Low Cost Housing is a new concept which deals with effective budgeting and following of techniques which help in reducing the cost construction through the use of locally available materials along with improved skills and technology without sacrificing the strength, performance and life of the structure.There is huge misconception that low cost housing is suitable for only sub standard works and they are constructed by utilizing cheap building materials of low quality.The fact is that Low cost housing is done by proper management of resources.Economy is also achieved by postponing finishing works or implementing them in phases.
Building Cost
The building construction cost can be divided into two parts namely:
Building material cost : 65 to 70 %
Labour cost : 65 to 70 %
Now in low cost housing, building material cost is less because we make use of the locally available materials and also the labour cost can be reduced by properly making the time schedule of our work. Cost of reduction is achieved by selection of more efficient material or by an improved design.
Areas from where cost can be reduced are:-
1) Reduce plinth area by using thinner wall concept.Ex.15 cms thick solid concrete block wall.
2) Use locally available material in an innovative form like soil cement blocks in place of burnt brick.
3) Use energy efficiency materials which consumes less energy like concrete block in place of burnt brick.
4) Use environmentally friendly materials which are substitute for conventional building components like use R.C.C. Door and window frames in place of wooden frames.
5) Preplan every component of a house and rationalize the design procedure for reducing the size of the component in the building.
6) By planning each and every component of a house the wastage of materials due to demolition of the unplanned component of the house can be avoided.
7) Each component of the house shall be checked whether if it’s necessary, if it is not necessary, then that component should not be used.
Cost reduction through adhoc methods
Foundation
Normally the foundation cost comes to about 10 to 15% of the total building and usually foundation depth of 3 to 4 ft. is adopted for single or double store building and also the concrete bed of 6″(15 Cms.) is used for the foundation which could be avoided.
It is recommended to adopt a foundation depth of 2 ft.(0.6m) for normal soil like gravely soil, red soils etc., and use the uncoursed rubble masonry with the bond stones and good packing. Similarly the foundation width is rationalized to 2 ft.(0.6m).To avoid cracks formation in foundation the masonry shall be thoroughly packed with cement mortar of 1:8 boulders and bond stones at regular intervals.
It is further suggested adopt arch foundation in ordinary soil for effecting reduction in construction cost up to 40%.This kind of foundation will help in bridging the loose pockets of soil which occurs along the foundation.
In the case black cotton and other soft soils it is recommend to use under ream pile foundation which saves about 20 to 25% in cost over the conventional method of construction.
Plinth
It is suggested to adopt 1 ft. height above ground level for the plinth and may be constructed with a cement mortar of 1:6. The plinth slab of 4 to 6″ which is normally adopted can be avoided and in its place brick on edge can be used for reducing the cost. By adopting this procedure the cost of plinth foundation can be reduced by about 35 to 50%.It is necessary to take precaution of providing impervious blanket like concrete slabs or stone slabs all round the building for enabling to reduce erosion of soil and thereby avoiding exposure of foundation surface and crack formation.
Walling
Wall thickness of 6 to 9″ is recommended for adoption in the construction of walls all-round the building and 41/2 ” for inside walls. It is suggested to use burnt bricks which are immersed in water for 24 hours and then shall be used for the walls
Rat – trap bond wall
It is a cavity wall construction with added advantage of thermal comfort and reduction in the quantity of bricks required for masonry work. By adopting this method of bonding of brick masonry compared to traditional English or Flemish bond masonry, it is possible to reduce in the material cost of bricks by 25% and about 10to 15% in the masonry cost. By adopting rat-trap bond method one can create aesthetically pleasing wall surface and plastering can be avoided.
Concrete block walling
In view of high energy consumption by burnt brick it is suggested to use concrete block (block hollow and solid) which consumes about only 1/3 of the energy of the burnt bricks in its production. By using concrete block masonry the wall thickness can be reduced from 20 cms to 15 Cms. Concrete block masonry saves mortar consumption, speedy construction of wall resulting in higher output of labour, plastering can be avoided thereby an overall saving of 10 to 25% can be achieved.
