High-rise Buildings N
Structural Systems for High-rise Buildings
( Lecture notes prepared by: Prof. Dr. JAHANGIR BAKHTERI )
Architect and the Engineer:
An architect usually deals with a design problem such as a high-rise building
in a comprehensive manner, and stresses an overall, rather than an elemental
approach to design thinking.
An engineer thinks in the reverse manner, starting with details, and without
sufficient regards for overall picture.
The principal form giver of a building is the architect, who must develop the
self-confidence to express the strength and inherent beauty of structure and
materials. He needs to understand the laws of nature as reflected by the play
of forces in the building assemblage.
Sears Tower, Chicago, USA Petronas Twin Tower, Malaysia
Taipei 101 Tower, Taiwan
Burj Dubai Tower:
Tallest structure in the world.
Total height 600 m, now
completed.
BOSTON -- Massachusetts Institute of Technology
is suing renowned architect Frank Gehry, alleging
serious design flaws in the Stata Center, a building
celebrated for its unconventional walls and radical
angles.
The school asserts the center, completed in spring
2004, has persistent leaks, drainage problems and
mold growing on its brick exterior. It says
accumulations of snow and ice have fallen
dangerously from window boxes and other areas of
its roofs, blocking emergency exits and causing
damage
MIT sues architect Gehry, flaws in building
The suit says MIT paid
Los Angeles-based Gehry
Partners $15 million to
design the Stata Center,
which cost $300 million
to build. It houses labs,
offices, classrooms
and meeting rooms.
"Gehry breached its
duties by providing
deficient design services
and drawings," according
to the suit, which also
names New Jersey-based
Beacon Skanska
Construction Co., now
known as Skanska USA
Building Inc. The suit,
filed Oct. 31, 2007 seeks
unspecified damages.
History of Development of High-rise Buildings:
Throughout history, man has always tried to express his greatness by
building monuments as high as he could practically achieve, with the
Egyptian pyramid as the best example.
- Egyptian pyramids made from masonry blocks are 481ft. high which is
equivalent to a 40-storey building.
- In 1907 the 47-storey Singer skyscraper in New York was built.
- For centuries there were two basic materials used in construction: wood
and masonry
- Development of new construction materials such as concrete & steel and
advanced technologies in construction made it possible to construct
high-rise buildings.
During the 1960s and 70s, F.R. Khan became the most prominent
innovator in the area of high-rise buildings, both in concrete and steel. The
methodology of shear wall-frame interaction in high-rise building was
developed by him in 1964, in which the stiffness of frame building could be
enhanced up to several times. Then, he initiated tubular design concept in
high-rise buildings and applied in design & construction of several building
structures. After that his next innovation was tube in tube concept in tall
buildings.
The innovations introduced by Khan not only improved the rigidity of tall
buildings, but also resulted in substantial economics over the cost of
buildings designed using traditional schemes.
Nowadays, the many efficient, large, sophisticated two and three
dimensional computer programs have made the analysis of complicated
structure much simpler, more economical, and accessible to the designers.
Contributing Factors to the Development High-Rise Buildings
(Assignment No. 1)
(1)There are four major factors which contributed to the development of
high-rise building which are:
(a) Development of high-strength materials
(b) Development of new design concepts
(c) Development of new structural systems
(d) Improved construction methods (Explain every one of them briefly)
Need for High-rise buildings:
High-rise buildings are closely related to the city; they are a natural
response to dense population concentration, shortage of land, and high
land costs. A tall building is an integral part of one large building
organism, the city, where the building or activity cells are interconnected
by multilevel movement systems.
High-rise buildings range in height from 10 to more than 100 stories.
Factors which governs/controls the height or the massing of a building
are: the client’s needs versus the land available, necessary services to
support the building and its inhabitants, the ecological impact of the
building,
What are the Problems caused by high-rise buildings?
Some impacts of skyscrapers on the city:
Example: 109 storey Sears Tower in Chicago, USA, The building’s
electrical system can serve a city of 147000 people and its air conditioning
complex can cool 6000 one-family houses. A total of 102 elevators
distribute about 16500 daily users to different parts of the building. Since
the building contains all necessary supports services facilities such as
shopping, entertainment, health, education, security, parking and other
utilities, therefore, people seldom leave it.
Load Action on High-Rise Buildings:
Loads acting on a structure are generated either directly by the forces of
nature or by man himself. In other words, there are two basic sources for
building loads: geophysical and man-made. Fig.1 represents sources of
building loads.
The geophysical forces are the result of continuous changes in nature,
which include gravitational loads, meteorological loads and seismological
forces. Gravitational loads include weight of the structure (Dead load), and
occupants of the building (Live load). Meteorological loads are wind loads,
rain & snow loads, temperature stresses. Seismological forces are
generated by the ground motion (i.e. earthquake forces).
Fig. 1 Sources of Building Loads
The structural facade is exposed to the controlled temperature of the interior
building environment and the daily and seasonal changes of weather. These
changes in temperature causes additional stress in the buildings. The
structure will have contraction for temperature drop and expansion for
temperature increase.
Creep is the time-dependent deformation that occurs in concrete for years
after the initial loading deformation. Creep in concrete members is
dependent on the magnitude of the stress, the length of time that the stress
is applied, and the age and strength of concrete
Shrinkage is a major cause for volume change in concrete and generally
characterized by a gradual loss of moisture within the concrete member. As
shrinkage stresses appear, additional restraint is required which, in turn,
places additional loads on the structure.
The man-made sources of loading are the shocks generated by cars,
elevators, machines and equipment. They can be caused by blasts and
impacts as well.
Locked-in stresses can be created by the prestressing of concrete. Forces
can also be locked into the structure during construction processes through
restrained volumes.
Loads and Forces in Building Structures
Loads acting in building structures are:
(a)Dead Loads: Dead loads include the weight of all permanent components of
the structure and any other immovable loads that are constant in magnitude and
permanently attached to the structure.
(b)Imposed Loads: Imposed loads are loads and forces that act on a structure by
character of use of the structure to the nature of use, activities due to people,
machinery installation, external forces etc. These loads are:
i) Live Loads
ii) Wind Loads
iii) Earthquake Loads
iv) Snow and Rain Loads
v) Soil and Hydrostatic Forces
vi) Erection Loads
vii) Other Forces; such as impact, vibration, temperature effects, shrinkage,
creep and forces due to the settlement of foundations.
Wind Effect on Tall Buildings
In old day’s skyscrapers, the enormous weight of masonry walls, relatively
thick slab and larger structural elements were such that wind action could
not overcome the locked-in-gravity forces.
The modern glass-walled skyscrapers with their optimum interior open
space have taken away the overall rigidity of the structure and wind action
has become a major problem for the designers of high-rise structures.
The wind loading on a building is influenced by the environment factors
such as large-scale roughness, shape and slenderness of the building and
wind speed.
The mean wind velocity and wind pressure in general increases with height
of the building as shown in Fig. 2
Fig. 2 Variations of wind velocity
and wind pressure on a building
Calculation of wind pressure on tall buildings:
Based on B.S(British Standard), the characteristics wind pressure on a
structure will be:
Wk = 0.613Vs
2
Where, Wk in N/m2
is characteristic wind pressure and Vs is design wind
speed in m/sec which can be calculated as follows:
Vs = VS1S2S3
Where, V = basic wind speed in m/sec
S1 = multiplying factor relating to topology
S2 = multiplying factor relating to height above ground and wind
braking.
S3 = multiplying factor related to life of structure
Note1: S1 may generally always be taken as unity except in the following
cases:
(a) On site adversely affected by very exposed hill slopes and crests where
wind acceleration is known to occur: S1 = 1.1
(b) On site in enclosed steep-sided valleys completely sheltered from wind:
S1 = 0.9
Note2: S3 is a probability factor relating the likelihood of the design wind speed
being exceeded to the probable life of the structure. A value of unity is
recommended for general use and corresponds to an excessive speed occurring
once in fifty years.
S2 for various heights of buildings can be taken from Table 1.
Notes:
h is height (in meters) above general level of terrain to top of structure or part
of structure.
Topographical factors
1. open country with no obstructions
2. open country with scattered wind-breaks
3. country with many wind-breaks; small towns; suburbs of large cities
4. city centers and other environments with large and frequent obstructions
Example of wind forces calculation
Assume an eight story framed office building shown below (Fig. D) is located in
Kabul City’s suburb (e.g. Gardanai Baghe Bala). Calculate the wind forces on the
structure assuming basic wind speed to be 40m/sec and the building is 22.5m by
22.5m in plan and having a 24m height above ground level.
Characteristic wind pressure is Wk = 0.613 Vs
2
, where,
Vs = design wind speed in m/sec = VS1S2S3
V = basic wind speed in m/sec
S1 = multiplying factor relating to topology
S2 = multiplying factor relating to height above ground and wind braking
S3 = multiplying factor related to life of structure
Assume basic wind speed of V = 40m/sec
In an hilly area the value for S1 = 1.1
The probability factor the life of structure S3 = 1
S2 the height factor for the structure with a height 24m and open country form with no
obstructions from the Table-1 using interpolation will be S2 = 1.026
Therefore, the design wind speed Vs = VS1S2S3 = 40 x 1.1 x 1 x 1.026 = 45.144 m/sec
Characteristic wind pressure = 0.613 x (45.144)2
= 1285.97 N/m2
= 1.28597kN/m2
Wind pressure on an interior frame will be Wki = 1.28597 x 7.5 = 9.645 kN/m
Wind concentrated force at each floor level will be Wp = 9.645 x 3 = 28.934 kN
Wind concentrated force at roof level will be Wkr = 9.645 x 1.5 = 14.467 kN
The concentrated forces due to wind pressure on the building in different floors and roof
Example of earthquake forces
Calculation:
Assume an eight stories framed office
building with live loads of 2.5 kN/m2
shown below (Fig. A) is located in
seismic zone III. Calculate the seismic
forces on the structure using seismic
coefficient method. Assume roof and
floor slabs thicknesses of 150mm,
beams section 250mm by 400mm,
columns section 400mm by 500mm
and story height is 3.0m.
As recommended by the codes assume
25% live loads are acting during
earthquake.
(a) Dead Loads
Weight of beams = 24 x 0.25 x 0.4 x 7.5 x 24 = 432.0 kN
Weight of columns = 16 x 0.4 x 0.5 x 3 x 24 = 230.4 kN
Weight of slabs = 22.5 x 22.5 x 0.15 x 24 = 1822.5 kN
Weight of walls = 22.5 x 4 x 3 x 0,12 x 20 = 648.0 kN
(b) Live loads
Live loads at all floors = 22.5 x 22.5 x 2.5 x 0.25 = 316.4 kN
(c)Lumped mass at floor level 1 = W1 = 432.0 + 230.4 + 1822.5 + 648.0 + 316.4
= 3449.3 kN
Similarly W1 = W2 = W3 = W4 = W5 = W6 = W7 = 3449.3 kN
Lumped mass at roof floor = 2693.7 kN ( no. L L)
(d) Base shear, VB = C αh W = C β I α0 W
where, C is a coefficient defining the flexibility of structure with the increase in
number of stories depending upon fundamental time period T and given in Fig. B,
αh = β I α0 and β is a coefficient depending upon soil – foundation system, I is
importance factor for the structure, and α0 is basic horizontal seismic coefficient.
W = Total gravity loads of the building = 7 x 3449.3 + 2693.7 = 26838.8 kN
Building is without bracing of shear walls therefore based on code, T = 0.1 x n
where, T is fundamental time period and n is the number of stores including
basement stores.
T = 0.1 x 8 = 0.8 seconds
Assume the building is located in zone III and rested on raft foundation then,
β = 1.0, I = 1.0, and
α0 = 0.04, αh = β I α0 = 1 x 1 x 0.04 = 0.04
Value of C from Fig. B below; for T = 0.8 seconds is C = 0.65 Therefore, the
base shear is
VB = 0.65 x 0.04 x 26838.8, = 697.8 kN
(e) Distribution of lateral seismic force induced along the height of the building
and shear distribution in the building are given by the formulas,
In which hi
is the height of ith
floor measured from the base of the building.
