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Determine the optimum block type for use in Saudi Arabia
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Determine the optimum block type for use in Saudi Arabia
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DETERMINE THE OPTIMUM BLOCK TYPE
FOR USE IN SAUDI ARABIA
by
Khaled Mohammed Al-Jammaz
A Thesis Presented to the
FACULTY OF THE SCHOOL OF ARCHITECTURE
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Fulfillment of the
Requirements for the Degree
MASTER OF BUILDING SCIENCE
May 2002
Copyright 2002 Khaled Mohammed Al-Jammaz
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UMI Number: 1411773
Copyright 2002 by
Al-Jammaz, Khaled Mohammed
All rights reserved.
®
UMI
UMI Microform 1411773
Copyright 2003 by ProQuest Information and Learning Company.
All rights reserved. This microform edition is protected against
unauthorized copying under Title 17, United States Code.
ProQuest Information and Learning Company
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P.O. Box 1346
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UNIVERSITY OF SOUTHERN CALIFORNIA
The Graduate School
University Park
LOS ANGELES, CALIFORNIA 90089-1695
This thesis, w ritten b y
U nder th e direction o f hls>.. Thesis
Com m ittee, an d approved b y a ll its members,
has been p resen ted to an d accepted b y The
Graduate School, in p a rtia l fulfillm ent o f
requirem ents fo r th e degree o f
< ^ > C- \ . J&-SA.C.
Dean o f Graduate Studies
D a te Aoc.t ij x»
THESIS COMMITTEE
s V L
Chairperson
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Acknowledgements
There are many people I need to thank in completing of my thesis, but there have
been some more involved then others. These are my parents and family that have
been patient throughout my academic career and supported all the paths I have taken
them on. The most important person that has stood by me and without whom would
have been lost during the last two years is Professor Marc Schiler, my thesis
advisor. Thank you truly Professor Schiler for your time and dedication in seeing me
succeed. Additional thanks goes to the thesis committee all of whom have been very
supportive. Professor G. Goetz Schierle, I would like to thank you for your advice
and guidance. Professor Glen Hickman I would like to thank you for your advice
and time. Professor Madhu Thangavelu I would like to thank you for taking the
time to help me. Special thanks goes out to: Dr. Rayed Al-Dekale, Dr. Naser Al-
Heemde, Dr. Ibraheem Al-Mofeez, Dr. Maher Omar, Dr. Abdulrahman Al-
Ibraheem, Dr. Saleh Al-Ajlan, Dr. Saleh Al-Siad, Dr. Mohammad AI-Hmoud,
Eng. Aftab Ahmad, Mr. Hayder Elawad, Arch. Rashed Al-Shaali, Arch. Kang-
Kye, Arch. Sreemathi Iyer, and All Master of Building Science students.
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Table of Contents
Acknowledgements..................................................................................................ii
List of Figures..........................................................................................................v
List of Tables..........................................................................................................vii
List of G raphs.......................................................................................................viii
Abstract.................................................................................................................. ix
1. Statement of Purpose.......................................................................................1
2. Introduction......................................................................................................2
3. Background.......................................................................................................3
4. Hypothesis........................................................................................................6
5. Material Specifications....................................................................................7
5.1. Red Clay Block.......................................................................................................................8
5.1.1. Components of Red Clay Block.............................................................................................8
5.1.2. Manufacturing Process of Red Clay Block............................................................................ 9
5.2. Concrete Block...................................................................................................................... 1 1
5.2.1. Components of Concrete Block............................................................................................. 1 1
5.2.2. Manufacturing Process of Concrete Block.............................................................................12
5.3. Volcanic Block......................................................................................................................13
5.3.1. Components of Volcanic Block.............................................................................................13
5.3.2. Manufacturing Process of Volcanic Block.............................................................................14
5.4. Siporex Block........................................................................................................................ 16
5.4.1. Components of Siporex Block............................................................................................... 16
5.4.2. Manufacturing Process of Siporex Block...............................................................................17
6. Testing of Materials for Thermal Resistance................................................ 18
6.1. Definition............................................................................................................................... 19
6.1.1. Definition of Thermal Conductivity................................................................................. 19
6.1.2. Definition of Thermal Conductance...................................................................................19
6.2. Testing Material for Each Block Type..................................................................................20
62.1. Theory.................................................................................................................................20
6.2.2. Test Preparation.................................................................................................................24
6.2.3. Results.................................................................................................................................26
6.2.4. Analysis...............................................................................................................................27
6.3. Testing Constructed Walls for Each Block Type................................................................28
6.3.1. Theory.................................................................................................................................28
6.3.2. Test Preparation..................................................................................................................32
6.3.3. Results.................................................................................................................................38
6.3.4. Analysis...............................................................................................................................39
iii
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7. Testing of Material for Compression Strength............................................. 40
7.1. Theory....................................................................................................................................41
1 2 . Test Preparation.....................................................................................................................42
7.3. Results...................................................................................................................................43
7.4. Analysis.................................................................................................................................45
8. Seismic Design Calculation.............................................................................46
8.1. Reason and Process of Seismic Design..................................................................................47
8.2. Sample Calculations..............................................................................................................48
8.3. Results...................................................................................................................................72
8.4. Analysis.................................................................................................................................73
9. Combined Analysis......................................................................................... 74
10. Conclusion.......................................................................................................78
11. Bibliography.................................................................................................... 80
12. Appendix..........................................................................................................81
12.1. Appendix A : Thermal Calculation....................................................................................... 81
12.2. Appendix B : Seismic Calculation ....................................................................................... 85
iv
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List of Figures
Figure 1: Ancient city in Riyadh, Saudi Arabia..................................................... 1
Figure 2: Present city of Riyadh, Saudi Arabia...................................................... 1
Figure 3: Central region, Riyadh, Saudi Arabia...................................................... 3
Figure 4: Southern region, Jazan, Saudi Arabia...................................................... 3
Figure 5: Northern region, Joufe, Saudi Arabia......................................................4
Figure 6: Eastern & Western region, Jeddah, Saudi Arabia.................................... 4
Figure 7: Internal Manufacturing City in Jubal, Saudi Arabia................................5
Figure 8: Red Clay Block...................................................................................... 7
Figure 9: Concrete Block....................................................................................... 7
Figure 10: Volcanic Block....................................................................................... 7
Figure 11: Siporex Block......................................................................................... 7
Figure 12: Large mixture machine...........................................................................9
Figure 13: Continuous block on the Conveyer belt.................................................. 9
Figure 14: Machine generates block size.................................................................. 9
Figure 15: Red clay laid to automated cart............................................................... 9
Figure 16: Evaporation room...................................................................................10
Figure 17: Kiln room.............................................................................................. 10
Figure 18: Recycling machine.................................................................................10
Figure 19: Packaging machine.................................................................................10
Figure 20: Mixing machine....................................................................................12
Figure 21: Forming machine...................................................................................12
Figure 22: Automatic cart........................................................................................12
Figure 23: Evaporating room...................................................................................12
Figure 24: Packaging area.......................................................................................12
Figure 25: Natural aggregate arrival site................................................................. 14
Figure 26: Mixture machine....................................................................................14
Figure 27: Forming machine...................................................................................15
Figure 28: Automated cart.......................................................................................15
Figure 29: Evaporating room...................................................................................15
Figure 30: Packaging area.......................................................................................15
Figure 31: Competed blocks on their way to construction site................................ 15
Figure 32: Frames where mixture is poured in to frames.........................................17
Figure 33: Size segmenting machine....................................................................... 17
Figure 34: Packaging area.......................................................................................17
Figure 35: Thermal Conductivity and Conductance.................................................19
Figure 36: Hot disk sensor...................................................................................... 20
Figure 37: the square curve indicates the temperature increase of the sensor itself and
the circular one shows how the temperature of the sample surface is increasing.. ..22
Figure 38: Hot disk sensor sandwiched between two slides of material...................25
Figure 39: Hot disk sensor sandwiched between two slides that have been glued 25
Figure 40: Masonry block wall................................................................................29
v
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Figure 41: Guarded hot plate................................................................................... 30
Figure 42: Dimensions of the wall sample.............................................................. 33
Figure 43: Measuring weight for samples............................................................... 33
Figure 44: Measuring sample dimensions............................................................... 34
Figure 45: Attaching thermocouple wires to sample................................................34
Figure 46: Taping hollow areas with tape............................................................... 35
Figure 47: Enclosing sample with a blanket............................................................ 35
Figure 48: Placing sample into guarded hot plate....................................................36
Figure 49: Placing steel cover over the guarded hot plate........................................36
Figure 50: Placing mesh material between sample and steel cover..........................37
Figure 51: Connecting thermocouple to the machine...............................................37
Figure 52: Grouting of blocks..................................................................................42
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)..................................... 48
Figure 54: Minimum shear wall length in each direction........................................ 49
Figure 55: Minimum shear wall length in each direction.........................................51
Figure 56: Minimum shear wall length in each direction.........................................53
Figure 57: Minimum shear wall length in each direction.........................................55
Figure 58: Minimum shear wall length in each direction.........................................57
Figure 59: Minimum shear wall length in each direction.........................................59
Figure 60: Minimum shear wall length in each direction.........................................61
Figure 61: Minimum shear wall length in each direction.........................................63
Figure 62: Minimum shear wall length in each direction.........................................65
Figure 63: Minimum shear wall length in each direction.........................................67
Figure 64: Minimum shear wall length in each direction.........................................69
Figure 65: Minimum shear wall length in each direction.............................. 71
VI
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List o f Tables
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the
Reinforcing Steel Designed to Carry the Entire Shear Load....................................48
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed
to Carry the Entire Shear Load................................................................................52
Table 3: Calculations of Minimum Shear Wall Length in Each Direction..............72
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List of Graphs
Graph 1: Red clay mixture...................................................................................... 8
Graph 2: Concrete mixture......................................................................................11
Graph 3: Volcanic mixture......................................................................................13
Graph 4: Siporex mixture....................................................................................... 16
Graph 5: Thermal Conductivity for block material................................................. 26
Graph 6: Thermal conductivity for each block wall sample.....................................38
Graph 7: Compression strength without grouting................................................... 43
Graph 8: Compression strength with grouting.........................................................44
Graph 9: Number of Factories and price per block in US Dollar.............................76
Graph 10: Over all price per block in US Dollar.................................................... 77
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Abstract
This thesis is to find the optimum block type for use in residential building within
Saudi Arabia. Four individual block types chosen for testing are red clay block,
concrete block, volcanic block, and Siporex block. Each of the block types chosen
for testing is based on current availability within Saudi Arabia. The optimum block
type was determined by testing for thermal resistance, structure, availability and cost.
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1. Statement of Purpose
The purpose of this study is to gain better knowledge of building materials used in
Saudi Arabia’s residential building. Materials are chosen based on what is available
currently in the country of Saudi Arabia. The four materials chosen are tested for
thermal resistance, compression strength and resistance to seismic force. Materials
used in the past were suitable for smaller residential construction (See Figure 1).
Since the country has advanced in the last three decades, building structures have
also evolved (See Figure 2). The evolution of residential construction is continuous.
The building materials chosen will be tested for use in larger residential and light
commercial structures. Their appropriateness for this form of structural building will
rest on their ability to respond to or more than just withstand extreme weather.
Figure 2: Present city of Riyadh, Saudi Arabia
(MffiR, 1999)
Figure 1: Ancient city in Riyadh, Saudi Arabia
(MIBR, 1999)
1
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2. Introduction
“Buildings in Saudi Arabia consume more than 65% of the total electric energy
generated in the country compared to 22% in the United Kingdom, 21% in the USA,
and 21% in Australia. Air conditioning equipment is the main consumer with the
yearly average of approximately 45% — 70% out of the 65% of the total electric
energy generated in the country” (MIE, 1997).
In Saudi Arabia, residential buildings utilize most of the country’s electricity due to
poor use of building material. Engineers and architects are looking for an innovative
way to reduce the electric use in residential buildings. Since residential building is
constructed from block types it is essential to determine which block type would
have the optimum characteristics to reduce electrical use, while having the optimum
properties of structure. There are many types of blocks available for use in
residential building, with each block type having unique characteristics.
It is important to identify which block type will have the optimum combination of
characteristics. Saudi Arabia is populated only in several disparate areas, with large
areas of desolate land. The natural environment is extremely hostile, both physically
and environmentally. The land lacks natural vegetation, has thermal extremes, and
the soil is extremely corrosive in particular areas. All this added to poor use of
building materials only increases use of electricity for residential applications.
2
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3. Background
In the past, materials used in the different regions of Saudi Arabia have consisted of
three basic types. Primarily, construction has consisted of stone, unfired mud-brick,
or a combination of the two materials. In the central region of Saudi Arabia,
residents reside in unfired mud-brick houses using stone foundations. This is
because the region is very hot and dry. The region is defined as desert. (See Figure 3)
Figure 3: Central region, Riyadh, Saudi Arabia (MIBR, 1999)
The southern region residents reside in houses constructed of mud or stone. This is a
mountainous region. The climate consists of heavy rain and colder weather. (See
Figure 4)
Figure 4: Southern region, Jazan, Saudi Arabia (MIBR, 1999)
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The northern region climate is very cold and rainy, and the residential construction is
also mud or stone. (See Figure 5)
Figure5: Northern region, Joufe, Saudi Arabia (MIBR, 1999)
In the eastern and western regions, housing construction is a combination of coral
aggregate block and wood. (See Figure 6) This combination is used because the
region is highly humid while the soil is high in salt content, due to the seas that
surround the country’s eastern and western borders.
