Structural Analysis I_Introduction_Reactions (2).ppt

structure architecture

1. Structural Analysis ( I )Structural Analysis ( I )

2. CONSTRUCTION ENGINEERING IS
THE ART OF USING MATERIALS
That Have Properties Which Can Only Be Estimated
TO BUILD REAL STRUCTURES
That Can Only Be Approximately Analyzed
TO WITHSTAND FORCES
That Are Not Accurately Known
SO THAT OUR RESPONSIBILITY WITH RESPECT
TO PUBLIC SAFETY IS SATISFIED.

3. Hours per week :
Lecture: 4 hours
Tutorial: 2 hours
Credit: 3
Prerequisite: BA 141


Class Performance/Attendance : 10%
Midterm 1/Assignment (7
th
week) : 30%
Midterm 2/Assignment (12
th
week): 20%
Final Exam : 40 %

4. Course Aim:
Calculation of the internal forces of each member of a structure
system
Course Description:
Definition of a structure, its support conditions and its various
structural forms in addition to various loading conditions that a
structure must support. Study the stability and determinacy of
structures. Calculation of reaction forces. Basic concepts of
structural analysis. Calculation of the internal forces (normal
forces, shear forces and bending moments) and its distribution on
statically determinate beams, frames and arches. Member forces
in trusses. Influence lines and its use to calculate the maximum
response functions in structures

5. Overall Course Contents:

1 -Introduction to structural analysis, scope, the definition of a
structure, its forms, supports and loads,
2 -Basic concepts of structural analysis, Study the stability and
determinacy, of structures. Equilibrium, Free-body diagram,
Reaction forces
3 -Internal Forces, sign convection, Relationships between load,
shear & bending moment. Methods of calculation of internal forces
4 -Internal forces in simple and compound beams
5 -Internal forces in simple, three-hinged, closed, multi-storey &
multi-bay frames
6 -Internal forces in arches
7 -Influence lines and its use to calculate the maximum response
functions in statically determinate beams and trusses .

6. Structural AnalysisStructural Analysis
• Structural analysis is a mathematical examination of a Structural analysis is a mathematical examination of a
complex structurecomplex structure
• Analysis breaks a complex system down to individual Analysis breaks a complex system down to individual
component partscomponent parts
• Uses geometry, trigonometry, algebra, and basic physicsUses geometry, trigonometry, algebra, and basic physics
““Used to calculate the forces in the Used to calculate the forces in the
structural elements and their structural elements and their
displacementsdisplacements””

7. Structural Elements and LoadsStructural Elements and Loads
What is a structureWhat is a structure??
Various components connected Various components connected
to carry different types of to carry different types of
loadsloads
The human skeleton is a structure which maintains the
shape of the body keeps the various organs and muscles
in the right place and transmits their weight and carried
loads down to the ground.

8. Examples of Civil StructuresExamples of Civil Structures
BuildingsBuildings

9. BridgesBridges
Examples of Civil StructuresExamples of Civil Structures
ShedsSheds

10. Structural Design ConsiderationStructural Design Consideration
•SafetySafety
•ServiceabilityServiceability
•Economics & Environmental Constraints Economics & Environmental Constraints
The design process requires a fundamental The design process requires a fundamental
knowledge of the followingsknowledge of the followings::
• Material PropertiesMaterial Properties
• Laws of mechanics which govern material responseLaws of mechanics which govern material response

11. Structural ElementsStructural Elements
Tie RodTie Rod “ structural “ structural
members subjected to tensile members subjected to tensile
force are often referred to as force are often referred to as
tie rods or bracing struts tie rods or bracing struts
almost used in Trusses”almost used in Trusses”
ColumnsColumns “ Vertical “ Vertical
structural members resist structural members resist
axial compressive loads” axial compressive loads”
Tie RodsTie Rods
ColumnColumn

12. BeamsBeams “ Straight horizontal members used to “ Straight horizontal members used to
primary carry vertical loads and are designed to primary carry vertical loads and are designed to
resist flexure or bending moments”resist flexure or bending moments”

13. Types of StructuresTypes of Structures
““The combination of structural elements and the The combination of structural elements and the
material from which they are composed is material from which they are composed is
referred to as a structural system”referred to as a structural system”
TrussesTrusses “ This type of structures is used when the “ This type of structures is used when the
span of a structure is large and its depth is not span of a structure is large and its depth is not
important criterion. Consists of slender elements important criterion. Consists of slender elements
usually arranged in Triangular fashion. It is used usually arranged in Triangular fashion. It is used
for Bridges, Roofs & Towersfor Bridges, Roofs & Towers ” ”

