Structures
Chapter 4
ME 108 - Statics
Outline
• Applications
• Simple truss
• Method of joints
• Method of section
Germany
Tacoma Narrows Bridge http://video.google.com/videoplay?docid=-323172185412005564&q=bruce+lee&pl=true
UK
Applications
Trusses are commonly used to
support a roof.
For a given truss geometry and
load, how can we determine the
forces in the truss members and
select their sizes?
A more challenging question is
that for a given load, how can
we design the trusses’ geometry
to minimize cost?
Applications
Trusses are also used in a variety
of structures like cranes and the
frames of aircraft or space
stations.
How can we design a light weight
structure that will meet load,
safety, and cost specifications?
Simple Trusses
• A truss is a structure composed of slender members
joined together at their end points.
• Joint connections are formed by bolting or welding the
ends of the members to a common plate, called a gusset
plate, or by simply passing a large bolt or pin through
each of the members
Simple Trusses
Planar Trusses
• Planar trusses lie on a single plane and are used to support
roofs and bridges
• The truss ABCD shows a typical roof-supporting truss
• Roof load is transmitted to the truss at joints by means of a
series of purlins, such as DD’
• The analysis of the forces developed in the truss members is 2D
Assumptions for Design
• 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 loadings are applied at the joint”
– Assumption true for most applications of bridge and
roof trusses
– Weight of the members neglected since forces
supported by the members are large in comparison
– If member’s weight is considered, apply it as a
vertical force, half of the magnitude applied at each
end of the member
Assumptions for Design
2.“The members are joined together by smooth pins”
– Assumption true when bolted or welded joints are
used, provided the center lines of the joining
members are concurrent
Assumptions for Design
• Each truss member acts as a two force member,
therefore the forces at the ends must be directed
along the axis of the member
• If the force tends to elongate the member, it is a
tensile force
• If the force tends to shorten the member, it is a
compressive force
• Important to state the nature of the force in the
actual design of a truss – tensile or compressive
• Compression members must be made thicker
than tensile member to account for the buckling
or column effect during compression
Simple Truss
• To prevent collapse, the form of a truss
must be rigid
• The four bar shape ABCD will collapse
unless a diagonal member AC is added
for support
• The simplest form that is rigid or stable is
a triangle
• A simple truss is constructed starting with
a basic triangular element such as ABC
and connecting two members (AD and
BD) to form an additional element. For
these trusses, the number of members
(M) and the number of joints (J) are
related by the equation M = 2 J – 3.
Method of Joints
Procedure for Analysis
Truss analysis by the method of joints: 1) draw FBDs for all joints in the structure.
2) apply equations of equilibrium to each joint.
3) solve for unknowns.
Method of Joints
• Before beginning, it is usually necessary to draw a
free-body diagram of the entire truss (i.e. treat the
truss as a single object) & determine the reactions
at its supports:
– E.g. consider the Warren truss which has
members 2 m in length & support loads at B & D
Method of Joints
– From the equilibrium equations:
Σ Fx = Ax = 0
Σ Fy = Ay + E 400 N 800 N = 0
Σ Mpoint A = (1 m)(400 N) (3 m)(800 N) + (4 m)E = 0
We obtain the reactions:
Ax = 0, Ay = 500 N & E = 700 N
• The next step is to choose a joint & draw its free-
body diagram:
– Isolate joint A by cutting members AB & AC
Method of Joints
Method of Joints
– The terms TAB & TAC are the axial forces in
members AB & AC, respectively
– Although the directions of the arrows
representing the unknown axial forces can be
chosen arbitrarily, notice that we have chosen
them so that a member is in tension if we
obtain a positive value for the axial force
– Consistently choosing the directions in this
way helps avoid errors
Method of Joints
– The equilibrium equations for joint A are:
Σ Fx = TAC + TAB cos 60°= 0
Σ Fy = TAB sin 60° + 500 N = 0
Solving these equations, we obtain the axial
force TAB = 577 N & TAC = 289 N
– Member AB is in compression & member AC
is in tension
Method of Joints
• Next, obtain a free-body diagram of joint
B by cutting members AB, BC & BD:
– From the equilibrium equations for joint B:
Σ Fx = TBD + TBC cos 60° + 577 cos 60° = 0
Σ Fy = 400 N + 577 sin 60° TBC sin 60° = 0
We obtain TBC = 115 N & TBD = 346 N
– Member BC is in tension & member BD is in compression
Method of Joints
• By continuing to draw free-body diagrams of the
joints, we can determine the axial forces of all
the members
• In 2 dimensions, you can obtain only 2
independent equilibrium equations from the free-
body diagram of a joint
• Summing the moments about a point does not
result in an additional independent equation
because the forces are concurrent
Method of Joints
• Therefore when applying the method of joints, you should choose joints to analyze that are subjected to no more than 2 unknown forces:
– In our example, we analyzed joint A first because it was subjected to the known reaction exerted by the pin support & 2 unknown forces, the axial forces TAB & TAC
– We could then analyze joint B because it was subjected to 2 known forces & 2 unknown forces, TBC & TBD
– If we had attempted to analyze joint B first, there would have been 3 unknown forces
Method of Joints
• When determining the axial forces in the
members of a truss, it will be simpler if you are
familiar with 3 particular types of joints:
1.Truss joints with 2 collinear members & no
load: the sum of the forces must equal zero,
T1 = T2. The axial forces are equal.
