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Lecture notes, lectures 1-4
Structural Analysis I (University of Sheffield)
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Scope of Structures 1
Topics
Section I: Basic structural analysis
Section I (Week 1 to Week 7)
Types of structures and loadings
Basic concepts and analysis techniques
Analysis of Trusses
Analysis of beams and frames
Section II: Deflection of structures
Deflection of beams
Integration method
Moment area method
Deflection of structures using energy
methods
Principle of virtual work
Lecture 1
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Lecture 1
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Why we study structural analysis?
Structural Engineering is about the conception, design and construction of
the structural systems that are needed in support of human activities. Civil
engineering projects include bridges, building, dams, storage facilities etc.
involves structural engineering.
Lecture 1
Answer: To prevent this to happen!
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Compassvale School, Singapore
•Structural analysis is rarely an end in
itself. Rather, it is a tool used by structural
designers to assist in the creation of
design concepts and to demonstrate that
these
concepts
satisfy
project
requirements.
Lecture 1
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Lecture 1
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Introduction
Structural forms and classifications
What is a structure?
• Structural forms and classifications
• Structures refer to a system of connected parts used to
support a load (e.g. self-weight, occupants/furniture for
building, traffics for bridges)
– Structural members and their characteristics
– Different forms of structures and their
properties
• Factors to consider
– Safety (most important!)
– Serviceability (e.g. extensive deflection or vibration is
not allowed)
– Esthetics
– Economic and environmental constraints (cost to build
and use, energy consumption etc.)
• Loads
– Different loadings type and their characteristics
• Structural analysis and design
– An introduction to the basic concepts
Lecture 1
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Structural members and their
characteristics
Tie Rod and Struts
• Tie rods normally refer to
structural members subjected
to tensile force only
• Normally tie rods are rather
slender
• Struts are members that could
subject to both tensile and
compressive forces
• They are commonly used in
trusses and bridges
• For the ease of design and construction, structures
are often composing of a number of structural
elements or members
• In this course, the following commonly encountered
member types shall be discussed:
– Tie rods
– Struts
– Beams
– Columns
More on tie rod in Wikipedia: http://en.wikipedia.org/wiki/Tie_rod
More on tie strut in Wikipedia: http://en.wikipedia.org/wiki/Strut
Lecture 1
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•Compression
Structures Strut
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•Force transfer mechanism
in an arch
•Stable 3-hinge arch
continues to support the load
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•Tension
Structures
Beams
Columns
• Beams are usually straight
horizontal members used
primarily to carry vertical
loads
• They are classified to how
they are supported
• Design to resist bending
moment and shear forces
• In steel beam, wide flange
cross section are commonly
used
• Columns usually
straight vertical
members to resist
vertical compression
• Tubes and wide flange
cross section are often
used
• A column is called a
beam-column if it is
subjected to both
vertical compression and
bending
More about beam in Wikipedia:
http://en.wikipedia.org/wiki/Beam_%
28structure%29
Lecture 1
More about column in Wikipedia: http://en.wikipedia.org/wiki/Column
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Structural forms and classifications
Trusses
• Used when long span is
needed with not much
depth constraints
• Consist of slender
struts/ties arranged in
triangular or other
fashions
• Planer trusses (all
members lie in same
plane) used in bridges
and roof
• Space truss (members lie
in different planes) used
in derricks and towers
• The combination of structural elements
(rods, strut, beam, column) and materials
(concrete, steel, timber) generate different
structural forms or structural systems
• In this course, focus shall be given to the
following two most popular types of skeleton
structural systems
– Trusses
– Frames (including simple beams)
Lecture 1
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Structural Analysis and Design
Selection of structural form
(trusses, frames) and material
(structural steel, concrete)
Find out the loads on
structures
Rechecking
Main study
objective of
this course
Structural analysis
(calculate the forces
in the structure)
Rechecking
Design (to provide
strong enough
members to resist the
forces
Build (to construct the
structures)
Lecture 1
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Next Lecture
• Some essential tools for structural analysis
– Principle of superposition
– Equilibrium
– Internal forces
– Free body diagrams
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Idealization structures
• In real sense, an exact analysis is almost impossible due
to uncertainty in members dimensions and loading
location (especially before the structure was built!)
