Design of Domes - Forgotten Books
33
Transcript of Design of Domes - Forgotten Books
DESIGN OF DOMES
! . s . TERRINGTONB.Sc.
upmo FROM
CONCRETE ammo
P! BLISHED BY
CONCRETE P! BLICATIONS LIMITED
14 DARTMO! TH STREET, LONDON, S .W. I
P! BLISHER’S NOTICE .
MANY aspects of reinforced concrete design cannot be adequatelydealt with in text-books due to the lengthy treatment they requireif they are to be really comprehensive and useful . It is believedthat CONCRETE AND CONSTR! CTIONAL ENGINEERING is the onlyperiodi cal in the English language which publishes complete treatiseson such subj ects
,but the articles necessarily appear in serial form
,
and for that reason they are cumbersome to use . Where their importance warrants re-publication
,such articles will in future be issued
as reprints in this form and published at a low price . A list of otherConcrete Series books is given on page 32 .
DESIGN OF DOMES
THE dome or domed shell commends itself from the architectural and astheticpoints of view and would be a comparatively inexpensive form of constructionbut for the practical difficulties due to the accurate erection of the curved shuttering (on both faces Where the inclination to the horizontal is greater than about30 deg . ) and the difiiculty of placing the concrete . In any structure a saving ofmaterials in the finished work due to an otherwise excellent design may be aecompanied by prohibitive erection costs ; in domes these cannot be ignored sincethe curved shuttering is expensive to make and erect and the costs of placingthe concrete and steel
,often at great heights , are particularly heavy .
From other points of view, concrete , suitably reinforced , is advantageoussince
,once the erection difficulties have been solved , the wet concrete readily
takes up the curvatures required in the two directions , thus avoiding the production of a series of flat surfaces which in most other materials only approximateto the curvatures of the shell . Moreover , the compression stresses in domesare small so that , particularly with the high- strength concretes now obtainable
,
the shell may be cast very thin,resulting in a light-weight construction . Its
adaptability for receiving fixings to hold an outer covering or plaster falseworkand the fact that the construction is fireproof are other advantages of concrete .
In practice , domes are usually spherical and are essentially sur faces ofrevolution about a vert ical axis . A vertical section through this axis in anydirection is , as a rule , an arc of a circle . Another possible form is the conicalsurface of a right cone with a vertical axis giving a triangular section throughits axis of revolution . Other shapes , such as the spheroid giving an ellipticalsection through its axis of revolution
,have not been adopted in this country .
By employing surfaces of revolution with vert ical axes and ensuring that allloading is symmetrical about these axes in every direction ,
only direct compression forces in the shell are required to maintain stability ,
provided the shell issupported in a plane at right angles to its axis of revolution
,the supports being
level all round the edge . These ar e the conditions in the des ign of domes ofnormal proport ions .
General Condit i ons for S tabil i ty .
The first point to be noted regarding the stability of the dome is that th eforces maintaining equilibrium act in two directions . In geographical termsthese directions are those of latitude and longitude . The second point to observeis that , in the case of reinforced concrete arches
,their transverse stiffness or
strength in bending afforded by the reinforcement contributes to their strengthand stability . In regard to domes , however , their stifiness is ignored and stabilityis attained by direct tension or compression in the planes previously mentioned
.
Additional strength may be obtained by reinforcing the shell on both faces,thus
enabling any section to resist bending as wellas direct stresses,but this isunneces
sary as the analyses which follow later will show . Hence one layer only offorcement is provided .
Before any analysis of the forces is attempted it is as well to make a brief
j. s . TERRINGTON .
survey of the nature of these forces which contribute to stability, so that the
obj ect of the mathematical investigation is clear . An analogy with the voussoirsin a stone arch may be drawn . I f , for example , a hemispherical domed shell isdivided into a number of sectors (Fig . I ) by j oints formed along lines of longitude ,all crossing at the crown (that is the point corresponding to the north pole of theterrestial sphere) , it is found that the shell remains in equilibrium by virtue ofthe top part of the wedge—shape sides of each sector pressing against one another .
The top of each sector tends to fall inwards and is supported by the equal andOpposite lateral reaction of the opposite sector . In addition
,in order to maintain
equilibrium the sectors must be tied by a circumferential band near the bottom .
fi ns /or! 50ndor 778 Fig . 1 .
Thus the lateral reactions near the crown exert horizontal circumferential compressions , and towards the support the lateral reactions , which have in effectbecome circumferential tensions
,are supplied by the band or circular tie near
the bottom . In other words,the lateral reactions near the crown
,called the
circumferential or hoop compression,which lower down the sides gradually change
to circumferential or hoop tension,are all in the horizontal plane such that the
lines of force are horizontal circles of latitude ar ound the surface of the shell .The whole structure must be symmetrical about the vertical axis of revolution ,
otherwise these forces will not be equal and neutralizeone another in the maintenance of equilibrium . In addition to these horizontal circular forces there aredirect compressions acting down the lines of longitude (Fig. At any particularlevel these longitudinal or meridional forces are proportional to the weight ofthe segment of shell and any other load supported above this level . The
DES IGN OF DOMES .
meridional stress is accordingly zero at the crown and a maximum at the support
.Hence it follows th at in a dome , unlike an arch , there is no thrust at the
crown and a circular port ion may b e removed from the shell at the top to formthe lantern without j eopardising the stability of the structure . In the mannerdescribed
,the forces in the two directions of curvature keep the shell together
provided the material ofwhich it i s made will withstand the compress ions induced(incidentally the stresses are quite small) and that the material is able , as in thecase of concrete reinforced with steel bars , to resist the tension stresses developed .
