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Structural analysis basics
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Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
1
Part I Structural Analysis II
1) Introduction on Structural Design
(Except from “Aims of Structural Design” published by the Institution of Structural
Engineers, August 1969.)
a) Factors in Design
Function : The structure must fulfill its intended function. This function may be
complex, concerned not only with the support of load, but with water tightness,
thermal and sound insulation, fire resistance, and chemical resistance etc. Load itself
may be complex and show not only a single design load, but a wide spectrum of loads
and their combinations occurring with varying frequency. Moreover, as soon as a
structure is built, the influence of weather, ageing, fatigue, use, ill-use, accident and
other causes of deterioration commence to act upon it. The designer seeks to ensure
the continued service ability of his structure during its intended life.
Safety : The structure must be safe. Most structures involve the public; there are few
to whose strength someone does not at some tie entrust his life. The consequences of
collapse are grave and the possibility of collapse must be made remote.
The designer’s estimate of adequate safety is partly analytical and partly a matter of
judgment-the accident-prone structure may be difficult to define, but can be
recognized by the experienced.
Economy : The structure must be of least cost. The client having decided, doubtless
in consultation with his designer, on the standard of strength, durability and other
qualities that he requires, the design will proceed with this intention; the designer is,
in fact, in the market place looking for the best value for his client’s money. Design
decisions, always implicitly and often explicitly, are economic decisions, right down
to the choice of the smallest detail.
The designer is all too seldom able to consider the balance of first cost and
maintenance costs. Clearly, each can be reduced at the expense of each other;
somewhere there must be a minimum total cost. Most owners of structures seem to be
so organized, however, that the two are paid for from different funds and the effect of
one on the other is not revealed. It must be assumed that owners have fiscal and
financial reasons for this.
Other factors to be considered: -
a) statutory requirements to be satisfied
b) physical constraints which may affect the method of construction
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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c) time constraints; structures may need to be completed within a very short time
d) environmental impacts to be considered
b) Process of Design
Appreciation of requirements : Design starts with the appreciation of the client’s
requirements; this is the critical stage in the process and calls for the closest
collaboration between the client and the designer.
Formulation of schemes : As many facts and factors as possible are collected and
assimilated; the mind concentrates upon them. By some process of the intellect or
imagination, often in moments of relaxation, concepts of the general structure suggest
themselves, sometimes vague, sometimes surprisingly complete.
Appraisal of schemes : These concepts are then developed approximately and
appraised by the designer. The amount of mathematical analysis at this stage will
depend on his experience; often no more is done than is enough to determine general
dimensions and the designer is likely to concentrate more on functional suitability,
ease of erection and economy.
At the end of this appraisal, the designer returns to the original requirements, perhaps
to the client. The process may have thrown up unexpected possibilities that modify
the requirements; alternatively what was thought feasible may be found to be not so.
The economic aspect may need rethinking.
The most economic solution is usually the one with a straight-forward, clear-cut
method of erection with a minimum of separate process following one another
without mutual interference. In considering cost, it is more useful to concentrate on
operations than on the minutiae of quantities and rates whose apparent precision may
mislead.
Analysis of final scheme : It is now that a full structural analysis is made. This is
frequently referred to as “design”; in fact, the significant design decision have already
been taken and the operation of analysis is much more a process of checking the
scheme for soundness in service and during erection.
Procedure of structural design :
a) Preliminary and feasibility study:
1. Analysis of functional requirements and constraints
2. Planning and scheming; appraisal of different schemes and preliminary design
b) Detailed design:
1. evaluation loads
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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2. structural design (size, shape, quality etc)
3. detailing and production of drawings
4. preparation of specification documents
Details and Specifications : Detailing must be seen as an integral and important part
of design; one cannot afford to lose control of the detailed design because:
− bad detailing can leading to disaster-in fact, it is much the more frequent
cause of trouble;
− bad detailing can greatly increase cost;
− it is easy without close supervision for details to creep in which are not
consistent with the rest of the structure and which can render an otherwise
carefully considered erection scheme impossible.
Specifications are part of the design, complementary to the detail drawings, and not
merely a means of regulating contractual relations.
Supervision of Construction : Supervision should be under the designer’s control;
his task is not yet complete when he hands over the bundle of design documents.
Construction is rarely without incident requiring decision, the consequences of which
only the designer can appreciate.
Other follow-up work : Preparation of contract documents, site supervision, contract
administration, sometimes re-designing parts of the works, preparation of as-built
drawing etc.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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c) Loading
Dead Loads
When considering a building as a whole at the preliminary design stage, it is best
and easiest to estimate the total weight of the building in terms of average kPa of
floor area.
Although expressed in terms of floor area, these overall dead-load values include
the weight of the structural floor itself, the roof, walls, shafts, columns, and
possibly the floor surfacing, ceilings, and partitions. However, it should be clear
that they are only rough approximations. They can be varied greatly depending
upon the designer’s skill and his estimate of the materials to be used for the basic
structural system, the floors themselves, the walls, and other permanent portions.
Table : Approximate Weight of Materials (lb/cu ft)
Masonry Water, Snow
Marble, granite 140-165 Water 62.5
Brick masonry 100-150 Snow, fresh 5
Concrete, normal 150 Snow, packed 10 and up
Concrete, lightweight 90-120 Snow, wet 40-50
Metals Miscellaneous
Steel 480 Sand 100-120
Aluminum 165 Glass 160
Brass 530 Asphalt 80-100
Mortar 100
Timber
Redwood 26 Pine 35-40
Douglas fir 32 Oak 54
Live Loads
The actual live load on the floor of a building varies greatly. It is possible to
concentrate a heavy loading such as a safe box or moving equipment over a
rather small area. Then it is necessary to design that small area for a heavy
concentration. However, when there is a large area support by a primary
structural component, the significance of that concentration as compared with the
overall load will be reduced correspondingly.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Since the actual live load over the floor area of a building can vary greatly with
time of day and occupancy type, it becomes quite difficult to determine a precise
live load in kPa to a specific floor design. Nevertheless, an average design load
value can be assigned when actual or probable type of building occupancy is
known. This would be classified as the basic live load for application when
considering the larger tributary areas.