• Hence, in practice, in order to carry out force analysis,
structural engineers must know how to idealize a
structures to a simple form/geometry for analysis
• Structural prismatic elements like rod/beam/column
members are normally idealized as “line elements” with
appropriate support conditions
• At the same time an appropriate idealized loading and
supports and connections should also be applied to the
structure
Lecture 2
•Idealization structures
Geometry
Loading
Supports and connections
•Principle of superposition
•Equilibrium
•Internal forces
•Free body diagrams
Lecture 2
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•Simplification
1
the purpose of analysis
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•All structures have to be simplified in a convenient form to carry out the structural
analysis; actual structures often have complicated details
•Experience and good judgement of structural behaviour are required in carrying
out such simplification
Support connections
• Real and actual supports/connections could come with many
different details
• In structural analysis, they are idealized into a few supports and
connection types
Pin connection
Lecture 2
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Fixed connection
Lecture 2
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Idealized connection types
Idealized structures
Roller support: no relative vertical movement, allows
some freedoms for slight rotation
Pin connection: no relative horizontal and vertical movement,
allows some freedoms for slight rotation
Fixed joint: no relative horizontal, vertical movement and rotation
• When selecting the model for each support, the
engineer must be aware how the assumptions will
affect the actual performance
• The assumptions must be reasonable for the
structural design
• The simplified loadings should give results that
closely approximate the actual loadings
• Common types of connections on coplanar
structures are given in Table 2.1 of Textbook (Page
37)
90
Roller support
Pin support
Pin connection
=90
Fixed connection
Lecture 2
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•Reactions
•Different types of idealised support conditions for planar structures:
Pin support
with Fx and
Fy
Tie support
with Fn
Generally
known as
roller
support with
only one
reaction
force Fy
Partially
fixed or fixed
support with
Fx , Fy and
M
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1 or a combination of bodies is to be analysed and it is
Quite often,
a body
isolated as a single body from the surrounding bodies.
Hence, a free body diagram is a diagrammatic representation of the
isolated body or combination of bodies treated as a single body, showing
ALL forces applied to it and by mechanical contact with other bodies that
are imagined to be removed.
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•Example: Draw the
free body diagram
for the beam.
Neglect the weight of
the beam.
Idealized structures: An example
Lecture 2
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•Example: A cylinder is
supported on a smooth
inclined surface by a two-bar
frame . Draw the free body
diagram for the cylinder, the
two bar frame and the pin at
C.
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•Example: Draw the free body diagram for the
pulley, the post and the beam CD.
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Equations of Equilibrium
• For the general 3D cases, six equations:
 Fx  0
 M x 0
 Fy  0
 M y 0
 Fz  0
 M z 0
• For planar structures (and problems appear in
this course), reduces to 3 equations:
 Fx  0 ,  F y  0 ,  M o  0
Lecture 2
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Principle of Superposition
Principle of Superposition
• Forms the basis for much theory of structural
analysis in this course
• The total displacements or internal loadings
P
(stress) at a point in a structure subjected to
several external loadings can be determined by
adding together the displacements or internal
loadings (stress) caused by each of the external
loads acting separately
• Linear relationship exist among loads, stresses and
displacements
Lecture 2
2P

2
• 2 requirements for the principle to apply:
– Material must behave in a linear-elastic manner,
Hooke’s Law is valid
– The geometry of the structure must not undergo
significant change when the loads are applied,
small displacement theory
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More examples on FBD
D
More examples on FBD
P
P
1
C
F
2
y
x
1
FxB
FxA
B
A
FyA
2
P
FyB
FBD for the whole structure
FxA
Fxc
FyA
Fxc
M
Fyc
Fyc
FxA + 0 = 0
FyA – F = 0
Moment above A: M + F3 = 0
FBD for ACD
FxA
Lecture 2
FyB
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•12kN
•18kN.m
•K
•1m •1m
Lecture 2
•18kN.m
•8kN/m
•10kN
•22kN.m
•18kN.m
•12kN
•8kN/m
•10kN
•22kN.m
•K
•4m
•1m•1m
•2m
•29kN
•8kN/m
•22kN.m
•10kN
•25kN
•17
•
•
•x=17/8
•K
•29kN
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•29
•12kN
FxB
FBD for CB
FyA
•15
•18
•25kN
•11
•28
•46.0625 •32
•17
•10
•
•Q(kN)
•20
•M(kN·m)
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•80kN.m
•160kN
•1m •1m
•40kN/m
•2m
•80kN.m
•40kN
•130
•80kN.m
•160kN
•40kN/m
•310kN
•120
•
•30
•130
•130kN
•310kN
•
•40kN
•
•210
•40kN
•40kN/m
•130kN
•2m
•4m
•160kN
•340
•
•190
•160
•Q(kN)
•M(kN·m)
•280
22 / 47
哈工大 土木工程学院
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Some hints on drawing FBD and
equilibrium equations (EE)
Internal forces
•
FBD should never contain any support
•
When removing the supports to the structures, replaced them by the
corresponding reactions
•
The number of reactions should always equal to the numbers of constrains
removed.