Figure6: Eastern & Western region, Jeddah, Saudi Arabia (MIBR, 1999)
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These different forms of construction have evolved over the years in Saudi Arabia
based on the building materials available in the country. Over the last decades
residential building has increased in the different regions since oil discovery in the
late 1960’s. The increased wealth from the new commodity advanced the country
during King Abdulaziz reign. King Abdulaziz during this time wanted to advance
the country through two means. First, was the education of the population and
second, was the construction of internal manufacturing facilities (See Figure 7).
Subsidies, land and interest free loans were given to manufacturing facilities of all
different types. The country’s manufacturing industry was advancing and
flourishing at a high rate during the 1970’s. Saudi Arabia was advancing and the
manufacturing industry was very successful during this time but there was one
problem that neither government nor individual manufacturers had foreseen. There
is no standard specification that is being used by the different manufacturers of the
same product. A particular manufacturer of one product will rely on the British
standards while another may use American or Italian standards (MIBR, 1999).
Figure 7: Internal Manufacturing City in Jubal, Saudi Arabia (MIBR, 1999)
5
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4. Hypothesis
Determination of the block types to be used in the Kingdom of Saudi Arabia will
require examination of several testing criteria due to environmental conditions
prevalent in the country. The environmental conditions to be considered include
extreme temperature, physically remote area of habitation, and scarcity of many
common building material resources.
It is hypothesized that the optimum building block for use in the Kingdom
of Saudi Arabia’ s unique environmental conditions and structural necessities is
red clay block.
Red clay is a natural material found in abundance within the country, it is easy to
excavate, and has low thermal conductivity. Block types to be tested include red
clay block, volcanic block, concrete block and siporex block. Each block type will
be tested for thermal resistance, compression strength, seismic force, availability,
and relative cost.
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5. Material Specifications
There are four m ain building blocks that will be tested. All four blocks are currently
manufactured and used in Saudi Arabia. The four blocks are red clay (See Figure 8),
concrete (See Figure 9), volcanic blocks (See Figure 10), and siporex (See Figure
11).
Figure 8: Red Clay Block Figure 9: Concrete Block
Figure 10: Volcanic Block Figure 11: Siporex Block
Each of the blocks has common elements but differs greatly in the main elements
used in manufacturing.
7
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5.1. Red Clay Block
Red clay block is a natural material that is easy to excavate and is found in
abundance within the Kingdom of Saudi Arabia. The most famous factory for red
clay block production is the Al-Yamamah factory in Truma, Saudi Arabia. The Al-
Yamamah factory produces an array of sizes of red clay block. The sample size
chosen for testing is the standard size which is: 20 cm x 20 cm x 40 cm, weighing at
16.690 Kilograms (Al-Yamamah factory, 1999).
5.1.1. Components of Red Clay Block
The red clay block is very simple in composition comprising of clay and sand. Clay
comprises eighty percent of the total mixture, with sand consisting of the remaining
twenty percent (See graph 1).
RED BLOCK MIXTURE
Sand
20%
Clay
80%
Graph I: Red clay mixture
8
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5.1.2. Manufacturing Process of Red Clay Block
The red clay is excavated from the ground from north Truma, Saudi Arabia and
transported as a natural component to Al-Yamamah factory. At arrival the natural
component is mixed with sand in a revolving mixing machine (See Figure 12) based
on the above graph. The mixture is released from the mixture machine to a second
machine that will extrude the mixture into a continuous block on the conveyer belt
(See Figure 13), sending the block to a third machine that will generate the desired
block size (See Figure 14). Once the desired block is generated, the blocks are laid
onto an automated cart (See Figure 15) that takes the blocks to the evaporation room
(See Figure 16). After evaporation the blocks are taken to the kiln for firing (See
Figure 17).
Figure 13: Continuous block on the Conveyer
belt
Figure 12: Large mixing machine
Figure 14: Machine generates block size Figure 15: Red clay laid to automated cart
9
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Figure 16: Evaporation room Figure 17: Kiln room
At the end of the firing process, the blocks are removed from the kiln and inspected.
Approximately ten to twenty percent of each batch will not pass inspection and will
be recycled (See Figure 18). The blocks that remain are packaged into desired
quantities (See Figure 19) and placed on trucks that will deliver the finished red clay
blocks to construction sites.
Figure 19: Packaging machine Figure 18: Recycling machine
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5.2. Concrete Block
Concrete block consists of many components that are chemically produced. There
are a hundred and fifty one factories that produce concrete blocks within the
Kingdom of Saudi Arabia. The most famous factory for concrete block
manufacturing is the Al-Modafer factory in Al-Kaije, Saudi Arabia. Al-Modafer
factory produces an array of sizes of concrete block. The sample size chosen for
testing is the standard size which is: 20 cm x 20 cm x 40 cm, weighing at 22.070
Kilograms (Al-Modafer factory, 2001).
5.2.1. Components of Concrete Block
The concrete block is comprised of cement, rough sand, sand, water, and chemicals.
The majority of the mixture is the cement component which itself is a chemical
mixture. Cement comprises fifty percent of the total mixture, rough sand consisting
eighteen percent, sand eighteen percent, water ten percent and chemical mixture only
four percent (See graph 2).
CONCRETE MIXTURE
Chemical
4% j
I
Water j
Cement
50%
Rough Sand
18%
Graph 2: Concrete mixture
1 1
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5.2.2. Manufacturing Process of Concrete Block
Cement, rough sand and sand are initially mixed together in the mixture machine
(See Figure 20), sent on a conveyer belt as a continuous line of material to the the
forming machine (See Figure 21). Within the forming machine the initial mixture
drops and is mixed with water and chemical mixture. Once the process of formation
is complete the final mixture is released onto automated wood panels that once full
are placed on automated carts (See Figure 22). Once the automated cart is full it
moves along the factory to the evaporation room (See Figure 23). At the end of the
evaporation process, the blocks are removed from the room and packaged into
desired quantities (See Figure 24). The blocks are then stored or placed on trucks
that will deliver the finished concrete blocks to construction sites.
Figure 20: Mixture machine
Figure 21: Forming machine
Figure 24: Packaging area
Figure 22: Automatic cart Figure 23: Evaporating room
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5.3. Volcanic Block
Volcanic ash is a natural material that is easy to excavate and is found in abundance
within the Kingdom of Saudi Arabia. The most famous factory for volcanic block
production is the Al-Borkane factory in Jeddah, Saudi Arabia. The Al-Borkane
factory produces an array of sizes of volcanic block. The sample size chosen for
testing is the standard size which is: 20 cm x 20 cm x 40 cm, weighing at 13.970
Kilograms (Al-Borkane factory, 2000).
5.3.1. Components of Volcanic Block
The volcanic block is composed of large volcanic ash, small volcanic ash, cement,
and water. The majority of the mixture is the volcanic ash component. Volcanic
blocks are comprised of a total of sixty five percent of volcanic ash (large and small
granules). The remaining mixture consists of twenty percent cement and fifteen
percent water (See graph 3).
VOLCANIC MIXTURE
Cement
20%
Graph 3: Volcanic mixture
13
Water
15%
Large
Volcanic ash
35%
Samll
Volcanic ash
30%
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5.3.2. Manufacturing Process of Volcanic Block
The volcanic ash is excavated from the southern region of Al-Medina, Saudi Arabia
and transported as a natural component to Al-Borkane factory. At arrival the ash is
separated into granule size (large and small) (See Figure 25) at which the natural
component is mixed with cement based on the percentages in the above graph. The
initial mixture is released onto a conveyer belt as a continuous line of material to the
(See Figure 26) second machine (the forming machine) (See Figure 27). Within the
forming machine the initial mixture drops and is mixed with water. Once the process
of formation is complete the final mixture is released onto automated fiberglass
panels that once full are placed on automated carts (See Figure 28). Once the
automated cart is full it moves along the factory to the evaporation room (See Figure
29). At the end of the evaporation process, the blocks are removed from the room
and packaged into desired quantities (See Figure 30). The blocks are then stored or
placed on trucks that will deliver the finished volcanic blocks to construction sites
(See Figure 31).
Figure 25: Natural aggregate arrival site Figure 26: Mixture machine
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Figure 28: Automated cart Figure 27: Forming machine
Figure 30: Packaging area Figure 29: Evaporating room
Figure 31: Competed blocks on there way to
construction site
15
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5.4. Siporex Block
Siporex block consists of many components that are primarily chemical based. The
manufacturing process is very complex and there is only one manufacturing factory
within the Kingdom of Saudi Arabia. The siporex factory located in Al-Kajje, Saudi
Arabia is the sole manufacture of the siporex block. The block is also known as
Hepel, Ytong or Clecon block and is produced in one size which is: 20 cm x 25 cm x
40 cm, weighing at 13.860 Kilograms (Siporex factory, 2001).
5.4.1 Components of Siporex Block
The siproex block consists of aluminum powder, sand, cement, water and chemical
mixture. The majority of the mixture is the alumna powder comprising thirty-six
percent of the total mixture, sand comprising thirty-five percent, cement comprising
nineteen percent, with water and chemical mixture comprising of the last ten percent
(See graph 4).
SIPOREX MIXTURE
Lye, Gypsum,
Magnesite,
Sol.Oil, Lime,
W ater Cement
1 0% 19%
35%
Graph 4: Siporex mixture
16
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5.4.2. Manufacturing Process of Siporex Block
The production process of the siporex block is simple yet complex due to the
complexity of the reaction process that takes place to produce the block. Initially the
components are mixed together based on the percentages in the above graph. The
mixture is then placed onto a truck that will pour the mixture into frames ( See
Figure 32) . The frames are filled approximately halfway, and left for fifteen to
twenty minutes. During this time a chemical reaction takes place, the mixture begins
to expand. Once the reaction process ends, the fiames with the expanded blocks are
removed and taken to a machine that will cut the large block into individual blocks
based on the standard size previously discussed (See Figure 33). The blocks are then
taken to a furnace room for baking. At the end of the evaporating process, the blocks
are packaged into desired quantities (See Figure 34). The blocks are then stored or
placed on trucks that will deliver the finished siporex blocks to construction sites.
Figure 32: Frames where mixture is poured in
to fiames m i ■ «
Figure 33: Size segmenting machine
Figure 34: Packaging area
17
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6. Testing of Materials for Thermal Resistance
With a variety of block types manufactured, engineers and architects began using
additional materials to insulate building structures. The use of insulation materials
increased the cost of building structures. If a particular block type was found for use
without the additional insulation, it would be both cost and time effective. Testing
the four blocks chosen will yield the best block for use in terms of thermal
conductivity.
In testing for thermal conductivity there were two testing methods used. The first
testing method tests the material for each block type. The second testing method
consists of constructing a wall for each block type, then the entire assembly is tested
for thermal conductance.
18
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6.1. Definition
6.1.1. Definition of Thermal Conductivity
Heat flows through every solid at characteristic rate which is known as conductivity.
Conductivity within the System International (SI) is the rate at which one Watt
temperature drop through the material is IK or 1C°, under steady heat flow
conditions. Thus, SI units for conductivity are W/m K or W/m C (Benjamin, 1999).
6.1.2. Definition of Thermal Conductance
Common building materials (i.e. brick) are usually available in standard thicknesses.
For such building materials it is beneficial to know the rate at which heat flows for
the standard thickness rather then per inch rate. Conductance within the System
International (SI) is designated as C, stands for Watt per meter squared (m2 ) of a
given thickness of material when temperature difference is IK. Thus, SI units for
conductance are W/m2 K (Benjamin, 1999).
moves through one square meter (m2 ) of material, which is one meter thick; with a
(4 ,n. m this
' example)
/Conductivity jg = 12 Stahl
Resistance R = 4 = 0.083
Figure 35: Thermal Conductivity and Conductance in English units (Benjamin, 1999)
19
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
6.2. Testing Material for Each Block Type
Testing of material for each block type took place at King Abdulaziz City of Science
and Technology (KACST). KACST had the equipment needed to conduct the
thermal conductivity test of material for each block type. The Hot Disk Thermal
Constants Analyzer, which has the capability of testing the material for each block
type was used.
6.2.1. Theory
Based on the theory of the Transient Plane Source technique, the Hot Disk Thermal
Constants Analyzer uses a sensor element in the shape of a double spiral( See Figure
36). This Hot Disk sensor acts both as a heat source for increasing the temperature
of the sample and a “resistance thermometer” for recording the time dependent
temperature increase. This spiral is supported by a material to protect its shape, give
it mechanical strength and keep it electrically insulated.
Figure 36: Hot disk sensor
20
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The spiral sensor is sandwiched between two halves of the sample (solid samples).
