STATICS OF RIGID BODIES
CHAPTER 1 FUNDAMENTAL CONCEPTS
PRINCIPLES OF STATICS
Mechanics may be defined as the physical science which
describes and predicts the conditions of rest or motion of bodies
under the action of force systems. In other words, where there is
motion or force, there is mechanics.
PRINCIPLES OF STATICS
ENGINEERING MECHANICS
It is defined as the science which considers the effects of forces
on rigid bodies.
RIGID BODY – It is defined as a definite amount of matter the parts of which
are fixed in position relative to each other.
 Rigid body – A body whose relative movement between its parts are negligible
relative to the gross motion of the body.
 Non-rigid body – A body whose relative movement between its parts are
significant relative to the gross motion of the body. Mechanics of the
deformable material.
PRINCIPLES OF STATICS
In engineering, mechanics is generally based on
Newton’s Laws and is often called Newtonian (or
Classical) Mechanics after the English scientist Sir Isaac
Newton (1642-1727).
ENGINEERING
MECHANICS
STATICS
FORCE SYSTEMS
DYNAMICS
APPLICATIONS
KINEMATICS
KINETICS
CONCURRENT
TRUSSES
TRANSLATION
TRANSLATION
PARALLEL
CENTROIDS
ROTATION
ROTATION
NON-CONCURRENT
FRICTION
PLANE MOTION
PLANE MOTION
PRINCIPLES OF STATICS
STATICS – It is the study of materials at rest. The actions of
all external forces acting on such materials are exactly
counterbalanced and there is a zero net force effect on
the material: such materials are said to be in a state of
static equilibrium.
DEFINITION OF TERMS
 RIGID BODY – rigid body is one which does not deform, in other words the
distance between the individual particles making up the rigid body remains
unchanged under the action of external forces.
 FORCE – A force represents the action of one body on another. Force can
be generated either by the direct contact of bodies or by their effect at a
distance. Forces always occur in pairs. Forces of a pair are always equal in
magnitude and opposite in direction. >units: kip-1000 pounds or ton- 2000 pounds
 Characteristics of forces:
Magnitude
Position of its line of action
Direction in which the force acts along its line of action
DEFINITION OF TERMS
 Principle of Transmissibility – states that the external effect of a
force on a body is the same for all points of application along its
line of action; it is independent of the point of application. While,
internal effect of a force, however, is definitely dependent on its
point of application.
 Space – The region occupied by bodies. Their positions and
orientations can be described by linear and angular measurements
relative to a specified coordinate system.
DEFINITION OF TERMS
 Time – The measure of the succession of events. Often, the
change in the physical quantities are described with respect
to time
 Mass – The measure of the inertia of a body, which indicates
the resistance to a change in its velocity.
 Particle – A body of negligible dimensions. This is a relative
matter to the surrounding effect. Hence, rotation effect is
insignificant.
STATICS
FORCE SYSTEMS
APPLICATIONS
CONCURRENT,
PARALLEL, NONCONCURRENT
TRUSSES, CENTROIDS,
FRICTION
FORCE SYSTEMS
 Force system is any arrangement where two or more forces act on a
body or on a group of related bodies. When the lines of action of all the
forces in a force system line in one plane, they referred to as being
coplanar; otherwise they are non-coplanar. These are classified
according to their lines of action;
CONCURRENT – forces whose lines of action pass through a common
point
PARALLEL – those in which the lines of action are parallel
NON-CONCURRENT – those in which the lines of action neither are
parallel nor intersect in a common point
FREE BODY DIAGRAMS
 It is one of the most important concepts in mechanics.
 It is introduced to help the student distinguish between action and
reaction forces.
 To do so, it is necessary to isolate the body being considered.
 A sketch of the isolated body which shows only the forces acting
upon the body is defined as a free-body diagrams.
SCALAR & VECTOR QUANTITY
 SCALAR QUANTITY
Physical quantities that can be completely described by a single
real number such as length, area, volume and mass.
