2. WEEK 1
• INTRODUCTION:
• Introduction to structural engineering, the analysis and design process: structural
forms, nature of loads, building materials;
– Types of supports, support reactions, degrees of static and kinematic indeterminacy for
planar and spatial structures, degrees of stability and redundancy members.
• Concept of axial shear, torsional, flexural rigidities and stiffness of structural members
• LEARNING OUTCOMES:
• To understand the concept and principles of structural analysis.
• To indentify and recognize the different types of loads, , supports and degree of
determinacy of structures.
• Familiarize with the concept of axial shear, torsional, flexural rigidities and stiffness of
structural members
4. Final Average Final Grade Final Average Final Grade
96≤ 𝑥 < 100 1.0 71≤ 𝑥 < 76 2.25
91≤ 𝑥 <96 1.25 61≤ 𝑥 < 71 2.5
86 ≤ 𝑥 <91 1.5 56≤ 𝑥 < 61 2.75
81 ≤ 𝑥 < 86 1.75 50≤ 𝑥 < 56 3.0
76≤ 𝑥 < 81 2.0 Below 50 Failed
Inc incomplete UD Unofficially Dropped
The final grade will correspond to the weighted average
scores shown below:
5. • Structural Theory – it is the method or tool by which we find out a
structure or a member of a structure behaves when subjected to certain
excitation. In other words finding out internal forces ( axial force, shear
force, moment), stress, strain, deflection, etc in a structure under applied
load conditions.
https://temple.manifoldapp.org/read/structural-
analysis/section/4681b7f4-641d-4afb-9e20-7558a
6. TYPES OF STRUCTURES
• 1. Beams- is the simplest of all structures which is a straight member
subjected to a transverse loads.( simply supported, cantilever, fixed
support, continuous beams)
•
7. 2. Planar and Space Trusses – For longer spans trusses are often used
instead of beams. Trusses support the loads primarily by axial forces
in the members. Triangular arrangements of members are used in
trusses.
8. • 3. . Arches and Cables- for longer span another type of structure is the
arch. It is characterized by low bending moment and large compressive
forces with supports capable of resisting horizontal forces due to the
tendency of the arch to flatten out under load. (suspension bridges).
Basically, a cable acts as tension and tends to lengthen under load.
•
9. .4. Rigid Frames – are structures used for buildings. They are assembled
for moment members joined at the ends by moment resisting
connections. From a structural standpoint rigid frames are often highly
statically indeterminate.
•
10. • 5. Composite Structures – Structures in which some of the members are
subjected to axial forces only, while some of the members are subjected
to bending are called composite structures.
11. BEAMS
• Members that are slender and support loads applied perpendicular to
their longitudinal axis.
Span, L
Distributed Load, w(x) Concentrated Load, P
Longitudinal
Axis
12. TYPES OF BEAMS
• Depends on the support configuration
M
Fv
FH
Fixed
FV
FV
FH
Pin
Roller
Pin
Roller
FV
FV
FH
14. • Different types of beams can be classified based on the type of support.
The four different types of beams are:
• Simply Supported Beam
• Fixed Beam
• Cantilever Beam
• Continuously Supported Beam
• Simply Supported Beam
• https://www.youtube.com/watch?v=hVFjJJLJizA
If the ends of a beam are made to rest freely on supports beam, it is
called a simple (freely) supported beam.
15. • Fixed Beam
If a beam is fixed at both ends it is free called fixed beam. Its another name is a built-in
beam or encase beam
Cantilever Beam
If a beam is fixed at one end while the other end is free, it is called cantilever beam.
16. • Continuously Supported Beam
If more than two supports are provided to the beam, it is called continuously
supported beam.
• Continuously Supported Beam
If more than two supports are provided to the beam, it is called
continuously supported beam.
17. • An overhanging beam, illustrated in Fig.4.1(c), is supported by a pin and a roller
support, with one or both ends of the beam extending beyond the supports.
• Overhanging Beam
18. • BUILDING LOADS
• The loads to be considered to be considered in designing a building are
the dead and live loads.
• Dead loads are static forces that are relatively constant for an extended
• time. They can be in tension or compression. Any part of a building
which is permanently installed heating to the total dead loads.(floor and
roof materials, beams, girders and columns, including the footings,
ceiling materials, including heating and air conditioning ducts and
electrical power installations, mechanical equipment, including elevators
and escalators.
• Live Loads- are transient and movable loads within the buildings. Wind,
earthquake, snow loads are also considered as live loads. (People in
class rooms, auditoriums, theater, and other buildings where people
assemble, office equipment and machines, furniture partitions which are
move occasionally, products in a warehouse & storage buildings.
19.
20. • Recommended values of live load are usually specified in building
codes. Reduction in the tabulated live loads is allowed by most building
codes when the area involved is very large. The reduction is the least of
the following values of R.
• R = 0.0086 x area ( when the area exceeds 11.2 𝑚2)
• R =
𝐷+𝐿
4.33𝐿
R≤ ( 0.40 for beams and girders, 0.60 for vertical
• members)
• Where: R = reduction factor
• D = dead loads, kPa
• L = Live load, kPa
• Live load reductions are not allowed for places of public assembly,
warehouses, garages and roofs.
21. • Wind loads
• Due to frequent occurrence of typhoons, wind loads must be included in
the analysis and design of buildings, especially for high rise structures.
Wind pressures can be computed approximately as:
• q = 0.0000473𝑉2
kPa ( V in km/hr)
• q = 0.00256𝑉2 psf (V in miles/hr)
• Earthquake loads
• The Phils. Falls within one of the most active earthquake zonesin the
world; so it is nec. To consider seismic forces in the design of buildings,
bridges and other structures. An earthquake causes an acceleration of
the ground surface. This acceleration has both vertical and horizontal
components. The vertical components is usually assumed to be
negligible, but the horizontal component is one of that can cause
considerable damaged and destruction.
• The minimum lateral earthquake force for which a building is to be
designed can be computed from the formula:
22. C factor is based on the fundamental building period T:
C =
1
15√𝑇
≤ 0.12
And the product of the factors C and S is to be
CS ≤ 0.14
For SI units, the building period T is computed byType equation here.