During a pre-set time, 200 resistance recordings are taken and from these the relation
between temperature and time is established. A few parameters, like the “Output of
Power” to increase the temperature of the spiral, the “Measuring Time” for recording
200 points and the size of the sensor are used to optimize the settings for the
experiment so that thermal conductivities from 0.01 W/mK to 400 W/mK can be
measured.
To theoretically describe how the Hot Disk behaves the thermal conductivity
equation has been solved assuming that the Hot Disk consists of a certain number of
concentric ring heat sources located in an infinitely large sample.
If the Hot Disk is electrically heated, the resistance increase as a function of time can
be given as:
R(t) = i?o{l + a • [AT/ + AIL(r)]}
R q is the resistance of the disk just before it is being heated or at time t = 0, < X is the
Temperature Coefficient of the Resistivity (TCR), AT) is the constant temperature
difference that develops almost momentarily over the thin insulating layers which are
covering the two sides of the Hot Disk sensor material (Nickel) and which make the
Hot Disk a convenient sensor. k T a ve (r) is the temperature increase of the sample
surface on the other side of the insulating layer and facing the Hot Disk sensor
(double spiral).
21
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From equation (1) we get the temperature increase recorded by the sensor:
A TaveT + A T i — — •
a
m
Ro
- 1
Here A7} is a measure of the “thermal contact” between the sensor and the sample
surface with A 7} = 0 representing perfect “thermal contact” closely realized by a
deposited (PVD or CVD) thin film or an electrically insulating sample.
Sensor
T
Sample t
Figure 37: the square curve indicates the temperature increase of the sensor itself and the circular
shows how the temperature of the sample surface is increasing
22
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
AT; becomes constant after a very short time A t; which can be estimated as:
M i = —
K i
Where S is the thickness of the insulating layer and Ki is the thermal diffusivity of
the layer material.
The time dependent temperature increase is given by the theory as:
M a v e ( T ) = P ° ■D(V)
7ry2-a-A
where Po is the total output of power from the sensor, a is the overall radius of the
disk, A is the thermal conductivity of the sample that is being tested and D(t) is a
dimensionless time dependent function with:
In this equation t is the time measured from the start of the transient recording, 0 is
the characteristic time defined as:
a2
©
K
where K is the thermal diffusivity of the sample.
2 3
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Now by making a computational plot of the recorded temperature increase versus
D(r), we get a straight line, the intercept of which is A 2} and the slope is
Since K and by that 0 are not known before the experiment the final straight line
from which the thermal conductivity is calculated, is obtained through a process of
iteration. In this way it is possible to determine both the thermal conductivity and
the thermal diffusivity from one single transient recording (KACTS, 2000).
6.2.2. Test Preparation
Samples were prepared individually using the following description. A sample block
was taken, using an automatic saw to slice off the surface areas of the samples,
leaving the center of each sample exposed. The new sample was then cut into two
equal dimension slides using the automatic saw. At this point, the Hot Disk sensor
was placed in between the two slides (See Figure 38).
The first tests were taken in the summer of 2001. There was fear of incorrect
configurations due to air gaps that remained in between the two slides, so a second
test was conducted. The second test took place in the winter of 2002, this time
gluing around the edges of the two slides after the sensor was placed in between (See
Figure 39). The second tests were done to assure the accuracy of the test and to
eliminate any inaccuracies due to air gaps between the sample slides. Testing was
done using the Hot Disk Thermal Constants Analyzer.
nYi-a-K
using experimental times much longer than AU .
2 4
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f i ?
\ ' . z .£ :2 J U * £ L
% -''3 'j> * r';,S 7
3 S X '.« ^
Figure 38: Hot disk sensor sandwiched between two slides of material
Figure 39: Hot disk sensor sandwiched between two slides that have been glued
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
6.2.3. Results
Upon completion of the testing done in the winter o f2002, the following results were
found for material of each block type (See graph 5). In the following graph, the x-
axis is diagramed with each of the materials, and the y-axis diagrams the thermal
conductivity (W/m.K) increasing in 0.2 intervals.
Thermal Conductivity for Block Material W/m.K
1.4
1.2
1
Thermal q.8
Conductivity
W/m.K 06
0.4
0.2 -
V ;
.
-
|
XXH
i* * * * ’ ..« '-Kt
**-
T * ^
s v -
u-
Red Block
Concrete
Block
Volcanic
Block
Siporex
Block
□ Thermal
Conductivity W/m.K
0.605 1.347 0.4201 0.1788
Graph 5: Thermal Conductivity for block material
Results of the thermal conductivity testing for material of each block type were
significantly varied from one another. The red clay block material indicated 0.6050
W/m.K. The concrete block material indicated 1.3470 W/m.K. The volcanic block
material indicated 0.4201 W/m.K. The siporex block material indicated 0.1788
W/m.K.
2 6
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6.2.4. Analysis
The thermal conductivity graph shows the output of the testing results, in regards to
thermal conductivity of material the concrete block shows the highest result “worst”.
Next is the red clay block material, which shows forty-four percent of the concrete
block material, though not the best result in comparison to block materials that
shows lower thermal conductivity. The second lowest result for thermal
conductivity is the volcanic block material, having lower results than the red clay
and concrete block material. Though, in comparison to the three block types
mentioned, the Siporex block material indicated the lowest thermal conductivity.
Therefore the Siporex block material would be the best choice for thermal
conductivity.
2 7
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6.3. Testing Constructed Walls for Each Block Type
The testing of constructed walls for each block type took place at King Fahd
University of Petroleum and Minerals (KFUPM). KFUPM had the equipment
needed to conduct the thermal conductivity test of the wall samples. The Guarded
Hot Plate was used to test the sample walls for thermal conductivity. The Guarded
Hot Plate conforms to American Society for Testing and Materials (ASTM) C 177-
85.
6.3.1. Theory
Thermal conductivity was measured under steady-state conditions using a guarded
hot plate. The test samples, 61 cm x 61 cm x 20 cm, were prepared for each of the
four types of masonry block (See Figure 40). In testing the samples with the guarded
hot plate, exposure to different temperatures from two opposite sides will result a
combination of conduction, convection, and radiation within the sample wall.
The guarded hot plate thermal conductance measuring system requires two samples
to be installed into the machine for measurement of thermal conductivity. The
samples are placed horizontally onto the guarded hot plate. The horizontal flat plate
heater arrangement is a 30.5 cm square center main heater with separately controlled
61 cm guarded heater (See figure 41). The guarded heater is used to eliminate lateral
heat flow to or from the main heater, allowing all heat produced from the main
heater to flow through the two test samples.
28
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The final temperature of the guarded hot plate depends on the power input of the
heater, the thermal conductance of the two test samples and the temperature of the
opposing heat sinks (See figure 41). The auxiliary heaters heat the incoming coolant
as it enters the lower and upper heat sinks. The heat sinks are used to produce a
constant and consistent temperature on the two far sides of the samples.
In the present testing, two identical test samples were not used. Instead, one of the
four prepared test wall samples was used to obtain the thermal conductivity of the
test samples. The second sample used was a dummy sample of known thermal
conductivity (calibrated), k\ (ASTM C177,2001).
Figure 40: Masonry block wall
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Coolant
Upper Heat Sink
Rubber Sheet
Dummy or
Test Sample SSi
Guard
Heater
Guard
Heater
Main Heater
Thermocouple
Test Sample
Soft Insulation*
Lower Heat Sink
Coolant
Figure 41: Guarded hot plate
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Once the testing sample and dummy sample were properly installed, the machine
was switched on. The sample temperatures were monitored until steady-state
conditions were obtained. The average temperature of the sample under steady-state
conditions were maintained at about 35± 2°C. The following measurements were
taken:
The power input at steady state condition, Q=N x V x I, where V=Main heater
voltage, l=Main heater current, and N=Correction factor of the equipment
The temperature difference across the dummy sample 1, ATi
The temperature difference across the test sample 2, AT 2
The thickness of the dummy sample 1, di
The thickness of the test sample 2, d2
Main heater surface area, A
The heat flow through the dummy sample was calculated by:
The heat flow, Q2, through the test sample was calculated by:
The thermal conductivity, k2, for test sample was calculated by:
31
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6.3.2. Test Preparation
Sample walls were prepared individually using the following description. Sample
blocks were used to build sample walls of 61 cm x 61 cm x 20 cm. Using mortar, the
individual blocks were connected into the specified dimensions of the wall (See
Figure 42). Once the wall was constructed, it was left for ten days to completely dry
and set. The walls were then shipped to the laboratory for testing.
The process of testing was first to measure the weight of the samples (See Figure 43)
and measure the external dimensions (See Figure 44). Because of the rough surface
of the sample, the thermocouple wires were fixed on both sides of the sample (See
Figure 45) and the surface temperatures were monitored by the automatic data
acquisition system. Top and lower surfaces of the samples were then covered with
tape to close off the hollow areas of the wall (See Figure 46). Then the samples were
covered with a blanket on both sides to have smooth contact with the plate surface
(See Figure 47).
The complete sample for testing was placed within the guarded hot plate (See Figure
48). Then a steel four sided covering was placed around the four sides of the
guarded hot plate, only exposing the top and bottom surface (See Figure 49). Once
the steel sided covering was placed around the guarded hot plate, gaps remained
around the four sides. At this time, mesh material was used to eliminate the gaps
between the guarded hot plate and steal covering (See Figure 50). The final step
required connection of the thermal cables to the machine (See Figure 51) and
running the test.
32
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61 cm
Cement
Mortar joint
61 cm
20 Cm
Figure 42: Dimensions of the wall sample
Figure 43: Measuring weight for samples
3 3
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Figure 44: Measuring sample dimensions
Figure 45: Attaching thermocouple wires to sample
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Figure 46: Taping hollow areas with tape
Figure 47: Enclosing sample with a blanket
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Figure 48: Placing sample into guarded hot plate
Figure 49: Placing steel cover over the guarded hot plate
3 6
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Figure 50: Placing mesh material between sample and steel cover
Figure 51: Connecting thermocouple to the machine
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
6.3.3. Results
Upon completion of the testing, the following results were found for the wall
samples of each block type (See graph 6). In the following graph, the x-axis is
diagramed with each of the block types, and the y-axis diagrams the thermal
conductivity (W/m.K) increasing in 0.2 intervals.
Thermal Conductivity for Each Block Type W/m.K
1.2-i/
1 -
0 .8 -
./I
Thermal
Conductivity 0.6
W/m.K
0.4
0.2 -
0 3 8 2a
A M
1
! ----
» . v
n
i ____ i
1 ' t * V
Kl ,v
V *.*/.
n
“ £
> r w X
u
Red Block
Concrete
Block
Volcanic
Block
Siporex
Block
□ Thermal
Conductivity W/m.K
0.382 1.025 0.411 0.325
Graph 6: Thermal conductivity for each block wall sample
Results of the thermal conductivity testing for the wall sample of each block type
were significantly varied from one another. The red clay block material indicates
0.382 W/m.K. The concrete block material indicates 1.025 W/m.K. The volcanic
block material indicates 0.411 W/m.K. The siporex block material indicates 0.325
W/m.K.
3 8
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63. 4. Analysis
Thermal conductivity for wall block type samples is shown in the graph with output
of the testing results, in regards to thermal conductivity. The concrete wall block
indicates the highest result “worst”. There is a significant drop to the next three
materials, whose values are closely grouped. The siporex is the best, followed
closely by the red clay block and then the volcanic block wall.
39
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7. Testing Material for Compression Strength
The variety of manufactured building block have different levels of compression
characteristics depending on block material components. Engineers and architects
traditionally specify building blocks that were found best suited for compression
only, for use as building materials, without regard for the long-term effectiveness of
thermal resistance. This practice, in the long-run effects the longevity of building
structures as well as energy usage and costs.
The testing of each block, including grouting and not grouting will provide a better
understanding of long-term strength and durability for each block type. The testing
for compression strength took place at King Saud University (KSU). KSU had the
equipment required to conduct the compression strength testing for each block type.
The Compressive Axial Load equipment will enable the testing of each block type
for compression strength, conforming to American Society for Testing and Materials
(ASTM) C 39.
4 0
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7.1. Theory
Compressive strength of material can be measured using several testing protocols.
Commonly, the compressive axial load test used to test concrete is used to test the
compressive strength of different block types. The Compressive Axial Load test can
be easily constructed and tests for endurance, pressure and longevity of each block
type.
The compressive strength method consists of applying a compressive axial load
either on a cylindrical concrete specimen or on a concrete block as specified by
ASTM C 39, at a continuous rate without shock and until a prescribed range of
failure occurs. The compressive strength is calculated by dividing the maximum load
attained during the test by the net cross-sectional area of the specimen (ASTM C39,
2001).
4 1
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7.2. Test Preparation
The first tests were taken in the summer of 2001 where each block type was tested
individually without filling the hollow gaps that are produced in manufacturing.
Since test results change if the hollow gaps are filled with grout, a second test was
required. The second test took place in the winter of 2002; this time filling the
hollow gaps with concrete grout (See Figure 52). The second tests were done to
observe the difference in the testing results considering gross areas, including
grouting. On both occasions, the Compressive Axial Load testing method was
used.