These quantities are describable by giving only a magnitude.
 VECTOR QUANTITY
Physical quantities such as displacement, velocity, force and
acceleration require both a magnitude and a direction are
called vector quantities.
SCALAR & VECTOR QUANTITY
Determine whether a scalar quantity, a vector quantity or neither
would be appropriate to describe each of the following situations.
a. The outside temperature is 15º C.
b. A truck is traveling at 60 km/hr.
c. The water is flowing due north at 5 km/hr.
d. The wind is blowing from the south.
e. A vertically upwards force of 10 Newton is applied to a rock.
f. The rock has a mass of 5 kilograms.
SCALAR & VECTOR QUANTITY
g. The box has a volume of .25 m³.
h. A car is speeding eastward.
i. The rock has a density of 5 gm/cm³.
j. A bulldozer moves the rock eastward 15m.
k. The wind is blowing at 20 km/hr from the south.
l. A stone dropped into a pond is sinking at the rate of 30 cm/sec.
SCALAR & VECTOR QUANTITY
 Vectors are determined by both magnitude and direction, they are
represented geometrically in 2 or 3 dimensional space as directed
line segments or arrows. The length of the arrow corresponds to the
magnitude of the vector while the direction of the arrow
corresponds to the direction of the vector.
PARALLELOGRAM LAW
 The method of vector addition is based on what is known as the
parallelogram law.
 The parallelogram law cannot be proved; it can only be
demonstrated by experiment.
TRIANGLE LAW
 If two forces are represented by their free vectors placed tip to tail,
their resultant vector is the third side of the triangle, the direction of
the resultant being from the tail of the first vector to the tip of the
last vector.
SOLUTION OF PROBLEMS (TECHNIQUE)
1. You should acquire the ability to organize your work in a neat and orderly
manner.
2. Identify or understand the problem then, start constructing a neat diagram
of the quantities involved.
3. Write the given legibly and other information required to answer the
problem.
4. Write out the equations you need to use before substituting.
5. If you were not able to come up with a correct solution, try to do dimensional
checking and conversion of units if applicable.
-dimensional checks means that the units on each side must be on the
same form
UNITS OF MEASUREMENT
 We’ll stick to MKS (SI) units in this course
MKS: meters; kilograms; seconds
As opposed to cgs: centimeter; gram; seconds
 Distance in meters (m)
1 meter is close to 40 inches
 Mass in kilograms (kg)
1 kg is about 2.2 pounds
 Time in seconds (s)
STATICS OF RIGID BODIES
CHAPTER 2 RESULTANT OF FORCE SYSTEMS
SIGN CONVENTION
Sign of Fx
Sign of Fy
Direction of F with
Respect to Origin O
Diagram
Y
+
+
Up to right
F
𝜽𝑿
O
X
O
+
-
-
+
Down to right
Up to left
Y
-
F
Y
F
𝜽𝑿
X
-
X
𝜽𝑿
𝜽𝑿
Down to left
F
O
O
Y
X
RESULTANT OF CONCURRENT FORCE
SYSTEM
Resultant of a force system is a force or a couple that will have the same
effect to the body, both in translation and rotation, if all the forces are
removed and replaced by the resultant.
The equation involving the resultant of force system are the following
1. Rx = ΣFx = F𝑥1 +F𝑥2 + F𝑥3 +...Rx = ΣFx = F𝑥1 +F𝑥2 + F𝑥3 +...
The x-component of the resultant is equal to the summation of
forces in the x-direction.
RESULTANT OF CONCURRENT FORCE
SYSTEM
2. Ry = Σfy = F𝑦1 +F𝑦2 + F𝑦3 +… Ry = Σfy = F𝑦1 +F𝑦2 + F𝑦3 +...
The y-component of the resultant is equal to the summation of
forces in the y-direction.