T =
0.91 𝐻
√𝐷√
(seconds)
Where H = height of the building, meters
D = dimension of the building in the direction being analyzed, meterspWhen the
building frame is made up of ductile material such as steel, the approximate building
period maybe calculated by
T = 0.10 x number of stories above the base
The lateral force from the eqn 1 is to be distributed as follows
F top = 0.07 TF ≤ 0.25𝐹
But F top is zero when T = 0.70 second. For any story at a height hn from ground
Fn = (F – F top) =
𝑊ℎ ℎ𝑛
∑𝑊𝑛 ℎ𝑛
F = ZIKCSW
23. • The terms in the preceding equations not previously defined are:
I = Occupancy importance factor
1.5 for essential facilities
1.25 for assembly halls
1.0 for all other occupancy
K = lateral force coefficient based on the building tyoe
S = numerical coefficient for the site resonance
1.5 unless site geothecnical data are available
W = total dead load + 25% of live loads for storage or warehouse
buildings
Wn = load on the nth floor
Z = earthquake zone coefficient
24. • Load combinations
• Knowing the different source of loading of a structure, the designer
must consider the different combinations of loads which have the most
unfavorable effect on the structure. Some load combinations are:
• 1. Dead load + Live load
• 2. Dead Load + Live Load + Wind Load
• 3. Dead Load + Live Load + Earthquake Load
• Conversion Between SI and U.S. Customary Units
• Length: The basic unit of length in SI system is m
• 1 ft = 0.3048 m 1 m = 3.281 ft
• 1 in = 25.40 mm 1mm = 0.03937 in
• Mass: The basic unit of mass in SI is Kilogram (kg). Lbm = pound-
mass
• 1 lbm = 0.4536kg 1kg = 2.205 lbm
25. • Force and Mass
– Weight is defined as the force required to restrain a body against the
acceleration of gravity. The unit of force in Si system is Newton (N). A
newton is defined as the force that gives an acceleration of
1m/𝑠2
𝑡𝑜 𝑎 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑘𝑔.
The expression for the weight of 1kg mass using the standard value for the
acceleration of gravity g = 9.807 m/𝑠2
is
W = mg = (1kg) (9.807m/𝑠2
) = 9.807 𝑁
1 lbf = (0.4536 kg) (9.807m/𝑠2) = 4.448 N
The US Customary unit kilopound (kip) is equal to a force of 1000 lb.
1 kip = 4.448 x 103 N
Stress
The unit of stress in the unit system is the Pascal (Pa) which is defined as
1 N/𝑚2
.
26. • ANALYSIS OF ROOF TRUSSES
• In building design wherein an open unobstructed space with a
width is to be of more than 40 f(12.2m)t and 50 ft (15.24m)is to be
provided, the roof is commonly supported by roof trusses spaced
from about 15 to 25 ft (4.57 to 7.62m) apart . These trusses may
rest on columns or on masonry walls along the sides of a building.
A roof truss attached to its supporting columns is commonly called
a bent, the analysis of which may be treated in the next chapter.
The discussion in the present chapter will be limited to roof trusses
supported on masonry walls. If the span of the roof trusses is small
say less than 40 or 50 ft (12.2 or 15.24m) , the truss may usually be
anchored to the wall at both ends, or the the anchor bolts may pass
through slotted holes in the bearing plate to make some provision
for expansion or contraction due to temperature changes. For
longer spans, trusses should be one end and supported on rollers or
rockers at the other end.
27. • A sketch showing the typical roof construction is shown. Here
the roof truss are supported on continuous masonry walls. The
distance between the adjacent roof trusses is called the bay. The
purlins are longitudinal beams which rest on the top chord, and
preferably at the joints of the truss. Unless the purlins are placed
at the joints, the top will be subjected to bending and direct
stresses the roof covering ( with or without sheathing) may rest
directly on the purlins or on the rafters which is in turn supported
by the purlins. A typical interior roof truss receives purlin loads
from both sides; so it supports roof loads on the equivalent of one
whole bay length. The span of the roof truss is the horizontal
distance between the supports; the rise is the vertical distance
from the reaves to eaves; the pitch is the ratio of the rise to the
span. Although the walls and the purlins assist in maintaining
longitudinal stability, additional bracing is still necessary. The
28. • Of the top chord, the bottom chord ; or both thus the names
top-chord bracing or bottom chord bracing. Although
approximate stress calculations maybe made , actual design
of the bracing system is largely a matter of experience and
judgement.
• The selection of the type of the roof truss to be used
generally depends on the length of span, the amount of
loading and the kind of materials to be used. Or vertical
members of the HOWE truss or the diagonal members of the
PRATT truss are normally in tension and may economically be
rods if the rest of the truss is built of wood. The top chord of
Howe, Pratt and Warren trusses may be horizontal (or with
sufficient slope to provide drainage) or with pronounced
slopes as indicated. The Fink and the Fan trusses, with or
29. • Light from windows facing to the north is desired. The
scissors truss is often used in church structures.. Three hinged
aarches with or without the rods, are often used in buildings
with long spans, such as armories or gymnasiums.
• DEADLOADS, SNOWLOADS AND WINDLOADS.
• A truss carries its own weight, the weight of the bracing
and ceiling and other suspended loads from the purlins. One
of the requirements of true truss action is that the loads be
applied at the joints only. Sometimes purlins are placed on
the top chord between panel joints. In this case the loads are
distributed to the adjacent joints during the truss analysis, but
both the direct stresses as determined from the truss analysis
and the bending stresses due to the intermediate purlin loads
between panel points must be considered in the design of the
30. • And the bracing system may be assumed to be divided
among the joints on the top chord, while ceiling and
suspended loads are assumed to be carried by the
appropriate lower chord joints.
• The loads on the trusses generally consist of dead , snow
and wind loads. The dead load includes 1) the weight of the
roof covering(with or without sheathing), rafters if any, and
purlins; (2) the weight of the bracing system; (3) the weight of
the truss itself; and (4) ceiling and other suspended loads.. If
desired, 1 and 4 may be ascertained before the beginning of
the truss analysis; items 2 and 3 must be first assumed and
then reviewed after design calculations have been made.
Fortunately items 2 and 3 are usually a small part of the total
load; so a rather large error in their assumed values may have
31. • Asphalt or asbestos, various types of shingles, tiles slates or
thin concrete slabs and tar and gravel. Sheathing or rafters
may or may not be used, depending on whether the roofing
material is self supporting. The weight of item 1 above can
easily vary from 5 to 25 lb per square foot(0.240 KPa to
1.2KPa) of roof surface. The weight of a plastered ceiling
maybe 8 to 10 lb(.0.384KPa to 0.480KPa) per square foot of
horizontal surface. The weight of the bracing system may
vary from 0.5 to 1.5 lb per square foot(0.024 to 0.72KPa) of
roof surface.
• The weight of the roof truss is usually estimated by use of
an appropriate empirical formula. Two of these are given
here, others are given in handbooks.