•
Away fix the global x and y axes before writing down the EE, this shall fix the
positive directions of forces
•
In general, one could always assume that all reactions are acting in the
positive direction
•
When writing down the EE, always use the condition of “Sum-To-Zero”.
That is, instead of writing:
FxB = -Pcos(45)
one shall write:
• Internal forces are forces developed inside a member
under the action of external loadings and reactions from
the supports
• In structural analysis, to find out the internal forces of
members are very important as such member forces
information are critical to the design of the structures so
that engineers could design a member section size strong
enough to resist the internal forces
• Hence, it is very important that in structural analysis all
the member forces are calculated correctly. Otherwise, it
could lead to error in design and eventually failure of
members and structures
FxB + Pcos(45) = 0
Lecture 2
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Next Lecture
• To obtain member force, one needs to make a
“cut” to the member under concern and expose
the corresponding internal forces
P
• Determinacy and stability
• Simple applications of equilibrium
equations
P
B
A
A
C
A simple beam ACB
N
N
M
M
V
B
V C
A cut before C exposes the
internal forces N, V and M
P
FxA
A
N
N
M
M
V
FyA
B
V C
FxB
FyB
FBD for RHS
FBD for LHS
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Determinacy
• Equilibrium equations (EE) provide sufficient conditions for
equilibrium
• When all forces (internal and reactions) can be determined
strictly from the EE, the structure is referred as statically
determine
• If the structure has more unknown forces than the numbers
of EE , the structure is a statically indeterminate structure
• If a structure is statically indeterminate, addition equations,
known as compatibility equations are needed to solve for the
forces
• In this course, we shall only work on statically determine
planar (2D) structures
Lecture 3
• Determinacy
• Stability
• Simple application of equilibrium
equations
•For stability, a minimum of 3
reaction components is needed:
1
Lecture 3
2
3
Lecture 3
4
Stability
• To ensure equilibrium of a structure or its members:
– Must satisfy EEs
– Members must be properly “held” or constrained
by their supports
• If the structure is not properly supported, instability
may be resulted and lead to collapse of the (partially
or whole) structure
• Instability could be caused by
– Partial constraints (i.e. no enough numbers of constraints
/ support reactions)
– Improper Constraints (i.e. enough numbers of constraints
but put in wrong positions/ directions)
Lecture 3
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•Examples of external statical classification:
•Computation of reactions using Equation of Equilibrium
•Example: Determine the reactions for the structure.
Lecture 3
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•Example: A pin-connected two-bar frame is supported and loaded as shown
below. Determine the reactions at supports A and B.
•And they must be properly arranged:
In some cases, the number of reaction forces may equal EEs.
