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Scranton, Pa.
E l em en t s o f
M a son r y D esi g n PREPARED ESPECIALLY FOR HOME STUDY
By
I. C. S. STAFF
IN COLLABORATION WITH
C. E. OROURKE, C. E.PROFESSOR OF STRUCTURAL ENGINEERING, CORNELL UNIVERSITY
3404 EDITION 1
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Copyright, 1936, by I n t e r n a t i o n a l T e x t b o o k Co m pa n y . Copyright in Great
Britain. All rights reserved
Printed in U. S. A.3404
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ELEMENTS OF MASONRY DESIGNSerial 3404 Edition 1
STRESSES IN MASONRY
INTRODUCTION
1. Classes of Masonry.A structure that is built either of
stone or of manufactured stone-like materials, such as concrete,
brick, or hollow clay tile, is known as a masonry structure.
Thus, masonry may be classified according to the materials
used as stone masonry, concrete masonry, brick masonry, or
hollow-tile masonry.
In stone masonry, blocks of stone of comparatively large
size are fitted in place and are usually held together by port-
land-cement mortar or cement-lime mortar. Although the
blocks used in stone masonry have a wide variety of shapes
and sizes, all stone masonry may be divided into two general
classes, namely, ashlar and rubble. The masonry is classed
as ashlar when the stones are cut so accurately and are laid so
carefully that the mortar joints are not more than \inch thick.
The stone blocks in rubble masonry either are of irregular
shape or are not laid with sufficient accuracy to meet the
requirements of ashlar.
Concrete consists of comparatively small particles of crushed
stone, gravel, or blast-furnace slag that are bound together by
cement mortar. Concrete may be either plain or reinforced
with steel bars or other shapes.
Brick masonry, or brickwork, is similar to ashlar masonry,
but the bricks are much smaller than the stone blocks and the
thickness of the mortar joints usually does not exceed J inch.
Also, in brick masonry, lime mortar, as well as portland-
C O P Y R I G H T E D B Y I N T E R N A T IO N A L T E X T B O O K C O M P A N Y . A L L R IG H T S R E S E RV E D
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2 ELEMENTS OF MASONRY DESIGN
cement and cement-lime mortar, is employed. Brick masonry
may be reinforced with steel bars, in which case it is called
reinforced brick masonry.
Concrete blocks or clay-tile blocks laid in mortar are frequently used for walls in buildings. Such blocks are made
hollow in order to reduce their weight and to provide an air
space in the walls. The faces of the blocks are essentially
plane and rectangular, but the surfaces may be grooved to
improve the adhesion of the mortar.
2. Comparison of Masonry With Other Structural Mate
rials. It is comparatively easy to give a masonry structure a
pleasing appearance. Masonry structures that are built of
good material, and are properly designed and erected, are very
durable and fairly fireproof. On the other hand, metal struc
tures are subject to rust and timber structures are subject to
decay, and structures of either material are seriously affected
by fire. Masonry, unlike steel and timber, is not suited for
resisting tensile stresses and should preferably be subjected
only to direct compression. Also, to carry a given compressive
stress safely, a masonry member must be considerably larger
than a metal one and somewhat larger than a timber one.
3. Uses of Different Kinds of Masonry.Concrete is
usually the most economical kind of masonry for retaining
walls, bridge piers, foundations, and tunnels. The facilitywith which concrete is placed accounts to a large extent for its
extensive usage. Stone masonry is employed where special
architectural treatment is desired or where permanence is of
vital importance. In localities near stone quarries where
suitable material for stone masonry is available, the cost of
preparing and laying large blocks of stone is often less than the
cost of making and placing concrete. Brickwork is usedextensively for the walls of buildings and for special purposes,
such as the lining of large sewers. Concrete blocks and terra
cotta blocks are especially adapted to wall and floor construc
tion.
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ELEMENTS OF MASONRY DESIGN 3
WORKING STRESSES FOR MASONRY
4. Introductory Remarks.In the design of any masonry
member, it is assumed that the masonry is monolithic; that is,the entire member is treated as a single piece of homogeneous
material. The same general method of procedure is followed
for all classes of masonry without reinforcement. Except for
the explanations dealing with reinforced brick masonry at the
end of this text, the general principles that are here presented
apply to plain concrete, stone masonry, unreinforced brick
work, and concrete-block and terra-cotta masonry. Rein-
forced-concrete design is taken up in separate texts.
5. General Characteristics of Masonry. Masonry is fairly
strong in compression, but is weak in shear and has practically
no strength in tension. Nevertheless, masonry is sometimes
used in walls subjected to horizontal thrust and has to resist
shear, or it is employed to span an opening and is subjected toboth flexural and shearing stresses. When masonry is to
resist flexural stresses, it should be laid with special care. It
should not be subjected to direct tension.
The strength of masonry depends on the quality of the
materials used and on the workmanship, or the care with which
the materials are handled and placed. Ordinarily, the blocks in
masonryincluding the coarse aggregate in concretehavemore strength than the mortar, and therefore the strength of
the mortar and the adhesion between the blocks and the mortar
are important factors. The quality of concrete is generally
specified by its ultimate compressive strength at the age of 28
days. For example, concrete which has a crushing strength of
2,000 pounds per square inch at the age of 28 days is called
2,000-pound concrete.
6. Allowable Compressive Strength of Masonry.Average
values of the strength of masonry under various kinds of stress
are merely rough approximations and may differ considerably
from the values obtained in a test of a particular specimen.
In most of the larger cities in the United States, the allowable
pressures on masonry are specified by the local building laws,
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4 ELEMENTS OF MASONRY DESIGN
the values differing widely in the various cities. When the
design is not governed by any law and there are no definite
specifications concerning the strength of the masonry to be
used, the working unit pressures given in Table I may be
employed. These values will be used in all problems in this
text.
TABLE I
WEIGHTS AND ALLOWABLE PRESSURES FOR MASONRY
Type of Masonry
Allowable Pressure, in Poundsper Square Inch Weight
inPounds
perCubicFoot
Portland-CementMortar
Cement-Lime
Mortar
LimeMortar
Brickwork....................................... 175 140 75 120Ashlar stonework:
Granite........................................ 800 640 400 168
Limestone...................................
500 400 250 168Sandstone................................... 400 320 160 144
Rubble stonework:
Granite........................................ 140 100 156
Limestone................................... 140 100 156
Sandstone................................... 140 100 132
Hollow blocks or tiles:
Concrete...................................... 80 70 72
Terra-cotta.................................
80 70. - .
60Concrete, 2,000-pound at
28 days........................................ 500 144
7. Allowable Shearing and Tensile Strength of Masonry.
For shear without diagonal tension, the allowable shearing unit
stress in stone masonry may be assumed to be about one-
quarter of the allowable compressive unit stress. The safe
strength of concrete in shear is generally taken as about 6per
cent of the ultimate compressive strength at the age of 28 days.
In determining the resistance to diagonal tension, the safe shear
ing strength of stone masonry may be assumed as about 30
pounds per square inch, and that of concrete as 2 per cent of
the ultimate 28-day compressive strength.
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ELEMENTS OF MASONRY DESIGN 5
The tensile strength of stone masonry should not be taken
as more than 15 pounds per square inch when portland-cement
mortar is used nor more than 5 pounds when lime mortar is
employed. However, many engineers disregard entirely the
tensile strength of stone masonry, and use a solid block of
stone or concrete whenever transverse stresses are to be resisted,
as in spanning openings. The tensile strength of 2,000-pound
concrete may be taken as 40 pounds per square inch.
TABLE II
ALLOWABLE STRESSES IN STONE BLOCKS
Allowable Stresses, in Pounds per Square Inch
MaterialBearing Flexure Shear
Granite........... 1,000 150 200
Limestone___ 700 125 150
Sandstone.... 400 75 150
8. Allowable Stresses for Solid Blocks of Stone. A
masonry member that is subjected primarily to flexural stresses
generally consists of a single block of stone or concrete. In
some cases, the bearing strength of the block is also important.
Average values of the allowable stresses for single blocks of
stone are given in Table II; those for concrete have been
specified in the preceding articles.
MASONRY PIERS
PRACTICAL CONSIDERATIONS
9. Limitations of Masonry Piers. Brickwork, stone
masonry, plain concrete, or hollow-block masonry, or a com
bination of two or more of these materials, is frequently used
for the construction of short masonry piers that carry central
loads. When the load on a masonry pier has considerable
eccentricity or the height of the pier is comparatively great,
reinforced masonry should preferably be used.
Brick is the material that is generally used in buildings for
masonry piers where metal reinforcement is not required.
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6 ELEMENTS OF MASONRY DESIGN
Concrete is just as satisfactory as brick from the standpoint of
service, but the cost of forms often makes the use of plain con
crete uneconomical for piers. It is commonly specified that
isolated piers in the interior of a building should not be builtof stone masonry, because of the tendency of natural stone to
Spall under great heat; and that in fireproof or semi-fireproof
construction isolated piers should not be built of hollow-block
masonry.