• For wooden roof truss: w = 0.5 + 0.075L (H. S. Jacoby)
• For steel truss : w = 0.4 + 0.04L (C. E. Fowler)
32. • In the above formulas, w is the weight of the truss in lbs per
square foot of horizontal surface, and L is the span in feet. It
should be noted that any empirical formula should be used
with discretion, or with adapting modifications. Some
preliminary are often advisable when empirical formulas are
used.
• The snow load which may come to the roof depends in the
climate of the locality and on the pitch of the roof truss. The
density of dry snow maybe taken at 8lbs per cu.ft (0.128KPa,
and that of wet snow at 12 lb per cu ft (0.192KPa). Snow tends
to inclined roof surface maybe assumed as (1 – θ /60) times
the estimated load per square foot of flat surface of flat surface,
wherein θ is the angle in degrees between the indicated roof
surface and the horizontal. If appropriate for the climate , the
snow load maybe assumed to be 15, 20 or 25 square foot(0.719,
33. The wind load on a roof surface depends on the pitch of
the roof truss and on the velocity of wind, which in turn is a
function of the height of the building. The wind pressure p in pounds per
square foot of vertical surface due to a wind velocity V in miles per hour is
usually assumed to vary from
P = 0.003𝑉2 to p = 0.004𝑉2 . Thus a provision for wind pressures of 20,
25, or 30 lb per square foot(.0.9576, 0.1.198, 1.44 Kpa)( of vertical surface
may provide for wind velocities of 75, 85, or 95 mph. The wind pressure
normal to an inclined roof is usually found by use of the Duchemin
empirical formula (1829) .
Pn = p
2 sin θ
1+ 𝑠𝑖𝑛2 θ
In which p, is the normal pressure on an inclined roof surface at an angle
θ 𝑤𝑖𝑡ℎ ℎ𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙, and p is the assumed pressure on the vertical surface.
Recent investigations have shown that wind may not only exert
leeward side; it may actually exert suction at the leeward side. Suction is
34. • Anchorage to prevent lifting of the truss. Also, if the windows on the
windward or the leeward side are open or broken, pressure or suction
may come to the to the inside of the roof. Although the use of
Duchemin formula as discussed above is generally considered to be
conservative, these newer conception are noteworthy.
• WIND LOADS AS RECOMMENDED IN THE 1940 ASCE Final Report.
Sub committee 31 of the American Society of Civil Engineers made
some definite recommendations in regards to wind forces in its 1940
final report . Although the report prescribes wind forces for both plane
and round roof surfaces, only those for plane surfaces have been
abstracted from the above mentioned source as follows.
• 1. A uniformly distributed force of 20 psf (.9576KPa)for the first 300
ft(91.46) above ground level, increased above this level by 2.5 psf(0.119
Kpa) for additional 100 ft(30.49m) of height, is recommended as
astandard wind load for the United states and Canada.
• 2. For plane surfaces inclined and to the wind and not more than
300 ft(91.46) above the ground, the external wind force maybe pressure
or suction depending on its exposure and slope. For a windward slope
35. • To the horizontal, a suction of 12 psf(0.575) is recommended; for a slopes
between 20 and 30 degrees, a suction uniformly diminishing from 12 psf(0.575
Kpa) to 0. ( p= 1.2∝ −36)𝑜𝑟 (47.55 ∝ −1726.4); and for slopes between 30 to
60 degrees, a pressure increasing uniformly from 0 to 9 psf (0.431KPa), (p =
.30∝ - 9 )or(14.288∝ −431.604). On the leeward slope, for all inclinations in
excess of zero, a suction of 9 psf(.431 KPa is recommended.
• 3. For a flat roof a normal external suction of not less than 12 psf (.575KPa)
should be considered as applied to the entire roof surface.
• 4. For buildings that are normally airtight an internal pressure or suction of
4.5 psf (0.2155 Kpa) should be considered as acting normal to the walls and
the roof. For buildings with 30 % or more of the wall surfaces open, an internal
pressure of 12 psf(0.575 Kpa) or an internal suction of 9 psf (.431KPa), is
recommended.; for buildings with wall opening varying from 0 to 30 % of the
wall space, an internal pressure varying
• Uniformly from 4.5(0.2155KPa) from to 12 psf(0.575KPa), (p = 4.5 + .25 n)or
(215.8155 +11.98n), n = percentage of opening , or an internal suction varying
uniformly from 4.5(0.2155KPa) to 9 psf(0.431KPa) ( p = 4.5 + 0.15n)/ (p =
215.8155 + 7.2n) is recommended.
36. • 5. The design wind force applied to any surface of a building is to be
a combination of the afore-mentioned appropriate external and internal
wind forces.
• 6. When wind forces are more than 300 ft(91.46m) above the
ground, the external and internal wind forces should be scaled up in the
proportioned that the prescribed wind force on the plane surface
normal to the wind at the level under consideration bears to 20 psf
(0.959KPa).
• The external and internal wind force on inclined plane surface of not
more than 300 ft (91.46m) above the ground as described in items 2 , 3
and 4.
37. • COMBINATIONS OF LOADS. The function of stress analysis is to
provide
• The designer with the most probable maximum or minimum(reversal)
stresses to which any truss member maybe subjected; consequently
consideration must be given to the combinations of dead, snow and
wind loads which are to be accommodated in the design. The usual
combinations are: 1) dead plus full snow on both sides, 2) dead plus
wind on either side, 3) dead plus half snow on both sides plus wind
from either side, 4) dead plus full snow on the leeward side and wind on
the windward side, 5) dead plus ice(which maybe 5 to 10 lbs per square
foot(0.239 to 0.478 KPa) of roof surface (on both sides plus wind on
either side , and 6) dead plus ice on both sides plus full snow on the
leeward side plus wind on the windward side. Much depends upon the
judgement of the designer as to which load combinations should be
used.
• Because maximum winds may come only occasional and are usually
of short duration, most specifications, most specifications allow a 33.5 %
percent increase in the unit working stress in cases where effect is
38. • Any other load combination in which wind load is included.
• Ordinarily a consideration of load combinations (1), (2), and 3 will
provide adequate design data. If there maybe a reversal of stress in any
one member, it must be caused by wind from the opposite side. The
minimum stress or maximum stress opposite in sign to that od dead
load, if any, must be due to the load combination (2) and or load
combination (3), because the inclusion of half snow in both sides of load
combination (3) will not nullify some of the reverse stress due to wind.