Cal culati on of Compres s ions and Tens i ons due to th e ! niformLoad .
The following analysis shows how to obtain the magnitude of the hoop compressions and tensions and the meridional compression at any point on the shell .
Fig . 2 .
In practice,the loading is often complicated by the addition of a point load
at the top due to the weight of a lantern or ornament . First , however , a simplethin spherical shell of uniform thicknes s will be dealt with and the effect of thepoint load will be temporarily deferred . Certain other loadings and stresses haveto be taken into account . Additional to the weight of the structural concretethere is the weight of the covering (if any) , for example , tiles , sheet metal , orasphalt
,and a suspended ceiling if provided . The weight of the covering is in
proport ion to the surface area of the shell , whereas the weight of the ceilingmay vary in proportion to that of the proj ected area . The sum of these per unitarea (which may be about 20 lb . per squar e foot ) when added to the self weightof the concrete shell per unit area is the load to be used in the design . Theeffects due to wind ,
temperature,and shrinkage are rather more complex and
difficult to estimate , but by reference to the example on arches and by comparison ,
it will be seen that the stresses involved are small as the dead load stresses areSee “Design of Arch Roofs, b y th eauthor . Published b y Concrete Publications Limited. Price 35.
j. s . TERRINGTON .
themselves small. It would therefore appear that horizontal pressure due towind from one direction would tend to cant the otherwise horizontal circumferential lines of force round the shell to a small angle with the horizontal
,and
at the same time slightly increase the meridional thrust on the leeward side .
These wind stresses,however
, can be amply covered by the superload due tosnow ,
etc. Temperature and shrinkage stresses are accounted for by the additionof reinforcement in both directions , regardless of whether the shell at any pointis in tension or compression , to the extent of about 02 per cent . of the crosssectional area .
A section through the vertical axis of revolution of a thin spherical shell isshown in Fig . 3 . Now consider the equilibrium of an elemental ring or zonebounded by two horizontal planes AB and CD close together
, ofwhich the positionsare defined by the angles (15 and (g!) d¢) in the illustration . The forces acting on
Fig . 3 .
this elemental ring are (I ) a series of thrusts T per unit length of the circle oflatitude AB down the lines of longitude tangential to the surface (2 ) a similarseries of thrusts (T dT) per unit length acting up the lines of longitude alsotangential to the surface round the circle of latitude CD ; and (3 ) the weightof the ring acting vertically downward . The two thrusts cannot be equal inmagnitude
,for they act at a small angle q to one another and between them
there is the weight of the ring acting at varying angles to them according to thelevel at which the ring is chosen . Hence the lower reaction is larger than thedownward thrust T by some quantity , say ,
dT. Clearly the thrust T is causedby the weight of the domed shell ANB ,
whose surface area is 27W x EN . Hence ,if w is its weight per unit area , its total weight is w x 27W x EN and ,
as EN r(r cos cfi) , the weight of the dome above the circle AB is
10 X 27t72(I cos As indicated , the sumof the vertical components of all
the thrusts T per unit length acting round the circumference of the circle ABis equal in magnitude to the weight of the portion of the dome above . Hence
T x 276 x EB cos (90°
96) w.2nr2(r cos qS)T x 2n(r sin cos (90
°w.2mz(r cos cfi)
T x 27W x sin2 95 w 2n72(1 cos qS)cos
sin2 95
DES IGN OF DOMES .
The difierence of the opposing thrusts T and (T +‘
dT) acting at the angles95 and (96 q) respectively to the vertical causes the ring or hoop force in thefollowing manner . Let H be the hoop force per unit length of sur facemeasured on a great-circle arc , that is , a line of longitude through thecrown of the shell. The breadth of the elemental ring in F ig . 3 , measuredon any such great circle arc is 17143 and by definition the hoop force on
the ring is H Also the component of T acting horizontally and radiallyfrom the vertical axis is T cos 93. This component in turn , as with any radialpressure (for example , in cylindrical water tanks) , causes a hoop tensionT cos ¢ x radius which radius at this level is 7 sin Hence the hoop tensiondue to T alone is T eos ¢ X f sin At the same time the thrusts (T £17)have a similar effect (T
’
+ dT) cos (95 x r sin (146) but in the oppositedirecti on causing hoop compression , and the difference of the two effects causesthe actual hoop compression or tension (whichever is greater) on the elementalring . It is therefore necessary to express thes e forces algebraically so that theirdifference may be obtained . Here the use of the differential calculus is invaluableand will be adopted .