Soil cement block technology
It is an alternative method of construction of walls using soil cement blocks in place of burnt bricks masonry. It is an energy efficient method of construction where soil mixed with 5% and above cement and pressed in hand operated machine and cured well and then used in the masonry. This masonry doesn’t require plastering on both sides of the wall. The overall economy that could be achieved with the soil cement technology is about 15 to 20% compared to conventional method of construction.
Doors and windows
It is suggested not to use wood for doors and windows and in its place concrete or steel section frames shall be used for achieving saving in cost up to 30 to 40%.Similiarly for shutters commercially available block boards, fibre or wooden practical boards etc., shall be used for reducing the cost by about 25%.By adopting brick jelly work and precast components effective ventilation could be provided to the building and also the construction cost could be saved up to 50% over the window components.
Lintals and Chajjas
The traditional R.C.C. lintels which are costly can be replaced by brick arches for small spans and save construction cost up to 30 to 40% over the traditional method of construction. By adopting arches of different shapes a good architectural pleasing appearance can be given to the external wall surfaces of the brick masonry.
Roofing
Normally 5″(12.5 cms) thick R.C.C. slabs is used for roofing of residential buildings. By adopting rationally designed insitu construction practices like filler slab and precast elements the construction cost of roofing can be reduced by about 20 to 25%.
Filler slabs
They are normal RCC slabs where bottom half (tension) concrete portions are replaced by filler materials such as bricks, tiles, cellular concrete blocks, etc.These filler materials are so placed as not to compromise structural strength, result in replacing unwanted and nonfunctional tension concrete, thus resulting in economy. These are safe, sound and provide aesthetically pleasing pattern ceilings and also need no plaster.
Jack arch roof/floor
They are easy to construct, save on cement and steel, are more appropriate in hot climates. These can be constructed using compressed earth blocks also as alternative to bricks for further economy.
Ferrocement channel/shell unit
Provide an economic solution to RCC slab by providing 30 to 40% cost reduction on floor/roof unit over RCC slabs without compromising the strength. These being precast, construction is speedy, economical due to avoidance of shuttering and facilitate quality control.
Finishing Work
The cost of finishing items like sanitary, electricity, painting etc., varies depending upon the type and quality of products used in the building and its cost reduction is left to the individual choice and liking.
Conclusion
The above list of suggestion for reducing construction cost is of general nature and it varies depending upon the nature of the building to be constructed, budget of the owner, geographical location where the house is to be constructed, availability of the building material, good construction management practices etc. However it is necessary that good planning and design methods shall be adopted by utilizing the services of an experienced engineer or an architect for supervising the work, thereby achieving overall cost effectiveness to the extent of 25% in actual practice.
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PONDING CONSIDERATIONS IN BUILDINGS
Flat roofs on which water may accumulate may require analysis to ensure that they are stable under ponding conditions. A flat roof may be considered stable and an analysis does not need to be made if both of the following two equations are satisfied:
C p + 0.9 C s £ 0.25
I d ³ 25S 4 / 10 6
Where C p = 32 L s L 4 p / 10 7 I p
C s = 32 SL 4 s / 10 7 s
L p = length, ft (m), of primary member or girder
L s = length, ft (m), of secondary member or purlin
S = spacing, ft (m), of secondary members
I p = moment of inertia of primary member, in 4
(mm 4 )
I s = moment of inertia of secondary member, in 4
(mm 4 )
I d = moment of inertia of steel deck supported on secondary members, in 4 /ft (mm 4 /m)
For trusses and other open-web members, I s should be decreased 15 percent. The total bending stress due to dead loads, gravity live loads, and ponding should not exceed 0.80F y , where F y is the minimum specified yield stress for the steel.