Therefore, h1
= 3m,
h2
= 6m, h3
= 9m, h4
= 12m, h5
= 15m, h6
= 18m, h7
= 21, and h8
= 24m,
With these forces Q1
to Q8
are worked out and shear force in the various stories are
computed as
Note: The seismic shear force for which the building is to be designed is indicated
in the last column of the Table Q and shown in the Fig. C .
Serviceability Criteria:
(a) With respect to wind design, the following aspects have to be considered to
ensure the satisfactory performance of a high-rise building structure under
service conditions:
Lateral deflection (drift) of the structure, particularly as this effects the stability
of the structure and the cracking of nonstructural elements and structural
members. Drift is the magnitude of displacements at the top of a building
relative to its base. Maximum allowable drift = H/500, where H is total height of
the building.
(b) Motion of the structure, as this affects comfort of the occupants.
Appropriate Shapes for Lateral Load Resisting High-rise Buildings:
Unsymmetrical buildings usually develop torsion due to lateral loads such as
wind and seismic forces. Hence, buildings should have a simple plan and need
to be symmetric both with respect to mass and rigidity. Therefore, the center of
mass and center of rigidity of the building either coincide with each other or
should be very close to each other to minimize the torsion. Irregular shape of
high-rise buildings may be designed as a combination of few regular shapes
with suitable movement joints.
In tall buildings, the length to width ratio in the plan of a building should not
exceed three. The figure below illustrates simple rules for layout plans for high-
rise buildings.
Do Don’t
Fig. 3 Guidelines for planning of lateral load resisting high-rise buildings
Residential High-rise Buildings Systems:
Residential buildings (apartment buildings, hotels, and dormitories) are
characterized by the presence of partitions that are designed during the
planning stage, are constructed with the progress of structure as shown in the
following figures. The presence of permanent partitions allows the columns’
layout to correspond to the architectural plan. The lateral resistance of
residential buildings is provided mostly by the frames in the buildings.
In residential buildings above 20 stories, a shear wall-frame interactive system
for resisting lateral loads is more appropriate. The major advantages of using
shear wall structures lie in the speed of construction, low reinforcing steel
content and acoustical privacy.
Office Buildings and their Structural Systems:
Office buildings are characterized by the absence of partitions during design
and construction, since office space is designed rather than offices, with
subsequent partitioning to accommodate the needs of a particular tenant as
shown in the next figure. With the increasing height of buildings in recent
years, new concepts evolved to economically provide resistance to lateral
forces due to wind and earthquake. This evaluation is shown in Fig. 4.
Residential building typical floor plan
Office building typical floor plan
As can be seen in Fig. 4, in buildings up to 20 stories, frame action is
sufficient to provide lateral resistance.
In buildings higher than 20 stories, the rigidity of frame is mostly insufficient,
and sway due to wind may begin to control the design. Introduction to shear
walls which interact with frames will enhance the total rigidity of the building.
Fig. 4 Structural System for Office Buildings with Different Heights
Rigid Frames:
Rigid joints are used between an assembly of linear elements to form
vertical and horizontal planes. The vertical planes consist of columns and
girders mostly on a rectangle grid; and horizontal planes consist of slabs,
beams and girders.
Types of Rigid Frames:
Rigid frames can be used in two forms, plane frame and space frame.
Plane Frames:
Plane frame is a two-dimensional element in a building which represents
the local deflection of the building in two dimensions only. Plane frame
carries the loads applied in its own plane and the loads from its
approximate tributary areas.
Space Frames:
Space frame is a three-dimensional element and carries the exact applied
loads.
The whole structure of a building having a regular plan can be represented
by a single space frame. The structural analysis of a building represented by
space frame will give the actual three-dimensional deflections in different
parts of the building.
Rigid Frames and Core Structure (shear walls):
Shear Core Structure:
In high-rise buildings, a common solution is to gather vertical transportation
and energy distribution system (e.g. elevators, stairs, toilets, mechanical
shafts) to form a core or cores. These cores are utilized as shear wall
systems to provide the necessary lateral stability for the buildings
When the resultant of lateral forces does not act through the centroid of the
building’s structural system, torsion will be developed in the structure.
Optimal torsional resistance is obtained with closed core sections.
A combination of rigid frame with shear core structure (shear walls) will
significantly increase the lateral resistance of the building as a result of the
core and frame interaction.
Framed Tube:
The framed tube consists of a closely spaced grid of exterior columns,
connected with beams without interior columns as shown in Fig. (i). The
efficiency of this system is derived from the great number of rigid joints
acting along the periphery.
Tube in Tube:
The exterior columns and beams are spaced so closely that the facade has
the appearance of a wall with perforated window openings. The entire
building acts as a hollow tube cantilevering out of the ground. The interior
core (tube) increases the stiffness of the building by sharing the loads with
the facade tube as shown in Fig. (ii).
Multi-cell Framed Tube:
The multi-cell framed tube can be visualized as an assemblage of individual
tubes resulting in a multiple-cell tube as shown in Fig. (iii). The increase in
stiffness is apparent. The system allows for the greatest height and the
most floor area.
The Importance and efficiency of the framed tube system:
The framed tube, which was first introduced in the sixties of last century, consists of
closely spaced grid of exterior columns, connected with beams, It is an efficient system
to provide lateral resistance without interior columns. The efficiency of this system is
derived from the great number of rigid joints acting along the periphery, creating large
tube. The framed tube represents a logical evolution of the conventional framed
structure, possessing the necessary lateral stiffness with excellent torsional qualities
while retaining the planning flexibility of interior column free space
To visualize the action of a framed tube, it is simple to start with a solid peripheral wall
(a solid tube) which obviously will act as a cantilever with a moment deflection. When
the wall is penetrated with small round opening (windows), it will still behave as a
cantilever. When the openings become larger and rectangular, instead of round, part of
the lateral forces are resisted by shear distortion of the columns and beams, and only the
rest by the moment (cantilever) deflection of the tube.
The ratio of the moment deflection to shear deflection depends on the stiffness
relationship between beams and columns. A typical distribution of column axial forces
in such a structure is shown in the figure below. With increasing beam stiffness and
increasing number of stories, a higher participation of the windward and leeward sides
may bring the framed tube closer to a rigid tube and high rotational resistance.
In addition, the framed tube has an unusually high torsional resistance owing to the
location of stiffness around the periphery. When the sway or wind stresses being
controlling the design (it may be around 40 stories), the framed tube is
supplemented by a core, to create the tube-in-tube system. The tallest tube-in-tube is
the 714-ft-tall, 52-story Shell Oil Building in Houston, Texas shown in Fig.(1).
When the height reach 70-80 stories, the tube in tube may no longer be a sufficiently
rigid system, & the exterior tube must be made rigid to act in 100% flexure as a
cantilever. Either a set of diagonal members within the peripheral tube creates an
exterior cantilevered truss system or interior connecting shear walls act as webs to
tie the opposite faces of the tube into a single unit.
The use of diagonals in the exterior in a reinforced concrete building was
investigated in a research project of the Illinois Institute of Technology in a 115-
story building (1450ft high).
The thorough study investigated all the architectural, structural, mechanical,
construction that the building is technically feasible at an estimated cost lower than
that of the 100-story John Hankook Building, its steel counterpart.
In addition, the framed tube has an unusually high torsional resistance owing to the
location of stiffness around the periphery. When the sway or wind stresses being
controlling the design (it may be around 40 stories), the framed tube is supplemented
by a core, to create the tube-in-tube system. The tallest tube-in-tube is the 714-ft-tall,
52-story Shell Oil Building in Houston, Texas shown in Fig.(1).
When the height reach 70-80 stories, the tube in tube may no longer be a sufficiently
rigid system, & the exterior tube must be made rigid to act in 100% flexure as a
cantilever. Either a set of diagonal members within the peripheral tube creates an
exterior cantilevered truss system or interior connecting shear walls act as webs to tie
the opposite faces of the tube into a single unit.
The alternate method of creating a rigid
tube is a multi-cell arrangement as shown
in Fig.(2) through the use of cross shear
walls in one or both directions.
Fig.(1)The tallest tube-in-tube 714-ft tall,
52-story Shell Oil Building in Houston, Texas
Fig.(2). Multi-cell framed tubes (plan)
Fig. (i) Framed Tube
Fig. (ii) Tube in Tube
Fig. (iii) Tube in Tube
Resistance to Lateral Loads
With the increasing use of light curtain walls, dry-wall partitions, and high-strength
concrete and steel reinforcement in tall buildings, the effect of lateral loads
especially wind load have become more significant.
Figure below shows the schematic plot of the lateral resistance (as represented by
the lateral load versus story deflection) in modern building. As a result, the lateral
resistance of the building is only slightly larger than that of frame alone, and no
reserve capacity against lateral load is available.
Lateral Loads Resisting Systems:
(a) Frame Structures:
The term “frame” denotes a structure that derives its resistance to lateral
loading from the rigidity of the connections between columns and beams or
slabs. Frames deform in a predominantly shear mode where relative story
deflections depend on the shear applied at the storey level as shown in
Fig. 5(a).
(b) Shear Wall Structures:
The term shear wall is actually a misnomer as far as high-rise building are
concerned, since a slender shear wall subjected to lateral forces has
predominantly moment deflections (i.e. bending mode) and only very
insignificant shear distortions as shown in Fig. 5(b).
A shear wall will share the lateral load proportional to its stiffness. The
distinguishing factor of a shear wall is its much higher moment of inertia
than a column. The calculation of lateral stiffness of the wall is simple, and
stresses in a shear wall without openings involve simple bending theory
only.
(c) Shear Wall-Frame Buildings:
The term shear wall-frame structure is used to denote any combination of frames
and shear walls. In this system, shear walls supplement frames that, if unaided,
frames often cannot be efficiently designed to satisfy lateral load requirements.
The great majority of modern multistory buildings are in fact shear wall-frame
structures, since elevator shafts, stair-case walls, and central core units of tall
buildings are mostly treated as shear walls. Using only shear walls to respond to
lateral loads is impractical above 150m. To be sufficiently strong, cores have to
become too large and do not correspond to their functions for vertical transportation
and energy distribution. Furthermore, deflection may be large enough to cause
cracking of partitions and windows or even to evoke unpleasant psychological
reactions among the building’s occupants. The lateral rigidity is greatly improved by
using not only the shear wall system but also the rigid frame to resist lateral forces.
The interaction of shear wall and frame is shown in Fig. 5(c)
Fig. 5 Deformation Modes in Different
Lateral Load Resisting Systems
Coupled Shear Wall Structures:
Openings normally occur in vertical rows throughout the height of a wall and the
connection between the wall segments is provided by either connecting beams or
floor slabs, or a combination of both. These types of walls are called “coupled shear
walls” or “pierced shear walls” as shown in Fig. 6 below.
If the openings are very small, their effect on the overall state of stress in the shear
wall will be minor and hence, can be neglected in the design. If the openings are
large enough, this will result in a system in which typical frame action predominates.
Fig. 6 Coupled Shear Wall Structure
Coupled shear walls supported on exterior columns only:
Parking areas under high-rise buildings require larger clear spans in
comparison to the top floors. Therefore, the shear walls must be stopped
and supported on exterior columns, thus leaving the parking area clear
free from columns and walls. The lower portions of such walls act as a
deep beam spanning between the supporting columns as shown in Fig.7(a).
The computer study has shown that the second floor beam
(supporting the shear walls) acts like a tension member for the coupled
shear wall above. The lintels over the doors for the next five stories act as
compression struts as shown in Fig. 7(b)
Fig. 7 Coupled shear walls supported on exterior columns
Comparison of High-rise Structural Systems:
Low-to medium-rise buildings are normally designed for gravity loads, and
then checked for their ability to resist lateral loads. However, high-rise
buildings are much more susceptible to lateral load action. With respect to
gravity loads, the weight of the structure increases almost linearly with the
number of stories. However, the amount of material needed for the
resistance of lateral forces increases at a drastically accelerating rate.
Fig.8 compares most of the general concepts for high-rise buildings in steel
and concrete as suggested by Fazlur Khan. The structural systems given for
certain height should not be considered an absolute rule.
