engineering statics engr 2301 chapter 6 - hcc learning web
TRANSCRIPT
Engineering Statics
ENGR 2301
Chapter 6
Analysis of Structures
Application
6 - 2
Design of support structures requires
knowing the loads, or forces, that each
member of the structure will
experience.
TRUSSES
Application Of Truss
One of the most famous structures in the world, the Eiffel
Tower, is a truss – a series of inter-connected triangles.
Trusses are also used to construct bridges, roofs, floors,
and even the Statue of Liberty.
The Eiffel tower was designed by
Gustave Eiffel and the Statue of
Liberty was designed by Frédéric
Bartholdi, with the help of Gustave
Eiffel who designed the interior
structure.
Truss
A technical definition of a truss is, “A rigid framework
composed of members connected at joints and arranged
into a network of triangles.”
Steven Ressler, Ph.D., U.S. Military Academy at West Point
A truss in architecture and engineering is a timber or metal
structural member that is formed of one triangle or a series of
triangles in a single plane. A truss requires less material than a
solid beam in attaining long spans for carrying heavy loads,
making it especially useful in constructing bridges and roofs.
History Of Truss
Some historians have stated that trusses date back to
18th century Greece, and others argue that there are
13th century structures in Italy that contain trusses.
Introduction
6 - 7
• For the equilibrium of structures made of several
connected parts, the internal forces as well the external
forces are considered.
• In the interaction between connected parts, Newton’s 3rd
Law states that the forces of action and reaction
between bodies in contact have the same magnitude,
same line of action, and opposite sense.
• Three categories of engineering structures are considered:
a) Trusses: formed from two-force members, i.e.,
straight members with end point connections and
forces that act only at these end points.
b) Frames: contain at least one one multi-force
member, i.e., member acted upon by 3 or more
forces.
c) Machines: structures containing moving parts
designed to transmit and modify forces.
ANALYSIS and DESIGN ASSUMPTIONS
When designing both the member and the joints of a truss, first it is necessary
to determine the forces in each truss member. This is called the force analysis
of a truss. When doing this, two assumptions are made:
1. All loads are applied at the joints. The weight of the truss
members is often neglected as the weight is usually small as
compared to the forces supported by the members.
2. The members are joined together by smooth pins. This
assumption is satisfied in most practical cases where the joints
are formed by bolting or welding.
With these two assumptions, the members act as two-
force members. They are loaded in either tension or
compression. Often compressive members are made
thicker to prevent buckling.
Definition of a Truss
6 - 9
• A truss consists of straight members connected at
joints. No member is continuous through a joint.
• Bolted or welded connections are assumed to be
pinned together. Forces acting at the member ends
reduce to a single force and no couple. Only two-
force members are considered.
• Most structures are made of several trusses joined
together to form a space framework. Each truss
carries those loads which act in its plane and may
be treated as a two-dimensional structure.
• When forces tend to pull the member apart, it is in
tension. When the forces tend to compress the
member, it is in compression.
Definition of a Truss
6 - 10
Members of a truss are slender and not capable of
supporting large lateral loads. Loads must be applied at
the joints.
Definition of a Truss
6 - 11
Simple Trusses
6 - 12
• A rigid truss will not collapse under
the application of a load.
• A simple truss is constructed by
successively adding two members and
one connection to the basic triangular
truss.
Analysis of Trusses by the Method of Joints
6 - 13
• Dismember the truss and create a freebody
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, and
opposite sense.
• Forces exerted by a member on the pins or
joints at its ends are directed along the member
and equal and opposite.
• Conditions of equilibrium are used to solve for
2 unknown forces at each pin (or joint), giving a
total of 2n solutions, where n=number of joints.
Forces are found by solving for unknown forces
while moving from joint to joint sequentially.
• Conditions for equilibrium for the entire truss
can be used to solve for 3 support reactions.
ZERO-FORCE MEMBERS
If a joint has only two non-colinear
members and there is no external
load or support reaction at that joint,
then those two members are zero-
force members. In this example
members DE, CD, AF, and AB are
zero force members.