When smaller areas are considered, the effect of a concentrated live board should
be considered as a special case and may be expressed as a single concentrated
load or as a uniform load in kPa, but of a magnitude much higher than the specified
basic live load. Such concentrations are often stated or suggested in building code
specifications.
Although the basic live load does not apply to smaller areas, it also will not be
accurate to apply it without some modification to larger tributary areas. Consider
that, in most building, it would be unusual to load every square foot of a large
tributary area completely, as could be the case for a limited area. Therefore, the
size of a tributary area determines the probability of loading every square meter
to basic live load levels. To simulate this effect, the basic live load can be
reduced when considering large tributary areas.
Wind Loads
Wind pressure on a building surface depends primarily on its velocity, the slope
of the surface, the shape of the surface, the protection from wind offered by other
structures and, to a smaller degree, the density of the air, which decreases with
altitude and temperature, and the surface texture. All other factors remaining
unchanged, the pressure due to wind is proportionate to the square of the velocity
and the density of the air:
P = CD x Q = CD ( ½V2D)
Where P is the pressure on a surface, V is the velocity of the wind, and D is the
density of air, CD is the numerical shape coefficient (called drag coefficient) and
Q is the dynamic pressure moving air, equal to ( ½V2D). The velocity of the
wind assumed in design could be different for structures at seaside and at top of
hill.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Earthquake Loads
Earthquake loads are specified with the following basic objectives: -
a. To protect the public from loss of life and serious injury
b. To prevent building from collapse and dangerous damage under a
maximum-intensity earthquake.
c. To insure building against any but very minor damages under moderate to
heavy earthquakes.
Equivalent static loads are specified so that the above objectives can be attained
within reason and without excessive cost.
Earthquake resistance calls for energy absorption (or ductility) rather than
strength only. If a building is able to deflect horizontally by several times the
amount envisioned under the basic seismic design load and still maintain its
vertical load-carrying capacity it will be able to absorb earthquakes considerably
heavier than the design earthquake. If such ductility is present, collapse of the
building can be prevented even if the building is seriously damaged. Thus, in
addition to seismic load design, the ductility and plasticity of a building should
be given due consideration.
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Consequences of structural inadequacy
Inadequacy may be of two types: the structure may cease to be serviceable or it
may collapse. The acceptability of the risk will depend on its consequences.
Normally, the loss resulting from unservicability will be much less than that
from collapse.
The response of a structure to the failure of a member may be one of the
following:
(a) the structure will collapse immediately, as with a pin-jointed statically
determinate truss;
(b) the structure will not only collapse but entrain the collapse of adjacent
structures;
(c) the structure may continue to stand, though with reduced strength, due to
redistribution of load. This is the situation of a structure specifically designed to
be “fall safe” and sometimes, but not always, of a statically indeterminate
structure.
The practical consequences of any of these three types of failure will vary
widely and must be assessed in a given case and set against the risk of failure
assessed above.
Clearly, if the failure of a structural member leads to a widespread collapse, the
consequences will be graver than if the collapse were limited to the structure
elements involved. Possibilities of such chain action type progressive failure
must be made very remote.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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2) Relationship of Stress and Strain
a) Stress
Normal Stress
The simplest form of stress is normal stress, which is the stress perpendicular to
the surface on which it act.
where σ is the normal stress,
P is the centric axial load, and
A is the area of the section.
The normal stress is usually expressed in Pascal (Pa), where one Pascal is equal
to one Newton per square meter, that is, 1 Pa = 1 N/ m2
. A Pascal is a very small
unit of stress, so one can usually expect to see stresses expressed in kPa or Mpa.
Shear Stress
Shear stress is stress parallel to the surface on which it acts.
where τ is the shear stress acting on the surface,
V is the force acting parallel to the surface and
A is the area on which the shear stress acts.
The unit for shear stress is same as that for normal stress, ie. Pascal (Pa). However,
the stress direction is parallel to the surface under force. The above formula
assumes that the shear stress is uniformly distributed over the surface.
Force P σ =
Area =
A
Force V τ =
Area =
A
V V
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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NORMAL STRESS
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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SHEAR STRESS
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Example 1
A steel bar of rectangular cross-section, 3cm by 2cm, carries an axial load of
40kN. Estimate the average tensile stress over a normal cross-section of the bar.
Solution
The area of a normal cross-section of the bar is
A = 0.03 x 0.02 = 0.6 x 10-3
m2
The average tensile stress over this cross-section is then
Example 2
A steel bolt, 2.5cm in diameter, carries a tensile load of 40kN. Estimate the average
tensile stress at the section a and at the screwed b, where the diameter at the root of
the thread is 2.1cm.
Solution
The cross-sectional area of the bolt at the section a is
π Aa =
4 (0.025)2 = 0.491 x 10
-3 m
2
The average tensile stress at A is then
P 40 x 103
σa = Aa
= 0.491 x 10
-3
= 81.4MN/m2
The cross-sectional area at the root of the thread, section b, is
π Ab =
4 (0.021)2 = 0.346 x 10
-3 m
2
The average tensile stress over this section is
P 40 x 103
σb = Ab
= 0.346 x 10
-3
= 115.6MN/m2
P 40 x 103
σ = A
= 0.6 x 10
-3
= 66.67 MN/m2
2 cm
40kN
Area of normal cross-section = 6cm2
40kN
3 cm
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Example 3
Two pieces of plastic are joined by gluing overlapping areas of 50 by 70 mm, as
shown below. If a tensile force of 780N (V) applied parallel to the glued surfaces
causes the glue to fail, at what shear stress did the glue fail?
Solution
The shear stress acting on the glued surface:
780 τ =
50 x 70 x 10-6
= 222860 Pa
= 222.860 kPa
= 223 kPa
Note that if the stress is to be expressed in Pa, the area must be given in m2.
Force V τ =
Area =
A
V V
V τ
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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b) Strain
Normal Strain
A member carrying a tensile load will stretch. The stretch is usually called
deformation, and the symbol for deformation is ∆L. To compare one material to
another, or one sample of the same material to another, it is customary to use strain,
which is the deformation per unit of length.