The hoop force in the elemental ring is due to'
the change in the value of Twhen cfiis increased by a small amount d¢,
and when this increase tends to zeroH.r .d¢ d. [T cos qS f sin qS]
From equation (I ) T
whence
sin¢.cos 96
By applying the rules for the differentiation of a product and a quotient ,that is
,
d(uv) ad v v .du
(unlit —ad v)
we obtain
s r
— sm24)
- z sm
sinq— I —2 cos qS
.
2¢
j. s . TERRINGTON .
In this form the equation is indeterminate when 95 0 and it is best writtencos qS
cos <1)H = wr
s r
Two critical points can now be examined,namely
,the crown
,where 95 0
,
and the level at which the hoop compression changes to hoop tension,that is
,
where H 0 .
Now H 0 when
or when cos d) cl) 0
45 0-618
93 51 deg . 48min .
In other words , round the circle of latitude at which the angle made with theaxis of revolution is 51 deg . 48min . there is no hoop compression and no hooptension , and the plane through this circle is called the plane of rupture . Also
,
wras 95 decreases the hoop compression , H ,
increases until it IS a maximum ,
2
at the crown . Similarly , as 96 increases to angles greater than 51 d eg . 48min .
the hoop effect increases but is of opposite sign,namely
,tension . On the other
hand the meridional thrust T, as may be seen from equation (I ) , is g at2
the crown and increases to a maximum at the support .
S tresses Due to aConcentrated Load at the Crown.
Having regard to its common occurrence,the stresses produced by a point
load at the crown must now be considered . The calculation is independent ofthe stresses due to the weight of the shell
,which are computed as before and
added to the point - load stresses to get the resultant stresses .
The thin spherical shell of radius r with a point load W at the crown is shownin section through its axis of revolution in Fig . 4 . As before
,consider an ele
mental annular zone ABCD . Thrusts T per unit length tangential to the shellact round the circle of latitude AB and thrusts (T dT) per unit length tangential to the shell act up lines of longitude round the circle CD . The positionsof the circles AB and CD with reference to the load W are defined by the angles(15 and (qS dci) in the figure
,as before .
The sumof the vertical components of the thrusts T is equal in magnitudeto the weight above AB
,the load W in this case . Hence
T X 27W x singb . cos (90°
95) WT.27zr . sin2qS W
TW
(compression) (6)sm2 95
As described before , the thrusts T and (T dT) are unequal , not in thiscase because of the weight of the elemental ring which is here assumed to be zero ,
8
DES IGN OF DOMES .
but because they act at different radii from the axis of revolution and each actsat a different angle to the direction of W.
From equation (2)H.r .d¢ d. [T cos r sin 45]
T !E TsmqS
whence
H
[ I cot 2 96]
cosec 2 Q6 (tension)
Fig . 4 .
At any circle of latitude , therefore , this hoop compression or tension mustbe added to that due to the weight of the shell to obtain the total circumferentialforce acting . At the crown
,where 96 is zero , thi s additional hoop tension becomes
infinite so that the lantern load must be spread over an appreciable area andif necessary the dome must be thickened
,reinforced strongly
,or both .
It is instructive to observe how the coefficients of the hoop and meridionalstresses due to the thin spherical shell of uniform thickness and the constantload W at the crown vary as the value of qSvaries . Table I shows these coefficients .
9
j . s . TERRINGTON .
TABLE I .
THIN SPHER ICAL SHELL OF ! NIFORM THIC! NESS W ITH LOAD W AT CROW N SUPPORTED ONCIRCLE OF LATITUDE .
Shell ofuniformthickness
Meridional thrusts T. Hoop forces H.
Meridional thrusts TCoeflicients of wr Coefficients ofw Coefficients of !
In Table I and Fig . 5, the coefficients for the uniform load and a point loadat the crown are given separately ,
but for any example the effect of both loadingsmust be considered simultaneously at any level . Thus the level of zero hoop forceor the j oint of rupture will no longer occur at the angle of 51 deg . 48min .
, butwill depend also on the relative magnitude of the unit weight w of the shell andthe point load W,
and on the radius 7. In other words , to find the value of cfiat the j oint of rupture for the combined loading the whole expression for H ,
namely E]
cosec2 ¢ must be equated to zero .
From this it will be seen that the greater W is in proportion to w, depending
he shell,the smaller qS becomes for H 0 , and
the level of the j oint of rupture rises , with the result that the load at the crownmay readily cause the whole of the shell to be in ring tension .
For example,consider a dome of radius r 25 ft . and of uniform weight
10 60 lb . per square foot . Consider a j oint at which 95 40 deg . Then without any load at the crown ,
H 60 x 25 300 lb . (compression) per foot
IO
DES IGN OF DOMES .