You can easily prove these results by
applying the equations of
equilibrium to joints D and A.
Zero-force members can be
removed (as shown in the
figure) when analyzing the
truss.
ZERO – FORCE MEMBERS (continued)
If three members form a truss joint for
which two of the members are collinear
and there is no external load or reaction at
that joint, then the third non-collinear
member is a zero force member.
Again, this can easily be proven. One can
also remove the zero-force member, as
shown, on the left, for analyzing the truss
further.
Please note that zero-force members
are used to increase stability and
rigidity of the truss, and to provide
support for various different loading
conditions.
Joints Under Special Loading Conditions
6 - 16
• Forces in opposite members intersecting in
two straight lines at a joint are equal.
• The forces in two opposite members are
equal when a load is aligned with a third
member. The third member force is equal
to the load (including zero load).
• The forces in two members connected at a
joint are equal if the members are aligned
and zero otherwise.
• Recognition of joints under special loading
conditions simplifies a truss analysis.
Space Trusses
6 - 17
• An elementary space truss consists of 6 members
connected at 4 joints to form a tetrahedron.
• A simple space truss is formed and can be
extended when 3 new members and 1 joint are
added at the same time.
• Equilibrium for the entire truss provides 6
additional equations which are not independent of
the joint equations.
• In a simple space truss, m = 3n - 6 where m is the
number of members and n is the number of joints.
• Conditions of equilibrium for the joints provide 3n
equations. For a simple truss, 3n = m + 6 and the
equations can be solved for m member forces and
6 support reactions.
Sample Problem 6.1
6 - 18
Using the method of joints, determine
the force in each member of the truss.
SOLUTION:
• What’s the first step to solving this
problem? Think, then discuss this with a
neighbor.
• DRAW THE FREE BODY DIAGRAM
FOR THE ENTIRE TRUSS (always first)
and solve for the 3 support reactions
• Draw this FBD and compare your sketch
with a neighbor. Discuss with each other
any differences.
Sample Problem 6.1
6 - 19
SOLUTION:
• Based on a free body diagram of the entire truss,
solve the 3 equilibrium equations for the reactions
at E and C.
ft 6ft 12lb 1000ft 24lb 2000
0
E
MC
lb 000,10E
xx CF 0 0xC
yy CF lb 10,000 lb 1000 - lb 20000
lb 7000yC
• Looking at the FBD, which “sum of moments”
equation could you apply in order to find one of
the unknown reactions with just this one equation?
• Next, apply the remaining
equilibrium conditions to
find the remaining 2 support
reactions.
Sample Problem 6.1
6 - 20
534
lb 2000 ADAB FF
CF
TF
AD
AB
lb 2500
lb 1500
FDB FDA
FDE 235 FDA CF
TF
DE
DB
lb 3000
lb 2500
• We now solve the problem by moving
sequentially from joint to joint and solving
the associated FBD for the unknown forces.
• Which joint should you start with, and why?
Think, then discuss with a neighbor.
• Joints A or C are equally good because each
has only 2 unknown forces. Use joint A and
draw its FBD and find the unknown forces.
• Which joint should you move to next, and why? Discuss.
• Joint D, since it has 2 unknowns remaining
(joint B has 3). Draw the FBD and solve.
Sample Problem 6.1
6 - 21
• There are now only two unknown member
forces at joint B. Assume both are in tension.
lb 3750
25001000054
54
BE
BEy
F
FF
CFBE lb 3750
lb 5250
375025001500053
53
BC
BCx
F
FF
TFBC lb 5250
• There is one remaining unknown member
force at joint E (or C). Use joint E and
assume the member is in tension.
lb 8750
37503000053
53
EC
ECx
F
FF
CFEC lb 8750
Sample Problem 6.1
6 - 22
• All member forces and support reactions are
known at joint C. However, the joint equilibrium
requirements may be applied to check the results.
checks 087507000
checks 087505250
54
53
y
x
F
F