Method of Joints
2.Truss joints with 2 noncollinear members
& no load: because the sum of the forces in
the x direction must equal zero, T2 = 0.
therefore T1 must also equal zero. The axial
forces are zero.
Method of Joints
3.Truss joints with 3 members, 2 of which
are collinear & no load: because the sum of
the forces in the x direction must equal zero,
T3 = 0. The sum of the forces in the y direction
must equal zero, so T1 = T2. The axial forces in
the collinear members are equal & the axial
force in the 3rd member is zero.
A member that supports no force is called a zero-
force member.
Analysis of trusses is simplified if we can easily
identify (such as by inspection) any zero force
members.
Zero force members
zero force member
not a zero force
member
example: identify any zero force members.
A
B
D
F
H
J
L
C E G I K
A
B
D
F
H
J
L
C E G I K
example: identify any zero force members.
Example 1
Determine the axial forces in the members of the truss in Figure.
Strategy
1st, draw a free-body diagram of the entire truss, treating it as a single object & determine the reactions at the supports. Then apply the method of joints, simplifying the task by identifying any special joints
Example 1
Solution
Determine the Reactions at the Supports:
Draw the free-body diagram of the entire truss:
Example 1
Solution
From the equilibrium equations:
Σ Fx = Ax + B = 0
Σ Fy = Ay 2 kN = 0
Σ Mpoint B = (6 m) Ax (10 m)(2 kN) = 0
We obtain the reactions Ax = 3.33 kN, Ay = 2 kN &
B = 3.33 kN.
Example 1
Solution
Identify Special Joints:
Because joint C has 3 members, 2 of which are
collinear & no load, the axial force in member BC is
zero, TBC = 0 & the axial forces in the collinear
members AC & CD are equal, TAC = TCD.
Draw Free-Body Diagrams of the Joints:
We know the reaction exerted on joint A by the
support & joint A is subjected to only 2 unknown
forces, the axial forces in members AB & AC.
Example 1
Solution
Example 1
Solution
The angle = arctan (5/3) = 59.0°
The equilibrium equations for joint A are:
Σ Fx = TAC sin 3.33 kN = 0
Σ Fy = 2 kN TAB TAC cos = 0
Solving these equations, we obtain TAB = 0 &
TAC = 3.89 kN.
Example 1
Solution
Now draw the free-body diagram of joint B:
Example 1
Solution From the equilibrium equation:
Σ Fx = TBD + 3.33 kN = 0
We obtain TBD = 3.33 kN. The negative sign
indicates that member BD is in compression.
The axial forces in the members are:
AB: 0
AC: 3.89 kN in tension (T)
BC: 0
BD: 3.33 kN in compression (C)
CD: 3.89 kN in tension (T)
Example 1
Critical Thinking
• Observe how our solution was simplified by
recognizing that joint C is the type of special joint
with 3 members, 2 of which are collinear & no
load:
– This allowed us to determine the axial forces
in all members of the truss by analyzing only
2 joints
Example 2
Each member of the truss in Figure will safely
support a tensile force of 10 kN & a compressive
force of 2 kN. What is the largest downward load F
that the truss will safely support?
Example 2
Strategy
This truss is identical to the one we analyzed in
Example 1. By applying the method of joints in the
same way, the axial forces in the members can be
determined in terms of the load F. The smallest
value of F that will cause a tensile force of 10 kN or
a compressive force of 2 kN in any of the members
is the largest value of F that the truss will support.
Example 2
Solution By using the method of joints in the same way as
in Example 1, we obtain the axial forces:
AB: 0
AC: 1.94F (T)
BC: 0
BD: 1.67F (C)
CD: 1.94F (T)
Example 2
Solution
For a given load F, the largest tensile force is 1.94F
(in members AC & CD) & the largest compressive
force is 1.67F (in member BD).
The largest safe tensile force would occur when
1.94F = 10 kN or when F = 5.14 kN.
The largest safe compressive force would occur
when 1.67F = 2 kN or when F = 1.20 kN.
Therefore, the largest load F that the truss will safely
support is 1.20 kN.