Fig. 8 Comparison of the
Concepts for Concrete and Steel
High-rise Buildings
From the structural point of view, the below-grade function outside the tower
should be for parking with longer spans; but from the ground level up, the first
segment should be for hotel, while the next segment should be for residential
floors. The topmost segment should be used for commercial and office space.
However, from the architectural/business point of view, the office, commercial
and parking floors are placed below apartment floors as shown in the figure
below.
Vertical Locations of Functions in a Multi - use Tall
Building
The Structural Flexibility in Column Spacing in High-rise Buildings :
- Closest column spacing should be at the lower level (base).
- Along the height of the building some intermediate columns can be dropped.
This type of arrangement of columns will avoid the use of complicated
heavy transfer beams. Figure below represents the floor plan of a
multistory building at different levels
Internal and External Movement of Structures:
Internal and external movements in the structure are caused by the expansion
and contraction due to the temperature changes. In order to get rid of forces
caused by the above phenomenon, expansion and contraction joints should
be provided in the structure. Differential settlements in foundations also create
forces in the structure, which should be considered in the design.
For the above reasons, movement joints must be provided in the buildings
(Fig.9). Movement joints may also be required when the structure changes in
size / shape, height, and foundation system. BS Codes recommend a
movement joint of 25mm in reinforced concrete frame structures at
approximately 50m length.
Fig.9 Plans Showing the Position of Movement Joints in Buildings.
Response of Structure:
Under the action of applied loads, the structure deflects vertically and
horizontally. The stresses and the deflections in the structure should be
within allowable limits.
Fig.10 indicates a measure of the response of the structure to the loadings.
Fig. 10 Life history of structure
Height-to-Width Ratio of a Building
With the increase of height-to width ratio, the
stiffness of the building should also increase.
Stiffness K is a function of size and number
of bays, structural system, rigidity and
Member connections.
For a plane frame structure the height-to-width
ratio is 5 to 7, or h / w = 5 ~ 7
Where, h is height of the building and w is the
width of the building.
Design of a shear wall subjected to wind and gravity loads:
The reinforced concrete wall shown in Fig.11 is 6m wide, 300mm thick and 36m
high, subjected to 1782kN vertical load. The wind pressure exerted on the wall is
11.69kN per linear meter of vertical height. Calculate the bending stresses and
shearing stresses in the wall.
P = 1782kN (gravity loads)
Vmax = wl = 11.69x36 = 421kN
Mmax = (wl2
) / 2 = 11.69x(36)2
/ 2 = 7575kN-m
Moment of inertia of the section will be:
I = tB3
/ 12 = [(0.3)x(6)3
] / 12 = 5.4m4
(i) Approximate Method
Continued ….
Fig.11 Design of a shear wall system
(ii) Design Based on Moment of Inertia
The maximum bending stress fmax
= Mc / I = 7575x3 / 5.4
= ± 4209kN/m2
= ± 4.209N/mm2
Average shear = v = V/A = 421 / (0.3x6) = 234kN/m2
or v = 0.234N/mm2
Compressive stress due to vertical load = P /A = -1782 /(6x0.3)
= -990kN / m2
= - 0.99N/mm2
Total stresses at base = -0.99 ± 4.209
Maximum compressive stress = - 5.199N/mm2
Maximum Tensile stress = 3.219N/mm2
The maximum compressive stress in concrete is - 5.199N/mm2
which is much lower than the allowable value.
The maximum tensile stress in concrete =3.219N/mm2
.
Therefore, reinforcement is needed.
In the figure shown, x=3.219x6/(3.219+5.199)= 2.29m
Total tensile force in the wall at base will be:
T = (3.219x2290x300)/2 = 1105727N
In case of the design for wind load/earthquake load, the allowable stress in
steel will be enhanced by 33%,
Therefore steel area As can be calculated as:
As = 1105727 / (0.87fyx4/3) = 1105727 / (0.87x460x4/3)
As = 2077.4mm2
Where, fy = 460N/mm2
yield strength of the high strength steel
The reinforcement should be provided at both ends of the wall
section, since wind and earthquake can act at any direction.
The foundation system should be designed to carry the tensile up lift.
Shafts:
Shafts are space structures which are stiff and strong in any direction.
Vertical shafts or tube structures in buildings are used to resist horizontal
and vertical loads. Their cross-section can be rectangular, square or
circular. In case when there is only one shaft in the building, it is usually
located central to a plan. In case of more than one shaft they should be
arranged in various symmetric locations. In case of preliminary design, a
tube with holes and openings less than 30% may be assumed as solid (i.e.
openings neglected). If more than 60% of the shaft surface is open, the
action will be assumed as a frame tube with a reduced strength and
stiffness.
When a shaft is relatively short and has an aspect ratio below 3, the
structural action will be shear –resisting. When the aspect ratio is 3 to 5
then shear dose not control the design.
For more slender shafts with aspect ratios above 5, bending action will the
dominant. Shafts with aspect ratios higher than 7 require tying of two or
more shafts together.
Design of a Tube System:
(Calculation of wall stresses)
The concrete shaft shown in Fig.12 is 6m square, 36m high, and 300mm
thick, is subjected to 11.69kN per linear meter of vertical height. Total
vertical load acting at the base of four walls is 7130kN i.e. each wall carries
1782.5kN.
Total shear at base = wl = 11.69x36 = 421kN.
Maximum moment = wl2
/2 = (11.69x362
) / 2 = 7575kNm
Approximately 3/4 of the moment is resisted by flange action,
i.e. Mf = (3/4)x7575 = 5681kNm
If the lever arm between the flange centroids is 5.7m,
then each flange carries a force of (5681)/(5.7) = ± 996.67kN
The stress in the flanges will be ±996.67/(5700x300) = ±0.583N/mm2
.
Therefore, the stress in the web due to moment will be ±0.583N/mm2
which
will decrease to zero at neutral axis
Fig.12 Approximate loads, moment, and stress in a tube structure.
A more accurate value of the stress can be obtained using the moment of
inertia of the tube as follows:
Moment of inertia of the tube is I = bh3
/12
= (6x63
) /12 – [5.4x(5.4)3
]/12 = 37.14m4
Stress in the shaft f = ± (Mc) / I = (7575x3) / (37.14) = ± 611.85 kN/m2
or the stress due to moment = ± 0.6118N/mm2
The average shear stress in the web is approximately given by:
v = V/A = 421/(2x6x0.3) = 116.94kN/m2
= 0.1169N/mm2
The vertical load of 7130kN produces a compressive stress of
7130/(4x5.7x0.3) = 1042.39kN/m2
= 1.0423N/mm2
Therefore, the combined stress from vertical compression and bending will
be as follows:
At the end where lateral force is applied the stress f = 0.6118 – 1.0423
= -0.4306N/mm2
(compressive)
At the other end f = -0.6118 -1.0423 = -1.6541N/mm2
(compressive).
Therefore, there is no tension in the shaft, which indicates the superiority
of the shaft systems.
Although there is no tensile stress in the shaft, but the designers provide
reinforcement in the walls so that greater strength and energy absorption
is provided.
As a conservative approximation, the designers neglect the effect of axial
load on the shaft and reinforcement provided to carry the entire moment
such that,
T = M/z = 7575/5.7 = 1328.95kN
T = 1328.95kN, if fy = 460N/mm2
(strength of the steel)
As = 1328950[(0.87x460) x 4/3] = 2490.53mm2
Therefore each flange wall needs 2490.53mm2
steel area.
Approximate Lateral Deflections for Vertical Elements:
It is necessary to calculate the lateral deflection or horizontal drift of the
building in the preliminary design stages. BS 8110 has recommended a
maximum value of Δ = h / 500, where h is the storey height or building height.
(a) A short solid wall : shear action, Fig.13
The deflection Δ will be,
Δ =(1.2Vh) / GA
where, G = (2/5)E for steel and concrete, E is modulus of elasticity
(b) A Tall solid wall or Tube deflect more by flexure, Fig.14
Δ = wh4
/(8EI), where, I = bd3
/12, b is width of the wall and d is its
thickness
Moment of inertia for Tube:
About axis 1-1: I = (1/12)[b1d3
1 - b2d3
2]
High-rise Building Systems:
Low – rise building ranges form 1 to 3 storey
Medium – rise building ranges form 4 to10 storey
High – rise building has 10 storey or more.
In high – rise building systems the vertical subsystems become the
controlling problem for two reasons:
(a) Higher vertical loads require larger columns, walls and shaft.
(b) Overturning moment, shear force and deflection produced by lateral
forces are much larger and need to be carefully tackled.
Low – or medium - rise buildings have shear resistance mode.
Filling of panels avoids the increase in size of beams and columns.
In High – rise buildings, the mode of resistance is of moment and deflection
rather than shear alone. Special structural arrangements will be needed.
They can be shear wall systems, tube/shaft systems, rigid-frame systems,
or a combination of them.
Basic Requirements for Providing Additional Resistance to Lateral
Forces and Deflection in High-rise Buildings:
(1)Increase the effective width of moment – resisting systems as shown in
Fig.15 below.
Fig. 15 Effective width in high-rise buildings
(2) Usage of truss system with chords and diagonals to increase the
efficiency of the system.
(3) Increase in the size of the structural elements in the lower
levels.
(4) Arrange the plan such that the overturn – resisting components
carry greater vertical loads.
(5) Usage of diagonal members in vertical systems.
(6) Providing horizontal diaphragm action by floors.
(7) Creating mega-frames by joining large vertical and horizontal
components such as two or more elevator shafts at multistory
intervals with a heavy floor systems or deep girder trusses.
Asymmetry in Building Forms:
When the building elevation is asymmetric, or when the support system
resultant is not axial with the building mass, overall bending will occur,
as shown in Fig.16 below.
Vertical Asymmetry: The vertical eccentricity between the resultant of
the gravity loads and the center line of the support system will produce
an overturning moment
Fig. 16 Eccentricity between the resultant of gravity loads and the support
system due to vertical asymmetry.
If the action by earthquake or wind is against the moment caused by the
eccentric gravity loads it will be beneficial. However, if the action of earthquake
or wind force is in the same direction of gravity load, then, it will make the
overturn problem worse.
Horizontal Asymmetry:
Horizontal asymmetry can also exist between the resultant of wind or
earthquake loads and that of shear resistance. This can produce
horizontal twisting (torsion) as illustrated in Fig.17
Fig. 17 Horizontal eccentricity due to horizontal asymmetry between the
shear resisting system and applied lateral load.
Wall System:
Walls are very rigid systems in their plane and can carry vertical &
horizontal loads. Walls can be concrete walls, precast concrete walls,
masonry walls and steel. When walls are braced, they can provide
excellent resistance to the horizontal loads in their plane. The potential for
resistance to lateral forces is high along the length of the wall but quite
weak across its thickness. The transverse resistance of wall to horizontal
forces is usually neglected. Two or more walls must be aligned
orthogonally i.e. at right angles to provide resistance to all lateral loads.
Fig.18 shows few layouts indicating the arrangement of various shear wall
systems.
Fig. 18 Plan views of shear wall
layouts
Torsional Moment Consideration in Wall Systems:
Torsion in shear wall systems can be neglected /
reduced when the center of orthogonal shear resistance
coincide or is close to the centriod of lateral loads. In
case the centers are apart from each other, then there
will be a horizontal moment (torsion) design problem.
Fig. 18 (a) and (b) represent unstable arrangement of
walls to resist horizontal forces. Because in Fig.18(a) the
walls supply no stiffness in the x-direction, while in
Fig. 18 (b) the centriod of resistance does not coincide
with the center of load application and there is almost no
stiffness against torsional rotation.
Fig.18 (c) represent a satisfactory arrangement. In the
arrangement shown in Fig.18 (d), there is a horizontal
torsion produced by load in x-direction, but the two walls in y-
direction form a couple that can provide torsional / rotational
resistance.
In Fig.18 (e), the tube / shaft form offers excellent resistance for
horizontal loads in any direction.
The layout shown in Fig.18 (f) is not only satisfactory with respect
to horizontal and rotation resistance, but also has additional
advantage of providing relief to the movement / deformation
caused by the temperature changes, creep and shrinkage.
The arrangement shown in Fig.18 (g) has sufficient shear
resistance but weak in torsional resistance.
Curved walls as shown in Fig.18 (h) offer good lateral resistance
due to the shall action, where floors serve as diaphragms.