3562555 0 /0 529 60mmao zmuo zzow w w w vo /aoFig . 5 .
height measured on any great-circle arc through the crown . Now if a load oflb . due to a lantern or ornament is placed at the top
,the hoop tension
2Induced at this level Is 5 000
£ 0$€Cz40
025 000
z .a.z 5
A net tension of 83 lb . per foot is therefore exerted . Thus an area which wouldotherwise b e in hoop compression is by reason of the concentrated load subj ectedto hoop tension .
x 2 41 383 lb . (tension) .
Example .
An example of a hemispherical domed shell of uniform thickness of 5 in .
with a radius of 37 ft . 6 in . will be used to illustrate the method of calculatingthe stresses and reinforcement required at any point . The weights of the shell
,
covering, and wind and snow load may be taken as 60 20 20 1 00 lb .
per square foot .
The meridional thrust at any point is T wr
I I
j. s . TERRINGTON .
The hoop compression or tension at any point is
(1 cos O)Select any three levels such that O 20 deg .
, 70 deg .,and 90 deg . (Fig .
and use Table I to find the coefficients in the expressions for T and H .
Fig . 6 .
At the level O 20 deg . the height above the springing is (see also p . 23 )375 COS 20°
375 X 0-
94 352 5 it .
T wrI COS 20
1 00 x X 05 16 lb . per foot (compression) .
1 cos 20°
CH wr 1 00 x 375 x 0 424I COS 200
lb . per foot (compression) .As the shell is 5 in . thick the meridional stress is
5 x 1 2
and the circumferential stress is
32 lb . per square inch (compression)
5 x 1 2
At the level O 70 deg . the height above the springing is (see375 COS 70
°
375 x 03 4 1 283 ft .
27 lb . per square inch (compression) .
T wr 1 00 x 37-5 x 0747 lb . per foot (compression) .
1 cH W 1 00 x 375 x 0-402
lb . per foot (tension) .
The meridional stress is 47 lb . per square inch (compression) .
The circumferential stress is 25 lb . per square inch (tension) .
Actually,of course , the concrete is not considered to take tension ,
andtherefore at a working stress of lb . per square inch the area of steel re
quired for hoop tension alone is 0 08 sq. in . per foot .
DES IGN OF DOMES .
At the springing level O 90 deg .
1 00 x 37-
5 X 1 -0 lb . (compression)
which is also the vertical reaction per foot run of circumference on the supports .
01 00 x 375 x 1 -0 lb . (tension) .
cos go
The meridional stress is 62 -5 lb . per square inch (compression) .
The circumferential stress is also 62 -
5 lb . per square inch (tens ion) .
To resist the latter tensile stress the reinforcement required is
sq. in . per foot .
The whole of the surface area requires reinforcement in both directions toresist temperature and shrinkage stresses in addition to the steel required overthose areas where hoop tension is developed . For this , about 0-2 per cent . ofthe cross- sectional area should be added , which in this example amounts to0-2
x 5 x 1 2 0-1 2 sq . m . per foot .
At the springing, therefore , where the greatest h00p tension is exerted , theCircumferential steel necessary is 0-208 0 -1 2 0-
328 sq. in . per foot , andfi- in. bar s at 4-in. centres providing 03 3 sq . in . would be suitable . At rightangles to these down the lines of longitude only temperature and shrinkage steelneed be provided , that is , 0-1 2 sq . in . per foot , and for this g—in. bars at 9-in.
centres are suffi cient .
The horizontal steel along the lines of latitude around the dome may bereduced slightly towards the crown as th e calculations show, particularly overthe zone of hoop compression , but a minimum of é-in. bars at 9—in. centres inboth directions should be placed at any point to allow for wind , temperature ,and shrinkage variations .
I t is as well now to see how the stresses are affected by a load at the crown .
Where this load is caused by a lantern , a circular portion of the dome from5 ft . to 20 ft . in diameter may be required to remain Open ,
but as has been explained the stability of th e whole structure is not impaired . Let the weightof the lantern less the self weight of the shell over an area 1 0 ft . in diameter be
lb . At the periphery of the lantern the angle made with the vertical axis
15
5% xI
f: 7 deg . 38min . Around this periphery the meridional thrust
W
2msin2O 23! x 375 x sin2 7°
38'
This is also the value of the circumferential tension due to this weight at
f5 x 1 2
Both these stresses are comparatively high for a dome and indicate that at
this level and each causes a stress 0 80 lb . per square inch .
IS
j. s . TERRINGTON .
this point the shell Should be thickened and specially reinforced to spread theload over a larger area .
At the other levels,where O 20 deg . , 70 deg . , and 90 deg ,
the stressesdue to this load are as follows :Where O 20 deg .
2ax 37-
5 sin2 20°
2 x n x 37-
5 0-342
247 lb . per foot (compression) ,W
2717 sin2OTherefore the total meridional thrust due to self weight
,superload
,and
lantern is 247 lb . per foot,giving an intensity of compression
247 lb . per foot (tension) .
f 6 lb . h .o
5 X 1 23 per square Inc
The total hoop thrust is 247 lb . per foot , giving an intensity
of compression of 22 lb . per square Inch .