Because both ∆ and L have the same units, either mm or m, ε will be unitless although
it is customary to express the units for as m/m.
Shear Strain
If a small square element is cut out of a body where a shear stress acts on one surface,
to maintain equilibrium in the y direction, there must also be an equal shear stress on
the opposite face.
The stresses would form a couple, and in order to maintain rotational equilibrium, an
equal and opposite couple must exist. The combination of forces due to the shear
stress on each face would cause the element to deform as shown:
The shearing strain is
But for small angles, measured in radians, tan γ = γ . Thus,
* Note that ∆s /L has no units and therefore γ has no units.
Normal Deformation, ∆L ∆L Normal Strain, ε =
Original Length, L =
L
∆s
L = tan γ
∆s
L = γ
∆s
L γ
P
∆L L
P
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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EXAMPLE 1
A cylindrical block is 40 cm long (L) and has a circular cross-section 10 cm (d) in
diameter. It carries a total compressive load of 90kN (P), and under this load contracts
0.04 cm (∆L). What is the average compressive stress over a normal cross-section and
the compressive strain ?
SOLUTION
The area if a normal cross-section is
π Aa =
4 (0.10)2 = 7.85 x 10
-3 m
2
The average compressive stress over this cross-section is then
P 90 x 103
σ = Aa
= 7.85 x 10
-3
= 11.46MN/m2
The average compressive strain over the length of the cylinder is
∆L 0.04 x 10-2
ε =
L =
40 x 10-2
= 0.01
90kN
90kN
40 cm long
Diameter = 10 cm
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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c) Stress – Strain Curves
In a testing machine, incremental force is applied to the specimen. The load and
change of length of the specimen are recorded at the different stages of loading. Stress,
Ơ, and strain, ε, are then calculated from
where P is the applied load and A is the area of the cross-section of the specimen. The
change of the area of the cross-section is usually ignored and the original area is used.
∆ L and L are the change of length and the original length of the specimen.
∆L ε =
L
P σ =
Aa and
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Mechanical Properties of Materials
The stress-strain curves shown are for three different kinds of material;
1. Low-carbon steel, a ductile material with a yield point.
2. A ductile material, such as aluminum alloy, which does not have a yield point.
3. A brittle material, such as cast iron or concrete in compression.
There are several points of interest which can be identified on the curves as follows:
1. Proportional limit: the maximum stress for which stress is proportional to strain.
(Stress at point P.)
2. Yield point: stress for which the strain increases without an increase in stress.
(Horizontal portion of the curve ab. Stress at point Y.)
3. Yield strength: the stress that will cause the material to undergo a certain specified
amount of permanent strain after unloading. (Usual permanent strain ε1 = 0.2
percent. Stress at point YS.)
4. Ultimate strength: maximum stress material can support up to failure. (Stress at
point U.)
5. Breaking strength: stress in the material based on original cross-sectional area at
the time it breaks. Also called fracture or rupture strength. (Stress at point B.)
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Tension test and Stress-Strain Curves
The mechanical properties of materials used in engineering are determined by tests
performed on small specimens of the material. The tests are conducted in
materials-testing laboratories equipped with testing machines capable of loading the
specimens in a variety of ways, including static and dynamic loading in tension and
compression. A tensile-test machine is shown.
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d) Hooke’s Law & Poisson’s Ratio
HOOKE’S LAW
For many materials, the lower end of the stress-strain curve is a straight line. This
behavior was recognized by Robert Hooke (1635-1703) and stated as Hooke’s Law:
“For an elastic body, stress is proportional to strain.”
σ = E ε
Hooke’s Law applied only up to the proportional limit of the material.
The constant E is called the elastic modulus, modulus of elasticity, or Young’s modulus.
E is equal to the slope of the stress-strain curve.
Stress σ E =
Strain =
ε
Since strain is dimensionless, E has the same units as stress, e.g. Pa, MPa. The value
of E for a given material is a constant. Materials with a high modulus of elasticity
have a high resistance to elastic deformation, and are said to be stiff.
HOOKE’S LAW FOR SHEAR
Experiments show that if the stress does not exceed the proportional limit, shearing
stress is also proportional to shearing strain.
Hooke’s law for shear may be expressed as
τ = G γ
Where G is the shear modulus (or modulus of
rigidity) of the material. G has the same units as
stress, e.g. Pa, Mpa.
The shear modulus, G, is another elastic constant for a given material. It measures the
resistance of a material to elastic shearing deformation.
Stress σ
Strain ε
E
1
τ
γ
G
1
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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EXAMPLE 1
An aluminum alloy sample is tested in tension. When the stress is 121 MPa the
normal strain is 1.7 x 10-3
m/m. Calculate the modulus of elasticity for this alloy.
SOLUTION
The modulus of elasticity:
σ
E =
ε
121 x 106
=
1.7 x 10-3
= 71.2 GPa
EXAMPLE 2
A 2 m long round bar of polystyrene plastic with a diameter of 25 mm carries a 4kN
tensile load. If the modulus of elasticity of the polystyrene is 3.0GPa, calculate the
longitudinal deformation in the bar.
SOLUTION
The deformation is
∆ = ε L
σ =
E L
P L =
A E
4000 x 2 =
(π / 4 x 252)x 10
–6 x 3 x 10
9
= 5.43 x 10-3
mm
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Poisson’s Ratio
When a load is applied along the axis of a bar, axial strain is produced. At the same
time, a lateral (perpendicular to the axis) strain is also produced. If the axial force is in
tension, the length of the bar increases and the cross-section contracts or decreases.
That is, a positive axial stress produces a positive axial strain and a negative lateral
strain. For a negative axial stress, the axial strain is negative and the lateral strain is
positive.
The ratio of lateral strain to axial strain is called Poisson’s ratio. It is constant for a
given material provided that the material is not stressed above the proportional limit,
is homogeneous, and has the same physical properties in all directions. Poisson’s ratio,
represented by the Greek lowercase letter v (nu), is defined by the equation
The negative sign ensures that Poisson’s ratio is a positive number.
The value of Poisson’s ratio, v, varies from 0.25 to 0.35 for different metals.