The provisions of the various codes in regard to the design
of masonry piers differ greatly, but the recommendations in the
following articles may be considered to represent good practice.
10. Causes of Failure of Masonry Piers.If a very short
block of plain concrete is subjected to sufficient pressure, it may
fail by crushing, but a plain-concrete pier whose height exceeds
its width usually fails by shearing along an inclined plane. A
brick pier generally fails by longitudinal splitting and sidewise
bulging of the masonry, as a result of failure of the individual
bricks either in tension or in flexure and the pulling apart of
the joints; neither the brick nor the mortar is crushed. For
best results, the mortar joints should be as thin as possible and
of uniform thickness. Also, the bricks should preferably be
laid on edge and the vertical joints should be continuous
through several courses instead of being staggered in each two
adjacent courses. No experimental data are available on the
crushing strength of stone masonry under the conditions
existing in an actual structure. However, the mortar is known
to be the weakest element in good ashlar masonry, and the
strongest stonework is obtained by using large blocks and thin
joints.
11. Brick Piers.A brick pier whose height is not more
than six times its least width may be considered a short post;
therefore, no reduction in the allowable compressive unit
stress, given in Table I, need be made because of the effect of
the height. For piers whose height exceeds six times the
width, the allowable unit stress should be reduced. A good
formula for determining the reduced unit stress is
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ELEMENTS OF MASONRY DESIGN 7
f - f ' (125 -T o V
in which / = allowable unit stress for high pier, in pounds per
square inch;
/i = allowable unit stress for masonry, as given in
Table I, in pounds per square inch;
h = height of pier, in inches;
6= least width of pier, in inches.
The least width of a brick pier should never be less than
one-twelfth of the height; in the case of an isolated pier, it
should not be less than one-tenth
of the height. Also, no pier over
8 feet high should be less than
12 inches square in cross-section.
Standard bricks are 8 inches long
and 3f inches wide and the mortar
joints between bricks are generally
about \ inch thick, but the width
of brickwork varies somewhat. In
estimating the strength of brick
work, it is best to assume that the width is a multiple of 4 inches.
But, in considering the weight of brickwork or the space occu
pied by it, the actual width should be determined as closely
as possible.
B a sem en ? F l o o r
F i g. 1
Ex a mp l e. If a pier constructed of brick laid in portland-cement
mortar is to be 20 inches square in cross-section and is to have a height
of 12 feet, what allowable unit stress should be used?
h 12X12So l u t io n . The ratio of the height to the width is - = , = 7.2.
Since this is greater than 6, the allowable unit stress should be less than
the value 175 lb. per sq. in. given in Table I for this class of masonry.
Thus,
/= / i ( l .2 5 gg| ) =175X ( l.2 5 ^ X 7 .2 ) =156 lb. per sq. in. Ans.
12. Plain-Concrete Piers. In Fig. 1 is illustrated a con
dition in building construction for which a plain-concrete pier
is commonly used. Here, the footing is some distance below
the basement floor, and the plain-concrete pier, or pedestal, p
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8 ELEMENTS OF MASONRY DESIGN
is introduced between the footing and the reinforced-concrete
column that is above the floor. The least width of such a
pedestal should not be less than one-third of its depth or height,
because the pedestal transfers the total column load to the
footing and is therefore heavily loaded.
Where a plain-concrete pier is used for any other purpose,
the least width should not be less than one-quarter of the
height.
13. Stone Piers. Stone piers are satisfactory for the sup
port of heavy loads when good strong stone is laid in courses
covering the entire cross-section. The top and bottom courses
should be bedded true and even, the thickness of the joints
should not be more than inch, and the mortar should be kept
back 1 inch from the face of the pier to prevent spalling, or
breaking off, of the edges of the blocks.
The least width of a pier of ashlar masonry should not be
less than one-tenth of the height; and, where the height exceeds
six times the width, the allowable unit stress should be deter
mined by the formula of Art. 11. The width of a rubble pier
should be not less than one-fifth of its height nor less than 20
inches.PIERS WITH CENTRAL LOADS
14. Unit Stress for Central Load.When the resultant loadon a pier is applied centrally, the stress in the pier is assumed
to be uniformly distributed over the area of the cross-section.
Thus, p
H r (I)
in which/=unit stress, in pounds per square inch;
P = resultant load on pier, in pounds;
A = cross-sectional area of pier, in square inches.
Piers of brick, stone, or hollow-block masonry are usually
square or rectangular in section, and plain-concrete piers are
square, rectangular, or circular. If the section of the pier is a
rectangle whose dimensions in inches are denoted by b and d,
the area is bdand p
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ELEMENTS OF MASONRY DESIGN 9
When the actual unit stress in a pier is less than the allow
able value for the material, the pier is considered to be safe.
Ex a mp l e. A vertical load of 350,000 pounds, in addition to theweight of the pier itself, is to be applied at the center of a 20 " X 24" rectan
gular pier 8 feet high that is built of granite ashlar laid in portland-cement
mortar. If the allowable unit stress is as given in Table I, is the pier
satisfactory?
So l u t io n . The unit stress due to the load o f 350,000 lb. is, by
formula 2,* P 350,000 0 1U
20X24 = 729lb- perSq- m'The unit stress, in lb. per sq. in., produced at the base of the pier by its
own weight is the weight o f a prism of masonry 1 sq. in., or sq. ft. , in
cross-section and 8 ft. high. This stress is
-j-gX 8X 16 8 = 9 lb. per sq. in.
and the total stress is 729+9 = 738 lb. per sq. in.
8X12The ratio of the height to the least width is =4.8, which is less
than 6, and the safe unit stress is as given in Table I, or 800 lb. per
sq. in. Since this is greater than the actual unit stress, the pier is satis
factory. Ans.
15. Safe Central Load on Pier.The total central load
that can be supported safely by a pier is equal to the product
of the allowable unit stress for the material and the cross-
sectional area of the pier. If / represents the allowable unit
stress, and the other letters have the same meanings as in the
preceding article,P = fA (1)
or, for a rectangular section,
P=fbd (2)
Ex a mp l e.A pier built of brick laid in lime mortar is 16 inches square
and 10 feet high. What central load can the pier safely support, in
addition to its own weight, if the allowable stress for brickwork is as
specified in Table I?
h 10 VI2So l u t io n . The ratio o f the height to the width is ^ =7 .5,
and the allowable unit stress should be found by the formula of Art. 11.
From Table I, the value o f / i for brick in lime mortar is 75 lb. per sq. in.
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10 ELEMENTS OF MASONRY DESIGN
Hence,
/ = / i ( 1 - 2 5 - ^ f ) = 75 X ( l .2 6 - ^ X 7 .5 ) =66 lb. per sq. in.
The unit stress produced by the weight of the pier itself is j^ -X 1 0
X120 = 8 lb. per sq. in., and the remaining unit stress is 66 8 = 58 lb.
per sq. in.
In formula 2, / = 58 lb. per sq. in. and b = d 16 in.; hence, the safe
superimposed load on the pier is
P = /M = 58X 16X 16 = 14,850 lb. Ans.
16. Design of Pier Carrying Central Load. The required
cross-sectional area of a pier carrying a central load is found by
dividing the load by the safe unit stress. Thus,
The ratio of the height to the least width of the pier must not
be excessive. Also, for a brick pier, the dimensions must bemultiples of 4 inches. In case the height of a brick or stone
pier is found to exceed six times its width, allowance must be
made for the reduced unit stress. Since the value of / in the
formula of Art. 11 depends on the width b,which is not known
until the required area is established, it is necessary first to
determine the area and dimensions on the assumption that the
ratio of the height to the width is less than 6. However, thevalue of / used for the first trial should be taken as the differ
ence between the allowable unit stress for a short pier and the
unit stress produced by the weight of the pier itself. Then, if
hthe ratio 7- exceeds 6, the corrected total allowable unit stress
b
is computed by the formula of Art. 11, and this value is reduced
to allow for the weight of the pier. The area is then determined for the net value thus established and, if necessary, the
section is redesigned. If the least width is changed, the cal
culations for the allowable unit stress / and the required
cross-sectional area Ashould be repeated.
E x a m p l e 1.A circular pier o f 2,000-pound concrete, 6 feet high, isto carry a central load of 180,000 pounds. What is the required diameter,
to the next larger inch?
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ELEMENTS OF MASONRY DESIGN 11
So l u t io n .-From Table I, the allowable unit stress is 500 lb. per sq.
in., and the stress due to the weight of the pier is jp -X 6 X 1 4 4 = 6 lb. per
sq. in. Then, the first trial area of the cross-section is taken as
, P 180,000A= -r = - =364 sq. in.
/ 494 n
V 364jjygg = 21.5, say 22 in. For this size, the6X12
ratio of the height to the diameter is = 3.3, which is less than 4.
Therefore, the diameter of 22 in. is satisfactory. Ans.