Precast Concrete Construction (Prefabrication)
In comparison to conventional reinforced concrete, Precast concrete
has the following advantages and disadvantages:
Advantages:
(i) Quality control / Factory production
(ii) Quicker Construction / Time effective
(iii) Less congested construction site (less wastage)
(iv) No affect of weather on construction
Disadvantages:
(i) Expensive & difficult jointing
(ii) Limited choices of sizing for the architect (i.e. modular system)
(iii) Heavy transportation cost with machinery and equipment need
for the erection
(iv) Construction handling loads consideration required
Long Span Structures:
Space Truss Horizontal systems:
Space truss systems are used to cover large area (100ft.
spans or more) in a flat floor or roof. They can be used
running in one direction and supported by main trusses
which in turn are supported on columns as shown in
Fig. 19. For truss design, the moment arms for the
resisting coupled will approximately be the overall depth
of the truss.
Fig. 19 Space Truss Horizontal
System
Arch Structures:
When the span of a structure exceeds 100ft.(30m), it is often more
economical to build a system made up of curved members, in the
form of arches, suspension cables, and thin shells. Fig 20
illustrates the difference between the action of an arch, suspension
structure, and a beam.
Theoretically, the main forces in the arches are direct axial forces
therefore; they need smaller sections in comparison to flexural
members (beams).
Fig.20 Force resisting systems in an arch, a suspension member, and a beam
Since H varies inversely to the rise h, it is desirable to use as high a rise as
possible. Considering aesthetic & practical requirement, a span / rise ratio
ranging from 5 to 8 and sometimes up to 12 are adopted.
Suspension Systems
A suspension cable is the reserve of an arch rib, using materials
under tension instead of compression. In suspension systems
there
is no buckling and an overall span -to-depth ratio of 10 is possible.
Suspension system are used in bridges and building roofs.
Fig.21 below represents a typical suspension bridge system.
Fig.21 Suspension bridge
system
Foundation (Substructure) Systems:
A foundation is the interface between a building and the earth.
Foundation System serve to transmit loads from the vertical
subsystems of a building to the earth. Fig.22 represents some typical
examples of the foundations system. The bearing capacity of the earth
(soil) is one of the important factor which controls the type of a
foundation to be selected for a specific structure.
Fig.22 Some Foundation
Systems
Classification of Foundations:
Shallow foundation:
Generally they are those footings where D/B < 1 where, D = depth of
footing below ground level and B = width of the footing as shown in
Fig. 23(a). They include spread footing, combined footing, mat
foundation, strip footing etc.
Deep Foundations:
They are those foundations where D/B > 4 as shown in Fig.23(b).
These types of foundations include piles, drilled piles, caissons etc.
Fig. 23 Classification of
Foundations
Selection of Foundations Type:
As a preliminary assessment, the total approximate weight of a
Multistory building should be calculated and considered as uniformly
spread over the entire plan area of the building i.e. W/A, where W is
total vertical load from building and A is the total horizontal area of
building at foundation level. If the allowable bearing capacity of the soil
q ≥ 4W/A then, an individual footing would be more economical than
that of mat foundation. If q < 2W/A then, it is preferable to use mat
foundation. Differential settling or lateral movement of components
should be avoided.
Line / Strip Footings:
Structural walls are usually supported on a line footing. Line footing is
placed directly beneath the wall as shown in Fig.24 (a). If P is the load
per unit length from the wall then, the earth pressure will be p = P/b
(force/unit area)
If the footing is subjected to additional overturning moment or eccentric
loading then the pressure distribution under the footing will no longer
be
uniform as shown in Fig.24 (b).
It is preferred to not have an eccentricity e > b/6, should it happen, then
part of the footing will tend to be lifted from the earth and there will be
no pressure on that part.
Fig.24 Line Footing
Mat Foundations:
If the weight of the entire building W, coincides with the centriod of
the mat foundation, then the uniform pressure under the
foundations is q = W/(ab), where ab is the area of the mat
foundations as shown in Fig.25.
In case of a moment M, due to the lateral force, the effect on the
earth pressure can be taken as W with an eccentricity, e = M / W.
Fig.25 Mat
Foundation
Spread Footings:
The spread footing support concentrated load from a column. If the
column’s load is P then the pressure on the earth is p = P/ab,
where ab is the area of the footing (square/rectangular).
Generally the punching shear produced by the column on the
footing is shown as a truncated cone or pyramid surface with slope
of 1:2 as shown in Fig.26. The obtained value for punching shear
should be less than the allowable value.
Fig.26 Punching shear in a spread footing.
Combined Footing:
For some reasons, one footing that can support two or more columns
may be designed which is called combined footing Fig.27. The reasons
can be the limitation by
adjoining property line or overlapping of two spread footing. The ideal
case of a combined footing is when the resultant of the column loads
coincide with the center of gravity of the combined footing.
Fig.27 Combined footings
Pile and Caisson Foundations:
When surface soil is too weak or the load too heavy, it is possible to
transfer the load to lower strata by means of piles or caissons.
Piles Foundations:
Piles are slender columns that support the loads through bearing at the top and
then transfer these loads through friction along the length of the pile due to
adhesion to the soil and through direct end bearing to the rock or some strong
stratum. Piles are used in groups or they can be used in a row underneath a
strip footing or beam footing. There are various piles which are used in the
construction industry. They are timber piles, cast in situ concrete piles, precast
concrete and precast prestressed piles(driven piles).
Caisson Foundations:
Caissons are large hollow piles or large cylindrical enclosures that are
sunk into the ground. They are sunk in soft soil with the help of their
cutting edges. The hole in the caisson can be in the range of one meter
to 3 meters.
Moment of inertia of geometric sections:
Moments of inertias of some frequently used cross-sections in the structural
engineering are given in the following table.
LOAD BEARING WALLS
Walls is a vertical load-bearing member whose length exceeds four times its
thickness. If the primary function of a wall is to support vertical loads, it can be
defined as a load bearing wall. The loads should act at one-third the depth of the
bearing area from the loaded face as shown in Fig.12 below.
Types of Load Bearing Walls
Generally there are two types of load bearing walls which are:
(a) Concrete walls
(b) Brick masonry walls
Reinforced Concrete Load Bearing Walls
BS 8110 recommends that it should contain at least 0.004bh of vertical
reinforcement and 0.0025bh of horizontal reinforcement. If tension occurs
across the section of the wall, a layer of steel must be provided near each face of
the wall.
Types of Reinforced Concrete Load Bearing Walls
Reinforced concrete load bearing walls may be short (stocky) or slender and
braced or unbraced.
BS 8110 Recommendation:
(i) Short / Stocky Wall: A wall where the effective height divided by the
thickness (le / h) does not exceed 15 (braced) or 10 (unbraced)
(ii) Slender Wall : When the above limits exceed
(iii) Unbraced Walls : A wall providing its own lateral stability is
called unbraced wall.
(iv) Braced Wall : A wall where the reactions to lateral forces ( i.e. forces
perpendicular to the plane of wall) are provided by lateral
supports.
Brick Masonry Walls
Clay Bricks
Clay bricks were used about 10000 years ago which were called ‘adobe’ a
Spanish word based on the Arabic word of ‘atob’ meaning sun dried
brick. Later on it was discovered that by baking or firing the clay bricks,
their strength and durability increased significantly which are used
commonly nowadays.
Calcium Silicate Bricks
Calcium silicate (sand-lime) bricks are made by molding lime mortar into
brick shapes. These bricks are manufactured by using steam under high
pressure.
Concrete Units
Modern concrete blocks are manufactured by vibrating a mixture of
Portland cement, sand and aggregate in a mold under pressure, curing with
pressurized steam.
Mortar
Mortar is the important component of brickwork which for load bearing
brickwork can be cement: lime: sand mix. The proportion varies on the
basis of strength requirement. The thickness of mortar joint recommended
by BS 5628 is 10 ~ 12mm.
Load Bearing Brickwork (LBB) Method and its Design Consideration
Load bearing brickwork construction is most appropriately used in
buildings where the floor areas are subdivided into a relatively large
number of rooms of small to medium sizes and in which the floor plan is
repeated on each storey throughout the height of the building. This system
avoids a heavy concentration of vertical loads at foundation level . The
LBB method differs from conventional RC method where beams and
columns support the loads while in LBB the walls act as structural
elements which provide support and stability for the buildings
Robustness in Load Bearing Brickwork Construction
Incorporating all safety factors recommended by the relevant codes of
practice, the factor of safety in LBB is much higher than that of RC
system. Even though considerable attention is needed to ensure the
robustness of the building. Because due to the abnormal loading such as
high wind pressure, earthquake force, gas explosion or vehicle impact in
the structure it is possible that the structure may suffer severe damage or
total collapse.
Fig. 14 shows three different layouts of LBB construction
Fig. 14
(a)
Partial Safety Factor m for Accidental Loads
Considering the probable effects of misuse or accident,
BS 5628-I recommends that the partial safety factors shown
in Table 2 ( Table 4 of BS 5628-I reproduced) should be
used in the design.
Table 2 : Partial Safety m Factors for Material
Fig 15 shows the effect of mortar mix proportions on the crushing strength of
brickwork built with medium strength bricks.
Compressive strength of LBB based on BS 5628-I
Table 3 ( Table 1 of BS 5628-I reproduced ) represents different mortar
designations with their constituent proportion by volume and mean
compressive strength at 28 days.
Table 3 : Requirements for mortar (Table 1 of BS 5628-I)
Table 4 ( Table 2(a) of BS 5628-I reproduced ) represents characteristic
compressive strength of masonry, fk made from various units using
different mortar designation.
Table 4 : Characteristic compressive strength of masonry, f k in N/mm2
REFERENCES:
1. Lin, T. Y. & Stotesbury, S.D., Structural concepts and systems for architects and
egineers, Van Nostrand Reinhold Company, USA, 1988
2. Fintel, M., Handbook of concrete engineering, Van Nostrand Reinhold Company,
USA, 1986
3. Beedle, L. S., Development in tall buildings 1983, Van Nostrand Reinhold
company, USA, 1983
4. Kowalczyk, R.M., Sinn, R. & Kilmister, M.B., Structural system for tall
buildings, McGraw-Hill, Inc., USA, 1995
5. Schueller, W., High-rise building structures, John Wiley & Sons, USA, 1977
(Assignment No. 1)
(Q#1) There are four major factors which contributed to the development of
high-rise building which are:
(a) Development of high-strength materials
(b) Development of new design concepts
(c) Development of new structural systems
(d) Improved construction methods ( Including Top-down construction method)
(Explain every one of them briefly)
(Q#2) Explain briefly:
(a) The advantages of pre-fabricated constructions
(b) The disadvantages of pre-fabricated constructions
(Q#3) Optional
Explain briefly the usage and construction technique of earth retaining diaphragm
wall in top-down construction of tall buildings.
Contributing Factors to the Development of High - Rise Buildings:
(a) Development of high-strength materials
- Some 50 years ago 3000 psi (20.7 MPa) concrete were used only.
- Presently as high as 20000 psi (138 MPa) concrete is used.
- 40 ksi (276 MPa) steel changed to 70 ksi (483 MPa).
- Plain steel bars changed to high strength deformed bars.
- Plain bars has fy= 250 MPa, High strength deformed bars fy = 460MPa
- WWF(welded wire fabric) has high strength and easy to use
which has a strength of 485 MPa (Malaysia & UK).
(b) Development of New Design Concepts
- The development of working stress design, Ultimate Strength Design
and Limit State Design.
- Considering inelastic behavior of the materials / structural elements
which results into economy and efficiency.
(c) Development of New Structural Systems
The development of flat plate system
 Smooth beamless ceiling,
 Reduced storey height, when used permits an additional storey in
every 10 storey height.
-Introduction of shear walls as bracing elements
-Shear wall – Frame systems
-Frame tube systems
-Tube in tube systems
(d) Improved Construction Methods
- Improved construction methods provide better technical solutions to
many problems and reduced construction costs
- Life-slab system
- Other prefabricated concrete systems
- Post-tensioned and pre-tensioned concrete construction
- Modern construction equipment
- New construction methods such as top-down construction technique
Plan of load bearing link houses
Fig.(1)The tallest tube-in-tube 714-ft tall, 52-story Shell Oil Building in Houston, Texas

High raise building presentation 2 for phd.ppt

  • 1.