For concrete it may be as low as v = 0.1 and for rubber as high as v = 0.5.
E, G, and v Relationships
It can be shown that the three elastic constants E, G, and v are not independent of each
other for an isotropic material. An isotropic material has the same properties in all
directions. The relationship between elastic constants is given by the equation
E G =
2(1 + v)
For most materials, v is in the neighborhood of ¼.
lateral strain lateral strain v =
axial strain = -
axial strain
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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MECHANICAL PROPERTIES OF SELECTED MATERIALS
Material Yield Strength
MPa
Ultimate
Strength
MPa
Young’s
Modulus
MPa
Shear
Modulus
MPa
Poisson’s
Ratio
METALLIC MATERIALS
Aluminum
2014-T6
6061-T6
Copper – Hard
Cast Iron
Gray
Malleable
Magnesium – Extruded
Steels
0.2% Carbon – Hot-rolled
0.8% Carbon – Hot-rolled
Stainless 18-8
Structural – 230G
Structural – 400G
NONMETALLIC MATERIALS
ABS Plastic
Concrete
Low Strength
High Strength
Vinyl – Rigid
Wood – Parallel to Grain
Douglas Fir – Dry
Red Oak - Dry
365
240
300
50
220
240
240
480
1140
230
400
-8
56
-44
58
-32
415
260
330
170
-690
340
340
410
830
1300
400
600
41
-21
-45
48
-51
-48
73
69
110
100
170
45
207
207
190
200
200
2.1
21
34
3.0
13
13
12
12
28
26
40
40
17
80
80
86
77
77
0.34
0.34
0.35
0.25
0.32
0.29
0.29
0.30
0.30
0.20
0.20
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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EXAMPLE 1
Determine the value for G for tin if E is known to be 20GPa and v is 0.36.
SOLUTION
E
G =
2(1 + v)
20 x 109
=
2(1 + 0.36)
= 7.35 x 109 Pa
EXAMPLE 2
A table of mechanical properties gives E = 96 GPa and G = 36 GPa for a titanium
alloy, but does not give the value for v.
Calculate v.
SOLUTION
E
v =
- 1
2G
96 x 109
=
- 1
2 x 36 x 109
= 1.3333 - 1
= 0.3333
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Axially loaded members
From the definition for strain,
δ
ε =
or δ = εL
L
From Hooke’s law,
σ = E ε
When the stress and strain are caused by axial loads, we have
P δ
= E
A L
PL
δ =
AE
δ
L
P
P
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Shear Stress in Bending
Consider a simply beam with a concentrated load at mid-span. If we cut the beam at
any transverse cross-section, a shear force V exists at the section to maintain
equilibrium. The shear force V is distributed in the form of vertical shear stresses
acting over the face of the section.
The vertical shear stresses are not uniform in magnitude over the face of the section.
It can be shown that the shear stresses:
(a) are highest at the neutral axis of the beam,
(b) are zero at the free surfaces (i.e. the top and bottom surfaces of the beam),
and
(c) varies with the distance from the neutral axis.
An important feature of the vertical shear stresses is that they give rise to
complementary horizontal shear stresses. At any point in a beam, the horizontal and
vertical shear stresses are always numerically equal in magnitude.
Vertical shear stresses developed at the section and acting over the face of the section
provide the internal resisting shear, V
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Horizontal shear stresses in a beam
The numerical magnitudes of the horizontal and vertical shear stresses at any point in
a beam are always equal.
(a) Loaded beam (b) Stresses acting on a portion of the beam.
(c) Horizontal shear stresses acting
on the upper and lower faces of an
elemental area in the beam. The
stresses produce a rotational moment.
(d) The rotational moment produced by the
horizontal shear stresses is exactly balanced by
a rotational moment associated with vertical
shear stresses. The rotational moment, hence
stresses, must be equal at the point.
(e) Horizontal shear-stress distribution
in a rectangular beam.
(f) Vertical shear-stress distribution in
rectangular beams.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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GENERAL SHEAR STRESS FORMULA FOR BEAMS
The shear stress (at the “imaginary cut”) is given by
VQ
τ =
It
where V = Shear Force at the section
I = Moment of inertia of the entire cross-section
t = Width of the section (at the “imaginary cut”)
and Q = Apy
Ap= The cross-sectional area above the “imaginary cut”
(the shaded area fghj in the diagram)
y = Distance from Neutral Axis to Centroid of the shaded area Ap
Note:
(a) The vertical shear stress at a point is equal to the horizontal shear stress.
(b) Using the area below the “imaginary cut” for Ap should give the same result.
For Rectangular Section (t = b), the maximum shear stress (at the Neutral Axis) is
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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EXAMPLE 1
If the maximum transverse shear force in a
beam with a section 40 by 90 mm deep
is 18kN, determine the value for the
maximum longitudinal and transverse shear
stresses in the beam.
SOLUTION
Both the longitudinal and transverse shear stresses have the same value at any point in
the beam. For a beam with a rectangular section, the maximum value occurs at the
neutral axis.
b h3
I = 12
40 x 903
= 12
= 2.430 x 106 mm
4
Q is the first moment about the neutral axis of the area from the plane where the stress
is evaluated to the outside fibers, and is shown shaded.
Q = y A
45 =
2 x 40 x 45
= 40 500 m3
V Q τ =
I t
18 000 x 40 500 x 10-9
=
2.430 x 106 x 10
-12 x 40 x 10
-3
= 7.5000 x 106 Pa
= 7.5000 MPa
40
45
45
y
N. A.
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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3) Analysis of Structure
a) Forces system (Force & Moment)
Force has been defined as the action of one body on another. It is a vector
quantity because its effect depends on the direction and on the magnitude of the
action and because it may be combined according to the parallelogram law of
vector combination.
Moment of a force is the rotational tendency of the force to move a body in the
direction of the force application. The magnitude of moment of a force to rotate a
body above an axis normal to the plane of the body is proportional to both the
magnitude of the force and to the moment arm, which is the perpendicular
distance from the axis to the line of action of the force.