It has been found, however, that with the exception of cases involving
small dead load and large wind load, there are very few cases of stress
reversal in roof trusses supported on masonry walls. In the usual cases
then, it seems desirable to device some sort of “ equivalent “ vertical
loading should be either that of full snow only or half snow plus a
certain fraction of the wind pressure normal to the roof surface. Hence
again, only experience and judgement can help to decide what to use as
“equivalent” loading for snow and wind.
39. Reactions for fixed Loads
General principles
A force is completely characterized by 3 properties . These are its
magnitude, direction and line of action or point of application.
For a body at rest to forces acting on a plane , there are 3 conditions
of equilibrium which must be satisfied.
∑Fx = 0 ∑Fy = 0 ∑M = 0
Special Cases of Equilibrium
1. Two forces: For 2 forces in equilibrium, they must be coplanar.,
collinear, equal in magnitude, and acting in opposite direction.
2. Three Forces: For 3 forces in equilibrium, a nec condition is that
they must be coplanar and concurrent.. If they are not coplanar
summation of forces perpendicular to the plane of any 2 of the 3 forces
will not be zero. Similarly,if they are not concurrent, moment about the
intersection of any of the 2 forces will not vanish.
40. • 3. Several Forces All Coplanar except One: If all forces
acting on a body are coplanar with the exception of one force,
equilibrium is not possible for there is no other force to
balance the non coplanar one.
• Types of Support
• Structures may be supported by rollers, hinge, links or fixed
end.
• Free Body diagram – A sketch indicating all external forces
and reactions representing the action exerted on the body.
• Sign Convention
• Principle of superposition – In many problems encountered in
the study of structural analysis, the structure is subjected to
41. • When the linear relationship is assumed to exist between the applied
loads and the resulting displacements in a structure, the forces acting
maybe considered separately and the structure analyzed for the
separate cases. The final result may be the algebraic sum of the
individual results.
• Stability and Determinacy of Structures
• In calculating the reactions for a structure, it is wise to visualize how
the loads are transmitted to the supports. The analyst must assess
whether the reactions can be determined by the equations of statics and
whether the structure will collapse and experience a change in
geometry due to loads.
• a) Stability and Determinacy of Beams.
• 1. Statically Determinate –The beam is statically determinate with 3
independent reactive force which can be calculated by 3 available
equations of static equilibrium.
• 2. Statically Indeterminate – It is statically indeterminate with one
42. • The number of independent reactive force and the number of
equations of static equilibrium.
• 3. Statically unstable (mechanism)- If there is no force to balance the
system.
•
• P P
• A B RAh A B
• P P
• A B RAv RBv
• RAh A B RBh
• P P
• A B RAv B RBv
• A B
• Rav RBv
43. • Determinacy and Stability
• A stable structure remains stable for any imaginable system of loads. Therefore, the
types of loads, their number and their points of application are not considered 4
• when deciding the stability or determinacy of the structure. A given structure
considered externally determinate if the total number of reaction components is equal
to the equations of equilibrium available. In other words:
• Determinacy:
45. • Determine the determinacy of the beam shown
• r = 5
• n = 1
• r > 3n
• 5 > 3(1) indeterminate
• Degree of determinacy
• 5-3(1)=2
• r -3n = 2( 2nd
degree)
46. • P2 MA P1 P2
• P1 RCh
• A B C Rah B RCv
• MC
• P1 P2 MA P1 P2
• Rah RCh
• A B C Rav B RCc
• P1 P2
• P1 P2 RCh
• A B C RAh B RCv
• Rav
• r = 6 n = 2 r = 3n 6 =3 (2) = 6 statically determinate
• r =7 n = 2 r>3n , 7 > 6 statically indeterminate 1st degree
• r = 5 n = 2 r < 3n 5 < 6 unstable
•
47.
48. • SUMMARY
• 1) If r < 3n the structure is unstable
• 2) If r= 3n the structure is determinate and stable UNLESS: - concurrent reactions -
parallel reactions
• 3) If r > 3n the structure is indeterminate and stable UNLESS: - concurrent reactions -
parallel reactions Degree of indeterminacy = r-3n = number of redundant reactions
• Note: If the structure is unstable, it does not matter if it is statically determinate or
indeterminat
49. • Indeterminate Structures
• A structure is termed as statically indeterminate, if it can not be analysed
from principles of statics alone, i.e. . A statically indeterminate structure
may be classified as:
1.Externally indeterminate, (example: continuous beams and frames
shown in figure-1(a) and (b)).
2.Internally indeterminate, (example: trusses shown in figure-1(c) and
(d)).
3.Both externally and internally indeterminate, (example: trussed beams,
continuous trusses shown in figure-1 (e) and (f)).
51. • Externally Indeterminate Structures
A structure is usually externally indeterminate or redundant if the reactions at the
supports can not be determined by using three equations of equilibrium, i.e. . In the
case of beams subjected to vertical loads only, two reactions can be determined by
conditions of equilibrium.
Therefore, simply supported cantilever and overhanging beams shown in figure 2 are
statically determinate structures
52. Beams. In regards to beams, if the reaction forces can be calculated
using equilibrium equations alone, they are statically determinate. On
On the other hand, if the reaction force can't be determined using
equilibrium equations only, other methods have to be used, and
the structure is said to be statically indeterminate
53.
54. • Stability and Determinacy of frame
• If 3m + r = 3J +c , the frame is statically determinate
• If 3m + r > 3j + c, the frame is statically indeterminate
• If 3m + r < 3j + c, the frame is unstable
. If r is the number of independent unknown reactive forces and c is the
numbers of equations of condition (c = 0 for a beam without internal
construction details or connections; c = 1 for a hinge; c = 2 for a roller);
we have , m is the no. of members and j is the no. of joints
Degree determinacy =( 3m +r) – 3j + c
55. • Determine the stability and determinacy of the frame:
• 2
• 2 internal hinge m = no. of members= 3
• j = no. of joints = 4
• 1 3 r = no. of support members = 3
• 4 c = internal hinge = 1
•
• 1
• 1
• 3m + r ? 3J + c
• 3(3) + 3 < 3( 4) + 1
• 12 < 13 therefore unstable
• r < 3n r = 5 3n = 3 (2) = 6
• 5 < 6 // unstable
•
56. • Determine the stability and determinacy of the structure
• hinge hinge
• r = 5, j= 6, m= 5, c = 2 r = 9
• 3m + r ? 3j + c n = 3
• 3(5) + 5 ? 3(6) + 2 r ? 3n
• 20 = 20 , therefore determinate 9 = 3(3)
• 9 =
9//determinate
57. • Determine the stability and determinacy of the frame shown.