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NUMBER OF CONNECTORS REQUIRED FOR BUILDING CONSTRUCTION
The total number of connectors to resist V h is computed from V h / q, where q is the allowable shear for one connector, kip (kN). Values of q for connectors in buildings are given in structural design guides.
The required number of shear connectors may be spaced uniformly between the sections of maximum and zero moment. Shear connectors should have at least 1 in (25.4 mm) of concrete cover in all directions; and unless studs are located directly over the web, stud diameters may not exceed 2.5 times the beam-flange thickness.
With heavy concentrated loads, the uniform spacing of shear connectors may not be sufficient between a concentrated load and the nearest point of zero moment. The number of shear connectors in this region should be at least
N 2 = N 1 [ ( M b / M m a x ) - 1] / ( b – 1 )
where M = moment at concentrated load, ft kip (kN-m)
M max = maximum moment in span, ft kip (kN-m)
N 1 = number of shear connectors required between M m a x and zero moment
b = S t r / S s or S e f f / S s , as applicable
S e f f = effective section modulus for partial composite action, in 3 (mm 3 )
Shear on Connectors
The total horizontal shear to be resisted by the shear connectors in building construction is taken as the smaller of the values given by the following two equations:
V h = 0.85 f ‘ c A c / 2
V h = A s F y / 2
where V h = total horizontal shear, kip (kN), between maximum positive moment and each end of steel beams (or between point of maximum positive moment and point of contraflexure in continuous beam)
f ‘ c = specified compressive strength of concrete at 28 days, ksi (MPa)
A c = actual area of effective concrete flange, in 2 (mm 2 )
A s = area of steel beam, in 2 (mm 2 )
In continuous composite construction, longitudinal reinforcing steel may be considered to act compositely with the steel beam in negative-moment regions. In this case, the total horizontal shear, kip (kN), between an interior support and each adjacent point of contraflexure should be taken as
V h = A s r F y r / 2
where A s r = area of longitudinal reinforcement at support within effective area, in 2 (mm 2 ); and F y r = specified minimum yield stress of longitudinal reinforcement, ksi (MPa).
COMPOSITE CONSTRUCTION
In composite construction, steel beams and a concrete slab are connected so that they act together to resist the load on the beam. The slab, in effect, serves as a cover plate. As a result, a lighter steel section may be used.
Construction In Buildings
There are two basic methods of composite construction.
Method 1. The steel beam is entirely encased in the concrete. Composite action in this case depends on the steel-concrete bond alone. Because the beam is completely braced laterally, the allowable stress in the flanges is 0.66F y , where F y is the yield strength, ksi (MPa), of the steel. Assuming the steel to carry the full dead load and the composite section to carry the live load, the maximum unit stress, ksi (MPa), in the steel is
F s = ( M D / S S ) + ( M L / S t r ) ? 0.66F y
where M D = dead-load moment, in-kip (kN-mm)
M L = live-load moment, in-kip (kN-mm)
S s = section modulus of steel beam, in 3 (mm 3 )
S t r = section modulus of transformed composite section, in 3 (mm 3 )
An alternative, shortcut method is permitted by the AISC specification. It assumes that the steel beam carries both live and dead loads and compensates for this by permitting
a higher stress in the steel:
f s = M D + M L / S s ? 0.76 F y
Method 2. The steel beam is connected to the concrete slab by shear connectors. Design is based on ultimate load and is independent of the use of temporary shores to support the steel until the concrete hardens. The maximum stress in the bottom flange is
F s = M D + M L / S t r £ 0.66 F y
To obtain the transformed composite section, treat the concrete above the neutral axis as an equivalent steel area by dividing the concrete area by n, the ratio of modulus of elasticity of steel to that of the concrete. In determination of the transformed section, only a portion of the concrete slab over the beam may be considered effective in resisting compressive flexural stresses (positive-moment regions). The width of slab on either side of the beam centerline that may be considered effective should not exceed any of the following:
1. One-eighth of the beam span between centers of sup- ports
2. Half the distance to the centerline of the adjacent beam
3. The distance from beam centerline to edge of slab
FASTENERS IN BUILDINGS
The AISC specification for allowable stresses for buildings specifies allowable unit tension and shear stresses on the cross-sectional area on the unthreaded body area of bolts and threaded parts. (Generally, rivets should not be used in direct tension.) When wind or seismic load are combined with gravity loads, the allowable stresses may be increased one-third.