Forces and moments are vector quantities and may be resolved into components;
that is to say, a force or a moment of a certain magnitude and direction may be
replaced and exactly represented by two or more components of different
magnitudes and in different directions.
Considering firstly the two-dimensional case shown below, the force F may be
replaced by the two components Fx and Fy provided that :
Fx = F cos θ and Fy = F cos θ
y
x
Fy
Fx
F
θ
Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials
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b) Types of structures and Support conditions
Type of structures :
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Support conditions :
Structures are either partially or completely restrained so that they cannot move freely
in space. The restraints are provided by supports that connect the structure to some
stationery body, such as the ground or another structure. There are generally three
types of supports and they are : -
b) Pin or hinged support
c) Roller support
d) Fixed support
Pin or Hinged support:-
A hinge support is represented by the symbol below :
It can resist force in any direction but cannot resist a moment.
Roller support :-
A roller is represented by the symbol below :
It can resist only force normal to the supporting surface through the center of the
connecting pin.
Fixed end support :-
A hinge support is represented by the symbol below :
It can resist force in any direction and moment about the fixed end. Therefore, it can
prevent the end from both translation and rotation.
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c) Equilibriums
A structure is considered to be in equilibrium if it remains at rest when subjected to a
system of forces and moments. If a structure is in equilibrium, then all its members
and parts are also in equilibrium.
For a structure to be in equilibrium, all the forces and moments (including support
reactions) acting on it must balance each other. For a plane structure subjected to
forces in its own plane, the conditions for equilibrium can be expressed by the
following equations of equilibrium :
Σ Fx = 0 Σ Fy = 0 Σ Mz = 0
The third equation above states that the sum of moments of all forces about any point
in the plane of the structure is zero.
Equations of Condition
Sometimes internal hinges are present within a structure. An internal hinge cannot
transmit moment. Therefore, the bending moment at a hinge is zero.
The condition that the moment is zero at a hinge provides an additional equation for
analyzing the structure. Such equations are commonly called equations of condition.
Determinate and Indeterminate Structures
A structure is externally determinate if all the support reactions can be obtained by
static, i.e.
No. of Support Reactions = No. of Equations (includes Equilibrium & Conditions)
A structure which is not determinate is called indeterminate.
Degree of Statically Indeterminacy = No. of Support Reactions – No. of Equations
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d) Free-Body Diagrams
1. Free-body diagrams make use of the concept that if a whole structure is in
equilibrium, any part of the structure is also in equilibrium.
2. Free-body diagrams can be plotted for various parts of a structure and also
for the entire structure.
3. When plotting a free-body diagram, it is important to indicate all the
possible forces and moments acting on the “cut” structure concerned.
4. Internal forces common to two free-bodies (on opposite sides of the “cut”)
should be denoted as equal in magnitudes but in opposite direction.
Free-body diagrams are very useful in finding the support reactions and determining
the internal forces in structures. The use of free-body diagrams is an important tool in
structural analysis and stress analysis.
D B C
E A
D B
C
E A
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Internal Forces at Cut Section of a Structure
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e) Example
EXAMPLE 1
A beam ABCDE has a pinned support at A and a roller support at D. It carries three
concentrated loads as shown. Determine the reactions.
SOLUTION
Support A can provide two components of reaction (VA and HA). Support D, being a
roller, provides only one reaction (VD), which will act vertically (that is, at right
angles to the direction of motion of the rollers). These reactions are shown on the free
body diagram.
(1)To determine HA
(∑H = 0) There are no horizontal loads; thus,
HA= 0
(2)To determine VD
Take moments about A:
(∑MA= 0)
+ (10 × 1) + (20 × 3.5) + (5 × 5.3) – (VD × 4.5) = 0
∴(VD = +23.67 kN (i.e. 23.67 kN upwards)
A B
10kN 5 kN 20kN
1m 2.5m 1m 0.8m
VA VD
10kN 5 kN 20kN
HA
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A B
C
D
(3)To determine VA
(∑V=0)
+VA + VD = 10 + 20 + 5
+ VA+ (+23.67) – 35 = 0
∴VA = + 11.33kN
(4)Check by taking moments about D:
∑MD = + (VA × 4.5) – (10 × 3.5) − (20 × 1) + (5 × 0.8)
= + (+11.33 × 4.5) – 51 = 51 – 51 = 0
∴correct
This last calculation provides a useful check on the mathematical accuracy of the
previous two calculations.
EXAMPLE 2
Beam ABCD has a pinned support at A and a roller support at C. It carries two
concentrated loads of 30kN each and a uniformly distributed load of 5 kN/m over the
right-hand half, as shown. Determine the reactions.
30kN
5 kN/m
30kN
1m
2m 2m 2m
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30kN
5 kN/m
30kN
A B
C
D
SOLUTION
Free body diagram,
(1)To determine HA
(∑H=0) There are no horizontal loads.
∴HA = 0
(2)To determine VC
Take moments about A:
Note that the moment of the U.D. load is the resultant total U.D. load
(5 ×××× 3 = 15 kN) multiplied by the distance from A to the line of action of that
resultant (i.e. 6 – 1.5 = 4.5 m)
(∑MA = 0)
+ (30 × 2) − (VC × 4) + (5 × 3 × 4.5 ) + (30 × 6)= 0
∴VC = + 76.875 kN
(3)To determine VA
(∑V =0)
+VA − 30 + VC − (5 × 3) − 15= 0
+VA − 30 + (+76.875) − 15 − 30 = 0
∴VA= − 1.875 kN (i.e. 1.875kN downwards)
(4)Check by taking moments about C:
∑MC= + (VA × 4) − (30 × 2) + (5 ×3 ×0.5) + (30 ×2)
= + (−1.875 × 4) −60 + 7.5 + 60 = 0
∴correct
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2 kN/m
2 kN/m
5 kN/m
15m
9m
2 kN/m
2 kN/m
5 kN/m
EXAMPLE 3
Determine the reactions at the support for the frame shown.
SOLUTION
Free-Body Diagram The free-body diagram of the frame is shown in the right hand
side above. Note that the trapezoidal loading distribution has been divided into two
simpler, uniform, and triangular, distributions whose areas and centroids are easier to
compute.