•
• r = 5, m= 3, j = 4 , c -= 0 r = 5, n = 1
• 3m + r > 3j + c r > 3n
• 3(3) + 5 > 3(4) 5 > 3(1)
• 14 > 12, therefore, indeterminate, 2nd degree 5> 3 // indeterminate
2°
•
58. • Determine the stability and determinacy of the frame shown
• hinge
•
• r = 3, m = 6, j = 7 c = 1 r = 5, n = 2
• 3m + r < 3j + c r < 3n
• 3(6) + 3 < 3(7) + 1 5< 6 // unstable
• 21 < 22 // unstable
59. • Determine the stability and determinacy of the frame:
• hinge
• r = 6 m = 7, j = 7, c = 1
• 3m + r > 3j + c
• 3(7) + 6 > 3 (7) + 1
• 27 > 22 // indeterminate 5th
degree
•
•
•
•
60. • Determine the stability and determinacy of the frame:
• hinge hinge
•
• m = 5
• J = 6
• c = 2
• r = 4
• 3m + r > 3j + c
• 3(5) + 4 < 3(6) + 2 19< 20 // unstable
61. • Determine the stability and determinacy of the frame shown
• r = 9, m = 20, j = 15, c = 0
• 3m + r > 3j + c
• 3(20) + 9 > 3 (15)
• 69 > 45 // indeterminate to the 24th degree
62. • Stability and Determinacy of Rigid Frames
• In each of the following frames is determinate or indeterminate. If statically indeterminate, what is the
number of degree of indeterminacy?
• A rigid frame in structural engineering is the load-resisting skeleton constructed with straight or curved
members interconnected by mostly rigid connections which resist movements induced at the joints of
members. Its members can take bending moment, shear, and axial loads
• 1) m = 6, j= 6, r = 3, c = 0 2) m = 7, r = 9, j = 8, c = 0
• 3m + r > 3j + c 3m + r > 3j + c
• 3(6) + 3 > 3(6) 3(7) + 9 > 3(8)
21 > 18, therefore indeterminate 30 > 24, therefore indeterminate
to the 3rd degree to the 6th degree
•
63.
64. • STABILITY AND DETERMINANCY OF TRUSSES
• Truss Structures
• A truss can be defined as a structure that is composed of links or bars,
assumed to be connected by frictionless pins at the joints, and arranged
so that the area enclosed within the boundaries of the structure is
subdivided by the bars into geometrical figures which are normally
triangles.
• Internal determinacy is a type of indeterminacy that is associated to
trusses. The basic form of the truss is a triangle. To make the truss, add
two members and one joint, and repeat.
65. • STABILITY AND DETERMINANCY OF TRUSSES
• A simple rigid truss is formed by starting with a triangle
consisting of three members and three joints and subsequently
adding two members and a joint to form another triangle. This
addition of 2 members and a joint is continued until the final truss
configuration is attained. A truss with j joints will require(j – 3)
additional joints to expand the structure beyond the original three
joint triangle. Since each additional joint is formed by the
intersection of 2 members , the total number of members m to
force a simple rigid truss
• m = 2(j – 3 ) + 3 = 2j – 3
•
66. • To develop a criterion for the stability and determinacy of trusses,
consider the equilibrium of a simple truss. At each joint of a truss
is a concurrent force system consisting of the forces in two members
plus any applied loads and reactions. The equations ∑𝐹ℎ = 0
and ∑𝐹𝑣 = 0 are sufficient to determine the unknown forces in the two
members acting at a point. For a truss with j joints there are 2j
independent equations of equilibrium which involve ( m + r ) unknowns,
where r is the number of reaction components. The criterion used for
classification is summarized as
• 1) If m + r = 2j, the truss is statically determinate
• 2) If m + r > 2j, the truss is statically indeterminate
• 3) If m + r < 2j, the truss is unstable (a mechanism)
Degree of determinacy is (m +r) – 2j
67. •
• m = 9
• r = 3
• j = 6
•
• a) m + r = 2j
• 9 + 3 = 2(6)
• 12 = > 12 , Therefore it is unstable bec all reactions are parallel
• b) No. of unknown forces = 12
• No. of available equations = 2(6) =12 // unstable
https://www.youtube.com/watch?v=gcPP8DCa2sw
https://www.youtube.com/watch?v=9nzbZi9khAo
68. • For the trusses shown, determine its stability and determinacy.
a) Determinate truss
m = 17, j = 10, and r = 3. m + r = 2j 17 + 3 = 2 (10) , 20= 20 //
So, degree of static indeterminacy = 0, that means it is a statically
determinate system
b) (Internally) indeterminate truss
m = 18, j= 10, and r = 3. 18 + 3 > 2(10), // 21> 20
So, degree of static indeterminacy = 1.
69. (Externally) indeterminate truss
m = 17, j = 10, and r = 4. m + r = 2j ; 21> 20
So, degree of static indeterminacy = 1.
It should be noted that in case b, we have one member more than what
is needed for a determinate system
a), where as c) has one unknown reaction component more than what is
needed for a determinate system. Sometimes, these two different types of
redundancy are treated differently; as internal indeterminacy and
system. Sometimes, these two different types of redundancy are treated
differently; as internal indeterminacy and external indeterminacy . Note
that a structure can be indeterminate either externally or internally or
both externally
and internally.
.
70. Classify the trusses shown as stable, determinate, or
indeterminate, and state the degree of indeterminacy
when necessary.
r = 3, m = 9, j = 6. From equation , m + r = 2j
9 + 3 = 2(6);
12 = 12
//Statically determinate
71. r = 3, m = 10, j = 6. From equation
m + r = 2j; 10 + 3 > 2(6)
13> 12 //.
Statically indeterminate to 1°(internally )
72. r = 3, m = 9, j = 6. From equation ,
m + r = 2j;
9 + 3 = 2(6). Statically determinate.
73. r = 3, m = 24, j = 14.
From equation , 24 + 3 < 2(14).;
27 < 28; Statically unstable.
74. m = 11, r = 5, j= 7, m+ r > 2j
11 + 5 = 2 (7)
16 > 14 // indeterminate to the degree
75. 1/3 Concept of axial shear, torsional, flexural rigidities and stiffness of
structural members.
Shear force is the force in the beam acting perpendicular to its longitudinal
(x) axis
Axial force is the force in the beam acting parallel to the longitudinal axis
76. Flexural rigidity is defined as the force couple required to bend a fixed non-rigid
structure in one unit of curvature or it can be defined as the resistance offered by a
structure while undergoing bending
Torsion is the twisting of an object due to an applied torque.