Figure52: Grouting of blocks
4 2
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7.3. Results
Testing without filling the hollow gaps of each block type yielded the following
results (See graph 7). The x-axis on the graph contains the different block types,
while the y-axis is the compressive strength (psi) increasing in 500 psi intervals. In
the Results of the initial test, the red clay block had the highest compression strength,
which reached 1534 psi. While the concrete block reached 720 psi, the volcanic
block reached 686 psi, and the Siporex block with the lowest compression strength,
reached 471 psi.
_ 3000
'3 5
3 2500
s z
c 2000
0 )
k -
* 1500
O
>
$ 1000
O
w
| 500
o
o
0
Compression strength without grouting
1534
v
/2U
686
471
- ^ . . -■ — --V
- ^
I n *-"
*->
Red Clay Concrete Volcanic Siporex
Block
Block Type
Graph 7: Compression strength without grouting
4 3
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The results of testing the blocks with grouting in the winter of 2002 are recorded in
graph 8. The x-axis shows the block type and the y-axis the strength in intervals of
500 psi. The grouted test results vary greatly from the un-grouted test, the concrete
block reached the highest compression strength at 2548 psi. While the red clay block
reached 2039 psi, coming in at second and the volcanic block reached 1109 psi. The
results changed for the red clay, concrete and volcanic blocks, but the Siporex block
remained the same at 471 psi.
C o m p ressio n stren g th w ith grouting
3000
1 2 5 0 0
g> 2 0 0 0
o
f f i
>
U >
0 )
< D
a
E
o
o
1500
1000
500
0
2548
2039
V
r
1109
471
Red Clay Concrete
Block
Volcanic Siporex
Block Type
Graph 8: Compression strength with grouting
4 4
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7.4. Analysis
Compression strength testing for the block types conducted in the summer of 2001
(un-grouted) and winter 2002 (grouted) resulted in major differences. The un
grouted test indicates that the red clay block has the highest compression strength.
The grouted test shows that the red clay block is in second place. In the second
testing, the concrete block has the highest compression strength. In both tests the
volcanic block came in third place.
The winter 2002 conclusion of testing is that compression strength for three of the
four blocks increased dramatically from the un-grouted test results. The graphs
display the figures of all the block types, outlining the significant change in
compression strength, except for Siporex block that shows the same strength in both
tests. The reason behind the unchanged compression strength for Siporex is that the
block is manufactured as a solid block.
4 5
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8. Seismic Design Calculation
“O you who believe! Take your precautions.. (Qur’an)
The above quote comes from an aye in the Holy Qur’an. Stating that every person
needs to be cautious is based on historical occurrences. Saudi Arabia is not thought
of as a country that experiences seismic activity. Though, centuries ago, Saudi
Arabia experienced numerous seismic activities. In the past few centuries, the
country had not experienced seismic activity, until 1998, when the north-west region
of the country experienced a 4.0 on the Richter scale.
The outcome of the 1998 earthquake was damage and destruction to the region’s
structures. Damage could have been m inim ized if architects and engineers had
designed for seismic activity. There were two problems that resulted in the recorded
damage, first the materials used and second the seismic structural design. Building
block had been tested for compression strength but never for seismic shear.
4 6
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8.1. Reason and Process of Seismic Design
To design for seismic resistance, the result of compression strength for each block
type is required. Seismic calculations provide the minimum length for reinforced
walls in each direction. The minimum length for reinforced shear walls is based on a
building’s ability to resist seismic forces. The following study includes two
examples:
• A one-story building of 1000 square feet floor area
• A two-story building of 1000 square feet floor area per story
Each of the two examples includes two design methods and two rebar sizes:
• Method 1 assume rebars to resist all shear, using # 4 and # 5 rebars
• Method 2 assume masonry to resist all shear, using # 4 and # 5 rebars
The examples and methods in the following calculations used in this section came
from Reinforced Masonry Engineering Handbook by James Ambrose, SE (Amrhein,
2000).
4 7
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8.2. Sample Calculations
8.2.1. One-story building - Red Clay Block
8.2.1.1. Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 7 5/8”
8” block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
b = 7.625”
Red clay block specified com p ressiv e stren g th
Fs = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See Figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llowable sh e a r s tre s s
34 psi (1.33) Fy = 45.2 psi
From Table 1: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerUBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
' »
U/Vd
(psi)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1J0+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2500 50 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 58 53 51 49 47 44 42 40 38
4000+ 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel Designed
to Carry the Entire Shear Load
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Rebar size (#4 bars, < f > V 2 ”)
Av = t t r2 = t t (0.5/2)2 Av = .2 in2
Rebar spacing
s = Av F s / (Fy b)
w here
s = re b a r sp a c in g (in)
Av = reb ar c ro s s sectio n area (in2)
Fs = allow able reb ar s tr e s s (psi)
Fv = allow able m aso n ry sh e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.2 (24000)/[45.2(7.625)] = 14”
U se m ultiple o f 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12”
2 w alls each direction: d = d ’/2 = 17.3/2
Wall length L=d’+4” = 8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”)
The minimum length, for the reinforced wall within each direction, is 9’-4”,
at 8” rebar spacing.
s = 8”
d ’= 17.3’
d = 8.6’
L = 9’-4”
L=9’-4”
L= 9’-4”
/n
L=9’-4”
L=9’-4”
Figure 54: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.I.2. Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 7 5/8”
8” block thickness with mortar
G rade 60 ste e l (Fy = 60,000 psi)
Red clay block S pecified co m p ressiv e stren g th
b = 7.625”
F s = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allow able s h e a r s tr e s s
34 psi (1.33) Fv = 45.2 psi
From Tablel: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerUBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
&
M/Ifd
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 14)+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2500 S O 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 56 53 51 49 47 44 42 40 38
4000+ 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel
Designed to Carry the Entire Shear Load
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Rebar size (# 5bars, < j > 5/8”)
Av = t t r2 = t t (0.625/2)2 Av=.3in2
Rebar spacing
s = Av Fs / (Fv b)
S= reb ar sp acin g (in)
Av = reb ar c ro s s sectio n area (inz)
Fs = allow able re b a r s tr e s s (psi)
F„ = allow able m aso n ry s h e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
Use m ultiple o f 8” m odule s = 16”
R e^uirad effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12” d - 1 7 . 3 ’
2 w alls each direction: d = d ’/2 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”) L = 9’-4”
The minimum length, for the reinforced wall within each direction, is 9’-4”,
at 16” rebar spacing
L=9’_4”
L=9’-4”
L=9
Figure 55: Minimum shear wall length in each direction
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8.2.I.3. Method Two: masonry to resist all shear, using # 4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
R ed clay block Specified com pressive stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allow able s h e a r s tre s s
18 psi (1.33) Fv = 24 p si
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
(P*0
U/Vd
0 0 0.1 O J Z 0 3 04 0 5 0:6 0 7 0.8 04 1.0*
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3500 39 36 36 33 3 1 29 27 24 22 20 18
4000* 40 38 36 33 3 1 29 27 24 22 20 18
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed to Cany the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (#4 bars, « J > %”)
Av = t t r2 = t t (0.5/2)2 Av = .2 in2
Rebar spacing
s = Av Fs / (Fv b)
where
s= reb ar sp a c in g (in)
Av = reb ar c ro s s sectio n a re a (in2)
Fs = allow able re b a r s tre s s (psi)
Fv = allow able m aso n ry s h e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
Required effective wall length
d’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d’ = 393”/12” d ’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 24 (8”) L = 16’-8”
The minimum length, for the reinforced wall within each direction, is 16’-8”,
at 24” rebar spacing
L= 16’-8”
L= 16’-8’
L= 16’-8’
L= 16’-8’
Figure 56: Minimum shear wall length in each direction
5 3
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.I.4. Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
8” wall or block thickness with mortar
G rade 60 ste e l (Fy = 60,000 psi)
Red clay block Specified co m p ressiv e stren g th
b = 7.625”
F s = 24,000 psi
fm = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 psi
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
v m
(P*0
UjVd
O J O 0.1 0.2 0.3 0.4 (U S 0:6 0.7 0.8 0.9 1.0+
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3500 39 38 36 33 31 29 27 24 22 20 18
4000i. 40 38 36 33 31 29 27 24 22 20 18
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonty Designed to Cany the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Av = .3 in2
Rebar size (#5 bars, < ( > 5/8”)
Av = Trr2 = TT (0.625/2)2
Rebar spacing
s = Av F s / (Fv b)
where
s= reb ar sp a c in g (in)
Av = reb ar c ro s s sectio n area (in2 )
Fs = allow able reb ar s tr e s s (psi)
Fv = allow able m aso n ry s h e a r s tre s s (psi)
b = wail th ic k n e ss (in)
s = 0.3 (24000)/[24(7.625)] = 39”
U se m ultiple o f 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393712”
2 w alls e a c h direction: d = d ’/2 = 32.8/2
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 40 (8”)
The m inim um length, for the reinforced wall within each direction, is 16’-8”,
at 40” rebar spacing
L= 16’-8”
s = 40”
d ’= 32.8’
d = 16.4’
L = 16’-8”
L= 16’-8”
L= 16’-8”
Figure 57: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.2. Two-story building Ground Floor - Red Clay Block
8.2.2.I. Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
Red clay block Specified com pressive stren g th
b = 7.625”
Fs = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llowable sh e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
From Tablel: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
r m
U/Vd
ip* )
0.0 0.1 0.2 0.3 0.4 OS 0.6 0.7 0 3 0 3 1J0+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2500 S O 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 56 53 51 49 47 44 42 40 38
400Of 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel
Designed to Carry the Entire Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (#4 bars, $ 1 /2 ”)
Av = t t r2 = t t (0.512)2 Av = .2 in2
Rebar spacing
s = Av Fs / (Fv b)
w here
s= re b a r sp a c in g (in)
Av = re b a r c ro s s sectio n area (in2)
Fs = allow able reb ar s tr e s s (psi)
Fv = allow able m aso n ry sh e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.2 (24000)/[45.2(7.625)] = 14”
U se m ultiple of 8” m odule s = 16”
R equired effective waii length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)] = 331”
d ’ = 331”/12” d ’= 27.5’
2 w alls e a c h direction: d = d’/2 = 27.5/2 d = 13.75’
Wall length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 16 (8”) L = 14’-8”
The minimum length, for the reinforced wall within each direction, is 14’-8”,
at 16” rebar spacing
L= 14’-8”
L= 14’-8”
V !