5 x 1 2
Where O 70 degW
z irr sinzO 271 x 37-5 sin2 70°
2715 x 375 x 0884
96 lb . per foot (compression) ,
96 lb . per foot (tension) ,95
and so the total meridional thrust at this level is 96 lb . per
5 x 1 2
The total hoop tension is 96 lb . per foot , which requires
foot,giving an intensity of compression of 49 lb . per square inch .
0-1 2 (for temperature , etc. ) 0 2 1 sq. in . of reinforcement per foot .
Where O 90 deg . (at the springing)W
2317 5s 27x x 37-5 sin290
°
271 x 37-5 x 1
85 lb . per foot (compression) ,
m—Zfi ? 85 lb . per foot (tension) .
Therefore the total meridional thrust at this level is 85 lb .
per foot , giving an intensity of compression of5
3
53
32 64 lb . per square inch .
The total hoop tension is 85 lb . per foot , which requires
In this example the level of the j oint of rupture is such that O has the following value
0-1 2 (for temperature , etc. ) 0-33 sq . in . of steel per foot .
I4
DES IGN OF DOMES .
cosO coszO) (1 cosO) 85 o
1—2 cos O cos 3 O 0-0226 0
cos O 0-65
49 deg 30minThus the load from the lantern has raised the level of the j oint of rupture
from 375 X cos 51°
48’
37-
5 X 0-618 23-2 ft . above the springing , as it
would be without this load , to a level of 37-5 X cos 49°
30'
375 X 06 5 24-
4ft . above the springing .
The results obtained in this example may therefore be summarised as follows .Where there is hoop thrust the point load at th e crown reduces the compressivestress
,and where there is hoop tension the tensile stress is increased . Fur ther,
if the crown load is sufficiently heavy , circumferential tension may exist at alllevels . When the dome is hemispherical , the meridional force , and hence thesupport reactions
,is vertical and no inclined reaction around the edge is required
to maintain equilibrium . The circumferential tension,wherever it occurs, is
absorbed by the reinforcement which has been designed to resist it .
IS
J . s. TERRINGTON .
THIN SPHERICAL DOMES OF ! ARYING THIC ! NES S .
S INCE the meridional thrust governs the thickness of the shell,the thickness
may be graded from a minimum at the crown to a maximum at the supportsto secure economy in large domes .
In Fig . 7 a section is shown through the centre of a spherical dome of weightw per unit area at the crown increasing by w’per unit area per unit of the angleO . Ignoring for the moment any load at the crown
,the effect of which may be
calculated on the lines illustrated in the foregoing example,the obj ect is to obtain
expressions giving the values of the meridional thrust T and the hoop force Hat any level at which the circumference of the circle of latitude makes an angle Owith the vertical axis , taking into account the fact that the thickness of theshell varies .
Fig . 7.
As before the equilibrium of an elemental ring ABCD must be consideredand for this the weight of the dome above the circle AB must first be obtained .
I f abcd is an elemental ring of this portion,its weight is (w w
'
0)2m' sinand the weight of the portion of the dome above AB is
(w w’9)2:frr
2 Sin 9 .d6
4>
(w sin 6 w’9 sin 6)d9
0
27r72 — w c05 6 + w
’sin9 — w
'
0cos t9
27172[ w cos O w
’Sin O m’Ocos O w]27rr
2[w(1 cos O) w
’
(sinO O cosO)]
As described for the dome of uniform thickness , a series of thrusts T perunit length acts tangentially to the surface round the circumference AB , andin a similar fashion thrusts (T dT) act round the circumference CD but in theopposite sense . Also the sum of the vertical components of all the thrusts T
1 6
! . S . TERRINGTON .
1—cosO—cos2 O1+cos O
smO cos2O—O2 sin2 O cosO—Ocos 2 O cos Osin 2 O
—sin3O—sin O cos2O+Ocos O(2sin2O+coszO)
sin2 O
1 cos O sin2 O(9)
The coefficients of wr and w'
r may be obtained from the expressions (8)and (9) for all values of O and are given in Tab le I I . From these the meridionalthrust and the hoop compression or tension may be calculated at any level .
TABLE II .
SPHER ICAL SHELL OF THIC! NESS ! ARY ING UNIFORMLY FROM CROWN To SPRINGING SUPPORTEDAROUND A CIRCLE OF LATITUDE
I
w wt . per unit areaat crown. w increase in w per radian as angle Oincreases .
Meridional thrust (T) Hoop force (H)
Coefficients ofwr Coefficients ofw'
r Coefficients of wr Coeffi cients of
For a shell of varying thickness the j oint of rupture will not necessarily beat a level such that O 51 deg . 48min .
,but will depend on the relativemagm
tude of w and w'
. For usual variations in thickness the stresses in the shell are
18
DES IGN OF DOMES .
not altered very much,as may be seen from the example given later . Als o the
additional meridional thrust (T) and hoop tension (H) due to a load at the crownwill be the same as for a shell of uniform thickness , the only difference beingthat the available ar ea of concrete in section will be changed and hence thecompressive stres ses induced .