Torsion is expressed in either the Pascal (Pa), an SI unit for newtons per square
metre, or in pounds per square inch (psi) while torque is expressed in newton metres
(N·m) or foot-pound force (ft·lbf).
77. In structural engineering, the term 'stiffness' refers to the rigidity of a structural element. In general
terms, this means the extent to which the element is able to resist deformation or deflection under the
action of an applied force.
Rigidity, also called stiffness, is a measure of elasticity, and represents a material's resistance to
permanent deformation. ... Rigidity is a material's resistance to bending, whereas strength is a
material's resistance to breakage. Rigidity is measured by finding the Young's modulus of a
particular material.
Difference between stiffness and rigidity
78.
79.
80. WEEK 2
•
• 2. Analysis for external reactions and internal stress resultants of statically
determinate structure
– Beams
• Ability to analyze external reactions and internal stress reactions of statically
81. •SHEAR AND MOMENT IN BEAMS
• Whenever a beam is subjected to bending loads, there will be induced
certain internal loadings effects which vary from section to section of
the beam. These loading effects are the shearing force and bending
moment, often referred to shear and moment, respectively. This chapter
is concerned with the variations in shear and bending moment in beams
supporting various combinations of loading under different conditions
of support. Maximum values which are needed in the design will also be
determined. The curve which the longitudinal axis of the beam will form
in the loaded , called the elastic curve, will also be drawn to indicate the
changes in the concavity of beam segments.
• Shear and Moment
• Shear is defined as the algebraic sum of the external loads to the left or
to the right of the section that are perpendicular to the axis of the beam.
Shear denoted by V, may be expressed mathematically as:
•
82. • V = (∑ LOADS)L OR
• V = (∑ LOADS)R
• BENDING MOMENT
• Is defined as the algebraic sums of the moments about the centroidal
axis of the of a beam section of all loads acting either to the left or to
the right of the section . This is expressed as:
• M = (∑ M) L = (∑ M) R P2
• P1 B a C
• B C
• A x a L – x D
• R1 L R2
83. •
• P1 P2
• a x - a M M
• x L – x
• R1 V R2
• VAC = R1- P1 MAC = R1x – P1(x – a)
• RULES OF SIGNS FOR LOAD, SHEAR AND MOMENT
• Loads acting upward are considered positive, and negative for
loads directed downward. For shear, a positive shearing force
tends to move the left segment upward relative to the right
segment, and vice versa.
84. •
• Positive Shear Negative Shear
• Bending Moment is considered positive if it causes the beam to bend
concave upward, and negative otherwise.
•
• Positive Bending Negative Bending
86. • b. Procedure for determining shear force and bending moment
• diagrams
• ٠ Compute the support reactions from the free-body diagram
• (FBD) of the entire beam.
• ٠ Divide the beam into segment so that the loading within each
• segment is continuous. Thus, the end-points of the segments are
• discontinuities of loading, including concentrated loads and
• couples.
• Perform the following steps for each segment of the beam:
• Introduce an imaginary cutting plane within the segment, located
• at a distance x from the left end of the beam, that cuts the beam
• into two parts.
• Draw a FBD for the part of the beam lying either to the left or to
• the right of the cutting plane, whichever is more convenient. At
• the cut section, show V and M acting in their positive directions.
87. • Determine the expressions for V and M from the equilibrium
• equations obtainable from the FBD. These expressions, which
• are usually functions of x, are the shear force and bending
• moment equations for the segment.
• Plot the expressions for V and M for the segment. It is visually
• desirable to draw the V-diagram below the FBD of the entire
• beam, and then draw the M- diagram below the V-diagram.
• The bending moment and shear force diagrams of the beam are
• composites of the V and M diagrams of the segments. These
• diagrams are usually discontinuous, or have discontinuous
• slopes. At the end-points of the segments due to discontinuities
• in loading.
88. INTERNAL REACTIONS IN BEAMS
• At any cut in a beam, there are 3 possible internal reactions required
for equilibrium:
– normal force,
– shear force,
– bending moment.
Pb/L
x
Left Side of Cut
V
M
N
Positive Directions
Shown!!!
89. INTERNAL REACTIONS IN BEAMS
• At any cut in a beam, there are 3 possible internal reactions required
for equilibrium:
– normal force,
– shear force,
– bending moment.
Pa/L
L - x
Right Side of Cut
V
M
N
Positive Directions
Shown!!!
90. FINDING INTERNAL REACTIONS
• Pick left side of the cut:
– Find the sum of all the vertical forces to the left of the cut, including V.
Solve for shear, V.
– Find the sum of all the horizontal forces to the left of the cut, including
N. Solve for axial force, N. It’s usually, but not always, 0.
– Sum the moments of all the forces to the left of the cut about the point
of the cut. Include M. Solve for bending moment, M
• Pick the right side of the cut:
– Same as above, except to the right of the cut.
91. DRAW SOME CONCLUSIONS
• The magnitude of the shear at a point equals the slope of the moment diagram at that
point.
• The area under the shear diagram between two points equals the change in moments
between those two points.
• At points where the shear is zero, the moment is a local maximum or minimum.
96. • Write the shear and moment equations, shear and moment values and draw the shear
and moment diagrams of the beams shown.
• 3Kn + ∑ 𝑀 𝐶 = 0
• A 2m B 4m C R1 (6) – 3(4) = 0
• R1 = 2 Kn; R2 = 1Kn
• R1 1 2 R2 Pass a section between A & B
• section 1 V & M Equations
• x 0<x<2, x = 2
• A M VAB = 2
• R1=2 V MAB = 2x linear eqn
• 3Kn 2<x<6
• B 2 x-2 M VBC = 2- 3 = -1
• R1= 2 x V MBC = 2x – 3(x-2) linear
•
97. • V and M Values
• V Values M Values
• VA = 2 Kn MA = 0
• VAB = 2 MAB = 2x = 2 (2) = 4 Kn – m x = 2
• VB = 2-3 = - 1 MBC = 2x – 3(x – 2 ) x = 6
• VBC = - 1 = 2(6) – 3(6 – 2)
• VC = 1-1 = 0 K (Checking) = 0 k (checking)
•
98. • V & M Diagrams
• 3kn
• V max = 2Kn ; Mmax = 4 Kn-m
•
• 2
• -1 -1 shear
• 4
• moment
99. • Write the shear and moment equations, shear and moment values and draw the shear
and moment diagrams of the beams shown.