Most building construction is done with bearing-type connections. Allowable bearing stresses apply to both bearing-type and slip-critical connections. In buildings, the allowable bearing stress F p , ksi (MPa), on projected area of fasteners is
F p = 1.2 F
where F u is the tensile strength of the connected part, ksi (MPa). Distance measured in the line of force to the nearest edge of the connected part (end distance) should be at least 1.5d, where d is the fastener diameter. The center-to-center spacing of fasteners should be at least 3d.
Design of stiffners under load
AISC requires that fasteners or welds for end connections of beams, girders, and trusses be designed for the combined effect of forces resulting from moment and shear induced by the rigidity of the connection. When flanges or moment connection plates for end connections of beams and girders are welded to the flange of an I- or H-shape column, a pair of column-web stiffeners having a combined cross-sectional area A s t not less than that calculated from the following equations must be provided whenever the calculated value of A s t is positive:
A s t = ( P b f – F y c t w c ( t b + 5k ) ) / F y s t
where F y c = column yield stress, ksi (MPa)
F y s t = stiffener yield stress, ksi (MPa)
K = distance, in (mm), between outer face of column flange and web toe of its fillet, if column is rolled shape, or equivalent distance if column is welded shape
P b f = computed force, kip (kN), delivered by flange of moment-connection plate multi plied by 5/3 , when computed force is due to live and dead load only, or by 4/3, when computed force is due to live and dead load in conjunction with wind or earthquake forces
T w c = thickness of column web, in (mm)
T b = thickness of flange or moment-connection plate delivering concentrated force, in (mm)
Notwithstanding the preceding requirements, a stiffener or a pair of stiffeners must be provided opposite the beam compression flange when the column-web depth clear of fillets d c is greater than
d c = ( 4100 t 3 w c ( F y e ) ½ ) / P b f
and a pair of stiffeners should be provided opposite the tension flange when the thickness of the column flange t f is less than
t f = 0.4 ( P b f ) ½ / F y c
Stiffeners required by the preceding equations should comply with the following additional criteria:
1. The width of each stiffener plus half the thickness of the column web should not be less than one-third the width of the flange or moment-connection plate delivering the concentrated force.
2. The thickness of stiffeners should not be less than t b /2.
3. The weld-joining stiffeners to the column web must be sized to carry the force in the stiffener caused by unbalanced moments on opposite sides of the column.
WEBS UNDER CONCENTRATED LOADS
Criteria for Buildings
The AISC specification for ASD for buildings places a limit on compressive stress in webs to prevent local web yielding. For a rolled beam, bearing stiffeners are required at a concentrated load if the stress f a , ksi (MPa), at the toe of the web fillet exceeds F a = 0.66F y w , where F y w is the minimum specified yield stress of the web steel, ksi (MPa). In the calculation of the stressed area, the load may be assumed distributed over the distance indicated in Fig. 9.4.
For a concentrated load applied at a distance larger than the depth of the beam from the end of the beam:
F a = R / f w ( N + 5K )
where R = concentrated load of reaction, kip (kN)
t w = web thickness, in (mm)
N = length of bearing, in (mm), (for end reaction, not less than k)
K = distance, in (mm), from outer face of flange to web toe of fillet
For a concentrated load applied close to the beam end:
f a = R / t w ( N + 2.5 k )
To prevent web crippling, the AISC specification requires that bearing stiffeners be provided on webs where concentrated loads occur when the compressive force
exceeds R, kip (kN), computed from the following:
For a concentrated load applied at a distance from the beam end of at least d/2, where d is the depth of beam:
R = 67.5 t 2 w [ 1 + 3 ( N / d ) ( t w / t f ) 1.5 ] (F y w t f / t w ) ½
where t f = flange thickness, in (mm)
For a concentrated load applied closer than d/2 from the beam end:
R = 34r 2 w [ 1 + 3 ( N / d ) ( t w / t f ) 1.5 ] ( F y w t f / t w ) ½
If stiffeners are provided and extend at least one-half of the web, R need not be computed.