Ex a mpl e 2.A pier of brick laid in portland-cement mortar is to
carry a central load of 32,000 pounds. If the height of the pier is to be
9 feet 6 inches and its cross-section is to be square or nearly square, what
should be the dimensions of the cross-section?
So l u t io n .First, the approximate cross-sectional area is determined
by assuming that the allowable unit stress is 175 r r jX 9.5X120 = 175
8 = 167 lb. per sq. in. Then,
. P 32,000 lnoA = j = ~ - = lQ2 sq. in.
h 114A 12 "X 1 6 " pier has an area of 192 sq. in., but for this size - = ^ - = 9.5
and, by the formula of Art. 11,
/ = / i ( l . 2 5 - i D =175X ( l .2 5 -| jX 9 .5 ) = 136 lb. per sq. in.
A more accurate value of the required area is
. 32,000 ^ = T M --ir250sq- in-and a 1 2"X 16" pier is too small. A 16 "X 16" pier will obviously be
hsatisfactory, as its area is 256 sq. in., and the ratio r will be reduced to
114 -j jr = 7.1 and the tequired area will be much less than 250 sq. in. Hence,
the pier should be 16 in. square. Ans.
EXAMPLES FOR PRACTICE
1. A pedestal of 2,000-pound concrete that is 3 feet high and 24 inches
square in cross-section is required to support a central load of 280,000
pounds in addition to its own weight, (a) What unit stress is produced?
(5) Is the pier safe according to Table I? , f (a) 489 lb. per sq. in.
1(6) Yes
2. What superimposed central load can safely be placed on a 20-inch
square pier of limestone ashlar laid in portland-cement mortar, if the
height o f the pier is 12 feet? Ans. 172,400 lb.
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12 ELEMENTS OF MASONRY DESIGN
3. A rectangular pier of 2,000-pound concrete is to be 8 feet high and
30 inches wide. What should be its other dimension if it has to support
a central load of 500,000 pounds? Ans. 34 in.
4. A square pier built of brick laid in cement-lime mortar is to carry
a central load of 15,000 pounds. If the height of the pier is to be 8 feet
8 inches, what should be the dimensions of its cross-section?
Ans. 12 in. X 12 in.
PIERS WITH ECCENTRIC VERTICAL LOADS
17. Investigation for Unit Stress. A masonry pier is some
times subjected to an eccentric load, and the stress on the pieris then not uniform. In Fig. 2 (a) is shown a part elevation,
and in view (b) a cross-section, of a rectangular pier subjected
to a load P that is eccentric in only one direction. As indi
cated in view (b), the load acts along the center line that is
parallel to the face AB of the
pier but has an eccentricity e
with respect to the center lineparallel to AD. The stress
produced in the pier by such
loading is the same that would
be produced by an imaginary
central load P and a bending
moment Pe.
a
a2 ,
a2
F ig. 2
If the dimensions of the pier section are band d,respectively,
the compressive unit stress due to the imaginary central loadp
would be This stress would be uniformly distributed over
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ELEMENTS OF MASONRY DESIGN 13
the section, as represented by the area mnopin view (a). The
assumed bending moment Pe would produce flexural unit
stresses varying at a uniform rate from a maximum compressive unit stress at the edge BCto a maximum tensile unit stress
at the edge AD of the cross-section. These flexural unit
stresses are represented by the trianglesfgoand hgpin view (a).
Since the section modulus of a rectangle is fed2, the flexural
unit stresses at the edges BCand AD are each equal to Pe-r-
6PbA bd2= -n f- The total unit stress at any point in the section
farP
is the resultant of the uniformly distributed stress and the
flexural unit stress at that point. Hence, the eccentric load
P produces at the edge BC,which is nearer to the load, a maxi-
Similarly, the minimum resultant stress which occurs at the
edge AD farther from the load, is
In these formulas/ = maximum unit stress, in pounds persquare inch, at edge of pier that is
nearer the load and perpendicular to
the direction of eccentricity;
P = eccentric load, in pounds;
b = dimension of pier section, in inches,
measured at right angles to eccentricity
of load;d = dimension of pier section, in inches,
measured in direction of eccentricity;
e = eccentricity of load, in inches;
/ ' = minimum unit stress, in pounds per
square inch, at edge of pier that is
farther from the load and perpendicular
to the direction of eccentricity.
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14 ELEMENTS OF MASONRY DESIGN
If the value of / ' for a particular loading is positive, the stress
throughout the pier is compressive, and only the stress / is of
interest in considering the strength of the pier. However, iff is negative, the resultant stress at the edge of the section
farther from the load is tensile and, since masonry is much
stronger in compression than in tension, the stresses at both
edges of the section are important.
Ex a m p l e. A masonry pier that is 20 in. X 24 in. in cross-section
supports a total resultant load, including its own weight, of 160,000
pounds, which acts midway between the 24-inch faces and at a distance of
9 inches from one of the 20-inch faces, or with an eccentricity of 3 inches
measured in the direction of the 24-inch dimension. Compute the maxi
mum and minimum unit stresses in the pier.
S o l u t i o n . Since the eccentricity is parallel to the 24-inch dimension,
d= 24 in. and 6 = 20 in. Also, P = 160,000 lb. and e= 3 in. Then, by for
mula 1, the maximum unit stress is
, P /, . Ge\ 160,000^ / , . 6X3\ _0 0 ., .
f= Vd { 1 + d) = 20X24 X ( 1 + i r ) = 583 lb - per 0nly Sq- m-By formula 2, the minimum unit stress is
160,000 / 6 X3\
20 X24 X { 24 )= 83 lb. per sq. in.
Ans.
Ans.
In this case, the value off is positive and, therefore, the minimum unit
Stress is compressive.
F i g. 3
18. Under some conditions, the load on a pier may be
eccentric in both directions, as indicated in Fig. 3 (a), where
the rectangle ABCD represents the cross-section of a pier and
the load P has an eccentricitye\ with respect to the center line
parallel to the edge AD and an eccentricity e2 with respect to
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ELEMENTS OF MASONRY DESIGN 15
the center line parallel to AB. The loading in this case is
equivalent to a central load P, a bending moment Pe1with
respect to the center line parallel to the faceAD of the pier, and
a bending moment Pezwith respect to the perpendicular centerline.
The central load would cause a uniformly distributed com-
Ppressive unit stress equal to . Because of the bending
moment Pei, there is a compressive unit stress of -r-jr-at thebal
edge BC and a tensile unit stress of equal amount at the edgeAD. Similarly, the bending moment Pe2produces a compres-
sive unit stress of ,,2at the edge AB and a tensile unit stressdtr
of like amount at the edge CD. The maximum resultant unit
stress in the section evidently occurs at the comer B, where
the stresses due to both bending moments are compressive.
The value of this resultant unit stress is or
in which / = maximum unit stress, in pounds per square inch,
at corner of pier that is nearest to load;
P = eccentric load, in pounds;
5 = one dimension of pier cross-section, in inches;d = other dimension of pier cross-section, in inches;
ei = eccentricity of load, in inches, measured in direc
tion of dimension d;
ei = eccentricity of load, in inches, measured in direc
tion of dimension b.
The minimum resultant unit stressf occurs at the corner
A, where both bending moments produce tension, and thisstress may be computed by the formula
E x a m p l e .A 30-inch square pier supports a total load of 200,000
pounds applied 9 inches from one face and 14 inches from an adjacent
face. Determine the maximum and minimum unit stresses in the pier.
(1)
(2)
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16 ELEMENTS OF MASONRY DESIGN
So l u t i o n. In formulas 1 and 2,P is 200,000 lb. and band dare each
30 in. The distance from any face of the pier to the parallel center line
of the cross-section is 15 in., and the eccentricities o f the load in directions
parallel to the faces of the pier are el = 15 9 = 6 in. and es=15 14 = 1
in. Therefore, the maximum unit stress is
The minimum unit stress is
P / 1_ 6 e ,_ 6 f?\ 200,000 / 6X 6 6X1\
1 bd \ d b ) 30X3 0 X \ 30 30 /
= 89 lb. per sq. in.
and this stress is 89 lb. per sq. in., tension. Ans.
19. Principle of Middle Third.Whenever the eccentricity
ein formula 2, Art. 17, is less than the value of is less thano a
unity and the minimum unit stressf is then positive. Con
versely, when eis greater than the value of ^ is greater than
unity and f is negative. Thus, where the load on a pier is
eccentric in only one direction, the pier will be entirely in com
pression if the eccentricity is less than and will be partly in
d
tension if the eccentricity is greater than A point that is atd ^
a distance of 7-. from the center line of the cross-section is at theo
edge of the middle third of the section, and the principle just
developed may be expressed as follows:
When a load on a rectangular pier is eccentric in only one
direction and lies within the middle third of the cross-section, the
stresses are entirely compressive; if the load is applied outside themiddle third, there is tension near the face of the pier that is
farther from the load and perpendicular to the direction of the
eccentricity.
= 533 lb. per sq. in., compression. Ans.