    High-rise Buildings N StructuralSystems for High-rise Buildings ( Lecture notes prepared by: Prof. Dr. JAHANGIR BAKHTERI ) Architect and the Engineer: An architect usually deals with a design problem such as a high-rise building in a comprehensive manner, and stresses an overall, rather than an elemental approach to design thinking. An engineer thinks in the reverse manner, starting with details, and without sufficient regards for overall picture. The principal form giver of a building is the architect, who must develop the self-confidence to express the strength and inherent beauty of structure and materials. He needs to understand the laws of nature as reflected by the play of forces in the building assemblage.
  • 2.
    Sears Tower, Chicago,USA Petronas Twin Tower, Malaysia
  • 3.
  • 4.
    Burj Dubai Tower: Talleststructure in the world. Total height 600 m, now completed.
  • 5.
    BOSTON -- MassachusettsInstitute of Technology is suing renowned architect Frank Gehry, alleging serious design flaws in the Stata Center, a building celebrated for its unconventional walls and radical angles. The school asserts the center, completed in spring 2004, has persistent leaks, drainage problems and mold growing on its brick exterior. It says accumulations of snow and ice have fallen dangerously from window boxes and other areas of its roofs, blocking emergency exits and causing damage MIT sues architect Gehry, flaws in building
  • 6.
    The suit saysMIT paid Los Angeles-based Gehry Partners $15 million to design the Stata Center, which cost $300 million to build. It houses labs, offices, classrooms and meeting rooms. "Gehry breached its duties by providing deficient design services and drawings," according to the suit, which also names New Jersey-based Beacon Skanska Construction Co., now known as Skanska USA Building Inc. The suit, filed Oct. 31, 2007 seeks unspecified damages.
  • 7.
    History of Developmentof High-rise Buildings: Throughout history, man has always tried to express his greatness by building monuments as high as he could practically achieve, with the Egyptian pyramid as the best example. - Egyptian pyramids made from masonry blocks are 481ft. high which is equivalent to a 40-storey building. - In 1907 the 47-storey Singer skyscraper in New York was built. - For centuries there were two basic materials used in construction: wood and masonry - Development of new construction materials such as concrete & steel and advanced technologies in construction made it possible to construct high-rise buildings. During the 1960s and 70s, F.R. Khan became the most prominent innovator in the area of high-rise buildings, both in concrete and steel. The methodology of shear wall-frame interaction in high-rise building was developed by him in 1964, in which the stiffness of frame building could be enhanced up to several times. Then, he initiated tubular design concept in high-rise buildings and applied in design & construction of several building structures. After that his next innovation was tube in tube concept in tall buildings.
  • 8.
    The innovations introducedby Khan not only improved the rigidity of tall buildings, but also resulted in substantial economics over the cost of buildings designed using traditional schemes. Nowadays, the many efficient, large, sophisticated two and three dimensional computer programs have made the analysis of complicated structure much simpler, more economical, and accessible to the designers.
  • 9.
    Contributing Factors tothe Development High-Rise Buildings (Assignment No. 1) (1)There are four major factors which contributed to the development of high-rise building which are: (a) Development of high-strength materials (b) Development of new design concepts (c) Development of new structural systems (d) Improved construction methods (Explain every one of them briefly) Need for High-rise buildings: High-rise buildings are closely related to the city; they are a natural response to dense population concentration, shortage of land, and high land costs. A tall building is an integral part of one large building organism, the city, where the building or activity cells are interconnected by multilevel movement systems. High-rise buildings range in height from 10 to more than 100 stories. Factors which governs/controls the height or the massing of a building are: the client’s needs versus the land available, necessary services to support the building and its inhabitants, the ecological impact of the building, What are the Problems caused by high-rise buildings?
  • 10.
    Some impacts ofskyscrapers on the city: Example: 109 storey Sears Tower in Chicago, USA, The building’s electrical system can serve a city of 147000 people and its air conditioning complex can cool 6000 one-family houses. A total of 102 elevators distribute about 16500 daily users to different parts of the building. Since the building contains all necessary supports services facilities such as shopping, entertainment, health, education, security, parking and other utilities, therefore, people seldom leave it. Load Action on High-Rise Buildings: Loads acting on a structure are generated either directly by the forces of nature or by man himself. In other words, there are two basic sources for building loads: geophysical and man-made. Fig.1 represents sources of building loads. The geophysical forces are the result of continuous changes in nature, which include gravitational loads, meteorological loads and seismological forces. Gravitational loads include weight of the structure (Dead load), and occupants of the building (Live load). Meteorological loads are wind loads, rain & snow loads, temperature stresses. Seismological forces are generated by the ground motion (i.e. earthquake forces).
  • 11.
    Fig. 1 Sourcesof Building Loads
  • 12.
    The structural facadeis exposed to the controlled temperature of the interior building environment and the daily and seasonal changes of weather. These changes in temperature causes additional stress in the buildings. The structure will have contraction for temperature drop and expansion for temperature increase. Creep is the time-dependent deformation that occurs in concrete for years after the initial loading deformation. Creep in concrete members is dependent on the magnitude of the stress, the length of time that the stress is applied, and the age and strength of concrete Shrinkage is a major cause for volume change in concrete and generally characterized by a gradual loss of moisture within the concrete member. As shrinkage stresses appear, additional restraint is required which, in turn, places additional loads on the structure. The man-made sources of loading are the shocks generated by cars, elevators, machines and equipment. They can be caused by blasts and impacts as well. Locked-in stresses can be created by the prestressing of concrete. Forces can also be locked into the structure during construction processes through restrained volumes.
  • 13.
    Loads and Forcesin Building Structures Loads acting in building structures are: (a)Dead Loads: Dead loads include the weight of all permanent components of the structure and any other immovable loads that are constant in magnitude and permanently attached to the structure. (b)Imposed Loads: Imposed loads are loads and forces that act on a structure by character of use of the structure to the nature of use, activities due to people, machinery installation, external forces etc. These loads are: i) Live Loads ii) Wind Loads iii) Earthquake Loads iv) Snow and Rain Loads v) Soil and Hydrostatic Forces vi) Erection Loads vii) Other Forces; such as impact, vibration, temperature effects, shrinkage, creep and forces due to the settlement of foundations.
  • 14.
    Wind Effect onTall Buildings In old day’s skyscrapers, the enormous weight of masonry walls, relatively thick slab and larger structural elements were such that wind action could not overcome the locked-in-gravity forces. The modern glass-walled skyscrapers with their optimum interior open space have taken away the overall rigidity of the structure and wind action has become a major problem for the designers of high-rise structures. The wind loading on a building is influenced by the environment factors such as large-scale roughness, shape and slenderness of the building and wind speed. The mean wind velocity and wind pressure in general increases with height of the building as shown in Fig. 2 Fig. 2 Variations of wind velocity and wind pressure on a building
  • 15.
    Calculation of windpressure on tall buildings: Based on B.S(British Standard), the characteristics wind pressure on a structure will be: Wk = 0.613Vs 2 Where, Wk in N/m2 is characteristic wind pressure and Vs is design wind speed in m/sec which can be calculated as follows: Vs = VS1S2S3 Where, V = basic wind speed in m/sec S1 = multiplying factor relating to topology S2 = multiplying factor relating to height above ground and wind braking. S3 = multiplying factor related to life of structure Note1: S1 may generally always be taken as unity except in the following cases: (a) On site adversely affected by very exposed hill slopes and crests where wind acceleration is known to occur: S1 = 1.1 (b) On site in enclosed steep-sided valleys completely sheltered from wind: S1 = 0.9
  • 16.
    Note2: S3 isa probability factor relating the likelihood of the design wind speed being exceeded to the probable life of the structure. A value of unity is recommended for general use and corresponds to an excessive speed occurring once in fifty years. S2 for various heights of buildings can be taken from Table 1.
  • 17.
    Notes: h is height(in meters) above general level of terrain to top of structure or part of structure. Topographical factors 1. open country with no obstructions 2. open country with scattered wind-breaks 3. country with many wind-breaks; small towns; suburbs of large cities 4. city centers and other environments with large and frequent obstructions
  • 18.
    Example of windforces calculation Assume an eight story framed office building shown below (Fig. D) is located in Kabul City’s suburb (e.g. Gardanai Baghe Bala). Calculate the wind forces on the structure assuming basic wind speed to be 40m/sec and the building is 22.5m by 22.5m in plan and having a 24m height above ground level.
  • 19.
    Characteristic wind pressureis Wk = 0.613 Vs 2 , where, Vs = design wind speed in m/sec = VS1S2S3 V = basic wind speed in m/sec S1 = multiplying factor relating to topology S2 = multiplying factor relating to height above ground and wind braking S3 = multiplying factor related to life of structure Assume basic wind speed of V = 40m/sec In an hilly area the value for S1 = 1.1 The probability factor the life of structure S3 = 1 S2 the height factor for the structure with a height 24m and open country form with no obstructions from the Table-1 using interpolation will be S2 = 1.026 Therefore, the design wind speed Vs = VS1S2S3 = 40 x 1.1 x 1 x 1.026 = 45.144 m/sec Characteristic wind pressure = 0.613 x (45.144)2 = 1285.97 N/m2 = 1.28597kN/m2 Wind pressure on an interior frame will be Wki = 1.28597 x 7.5 = 9.645 kN/m Wind concentrated force at each floor level will be Wp = 9.645 x 3 = 28.934 kN Wind concentrated force at roof level will be Wkr = 9.645 x 1.5 = 14.467 kN The concentrated forces due to wind pressure on the building in different floors and roof
  • 21.
    Example of earthquakeforces Calculation: Assume an eight stories framed office building with live loads of 2.5 kN/m2 shown below (Fig. A) is located in seismic zone III. Calculate the seismic forces on the structure using seismic coefficient method. Assume roof and floor slabs thicknesses of 150mm, beams section 250mm by 400mm, columns section 400mm by 500mm and story height is 3.0m. As recommended by the codes assume 25% live loads are acting during earthquake.
  • 22.
    (a) Dead Loads Weightof beams = 24 x 0.25 x 0.4 x 7.5 x 24 = 432.0 kN Weight of columns = 16 x 0.4 x 0.5 x 3 x 24 = 230.4 kN Weight of slabs = 22.5 x 22.5 x 0.15 x 24 = 1822.5 kN Weight of walls = 22.5 x 4 x 3 x 0,12 x 20 = 648.0 kN (b) Live loads Live loads at all floors = 22.5 x 22.5 x 2.5 x 0.25 = 316.4 kN (c)Lumped mass at floor level 1 = W1 = 432.0 + 230.4 + 1822.5 + 648.0 + 316.4 = 3449.3 kN Similarly W1 = W2 = W3 = W4 = W5 = W6 = W7 = 3449.3 kN Lumped mass at roof floor = 2693.7 kN ( no. L L) (d) Base shear, VB = C αh W = C β I α0 W where, C is a coefficient defining the flexibility of structure with the increase in number of stories depending upon fundamental time period T and given in Fig. B, αh = β I α0 and β is a coefficient depending upon soil – foundation system, I is importance factor for the structure, and α0 is basic horizontal seismic coefficient.
  • 23.
    W = Totalgravity loads of the building = 7 x 3449.3 + 2693.7 = 26838.8 kN Building is without bracing of shear walls therefore based on code, T = 0.1 x n where, T is fundamental time period and n is the number of stores including basement stores. T = 0.1 x 8 = 0.8 seconds Assume the building is located in zone III and rested on raft foundation then, β = 1.0, I = 1.0, and α0 = 0.04, αh = β I α0 = 1 x 1 x 0.04 = 0.04 Value of C from Fig. B below; for T = 0.8 seconds is C = 0.65 Therefore, the base shear is VB = 0.65 x 0.04 x 26838.8, = 697.8 kN (e) Distribution of lateral seismic force induced along the height of the building and shear distribution in the building are given by the formulas,
  • 24.
    In which hi isthe height of ith floor measured from the base of the building. Therefore, h1 = 3m, h2 = 6m, h3 = 9m, h4 = 12m, h5 = 15m, h6 = 18m, h7 = 21, and h8 = 24m, With these forces Q1 to Q8 are worked out and shear force in the various stories are computed as
  • 26.