Another consideration is prevention of sidesway web buckling. The AISC specification requires bearing stiffeners when the compressive force from a concentrated load
exceeds limits that depend on the relative slenderness of web and flange r w f and whether or not the loaded flange is restrained against rotation:
r w f = ( d c / t w ) / ( l / b f )
where l = largest unbraced length, in (mm), along either top or bottom flange at point of application of load
b = flange width, in (mm)
d c = web depth clear of fillets = d – 2k
Stiffeners are required if the concentrated load exceeds R, kip (kN), computed from
R = 6800 t 3 w / h ( 1 + 0.4 r 3 w f )
where h = clear distance, in (mm), between flanges, and r w f
is less than 2.3 when the loaded flange is restrained against rotation. If the loaded flange is not restrained and r w f is less than 1.7,
R = 0.4 r 3 w f ( 6800 t 3 w / h )
R need not be computed for larger values of r w f ..
Combined Axial Compression Or Tension And Bending
The AISC specification for allowable stress design for buildings includes three interaction formulas for combined axial compression and bending.
When the ratio of computed axial stress to allowable axial stress f /F a exceeds 0.15, both of the following equations must be satisfied:
( f a / F a ) + ( C m x f b x ) / (1– f a /F ‘ e x ) F b x + C m y f b y / (1 – f a / F ‘ e y ) F b y ? 1
f a / 0.60F y + f b x /F b x + f b y / F b y ? 1
when f a /F a ? 0.15, the following equation may be used instead of the preceding two:
f a / F a + f b x / F b x + f b y / F b y ? 1
In the preceding equations, subscripts x and y indicate the axis of bending about which the stress occurs, and
In the preceding equations, subscripts x and y indicate the axis of bending about which the stress occurs, and
F a = axial stress that would be permitted if axial force alone existed, ksi (MPa)
F b = compressive bending stress that would be permitted if bending moment alone existed, ksi (MPa)
F ‘ e = 149,000 / ( Kl b / r b ) 2 , ksi (MPa); as for F a , F b , and 0.6F y , F ‘ e may be increased one-third for wind and seismic loads
Lb = actual unbraced length in plane of bending, in (mm)
r b = radius of gyration about bending axis, in (mm)
K = effective-length factor in plane of bending
f a = computed axial stress, ksi (MPa)
f b = computed compressive bending stress at point under consideration, ksi (MPa)
C m = adjustment coefficient
Deflections of Bents and Shear Walls
Horizontal deflections in the planes of bents and shear walls can be computed on the assumption that they act as cantilevers. Deflections of braced bents can be calculated
by the dummy-unit-load method or a matrix method. Deflections of rigid frames can be computed by adding the drifts of the stories, as determined by moment distribution
or a matrix method.
For a shear wall (Fig) the deflection in its plane induced by a load in its plane is the sum of the flexural
Figure showing Building frame resists lateral forces with (a) wind bents or (g) shear walls or a combination of the two. Bents may be braced in any of several ways, including (b) X bracing, (c) K bracing, (d) inverted V bracing, (e) knee bracing, and (f) rigid connections.
deflection as a cantilever and the deflection due to shear. Thus, for a wall with solid rectangular cross section, the deflection at the top due to uniform load is
? = 1.5 wH / Et [ ( H / L ) 3 + H / L]
where w = uniform lateral load
H = height of the wall
E = modulus of elasticity of the wall material
t = wall thickness
L = length of wall
For a shear wall with a concentrated load P at the top, the deflection at the top is
? c = ( 4 P / Et ) [ ( H / L ) 3 + 0.75 H / L ]
If the wall is fixed against rotation at the top, however, the deflection is
? f = ( P / Et ) [ ( H / L ) 3 + 3 H / L ]
Units used in these equations are those commonly applied in United States Customary System (USCS) and the System International (SI) measurements, that is, kip (kN), lb /in 2 (MPa), ft (m), and in (mm).