20. Kern. In case the load on a pier is eccentric in both
directions, there will be no tension in the pier if the load lies
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ELEMENTS OF MASONRY DESIGN 17
within the so-called kern of the section, which is a certain area
near the axis. For the rectangular section A BCD in Fig. 3
(6), the kern is the shaded rhombusEFGH,
whose vertexesE
and Gare at the ends of the middle third of the center line that
is parallel to the face AB,and whose vertexes FandH are at
the ends of the middle third of the perpendicular center line.
If the load acting on the section is applied at any point within
the kern, the entire section will be in compression; but, if the
load is applied outside the kern, there will be tension in some
parts of the section.
21. Limitation of Tensile Stress. The tensile strength of
the masonry is usually neglected in the design of masonry
members, such as piers, that are intended primarily to resist
compressive stresses. Therefore, it is usually considered good
practice to have the resultant load on a masonry pier lie within
the kern of the cross-section.
In the case of a rectangular pier carrying a load that is
eccentric in only one direction, the load should preferably be
applied within the middle third of the cross-section. When
such a load is applied exactly at the edge of the middle third,
the eccentricity e is equal to J d. If this value for eis substi
tuted in formula 2, Art. 17, it is found that the minimum unit
stress / ' is equal to zero; also, from formula 1, Art. 17, the
maximum unit stress is
Hence, the maximum unit stress for a load acting at the edge
of the middle third is just twice as great as the stress ior an
equal central load.
22. Design of Eccentrically Loaded Pier.The usual procedure in the design of a masonry pier that is to carry an eccen
tric load is to assume a probable section and then to investigate
its suitability and, if necessary, to make the required changes
in the assumed dimensions. If it is desired to avoid tension in
the pier, the load should lie within the middle third of the
section, and the dimension that is in the direction of the eccen
tricity should not be less than six times the eccentricity. Experi
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18 ELEMENTS OF MASONRY DESIGN
ence in masonry design and familiarity with designs used under
similar conditions are valuable aids in selecting the first section
to be investigated. However, when information is not avail
able as a guide, it is best to compute first the required area
for a central load and to assume for the first investigation a
section of somewhat larger area, as illustrated in the following
examples.
Ex a mpl e 1.A square pier o f limestone ashlar laid in portland-
cement mortar is 12 feet high and is to support a superimposed load of
200,000 pounds applied 4 inches from the axis of the pier and on a center
line perpendicular to one face of the pier. Determine the required width,
so that there will be no tension in the masonry.
So l u t io n .As stated in Art. 13, the width of an ashlar pier should
not be less than one-tenth of the height, or ~ X 1 2 = 1.2 ft., say 15 in.
Also, to avoid tension in the masonry, the width should not be less than
six times the eccentricity, or 6X 4 = 24 in.
The allowable unit stress, exclusive of that due to the weight of the
pier, is 500 j jX12X168 = 480 lb. per sq. in. and the required area
for a central load of 200,000 lb. is
A P
f
200,000
486= 412 sq. in.
For a load applied at the edge of the middle third of the section, the
maximum unit stress is twice the average value; therefore, in the limiting
case for no tension in the pier, the required area is 2X 412 = 824 sq. in.,which can be provided by a 29 "X 29" section. As the 29-in. width of this
section is somewhat greater than the width of 24 in., for which the load
would be applied at the edge of the middle third, the required area will
be a little less than 824 sq. in. and a 28 "X 28" pier will be tried. From
formula 1, Art. 17, the maximum unit stress for this section would be
, f 6e\ 200,000 / 1 . 6X4\
f ~b d ( 1 + j ) = T8X28 X ( 1 + ^ 8 - ) = 474 lb< per Sq- m-
h 12X12Also, -r= jrs = 5.1, which is less than 6, and no reduction in the allow-b 28
able unit stress of 486 lb. per sq. in. need be made. It will be found that
a pier with a section 27 in. square is too small, and a 28-in. square pier
would be used. Ans.
Ex a m p l e 2.A pier of brick laid in cement-lime mortar is to be 16
feet high and 32 inches wide. If it is to carry a vertical load of 60,000
pounds applied 2 inches from its axis, the eccentricity being measured at
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ELEMENTS OF MASONRY DESIGN 19
right angles to the 32-inch dimension, what should be the other dimension,
or thickness, of the pier?
So l u t i o n.The reduced allowable unit stress in this case is 140r^
X16X120 = 14013 = 127 lb. per sq. in. For a central load o f 60,000 lb.
and a pier for which ^ is less than 6, the required cross-sectional area of
the pier would be p
A = 7 = ' X27~ = 472 sq. in.
If 50 per cent is added as an allowance for the eccentricity of the load
and the probable reduction in the allowable unit stress because the height
may exceed six times the least width, the estimated area would be 1.5X472
= 708 sq. in. The thickness corresponding to this area is 708-^32=22.1,
and a tentative dimension of 24 in. will be tried. For this dimension,
h 16X12^ = = 8, or more than 6; hence, the allowable unit stress should be
reduced in accordance with the formula of Art. 11. Thus,
/ = / , ( l . 2 5 - i 0 =140X ( l . 2 5 - ^ X 8 ) =119 lb. per sq. in.
and the safe unit stress, exclusive of the stress due to the weight of the
pier, is 11913 = 106 lb. per sq. in. Also, by formula 1, Art. 17, in which
P = 60,000 lb., 6 = 32 in., d = 24 in., and e = 2 in., the actual unit stress due
to the superimposed load is
, F / . , 6 e \ _ 60 ,0 00 ^ / , , 6 X 2 \
* bd \1 + d ) 32X24X ( 1 + 24 / 117 lb- Per s
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20 ELEMENTS OF MASONRY DESIGN
MASONRY WALLS
WALLS OF UNIFORM CROSS-SECTION
MINIMUM ALLOWABLE THICKNESSES OF BUILDING WALLS
23. Introduction. The thickness of a masonry wall
should in any case be sufficient to resist safely the stresses dueto the dead and live loads on the structure. The exterior
walls of buildings must also provide resistance to fire and must
not be too thin in proportion to the height. Minimum allow
able thicknesses of masonry walls under various conditions are
specified by the building codes of cities and of organizations
such as the United States Department of Commerce and the
National Board of Fire Underwriters. The requirements ofthe various codes are fairly similar, although there are some
notable differences. Under ordinary circumstances, the mini
mum allowable thickness of a wall can be adopted with safety,
but it may sometimes be necessary to use a greater thickness
in order to bring the unit stresses below the permissible limits.
The recommendations in the following articles are based
largely on a report of the Building Code Committee of theDepartment of Commerce entitled Recommended Minimum
Requirements for Masonry Wall Construction.
24. Exterior and Party Walls in Skeleton Construction.
From an engineering standpoint, buildings may be divided into
two types, known as theskeleton typeand the wall-bearing type.
In skeleton construction, the columns and the floors and roof
form a self-sustaining framework, or skeleton, the floors andthe roof being supported by the columns; and the walls and
other parts of the structure are supported by the skeleton.
Thus, the outside walls serve merely to enclose the building,
and are supported at each floor by beams running between
columns. Where there is a party wall, which is a wall between
two buildings and adapted for the joint use of both, it is like
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ELEMENTS OF MASONRY DESIGN 21
wise supported by beams and columns that are common to the
two buildings. Exterior and party walls are generally of solid
brickwork but, in buildings with reinforced-concrete skeletons,the enclosure walls are sometimes of reinforced concrete.
Exterior walls of unreinforced masonry in skeleton construc
tion should be at least 8 inches thick. In the case of a party
wall, the minimum thickness should be 12inches.
25. Bearing and Non-Bearing Walls in Wall-Bearing Con
struction. In wall-bearing construction, the walls rest on
suitable foundations, and the floors and roof are supportedeither entirely by the walls or by the walls and interior columns.
The walls are usually of brick, but in buildings that are only
one or two stories high, hollow blocks are often used in the
walls. Party walls of brick masonry should be laid in cement
mortar or cement-lime mortar. Wall-bearing construction is
not recommended for buildings of more than ten or twelve
stories.Even in wall-bearing construction, a distinction must be
made between so-called bearing walls and non-bearing walls.
A bearing wall is one that supports a vertical load in addition
to its own weight. Thus, in dwellings, the floor and roof loads
are generally carried to the side walls by closely-spaced wooden
joists. Also, in structures such as armories and gymnasiums,
the walls frequently support balconies and roof trusses. Nonbearing walls, which support only their own weight, are used
along the ends of buildings. These walls run parallel to the
floor joists and do not receive any load from the joists.