L= 14’-8”
L= 14’-8”
Figure 58: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.2.2. Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
8” wall or block thickness with mortar
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Red clay block Specified co m p ressiv e stre n g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
h
□ □rr
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allow able s h e a r s tr e s s
34 p si (1.33) Fv = 45.2 psi
From Tablet: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerlJBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
I ' m
U/Vd
CP*/
0:0 0.1 0.2 0.3 0.4 O J S 0.6 0.7 0.8 OS 1J0+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2S00 50 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 56 53 51 49 47 44 42 40 38
4000+ 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel
Designed to Carry the Entire Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Av = .3 in2
Rebar size (# 5bars, < j > 5/8”)
Av = t t r2 = t t (0.625/2)2
R e b a r s p a c in g
s = Av Fs / (Fv b)
w here
s= re b a r sp acin g (in)
Av = re b a r c ro s s se c tio n area (in2)
Fs = allow able re b a r s tr e s s (psi)
Fv = allow able m aso n ry sh e a r s tr e s s (psi)
b = wall th ic k n e ss (in)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple of 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)] = 331”
d ’ = 331”/12”
2 w alls each direction: d = d ’/2 = 27.5/2
Wall length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 24 (8”)
The minimum length, for the reinforced wall within each direction, is 14’-8”,
at 24” rebar spacing
L= I4’-8”
s = 24”
d’= 27.5’
d = 13.75’
L = 14’-8”
L= 14’-8”
L= 14’-8”
Figure 59: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.2.3. Method Two: masonry to resist all shear, using # 4 rebars
A ssum e:
8” (nominal) wall = 75/8”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
Red clay block Specified com pressive stren g th
b = 7.625”
F s = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
/"TPX
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allowable s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerUBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
'»
(psi)
U/Vd
0J > 0.1 0.2 03 0 A 0 5 QA 0.7 0 8 0.9 1.0*
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3500 39 38 36 33 31 29 27 24 22 20 18
4000* 40 38 36 33 3 1 29 27 24 22 20 18
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed to Carry the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (#4 bars, i j > %”)
Av = t t r2 = t t (0.5/2)2 Av = .2 in2
R e b a r sp a c in g
s = Av Fs / (Fv b)
w here
s= reb ar sp a c in g (in)
Av = reb ar c ro s s section area (in2 )
Fs = allow able reb ar s tre s s (psi)
Fv = allow able m asonry sh e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple o f 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623712” d ’= 51.9’
2 w alls each direction: d = 6’ 1 2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 24 (8”) L = 26’-8”
The minimum length, for the reinforced wall within each direction, is 26’-8”,
at 24” rebar spacing
L= 26’-8’
L=26’-8’
L= 26’-8'
L=26’-8’
Figure 60: Minimum shear wall length in each direction
61
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.2.4. Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
Red clay block S pecified co m p ressiv e stre n g th
b = 7.625”
F s = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
T
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 psi
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
(w
(P»0
U/Vd
0.0 0.1 0.2 QJ 0A O JS 0.6 0.7 0.8 0.9 1.0*
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3500 39 38 36 33 31 29 27 24 22 20 18
4000+ 40 38 36 33 31 29 27 24 22 20 18
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed to Carry the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (#5 bars, < j > 5/8”)
Av = t t r2 = t t (0.625/2)2 Av = .3 in2
R e b a r sp a c in g
s = Av Fs / (Fv b)
w here
s= re b a r sp acin g (in)
Av = re b a r c ro s s se c tio n area (in2)
Fs = allow able re b a r s tre s s (psi)
Fv = allow able m aso n ry s h e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.3 (24000)/[24(7.625)] = 39”
U se m ultiple of 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d’ = 623”/12” d ’= 51.9’
2 w alls each direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 32 (8”) L = 26’-8’
The minimum length, for the reinforced wall within each direction, is 26’-8”,
at 32” rebar spacing
L= 26’-8’
L= 26’-8’
L=26’-8’
I
r
L= 26’-8’
Figure 61 Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.3. T w o -sto ry b u ild in g F ir s t F lo o r - R e d C la y B lo c k
8 .2 3 .1 . M e th o d O n e : R e b a rs to re s is t a ll s h e a r, u sin g # 4 r e b a r s
A ssum e:
8” (nominal) wall = 75/8”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
Red clay block Specified com p ressiv e stren g th
b = 7.625”
Fs = 24,000 psi
f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allowable s h e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
From Table 1: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
£
U/Vd
tP*9
0J O 0.1 02 O J 0A O S 0.6 0.7 0£ 0.9 1.0+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2500 50 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 56 53 51 49 47 44 42 40 38
4000+ 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel
Designed to Carry the Entire Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (#4 bars, < | > Vi”)
Av = t t r2 = t t (0.5/2)2 A v = .2 in2
Rebar spacing
s = Ay Fs I (Fy b)
where
s= rebar spacing (in)
Av = rebar cross section area (in2 )
Fs = allowable rebar stress (psi)
Fv = allowable masonry shear stress (psi)
b = wall thickness (in)
s = 0.2 (24000)/[45.2(7.625)] = 14”
Use multiple of 8” module s = 16”
Required effective wall length
d’ = V/(Fv b) = 65715# /[45.2(7.625)] = 190”
d’ = 190”/12” d’= 15.8’
2 walls each direction: d = d’/2 = 7.9/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23’
Round to 8” module: 16 (8”) L = 8’-8”
The minimum length, for the reinforced wall within each direction, is 8’-8”,
at 16” rebar spacing
L= 8’-8”
L= 8’-8”
L= 8’-8”
L= 8’-8”
Figure 62: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.3.2. Method One: Rebars to resist all shear, using #5 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
8” wall or block thickness with mortar
G rade 60 ste e l (Fy = 60,000 psi) Fs = 24,000 psi
Red clay block S pecified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □ □
h
□ car
□ □ □
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
Allow able s h e a r s tr e s s
34 p si (1.33) Fv = 45.2 psi
From Tablel: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerUBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
C
U/Vd
(P*Q
04) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 O J B 0.9 1J0+
1500 39 38 37 36 35 34 33 32 31 30 29
2000 45 44 42 41 40 39 38 37 36 35 34
2500 50 49 48 46 45 44 43 41 40 39 38
3000 55 53 52 51 49 48 47 44 42 40 38
3500 59 58 56 53 51 49 47 44 42 40 38
4000+ 60 58 56 53 51 49 47 44 42 40 38
Table 1: Allowable Shear Wall Stresses (psi) Non-Inspected Masonry with the Reinforcing Steel
Designed to Carry the Entire Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Rebar size (# 5bars, 4 > 5/8”)
Av = t t r2 = t t (0.625/2)2 Av = .3 in2
Rebar spacing
s = Ay F s I (Fv b)
w here
s= re b a r sp a c in g (in)
Av = re b a r c ro s s sectio n a re a (in2)
Fs = allow able reb ar s tr e s s (psi)
Fv = allow able m aso n ry s h e a r stre s s (psi)
b = w all th ic k n e ss (in)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[45.2(7.625)] = 190”
d ’ = 190”/12” d ’= 15.8’
2 w alls each direction: d = d ’/2 = 15.8/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23
R ound to 8” m odule: 24 (8”) L = 8’-8”
The m inim um length, for the reinforced wall within each direction, is 8’-8”,
at 16” rebar spacing
L= 8’-8’
L=8’-8”
I
L= 8’-8’
L= 8’-8’
Figure 63: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.3.3. Method Two: masonry to resist all shear, using # 4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Red clay block Specified co m pressive stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llowable s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (PerUBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
£
( p « J)
y/vd
0.0 0.1 0.2 0.3 0 < 4 O S 0:6 0.7 0.8 0.9 1.0+
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3500 39 38 36 33 31 29 27 24 22 20 18
4000+ 40 38 36 33 31 29 27 24 22 20 18
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed to Carry the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
R ebar size (#4 b ars, $ V i”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R e b a r sp a c in g
s = Av Fs /(F v b)
w here
s= reb ar sp acin g (in)
Av = reb ar c ro s s sectio n area (in2)
Fs = allow able re b a r s tr e s s (psi)
Fv = allow able m aso n ry sh e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple o f 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)J = 359”
d ’ = 359”/12” d ’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 24 (8”) L = 15’-8””
The minimum length, for the reinforced wall within each direction, is 15’-8”,
at 24” rebar spacing
L= 15’-8”
L= 15’-8
L= 15’-8”
L= 15’-8”
Figure 64: Minimum shear wall length in each direction
6 9
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.2.3.4. Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) w all = 75/8”
8” wall or block thickness with mortar
G rade 60 steel (Fy = 60,000 psi)
R ed clay block S pecified co m p ressiv e stren g th
b = 7.625”
F s = 24,000 psi
f m - 2039 psi
C onservative h/d ratio h/d =>1
Conservative h/d ratio is from the following equation M/Vd = Vh/Vd = h/d (See figure 53)
□ □□
□ □□
Figure 53: Masonry shear wall h/d ratios (Schierle, 1997)
A llow able s h e a r s tr e s s
18 psi (1.33) Fv = 24 psi
From Table 2: Allowable shear stress based on compression strength
Allowable shear stress is multiplied by 1.33 (Per UBC section 1603.5 Stress Increases. Allowable
stresses and soil bearing values specified in this code for working stress design may be increased one
third when considering wind or earthquake forces either acting alone or when combined with vertical
loads. No increase will be allowed for vertical loads acting alone.)
(P»0
MfVd
0.0 0.1 0.2 03 0.4 OS 03 0.7 0 £ 03 1.0+
1500 26 25 25 24 23 23 22 21 21 20 18
2000 30 29 28 28 27 26 25 24 22 20 18
2500 33 33 32 31 30 29 27 24 22 20 18
3000 37 36 35 33 31 29 27 24 22 20 18
3S00 39 38 36 33 31 29 27 24 22 20 18
4000+ 40 38 36 33 31 29 27 24 22 20 1S
Table 2: Allowable Shear Wall Stresses (psi) for Non-Inspected Masonry Designed to Carry the Entire
Shear Load
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Av = .3 in2
Rebar size (#5 bars, i f > 5/8”)
Av = t t r2 = t t (0.625/2)2
Rebar spacing
s = Av F s / (Fv b)
where
s= reb ar sp acin g (in)
Av = reb ar c ro s s sectio n area (in2)
Fs = allow able reb ar s tre s s (psi)
Fv = allow able m asonry s h e a r s tre s s (psi)
b = wall th ic k n e ss (in)
s = 0.3 (24000)/[24(7.625)] = 39”
U se m ultiple o f 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)] = 359”
d ’ = 359”/12”
2 w alls each direction: d = d ’/2 = 30/2
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 40 (8”)
The m inim um length, for the reinforced wall within each direction, is 15’-8”,
at 40” rebar spacing
L= 15’-8”
L= 15’-8”
s = 40”
d’= 30’
d = 15’
L = 15’-8”
L= 15’-8”
L= 15’-8”
Figure 65: Minimum shear wall length in each direction
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
8.3. Results
Table 3 gives results of the forty-eight different calculations based on the above
sample study. Each of 48 the calculations can be found in Appendix B starting on
page 84.
Block
Type
Rebar
Size
Rebars to resist all shear Minimum
shear wall
length in
each
direction
Masonry to resist all shear Minimum
shear wall
length in
each
direction
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26’-8”
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26’-8”
Table 3: Calculations of Minimum Shear Wall Length in Each Direction
72
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8.4. Analysis
Comparing the two studies, the findings show that using rebars to resist all shears
uses m inim um shear wall length in each direction; whereas, the second study which
uses masonry to resist all shear, uses greater shear wall length. However, when
using only masonry to resist all shears, the block type makes no difference.
Using #4 rebars are better then #5 rebars. Rebars #4require less shear wall length
than #5 rebars. However, #4 rebars require closer bar spacing than #5 rebars (16” vs.
24”).
Finally comparing overall block types, it was found that concrete block requires the
shortest shear wall length. While on the other end of the spectrum, Siporex and
volcanic block required the greatest length of reinforced wall.
73
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9. Combined Analysis
Before starting on the quest of finding the best block type for building in Saudi
Arabia, it was determined that a knowledge of manufacturing processes was
important. Looking at the individual block construction process, it was concluded
that there was simplicity and complexity to manufacturing the different block types.
The process of each block type was related to the number of manufacturers that
produced it.
The first manufacturing process analyzed was the red clay block. In comparison to
the other three block types, the red clay block manufacturing process was the
simplest. The second easiest block type to manufacture is the concrete block. The
concrete manufacturing process is the same as the process that produces the volcanic
block. Both are less complex in comparison to Siporex block manufacturing. Each
of the block types undergoes steps before completion, but the Siporex block demands
extra needs in the mixture and completion process.
The red clay block manufacturing process is the simplest, but there are other factors
that make the red clay block desirable. The additional factors are that the material
mixture is also the simplest and the main material in the mixture are clay, sand and
water making the mixture environmentally friendly. Also, production of the red clay
block uses the lowest amount of electricity because the firing process depends on the
use of inexpensive oil to fire the completed blocks. This is perhaps unique to Saudi
Arabia. Lastly, any damaged blocks are ground up and recycled into the next batch
74
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of red clay blocks, which is also, very environmentally sound. However, throughout
the testing and calculation process red clay block usually came in second, in
comparison to its competitors. But it is the final comparison which determines the
optimum block type. Red clay block scored second for thermal conductivity for
wall, compression strength and seismic, and third in thermal conductivity for
material. Furthermore, in the thermal cases there was very little difference between
the three best performers.
Comparing the red clay block to the concrete block, it is found that concrete block
has strength in particular testing. The concrete took first place in seismic and
compression strength. But coming in first does not automatically mean that the
concrete block is the optimum block. Since the concrete block, had the highest
conductivity by a large margin, this placed the red clay block ahead.
There are other factors to be considered before choosing the optimum block type
which are cost and availability. Though the majority of block manufacturers (one
hundred and fifty-one) produce Concrete blocks, making it the lowest price at
$0.34/block, the other end of the spectrum is the Siporex block, which is produced
by only one manufacturer in Saudi Arabia and costs $1.52/block. The volcanic and
red clay blocks are in between. Volcanic block is produced by four manufacturers
costing $0.58/block, in comparison to the red clay block, that is produced by twelve
manufactures and costing $0.53/block (See graph 9).
75
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N um ber o f F acto ries & P rice p e r B lock in US D ollar
Number of Factories
Price per Block in US Dollar
Volcanic Siporex Red Clay Concrete
Block Type
Graph 9: Number of Factories and price per block in US Dollar
If demand for the red clay block were to rise and more manufactures were to produce
the block, it would drive the price of the red clay block down. This theory emerges
from the basic information above. The greater number of manufactures equaled a
lower per unit price and vice versa, the lower number of manufactures, the greater
the per unit price. There is a direct relationship between number of manufacturers
and the unit price of the block. Once there is a demand for a product, more
competition emerges, lowering the price of the product until it reflects only raw
material and process costs. If this was to occur, more manufacturers would produce
red clay blocks due to demand, then the unit price of the block would be lower. This
is in theory only, but it is possible, based on the historic use and popularity of cement
blocks (MFNE, 1999).
76
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Over all price per block in US Dollar
Red Clay Concrete Volcanic Siporex
Block Type
B Price per Block
B Price Per Block with grout, plaster and labor
B Price of Concrete block with Insulation_________
Graph 10: Over all price per block in US Dollar
The last comparison of the analyses is based on comparing price of manufactured
block to use of in construction. The use of the different bock types adds other cost
variables to using a particular block type in construction. The cost variables to be
considered are the cost of grout, plaster, labor and insulation “for concrete block
only”. Concluding that price of each unit of block is not the only variable to the cost
of each block type use in construction. As the above graph displays, when other cost
variables are accounted for, red clay block was the least expensive block type to use
in construction (See graph 10). Whereas the concrete block is cheaper in the per unit
price, is more expensive then either red clay or volcanic block when used in
construction. Concrete block use in construction increases cost because of the need
for additional insulation.