+ → ∑ Fx = 0
Ax + 2(15) = 0
Ax = − 30 kN
+ ↑ ∑ Fy = 0
Ay −2(9)− (3)(9) =0
AY =31.5 kN
+ ∑MA=0
MA – 2 x 15 x 15/2 – 2 x 9 x 9/2 – 3/2 x 9 x 9/3 x 9 = 0
MA = 387 kNm
Checking Computations
∑MB = -30 x 15 – 31.5 x 9 + 387 + 2 x 15 x 15/2 + 2 x 9 x 9/2 + 3/2 x 9 x 9/3 = 0
=> checked OK
1
2
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4 x 10m = 40m
30kN
30kN
15 m
HL
VL VR
L R
EXAMPLE 4
Determine the truss reaction forces.
SOLUTION
+ ∑ML =0 = 30(20) +30(30) – 40V R VR = 37.5N↑
→ + ∑H = 0 = HL HL =0
↑ + ∑V = 0 = VL + VR –30- 30 VL = 22.5 N↑
Check
+ ∑ Mc = 0 = V L (20) + 30(10) –VR (20)
= 750 - 750 = 0 = > checked OK
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4) Beams
a) Calculations of Beams Reactions
Since the forces applied to a beam are in one plane, there are three equilibrium
equations for determining the support reactions. The equations are
where A is any point in the plane of loading.
Once the support reactions are known, internal forces and stresses in the beam may be
calculated.
Equations of Condition
Sometimes hinges or pinned joints are introduced into beams and frames. The
condition of hinges or pinned joints can provide equations for determining the support
reactions. This additional equation is called an equation of condition.
Feature of hinges or pinned joints :
1.A hinge is capable of transmitting only horizontal and vertical forces.
2.No moment can be transmitted at a hinge. (ie moment above joint = zero)
3.The bending moment at a hinge is zero.
The hinge is a particularly convenient location for “separation” of the structure into
free-bodies for computing support reactions.
∑Fx = 0 ∑Fy = 0 ∑MA = 0
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F
L
L / 2
Hinge
C A B
a
F
F / 2 F / 2
A B
F / 2
F / 2
F a / 2
C B
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INTERNAL FORCES IN A BEAM
(Shear Force, Bending Moment and Axial Force)
Before a structural member can be designed, it is necessary to determine the internal
forces that act within the member. In general, the internal forces for a beam section
will consist of a shear force V, a bending moment M, and a normal force (axial force)
N. For beams with no axial loading, the axial force N is zero.
Sign Convention (considering a small segment of the member):
Shear Force V: Positive shear ends to rotate the segment clockwise
Moment M : Positive moment bends the segment concave upward
(so as to “hold water”)
Axial Force N : Tension is positive.
An important feature of the above sign convention (often called the beam convention)
is that it gives the same (positive or negative) results regardless of which side of the
section is used in computing the internal forces.
Positive sign convention
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Procedure for finding V, M and N at a beam section
1. Compute the Support Reactions
2. Draw a Free-Body Diagram of the beam segment.
Keep all internal loading on the member in their exact location.
Draw a free-body diagram of the beam segment to the left or right of the section.
(Although the left or right segment could equally be used, we should select the
segment that requires the least computation.)
Indicate at the section the unknowns V, M and N acting in their positive
directions.
3. Use the Equations of Equilibrium to determine V, M and N,.
If the solution gives a negative value for V, M or N, the actual force or moment
acts in the negative direction.
4. Check the calculations using the opposite beam segment.
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EXAMPLE 1
Determine the shear V and the moment M at section P of the overhanging beam as
shown below.
SOLUTION
(a) First to determine reactions RB and RC. Taking moments about point C
10RB = 8(12) + 4(10)(5) − 10(3)
RB = 26.6 kN
Taking moments about point B,
10RC = 4(10)(5) + 10(13) − 8(2)
RC = 31.4 kN
Check by taking the summation of vertical forces,
8 + 4(10) + 10 = 26.6 + 31.4
58 = 58 (Ok)
(b) V and M at P using left free body,
V= sum of upward forces on ABP
= + 26.6 − 8 − 4(4) = +2.6 kN
M=sum of clockwise moments of forces on ABP about P
= 26.6(4) − 8(6) −4(4)(2) = +26.4kN m
(c) V and M at P using right free body,
V=sum of downward forces on DCP
= +10 + 4(6) − 31.4 = +2.6 kN (check)
M=sum of counterclockwise moments of forces on DCP about P
= 31.4(6) − 10(9) − 4(6)(3) = +26.4 kN m (Check)
2 m 10 m 3 m 4 m
A B C D
P
8 kN 10 kN 4 kN/m
RB Rc
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EXAMPLE 2
Determine the shear V and the moment M at section P of the cantilever.
SOLUTION
(a) Determine reactions MA and RA.
Taking moments about point A,
MA = 36(2.5) + 4(5.5)(5.5/2) = 150.5 kNm
Taking the summation of vertical forces,
RA = 36 + 4(5.5) = 58 kN
A check may be made by taking moments about point C, although it does not seem to
add much to an independent check. If done, it is
10.5 +36(3) + 4(5.5)(5.5/2) = 58(5.5)
319 = 319 (OK)
(b) V and M using left free body,
V = sum of upward forces on ABP
= + 58 −36 − 4(3.5) = +8 kN
M = sum of clockwise moments of forces on ABP about P
= -105.5 + 58(3.5) −36(1) − 4(3.5)(3.5/2) = −8 kNm
(c) V and M using right free body.
V = sum of downward forces on CP
= + 4(2) = +8kN (Check)
M = sum of counterclockwise moments of forces on CP about P
= − 4(2)(1) = −8kN.m (Check)
A P
RA
B C
3.5 m
36 kN
4 kN/m 2.5 m
5.5 m
MA
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b) Shear and Bending Diagrams
By the methods discussed before, the magnitude and sign of the shear forces and
bending moments may be obtained at many sections of a beam. When these values are
plotted on a base line representing the length of a beam, the resulting diagrams are
called, respectively, the shear diagram and the bending moment diagram.