26. Thickness of Exterior and Party Walls in Wall-Bearing
Construction.Recommended minimum thicknesses for high
exterior and party walls of masonry in buildings of the wall
bearing type are indicated in Fig. 4. In view (a) is represented
an outside bearing wall; in view (b), a party wall; and in view
(c), an outside non-bearing wall. As shown in the illustration,
the thickness of a bearing walleither outside or party
should be not less than 12 inches for the uppermost 35 feet of
its height and should be increased by 4 inches for each successive
35 feet or fraction thereof measured downwards from the top
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22 ELEMENTS OF MASONRY DESIGN
of the wall. For a non-bearing wall, the thickness may be 4
inches less than the minimum recommended value for a bearing
wall, provided that it is not less than 12inches.The thickness of a masonry wall should not be less than one-
twentieth of the height between successive floors or other
substantial lateral supports, unless the wall is reinforced by
F i g. 4
cross-walls, piers, or buttresses at intervals not exceeding
twenty times the thickness. Where an exterior bearing wall
or a party wall is stiffened, at intervals of not more than 12feet, by cross-walls or by projections at least 2 feet deep, the
thickness need be only 12 inches for the uppermost 70 feet of
the height and should be increased by 4 inches for each succes
sive 70 feet or fraction thereof.
The preceding recommendations apply to most walls of solid
masonry and, in general, also to walls of hollow masonry.
However, the height of a hollow wall above the top of the
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ELEMENTS OF MASONRY DESIGN 23
foundation wall should never exceed 50 feet, and the thickness
of such a wall should not be less than one-eighteenth of the
vertical or horizontal distance between substantial lateral sup
ports. Also, walls of rubble masonry should be 4 inchesthicker than similar walls of other classes of solid masonry and
should never be less than 16 inches thick.
Each change in the thickness of a wall should be made at a
floor level and not between floors. Where a change in wall
thickness would theoretically occur between floor levels, the
greater thickness should be continued to the higher floor.
27. Exterior bearing and non-bearing walls or party walls
in wall-bearing construction may be only 8 inches thick under
the following conditions: (1) When they are in the top story
of a building not exceeding three stories or 40 feet in height or
in a one-story commercial or industrial building, provided the
unsupported height of the 8-inch wall does not exceed 12 feet
and the roof beams are horizontal. (2) When they are in a
one- or two-family dwelling and are either solid walls not more
than 30 feet high or hollow walls not more than 20 feet high.
If such a dwelling has a gable wall, which comes to a point at
the roof, the portion within 5 feet of the peak need not be
included in determining the height.
28. Foundation Walls. The foundation walls support the
vertical loads from the structure resting on them and usuallyalso resist lateral earth pressure. Thus, they generally act as
both bearing and retaining walls. They must, therefore,
possess adequate strength and stability and must be at least as
thick as the walls they support.
Foundation walls are usually of plain or reinforced concrete,
stone, or brick, but sometimes they are of hollow masonry. A
foundation wall of rubble stone should be not less than 18inches thick. If the wall is of concrete, brick, coursed stone,
or hollow masonry, its thickness should be not less than 12
inches. However, where the foundation walls are built in
trenches and the material enclosed by them is not excavated
to provide a basement, a minimum thickness of 8 inches may
be used if the entire wall including the foundation wall is
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24 ELEMENTS OF MASONRY DESIGN
not over 30 feet high for solid masonry or 20 feet high for hollow
masonry. Foundation walls of masonry should be laid in
cement mortar. Also, all foundation walls should extend
below the level of frost action.
29. Fire Walls and Partitions.A fire wall is a wall that
subdivides a building in order to restrict the spread of fire and
that extends continuously from the foundation to and above
the roof. Such walls should preferably be of solid brick
masonry, laid in cement mortar or cement-lime mortar, or of
reinforced concrete. When a fire wall is constructed of brick,
the requirements in regard to minimum thickness are the same
as those for party walls in wall-bearing construction, except
that in residential buildings a brick fire wall may be 8 inches
thick for the uppermost 20 feet of its height. Fire walls of
hollow masonry, other than hollow brick walls, should be not
less than 16 inches thick in any part, but a minimum thickness
of 12 inches may be used for all such walls in a residentialbuilding or for hollow walls of brick in all buildings.
Partitions may be used in the various stories of a building
merely to subdivide the floor space. Often they are provided
to restrict the spread of fire, and are then called fire partitions
or fire division walls. The thickness of fire partitions of solid
brick should not be less than 8inches. If such partitions are
of hollow masonry, they should never be less than 12 inchesthick. In buildings used for heavy manufacturing, the mini
mum thickness of fire partitions should be 12inches if of hollow
brick construction and 16 inches if of other types of hollow
masonry.
A partition that serves as a bearing wall, but not as a fire
division wall, should have a thickness of at least one-eighteenth
of the height between floors or floor beams. Non-bearing
partitions that do not serve as fire division walls are usually
made of hollow masonry. The thickness, exclusive of plaster,
should be not less than 2 inches for an unsupported height up
to 8feet, 3 inches for a height of 12 feet, 4 inches for a height
of 15 feet, 6 inches for a height of 20 feet, and 8 inches for a
height of 25 feet.
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ELEMENTS OP MASONRY DESIGN 25
30. Parapets. Exterior walls, party walls, and fire walls
of masonry should be carried above the roof wherever the roof
is not of fireproof construction and the building is within about
50 feet of other buildings. The portion of the wall projectingabove the roof is known as a parapet. In buildings in which
8-inch walls are permitted, the parapet should be at least 8
inches thick and 2 feet high. In other buildings, the parapet
should be not less than 12 inches thick and 3 feet high.
INVESTIGATION AND DESIGN
Roof400
600
600% jeL
600* -j / g f
6,080
600 -c
3,ZOO
31. Walls of Uniform Section With Uniform Vertical Loads.If masonry walls are built in conformity with the minimum
requirements specified in
the preceding articles, the
maximum compressive unit
stresses produced by the
actual loads will usually be
well within the allowablelimits. However, in some
cases, it may be necessary
to investigate the stresses
developed in a wall or to
design a wall to withstand
a certain loading.
The simplest type of wall
is one that is of uniform
cross-section and carries
uniformly-distributed loads.
Since all portions of the
wall are then alike, a foot
length of the wall is usually
considered in investigation
or design. In solving a
particular problem, it is first
necessary to determine the amount and position of the resultant
load and then to proceed as described for a pier.
In order that a wall may be considered safe, the unit stress
at any point must not exceed the allowable value for the kind
t
!6
0>)
F i g. 5
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26 ELEMENTS OF MASONRY DESIGN
of masonry used in the wall. Also, if tension in the masonry
is not permitted, the resultant of the loads must lie within the
middle third of the wall. For the purpose of investigation or
design, the load from closely spaced joists may be assumed
to be uniformly distributed.
Ex a mp l e. In Fig. 5 (a) is shown a proposed section of an outside
bearing wall of a dwelling. The wall is to be of brick laid in portland-
cement'mortar and is to support its own weight and joists from five floors
and a roof; the basement floor is to rest directly on the soil. The load
per linear foot from each floor, which may be considered uniform for the
entire length of the wall, is 600 pounds, and from the roof 400 pounds.
Each of these loads is applied 4 inches from the inside face of the wall.If the wind and earth pressures are disregarded, is the wall thick enough
just above the footing?
So l u t io n . In this case, a linear foot of wall is considered. It is
necessary first to find the weight of the various parts of the wall and then
to determine the amount and the line of action of the resultant of all the
loads. Since the brickwork weighs 120 lb. per cu. ft., the weights of the
parts of the wall per lin. ft. are as follows;
1912" wall, 2 7 X ^ X 1 2 0 = 3,240 lb.
16" wall, 38X j ^ X 120 = 6,080 lb
20" wall, 16 xf^ X 12 0 = 3,200 lb.
The positions of the loads acting on the wall are shown in Fig. 5 (6).
The resultant load and its distance from the outside face o f the 20-in. wall
are computed in the following manner:
LoadWeight, Arm, Moment,
in Lb. in In. in In.-Lb.
R oo f..................................... 400 8 3,200
5th floor.............................. 600 8 4,800
12" wall............................... 3,240 6 19,440
4th, 3d, 2d floors.............. 1,800 12 21,600
16" wall.............................. 6,080 8 48,640
1st floor..............................
600 16 9,600
20" w a ll ............................. 3,200 10 32,000
Total........................... 15,920 139,280
Hence, the resultant load, which amounts to 15,920 lb., acts at a distance
from the outside face of the wall equal to 139,280+15,920 = 8.75 in. The
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ELEMENTS OF MASONRY DESIGN 27
20eccentricity of the load is 8.75 = 1.25 in. and by formula 1, Art. 17,
the maximum unit pressure is
15,920 / 6 X 1 .25\
1 2 X2 0 * \ + 20 )
= 91.2 lb. per sq. in.
This is well below the allowable value of 175 lb. per sq. in., but the thick
nesses of the wall cannot be reduced because they are the smallest that
are permitted by the requirements in Fig. 4 (a).