Example 2
Critical Thinking
• This example demonstrates why engineers
analyze structures:
– By doing so, they can determine the loads
that an existing structure will support or
design a structure to support given loads
– In this example, the tensile & compressive
loads the members of the truss will support
are given
Example 2
Critical Thinking
– Information of that kind must be obtained by
applying the methods of mechanics of
materials to the individual members
– Then statics can be used, as we have done in
this examples, to determine the axial loads in
the members in terms of the external loads on
the structure
Example 3
Determine the force in each member of the
truss and indicate whether the members are
in tension or compression.
Example 4
Determine the force in each member of the
truss and indicate whether the members are
in tension or compression.
Example 5
Example
The Method of Sections
• When we need to know the axial forces only in certain members of a truss, we often can determine them more quickly using the method of sections than the method of joints
• E.g. consider the Warren truss we used for the method of joints:
– It supports loads at B & D & each member is 2 m in length
– Suppose we need to determine only the axial force in member BC
The Method of Sections
– Just as in the method of joints, we begin by
drawing a free-body diagram of the entire
truss & determining the reactions at the
supports:
– The next step is to cut the members AC, BC &
BD to obtain a free-body diagram of a part, or
a section, of the truss:
The Method of Sections
– Summing moments about point B,
the equilibrium equations for the
section are:
Σ Fx = TAC + TBD + TBC cos 60° = 0
Σ Fy = 500 N 400 N TBC sin 60° = 0
Σ Mpoint B = (2 sin 60° m)TAC
(2 cos 60° m)(500 N) = 0
Solving them, we obtain TAC = 289 N, TBC = 115 N & TBD = 346 N.
The Method of Sections
• Notice how similar this method is to the method of joints:
– Both methods involve cutting members to obtain free-body diagrams of parts of a truss
– In the method of joints, we move from joint to joint, drawing free-body diagrams of the joints & determining the axial forces in the members as we go
– In the method of sections, we try to obtain a single free-body diagram that allows us to determine the axial forces in specific members
The Method of Sections
– In our example, we obtained a free-body diagram by cutting 3 members, including the 1 (member BC) whose axial force we wanted to determine
• In contrast to the method of joints, the forces on the free-body diagrams used in the method of sections are not usually concurrent:
– As in our example, we can obtain 3 independent equilibrium equations
– Although there are exceptions, it is usually necessary to choose a section that requires cutting no more than 3 members, or there will be more unknown axial forces than equilibrium equations
Example
The truss in Figure supports a 100-kN load. The
horizontal members are each 1 m in length.
Determine the axial force in member CJ & state
whether it is in tension or compression.
Example
Strategy
We need to obtain a section by cutting members
that include member CJ. By cutting members CD,
CJ & IJ, we will obtain a free-body diagram with 3
unknown axial forces.
Solution
To obtain a section, we cut members CD, CJ & IJ
& draw the free-body diagram of the part of the
truss on the right side of the truss
Example
Solution
From the equilibrium
equation:
Σ Fy = TCJ sin 45° 100 kN
= 0
We obtain TCJ = 141.4 kN.
The axial force in member
CJ is 141.4 kN (T).
Example
Critical Thinking
• We designed this example to demonstrate that
the method of sections can be very
advantageous when you only need to determine
the axial forces in particular members of a truss
– Imagine calculating the axial force in member
CJ using the method of joints
• But in engineering applications it is usually
necessary to know the axial forces in all the
members of a truss & in that case the 2 methods
are comparable
Example
Determine the axial forces in members DG & BE &
CG of the truss in Figure.
Example
Strategy
We can’t obtain a section that involves cutting
members DG & BE without cutting more than 3
members. However, cutting members DG, BE, CD
& BC results in a section with which we can
determine the axial forces in members DG & BE.
Example
Solution
Determine the Reactions at the Supports:
Draw the free-body diagram of the entire truss:
Example
Solution
From the equilibrium equations:
Σ Fx = Ax = 0
Σ Fy = Ay + K F 2F F = 0
Σ Mpoint A = LF (2L)(2F) (3L)F + (4L)K = 0
We obtain the reactions Ax = 0, Ay = 2F & K = 2F.
Example
Solution Choose a Section:
We obtain a section by cutting
members DG, CD, BC & BE.
Because the lines of action of
TBE, TBC & TCD pass through point
B, we can determine TDG by
summing moments about B:
Σ Mpoint B = L(2F) (2L)TDG = 0
Example
Solution
The axial force TDG = F.
Then from the equilibrium equation:
Σ Fx = TDG + TBE = 0
We see that TBE = TDG = F.
Member DG is in compression & member BE is in
tension.
Example
Critical Thinking
• This is a clever example but not 1 that is typical
of problems faced in practice:
– The section used to solve it might not be
obvious even to a person with experience
analyzing structures
– Notice that the free-body diagram of the
section of the truss is statically indeterminate,
although it can be sued to determine the axial
forces in members DG & BE
Example Problem
For the truss shown, determine
the forces in members DC and FG.