Where shear walls contain openings, such as those for doors, corridors, or windows, computations for deflection and rigidity are more complicated. Approximate methods, however, may be used.
Load Distribution To Bents And Shear Walls
Provision should be made for all structures to transmit lateral loads, such as those from wind, earthquakes, and traction and braking of vehicles, to foundations and their supports
that have high resistance to displacement. For this purpose, various types of bracing may be used, including struts, tension ties, diaphragms, trusses, and shear walls.
Plate Girder in Buildings
For greatest resistance to bending, as much of a plate girder cross section as practicable should be concentrated in the flanges, at the greatest distance from the neutral axis. This
might require, however, a web so thin that the girder would fail by web buckling before it reached its bending capacity.
To preclude this, the AISC specification limits h/t.
For an unstiffened web, this ratio should not exceed.
h / t = 14,000 / (F Y (F y + 16.5) ) ½
where F y = yield strength of compression flange, ksi (MPa).
Larger values of h/t may be used, however, if the web is stiffened at appropriate intervals.
For this purpose, vertical angles may be fastened to the web or vertical plates welded to it. These transverse stiffeners are not required, though, when h/t is less than the value
computed from the preceding equation or Table 9.4.
Critical h/t for Plate Girders in Buildings should be taken from the codes
With transverse stiffeners spaced not more than 1.5 times the girder depth apart, the web clear-depth/thickness ratio may be as large as
h / t = 2000 / ( F y ) ½
If, however, the web depth/thickness ratio h/t exceeds 760 / (F b ) ½ , where F b , ksi (MPa), is the allowable bending stress in the compression flange that would ordinarily apply, this stress should be reduced to F ‘ b , given by the following equations:
F ‘ b = R P G R e F ¬b
R P G = [ 1 – 0.0005 ( A w / A f ) ( h/t – 760 / ( F b )) ½ ] ? 1.0
R e = [ ( 12 + ( A w /A f ) (3a – a 3 ) ) / ( 12 + 2 (A w / A f ) ] ? 1.0
Where A w = web area , in 2 (mm 2 )
A f = area of compression flange, in 2 (mm 2 )
a = 0.6 F y w / F b ? 1.0
F y w = minimum specified yield stress, ksi, (MPa), of web steel
In a hybrid girder, where the flange steel has a higher yield strength than the web, the preceding equation protects against excessive yielding of the lower strength web in the vicinity of the higher strength flanges. For nonhybrid girders, R e = 1.0.
Bearing Of Milled Surfaces
In building construction, allowable bearing stress for milled surfaces, including bearing stiffeners, and pins in reamed, drilled, or bored holes, is F p = 0.90F y , where F y is the yield strength of the steel, ksi (MPa)
For expansion rollers and rockers, the allowable bearing stress, kip/linear in (kN/mm), is
F p = ( ( F y – 13 ) / 20 ) 0.66d
where d is the diameter, in (mm), of the roller or rocker. When parts in contact have different yield strengths, F y is the smaller value.
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Column Base Plates
The area A 1 , in 2 (mm 2 ), required for a base plate under a column supported by concrete should be taken as the larger of the values calculated from the equation cited earlier, with R taken as the total column load, kip (kN), or
A 1 = R / 0.70 f ‘ c
Unless the projections of the plate beyond the column are small, the plate may be designed as a cantilever assumed to be fixed at the edges of a rectangle with sides equal to 0.80b and 0.95d, where b is the column flange width, in (mm), and d is the column depth, in (mm).