    Note: The seismicshear force for which the building is to be designed is indicated in the last column of the Table Q and shown in the Fig. C .
  • 27.
    Serviceability Criteria: (a) Withrespect to wind design, the following aspects have to be considered to ensure the satisfactory performance of a high-rise building structure under service conditions: Lateral deflection (drift) of the structure, particularly as this effects the stability of the structure and the cracking of nonstructural elements and structural members. Drift is the magnitude of displacements at the top of a building relative to its base. Maximum allowable drift = H/500, where H is total height of the building. (b) Motion of the structure, as this affects comfort of the occupants. Appropriate Shapes for Lateral Load Resisting High-rise Buildings: Unsymmetrical buildings usually develop torsion due to lateral loads such as wind and seismic forces. Hence, buildings should have a simple plan and need to be symmetric both with respect to mass and rigidity. Therefore, the center of mass and center of rigidity of the building either coincide with each other or should be very close to each other to minimize the torsion. Irregular shape of high-rise buildings may be designed as a combination of few regular shapes with suitable movement joints. In tall buildings, the length to width ratio in the plan of a building should not exceed three. The figure below illustrates simple rules for layout plans for high- rise buildings.
  • 28.
    Do Don’t Fig. 3Guidelines for planning of lateral load resisting high-rise buildings
  • 29.
    Residential High-rise BuildingsSystems: Residential buildings (apartment buildings, hotels, and dormitories) are characterized by the presence of partitions that are designed during the planning stage, are constructed with the progress of structure as shown in the following figures. The presence of permanent partitions allows the columns’ layout to correspond to the architectural plan. The lateral resistance of residential buildings is provided mostly by the frames in the buildings. In residential buildings above 20 stories, a shear wall-frame interactive system for resisting lateral loads is more appropriate. The major advantages of using shear wall structures lie in the speed of construction, low reinforcing steel content and acoustical privacy. Office Buildings and their Structural Systems: Office buildings are characterized by the absence of partitions during design and construction, since office space is designed rather than offices, with subsequent partitioning to accommodate the needs of a particular tenant as shown in the next figure. With the increasing height of buildings in recent years, new concepts evolved to economically provide resistance to lateral forces due to wind and earthquake. This evaluation is shown in Fig. 4.
  • 30.
    Residential building typicalfloor plan Office building typical floor plan
  • 31.
    As can beseen in Fig. 4, in buildings up to 20 stories, frame action is sufficient to provide lateral resistance. In buildings higher than 20 stories, the rigidity of frame is mostly insufficient, and sway due to wind may begin to control the design. Introduction to shear walls which interact with frames will enhance the total rigidity of the building. Fig. 4 Structural System for Office Buildings with Different Heights
  • 32.
    Rigid Frames: Rigid jointsare used between an assembly of linear elements to form vertical and horizontal planes. The vertical planes consist of columns and girders mostly on a rectangle grid; and horizontal planes consist of slabs, beams and girders. Types of Rigid Frames: Rigid frames can be used in two forms, plane frame and space frame. Plane Frames: Plane frame is a two-dimensional element in a building which represents the local deflection of the building in two dimensions only. Plane frame carries the loads applied in its own plane and the loads from its approximate tributary areas. Space Frames: Space frame is a three-dimensional element and carries the exact applied loads.
  • 33.
    The whole structureof a building having a regular plan can be represented by a single space frame. The structural analysis of a building represented by space frame will give the actual three-dimensional deflections in different parts of the building. Rigid Frames and Core Structure (shear walls): Shear Core Structure: In high-rise buildings, a common solution is to gather vertical transportation and energy distribution system (e.g. elevators, stairs, toilets, mechanical shafts) to form a core or cores. These cores are utilized as shear wall systems to provide the necessary lateral stability for the buildings When the resultant of lateral forces does not act through the centroid of the building’s structural system, torsion will be developed in the structure. Optimal torsional resistance is obtained with closed core sections. A combination of rigid frame with shear core structure (shear walls) will significantly increase the lateral resistance of the building as a result of the core and frame interaction.
  • 34.
    Framed Tube: The framedtube consists of a closely spaced grid of exterior columns, connected with beams without interior columns as shown in Fig. (i). The efficiency of this system is derived from the great number of rigid joints acting along the periphery. Tube in Tube: The exterior columns and beams are spaced so closely that the facade has the appearance of a wall with perforated window openings. The entire building acts as a hollow tube cantilevering out of the ground. The interior core (tube) increases the stiffness of the building by sharing the loads with the facade tube as shown in Fig. (ii). Multi-cell Framed Tube: The multi-cell framed tube can be visualized as an assemblage of individual tubes resulting in a multiple-cell tube as shown in Fig. (iii). The increase in stiffness is apparent. The system allows for the greatest height and the most floor area.
  • 35.
    The Importance andefficiency of the framed tube system: The framed tube, which was first introduced in the sixties of last century, consists of closely spaced grid of exterior columns, connected with beams, It is an efficient system to provide lateral resistance without interior columns. The efficiency of this system is derived from the great number of rigid joints acting along the periphery, creating large tube. The framed tube represents a logical evolution of the conventional framed structure, possessing the necessary lateral stiffness with excellent torsional qualities while retaining the planning flexibility of interior column free space To visualize the action of a framed tube, it is simple to start with a solid peripheral wall (a solid tube) which obviously will act as a cantilever with a moment deflection. When the wall is penetrated with small round opening (windows), it will still behave as a cantilever. When the openings become larger and rectangular, instead of round, part of the lateral forces are resisted by shear distortion of the columns and beams, and only the rest by the moment (cantilever) deflection of the tube. The ratio of the moment deflection to shear deflection depends on the stiffness relationship between beams and columns. A typical distribution of column axial forces in such a structure is shown in the figure below. With increasing beam stiffness and increasing number of stories, a higher participation of the windward and leeward sides may bring the framed tube closer to a rigid tube and high rotational resistance.
  • 38.
    In addition, theframed tube has an unusually high torsional resistance owing to the location of stiffness around the periphery. When the sway or wind stresses being controlling the design (it may be around 40 stories), the framed tube is supplemented by a core, to create the tube-in-tube system. The tallest tube-in-tube is the 714-ft-tall, 52-story Shell Oil Building in Houston, Texas shown in Fig.(1). When the height reach 70-80 stories, the tube in tube may no longer be a sufficiently rigid system, & the exterior tube must be made rigid to act in 100% flexure as a cantilever. Either a set of diagonal members within the peripheral tube creates an exterior cantilevered truss system or interior connecting shear walls act as webs to tie the opposite faces of the tube into a single unit. The use of diagonals in the exterior in a reinforced concrete building was investigated in a research project of the Illinois Institute of Technology in a 115- story building (1450ft high). The thorough study investigated all the architectural, structural, mechanical, construction that the building is technically feasible at an estimated cost lower than that of the 100-story John Hankook Building, its steel counterpart.
  • 39.
    In addition, theframed tube has an unusually high torsional resistance owing to the location of stiffness around the periphery. When the sway or wind stresses being controlling the design (it may be around 40 stories), the framed tube is supplemented by a core, to create the tube-in-tube system. The tallest tube-in-tube is the 714-ft-tall, 52-story Shell Oil Building in Houston, Texas shown in Fig.(1). When the height reach 70-80 stories, the tube in tube may no longer be a sufficiently rigid system, & the exterior tube must be made rigid to act in 100% flexure as a cantilever. Either a set of diagonal members within the peripheral tube creates an exterior cantilevered truss system or interior connecting shear walls act as webs to tie the opposite faces of the tube into a single unit. The alternate method of creating a rigid tube is a multi-cell arrangement as shown in Fig.(2) through the use of cross shear walls in one or both directions. Fig.(1)The tallest tube-in-tube 714-ft tall, 52-story Shell Oil Building in Houston, Texas
  • 40.
  • 41.
    Fig. (i) FramedTube Fig. (ii) Tube in Tube Fig. (iii) Tube in Tube
  • 42.
    Resistance to LateralLoads With the increasing use of light curtain walls, dry-wall partitions, and high-strength concrete and steel reinforcement in tall buildings, the effect of lateral loads especially wind load have become more significant. Figure below shows the schematic plot of the lateral resistance (as represented by the lateral load versus story deflection) in modern building. As a result, the lateral resistance of the building is only slightly larger than that of frame alone, and no reserve capacity against lateral load is available.
  • 43.
    Lateral Loads ResistingSystems: (a) Frame Structures: The term “frame” denotes a structure that derives its resistance to lateral loading from the rigidity of the connections between columns and beams or slabs. Frames deform in a predominantly shear mode where relative story deflections depend on the shear applied at the storey level as shown in Fig. 5(a). (b) Shear Wall Structures: The term shear wall is actually a misnomer as far as high-rise building are concerned, since a slender shear wall subjected to lateral forces has predominantly moment deflections (i.e. bending mode) and only very insignificant shear distortions as shown in Fig. 5(b). A shear wall will share the lateral load proportional to its stiffness. The distinguishing factor of a shear wall is its much higher moment of inertia than a column. The calculation of lateral stiffness of the wall is simple, and stresses in a shear wall without openings involve simple bending theory only.
  • 44.
    (c) Shear Wall-FrameBuildings: The term shear wall-frame structure is used to denote any combination of frames and shear walls. In this system, shear walls supplement frames that, if unaided, frames often cannot be efficiently designed to satisfy lateral load requirements. The great majority of modern multistory buildings are in fact shear wall-frame structures, since elevator shafts, stair-case walls, and central core units of tall buildings are mostly treated as shear walls. Using only shear walls to respond to lateral loads is impractical above 150m. To be sufficiently strong, cores have to become too large and do not correspond to their functions for vertical transportation and energy distribution. Furthermore, deflection may be large enough to cause cracking of partitions and windows or even to evoke unpleasant psychological reactions among the building’s occupants. The lateral rigidity is greatly improved by using not only the shear wall system but also the rigid frame to resist lateral forces. The interaction of shear wall and frame is shown in Fig. 5(c)
  • 45.
    Fig. 5 DeformationModes in Different Lateral Load Resisting Systems
  • 46.
    Coupled Shear WallStructures: Openings normally occur in vertical rows throughout the height of a wall and the connection between the wall segments is provided by either connecting beams or floor slabs, or a combination of both. These types of walls are called “coupled shear walls” or “pierced shear walls” as shown in Fig. 6 below. If the openings are very small, their effect on the overall state of stress in the shear wall will be minor and hence, can be neglected in the design. If the openings are large enough, this will result in a system in which typical frame action predominates. Fig. 6 Coupled Shear Wall Structure
  • 47.
    Coupled shear wallssupported on exterior columns only: Parking areas under high-rise buildings require larger clear spans in comparison to the top floors. Therefore, the shear walls must be stopped and supported on exterior columns, thus leaving the parking area clear free from columns and walls. The lower portions of such walls act as a deep beam spanning between the supporting columns as shown in Fig.7(a). The computer study has shown that the second floor beam (supporting the shear walls) acts like a tension member for the coupled shear wall above. The lintels over the doors for the next five stories act as compression struts as shown in Fig. 7(b)
  • 48.
    Fig. 7 Coupledshear walls supported on exterior columns
  • 49.
    Comparison of High-riseStructural Systems: Low-to medium-rise buildings are normally designed for gravity loads, and then checked for their ability to resist lateral loads. However, high-rise buildings are much more susceptible to lateral load action. With respect to gravity loads, the weight of the structure increases almost linearly with the number of stories. However, the amount of material needed for the resistance of lateral forces increases at a drastically accelerating rate. Fig.8 compares most of the general concepts for high-rise buildings in steel and concrete as suggested by Fazlur Khan. The structural systems given for certain height should not be considered an absolute rule.
  • 50.
    Fig. 8 Comparisonof the Concepts for Concrete and Steel High-rise Buildings
  • 51.
    From the structuralpoint of view, the below-grade function outside the tower should be for parking with longer spans; but from the ground level up, the first segment should be for hotel, while the next segment should be for residential floors. The topmost segment should be used for commercial and office space. However, from the architectural/business point of view, the office, commercial and parking floors are placed below apartment floors as shown in the figure below.