Shear and moment diagrams are very useful to a designer, as they allow him to see at
a glance the critical sections of the beam and the forces to design for Draftsman like
precision in drawing the shear and bending moment diagrams is usually not necessary,
as long as the significant numerical values are clearly marked on the diagrams.
The most fundamental approach in constructing the shear and bending moment
diagrams for a beam is to use the procedure of sectioning. With some experience, it is
not difficult to identify the sections at which the shear force and moments should be
calculated directly. The shape of the shear and bending moment diagrams between
these sections are also readily identified after some experience and can be sketched in.
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EXAMPLE 1
Draw the shear and moment diagrams for the beam shown.
∑Mc =0
−(AY × 8) + 2 × 8 × 4 + 6 × 3 = 0
∑Fy = 0
10.25 − 2 × 8 −6 + CY = 0
CY= −10.25 + 16.00 + 6.00 = 11.75MN
For 0 ≤ x ≤ 5 m:
∑Fy = 0
10.25−2 × x – V = 0
V =10.25− 2×(MN)
∑MO = 0
x −(10.25×x)+2×x×
2 +M =0
M =10.25x − x2(MNm)
For 5 ≤ x ≤ 8 m:
∑FY = 0
10.25 −2 × x − 6−V=0
V=10.25 − 6 −2x
=4.25− 2x(MN)
∑MO=0
x −(10.25×x)+2×x×
2 +6(x-5)+M =0
M =10.25x − 6(x−5) − x2(MNm)
64.00+18.00 Ay =
8
82.00 =
8 =10.25MN
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EXAMPLE 2
Draw the shear and moment diagrams for
the beam shown.
∑MC = 0
50×6−BY×4=0
50×6 By=
4
=75.000kN
∑FY=0
−50+75.000+Cy = 0
CY=50.000-75.000
= - 25.000kN
For 0 ≤ x ≤ 2 m
∑Fy=0
-50-V=0
V=-50.00(kN)
∑ MO = 0
50 × x + M = 0 M = -50.00x(kNm)
For 2 ≤ x ≤ 6 m
∑Fy= 0
-50+75.00-V=0
V=-50.00 + 75.00
=25.00(kN)
∑MO=0
50 × x - 75.00(x-2) + M=0
M = -50.00x + 75.00(x-2)(kNm)
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EXAMPLE 3
Sketch the Shear Force diagram and Bending Moment diagram for the cantilever
shown below.
SOLUTION
Even though there is no loading on AB and therefore no shear force or bending
moment, the distance x will still be measured from A rather than B.
Shear force
0<x<2 Q=0
2<x<5 Q=-5000-2000(x-2)
5<x<8 Q=-5000-6000-3000=-14000N
Bending moment
0<x<2 M = 0
(x-2) 2<x<5 M = -5000(x-2)-2000(x-2)
2
= -1000(x-2)(x+3)Nm
(parabolic distribution)
1 5<x<8 M = -5000(x-2)-6000(x-3
2 )-3000(x-5)
= -14000x+46000Nm (linear distribution)
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SHAPES OF SHEAR & BENDING MOMENT DIAGRAMS
A. Beams under Point Loads only
1. Shears are constant along sections between point loads.
2. The shear diagram consists of a series of horizontal lines.
3. Moment vary linearly between point loads
4. The moment diagram is composed of sloped lines.
B. Beams under Uniformly Distributed Loads (UDL) only
1. A Uniformly Distributed Lad produces linearly varying shear forces.
2. The shear diagram consists of sloped line or a series of sloped lines.
3. A UDL produces parabolically varying moments
4. The moment diagram is composed of 2nd –order parabolic curves.
Characteristic of shear force and bending moment diagrams of beams under the
following different loading cases:
1. Section with No Load:
Shear Diagram is a Horizontal Straight Line
Moment Diagram is a Sloping Straight Line
2. At a Point Load:
There is a Jump in the Shear Diagram
3. At a Point Moment :
There is a Jump in the Moment Diagram
4. Section under UDL:
Shear Diagram is Sloping Straight Line(1 st order)
Moment Diagram is a curve (2 nd order parabolic)
5. Section under Linearly Varying Load:
Shear Diagram is a Curve (2 nd order)
Moment Diagram is a curve (3 rd order)
The Curve of the Bending Moment Diagram is 1 order above the curve of the
Shear Diagram.
Maximum and Minimum Bending Moments occur where the Shear Diagram
Passes through the X-axis(i.e. at points of zero shear).
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Sharp of Shear and Bending Moment diagrams
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c) Deflections
The qualitative deflected shape (also called “elastic curve”) of a beam is simply an
approximate exaggerated sketch of the deformed beam due to the given loading. The
deflected shape is useful in understanding the structural behavior.
A sketch of the deflected shape can be made by making use of the bending moment
diagram:
Where the moment is positive, the deflected shape should be concave upward.
Where the moment is negative, the deflected shape should be concave downward.
Where the moment is zero, there is a point of zero curvature.
At a roller or pin support, the vertical deflection must be zero and the member may
rotate at that support.
At a fixed support, the rotation must also be zero
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Principle of Superposition :
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EXAMPLE 1
Use the methods of superposition to find the deflection at the free end of the
cantilever beam as shown.
SOLUTION
Two basic loadings are used as shown in diagrams (b) and (c) above.
The deflection at A is equal to the sum of the deflections under the Point Load and the
Uniform Distributed Load.
P L3 w L
4
δA = 3 EI
+ 8 EI
2 (9)3 0.5 (9)
4
δA = 3 EI
+ 8 EI
896 δA =
E I
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EXAMPLE 2
Use the method of superposition to find the deflection at the middle of the simply
supported beam shown below:
SOLUTION
Two basic loadings are used. The deflection at point M is equal to the sum of the
deflections due to the point load and UDL.
P a 5 w L4
∆M = 48 EI
(3L2 – 4a
2) +
384 EI
10 (3) 5 (1.5)(12)4
∆M = 48 EI
(3(12)2 – 4(3)
2) +
384 EI
652 ∆M =
E I
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5) TRUSSES
A truss is a structure composed of straight slender members joined together at their
end points. The members are usually formed into triangular patterns to produce an
efficient light-weight, load bearing structure.