3. Rz = ΣFz = F𝑧1 +F𝑧2 + F𝑧3 +... Rz = ΣFz = F𝑧1 +F𝑧2 + F𝑧3 +...
The z-component of the resultant is equal to the summation of
forces in the z-direction.
CONCURRENT FORCE SYSTEM
 The line of action of each forces in coplanar concurrent force system
are on the same plane. All of these forces meet at a common point,
thus concurrent. In x-y plane, the resultant can be found by the
following formulas:
RESULTANT OF SPATIAL CONCURRENT
FORCE SYSTEM
 Spatial concurrent forces (forces in 3-dimensional space) meet at a
common point but do not lie in a single plane. The resultant can be
found as follows:
DIRECTION COSINES
CONCURRENT FORCE SYSTEM
 EXAMPLE 1
 Three ropes are tied to a small metal ring. At the end of each rope
three students are pulling, each trying to move the ring in their
direction. Find the net force in the ring due to the three applied forces.
CONCURRENT FORCE SYSTEM
 EXAMPLE 2
Find the resultant vector of vectors A and B shown below.
CONCURRENT FORCE SYSTEM
 EXAMPLE 3
Three vectors A, B, and C are shown in the figure below. Find one
vector (magnitude and direction) that will have the same effect as
the three vectors shown below.
CONCURRENT FORCE SYSTEM
 EXAMPLE 4
P is directed at an angle α from x-axis and the 200 N force is acting at a slope of 5
vertical to 12 horizontal.
1. Find P and α if the resultant is 500 N to the right along the x-axis.
2. Find P and α if the resultant is 500 N upward to the right with a slope of 3 horizontal to 4
vertical.
CONCURRENT FORCE SYSTEMS
 EXAMPLE 5
Forces F, P, and T are concurrent and acting in the direction as shown
below
a.Find the value of F and 𝛼 if T = 450 N, P = 250
N, β = 30°, and the resultant is 300 N acting up
along the y-axis.
a.Find the value of F and 𝛼 if T = 450 N, P = 250
N, β = 30° and the resultant is zero.
SPATIAL FORCE SYSTEM
y
Resultant of three mutually
perpendicular forces.
𝐹=
Fy
𝜃y
𝜃x
O
Fz
z
F
𝜃z
𝐹𝑋2 + 𝐹𝑌2 + 𝐹𝑍 ²
(1)
𝐹𝑋 = 𝐹𝑐𝑜𝑠𝜃𝑋
Fx
x
𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦
𝐹𝑧 = 𝐹𝑐𝑜𝑠𝜃𝑧
(2)
SPATIAL FORCE SYSTEMS
Y
We will use a right handed coordinate system to
develop the theory of vector algebra that follows.
X
Z
𝑦
Cartesian Vector Representation
𝐴𝑦
𝐴 = 𝐴𝑋 + 𝐴𝑌 + 𝐴𝑍
Magnitude of Cartesian
Vector
𝑌
𝐴𝑋
𝑥
𝐴=
𝐴𝑧
𝑗
𝑧
𝑘
𝑍
𝑋
𝑖
𝐴²𝑋 + 𝐴²𝑦 + 𝐴²𝑣
SPATIAL FORCE SYSTEM
(+)
y
Fz
Fy
Fx
B
F = 300 lb
5
Fy
Fx
(3,5,-3)
O
1
A (-2,1,3)
z
(+)
𝑥2 + 𝑦2 + 𝑧2
𝑑=
52 + 42 + 62
𝒅 = 𝟖. 𝟕𝟖
𝐹𝑋 𝐹𝑌 𝐹𝑍 𝐹
=
=
=
𝑥
𝑦
𝑧
𝑑
(3)
(4)
𝐹𝑋 𝐹𝑌 𝐹𝑍 300
=
=
=
5
4
6
8.78
3
(+)
2
Fz
3
3
𝑑=
x
𝑭𝑿 = 𝟏𝟕𝟏 𝒍𝒃;
The components of a force as determined by the
coordinates of two points on its line of action.