However, instability or movement of structure could still occur
if support reactions are
• Concurrent (i.e. pass through) at a point
• All parallel and acting along one direction only
Simple application of EEs
Simple application of EEs
Procedure of analysis using EEs
Some useful hints
(1) Identify all the supports and hinge connections (if
any)
(2) Remove the supports and re-apply the reactions
(3) Draw FBD for the whole structures or FBDs for
separate parts of the structure. Remember that 3
EEs could by applied to each separated FBD
(4) Apply EEs in turns (to different separate parts if
needed) to solve for the unknown reaction and
forces
(5) Always try to solve one unknown at a time
(1) There is no need to know in advance the directions
of the reaction before the analysis. One could
always assuming the reactions are acting in the
positive directions, if eventually the answer is
negative, it simply implies that the reaction is
acting in the opposite direction
(2) When apply the EEs, very often the solutions will
become very straightforward by take moment (i.e.
using the M=0 condition) at appropriate point
(3) Always write the EE in “sum-to-zero” form, this
could reduce the chance of making errors
Lecture 3
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Numerical examples
The 6kN/m udl is converted to
36kN at its centroid
Note: the 6.8kNm
moment at the free
end should be
included when taking
moment at A
  Fx  0; Ax  270 cos 600  0
Ax  135kN
Segment BC :
With anti - clockwise in the  direction,
With anti - clockwise in the  direction,
0
0
 M A  0;  270 sin 60 (3)  270 cos 60 (0.3)  B y (4.2)  6.8  0
 M c  0;  8  B y (4.5)  0  B y  1.78kN
B y  159kN
   Fy  0;  1.78  C y  0  C y  1.78kN
   Fy  0;  270 sin 600  159  Ay  0
  Fx  0; Bx  0
Ay  74.8kN
Lecture 3
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Lecture 3
18
•Example :
•Compute the reactions for the structure shown below.
Segment AB :
With anti - clockwise in the  direction,
 M A  0; M A  36(3)  (1.78)(6)  0
M A  97.3kN.m
   Fy  0; Ay  36  1.78  0
Ay  34.2kN
  Fx  0; Ax  0
Lecture 3
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•40kN/m
•A
•120kN
•B
•8m
•2m
•D
•K
•C
•3m
•3m
•120kN
•40kN/m
•145kN
•40kN/m
•A
•C
•K
•A
•(kN·m)
•
•180
•263
•60
•
•
•175
•20kN
•D
•120
•145
•3m
•10kN
•
•235kN
•Home Works
•120kN
•B
•60kN
•60kN
•60 (kN)
•3m
•B
•3m
•10kN
•E
•2m •2m
•5kN/m
•2m
•2m
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•2kN/m
•C
•2m
•20kN·m
•2m
•2m
•F
•2m
•D
•4m
•10kN
•2m
•5m
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•q
•B
•A
•B
•x
•D
•l
1
q( l  x )
2
•A
•D
•C
1
q( l  x ) 2
8
•B
q( l  x )
1
1
2
x  qx 2  ql  x 
2
2
8
•C 1 q( l  x )
•D
•C
•l - x
•l
1
1
q( l  x ) x  qx 2
2
2
•C
•l - x
•l
•A
•D
•l
•q
•A
•x
2
•B
x  (3  2 2)l  0.172l
•F
•q
P
•A
•D
•a
•B
•a
•a
•C
•E
•a
•FPa/2
•FP/4
•FQ
•F
•A
•C
•B
l
l
•FP
a
•FPa/4
•M
•a
•FP
•FP/2
•
0.5ql 2
•
0.125ql 2
•
0.5ql
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•M(kN·m)
0.5ql
•
•FQ(kN)
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Past Years Exam
Past Years Exam
Past Years Exam
Past Years Exam
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Past Years Exam
Some remarks
• More examples could be found in the textbook
• When udl is involved, one needs to convert it to the
equivalent point load acts at the centroid
• In all exercises, calculations are always very simple
but one need to draw FBDs and apply EEs
correctly to get the correct answers
• More exercises are given in Tutorial sheet 1
• More practice is useful to allow one to solve the
this type of question in a faster speed with less
chances to make error
Lecture 3
34
Next Lecture
• Introduction to planar truss
• Determinacy and stability of truss
Lecture 3
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Introduction to planar trusses
• A truss is a structure composed of slender members
joined together at their end points
• The joint connections are usually formed by bolting
or welding the ends of the members to a common
plate called gusset plate
• Planar trusses lie in a single plane and is often used
to support roof or bridges
Lecture 4
•
•
•
•
Introduction to planar truss
Roof and bridge trusses
Internal forces
Determinacy and stability of truss
Lecture 4
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Lecture 4
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Roof trusses
2
Different forms of roof trusses