The joint connections are usually formed by bolting or welding the ends of the
members to a common plate, called a gusset plate, or by passing a large bolt or pin
through each of the members. In the analysis of trusses, the members are commonly
assumed to be connected at the joints by frictionless pins. Since no moment can be
transferred through a frictionless pin joint, truss members are assumed to carry only
axial force.
Planar trusses lie in single plane and are often used to support roofs and bridges.
Space trusses, which are three-dimensional structures, are also in common use.
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CONVENTONAL TYPES OF BRIDGE AND ROOF TRUSSES
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Space Trusses
A space truss consists of members joined together at their ends to form a stable
three-dimensional structure.
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ANALYSIS OF TRUSSES
A truss is completely analyzed when the support reactions and the forces in all the
members are determined. The members forces may be analyzed by:
(1) the method of joints (considering the equilibrium of a joint),
(2) the method of sections (considering the equilibrium of a section of the truss),
or a combination of the two methods.
The analysis of a truss is based on the following assumptions:
1. Members are connected to joints by frictionless hinges.
2. Loads and reactions are applied only at the joints.
3. Members are straight and carry only axial load.
Although members in real trusses are commonly connected to gusset plates by welded
or bolted connections, the forces obtained under the above assumptions are reasonably
accurate if the members are relatively slender and the connections are such that the
centroidal axes of the members meeting at a joint are concurrent at a point.
Sign Convention
The convention is to label a tension force positive and a compression force negative.
Alternatively, we add a T after the numerical value of a force to indicate tension and a
C to indicate compression.
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(a) Methods for Analysis of Plane Trusses
The member forces in a statically determinate truss can be found by making use of the
Equations of equilibrium. The process is to consider different free-body diagrams of
Parts of the structure.
The two common techniques are:
(1) Method of Joints
In this method, a free-body of each joint is considered, one joint at a time.
Two independent equilibrium equations are available for each joint.
You should work each time with at a joint with only two unknown member
forces Once the unknown forces at one joint are determined, they become known forces
for other joints.
(2) Method of Sections
In the method, a larger part of the structure is taken as a free-body.
All three equations of equilibrium are available and three unknown bar forces
can be determined.
This method is particularly useful when only certain bar forces are required.
In practice, it is often convenient to use a combination of the two methods. The
key is to choose the most convenient free-body diagrams.
(a) Section 1.1 (b) Section 2.2
Method of Joints Method of Sections
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(1) The Method of Joints
If a truss is in equilibrium, then each of its joints must also be in equilibrium. Hence, the method of joints consists of satisfying the Equilibrium conditions :
ΣFx = 0 andΣFy = 0
for the forces exerted on the pin at each joint of the truss.
Procedure for Analysis:
First determine the reactions at the supports by considering the equilibrium of the
entire truss. Then draw the free-body diagram of a joint having at least one known
force and at most two unknown forces. (If this joint is at one of the supports, it may be
necessary to know the external reactions at the support.) By inspection, attempt to
show the unknown forces acting on the joint with the correct sense of direction. The x
and y axes should be oriented such that the forces on the free-body diagram can be
easily resolved into their x and y components, and then apply the two force
equilibrium equations ΣFx = 0 and ΣFy = 0. It should also be noted that if one of
the axes is oriented along one of the unknown forces, the other unknown can be
determined by summing forces along the other axis. Solve for the two unknown
member forces and verify their correct directional sense.
Continue to analyze each of the other joints, where again it is necessary to choose a
joint having at most two unknowns and at least one known force, realize that once the
force in a member is found from the analysis of a joint at one of its ends, the result
can be used to analyze the forces acting on the joint at its other end. Strict adherence
to the principle of action, equal but opposite force reaction, must, of course, be
observed. Remember, a member in compression” pushes” on the joint and a member
in tension “pulls “ on the joint.
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(2) The Method of Sections
If the forces in only a few members of a truss are to be found, the method of sections
generally provides the most direct means of obtaining these forces. The method of
sections consists of passing an imaginary section through the truss, thus cutting it into
two parts. Provided the entire truss is in equilibrium, each of two parts must also be in
equilibrium; and as a result, the three equations of equilibrium may be applied to
either one of these two parts to determine the member forces at the “cut section”
Procedure for Analysis
Free-Body Diagram Make a decision as to how to “cut “ or section the truss
through the members where forces are to be determined. Before isolating the
appropriate section, it may first be necessary to determine the truss’s external
reactions, so that the three equilibrium equations are used only to solve for member
forces at the cut section. Draw the free-body diagram of that part of the sectioned
truss which has the least number of forces on it. By inspection, attempt to show the
unknown member forces acting in the correct sense of direction.
Equations of Equilibrium Try to apply the three equations of equilibrium such that
simultaneous solution of equations is avoided. In this regard, moments should be
summed about a point that lies at the inter-section of the lines of action of two
unknown forces; in this way, the third unknown force is determined directly from the
moment equation. If two of the unknown forces are parallel, forces may be summed
perpendicular to the direction of these unknowns to determine directly the third
unknown force.
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Applying the Method of Sections
In applying the method of sections, two decisions are made:
(1) Choosing the free-body,
(2) Choosing the points for taking moments about.
In choosing the free-body, remember that the “cut” does not need to be a straight
line.
In choosing a point for taking moment, remember that it can be any point in the
plane. It does not have to be at a joint or a support.
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Zero-Force Members
Truss analysis using method of joints is greatly simplified if one is able to first
determine those members that support no loading. These Zero-force members may be
necessary for the stability of the truss during construction and to provide support if the
applied loading is changed. The zero-force members of a truss can generally be
determined by inspection of the joints.
CASE 1 If no external load is applied to a joint that consists of two bars, the
force in both bars must be zero.
CASE 2 If no external load acts at a joint composed of three bars --- Two of which are
collinear, the force in the bar that is not collinear is zero.
ΣFy = 0 requires F3 = 0
ΣFx = 0 requires F1 = F2
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Zero-Force Members
EXAMPLE
EXAMPLE
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EXAMPLE
EXAMPLE