𝑭𝒚 = 𝟏𝟑𝟕 𝒍𝒃;
𝑭𝒛 = −𝟐𝟎𝟓 𝒍𝒃
SAMPLE PROBLEM
y
 CE BOARD MAY 2014
A
A flag pole is supported by tension
wires AB, AC, and AD to resist an
uplift force of 570 N acting on the axis
of the pole. In this problem, a=3.5 m,
b=1.5m, c=4m, d=4.5m. L=6m.
D
L
a
B
b
z
c
O
d
C
x
Find the forces from AB, AC and AD
PARALLEL FORCE SYSTEMS
 Parallel forces can be in the same or in opposite directions. The sign
of the direction can be chosen arbitrarily, meaning, taking one
direction as positive makes the opposite direction negative. The
complete definition of the resultant is according to its magnitude,
direction,
and
line
of
action.
Moment of a Force
 The moment of a force about an axis line is the measure of its ability
to produce turning or twisting about the axis.
Y
F
d
O
The distance d is frequently called
the moment arm of the force.
Y
Units:
(lb-ft)²
lb-in
gram-cm
ton-ft
(+) clockwise rotation
(-) counterclockwise rotation
Varignon’s Theorem
 It was originally developed by the French mathematician Pierre
Varignon (1654-1722).
 It states that the moment of a force about a point is equal to the sum
of the moments of the components of the force about the point.
 This theorem can be proven easily using the vector cross product
since the cross product obeys the distributive law.
Since, F = F1 + F2
 We can use the principle of moments by
resolving the force into its rectangular
components and then determine the moment
using a scalar analysis.
COPLANAR PARALLEL FORCE SYSTEM
 EXAMPLE 1
A parallel force system acts on the lever shown in the figure.
Determine the magnitude and position of the resultant.
 SOLUTION:
𝑅 = σ𝐹
𝑅 = −30 − 60 + 20 − 40
𝑅 = 110
𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑
𝑀𝐴 = ෍ 𝑥 𝐹
= 30 2 + 60 5 − 20 7 + 40 11
= 660 𝑓𝑡 • 𝑙𝑏 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑀𝐴 = 𝑅𝑑
660 = 110𝑑
𝑑 = 6 𝑓𝑡
𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝐴
Thus, R = 110 lb downward at 6 ft to the right of A.
Applications of the Principle Moments
Y
Fy
x
Fy
𝜃
A
Fx
d
𝑦
𝑖𝑦
Fx
O
𝑖𝑥
Fx
+
B
Fy
C
F
 We can use the principle of
moments by resolving the
force into its rectangular
components
and
then
determine the moment using
a scalar analysis.
X
+ 𝑀𝑂 = 𝐹𝑋 ∙ 𝑖𝑦
+ 𝑀𝑂 = 𝐹𝑦 ∙ 𝑖𝑥
Sample Problem
 Assuming clockwise moments as positive, compute the moment of force F = 450
lb and of force P = 361 lb about points A, B, C and D.
A
F
C
1’
P
D
B
Applying the principle that the
moment of a force is equal to
the moment sum of its
components, we have
P
d
A
𝜃
C
B
𝑷𝑿
𝑷𝒀
Moment of a force in terms of its components
+ 𝑀𝑂 = 𝑃 ∙ 𝑑 = 𝑃𝑦 ∙ 𝐴𝐵
Resultant of Distributed Loads
The resultant of a distributed load is equal to the area of the load
diagram. It is acting at the centroid of that area as indicated.
Rectangular Load
𝑅 = 𝑤𝑜 𝐿
Triangular Load
1
𝑅 = 𝑤𝑜 𝐿
2
Trapezoidal Load
1
𝑅 = 𝑤𝑜1 𝐿 + (𝑤𝑜2 − 𝑤𝑜1 )𝐿
2
Sample Problem
Determine the moment of a force about point O.