32. Effect of Lateral Thrust.When a wall is subjected
to lateral thrust, as from wind or earth pressure, the resultant
load on the wall is inclined. In this case, it is most convenient
to determine first the amount and the line of action of the
resultant of the vertical loads only and
then to combine this resultant with
the horizontal or inclined force repre
senting the lateral pressure. The
resultant of the vertical forces is usu
ally found most easily by calculation,
but the graphic method is convenientfor locating the point where the resul
tant of all the forces cuts the base of
the wall. However, the horizontal
distance from any point on the base
to the resultant may also be found by
dividing the algebraic sum of the
moments of all the forces about that point by the vertical
component of the resultant.
When the resultant force acting on a wall is inclined, the
horizontal component tends to overturn the wall and to cause
the upper part of the wall to slide on the lower part, whereas
the vertical component resists overturning and sliding. The
conditions are indicated diagrammatically in Fig. 6, in which
Pv represents the vertical component of the resultant and Ph
represents the horizontal component. The overturning moment
about the edge a of the base of the wall is PhXab, and the
resisting moment isPvXac. If the resisting moment is greater
than the overturning moment, the wall is stable against over
turning; on the other hand, if the overturning moment exceeds
F i g. 6
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28 ELEMENTS OF MASONRY DESIGN
the resisting moment, the wall is unstable. However, the
factor of safety against overturning should preferably be at
least 2 ; that is, the quotient obtained by dividing the resisting
moment by the overturning moment should not be less than 2.
33. If a building wall were designed so that the vertical
loads acting on it would provide sufficient resistance to over
turning by wind and earth pressure, an excessive thickness of
wall would be required because the arm of the horizontal wind
pressure about the base of the wall is much greater than the
arm of the vertical load about the outer edge of the base.However, considerable lateral resistance to wind pressure is
furnished by the floors of the building, and it is customary to
make allowance for this resistance in designing the wall. Since
it is not possible to estimate even roughly the amount of the
resistance supplied by a particular floor, the required thickness
of the wall must be based on experience and empirical rules.
It may be taken for granted that exterior walls built in accordance with the requirements in Arts. 23 to 28, inclusive, will be
stable against overturning by external lateral pressure.
34. A wall will not slide if the ratio of the horizontal com
ponent of the resultant force acting on the wall to the vertical
component does not exceed the coefficient of friction for two
blocks of the material of which the wall is composed. The
value of this coefficient for any kind of masonry used in a wallmay be taken as 0.65. However, the factor of safety against
sliding should be about 1.5. Thus, if the product of the verti
cal component of the resultant force and the coefficient of
friction is divided by the horizontal component of the resultant,
the quotient should be at least 1.5. Even if the lateral resist
ance of the floors is neglected, a building wall that meets the
minimum requirements for stability against overturning willbe amply safe against sliding.
35. Distribution of Concentrated Loads.Loads are often
carried to walls by beams, girders, or trusses spaced relatively
far apart. Such loads are gradually distributed over the
masonry below, but the pressure directly under them is much
greater than elsewhere. Where the portion of the bottom of
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ELEMENTS OF MASONRY DESIGN 29
the beam or trass which bears on the wall is so small that the
unit pressure on the masonry would exceed the safe strength
of the material, a steel bearing plate must be provided under
the member to spread the load over a larger area of masonry.
In the investigation or design of a wall of
uniform cross-section that carries concen
trated loads, it is necessary to consider a
length of wall equal to the distance between
loads, but otherwise the procedure is similar
to that explained for uniform loads.
Ex a mpl e.A section through a proposed brick
wall for a storage warehouse is shown in Pig. 7.
The floor beams and roof beams are spaced 6 feet
apart horizontally; the load from each roof beam is
4,800 pounds and is assumed to be applied at a dis
tance of 3 inches from the inside face of the wall;
the load from each beam of the sixth, fifth, fourth,
third, and second floors is 18,000 pounds and is
applied 4 inches from the inside face of the wall;
and the load from each first-floor beam is 20,000
pounds and is applied 4 inches from the inside face
of the wall. What is the maximum unit pressure in
the wall just above the footing, for the vertical
loads alone?
So l u t io n .T he calculations for determining
the resultant load for a 6-ft. length of wall and its '
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30 ELEMENTS OF MASONRY DESIGN
The distance from the outside face of the 24-in. wall to the line o f action o f
the resultant load is 2,364,200-5-203,600 = 11.6 in., and the distance from
24the center o f the base of that wall to the resultant load is -jp 11.6 = 0.4
in. Hence, by formula 1, Art. 17, the maximum unit pressure is
203,600 w / 6X0.4\
72 X24 X l + 24 )= 130 lb. per sq. in. Ans.
This is well below the usual allowable value of 175 lb. per sq. in.
EXAMPLES FOR PRACTICE
1. An exterior bearing wall in an apartment house is of brickwork
and has a cross-section similar to the upper 121 feet of the wall shown in
Fig. 4 (a). The 12-inch, 16-inch, and 20-inch thicknesses each extend
for a height of 35 feet, as in the illustration, and the height of the 24-inch
portion is 16 feet. The load from the top tier of joists is treated as a
uniform load of 500 pounds per linear foot, and that from each of the lower
tiers of joists is a uniform load of 800 pounds per linear foot. The loads
from the three upper tiers of joists are applied 4 inches from the inside
face of the 12-inch portion of the wall; the loads from the next three tiers
are applied 4 inches from the inside face of the 16-inch portion; the loadsfrom the next three tiers, 4 inches from the inside face of the 20-inch
portion; and the load from one tier, 4 inches from the inside face of the
24-inch portion. What is the distance from the outside face of the wall
to the line of action of the resultant of the vertical loads acting on the
wall above the base of the 24-inch portion? Ans. 10.1 in.
2. Compute the maximum unit pressure in the wall of the preceding
example at the base of the 24-inch portion . Ans. 145 lb. per sq. in.
WALLS WITH PROJECTIONS
TYPES OF PROJECTIONS
36. Pilasters.When a concentrated load is applied to a
masonry wall, the thickness of the wall is often increased for
a short distance on each side of the load, as shown in the plan
in Fig. 8 at a. If the load is vertical, the purpose of the additional masonry is merely to increase the bearing area of the
wall directly under the load. These projections, which are
called pilasters, may be placed on the inside of the wall, on the
outside, or on both the inside and outside of the wall at each
load. Brick or stone pilasters should be well bonded to the
wall so that all the masonry will act as a unit in supporting the
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ELEMENTS OF MASONRY DESIGN 31
load. The width b of a pilaster should be between one-sixth
and one-third of its length c.
A pilaster and the part of the wall opposite the pilaster are
assumed to act together in supporting the concen
trated load. Thus, in the illustration, the pilaster a
and the portion of the wall between the dotted lines
are treated as a unit in computing the unit stress
produced by a given load or in determining the
dimensions band cthat are required for the support
of a certain load. The method of procedure in
either investigation or design is similar to that for a
masonry pier.
37. Buttresses. When a wall is required to sustain a
horizontal or an inclined load, such as the reaction of a roof
truss exposed to the wind, a heavy masonry projection is
provided to help prevent the wall from overturning. Such a
projection is usually located on the side of the wall toward
which the horizontal component of the load acts. The pro
jection is then subjected to compressive stress and is known as
a buttress. In order that buttresses will act in unison with the
rest of the wall, they must be well bonded into the wall.
38. Counterforts.When a projection that is provided to
help resist a horizontal or an inclined load is placed on the side
of the wall away from which the horizontal component of theload acts, the masonry in the projection is subjected to tension.
The projection is then called a counterfort. Since plain
masonry is very weak in tension, masonry counterforts must
be reinforced with steel.
STABILITY OF BUTTRESSED WALLS
39. Loads on Buttressed Walls. In Fig. 9 is shown a verti
cal section through a gymnasium, in which the steel roof
trusses rest upon masonry walls. Opposite each truss the
walls a and b are provided with buttresses c and
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32 ELEMENTS OF MASONRY DESIGN
is subjected to a vertical load from the truss and also to a
horizontal thrust from the wind load on the truss. The wind
also exerts a horizontal pressure on the walls, acting on thewall awhen it is blowing from the left and on the wall bwhen
it is blowing from the right. Each buttress carries, in addition
to its own weight and the roof load, a vertical load from the
spectators gallery and from each floor.
In order to determine the stability of the walls and but
tresses, it is necessary to consider several conditions. As the
wind may blow against either side, two independent sets of com
putations must be made. First, it should be assumed that
the wind blows from the right-hand side. The reactions from
both the dead loads and the vertical live loads on the floors
tend to resist overturning, but the live loads do not always
act and are disregarded in determining the location of theresultant of all loads acting on the wall and buttress. How
ever, in determining the maximum compressive unit stress at
the bottom of the buttress, the reactions from the maximum
live loads are included. After both walls a and b have been
investigated with the wind blowing from the right, other sets
of computations are made on the assumption that the wind is
blowing from the left.