These results are plotted in Fig . 8, enabling values between the anglesactually calculated to be interpolated .
Exampl e .
Consider the hemispherical domed shell of radius 37 ft . 6 in . as in the previons example
,but of thickness varying from 3 in . at the crown to 6 in . at the
supports . The weight of shell and superload at the crown is 36 40 76 lb .
per square foot and at the support 72 40 1 1 2 lb . per square foot . Therefore1 1 2 76
ft
2
For the values of O 20 deg ., 70 deg .
, and 90 deg . the magnitude of themeridional thrust and hoop force may be evaluated from the coefficients givenin Table I I .
121 76 lb . and w'
22 -
9 lb . per radian .
Tab . ) H (lb )
20 37 X 0 -5 1 6 2 2 -9 X 0 -1 0] 76 X 375 X 0 -425) 2 2 -9 x 37
-5 x -25)
(compress ion) 5 (compres s ion)70 37
-5I76 x 0 -747 2 2 -
9 x 0 -
59 ] 76 x 37 5 x 04 02 ) 2 2 -9 x 37 5 x
(compress ion) (tens ion)9 0 37 X 1 -0 2 2 -9 X 1 -01 76 X 3 75 X TO ) 2 2 -9 X 37
-5 X (+ I -o)
(compress ion) (tens ion)
O 20 deg . the thickness of the shell'
is 3 3 X 3-67 in .
The meridional compressive stress is36
1
75
20
1 235
-
4 lb . per square inch .
The hoop compress ive stress is3 6
1
74
31 2323 lb . per square inch .
O 70 deg . th e thickness of the shell is 3 3 X 3—8 5-
34 in .
The meridional compressive stress is5
_ 41-2 lb . per square inch .
0-2
X 1 2
1 005 34 X 0 20 sq m of steel
per foot , or, say ,—in. bars at 6-in. centres (0-22 sq . in . per foot) .
At O 90 deg. the thickness of the shell is 6 in .
6 X 1 2
The hoop tension requires
The meridional compressive stress is 51 5 lb . per square inch .
I9
J . s . TERRINGTON .
6 o
0 f 11 00
X X 1 2 0 35 sq m o stee
per foot , or , say , g—in. bars at 6-in. centres (0-393 sq. in . per foot)
The hoop tension requires
Graph ical Methods .
The results of the previous analyses may be arrived at by the followinggraphical construction .
Consider a hemispherical dome 5 in . thick and let the radius be 37 ft . 6 in .
as before . As shown in the first example the weight of the shell above anylevel is proportional to its height and therefore the weight of any annular zoneABCD is also proportional to its height . Draw the dome to scale and dividethe height into a convenient number of equal parts (the larger the number themore accurate the results) , and draw horizontal lines through the j oints .
”
The weight of each zone or part of the shell between two adj acent horizontal
Fig . 9 .
lines is equal to the total weight of the shell, superload , and finishes divided bythe number of zones . Draw a vertical line oy representing to scale the totalestimated load
,and divide this line into the same number of equal parts as there
are zones to obtain the weight of each zone .
Considering the j oint A (Fig. draw 0aparallel to the tangent to the shellat A
,and draw pahorizontally cutting 0ain a. Then oap is the triangle of
forces for the j oint A,op representing the weight of the portion of the dome
above A,outhe magnitude of the total meridional thrust round the j oint A,
and ap the horizontal force causing hoop compression .
For the j oint B,draw ob parallel to the tangent to the shell at B and draw
qb horizontally cutting 0b in b . From adraw ab ’ perpendicular to bq. Thenoqis the weight of the dome above B ,
0b is the magnitude of the total meridional
20
DES IGN OF DOMES .
thrus t round the j oint B ,and (bq—ap) b b
’ is the horizontal force causing hoopcompression .
For the j oint C , draw oc parallel to the tangent to the shell at C.
and draw re
horizontally,cutting 00 in 0. Draw the perpendicular bc' . Then or 15 the weight
of the dome above C ,00 is the magnitude of the total meridional thrust round
the j oint C,and (cr bq) cc
'
is the horizontal force causing hoop compression .
Lower down at a j oint G,by the same construction , 0g gives the magnitude
of the meridional thrust T,01) is the weight of the dome above G, and
force
(gv kw) g' is the horizontal force which is now in the opposite direction
and , in consequence , indicates hoop tension .
The intercepts up,b b
'
, cc'
, etc ., gg
'
,etc .
,Show the variation in
magnitude of the hoop forces from compression above the level O 51 deg . 48min .
to tension below the plane indicated by this angle ; these are proportional tothe hoop forces , so that for any zone GH,
the mean intercept X is a measure ofthe hoop tension in this zone . The exact values of the circumferential stressesare obtained from the values of ap,
b b'
, cc gg’
,etc
.,which are scaled
off the stress diagram . Any of these intercepts represents the horizontal radial
2 1
J . s . TERRINGTON .
force distributed round the circumference . The radial force per unit of circumXference 15 therefore where rah IS the horizontal radius of the zone considered .