Sample Problem
Determine the moment of the force about point O.
Cartesian Vector Formulation
y
y
z
z
x
z
Right Hand Rule for Moment
y
j
i
k
z
xx
jxk=i
k x j = -i
ixi=0
kxi=j
j x i = -k
jxj=0
ixj=k
i x k = -j
kxk=0
Resultant Moment of a System of Forces
𝒓𝒙 , 𝒓𝒚 , 𝒓𝒛 represent the x, y, z components of the position vector drawn from point O to any
point on the line of action of the force
F, F, F
x
y
z
represent the x, y, z components of the force vector
If a body is acted upon by a system of forces, the resultant moment of the
forces about point O can be determined by vector addition of the moment
of each force. This resultant can be written symbolically as,
(𝑴𝑹 )𝒐 = ෍( 𝒓 𝒙 𝑭)
Sample Problem
y
x
z
 Two forces act on the rod shown in the
figure. Determine the resultant moment
they create about the flange at O. Express
the result as a Cartesian vector.
Sample Problem
A 200 lb force F
passes
through
points A to point
B. Compute the
moment of force
F about each
coordinate axis.
y
2’
B
4’
4’
F = 200 lb
3’
x
C
6’
z
A
Sample Problem
y
A force of 100 lb is
directed from A
toward B in the
cube. Determine
the moment of the
force about each
of the coordinate.
F
A
C
4’
D
B
4’
O
z
E
4’
x
4’
Couples
A couple is defined as two parallel forces that have the
same magnitude, but opposite directions, and are
separated by a perpendicular distance d.
F
σ 𝑀𝐴 = 𝐹 ∙ 𝑑
d
A
a
B
σ 𝑀𝐴 = 𝐹 𝑑 + 𝑎 − 𝐹 ∙ 𝑎 = 𝐹 ∙ 𝑑
𝐶 = 𝐹 ∙ 𝑑
F
IMPORTANT POINTS
 A couple moment is produced by two non-collinear forces that are equal
in magnitude but opposite in direction. Its effect is to produce pure
rotation, or tendency for rotation in a specified direction.
 A couple moment is a free vector, and as a result it causes the same
rotational effect on a body regardless of where the couple moment is
applied to the body.
 The moment of the two couple forces can be determined about any
point. For convenience, this point is often chosen on the line of action of
one of the forces in order to eliminate the moment of this force about the
point.
ASSIGNMENT
80 lb
80 lb
100 lb
100 lb
100 lb
 Determine the resultant
moment about point A
of the system of forces
shown in the figure.
Each square is 1 ft on a
side.
200 lb
A
EQUILIBRIUM OF FORCE SYSTEM
EQUILIBRIUM is the term used to designate the
condition where the resultant of a system is zero.
The physical meaning of EQUILIBRIUM, as applied
to a body, is that the body either is at rest or is
moving in a straight line with constant velocity.
How to solve equations of equilibrium
successfully?
Identify all the known and unknown external forces
that act on the body.
Draw a free-body diagram.
A thorough understanding of how to draw a free-body
diagram is of primary importance for solving problems in
mechanics.
Support Reactions
A support prevents the translation of a body in a given
direction by exerting a force on the body in the opposite
direction.
A support prevents the rotation of a body in a given
direction by exerting a couple moment on the body in
the opposite direction.
Types of Support
Actual types of support
SAMPLE FBD
Weight and the Center of
Gravity.
When a body is within a
gravitational field, then
each of its particles has
a specified weight.
SAMPLE FBD
A bolted connection at A will allow
for any slight rotation that occurs
here when the load is applied, and
so a pin can be considered for this
support.
At B a roller can be considered
since this support offers no
resistance to horizontal movement.
SAMPLE PROBLEM
Draw the free-body diagram of the uniform beam shown
in the figure. The beam has a mass of 100 kg.