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ELEMENTS OF MASONRY DESIGN 33
40. Investigation of Walls With Buttresses.In the investi
gation of the stability of a buttressed wall, it is customary to
assume that the buttress and the portion of the wall directly
opposite the buttress act as a unit, and to neglect entirely any
resistance to overturning, sliding, or stress that may be fur
nished by the remainder of the wall. The problem may be
solved entirely by calculation, but it is usually convenient to
adopt the following method, which is partly graphic.
First, a cross-section of the wall and a buttress is drawn to
a convenient scale, the weight of the masonry resisting over
turning is calculated, and the center of gravity of that mass of
masonry is located. On the cross-section are located accu
rately the lines of action of the weight of the masonry, the
reaction from the roof truss, and any other forces that act on
the portion of the masonry under consideration. The position,
in the cross-section, of the line of action of the resultant force
acting on the masonry is then determined by any suitable
graphic method. If this resultant passes well within the base
of the section, the structure will be safe against overturning.
However, in order to avoid tension in the masonry, the resultant
force should cut the base of the section inside the middle third.
The resistance to sliding and to crushing of the masonry
should also be investigated. To determine the factor of safety
against sliding, the product of the vertical component of the
resultant force and the coefficient of friction for the masonryis divided by the horizontal component of the resultant force.
For safety against crushing, the maximum pressure, as deter
mined by formula 1, Art. 17, must not be greater than the
allowable value for the masonry.
Where the buttress and wall increase in thickness toward the
bottom, and the floor loads are applied at various heights, it
is often necessary to investigate the safety of the wall at eachchange in thickness, as well as at the bottom. In investigating
the safety of the wall at any level, only the forces applied to
the wall above that level are considered.
E x a m p l e .-In Fig. 10 (a) is shown an enlarged section through the
wall aand a buttress cin Fig. 9. The wall and buttress are of limestone
ashlar, and the width of the buttress, measured parallel to the wall, is
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34 ELEMENTS OF MASONRY DESIGN
F ig. 10
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ELEMENTS OF MASONRY DESIGN 35
4 feet below point p , Fig. 10 (a), and 3 feet above that point. The wind
is assumed to be blowing from the right, and the truss reaction for the
combined dead and wind loads has a vertical component of 54,000 pounds
and a horizontal component of 5,500 pounds. The dead load from the
gallery is carried almost entirely by the. truss, and the amount transmitted
directly to the wall may be neglected. The wall carries a dead load of
19,000 pounds from the first floor and an equal load from the locker-room
floor, each of these loads being applied 8inches from the inside face of the
wall. The load on the basement floor is carried directly by the soil. Find
the distance from the outside face of the buttress to the point where the
resultant load cuts (a) the top of the footing and (6) the section x-x at the
first-floor line.
So l u t i o n.(a) In order to determine the weight of the masonry and
the position of the center of gravity, the cross-section may be imagined
to be made up of the divisions numbered 1, 2, 3, 4, 5, 6, 7, 8, and 9 in Fig.
10 (a). In this case, the width of the masonry under consideration is not
uniform for the entire height of the buttress, this width being 4 ft. below
point p and 3 ft. above that point. Therefore, in locating the center of
gravity, it is necessary to use the volumes of masonry in the divisions and
not the areas. These volumes, their respective arms from the inside faceof the lower portion of the wall, and the moments about that face are as
follows:
Division Volume, in Cu. Ft. Arm, in Ft. Moment
1 1.5X6.75X3 = 30.4 2 61
2 1X2.25X3.5X3 = 11.8 3.5 41
3 0.25X2X3 1.5 2.63 4
4 2X 7.5X 3 45.0 1.5 68
5 2.5X8X 3 60.0 3.75 2256 2.5X13X3 97.5 1.25 122
7 4X10.5X3 126.0 4.5 567
8 6.5X2X3 39.0 3.25 127
9 7X23.5X 4 658.0 3.5 2,303
1,069.2, say 3,518, say
1,070 3,520
If the masonry is assumed to weigh 168 lb. per cu. ft., the total weight
of the masonry is 1,070X168 = 179,800 lb. The distance from the insideface of the wall to the center of gravity o f the wall and buttress is 3,520-f-
1,070 = 3.28 ft., or 3 ft. 3 in.
The forces acting on the buttress above the footing are shown in Fig.
10 (a). In order to present clearly the graphic method of locating the line
of action of the resultant of these loads, the outline of the cross-section
is here redrawn in view (5), but in practice it would not be necessary to
redraw the section. The first step is to locate the point a. where the
reaction from the truss is applied, and to establish graphically the line of
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36 ELEMENTS OF MASONRY DESIGN
action ab of the resultant of the two components of that reaction. This
is done by laying off, to some convenient scale, the length ac vertically
downwards to represent 54,000 lb. and the length cb horizontally toward
the left to represent 5,500 lb., and then joining aand bby a straight line.
The next step is to locate the line of action of the resultant deo f the truss
reaction and the vertical loads at the first floor and the locker-room floor.
The line abis prolonged upwards to intersect at d the vertical line repre
senting the common line of action of the two floor loads. The required
line of action de is determined by laying off from d the vertical distance
dfto represent the total load of 38,000 lb. from the two floors, making the
inclined distancef e parallel and equal to ab,and connecting dand eby a
straight line. The last step is to draw a vertical line at a distance o f 3.28ft. from the face of the wall to represent the line of action of the entire
weight of the masonry, which is 179,800 lb., and to extend the line de
until it intersects this vertical line of action at g. The final resultant gh
of all the forces is determined by making the vertical distance giequal to
179,800 lb., making illequal and parallel to de,and then drawing the line
from gto h. The resultant ghcuts the base of the buttress at a distance
of 3 ft. 4J in. from the outside face of the buttress, whereas the edge of
the middle third is 1 x 7 ' 0 " = 2 ft. 4 in. from that face. Ans.
(b) In the investigation for the section x-x, Fig. 10 (a), only divisions
1 to 7, inclusive, of the cross-section are included in determining the weight
of masonry that resists the truss reaction. The volume of this part of
the wall is 372.2, or say 372, cu. ft. and the sum of the moments about the
inside face of the wall is 1,088. Hence, the weight of the masonry
is 372X168 = 62,500 lb., and the distance from the inside face of the wall
to the center of gravity is 1,088-r-372 = 2.92 ft., or 2 ft. 11 in.
The resultant of the truss reaction and the weight of the masonry,which are the only forces acting above the first floor, is then located as in
Fig. 10 (c). The resultant abof the components of the truss reaction is
determined as in view (b). It is then prolonged to intersect the vertical
line of action of the weight of the masonry at j ; and the resultant jk of
the weight of the masonry and the truss reaction is determined by laying
off j l to represent the weight of 62,500 lb., making Ik parallel and equal
to ab,and joining the pointsj and k. This final resultant cuts the section
x-xat m. The distance from the outside face of the buttress to the intersection point m is 2ft. 11| in., whereas the edge of the middle third is at
a distance of J X6' 6" = 2 ft. 2 in. from that outside face. Ans.
41. Design of Buttressed Walls. The procedure in design
ing a buttressed wall consists in assuming a cross-section for
the wall and buttress, and investigating its stability against
overturning and sliding. Usually, it will be unnecessary to
compute the crushing stress, but under heavy loads that stress
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ELEMENTS OF MASONRY DESIGN 37
must also be investigated. In the selection of the trial section,
the required architectural effect will usually govern, but past
experience and good judgment are frequently helpful. If the
trial section is not thick enough, as indicated by the fact thatthe resultant of the forces acting on the wall passes outside the
middle third of the base or the maximum unit pressure exceeds
the allowable value, the wall or buttress should be made
heavier and the new section should be investigated. In case
the resultant lies quite near the center of the wall, and the unit
pressure is much lower than the allowable value, it is advis
able to reduce the size of the wall in order to lower the cost.Several sections may have to be tried and investigated before
the final design is adopted. Thus, in the case of the section
shown in Fig. 10 (a) and investigated in the example of the
preceding article, it is found that the resultant of the forces
acting on the masonry cuts the base of the section very close
to the center, and a smaller buttress would be tried.
EXAMPLE FOR PRACTICE
The buttress shown in section in Fig. 10 (a) and considered in the
example of Art. 40 is altered in the following manner: The thickness of
division 7 is reduced to 3 feet, the thickness of division 8 is made 5 feet 6
inches, and the thickness of division 9 is made 6 feet. Determine the
distance from the outside face of the buttress to the point where the
resultant of all the forces cuts the top of the footing of the new section.
Ans. 2 ft. 7% in.
WALLS WITH OPENINGS
42. Distribution of Pressure. When a wall contains an
opening, as for a window or a door, the weight of the masonry
over the opening and any concentrated or uniform load which
may be applied to that masonry must be carried by the part
of the wall on each side of the opening. Usually, the load
from the masonry over an opening is transmitted to the masonry
alongside the opening by a beam, called a lintel, or by an
arch. Hence, when a wall contains openings, the unit pressure
at a certain level in a portion of the wall alongside an opening
is greater than the pressure at a similar level in a solid wall of
the same material and thickness.