27”oh
As In cyhndrical tanks , thIs radial force produces a hoop tension or compressionX X .
X rh
in the circumferential ring GH. I f this force is a tension,as
27mm 27x
is actually the case for the zone GH, steel 15 required to resist it , and is providedat a definite spacing centre to centre and spread uniformly over the length 3 ,
the bars following circles of latitude round the shell (Fig. Further up theshell where this force is a compression ,
the cross section 3 X t must be sufficientlylarge so that the allowable working compressive stress of the material is notexceeded . In the same way the meridional thrust T—also scaled off the diagram—must be divided by 271m, to obtain the force per unit length of circumference .
This force in turn must be divided by the thickness to obtain themeridionalcompressive stress .
2 2
DES IGN OF DOMES .
GRAPHICAL CHEC ! OF FIRS T EXAMPLE .
As a direct comparison with the fir st example this method , or rather theprinciples outlined in it , may be used to check the results previously obtained .
The radius of the hemispherical dome is 37 ft . 6 in . and the thickness is 5 in .
The weight of the shell together with its covering and superload is 1 00 lb . persquare foot . The total weight of the dome is 1 00 X 231 X 37
-
52 lb .
If the shell is divided into ten zones , each weighs lb . Draw the dome toscale and draw the horizontal j oints at O 2 20 deg , 70 deg . and 90 deg . Thelengths of the radii are 37-5 X sin 20
°
1 2 -83 ft . ; 37-
5 X sin 70°
35-2 ft . ;
and 37-
5 X sin 90°
37-
5 ft . , and the heights above the springing are37
-
5 X cos 20°
35-2 ft . 375 X cos 70
°
1 2 -83 ft . , and 0 (see p .
With these loads draw the force diagram as in Fig . 1 1 . On this force diagram
draw horizontal lines at heights proportional t o 3—5o
—2and of the total weight .
37‘
5 375From the origin at the top draw diagonal lines to the points of intersection ofthese horizontal lines and the outer locus lines . The lengths of these diagonallines give the values of the total meridional thrust . At O 20 deg . this scales
lb .,and at 70 deg . it scales lb . The horizontal radii are 1 2 -83 ft .
and 35-2 ft . respectively . The meridional thrusts per foot are therefore
271 X 1 2 -83 231 X 35-2diagram may b e used to obtain the hoop forces .
Actually the diagram cannot be read directly for O 20 deg , but may b eso read for O 70 deg . Thus the intercept at the latter angle scales lb .
271X 4force diagrams for the upper and lower levels of a ring 1 234 ,
say, 3
-
75 ft . high,
symmetrically spaced about the ordinate for O 20 deg . the hoop compress ionmay be scaled off . This simply amounts to drawing the force diagram to alarger scale near the top . The height of the dome above the ring now Chosen is
37“
5The angle which the meridi onal thrust makes with the horizontal at this
4 37 075
375lb .
, and the angle which the meridional thrust makes
37“
5force at the higher level 13 cot 9
° X 63 14 lb . at thelower level it is cot 27 deg . 20min. X 1 -935 lb .
23
Where applicable this force
acting on an arc 4-0 ft . high , giving H lb . By drawing the
0-
425 ft . and its weight is X lb .
level is cos 9 deg . At the lower side of this ring the vertical load is
with the horizontal is cos“ 1
27 deg . 20 min . The total horizontal
J. s . TERRINGTON .
J. s . TERRINGTON .
DES IGN OF DOil/IES .
TABLE I II .
Levelof joint Weight of dome above each joint(4> deg ) ZW ’
EMI cos w’
(5in (b 45 cos ¢ )l
M x 37-
5’[76 x 0 0 1 5 2 2 -
9 x o I0 -06 0 0 1 4]
I 0 -1 34 o-cmo 2 34 0 1 03]o 3 57 o-zoé ]
E 0 5 0 3 4 1 ]I 0 -657 0 -
52 l0 -826 0 -
741 -0 1 -0
90 deg ,the scaled values of the total meridional thrusts (lb . ) and the total hori
zontal forces (lb . ) causing hoop stresses are :
Total horiz ontal forceover 1 0-deg . are.
The horizontal radii at these levels are37
-
5 sin 20°
1 2-83 ft . 37
-
5 sin 70°
35-25 ft . and 37-5 sin 90
°
375 ft .At all levels the length of are over which the hoop force acts is
1 0°
375 X180°
X 7! 65 5 ft .
At th e level O 20 deg . T271
2
35
1
0
2
0
8lb . per foot (compression)
2336
25
225 lb . per foot (compression)
At the level O 70 deg . T228
20
522 lb . per foot (compression)
“
5
20
225 lb . per foot (tension)
At the level O 90 deg . T-
5lb . per foot (compression)
2:15
:50
n lb . per foot (tension)
These results are practically identical with those obtained analytically .