Equilibrium of Concurrent Force System
2
𝑅=
(෍ 𝑋) + (෍ 𝑌)
෍𝑋 = 0
෍𝑌 = 0
2
SAMPLE PROBLEM
The cable and boom shown in the figure support a load
of 600 lb. Determine the tensile force T in the cable and
the compressive force C in the boom.
30°
45°
600 lb
SAMPLE PROBLEM
𝑊
𝑃
𝜃
30°
A 300 lb box is held at rest
on a smooth plane by a
force P inclined at an
angle 𝜃 with the plane as
shown in the figure. If
𝜃 =45°, determine the vale
of P and the normal
pressure N exerted by the
plane.
Equilibrium of bodies acted upon by
coplanar non-concurrent force system
෍𝑋 = 0
෍𝑌 = 0
෍ 𝑀𝑜 = 0
Sample Problem
Determine the horizontal
and vertical components
of reaction on the beam
caused by the pin at B
and the rocker at A as
shown in the figure.
Neglect the weight of the
beam.
Sample Problem
The member shown in
the
figure
is
pin
connected at A and
rests against a smooth
support at B. Determine
the
horizontal
and
vertical components of
reaction at the pin A.
Sample Problem
Compute the total
reaction at A and B
on the truss shown
in the figure.
2240 lb
2000 lb
90°
𝜃
20’
𝜃
20’
20’
1000 lb
20’
20’
2000 lb
3000 lb
Sample Problems
600 lb
240 lb
E
D
16’
12’
400 lb
B
A
C
12’
12’
The truss shown in the figure
is supported on rollers at A
and a hinge at B. Solve for
the components of the
reactions. Compute the
total reactions at A and B.
Sample Problem
Determine the horizontal
and vertical components
of
reaction
on
the
member at the pin A, and
the normal reaction at the
roller B in the figure.
Sample Problem
The box wrench in the
figure is used to tighten
the bolt at A. If the
wrench does not turn
when the load is applied
to the handle, determine
the torque or moment
applied to the bolt and
the force of the wrench
on the bolt.
Sample Problem
150 lb/ft
120 lb
60 lb/ft
3’
𝑅1
6’
Determine the reactions for
the beam loaded as shown in
the figure.
6’
𝑅2
ASSIGNMENT
1000 lb
2000 lb
The fink truss shown in
the
figure
is
supported by a roller
at A and a hinge at
B. the given loads are
normal
to
the
inclined
member.
Determine
the
reactions at A and B.
2000 lb
2000 lb
1000 lb
30°
60°
60°
60’
30°
ASSIGNMENT
Determine the amount
and direction of the
smallest force P required
to start the wheel in the
figure over the block.
What is the reaction at
the block?
P
6’’
𝛼
2’
2000 lb
30°
STATICS OF RIGID BODIES
CHAPTER 4
ANALYSIS OF STRUCTURES
CHAPTER OBJECTIVE
To show how to determine the forces in the
members of a truss using the method of joints
and the method of sections.
To analyze the forces acting on the
members
of
frames
and
machines
composed of pin-connected members.
ANALYSIS OF STRUCTURE
Analysis of structure is the process by which
we determine how the loads applied to a
structure are distributed throughout the
structure.
TYPES OF ELEMENTARY STRUTURES
Two types of structures will be studied;
Pin-connected trusses and,
Pin-connected frames.
WHAT IS A TRUSS?
A truss is a structure composed of slender
members joined together at their end points.
The
members
commonly
used
in
construction consist of wooden struts or
metal bars.
WHAT IS A TRUSS?
The purpose of a truss is to support a larger
load or span a greater distance than any
individual member from which the truss may
be built.
SIMPLE TRUSS
 A simple truss consists of
triangular
elements
connected
together
by
pinned joints. The forces
within its members can be
determined by assuming the
members are all two-force
members,
connected
concurrently at each joint.
The members are either in
tension or compression, or
carry no force.