77
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10. Conclusion
The best determinants for finding the optimum block type came from testing for
thermal resistance, structure, availability and cost. Using the end results of each
testing method and calculations, several conclusions can be made. Throughout the
testing process, it has been found that there may not be one correct option, but
several depending on the outcome of the different tests and region of Saudi Arabia
which the block would be the optimum choice for use in construction.
The final outcome of testing shows that siporex block was best in regards to thermal
resistance. Negative factors of the siporex block are cost, availability, compression
strength and seismic activity. Siporex block having the highest thermal resistance
does not alone make it an optimal option. Beneficial use of siporex block is within
the central region of the country, used as insulation can protect against the extremely
hot and dry weather of the desert climate.
The concrete block did come in first for compression strength, seismic activity, cost
and availability. Concrete block does not have insulation capabilities and would not
be a great choice for regions that have extreme weather. Though a concept that can
be looked at is combination of concrete for structure and siporex for insulation,
which would be the optimum decision for use in all regions within Saudi Axabia.
This was an idea of using two block types to get the optimum block. Considerations
would have to be made for the ability to combine two block types in building
78
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structure. Architects and engineers would focus one block type for structure with
another for insulation. The negative aspects are cost, time, availability, and expertise
to use the combined blocks in construction.
If combination of block types took place in manufacturing, this can be the
development of a new type of building block. Or if the manufacturing process was
to change the block unit design, it could produce new blocks that had higher
insulating abilities with greater compression strength that would allow for higher
seismic activity.
Another option is to consider the overall cost of using concrete block for structure
and a separate insulation material with appropriate finish (grout, plaster and labor).
This option is less cost’'- then use of siporex block for insulation. However, when
adding the cost of appropriate finish, it is still found that use of red clay block solely,
is the least expensive.
With the block types and current design available, the optimum choice would be the
red clay block. Since, each of the five regions within Saudi Arabia consists of at
least one red clay block manufacture. The red clay block is easy to manufacture, is
environmentally friendly, and has a very simple component mixture. Testing
concluded, red clay block did not place first, but came in a close second throughout
each of the testing processes. Finalizing the discussion and conclusion would mean
that it is the optimum block type in comparison to its testing competitors within the
different regions of Saudi Arabia.
79
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11. Bibliography
Al-Borkane factory (2000), Al-Borkane Catalog. Jeddah, Saudi Arabia
Al-Modafer factory (2001), Al-Modafer Catalog. Al-Karje, Saudi Arabia.
Al-Yamamah factory (1999), Al-Yamamah Catalog. Truma, Saudi Arabia.
Ambrose James (2000), Reinforced Masonry Engineering Handbook, Masonry
Institute of America. USA
ASTM C39 (2001), American Society fo r Testing and Materials C39. American
Society for Testing and Materials. Unites States.
ASTM C177 (2001), American Society fo r Testing and Materials C177, American
Society for Testing and Materials. Unites States.
Benjamin Stein., and John S. Reynolds (1999), Mechanical and Electrical
Equipment fo r Buildings, 9th Edition, published simultaneously in Canada.
Holy Qur'an, (1999), Chapter 120, verses 71, King Fahd Complex For The Printing
of The Holy Qur’an, Madinah, Saudi Arabia.
KACST (2000), Hot Disk Thermal Constants Analyzer Manual. King Abdulaziz
City of Science and Technology. Saudi Arabia.
KFUPM (2002), Testing ofFour Types o f Masonry Block to Determine Thermal
Conductivity. King Fahd University of Petroleum and Minerals. Saudi Arabia,
Dhran.
MFNE, Ministry of Finance and National Economy (1999), Guide to Saudi Arabia
Manufactured Products, 10th Edition, Saudi Arabia.
MIBR, Ministry of the indigenous businesses and the rural one (1999), The desert
brides, Riyadh, An Isbar studies, researches, and flags
MIE, Ministry of Industry & Electricity (1997), Electrical growth and development
in Kingdom ofSaudi Arabia up to the year 1997. Electrical Affairs Agency, Riyadh,
Kingdom of Saudi Arabia.
Schierle, G G (1997) Lecture Notes on Seismic Design, USC
Siporex factory (2001), Siporex Catalog. Al-Kaije, Saudi Arabia.
80
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12. Appendix
12.1 Appendix A : Thermal Calculation
OCER1125
Red Clay Block
Date of Test
29-Dec-01
Main Heater Voltage, V = 7.864
Main Heater Current, I = 0.7265
Thickness of the upper or dummy sample, d1 = 0.030209
Thickness of the lower or test sample, d2 = 0.19906
Surface area of the main heater, A = 0.0948
Correction factor of the instrument, N = 0.986
Dummy. Polystyrene
T2 = 46.780 C, T l= 29.463 C
T2-T1= 17.317 C, i.e. delta T1
T-avg = 38.122
For dummy sample (polystyrene), k = 1.864 E-04 * T(C) + 0.0329587
_kl. k-valuc of dummy sample at 38.12 is_________0.040065 W/m.K
Hot Cold
Brick 44.08 24.57
43.89 24.97
43.58 24.98 Brick Mortar
43.97 24.52 Area= 0.087216 0.00758
44.08 24.86
Avg. 43.92 24.78
Mortar 42.91 25.04
42.78 26.07
Avg. 42.85 25.56
W. Temp 43.83 24.84
Delta T of sample = 18.992
CALCULATIONS:
Total Heat Flow, Q = N*V*I = 5.633211
Heat flow thro'the dummy sample, Ql=kl*A*dcltaTl/dl = 2.177247
Heat flow thro’ thctcstsamplc, Q2 = Q-Q1 = 3.455965
k of test sample = (Q2*d2)/(A*delta T2) = 038210
k = 038210 W/m.k
T„.,„ = 3434 C
Obtained from weighted temp.
81
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OCER1125
Concrete Block
Main Heater Voltage, V = 11.541
Main Heater Current, 1 = 1.0636
Thickness of the upper or dummy sample, dl
0.030209
Thickness ofthc lower or test sample, d2 = 0.19587
Surface area of the main heater, A = 0.0948
Correction factor of the instrument, N = 0.986
Dummy. Polystyrene
T2 = 52.372 C, T1 = 29.415 C
T2-T1= 22.957 C, i.e. delta T
T-avg = 40.893
For dummy sample (polystyrene), k = 1.864 E-04 * T(C) +
0.0329587
_______ k l. k-valuc of dummy sample at 40.89 is________ 0.040581 W/mJC
Hot Cold
Test Sample Brick 47.48 28.04
47.62 26.52
45.70 28.25
46.21 27.05
45.04 28.31
Avg. 46.41 27.63
Mortar 43.70 28.69
44.89 29.13
Avg. 44.30 28.91
W. Temp 46.24 27.73
Delta T of sample = 18.504
Brick Mortar
Area= 0.087216 0.00758
CALCULATIONS:
Total Heat Flow, Q = N*V*I = 12.10316
Heat flow thro'the dummy sample, Ql=kl*A*dcltaTl/dl = 2.923601
Heat flow thro'the test sample, Q2 = Q - Q1 = 9.179556
k of test sample = (Q2*d2)/(A*delta T2) = 1.02498
k = 1.02498 W/m.k
T.jm .=36.99 C
Obtained from weighted temp.
8 2
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OCER1125
Volcanic Block
Main Heater Voltage, V = 8.300
Main Heater Current, I = 0.7667
Thickness of the upper or dummy sample, d 1 = 0.030209
Thickness o f the lower or test sample, d2 = 0.19732
Surface area of the main heater, A = 0.0948
Correction factor of the instrument, N = 0.986
Dummy. Polystyrene
T2= 48.018 C ,T 1 = 29.439 C
T2-T1= 18.579 C, i.e. delta T
T-avg = 38.729
For dummy sample (polystyrene), k = 1.864 E-04 * T(C) + 0.0329587
_______ k l, k-valuc of dummy sample at 38.73 is________ 0.040178 W/m.K
Hot Cold
Test Samnle Brick 44.68 24.42
44.84 25.16
44.90 25.26
45.41 25.17
45.63 2579
Avg. 45.09 25.06
Mortar 43.94 25.33
44.40 25.62
Avg. 44.17 25.48
W. Temp 45.02 25.09
Delta T of sample = 19.925
Area=
Brick Mortar
0.087216 0.00758
CALCULATIONS;
Total Heat Flow, Q = N*V*I = 6.274519
Heat flow thro’ the dummy sample, Ql= kl *A*dclta Tl/dl = 2.342534
Heat flow thro' the test sample, Q2 = Q - Q 1 = 3.931985
k of test sample = (Q2*d2)/(A*delta T2) =
k = 0.41075 W/m.k
T„..„ = 35.05 C
0.41075
Obtained from weighted temp.
83
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OCER1125 Siporex Block
(Solid)
Main Heater Voltage, V
Main Heater Current, 1
Thickness of the upper or dummy sample, dl
Thickness of the lower or test sample, d2
Surface area of the main heater, A
Correction factor of the instrument, N
7.313
0.6755
0.030209
0.19745
0.0948
0.986
Dummy. Polystyrene
T2 = 45323 C, T1 = 29390 C
T2-T1= 15.933 C,i.e. delta T1
T-avg = 37357
For dummy sample (polystyrene), k = 1.864 E-04 * T(C) + 0.0329587
_______ kl. k-valuc of dummy sample at 3736 is_________ 0.039922 W/m.K
Hot Cold
Test Sample Brick 43.55 25.46
43.93 24.47
43.15 24.11
43.12 25.19
43.51 24.49
Ave. 43.45 24.74
Mortar 40.47 25.88
42.27 26.51
Avg. 41.37 26.20
W. Temp 4339 24.86
Delta Tof sample = 18.425
Brick Mortar
Area= 0.087216 0.00758
CALCULATIONS:
Total Heat Flow, Q = N*V*I
Heat flow thro’ the dummy sample, Ql=kl*A*dclta Tl/dl
Heat flow thro’ the test sample, Q2 = Q - Q1
4.870772
1.996088
2.874684
k of test sample = (Q2*d2)/(A*deIta T2) =
k = 032495 W/m.k
T„.,„ = 34.07 C
032495
Obtained from weighted temp.