14. Howe Truss
Fink Truss
Pratt Truss
Warren Truss

15. Cables :Cables :
usually flexible, carry loads usually flexible, carry loads
in tension and supportin tension and support
bridges and building roofs as bridges and building roofs as
it have advantages over it have advantages over
beams and trusses for spans beams and trusses for spans
> 46 ms. It is use is limited by > 46 ms. It is use is limited by
their sag, weight and method their sag, weight and method
of anchorage.of anchorage.
Cables and Arches Cables and Arches “ These types of structures are used for “ These types of structures are used for
large span”large span”

16. Three-hinged arch
Hinge 1 Hinge 2
Hinge 3
Arches :Arches : Its strength in compression, since it has a Its strength in compression, since it has a
reverse curvature.reverse curvature. Frequently used in bridge structures, Frequently used in bridge structures,
dome roofs and for opening in masonry wall.dome roofs and for opening in masonry wall.

17. FramesFrames
•They are often used in buildings and are composed of beams They are often used in buildings and are composed of beams
and columns that are eight pin or fixed connected.and columns that are eight pin or fixed connected.
• It also extend in two or three dimensionsIt also extend in two or three dimensions
•The loadings of frames causes bending of its membersThe loadings of frames causes bending of its members
•The strength of such a frame is derived from the moment The strength of such a frame is derived from the moment
interactions between beams and columns at the rigid joint. interactions between beams and columns at the rigid joint.
10 kN
20 kN
20 kN
8 m
8 m
8 m
8 m8 m8 m

18. Surface structuresSurface structures
•
It is made of material having a very small thickness compared to It is made of material having a very small thickness compared to
its other dimensions. The material is acting as a membrane that is its other dimensions. The material is acting as a membrane that is
subjected to pure tension. subjected to pure tension.
• It may be shaped as folded plates, cylinders, made of material It may be shaped as folded plates, cylinders, made of material
such as reinforced concrete. These types of structures are referred such as reinforced concrete. These types of structures are referred
to as thin plates or shellsto as thin plates or shells

20. LoadsLoads

21. LoadsLoads
“ “ Designing loads is often specified in codes”Designing loads is often specified in codes”
Dead LoadsDead Loads “Consist of the weights of the various “Consist of the weights of the various
structural members and any other objects that are structural members and any other objects that are
permanently attached to the structure.” permanently attached to the structure.”
Example Example Given the material properties (density) and the size Given the material properties (density) and the size
of the various components of the structure, the dead load can of the various components of the structure, the dead load can
be easily calculated be easily calculated
W = density W = density  volume volume

22. Live LoadsLive Loads “Can vary both in their magnitude and “Can vary both in their magnitude and
location. They may be caused by the weights of location. They may be caused by the weights of
objects temporarily placed on a structure, moving objects temporarily placed on a structure, moving
vehicles or natural forces” vehicles or natural forces”
Example Example
BuildingsBuildings : furniture & people : furniture & people
BridgesBridges : Moving vehicles : Moving vehicles
Wind loadsWind loads: when the structure block the flow of wind, : when the structure block the flow of wind,
the wind’s kinetic energy is converted into potential the wind’s kinetic energy is converted into potential
energy of pressure which causes wind loadings. energy of pressure which causes wind loadings.

23. Idealized StructureIdealized Structure

24. Idealized StructureIdealized Structure
An exact analysis of a structure can not be An exact analysis of a structure can not be
carried out since estimates always have to carried out since estimates always have to
be made of the :be made of the :
Loadings Loadings
Strength of Material Strength of Material
Elements connectivityElements connectivity
It is important to develop a model or It is important to develop a model or
idealized structure so that idealized structure so that a practical force a practical force
analysis of the members can be performedanalysis of the members can be performed

25. Examples on Idealized StructuresExamples on Idealized Structures

26. Examples on Idealized Structures (cont.)Examples on Idealized Structures (cont.)

27. Elements ConnectionsElements Connections
The three types of joints The three types of joints
most often specified are most often specified are
 The Pin ConnectionThe Pin Connection
 Pin (Hinged) SupportPin (Hinged) Support
A pin connected joint and roller support allow A pin connected joint and roller support allow
some freedom for slight rotationsome freedom for slight rotation

28.  The Fixed ConnectionThe Fixed Connection
 Fixed SupportFixed Support
Fixed joints allow Fixed joints allow NONO relative rotation relative rotation
between the connected membersbetween the connected members
It is important to be able to recognize the symbols for
these connections and the kind of reactions they exerted
on their attached members by noting how the connection
prevents degree of freedom or displacement of the
member

29. ReactionsReactions
3kg
R
a = 3kg
Forces developed at structure Forces developed at structure
supports to maintain supports to maintain
equilibrium.equilibrium.
Ex: If a 3kg jug of water Ex: If a 3kg jug of water
rests on the ground, there is a rests on the ground, there is a
3kg reaction (R3kg reaction (R
aa) keeping the ) keeping the
bottle from going to the bottle from going to the
center of the earth.center of the earth.