In the case of a bridge, such as shown in the figure, the
load on the deck is first transmitted to stringers,
then to floor beams, and finally to the joints of the two
supporting side trusses.
Assumption for Design
All loadings are applied at the joints.
The members are joined together by smooth pins.
If the force tends to
elongate the member, it is
a tensile force (T)
whereas if it tends to
shorten the member, it is a
compressive force (C)
Frames and Machines
Frames and machines are structures that contain one or
more multi force members, that is, members with three or
more forces or couples acting on them. Frames are
designed to support loads, and machines transmit and
alter the effect of forces.
The forces acting at the joints of a
frame or machine can be determined
by drawing the free-body diagrams of
each of its members or parts.
NATURE OF SUPPORTS AND THEIR
REACTIONS
Statically determinate structures
These are structures that can be analyzed using statics alone. If
the number of unknown forces is equal or less than he number
of equations, the structure is said to be statically determinate.
Statically indeterminate structures
These structures indicate that there’s at least one more
unknown reaction force than there are equations of equilibrium,
meaning that the sum of the forces and moments in each
direction is equal to zero.
Structural Stability/ Instability
Static Indeterminacy of Structures – External and Internal
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠(𝑟) = 3
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠(𝑒) = 3
𝒓=𝒆
∆ 𝑺𝑻𝑨𝑻𝑰𝑪𝑨𝑳𝑳𝒀 𝑫𝑬𝑻𝑬𝑹𝑴𝑰𝑵𝑨𝑻𝑬 (𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍)
𝐼𝑓 𝒓 > 𝒆: 𝑺𝒕𝒂𝒕𝒊𝒄𝒂𝒍𝒍𝒚 𝑰𝒏𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒆
Structural Stability/ Instability
Static Indeterminacy of Structures – External and Internal
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑒𝑚𝑏𝑒𝑟(𝑚) = 9
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑜𝑖𝑛𝑡𝑠 (𝑗) = 6
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 (𝑅) = 3
∆ 𝒎 + 𝑹 = 𝟐𝒋
∆ 𝑺𝑻𝑨𝑻𝑰𝑪𝑨𝑳𝑳𝒀 𝑫𝑬𝑻𝑬𝑹𝑴𝑰𝑵𝑨𝑻𝑬 (𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍)
Structural Stability/ Instability
Static Indeterminacy of Structures – External and Internal
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑒𝑚𝑏𝑒𝑟(𝑚) = 10
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑜𝑖𝑛𝑡𝑠 (𝑗) = 6
 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 (𝑅) = 3
∆ 𝒎 + 𝑹 > 𝟐𝒋
∆ 𝑺𝑻𝑨𝑻𝑰𝑪𝑨𝑳𝑳𝒀 𝑰𝑵𝑫𝑬𝑻𝑬𝑹𝑴𝑰𝑵𝑨𝑻𝑬 (𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍)
Structural Stability/ Instability
Static Indeterminacy of Structures – External and Internal
m + R = 2j
Statically Determinate Internally
m + R > 2j
Statically Indeterminate Internally
m + R < 2j
Unstable Truss
Zero Force Members
 If
only
two
non-collinear
members form a truss joint and
no external load or support
reaction is applied to the joint,
the two members must be zero
force members.
Zero Force Members
If three members form a truss
joint for which two of the
members are collinear, the
third member is a zero-force
member provided no external
force or support reaction is
applied to the joint.
Analysis Method; METHOD OF JOINTS
 Start with any joint where at least one known load exists and
where not more than two unknown forces are present.
METHOD OF JOINTS
Is the truss statically
determinant externally?
Is the truss statically
determinant internally?
Are there any Zero Force
Members in the truss?
METHOD OF JOINTS
𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒𝑠;
CH, GH, CG, FG, CF
CD, DF, EF, DE
Assignment
Determine the force
in each member of
the Pratt truss, and
state if the members
are in tension or
compression.