.
=
fnor = f"nor nnor
#
0
Mct = 0
Mct
"
#
−fnor
0
0
−Mct =
−Mct
As with the slider, the force is strictly normal, which implies that there is only one force unknown.
(Note – we might also require here that the force must be compressive, as the contact clearly could not
contain a tensile load.) The constrained rotation implies that there is a sustained moment, giving one
moment unknown. (Remember that the moment must be equal and opposite as well.)
Fixed
Recall our discussion of “fixed joints” from the previous section.
" #
Mx
z
M = My
Mz
y
x
" #
fx
f = fy
fz
We recall that fixed contact implies the presence of two force unknowns and one moment unknown in
2D, and three force and three moment unknowns in 3D. Consider the following rigid body. We can
think of this body as being split in half, with a “fixed joint” in between them.
=
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+
9.6
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 9
solids.uccs.edu/teaching/mae2103
There are three things worth noting:
(1) Two bodies joined with a fixed joint are identical to a rigid body.
(2) A fixed joint adds the maximum number of unknowns, and splitting a body into two adds the
maximum number of unknowns.
(3) You can always find the internal connecting force/moment in a rigid body by splitting it in two
pieces and solving for the connecting forces.
We will make considerable use of this kind of technique in subsequent sections.
4.9
Multi-body equilibrium
Previously, we always had just one 2D/3D moment and one 2D/3D force equilibrium equation. We will always need at
least as many force/moment equilibrium eqs as we have rigid bodies, but we can be clever about how we choose them:
For instance, consider the following multi-body structure:
Here, we can do equilibrium for the two beams seperately, or we can do equilibrium for the entire combined structure
and one of the members. It generally doesn’t matter, and you can be clever about how you set your problems up.
4.10
Multi-body solution strategy
We will follow a methodical procedure for solving these problems, similarly to the procedure used for problems solved in
previous sections.
1. Determine unknowns: (requested quantities, reaction forces, connecting forces)p
2. Draw FBD for each rigid body
3. Write equations:
ˆ Force equilibrium for each member (and one for each multi-pinned joint, if necessary)
ˆ Moment equilibrium for each member
ˆ Other: Constitutive, geometric, etc.
4. Solve
5. Sanity check
6. Substitute values
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9.7
Lecture 10
4.11
Equilibrium of multiple rigid bodies
Multi-body examples
Example 4.6
Consider the following frame:
L1
L2
`1
f1
θ
`2
f2
Find all of the reaction and connection forces:
1. Unknowns: ffixx , ffixy , Mfixz , fpinx , fpiny , froly
2. FBD
L1
L2
ffix
Mfix
frol
fpin
`1
θ
f1
f2
`2
−fpin
where
"
ffix
ffixx
= ffixy
0
#
"
0
0
M=
Mfixz
#
"
fpinx
fpinx
= fpiny
0
#
"
0
#
"
f1 sin θ
f1 = −f1 cos θ
0
frol = froly
0
#
"
0
f2 = −f2
0
#
(4.58)
3. Equations:
forces:
"
X
# " #
ffixx + fpinx + f sin θ
0
f = ffixy + fpiny − f1 cos θ = 0
0
0
"
X
# " #
−fpinx
0
f = −fpiny − f2 + froly = 0
0
0
moments: take about the fixed joint and then about the pin:
distance vectors:
" #
" #
" #
`1
L1
`2
r1 = 0
rpin = 0
r2 = 0
0
0
0
rrol
(4.59)
" #
L2
= 0
0
(4.60)
moments:
"
M1 =
0
0
#
−`1 f1 cos θ
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"
Mpin
0
0
=
L1 fpiny
#
"
0
0
M2 =
−`2 f2
#
"
Mrol
0
0
=
L2 froly
#
(4.61)
10.1
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 10
solids.uccs.edu/teaching/mae2103
summing the moments for beam 1:
"
X
M = M1 + Mpin + Mfix
# " #
0
0
0
=
= 0
0
−`1 f1 cos θ + L1 fpiny + Mfix
(4.62)
for beam 2:
"
X
M = M2 + Mrol
# " #
0
0
0
=
= 0
0
−`2 f2 + L2 froly
(4.63)
4. Solve:
beam 2 moment z:
froly =
`2
f2
L2
(4.64)
beam 2 force x:
(4.65)
fpinx = 0
beam 2 force y:
−fpiny − f2 + froly = 0 =⇒ fpiny = f2
`2
−1
L2
(4.66)
beam 1 moment z:
−`1 f1 cos θ + L1 fpiny + Mfix = 0 =⇒ Mfix = `1 f1 cos θ − L1 f2
`2
−1
L2
(4.67)
beam 1 force x:
ffixx + fpinx + f1 sin θ = 0 =⇒ ffixx = −f1 sin θ
(4.68)
beam 1 force y:
ffixy + fpiny − f1 cos θ = 0 =⇒ ffixy = f1 cos θ − f2
`2
−1
L2
(4.69)
Example 4.7
Consider the following frame connected using small pulleys and pinned joints. Find all connecting and reaction
forces:
All content © 2016, Brandon Runnels
10.2
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 10
solids.uccs.edu/teaching/mae2103
L
L
b
a
H
c
w
d
e
c
c
c
c
c
c
1. Identify unknowns: fxa , fya , fxb , fyb , fb1x
, fb1y
, fb2x
, fb2y
, fpullx
, fpully
, fxe , fye – 12 unknowns.