  • 52.
    Vertical Locations ofFunctions in a Multi - use Tall Building The Structural Flexibility in Column Spacing in High-rise Buildings : - Closest column spacing should be at the lower level (base). - Along the height of the building some intermediate columns can be dropped. This type of arrangement of columns will avoid the use of complicated heavy transfer beams. Figure below represents the floor plan of a multistory building at different levels
  • 53.
    Internal and ExternalMovement of Structures: Internal and external movements in the structure are caused by the expansion and contraction due to the temperature changes. In order to get rid of forces caused by the above phenomenon, expansion and contraction joints should be provided in the structure. Differential settlements in foundations also create forces in the structure, which should be considered in the design. For the above reasons, movement joints must be provided in the buildings (Fig.9). Movement joints may also be required when the structure changes in size / shape, height, and foundation system. BS Codes recommend a movement joint of 25mm in reinforced concrete frame structures at approximately 50m length.
  • 54.
    Fig.9 Plans Showingthe Position of Movement Joints in Buildings. Response of Structure: Under the action of applied loads, the structure deflects vertically and horizontally. The stresses and the deflections in the structure should be within allowable limits. Fig.10 indicates a measure of the response of the structure to the loadings.
  • 55.
    Fig. 10 Lifehistory of structure
  • 56.
    Height-to-Width Ratio ofa Building With the increase of height-to width ratio, the stiffness of the building should also increase. Stiffness K is a function of size and number of bays, structural system, rigidity and Member connections. For a plane frame structure the height-to-width ratio is 5 to 7, or h / w = 5 ~ 7 Where, h is height of the building and w is the width of the building.
  • 57.
    Design of ashear wall subjected to wind and gravity loads: The reinforced concrete wall shown in Fig.11 is 6m wide, 300mm thick and 36m high, subjected to 1782kN vertical load. The wind pressure exerted on the wall is 11.69kN per linear meter of vertical height. Calculate the bending stresses and shearing stresses in the wall. P = 1782kN (gravity loads) Vmax = wl = 11.69x36 = 421kN Mmax = (wl2 ) / 2 = 11.69x(36)2 / 2 = 7575kN-m Moment of inertia of the section will be: I = tB3 / 12 = [(0.3)x(6)3 ] / 12 = 5.4m4 (i) Approximate Method
  • 58.
  • 59.
    Fig.11 Design ofa shear wall system (ii) Design Based on Moment of Inertia The maximum bending stress fmax = Mc / I = 7575x3 / 5.4 = ± 4209kN/m2 = ± 4.209N/mm2 Average shear = v = V/A = 421 / (0.3x6) = 234kN/m2 or v = 0.234N/mm2 Compressive stress due to vertical load = P /A = -1782 /(6x0.3) = -990kN / m2 = - 0.99N/mm2 Total stresses at base = -0.99 ± 4.209 Maximum compressive stress = - 5.199N/mm2 Maximum Tensile stress = 3.219N/mm2
  • 60.
    The maximum compressivestress in concrete is - 5.199N/mm2 which is much lower than the allowable value. The maximum tensile stress in concrete =3.219N/mm2 . Therefore, reinforcement is needed. In the figure shown, x=3.219x6/(3.219+5.199)= 2.29m Total tensile force in the wall at base will be: T = (3.219x2290x300)/2 = 1105727N In case of the design for wind load/earthquake load, the allowable stress in steel will be enhanced by 33%, Therefore steel area As can be calculated as: As = 1105727 / (0.87fyx4/3) = 1105727 / (0.87x460x4/3) As = 2077.4mm2 Where, fy = 460N/mm2 yield strength of the high strength steel
  • 61.
    The reinforcement shouldbe provided at both ends of the wall section, since wind and earthquake can act at any direction. The foundation system should be designed to carry the tensile up lift. Shafts: Shafts are space structures which are stiff and strong in any direction. Vertical shafts or tube structures in buildings are used to resist horizontal and vertical loads. Their cross-section can be rectangular, square or circular. In case when there is only one shaft in the building, it is usually located central to a plan. In case of more than one shaft they should be arranged in various symmetric locations. In case of preliminary design, a tube with holes and openings less than 30% may be assumed as solid (i.e. openings neglected). If more than 60% of the shaft surface is open, the action will be assumed as a frame tube with a reduced strength and stiffness. When a shaft is relatively short and has an aspect ratio below 3, the structural action will be shear –resisting. When the aspect ratio is 3 to 5 then shear dose not control the design.
  • 62.
    For more slendershafts with aspect ratios above 5, bending action will the dominant. Shafts with aspect ratios higher than 7 require tying of two or more shafts together. Design of a Tube System: (Calculation of wall stresses) The concrete shaft shown in Fig.12 is 6m square, 36m high, and 300mm thick, is subjected to 11.69kN per linear meter of vertical height. Total vertical load acting at the base of four walls is 7130kN i.e. each wall carries 1782.5kN. Total shear at base = wl = 11.69x36 = 421kN. Maximum moment = wl2 /2 = (11.69x362 ) / 2 = 7575kNm Approximately 3/4 of the moment is resisted by flange action, i.e. Mf = (3/4)x7575 = 5681kNm If the lever arm between the flange centroids is 5.7m, then each flange carries a force of (5681)/(5.7) = ± 996.67kN The stress in the flanges will be ±996.67/(5700x300) = ±0.583N/mm2 . Therefore, the stress in the web due to moment will be ±0.583N/mm2 which will decrease to zero at neutral axis
  • 63.
    Fig.12 Approximate loads,moment, and stress in a tube structure.
  • 64.
    A more accuratevalue of the stress can be obtained using the moment of inertia of the tube as follows: Moment of inertia of the tube is I = bh3 /12 = (6x63 ) /12 – [5.4x(5.4)3 ]/12 = 37.14m4 Stress in the shaft f = ± (Mc) / I = (7575x3) / (37.14) = ± 611.85 kN/m2 or the stress due to moment = ± 0.6118N/mm2 The average shear stress in the web is approximately given by: v = V/A = 421/(2x6x0.3) = 116.94kN/m2 = 0.1169N/mm2 The vertical load of 7130kN produces a compressive stress of 7130/(4x5.7x0.3) = 1042.39kN/m2 = 1.0423N/mm2 Therefore, the combined stress from vertical compression and bending will be as follows: At the end where lateral force is applied the stress f = 0.6118 – 1.0423 = -0.4306N/mm2 (compressive) At the other end f = -0.6118 -1.0423 = -1.6541N/mm2 (compressive).
  • 65.
    Therefore, there isno tension in the shaft, which indicates the superiority of the shaft systems. Although there is no tensile stress in the shaft, but the designers provide reinforcement in the walls so that greater strength and energy absorption is provided. As a conservative approximation, the designers neglect the effect of axial load on the shaft and reinforcement provided to carry the entire moment such that, T = M/z = 7575/5.7 = 1328.95kN T = 1328.95kN, if fy = 460N/mm2 (strength of the steel) As = 1328950[(0.87x460) x 4/3] = 2490.53mm2 Therefore each flange wall needs 2490.53mm2 steel area.
  • 66.
    Approximate Lateral Deflectionsfor Vertical Elements: It is necessary to calculate the lateral deflection or horizontal drift of the building in the preliminary design stages. BS 8110 has recommended a maximum value of Δ = h / 500, where h is the storey height or building height. (a) A short solid wall : shear action, Fig.13 The deflection Δ will be, Δ =(1.2Vh) / GA where, G = (2/5)E for steel and concrete, E is modulus of elasticity
  • 67.
    (b) A Tallsolid wall or Tube deflect more by flexure, Fig.14 Δ = wh4 /(8EI), where, I = bd3 /12, b is width of the wall and d is its thickness
  • 68.
    Moment of inertiafor Tube: About axis 1-1: I = (1/12)[b1d3 1 - b2d3 2]
  • 69.
    High-rise Building Systems: Low– rise building ranges form 1 to 3 storey Medium – rise building ranges form 4 to10 storey High – rise building has 10 storey or more. In high – rise building systems the vertical subsystems become the controlling problem for two reasons: (a) Higher vertical loads require larger columns, walls and shaft. (b) Overturning moment, shear force and deflection produced by lateral forces are much larger and need to be carefully tackled. Low – or medium - rise buildings have shear resistance mode. Filling of panels avoids the increase in size of beams and columns. In High – rise buildings, the mode of resistance is of moment and deflection rather than shear alone. Special structural arrangements will be needed. They can be shear wall systems, tube/shaft systems, rigid-frame systems, or a combination of them.
  • 70.
    Basic Requirements forProviding Additional Resistance to Lateral Forces and Deflection in High-rise Buildings: (1)Increase the effective width of moment – resisting systems as shown in Fig.15 below. Fig. 15 Effective width in high-rise buildings
  • 71.
    (2) Usage oftruss system with chords and diagonals to increase the efficiency of the system. (3) Increase in the size of the structural elements in the lower levels. (4) Arrange the plan such that the overturn – resisting components carry greater vertical loads. (5) Usage of diagonal members in vertical systems. (6) Providing horizontal diaphragm action by floors. (7) Creating mega-frames by joining large vertical and horizontal components such as two or more elevator shafts at multistory intervals with a heavy floor systems or deep girder trusses.
  • 72.
    Asymmetry in BuildingForms: When the building elevation is asymmetric, or when the support system resultant is not axial with the building mass, overall bending will occur, as shown in Fig.16 below. Vertical Asymmetry: The vertical eccentricity between the resultant of the gravity loads and the center line of the support system will produce an overturning moment
  • 73.
    Fig. 16 Eccentricitybetween the resultant of gravity loads and the support system due to vertical asymmetry.
  • 74.
    If the actionby earthquake or wind is against the moment caused by the eccentric gravity loads it will be beneficial. However, if the action of earthquake or wind force is in the same direction of gravity load, then, it will make the overturn problem worse. Horizontal Asymmetry: Horizontal asymmetry can also exist between the resultant of wind or earthquake loads and that of shear resistance. This can produce horizontal twisting (torsion) as illustrated in Fig.17
  • 75.
    Fig. 17 Horizontaleccentricity due to horizontal asymmetry between the shear resisting system and applied lateral load.
  • 76.
    Wall System: Walls arevery rigid systems in their plane and can carry vertical & horizontal loads. Walls can be concrete walls, precast concrete walls, masonry walls and steel. When walls are braced, they can provide excellent resistance to the horizontal loads in their plane. The potential for resistance to lateral forces is high along the length of the wall but quite weak across its thickness. The transverse resistance of wall to horizontal forces is usually neglected. Two or more walls must be aligned orthogonally i.e. at right angles to provide resistance to all lateral loads. Fig.18 shows few layouts indicating the arrangement of various shear wall systems.
  • 77.
    Fig. 18 Planviews of shear wall layouts
  • 78.
    Torsional Moment Considerationin Wall Systems: Torsion in shear wall systems can be neglected / reduced when the center of orthogonal shear resistance coincide or is close to the centriod of lateral loads. In case the centers are apart from each other, then there will be a horizontal moment (torsion) design problem. Fig. 18 (a) and (b) represent unstable arrangement of walls to resist horizontal forces. Because in Fig.18(a) the walls supply no stiffness in the x-direction, while in Fig. 18 (b) the centriod of resistance does not coincide with the center of load application and there is almost no stiffness against torsional rotation. Fig.18 (c) represent a satisfactory arrangement. In the arrangement shown in Fig.18 (d), there is a horizontal
  • 79.
    torsion produced byload in x-direction, but the two walls in y- direction form a couple that can provide torsional / rotational resistance. In Fig.18 (e), the tube / shaft form offers excellent resistance for horizontal loads in any direction. The layout shown in Fig.18 (f) is not only satisfactory with respect to horizontal and rotation resistance, but also has additional advantage of providing relief to the movement / deformation caused by the temperature changes, creep and shrinkage. The arrangement shown in Fig.18 (g) has sufficient shear resistance but weak in torsional resistance. Curved walls as shown in Fig.18 (h) offer good lateral resistance due to the shall action, where floors serve as diaphragms.
  • 80.