•
• w = 4 Kn /m RA = RB = wL / 2 = 4(4)/2 = 8 Kn
• A B Pass a section between A & B
• 4 m V & M Equations
• wL = 4x 0<x<4 x = 4
• x/2 x/2 VAB = 8 – 4x linear eqn
• M MAB = 8x -4x (x/2)= 8x - 2𝑥2
2nd degree
• x V values
• 8 V VA = 8 Kn
• VAB = 8 – 4x x= 4
• VAB = 8 – 4(4) = 6- 16 = - 8
• VB = - 8 + 8 = 0 ok checking
100. • Before solving for the moment values draw first the shear diagram to see if there is a
point of zero shear. The point of zero shear is the point on the shear diagram where
the locus intersects at the horizontal line. In which there lies a point of inflection it is
where the highest point on the locus or there is a maximum moment at that point.
• By symmetry the value of x = 2m
• A B Moment values
• 4 MA = 0
• x MAC = MAB ; x = 2m
• 8 MAC = 8x - 2𝑥2
= 8 (2)- 2(22
) = 8 Kn.m
• C MBC = MAB ; x= 4
• 4 MBC = 8(4) - 2(42) = 0 ok checking
• -8 Since the equation is 2nd degree the locus is
parabolic
101. • Write the shear and moment equations, shear and moment values and draw the shear
and moment diagrams of the beams shown.
• w = 4 Kn/m Solving for the reactions
• + ∑MB = 0
• R A(3) – ½(4)(3)1/3)(3)= 0
• A B RA = 2Kn ; RB = 6-2 = 4 Kn
RA = 2 3m RB = 4 Pass a section between AB
V & M Equations Since the value of y is no longer the same as w
0<x<3; x = 3m solve for y in terms of x by ratio and
proportion
¾ = x/y
• y = 4x/3
• y 4
• x
• L = 3
•
•
102. • Analyze the left section of the beam
• ½ xy
• 2/3 x 1/3x
• y = 4x/3
• A
•
• 2
• V and M equations
• 0 < x < 3
• VAB = 2 -1/2 (xy) = 2 – x/2 (4x/3) = 2 - 2𝑥2
/3
• MAB = 2x - 2𝑥2
/3(1/3) (x) = 2x - 2𝑥3
/9
• V- Values
• VA= 2 Kn
• VAB == 2 - 2𝑥2
/3= 2 – 2(32
)/3 = -4
• VB = -4 + 4 = 0 ok //
103. • Draw the shear diagram before solving for the moment values
• solving for the value of x using the shear equation
• VAB and equate t0 zero
• VAB = 2 -2𝑥2/3 = 0
• 3 x = 1.73m
• x Moment Values
• 2 3 - x MA = 0
• C MAC = MAB when x = 1.73
• MAC = 2x –( 2/9) 𝑥3 =2(1.73) –(2/9)1.733
• 3.31 -4 MAC = 2.31 Kn-m
• MCB = MBC when x = 3
• MBC = 2x –( 2/9) 𝑥3 =2(3) –(2/9)33 = 0//
104. Sample Problem 4.1
The simply supported beam in Fig. (a) carries two
concentrated
loads. (1) Derive the expressions for the shear
force and the bending
moment for each segment of the beam. (2) Sketch
the shear force
and bending moment diagrams. Neglect the weight
of the beam.
Note that the support reactions at A and D have
been computed and
are shown in Fig. (a).
Solution
∑MD = 0, RA(7) – 14 (5) – 28(2) = 0
RA = 18, RD= 24
Part 1
The determination of
the expressions for V
and M for each of the
three beam segments
(AB,BC, and CD) is
explained below.
105. • Segment AB (0＜x＜2 m) x = 2
• VAB = 18
MAB = 18 x
106. • Segment BC (2＜x＜5 m) x = 5
• VBC = 18 – 14 = 4
• MBC = 18x -14 (x-2)
• MBC = 18 x – 14x + 28 = 4x + 28
108. • V & M Values
• V values M values
• VA = 18 Kn MA = 0
• VAB = 18 MAB = 18x when x = 2
• VB = 18 – 14 = 4 = 18 (2) = 36 Kn -m
• VBC = 4 MBC = 4x + 28 when x = 5
• VC = 4 – 28 = -24 = 4(5) + 28 = 48 Kn- m
• VCD = -24 MCD = 160 – 24x when x = 7
• VD = -24 + 24 = 0 checking ok = 160 – 24(7) = 0
• MC = 0 ok
109. Part 2
The V-diagram reveals that
the largest shear force in
the beam is -24 kN :
segment CD
The M-diagram reveals
that the maximum bending
moment is +48 kN·m : the
28-kN load at C.
Note that at each
concentrated force the Vdiagram
“jumps” by an
amount equal to the force.
There is a discontinuity in
the slope of the M-diagram
at each concentrated force.
110. • The cantilever beam in
• Fig.(a) carries a triangular
• load. The intensity of
• which varies from zero at
• the left end to 360 lb/ft at
• the right end. In addition, a
• 1000-lb upward vertical
• load acts at the free end of
• the beam. (1) Derive the
• shear force and bending
• moment equations. And
• (2) draw the shear force
• and bending moment
• diagrams. Neglect the
• weight of the beam.)
Note that the triangular load has been
replaced by is resultant, which is the
force 0.5 (12) (360) = 2160 lb (area
under the loading diagram) acting at
the centroid of the loading diagram.
111. • Pass a section between AB
0 < x < 12 x=12
VAB =1000 – ½ xy = 1000 – ½ x (30x)
= 1000 – 15𝑥2
MAB = 1000x -15𝑥2
(1/3 x) = 1000x -5𝑥3
V Values
VA = 1000 lbs
VAB = 1000 – 15𝑥2
= 1000 – 15 (12)2
= - 1160
VB = - 1160 + 1160 = 0 K Checking
112. Part 2
The location of the
section where the
shear force is zero is
found from
V = 1000 – 15𝑥2
x = 8.165 ft
x = 8.165 ft .
M Values
MA = 0
MAC = MAB when x = 8.165
MAC = 1000x -5𝑥3
= 1000(8.165) - 5(8.165)3
= 5,443 ft – lbs
MCB = MBC when x + 12
MCB = 1000(12) – 5122
=1160
Checking MC is negative -1160 which is
equal but opposite in direction with the
moment in MCB ok//
116. • Area Method for Drawing Shear- Moment Diagrams
• Useful relationships between the loading, shear force, and
• bending moment can be derived from the equilibrium
• equations.
• These relationships enable us to plot the shear force diagram
• directly from the load diagram, and then construct the bending
• moment diagram from the shear force diagram. This technique,
• called the area method, allows us to draw the shear force and
• bending moment diagrams without having to derive the
• equations for V and M.