2. Free body diagram:
pulleys:
fb
"
0
−w
0
"
" #
w
0
0
−w
0
0
#
#
c
fpull
"
#
c
fb1
c
fb1x
c
= fb1x
0
0
−w
0
beam 1:
" a#
fx
f a = fya
0
"
−fxb
−f = −fyb
0
"
#
b
#
beam 2:
" #
0
fd = w
0
c
fb2
" e#
fx
e
f = fye
0
c
fb2x
c
f
= b2y
0
"
#
3. Write equations:
All content © 2016, Brandon Runnels
10.3
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 10
solids.uccs.edu/teaching/mae2103
force equilibrium – pulleys:
X
" b
#
fx + w
f = fyb − w = 0
0
#
c
fpullx
−w
c
−w =0
f = fpully
0
"
X
force equilibrium – beams:
" a
c #
fx − fxb + fb1x
X
c
fb1 = fya − fyb + fb2x
=0
0
"
X
fb2 =
fye
(4.70)
#
c
fxe + fb2x
c
+ w + fb2y = 0
0
(4.71)
force equilibrium – pinned joint:
#
c
c
c
fb1x
+ fb2x
+ fpullx
c
c
c
= fb1y + fb2y + fpully = 0
0
"
X
f=
c
fb1
+
c
fb2
+
c
fpull
(4.72)
moment equilibrium – pulleys:
=⇒ moment is trivially zero because tensions are equal.
moment equilibrium – beams:
=⇒ choose pivots for both, and compute distance vectors:
" #
" #
"
#
L
L
2L
b
c
d
r = 0
rb1 = 0
r = H/2
0
0
0
so the moments are:

 "
#
0
î
ĵ
k̂
b
0


L
0
0 =
M =
−Lfyb
−fxb −fyb 0
"
# " #
0
î
ĵ
k̂
d
M = L H/2 0 = 0
Lw
0
w
0
giving the equilibrium equations
"
#
0
X
0
Mb1 =
=0
c
2Lfb1y
− Lfyb
"
Mcb1
Mcb2
î
= 2L
c
fb1x
"
î
= 2L
c
fb2x
c
fb1y
ĵ
H
c
fb2y
Mb2 =
2L
= H
0
#
# "
#
0
k̂
0
0 =
c
2Lfb1y
0
# "
#
0
k̂
0
0 =
c
c
2Lfb2y
− Hfb2x
0
ĵ
0
"
X
"
c
rb2
0
0
(4.73)
(4.74)
(4.75)
#
c
c
Lw + 2Lfb2y
− Hfb2x
=0
(4.76)
4. Solve:
pulley force equilibrium:
fxb = −w
fyb = w
c
fpullx
=w
c
fpully
=w
(4.77)
beam 1 moment z:
c
c
2L
fb1y − L
(w ) = 0 =⇒ fb1y =
All content © 2016, Brandon Runnels
w
2
(4.78)
10.4
MAE2103 - Engineering Mechanics I
University of Colorado Colorado Springs
Course Notes - Lecture 10
solids.uccs.edu/teaching/mae2103
pinned joint force y:
c
c
(w /2) + fb2y
+ (w ) = 0 =⇒ fb2y
=−
3w
2
(4.79)
beam 2 moment z:
c
c
Lw + 2L(−3w /2) − Hfb2x
= 0 =⇒ fb2x
=−
2Lw
H
(4.80)
pinned joint force x:
c
fb1x
+
c
fb2x
c
+ fpullx
|{z}
= 0 =⇒
|{z}
−2Lw /H
c
fb1x
2L
=w
−1
H
(4.81)
w
beam 1 force x:
fxa − fxb +
|{z}
−w
c
fb1x
|{z}
= 0 =⇒ fxa = −
w 2L
H
(4.82)
w (2L/H−1)
beam 1 force y:
w
c
fya − fyb + fb2y
= 0 =⇒ fya =
2
|{z} |{z}
−w
(4.83)
−3w /2
beam 2 force x and y:
2Lw
c
fxe + fb2x
= 0 =⇒ fxe =
|{z}
H
−2Lw /H
All content © 2016, Brandon Runnels
w
c
fye + w + fb2y
= 0 =⇒ fye =
2
|{z}
(4.84)
−3w /2
10.5