    Precast Concrete Construction(Prefabrication) In comparison to conventional reinforced concrete, Precast concrete has the following advantages and disadvantages: Advantages: (i) Quality control / Factory production (ii) Quicker Construction / Time effective (iii) Less congested construction site (less wastage) (iv) No affect of weather on construction Disadvantages: (i) Expensive & difficult jointing (ii) Limited choices of sizing for the architect (i.e. modular system) (iii) Heavy transportation cost with machinery and equipment need for the erection (iv) Construction handling loads consideration required
  • 81.
    Long Span Structures: SpaceTruss Horizontal systems: Space truss systems are used to cover large area (100ft. spans or more) in a flat floor or roof. They can be used running in one direction and supported by main trusses which in turn are supported on columns as shown in Fig. 19. For truss design, the moment arms for the resisting coupled will approximately be the overall depth of the truss.
  • 82.
    Fig. 19 SpaceTruss Horizontal System
  • 83.
    Arch Structures: When thespan of a structure exceeds 100ft.(30m), it is often more economical to build a system made up of curved members, in the form of arches, suspension cables, and thin shells. Fig 20 illustrates the difference between the action of an arch, suspension structure, and a beam. Theoretically, the main forces in the arches are direct axial forces therefore; they need smaller sections in comparison to flexural members (beams). Fig.20 Force resisting systems in an arch, a suspension member, and a beam
  • 84.
    Since H variesinversely to the rise h, it is desirable to use as high a rise as possible. Considering aesthetic & practical requirement, a span / rise ratio ranging from 5 to 8 and sometimes up to 12 are adopted.
  • 85.
    Suspension Systems A suspensioncable is the reserve of an arch rib, using materials under tension instead of compression. In suspension systems there is no buckling and an overall span -to-depth ratio of 10 is possible. Suspension system are used in bridges and building roofs. Fig.21 below represents a typical suspension bridge system. Fig.21 Suspension bridge system
  • 86.
    Foundation (Substructure) Systems: Afoundation is the interface between a building and the earth. Foundation System serve to transmit loads from the vertical subsystems of a building to the earth. Fig.22 represents some typical examples of the foundations system. The bearing capacity of the earth (soil) is one of the important factor which controls the type of a foundation to be selected for a specific structure.
  • 88.
  • 89.
    Classification of Foundations: Shallowfoundation: Generally they are those footings where D/B < 1 where, D = depth of footing below ground level and B = width of the footing as shown in Fig. 23(a). They include spread footing, combined footing, mat foundation, strip footing etc. Deep Foundations: They are those foundations where D/B > 4 as shown in Fig.23(b). These types of foundations include piles, drilled piles, caissons etc.
  • 90.
    Fig. 23 Classificationof Foundations
  • 91.
    Selection of FoundationsType: As a preliminary assessment, the total approximate weight of a Multistory building should be calculated and considered as uniformly spread over the entire plan area of the building i.e. W/A, where W is total vertical load from building and A is the total horizontal area of building at foundation level. If the allowable bearing capacity of the soil q ≥ 4W/A then, an individual footing would be more economical than that of mat foundation. If q < 2W/A then, it is preferable to use mat foundation. Differential settling or lateral movement of components should be avoided.
  • 92.
    Line / StripFootings: Structural walls are usually supported on a line footing. Line footing is placed directly beneath the wall as shown in Fig.24 (a). If P is the load per unit length from the wall then, the earth pressure will be p = P/b (force/unit area) If the footing is subjected to additional overturning moment or eccentric loading then the pressure distribution under the footing will no longer be uniform as shown in Fig.24 (b). It is preferred to not have an eccentricity e > b/6, should it happen, then part of the footing will tend to be lifted from the earth and there will be no pressure on that part.
  • 93.
  • 94.
    Mat Foundations: If theweight of the entire building W, coincides with the centriod of the mat foundation, then the uniform pressure under the foundations is q = W/(ab), where ab is the area of the mat foundations as shown in Fig.25. In case of a moment M, due to the lateral force, the effect on the earth pressure can be taken as W with an eccentricity, e = M / W.
  • 95.
  • 96.
    Spread Footings: The spreadfooting support concentrated load from a column. If the column’s load is P then the pressure on the earth is p = P/ab, where ab is the area of the footing (square/rectangular). Generally the punching shear produced by the column on the footing is shown as a truncated cone or pyramid surface with slope of 1:2 as shown in Fig.26. The obtained value for punching shear should be less than the allowable value.
  • 97.
    Fig.26 Punching shearin a spread footing.
  • 98.
    Combined Footing: For somereasons, one footing that can support two or more columns may be designed which is called combined footing Fig.27. The reasons can be the limitation by adjoining property line or overlapping of two spread footing. The ideal case of a combined footing is when the resultant of the column loads coincide with the center of gravity of the combined footing.
  • 99.
  • 100.
    Pile and CaissonFoundations: When surface soil is too weak or the load too heavy, it is possible to transfer the load to lower strata by means of piles or caissons. Piles Foundations: Piles are slender columns that support the loads through bearing at the top and then transfer these loads through friction along the length of the pile due to adhesion to the soil and through direct end bearing to the rock or some strong stratum. Piles are used in groups or they can be used in a row underneath a strip footing or beam footing. There are various piles which are used in the construction industry. They are timber piles, cast in situ concrete piles, precast concrete and precast prestressed piles(driven piles).
  • 101.
    Caisson Foundations: Caissons arelarge hollow piles or large cylindrical enclosures that are sunk into the ground. They are sunk in soft soil with the help of their cutting edges. The hole in the caisson can be in the range of one meter to 3 meters.
  • 102.
    Moment of inertiaof geometric sections: Moments of inertias of some frequently used cross-sections in the structural engineering are given in the following table.
  • 105.
    LOAD BEARING WALLS Wallsis a vertical load-bearing member whose length exceeds four times its thickness. If the primary function of a wall is to support vertical loads, it can be defined as a load bearing wall. The loads should act at one-third the depth of the bearing area from the loaded face as shown in Fig.12 below. Types of Load Bearing Walls Generally there are two types of load bearing walls which are: (a) Concrete walls (b) Brick masonry walls
  • 106.
    Reinforced Concrete LoadBearing Walls BS 8110 recommends that it should contain at least 0.004bh of vertical reinforcement and 0.0025bh of horizontal reinforcement. If tension occurs across the section of the wall, a layer of steel must be provided near each face of the wall. Types of Reinforced Concrete Load Bearing Walls Reinforced concrete load bearing walls may be short (stocky) or slender and braced or unbraced. BS 8110 Recommendation: (i) Short / Stocky Wall: A wall where the effective height divided by the thickness (le / h) does not exceed 15 (braced) or 10 (unbraced) (ii) Slender Wall : When the above limits exceed (iii) Unbraced Walls : A wall providing its own lateral stability is called unbraced wall. (iv) Braced Wall : A wall where the reactions to lateral forces ( i.e. forces perpendicular to the plane of wall) are provided by lateral supports.
  • 107.
    Brick Masonry Walls ClayBricks Clay bricks were used about 10000 years ago which were called ‘adobe’ a Spanish word based on the Arabic word of ‘atob’ meaning sun dried brick. Later on it was discovered that by baking or firing the clay bricks, their strength and durability increased significantly which are used commonly nowadays. Calcium Silicate Bricks Calcium silicate (sand-lime) bricks are made by molding lime mortar into brick shapes. These bricks are manufactured by using steam under high pressure. Concrete Units Modern concrete blocks are manufactured by vibrating a mixture of Portland cement, sand and aggregate in a mold under pressure, curing with pressurized steam.
  • 108.
    Mortar Mortar is theimportant component of brickwork which for load bearing brickwork can be cement: lime: sand mix. The proportion varies on the basis of strength requirement. The thickness of mortar joint recommended by BS 5628 is 10 ~ 12mm. Load Bearing Brickwork (LBB) Method and its Design Consideration Load bearing brickwork construction is most appropriately used in buildings where the floor areas are subdivided into a relatively large number of rooms of small to medium sizes and in which the floor plan is repeated on each storey throughout the height of the building. This system avoids a heavy concentration of vertical loads at foundation level . The LBB method differs from conventional RC method where beams and columns support the loads while in LBB the walls act as structural elements which provide support and stability for the buildings
  • 109.
    Robustness in LoadBearing Brickwork Construction Incorporating all safety factors recommended by the relevant codes of practice, the factor of safety in LBB is much higher than that of RC system. Even though considerable attention is needed to ensure the robustness of the building. Because due to the abnormal loading such as high wind pressure, earthquake force, gas explosion or vehicle impact in the structure it is possible that the structure may suffer severe damage or total collapse. Fig. 14 shows three different layouts of LBB construction Fig. 14 (a)
  • 111.
    Partial Safety Factorm for Accidental Loads Considering the probable effects of misuse or accident, BS 5628-I recommends that the partial safety factors shown in Table 2 ( Table 4 of BS 5628-I reproduced) should be used in the design. Table 2 : Partial Safety m Factors for Material
  • 112.
    Fig 15 showsthe effect of mortar mix proportions on the crushing strength of brickwork built with medium strength bricks.
  • 113.
    Compressive strength ofLBB based on BS 5628-I Table 3 ( Table 1 of BS 5628-I reproduced ) represents different mortar designations with their constituent proportion by volume and mean compressive strength at 28 days. Table 3 : Requirements for mortar (Table 1 of BS 5628-I)
  • 114.
    Table 4 (Table 2(a) of BS 5628-I reproduced ) represents characteristic compressive strength of masonry, fk made from various units using different mortar designation. Table 4 : Characteristic compressive strength of masonry, f k in N/mm2
  • 115.
    REFERENCES: 1. Lin, T.Y. & Stotesbury, S.D., Structural concepts and systems for architects and egineers, Van Nostrand Reinhold Company, USA, 1988 2. Fintel, M., Handbook of concrete engineering, Van Nostrand Reinhold Company, USA, 1986 3. Beedle, L. S., Development in tall buildings 1983, Van Nostrand Reinhold company, USA, 1983 4. Kowalczyk, R.M., Sinn, R. & Kilmister, M.B., Structural system for tall buildings, McGraw-Hill, Inc., USA, 1995 5. Schueller, W., High-rise building structures, John Wiley & Sons, USA, 1977
  • 116.
    (Assignment No. 1) (Q#1)There are four major factors which contributed to the development of high-rise building which are: (a) Development of high-strength materials (b) Development of new design concepts (c) Development of new structural systems (d) Improved construction methods ( Including Top-down construction method) (Explain every one of them briefly) (Q#2) Explain briefly: (a) The advantages of pre-fabricated constructions (b) The disadvantages of pre-fabricated constructions (Q#3) Optional Explain briefly the usage and construction technique of earth retaining diaphragm wall in top-down construction of tall buildings.
  • 117.
    Contributing Factors tothe Development of High - Rise Buildings: (a) Development of high-strength materials - Some 50 years ago 3000 psi (20.7 MPa) concrete were used only. - Presently as high as 20000 psi (138 MPa) concrete is used. - 40 ksi (276 MPa) steel changed to 70 ksi (483 MPa). - Plain steel bars changed to high strength deformed bars. - Plain bars has fy= 250 MPa, High strength deformed bars fy = 460MPa - WWF(welded wire fabric) has high strength and easy to use which has a strength of 485 MPa (Malaysia & UK). (b) Development of New Design Concepts - The development of working stress design, Ultimate Strength Design and Limit State Design. - Considering inelastic behavior of the materials / structural elements which results into economy and efficiency.
  • 118.
    (c) Development ofNew Structural Systems The development of flat plate system  Smooth beamless ceiling,  Reduced storey height, when used permits an additional storey in every 10 storey height. -Introduction of shear walls as bracing elements -Shear wall – Frame systems -Frame tube systems -Tube in tube systems
  • 119.
    (d) Improved ConstructionMethods - Improved construction methods provide better technical solutions to many problems and reduced construction costs - Life-slab system - Other prefabricated concrete systems - Post-tensioned and pre-tensioned concrete construction - Modern construction equipment - New construction methods such as top-down construction technique
  • 124.
    Plan of loadbearing link houses
  • 129.
    Fig.(1)The tallest tube-in-tube714-ft tall, 52-story Shell Oil Building in Houston, Texas