30. A bridge across a river A bridge across a river
has a 200 lb man in the has a 200 lb man in the
center. What are the center. What are the
reactions at each end, reactions at each end,
assuming the bridge has assuming the bridge has
no weight?no weight?

31. Types of SupportsTypes of Supports
The four types of supports that can be found in structures The four types of supports that can be found in structures
are; roller, frictionless surface, pinned, and fixed. The type are; roller, frictionless surface, pinned, and fixed. The type
of support affects the forces and moments that are used to of support affects the forces and moments that are used to
represent these supports. represent these supports.

32. Structural elements carry their loading to other Structural elements carry their loading to other
elements or the ground through connections or elements or the ground through connections or
supports.supports.
It is important to realize that all of the graphical It is important to realize that all of the graphical
representations of supports are idealizations of a real representations of supports are idealizations of a real
connection connection
In order to be able to analyze a structure it is In order to be able to analyze a structure it is
necessary to be clear about the forces that can be necessary to be clear about the forces that can be
resisted at each support.resisted at each support.
In order to facilitate the analysis of a structure, it is In order to facilitate the analysis of a structure, it is
often necessary to idealize the behavior of a support.often necessary to idealize the behavior of a support.
It is expected that these representative forces and It is expected that these representative forces and
moments, if properly calculated, will bring about moments, if properly calculated, will bring about
equilibrium in the structural elementequilibrium in the structural element. .

33. ROLLER SUPPORTS
Roller supports are free to rotate and translate along the Roller supports are free to rotate and translate along the
surface upon which the roller rests.surface upon which the roller rests.
The surface can be horizontal, vertical, or sloped at any The surface can be horizontal, vertical, or sloped at any
angle.angle.
The resulting reaction force is always a single force that is The resulting reaction force is always a single force that is
perpendicular to, and away from, the surface.perpendicular to, and away from, the surface.
A roller support A roller support
cannot provide any cannot provide any
resistance to lateral resistance to lateral
forces forces

34. PINNED (HINGED ) SUPPORTSPINNED (HINGED ) SUPPORTS
A pinned (hinged) support can resist both vertical and A pinned (hinged) support can resist both vertical and
horizontal forces but not a moment. horizontal forces but not a moment.
They will allow the structural member to rotate, but not to They will allow the structural member to rotate, but not to
translate in any direction.translate in any direction.
A single pinned connection A single pinned connection
is usually not sufficient to is usually not sufficient to
make a structure stable. make a structure stable.
Another support must be Another support must be
provided at some point to provided at some point to
prevent any rotation of the prevent any rotation of the
structure.structure.

35. FIXED SUPPORTSFIXED SUPPORTS
Fixed supports can resist Fixed supports can resist
vertical and horizontal forces vertical and horizontal forces
as well as a moment. Since as well as a moment. Since
they restrain both rotation they restrain both rotation
and translation, they are also and translation, they are also
known as rigid supports. known as rigid supports.
This means that a structure This means that a structure
only needs one fixed support only needs one fixed support
in order to be stable in order to be stable

36. SUPPORTSSUPPORTS

37. Link MembersLink Members
Direction of
possible motion
RR
AA
BB
A link AB is shown symbolically, the direction of any possible
small displacement of point B is a long a circle with center A.
The reaction RR direction must be perpendicular to the
possible direction of motion. Consequently, the reaction
must coincides with the link member.

38. EquilibriumEquilibrium
Σ FΣ F
xx = 0 means “The sum of the forces in the x = 0 means “The sum of the forces in the x
direction is 0” direction is 0”
Σ FΣ F
yy = 0 means “The sum of the forces in the y = 0 means “The sum of the forces in the y
direction is 0”direction is 0”
ΣMΣM
OO = 0 means “The sum of the moments = 0 means “The sum of the moments
about an axis perpendicular to the xy plane for about an axis perpendicular to the xy plane for
all the forces acting on an object = 0”all the forces acting on an object = 0”

39. Procedures forProcedures for Calculation of External Calculation of External
Support ReactionsSupport Reactions
1. Disassemble the structure and draw the Free Disassemble the structure and draw the Free
Body Diagram for each member showing all Body Diagram for each member showing all
the support reactions and the idealized the support reactions and the idealized
applied loads. applied loads.
2.2.Apply the proper Equations of Equilibrium Apply the proper Equations of Equilibrium
sequentially to find the unknown support sequentially to find the unknown support
reactions.reactions.