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38 ELEMENTS OF MASONRY DESIGN
The masonry alongside an opening may be treated as a pier.
The load on such a pier consists of the weight of the masonry
directly above it, the weight of the masonry above half of eachadjacent opening, and any concentrated or uniform loads that
may be applied to the masonry whose weight comes on the pier.
(b)F i g. XI
Ex a mp l e. In Fig. 11 (a) and (b) is shown part of the front wall of a
factory building which is built of brick laid in portland-cement mortar.
The wall supports, in addition to its own weight, a uniform load of 1,000
pounds per linear foot from each floor and 300 pounds per linear foot from
the roof. What is the maximum unit pressure on the masonry?
So l u t io n . It is evident that the greatest unit pressure is produced
in any pier between two windows and at the level z-z,which is at the bot
toms of the windows in the first story. The load on one of these piersmay be taken as the sum of the weight of the masonry between the sections
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ELEMENTS OF MASONRY DESIGN 39
x-x and y-y, which weight may be assumed to be centrally applied, and
the roof and floor loads between the same sections. The volume of the
masonry above the levelz-zis as follows:
If the brickwork is assumed to weigh 120 lb. per cu. ft., the weight of the
masonry between the sections x-x and y-y is 491.3X120 = 58,960 lb.Since the roof load is 300 lb. per lin. ft. of wall, the total roof load
between sections x-x and y-y is 300 X8.5 = 2,550 lb. Also, the total load
from the fifth, fourth, third, and second floors is 1,000 X8.5 X 4 = 34,000
]b. The total load on the pier at the level z-zis
As this load is assumed to be centrally applied to the pier, the maximum
unit pressure may be found by formula 2, Art. 14. In this case,P = 95,510
lb., 6= 4.5 ft. or 54 in., and d = 20 in., and the required pressure is
43. Arrangement of Openings.When many openings are
to be provided in a masonry wall, their relative positions
should be determined by taking into consideration not only
convenience and appearance, but also the effect of the openings
on the strength of the wall. Walls with openings are morelikely to be cracked than solid walls, because of the fact that
the unit pressure on the mortar is not uniform at all points at
the same level. Therefore, special precautions should be
taken to have the parts of the wall well bonded. Also, when
the combined width of the openings exceeds one-third of the
total length of the wall, the thickness of the wall between
openings is increased by means of pilasters or buttresses.Whenever possible, the window openings in the different
stories should be placed directly over one another. If it is
absolutely necessary to place an opening under a pier, rigidity
should be attained by the use of a heavy steel beam over the
opening; the beam helps greatly to prevent the masonry from
cracking just above the opening.
Total................... 491.3 cu. ft.
58,960+2,550+34,000 = 95,510 lb.
P 95,510
J bd 54X20= 88.4 lb. per sq. in. Ans.
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40 ELEMENTS OF MASONRY DESIGN
MASONRY BEAMS
GENERAL FORMULAS
44. Usual Types of Masonry Beams.A masonry beam is
practically always either a single block of stone or a plain-
concrete member. Since neither stone nor plain concrete isable to resist shocks, masonry beams should not be used for
supporting live loads. In fact,
the use of masonry beams is
usually confined to the construc
tion of corbels and lintels. A
corbel is a short cantilever beam
that is partly embedded in a wall
and projects beyond the face of
the wall below to provide support
for an eccentric load, such as the
load from an overhanging wall
for a bay window or from the
end of a beam, girder, truss, orarch. In Fig. 12, the stone
corbel a supports the end of the
beam b. A masonry lintel may
be used over an opening where the span is short and the load
is light. A stone lintel over a window is shown in Fig. 13.
Masonry beams are almost always rectangular in cross-
section. However, the exposed end of a corbel is frequentlymade ornamental for the sake of appearance. When a beam
is of stratified stone, it should be set with the layers vertical
or on edge, as indicated in Fig. 13 at a.
45. Formulas for Investigation of Rectangular Beams.
For the usual masonry beams, which are rectangular in cross-
section at the points of maximum shear and maximum bendingmoment, it is simply necessary to apply the fundamental
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ELEMENTS OF MASONRY DESIGN 41
formulas for rectangular beams. Thus, if bdenotes the width
and d the depth of the cross-section, both expressed in inches,
bd?
the section modulus, in inches3, is S = . Also, if / i represents
the allowable flexural unit stress, in pounds per square inch,
the resisting moment Mi of the section, in inch-pounds, may
be found by the flexure formulaMx=fiS or
Mi = (1)
When it is required to find the load that can be carried
safely by a given masonry beam, the first step is to compute
the allowable bending moment Mi by applying formula 1.
Then the expression for the bending moment in terms of the
unknown load is equated to the allowable bending moment,
and the allowable load is calculated by solving that equation.
The shearing unit stress for the loading that develops the
safe resisting moment of the beam will generally be well below
the allowable value. However, in case there is doubt con
cerning the shearing strength of the beam, the maximum
shearing unit stress should be calculated by the formula
3 V
V 2 bd (2)
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42 ELEMENTS OF MASONRY DESIGN
in which v = shearing unit stress, in pounds per square inch;
V=maximum shear, in pounds.
When the value of v that is found by formula 2 is less than
the allowable shearing unit stress vu the load determined bythe bending moment is satisfactory. However, if that value
of v is greater than the safe stress vuthe safe load is calculated
as follows: First, the allowable total shear Fi is found by the
formula
Fi= vibd (3)
Then, the expression for the shear in terms of the unknown
load is equated to the allowable shear, and the required load
is determined by solving the equation thus obtained.
Ex a mp l e . What uniformly distributed load, in addition to its own
weight, can be safely carried by a granite beam that is 8inches wide and
11inches deep and is simply supported on a span of S feet?
So l u t i o n . From Table II, the allowable flexural unit stressfi is 150
lb. per sq. in. Then, by formula 1, the allowable bending moment is
= 1)0X | X .11L 24 ,200 in.-lb.6 6
The bending moment due to a uniform load of tolb. per ft. on a span lo f
5 ft. iswP t oX 52 25 to,, ,, . ,,
M = -g - = g = g - ft.- lb., or 37.5 tom.-lb.
Then, 37.5 to= 24,200
and to= 645 lb. per ft.
From Table I, granite weighs 168 lb. per cu. ft. Hence, the weight
8X11of the beam itself is X 168 = 103 lb. per ft., and the allowable super
imposed load is 645 103 = 542 lb. per ft.
The maximum shear is V=\wl = JX645X 5 = 1,613 lb. and the shear
ing unit stress is, by formula 2,
3 V 3 1,613
v ~ 2 M = 2X 8 x n =27 '5 lb- per Sq- ln
which is much below the allowable value of 200 lb. per sq. in. Hence,
the safe superimposed load is 542 lb. per ft. Ans.
46. Sometimes, it is desired to determine whether a given
beam of width b and depth d can safely support a specified
loading. In such a case, the first step is to compute the maxi-
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ELEMENTS OF MASONRY DESIGN 43
mum bending moment Mand the maximum shear Vthat will
be produced by the load. Then, the maximum flexural unit
stress / may be found by the formula
and the maximum shearing unit stress vmay be calculated by
formula 2 of the preceding article. The beam is safe if the
values of / and vthus obtained are less than the allowable unit
stresses in flexure and shear, respectively.
47. Formulas for Design of Rectangular Beams. In
designing a beam to carry a specified load, the first step is to
compute the maximum bending moment M. When a masonry
beam is used, the width bis generally fixed by the type of con
struction, and the allowable flexural unit stress /i is specified.
Then, the required depth dmay be found by the formula
When the shearing strength of the beam is in doubt, it
should be investigated by computing the maximum shear V,
substituting the computed value in formula 2, Art. 45, andcomparing the shearing unit stress v thus found with the
allowable shearing unit stress ?>i. If v is less than vu the pre
viously computed value of d may be adopted. On the other
hand, if v exceeds vu a larger beam is needed. The required
depth may then be found by the formula
Ex a m p l e . A sandstone beam 12 inches wide is to be subjected to a
bending moment of 5,000 inch-pounds. What depth is required to resist
this bending moment?
So l u t io n . Here, M=5,000 in.-lb., /i = 75 lb. per sq. in., and 6 = 12
in. Then, by formula 1,
(1)
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44 ELEMENTS OF MASONRY DESIGN
CORBELS
48. Forces Acting on Corbel.The corbel a, in Fig. 12,
serves to support the load P from the beam b. The overturn
ing of the corbel is prevented by the weight Wof the masonry
above it. Ordinarily, this weight may be assumed to be con
centrated at the center of the wall. The downward forces P
and Ware balanced by the upward reaction R,which is equal
and opposite to the resultant of P and W. In order to avoid
tension in the bed joint between the under surface of the cor
bel and the wall, the load W should be such that the reaction
Rwill lie within the middle third of the joint. Since the face
of the wall is liable to spall directly beneath the corbel, it is
customary to assume that the effective edge of the joint is \
inch from the face of the wall, as ind