• First consider beam subjected to distributed loading and then
• discuss concentrated forces and couples.
117. • Procedure for the Area Method
• Compute the support reactions force the free-body diagrams
• (FBD) of the entire beam.
• Draw the load diagram of the beam (which is essentially a
• FBD) showing the values of the loads, including the support
• reactions. Use the sign conventions in Fig. 4.3 to determine
• the correct sign of each load.
• Working from left to right, construct the V-and M-diagram
• for each segment of the beam using Eqs. (4.1)-(4.6).
• When reach the right end of the beam, check to see whether
• the computed values of V-and M are consistent with the end
• conditions. If they are not, you made an error in the
• computations.
118. • Sample Problem 4.4
• The simply supported beam in Fig. (a) supports 30-kN concentrated
• force at B and a 40-kN•m couple at D. Sketch the shear force and
• beading moment diagrams by the area method. Neglect the weight
• of the beam.
119. • Solution
• Load Diagram
• The load diagram for the beam is shown in Fig. (b). The reactions
• at A and E are found from equilibrium analysis. Indicating its sign
• as established by the sign conventions in Fig. 4.3.
120.
121. EXAMPLE 2
Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Figure
below , draw the shear and moment diagrams for the beams specified in the following
problems.
SOLUTION
122. • Sample Problem 4.5
• The overhanging beam in Fig. (a) carries two uniformly distributed
• loads and a concentrated load Using the area method. Draw the
• shear force and bending moment diagrams for the beam.
123. Load Diagram
The load
diagram for the
beam is given
in Fig. (b)
Shear Force
Diagram
The steps
required to
construct the
shear force
diagram in Fig.
(c) are now
detailed.
125. EXAMPLE
3
SOLUTION
Beam carrying the triangular loads shown in the Figure below, draw the shear and
moment diagrams for the beams specified in the following problems.
4. Load in AB is linear, thus, VAB is
second
degree or parabolic curve. The
load is from
0 at A to wo (wo is downward or -
wo) at B,
thus the slope of VAB is
decreasing.
5. VBC is also parabolic since the
load in BC is
linear. The magnitude of load in
BC is from -
wo to 0 or increasing, thus the
slope of
VBC is increasing.
126. PROBLEM
S
1. Write shear and moment equations for the
beams as shown below in the following problems.
In each problem, let x be the distance measured
from left end of the beam. Also, draw shear and
moment diagrams, specifying values at all change
of loading positions and at points of zero shear.
Neglect the mass of the beam in each problem.
2. Write shear and moment equations for the beams
as shown below in the following problems. In each
problem, let x be the distance measured from left end
of the beam. Also, draw shear and moment diagrams,
specifying values at all change of loading positions and
at points of zero shear. Neglect the mass of the beam
in each problem.
3. draw the shear and moment diagrams for the
beams specified in the following problems. Give
numerical values at all change of loading
positions and at all points of zero shear.
127.
128. EXAMPLE: FIND THE INTERNAL REACTIONS AT
POINTS INDICATED. ALL AXIAL FORCE
REACTIONS ARE ZERO. POINTS ARE 2-FT APART.
20
m
P = 20 Kn
12 Kn
8 Kn
12
m
1
7
10
6
2 3 9
4 5 8
Point 6 is just left of P and Point 7 is just right of P.
129. EXAMPLE: DRAW SHEAR & MOMENT
DIAGRAMS FOR THE FOLLOWING BEAM
3 m 1 m
1 m
12 kN 8 kN
A C
B
D
RA = 7 kN RC = 13 kN
130. 3 m 1 m
1 m
12 kN
A C
B
D
V
(kN)
M
(kN-m)
7
-5
8
8 kN
7
-15
8
7
-8
2.4 m
131. 20
m
P = 20 Kn
12 Kn
8 Kn
12
m
V
(Kn)
M
(KN-m)
8 Kn
-12 Kn
96 Kn-m
x
x
V & M Diagrams
a
b
c
What is the area of the
blue rectangle?
96 ft-kips
What is the area of
the green rectangle?
-96 Kn-m
132. • For the cantilever beam and loading shown, write the shear and
moment equations and draw the shear and moment diagrams and the
elastic curve.
• 2 kN/m 8 kN
• A 6m B 4m C
•
• For the beam shown, write the V & M equations and draw the V & M
diagrams and elastic curve.
• 8 kN 16 kN 12 kN
•
• A B C D E
• 3m 3m 5m 4m
•
136. • Equilibrium Structures, Support Reactions, Determinacy and Stability of Beams and
Frames
Engineering structures must remain in equilibrium both externally and internally when
subjected to a system of forces. The equilibrium requirements for structures in two and
three dimensions are stated below.
• 3.1.1 Equilibrium in Two Dimensions
• Procedure for Computation of Support Reactions
• •Sketch a free-body diagram of the structure, identifying all the unknown reactions
using an arrow diagram.
• •Check the stability and determinacy of the structure using equation 3.3 or 3.4. If the
structure is classified as determinate, proceed with the analysis.
• •Determine the unknown reactions by applying the three equations of equilibrium. If a
computed reaction results in a negative answer, the initially assumed direction of the
unknown reaction, as indicated by the arrow head on the free-body diagram, is wrong
and should be corrected to show the opposite direction. Once the correction is made,
the magnitude of the force should be indicated as a positive number in the corrected
arrow head on the free-body diagram
137. A cantilever beam is subjected to a uniformly distributed load and an inclined concentrated load,
as shown in figure 3.9a. Determine the reactions at support A.
138.
139.
140.
141. • A beam with an overhang is subjected to a varying load, as shown
in Figure 3.11a. Determine the reactions at supports A and B.
142. • A compound beam is subjected to the loads shown in Figure
3.13a. Find the support reactions at A and B of the beam.
143. • Computation of reactions. The analysis of a
compound structure must always begin
with the analysis of the complimentary
structure, as the complimentary structure
is supported by the primary structure.
Using the equations of equilibrium, the
support reactions of the beam are
determined as follows:
• Analysis of the complimentary
structure CB.
144.
145.
146. The negative sign implies that the
originally assumed direction of Ay was
not correct. Therefore, Ay acts
downward instead of upward as was
initially assumed. This should be
corrected in the subsequent analysis.
147. • .1 Classify the structures shown in Figure P3.1a to Figure P3.1p as statically
determinate or indeterminate, and statically stable or unstable. If indeterminate, state
the degree of indeterminacy.
148. • . Determine the support reactions for the beams shown in Figure P3.2 through
Figure P3.12.