40. Example 1Example 1
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

41. Example 1 (continue)Example 1 (continue)
Solution:Solution:

42. Example 2Example 2
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

43. Example 2 (continue)Example 2 (continue)
Solution:Solution:

44. Example 3Example 3
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

45. Example 3 (continue)Example 3 (continue)
Solution:Solution:

46. Example 4Example 4
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

47. Example 4 (continue)Example 4 (continue)
Solution:Solution:

48. Example 5Example 5
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

49. Example 5 (continue)Example 5 (continue)
Solution:Solution:

50. Example 6Example 6
Find the reactions of the supports of the following archFind the reactions of the supports of the following arch

51. Example 6 (continue)Example 6 (continue)
Solution:Solution:

52. Example 7Example 7
Find the reactions of the supports of the following beamFind the reactions of the supports of the following beam

53. Example 7 (continue)Example 7 (continue)
Solution:Solution:

54. Example 8Example 8
Find the reactions of the supports of the following frameFind the reactions of the supports of the following frame

55. Example 8 (continue)Example 8 (continue)
Solution:Solution:

56. Example 8 (continue)Example 8 (continue)
Solution:Solution:

57. StabilityStability
It is not sufficient to satisfy the equations of equilibrium to It is not sufficient to satisfy the equations of equilibrium to
ensure structure stability. The members must be properly ensure structure stability. The members must be properly
constrained by their supports.constrained by their supports.
A structure will be geometrically A structure will be geometrically unstableunstable that it will that it will
move slightly or collapse – if there are fewer reactive forces move slightly or collapse – if there are fewer reactive forces
than equations of equilibrium, i.e.than equations of equilibrium, i.e.

58. Improper ConstraintsImproper Constraints::
In some cases the number of unknown forces equals to the In some cases the number of unknown forces equals to the
number of equation of equilibrium. However, instability or number of equation of equilibrium. However, instability or
movement of a structure can be developed because of the movement of a structure can be developed because of the
improper constraining by the supportsimproper constraining by the supports..
• Support reactions are concurrent at a point.Support reactions are concurrent at a point.
• Support reactions are parallel.Support reactions are parallel.

59. DeterminacyDeterminacy
It is said that the structure is It is said that the structure is Statically Determinant;Statically Determinant; if if
all the forces in a structure can be determined using all the forces in a structure can be determined using
Equations of Equilibrium.Equations of Equilibrium.
R = NR = N
It is said that the structure is It is said that the structure is Statically Indeterminate;Statically Indeterminate; if if
the a structure has more unknown forces than the the a structure has more unknown forces than the
available equilibrium equations.available equilibrium equations.
R > NR > N
If a structure is statically indeterminate, the additional If a structure is statically indeterminate, the additional
equations needed to solve for unknown reactions are equations needed to solve for unknown reactions are
obtained by relating the applied loads and reactions to a obtained by relating the applied loads and reactions to a
known displacement or slope at different points on the known displacement or slope at different points on the
structure, these equations called structure, these equations called compatibility equationscompatibility equations

60. ExamplesExamples

61. ExamplesExamples

62. ExamplesExamples

63. W t/mW t/m
LL
WLWL
L/2L/2
L/3L/3
LL
W t/mW t/m
WL/2WL/2
WL/2WL/2
LL
W t/mW t/m
L/2L/2
Load IdealizationLoad Idealization

64. L
`L
`
WLWL``
L
`/2
L
`/2
LL
W t/mW t/m
LL
WLWLL/2L/2
W t/mW t/m

65. Application of the Equation of EquilibriumApplication of the Equation of Equilibrium
(1) Calculation of the External Reaction(1) Calculation of the External Reaction
(2) Calculation of the Internal Forces(2) Calculation of the Internal Forces
 Normal ForceNormal Force
Shearing ForceShearing Force
Bending MomentBending Moment
QQ
NN
QQ
NN
MM
MM

66. Procedure forProcedure for Calculation of Internal ForcesCalculation of Internal Forces
1.1.Calculate external support reactions as previously Calculate external support reactions as previously
mentioned. mentioned.
2.2.Pass an imaginary section through the member Pass an imaginary section through the member
perpendicular to its axis at the point where the perpendicular to its axis at the point where the
internal forces is determined. internal forces is determined.
3.3.Apply the proper Equation of Equilibrium to find the Apply the proper Equation of Equilibrium to find the
unknown internal forces N, Q and M.unknown internal forces N, Q and M.
4.4.If the equation of equilibrium yields a quantity If the equation of equilibrium yields a quantity
having a negative magnitude, so it should be reversed. having a negative magnitude, so it should be reversed.

67. Examples to be worked outExamples to be worked out

68. Examples to be worked outExamples to be worked out

69. Examples to be worked outExamples to be worked out