8 4
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12.2 Appendix B: Seismic Calculation
Seismic design
Assume: Z=0.4, Rw = 8,1= 1, C = 2.75
Dead load
First floor W = 250 x 40 x 50 = 250,000
Roof W = 170x40x50 = 170,000
Base shear
V = W Z IC / R w = (250,000+170,000) 0.4 x 1 x 2.75 / 6
V = 0.183 x (250,000+170,000) x 1.5 = 115,290
Vertical force distribution
Fx = (V-Ft)wxhx/S (w i hi)
Level wx hx wx hx wxhx/X w; hi Fx = V (wx hx / £ wj hj)
Roof 170,000# 18 306,000# 0.57 65,715.3#
Floor 250,000# 9 2,250,000# 0.42 48,421.8#
£ W i hi = 5,310,000# V= 115,290.1#
Shear per floor
First Floor 65,715.3#
Ground floor V = 48,421.8 + 65,715.3 = 114,137.1#
Horizontal Spaceing
Space horizontal rebars at multiple of 8" based on:
As = 0.0007 Area of masonry
where
As = area of horizontal steel (spaced to get 0.0007)
0.0007 from UBC
# 4 bars spaced 32"
# 5 bars spaced 48" (max. allowed by code)
85
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One-story building - Red Clay Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th fm = 2039 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tr e s s
34 psi (1.33) Fv = 45.2 psi
R ebar size (#4 bars, f V z”)
Av = t t r2 = t t (0.5/2)2 A v = .2 in2
R ebar sp acin g
s = Av F s / (Fv b)
s = 0.2 (24000)/[45.2(7.625)] = 14”
U se m ultiple o f 8” m odule s = 8”
R equired effective wall length
d ’ = V/( Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12” d ’= 17.3’
2 w alls each direction: d = d 72 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 8 (8”) L = 8’.93”=8-
11.16
One-story building - Red Clay Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) Fs = 24,000
psi
Specified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Allowable sh e a r s tr e s s
18 p si (1.33) Fv = 24 psi
R ebar size (#4 b ars, f %”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
Required effective wall length
d ’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393”/12” d ’= 32.8’
2 w alls each direction: d = d’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
Round to (8“ m odule: 24 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Red Clay Block
Method One: Rebars to resist all shear, using #5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
34 psi (1.33) Fv = 45.2 psi
R ebar siz e (# 5bars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs /(F v b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12” d ’= 17.3’
2 w alls each direction: d = d’/2 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”) L = 8’.93”= 8-
11.16
One-story building - Red Clay Block
Method Two: masonry to resist all shear, using # 5rebars
Assume:
8” (nominal) wall = 75/8” b = 7.625”
Grade 60 steel (Fy = 60,000 psi) Fs = 24,000 psi
Specified compressive strength f m = 2039 psi
Conservative h/d ratio h/d =>1
Allowable shear stress
18 psi (1.33) Fv = 24 psi
Rebar size (#5 bars, f 5/8”)
Av = t t r2 = t t (0.625/2)2 Av = .3 in2
Rebar spacing
s = Av Fs / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39”
Use multiple of 8” module s = 32”
Required effective wall length
d’ = V/( Fv b) = 72,000# /[24(7.625)] = 393”
d’ = 393”/12” d’= 32.8’
2 walls each direction: d = d’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
Round to (8“ module: 32 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Concrete Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
Specified co m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
38 p si (1.33)
R ebar siz e (#4 b ars, f V 2”)
Av = tt r2 = tt (0.5/2)2
R eb ar sp a c in g
s = Av Fs /(F v b)
s = 0.2 (24000)/[50.54(7.625)] = 12.45”
U se m ultiple o f 8” m odule s = 8”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[50.54(7.625)] = 186.8”
d ’ = 186.8712” d ’= 15.5’
2 w alls e a c h direction: d = d 72 = 15.5/2 d = 7.7’
Wall length L=d+4”=7.7’+.33’=8.03
R ound to 8” m odule: 8 (8”) L = 8’.03”=8-
0.36
b = 7.625”
F s = 24,000 p si
f m = 2548 psi
h/d =>1
Fv = 50.54 psi
Av = .2 in2
One-story building - Concrete Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 p si
S pecified co m p ressiv e stren g th f m = 2548 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 psi (1.33) Fv = 24 p si
R ebar siz e (#4 b ars, f V z” )
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R eb ar sp acin g
s = Av Fs /(F v b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393712” d’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 24 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission
One-story building - Concrete
Method One: Rebars to resist all shear, using #5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 2548 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tre s s
34 psi (1.33) Fv = 45.2 p si
R ebar size (#5 b ars, f 5/8”)
Av = tt r2 = i t (0.625/2)2 Av = .3 in2
R eb ar sp a c in g
s = Av Fs /(F v b)
s = 0.3 (24000)/[45.2(7.625)] = 20”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12” d ’= 17.3’
2 w alls e a c h direction: d = d’/2 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”) L = 8’.93”= 8-
11.16
One-story building - Concrete Block
Method Two: masonry to resist all shear, using # 5rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 2548 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tre s s
18 p si (1.33) Fv = 24 p si
R ebar siz e (#5 b ars, f 5/8”)
Av = t t r2 = t t (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs I (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393”/12” d ’= 32.8’
2 w alls e a c h direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 32 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Volcanic Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
29 psi (1.33) Fv = 38.57 psi
R ebar siz e (#4 bars, f V")
Av = t t r2 = t t (0.5/2)2 A v = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[38.57(7.625)] = 16”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[38.57(7.625)] = 244”
d ’ = 244712” d ’= 20’
2 w alls each direction: d = d ’/2 = 20/2 d = 10’
Wall length L=d+4”=10’+.33’=10.33
R ound to 8” m odule: 16 (8”) L = 10’.33”=10-
3.96
O n e -s to ry b u ild in g - V o lcan ic B lco k
M e th o d T w o : m a s o n ry to re s is t all s h e a r, u sin g # 4 r e b a r s
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
18 p si (1.33) Fv = 24 psi
R ebar siz e (#4 b ars, f %”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp acin g
s = Av F s / (Fy b)
s = 0.2 (24000)/[24(7.625)] = 26”
Use m ultiple of 8” m odule s = 24”
R equired effective wall length
d ’ = V/( Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393712" d ’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 24 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Volcanic Block
Method One: Rebars to resist all shear, using #5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = ir r 2 = it (0.625/2)2 Av = .3 in 2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)J = 208”
d ’ = 208”/12” d ’= 17.3’
2 w alls each direction: d = d ’/2 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”) L = 8’.93”=8-
11.16
One-story building - Volcanic Blcok
Method Two: masonry to resist all shear, using # 5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) Fs = 24,000psi
Specified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
18 psi (1.33) Fv = 24 p si
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs /(F v b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple of 8” m odule s = 32”
R equired effective wall length
d’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 393”/12” d ’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 32 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Siporex Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 p si
S pecified c o m p ressiv e stre n g th f m = 471 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
29 psi (1.33) Fv = 38.57 psi
R eb ar siz e (#4 b ars, f %")
Av = tt r2 = i t (0.5/2)2 Av = .2 in2
R eb ar sp a c in g
s = Av Fs /(F v b)
s = 0.2 (24000)/[38.7(7.625)] = 16.3”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[38.7(7.625)J = 244”
d ’ = 244”/12” d ’= 20.3’
2 w alls each direction: d = d ’/2 = 20.3/2 d = 10.2’
Wall length L=d+4”=10.2’+.33’=10.53
R ound to 8” m odule: 16 (8”) L = 10’-8”
One-story building - Siporex Block
Method Two: masonry to resist all shear, using # 4 rebars
A cqum A '
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
S pecified c o m p re ssiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
18 p si (1.33)
R ebar size (#4 b a rs, f Vi”)
Av = tt r2 = tt (0.5/2)2
R eb ar sp acin g
s = Av F s / (Fy b)
s = 0.2 (24000)/[24(7.625)] = 26 ”
U se m ultiple o f 8” m odule
R equired effective wall length
d ’ = V/( Fv b) = 72,0 0 0 # /[24(7.625)1 = 393”
d ’ = 393”/12” d ’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 24 (8”) L = 16’-8:
b = 7.625”
F s = 24,000 psi
f m = 471 psi
h/d =>1
Fy = 24 p si
Av = .2 in2
s = 24”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
One-story building - Siporex Block
Method One: Rebars to resist all shear, using #5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 471 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = t t r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[45.2(7.625)] = 208”
d ’ = 208”/12” d ’= 17.3’
2 w alls each direction: d = d ’/2 = 17.3/2 d = 8.6’
Wall length L=d+4”=8.6’+.33’=8.93
R ound to 8” m odule: 16 (8”) L = 8’.93”=8-
11.16
One-story building - Siporex Block
Method Two: masonry to resist all shear, using # 5rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stre n g th f m = 471 p si
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs /(F v b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple of 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 72,000# /[24(7.625)] = 393”
d ’ = 3 9 3712” d ’= 32.8’
2 w alls each direction: d = d ’/2 = 32.8/2 d = 16.4’
Wall length L = d+4” = 16.4+.33=16.73’
R ound to (8“ m odule: 32 (8”) L = 16’-8”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Red Clay Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nominal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi)
Specified co m p ressiv e stren g th
C onservative h/d ratio
Allowable s h e a r s tre s s
34 psi (1.33)
R ebar size (#4 b ars, f ’ /z”)
Av = tt r2 = tt (0.5/2)2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[45.2(7.625)] = 14
U se m ultiple of 8” m odule
R equired effective wall length
d ’ = V/( Fv b) = 114137# l[45.2(7.625)] = 331”
d’ = 331”/12” d ’= 27.5’
2 w alls each direction: d = d ’/2 = 27.5/2 d = 13.75’
Wall length L=d+4”=13.75’+.33'=14.08
R ound to 8” m odule: 8 (8”) L = 14.08
F s = 24,000 psi
f m = 2039 psi
h/d =>1
Fv = 45.2 psi
Av = .2 in2
s = 8”
Two-story building Ground Floor - Red Clay Block
Method Two: masonry to resist all shear, using # 4 rebars
A q q iii t ip "
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Allowable s h e a r stre s s
18 psi (1.33) Fv = 24 psi
R ebar size (#4 b ars, f Vt” )
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple o f 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623”/12” d ’= 51.9’
2 w alls each direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 24 (8”) L = 26.23”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Red Clay Block
Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Allowable s h e a r s tre s s
34 p si (1.33) Fv = 45.2 psi
R ebar size (# 5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R eb ar sp a c in g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)] = 331”
d ’ = 331”/12” d’= 27.5’
2 w alls each direction: d = d ’/2 = 27.5/2 d = 13.75’
Wall length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 16 (8”) L = 14.08
Two-story building Ground Floor - Red Clay Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssum e:
8” (nom inal) w all = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 2039 p si
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R eb ar siz e (#5 b ars, f 5/8”)
Av = T T r2 = TT (0.625/2)2 Av = .3 in 2
R eb ar sp a c in g
s = Av F s / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623712” d ’= 51.9’
2 w alls each direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 32 (8”) L = 26.23
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Concrete Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 steel (Fy = 60,000 psi)
S pecified co m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
38 p si (1.33)
R ebar siz e (#4 b ars, f V z” )
Av = t t r2 = tt (0.512)2
b = 7.625”
Fs = 24,000 psi
f m = 2548 psi
h/d =>1
Fy = 50.54 psi
Av = .2 in2
R ebar sp a c in g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[50.54(7.625)] = 12.45”
U se m ultiple of 8” m odule s = 8”
R equired effective w all length
d ’ = V/(Fv b) = 114137#/[50.54(7.625)] = 296”
d’ = 296”/12” d ’= 24.6’
2 w alls each direction: d = d ’/2 = 24.6/2 d = 12.3’
Wall length L=d+4”=12.3’+.33’=12.63
R ound to 8” m odule: 8 (8”) L = 12.63
Two-story building Ground Floor - Concrete Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified c o m p ressiv e stre n g th f m = 2548 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 p si
R eb ar size (#4 b ars, f V z” )
Av = i t r2 = tt (0.512)2 Av = .2 in2
R eb ar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623'
d> = 623”/12” d ’= 5 1 .9 ’
d = 25.9’ 2 w alls each direction: d = d ’/2 = 51.9/2
W all length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 24 (8”) L = 26.23’
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Concrete Block
Method One: Rebars to resist all shear, using #5 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
S pecified co m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tre s s
34 psi (1.33)
R ebar size (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20”
U se m ultiple of 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)] = 331
d ’ = 331”/12”
2 w alls each direction: d = d’/2 = 27.5/2
b = 7.625”
Fs = 24,000 psi
f m = 2548 psi
h/d =>1
Fv = 45.2 psi
Av = .3 in2
d ’= 27.5’
d =13.75’
s = 16”
Wall length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 16 (8”)
Two-story building Ground Floor - Concrete Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
R ebar sp acin g
s = Av F s I (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623”/12” d ’= 51.9’
2 w alls e a c h direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 32 (8”) L = 26.23
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
S pecified co m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
18 p si (1.33)
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2
b = 7.625”
F s = 24,000 psi
f m = 2548 psi
h/d =>1
Fv = 24 psi
Av = .3 in2
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Volcanic Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8”
G rade 60 steel (Fy = 60,000 psi)
Specified co m p ressiv e stren g th
C onservative h/d ratio
Allowable sh e a r s tre s s
29 psi (1.33)
R ebar size (#4 b ars, f V i”)
b = 7.625”
F s = 24,000 psi
fm = 1109 psi
h/d =>1
Fv = 38.57 psi
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[38.57(7.625)J = 16”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[38.57(7.625)] = 388”
d ’ — 388”/12” d ’= 32.3’
2 w alls each direction: d = d’/2 = 32.3/2 d = 16.15’
Wall length L=d+4”=16.15’+.33’=16.48
R ound to 8” m odule: 16 (8”) L = 16.48
Two-story building Ground Floor - Volcanic Block
Method Two: masonry to resist all shear, using # 4 rebars
R ebar sp acin g
s = Av F s I (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
R equired effective wall length
d’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623”/12” d ’= 5 1 .9 ’
2 w alls each direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
Round to (8“ m odule: 24 (8”) L = 26.23
A e e n n iP *
8” (nom inal) wall = 75/8”
G rade 60 steel (Fy = 60,000 psi)
Specified co m pressive stren g th
C onservative h/d ratio
A llowable sh e a r s tre s s
18 psi (1.33)
R ebar siz e (#4 b ars, f V i”)
Av = tt r2 = tt (0.5/2)2
b = 7.625”
Fs = 24,000 psi
fm = 1109 psi
h/d =>1