To minimize material requirements, the plate projections should be nearly equal. For this purpose, the plate length N, in (mm) (in the direction of d), may be taken as
N = ( A 1 ) ½ + 0.5 ( 0.95d – 0.80b )
The width B, in (mm), of the plate then may be calculated by dividing A1 by N. Both B and N may be selected in full inches (millimeters) so that BN ³ A 1 . In that case, the bearing pressure f p , ksi (MPa), may be determined from the preceding equation. Thickness of plate, determined by cantilever bending, is given by
t = 2p (f p / F y ) ½
where F y = minimum specified yield strength, ksi (MPa), of plate; and p = larger of 0.5 (N – 0.95d) and 0.5 ( B – 0.80b).
When the plate projections are small, the area A 2 should be taken as the maximum area of the portion of the supporting surface that is geometrically similar to and concentric with the loaded area. Thus, for an H-shaped column, the column load may be assumed distributed to the concrete over an H-shaped area with flange thickness L, in (mm), and web thickness 2L:
L = ( 1 / 4 ) (d + b) – 1 / 4 ( (d + b) 2 – 4R / F P ) ½
where F p = allowable bearing pressure, ksi (MPa), on support. (If L is an imaginary number, the loaded portion of the supporting surface may be assumed rectangular as discussed earlier.) Thickness of the base plate should be taken as the larger
of the values calculated from the preceding equation and
t = L (¬3 f p / F b ) ½
Bearing Plates
To resist a beam reaction, the minimum bearing length N in the direction of the beam span for a bearing plate is deter- mined by equations for prevention of local web yielding and web crippling. A larger N is generally desirable but is limited by the available wall thickness.
When the plate covers the full area of a concrete support, the area, in 2 (mm 2 ), required by the bearing plate is
A 1 = R / 0.35 f ‘ c
Where R = beam reaction, kip (kN), f ‘ c = specified compressive strength of the concrete, ksi (MPa). When the plate covers less than the full area of the concrete support, then, as determined from Table
Full area of concrete support 0.30 f 'c
Sandstone and limestone 0.40
Brick in cement mortar 0.25
♣
Units in MPa = 6.895 X ksi
where A 2 = full cross-sectional area of concrete support, in 2 (mm 2 ).
With N established, usually rounded to full inches (millimeters), the minimum width of plate B, in (mm), may be calculated by dividing A 1 by N and then rounded off to full inches (millimeters), so that BN ³ A 1 . Actual bearing pres- sure f p , ksi (MPa), under the plate then is
F p = R / BN
The plate thickness usually is determined with the assumption of cantilever bending of the plate:
t = ( ( 1 / 2) B – k ) ( 3 f p / F b ) ½
where t = minimum plate thickness, in (mm)
k = distance, in (mm), from beam bottom to top of web fillet
F b = allowable bending stress of plate, ksi (MPa)
Stresses In Thin Shells
Results of membrane and bending theories are expressed in terms of unit forces and unit moments, acting per unit of length over the thickness of the shell. To compute the unit stresses from these forces and moments, usual practice is to assume normal forces and shears to be uniformly distributed over the shell thickness and bending stresses to be linearly distributed.
Then, normal stresses can be computed from equations of the form:
f x = N x / t + ( M x / t 3 /12) z
where z = distance from middle surface
t = shell thickness
M x = unit bending moment about an axis parallel to direction of unit normal force N x .
Similarly, shearing stresses produced by central shears T and twisting moments D may be calculated from equations of the form:
V xy = T / t ± D / ( t 3 / 12 ) z
Normal shearing stresses may be computed on the assumption of a parabolic stress distribution over the shell thickness:
V xy = V / t 3 / 6 ( t 2 / 4 – z 2 )
where V = unit shear force normal to middle surface.