• They are often used as part
of an industrial building
frame
• Roof load is transmitted to the truss at the joints by
means of a series of purlins
• To keep the frame rigid and thereby capable of
resisting horizontal wind forces, knee braces are
sometimes used at the supporting column
Lecture 4
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Different forms of bridge trusses
Bridge trusses
• The load on the deck is first
transmitted to the stringers 
floor beams  joints of
supporting side truss
• The top and bottom cords of
these side trusses are
connected by top and bottom
lateral bracings resisting
lateral forces
• Additional stability is
provided by the portal and
sway bracing
• In the case of a long span
truss, a roller is provided at
one end for thermal expansion
Lecture 4
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Internal force of truss members
Assumptions of analysis and design
P1
1. The gusset plates provide limited amount of
moment resistance
2. The members are joined together by smooth pins
3. No moment at the two ends of members
4. All loadings are applied at the joints
5. All supports are also located at the joints
6. The above assumptions lead to the conclusion
that only axial forces (tension or compression)
are acting in the all members regarding of their
sizes, locations and dimensions
y
x
Set the x and y axes parallel and perpendicular
to the orientation of the member respectively
V1
Assume that the two forces at the two ends are
P1, V1 and P2 and V2 respectively
V2
P1
P2
y
x
V1=0
Using the conditions Fx=0 and Fy=0, the
forces must be equal and opposite at the two
ends
Now taking moment at one end since M=0
LV1=0
where L0 is the length of the member.
Hence, V1=0 and only the equal and opposite
axial force exist
V1 =0
P1
Lecture 4
Since both end of any truss members are
pinned end, there is no moment at the ends
Thus, all truss members are axial force
members
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CV2102: Structures 1
CV2102: Structures 1
Classification of coplanar trusses
Compound trusses
Simple Truss
• Simplest framework of
rigid (stable) with the
basic “stable” triangle
element (ADC)
• Addition members are
added in to connect with
the basic triangle
• Note that simple trusses
may not consist entirely
of triangles
Formed by joining two
or more simple trusses
together
Type 1, Figure (a) :
connected by a common
joint and bar
Type 2 , Figure (b) :
connected by three bars
Type 3, Figure (c):
substituted bars of a
truss by simple trusses
Lecture 4
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Lecture 4
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CV2102: Structures 1
Determinacy of planar trusses
Complex truss
A complex truss is one
that cannot be classified as
being either simple or
compound trusses
Lecture 4
•Overall classification
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CV2102: Structures 1
CV2102: Structures 1
•Condition equations
CV2102: Structures 1
CV2102: Structures 1
Internal stability of trusses
Internal stability of trusses
• To determine the internal stability
of a compound truss, it is necessary
to identify the way in which the
simple truss are connected together
• A truss is internally unstable if some of it
components forms a collapsible mechanism
• The internal stability can be checked by careful
inspection of the arrangement of its members
• If it can be determined that each joint is held
fixed so that it cannot move in a “rigid body”
sense with respect to the other joints, then the
truss will be stable
• However, if a truss is constructed so that it does
not hold its joints in a fixed position, it will be
unstable or have a “critical form”
• A simple truss will always be internally stable
• The truss at the top is unstable as
there is no restraint between joints
C and F or B and E
• The truss at the bottom unstable
since the inner simple truss ABC is
connected to DEF using 3 bars
which are concurrent at point O.
• An external load can be applied at
A, B or C and causes the truss to
rotate slightly
Lecture 4
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CV2102: Structures 1
CV2102: Structures 1
•Examples
CV2102: Structures 1
•Examples
CV2102: Structures 1
Next Lecture
• Analysis of planar trusses I: method of
joints
• Zero force members
Lecture 4
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