Fy = 24 psi
A v = .2 in2
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Volcanic Block
Method One: Rebars to resist all shear, using #5 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
Specified co m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
34 psi (1.33)
R ebar size (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2
b = 7.625”
F s = 24,000 psi
fm = 1109 psi
h/d =>1
Fv = 45.2 psi
Av = .3 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s s 0.3 (24000)/[45.2(7.625)] = 20”
Use m ultiple o f 8 ” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)J = 331”
d ’ = 331”/12” d ’= 27.5’
2 w alls each direction: d = d ’/2 = 27.5/2 d = 13.75’
Wall length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 16 (8”) L = 14.08
Two-story building Ground Floor - Volcanic Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000psi
Specified co m p ressiv e stren g th fm = 1109 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tre s s
18 p si (1.33) Fv = 2 4 p s i
R eb ar size (#5 b a rs, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in 2
R ebar sp acin g
s = Av F s I (Fy b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
a i o n j ..i. . _
U se m ultiple of 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d’ = 623”/12”
2 w alls each direction: d = d ’/2 = 51.9/2
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 32 (8”) L = 26.23
d’= 51.9’
d = 25.9’
s = 32
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Siporex Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
S pecified c o m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
b = 7.625”
Fs = 24,000 psi
f m = 471 psi
h/d =>1
29 psi (1.33)
R eb ar size (#4 b ars, f Vz")
Av = tt r2 = tt (0.5/2)2
Fy = 38.57 psi
Av = .2 in2
R ebar sp acin g
s = Av Fs /(F v b)
S = 0.2 (24000)/[38.7(7.625)] = 16.3”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 114137#/[38.7(7.625)] = 386”
d ’ = 386”/12” d ’= 32.16’
2 w alls each direction: d = d ’/2 = 32.16/2 d = 16.08’
Wall length L=d+4”=16.08’+.33’=16.41
R ound to 8” m odule: 16 (8”) L = 16.41
Two-story building Ground Floor - Siporex Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssu m e:
8” (nom inal) wall = 75/8”
G rade 60 ste e l (Fy = 60,000 psi)
S pecified c o m p ressiv e stren g th
C onservative h/d ratio
A llow able s h e a r s tr e s s
18 p si (1.33)
R ebar siz e (#4 b a rs, f Vz)
Av = tt r2 = tt (0.5/2)2
R ebar sp a c in g
s = Ay F s / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623”/12”
2 w alls each direction: d = d ’/2 = 51.9/2
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 24 (8”)
b = 7.625”
Fs = 24,000 psi
f m - 471 p si
h/d =>1
Fv = 24 psi
Av = .2 in2
L = 26.23
d ’= 51.9’
d = 25.9’
s = 24”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building Ground Floor - Siporex Block
Method One: Rebars to resist all shear, using #5 rebars
A ssu m e:
8” (nom inal) w all = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 471 p si
C onservative h/d ratio h/d =>1
A llow able s h e a r s tre s s
34 p si (1.33) Fv = 45.2 psi
R eb ar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in 2
R ebar sp acin g
s = Av Fs I (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 114137# /[45.2(7.625)] = 331”
d ’ = 331712” d ’= 27.5’
2 w alls each direction: d = d ’/2 = 27.5/2 d = 13.75’
W all length L=d+4”=13.75’+.33’=14.08
R ound to 8” m odule: 16 (8”) L = 14.08
Two-story building Ground Floor - Siporex Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nominal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) Fs = 24,000 psi
S pecified co m p ressiv e stren g th f m = 471 psi
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 psi (1.33) Fv = 24 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = Trr2 = T T (0.625/2)2 Av = .3in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/( Fv b) = 114137# /[24(7.625)] = 623”
d ’ = 623712” d ’= 5 1 .9 ’
2 w alls each direction: d = d ’/2 = 51.9/2 d = 25.9’
Wall length L = d+4” = 25.9+.33=26.23’
R ound to (8“ m odule: 32 (8”) L = 26.23
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Red Clay Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8”
G rade 60 steel (Fy = 60,000 psi)
Specified co m p ressiv e stren g th
C onservative h/d ratio
A llowable s h e a r s tre s s
34 psi (1.33)
R ebar size (#4 b ars, f V i”)
Av = t t r2 = t t (0.5/2)2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[45.2(7.625)] = 14”
U se m ultiple o f 8” m odule
R equired effective wall length
d ’ = V/(Fv b) = 65715#/[45.2(7.625)] = 190”
d ’ = 190”/12” d’= 15.8’
2 w alls each direction: d = d ’/2 = 7.9/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23’
R ound to 8” m odule: 16 (8”) L = 8.23
Two-story building First Floor - Red Clay Block
Method Two: masonry to resist all shear, using # 4 rebars
b = 7.625”
Fs = 24,000 psi
f m = 2039 psi
h/d =>1
Fv = 45.2 psi
Av = .2 in2
s = 16”
A ssum e:
8” (nominal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 2039 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R ebar size (#4 b a rs, f Vi”)
Av = t t r2 = t t (0.5/2)2 A v = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)] = 359”
d’ = 359”/12” d ’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 24 (8”) L = 15.33”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Red Clay Block
Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stre n g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Allowable s h e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
R ebar size (# 5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in 2
R e b a rsp a c in g
s = Av Fs /(F v b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple of 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 65715#/[45.2(7.625)] = 190”
d’ = 190”/12” d ’= 15.8’
2 w alls each direction: d = d ’/2 = 15.8/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23
R ound to 8” m odule: 24 (8”) L = 8.23
Two-story building First Floor - Red Clay Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stre n g th f m = 2039 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R ebar size (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 A v = .3 in 2
R ebar sp acin g
s = Ay Fs / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39”
U se m ultiple o f 8 ” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)J = 359”
d ’ = 359”/12” d ’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 32 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Concrete Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8 ” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
S pecified c o m p re ssiv e stren g th f m = 2548 psi
C o n serv ativ e h/d ratio h/d =>1
A llow able s h e a r s tr e s s
38 psi (1.33) Fv = 50.54 psi
R eb ar siz e (#4 b ars, f %”)
A v = t t r2 = t t (0.5/2)2 A v = .2 in2
R ebar sp a c in g
s = Av Fs /( F v b)
s = 0.2 (24000)/[50.54(7.625)] = 12.45”
U se m ultiple of 8” m odule s = 8”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[50.54(7.625)J = 170”
d ’ = 170”/12” d ’= 14.16’
2 w alls e a c h direction: d = d ’ /2 = 14.16/2 d = 7.08’
Wall length L=d+4”=7.08’+.33’=7.41
R ound to 8” m odule: 8 (8”) L = 7.41
Two-story building First Floor - Concrete Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th fm = 2548 psi
C o n servative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 p si
R ebar siz e (#4 b ars, f %”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp a c in g
s = Av F s / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple of 8 ” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)] = 359”
d ’ = 359”/12” d ’= 30’
2 w alls e a c h direction: d = d ’/2 = 30/2 d = 15’
Wall len g th L = d+4” = 15+.33=15.33
R ound to (8“ m odule: 24 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Concrete Block
Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) Fs = 24,000 psi
S pecified co m p ressiv e stren g th f m = 2548 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r stre s s
34 psi (1.33) Fv = 45.2 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in 2
R ebar sp a c in g
s = Av F s I (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20”
Use m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[45.2(7.625)] = 190”
d ’ = 190”/12” d ’= 15.8’
2 w alls e a c h direction: d = d 72 = 15.8/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23
R ound to 8” m odule: 16 (8”) L = 8.23
Two-story building First Floor - Concrete Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 p si
Specified co m p ressiv e stren g th f m = 2548 psi
C onservative h/d ratio h/d =>1
Allowable s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av F s / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple of 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)J = 359”
d’ = 359”/12” d ’= 30’
2 w alls each direction: d = d’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
Round to (8“ m odule: 32 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor — Volcanic Block
Method One: Rebars to resist all shear, using #4 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) Fs = 24,000 psi
Specified co m p ressiv e stre n g th f m = 1109 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tr e s s
29 psi (1.33) Fv = 38.57 psi
R ebar size (#4 b ars, f %”)
Av = tt r2 = tt (0.512)2 Av = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[38.57(7.625)] = 16”
U se m ultiple o f 8” m odule s = 16”
R equired effective wall length
d ’ = V/( Fv b) = 65715# /[38.57(7.625)] = 223”
d ’ = 223”/12” d ’= 18.58’
2 w alls each direction: d = d ’/2 = 18.58/2 d = 9.29’
Wall length L=d+4”=9.29’+.33’=9.62
R ound to 8” m odule: 16 (8”) L = 9.26
Two-story building First Floor - Volcanic Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssum e:
8” (nominal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) Fs = 24,000 psi
Specified co m p ressiv e stre n g th fm = 1109 psi
C onservative h/d ratio h/d =>1
A llowable s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R ebar siz e (#4 b ars, f 1 /z”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple o f 8” m odule s = 24”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /I24(7.625)] = 359”
d ’ = 359”/12” d’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 24 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Volcanic Block
Method One: Rebars to resist all shear, using #5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
34 psi (1.33) Fv = 45.2 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in 2
R eb ar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/( Fv b) = 65715# /[45.2(7.625)] = 190”
d ’ = 190”/12” d ’= 15.8’
2 w alls each direction: d = d ’/2 = 15.8/2 d = 7.9’
Wall length L=d+4”=15.8’+.33’=8.23
R ound to 8” m odule: 16 (8”) L = 8.23
Two-story building First Floor - Volcanic Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000psi
S pecified co m p ressiv e stren g th f m = 1109 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
18 psi (1.33) Fv = 24 p si
R ebar siz e (#5 b ars, f 5/8”)
Av = T T r2 = T T (0.625/2)2 Av = .3in2
R ebar sp acin g
s = Av Fs / (Fv b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)J = 359”
d’ = 359”/12” d ’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 32 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Siporex Block
Method One: Rebars to resist all shear, using #4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
Specified co m p ressiv e stren g th f m = 471 psi
C o nservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
29 p si (1.33) Fv = 38.57 psi
R ebar siz e (#4 b ars, f Vz” )
Av = t t r2 = u (0.5/2)2 A v = .2 in2
R eb ar sp a c in g
s = Av F s / (Fy b)
s = 0.2 (24000)/[38.7(7.625)J = 16.3”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d’ = V/(Fv b) = 65715# /[38.7(7.625)J = 222”
d ’ = 2 2 2 7 1 2 ” d ’= 1 8 .5 ’
2 w alls e a c h direction: d = d ’/2 = 18.5/2 d = 9.25’
Wall length L=d+4”=9.25’+.33’=9.58
R ound to 8 ” m odule: 16 (8”) L = 9.58
Two-story building First Floor - Siporex Block
Method Two: masonry to resist all shear, using # 4 rebars
A ssu m e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 ste e l (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 471 p si
C onservative h/d ratio h/d =>1
A llow able s h e a r s tr e s s
18 p si (1.33) Fv = 24 p si
R ebar siz e (#4 b ars, f %”)
Av = tt r2 = tt (0.5/2)2 Av = .2 in2
R ebar sp a c in g
s = Av Fs / (Fv b)
s = 0.2 (24000)/[24(7.625)] = 26”
U se m ultiple o f 8” m odule s = 24”
R equired effective w all length
d ’ = V/(Fv b) = 65715# /[24(7.625)] = 359”
d ’ = 3 5 9 7 1 2 ” d ’= 30’
2 w alls e a c h direction: d = d ’/2 = 30/2 d = 15’
Wall len g th L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 24 (8”) L = 15.33”
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Two-story building First Floor - Siporex Block
Method One: Rebars to resist all shear, using #5 rebars
Assume*
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m pressive stren g th f m = 471 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
34 psi (1.33) Fy = 45.2 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = rr r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs I (Fv b)
s = 0.3 (24000)/[45.2(7.625)] = 20.8”
U se m ultiple of 8” m odule s = 16”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[45.2(7.625)] = 190”
d ’ = 190”/12” d ’= 1 5.8’
2 w alls each direction: d = d ’/2 = 15.8/2 d = 7.9’
Wall length L=d+4”=7.9’+.33’=8.23
R ound to 8” m odule: 16 (8”) L = 8.23
Two-story building First Floor - Siporex Block
Method Two: masonry to resist all shear, using # 5 rebars
A ssum e:
8” (nom inal) wall = 75/8” b = 7.625”
G rade 60 steel (Fy = 60,000 psi) F s = 24,000 psi
S pecified co m p ressiv e stren g th f m = 471 psi
C onservative h/d ratio h/d =>1
Allow able s h e a r s tre s s
18 psi (1.33) Fv = 24 psi
R ebar siz e (#5 b ars, f 5/8”)
Av = tt r2 = tt (0.625/2)2 Av = .3 in2
R ebar sp acin g
s = Av Fs / (Fy b)
s = 0.3 (24000)/[24(7.625)] = 39.3”
U se m ultiple o f 8” m odule s = 32”
R equired effective wall length
d ’ = V/(Fv b) = 65715# /[24(7.625)] = 359”
d ’ = 359”/12” d’= 30’
2 w alls each direction: d = d ’/2 = 30/2 d = 15’
Wall length L = d+4” = 15+.33=15.33’
R ound to (8“ m odule: 32 (8”) L = 15.33
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
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University of Southern California Dissertations and Theses
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Asset Metadata
Creator
Al-Jammaz, Khaled Mohammed
(author)
Core Title
Determine the optimum block type for use in Saudi Arabia
School
School of Architecture
Degree
Master of Building Science / Master in Biomedical Sciences
Degree Program
Building Science
Publisher
University of Southern California
(original),
University of Southern California. Libraries
(digital)
Tag
Architecture,engineering, civil,engineering, materials science,OAI-PMH Harvest
Language
English
Contributor
Digitized by ProQuest
(provenance)
Advisor
Schiler, Marc E. (
committee chair
), Hickman, Glen (
committee member
), Schierle, G. Goetz (
committee member
), Thangavelu, Madhu (
committee member
)
Permanent Link (DOI)
https://doi.org/10.25549/usctheses-c16-293450
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UC11341020
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1411773.pdf (filename),usctheses-c16-293450 (legacy record id)
Legacy Identifier
1411773.pdf
Dmrecord
293450
Document Type
Thesis
Rights
Al-Jammaz, Khaled Mohammed
Type
texts
Source
University of Southern California
(contributing entity),
University of Southern California Dissertations and Theses
(collection)
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The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the au...
Repository Name
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Repository Location
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Tags
engineering, civil
engineering, materials science