From the force diagramin Fzg 9 it will be seen that th e total horizontalthrust causing hoop compression is represented by the large ordinate at the j ointof ruptureand is equal to the sum of the separate intercepts ap,
b b’
,cc
'
,etc . ,
downto this j oint . This ordinate also gives the total hoop tension for a hemisphericalshell, and in the same way the intercepts below the j oint of rupture ar e a measureof the horizontal force causing the hoop tension at the various levels
,the sum
27
J . s . TERRINGTON .
of which amounts to the aggregate hoop tension force . If the height of the domeis less than the radius this total hoop tension is not taken up completely at thevarious levels below the j oint of rupture
,with the result that the intercept at
the springing level is represented by the total ordinate . Thus in Fig. 9 , if thedome is only seven zones high
,that is
,h. z
17
07» then the intercepts for the twolowest j oints F and G are ff
’ and gv and the hoop tension which (in a hemispherical dome) would have been taken up gradually from G to ! has to beabsorbed completely at the level G and the hoop force at this level is gv . Hencea strong band of reinforcement
,determined by the magnitude of the ordinate
gv , is necessary at the bottom of this shallow dome in contrast with the hemispherical dome in which the hoop tension to be provided for varies more graduallydown -to the springing level .
Flat Domed RoofsThis problem of flat domes has frequently to be considered in the construction
of roofs and floors of water towers which are often constructed as flat domes .
Here the height is so small that there is no hoop tension in the dome,and the
hoop thrust must be absorbed in ring tension by a band of reinforcement which,
for a roof , can be accommodated in a continuous cornice around the periphery
Fig . 14 .
at the top of the tank . The calculation of the cross- sectional area of this bandof reinforcement is simple .
Let W the weight of the Shell and the load above it , which for a roof is anominal superload and for a floor is the weight of the water above . F ig . 14shows a section through the dome
,the ring beam supporting the latter being 2r
in diameter . At this level the shell makes an angle at with the horizontal .W WThen is the total meridional thrust and —
.cos or is the total radial thrust .
$111 or Slnat
This total radial thrust has a magnitude of per unit of circumference
and induces a circumferential tension Ww t or
X rW Cot onin the ring beam .
27W
This force has to be resisted by ring steel working at the usual stress ofor lb . per square inch
,ample laps and hooks being provided as the working
stress is constant at any section of the circumference .
Application to Piled Foundations .
Another application of the theory of the dome which has the advantage ofeconomy in reinforcement is in the design of circular piled foundations . In acircular piled b ase
,as in a chimney foundation , it is imperative to place the piles
28
DES IGN OF DOMES .
at as large a radius as possible to take up the additional thrust due to W ind ,
thereby lessening the maximum pile load and increasing the factor of safetyagainst overturning of the structure . In doing this the reaction from the outerpiles will not be transferred directly to the brickwork . The outer ring of piles
,
for instance , in Fig . 1 5 must be resisted in radial bending by inserting a mat ofreinforcement at the bottom of the concrete raft slab or alternatively by applyingthe principles of the dome in the following manner .
The thick concrete base or raft slab being stiff th e -dead weight
Fig . 15 .
from the chimney uni formly over the piles so that the pile load is the total deadweight divided by the number of piles .
The total dead load reaction from the outer ring of piles may b e transmittedto the brickwork by any inward inclined thrust T at a predetermined angle or,provided the horizontal radial component of this thrust T is balanced by aninward force supplied by circumferential tension . As in a domed sh ell
,this
ring steel may b e calculated in the following manner . If each pile reactiondue to the dead load is of magnitude P acting at a diameter 27 , then the totalreaction of 11 2 piles in the outer ring is n,P. Hence T Sinai ngp, and theoutward radial thrust is n cos a. The radial thrust per unit of circum
ference is cot aand the ring tension is As in the ring beam at the
29
J . 5 . TERRINGTON .
bottom of a flat dome , provision must be made for this total circumferentialforce
,at the usual allowable working stress of steel in tension
,in the form of a
continuous ring at the edge of the base with suitable secondary reinforcement .
The vertical reactions due to dead load from the outside piles are therebytransmitted directly into the brickwork without cantilever steel .
The additional load on the piles due to wind,and particularly those in the
outer ring on which this load is greatest , must be provided for in cantileverbending in the more orthodox fashion since the pile reactions from this causevary round the ring according to the direction in which the wind is blowing andthe diagonal or inclined thrusts are therefore not constant at any moment roundthe circumference . Moreover
,the additional forces are only ’ occasional and
intermittent . The method of obtaining the pile reaction due to the wind doesnot come within the scope of these articles . An outline of the method
,however
,
is as follows . Let n1 be the number of piles at a radius 71 and ii 2 the numberof piles at a radius 72 , etc. Then I
nIW
moment of inertia of the pilesabout a diameter %(n171
2 nzr z2 I. The distance of the farthest
Fig . 16 .
pile from the centre is r z . Now if Mmis the total overturning moment due towind ,
then the additional load on any pile at a distance x, from a diameter is
M M 2
,and the
maximum extra load is on any outside leeward pile when the wind is blowing in theMw X 72
and on any pile at distance si z from a diameter is
direction of the arrow and is These additional loads on the outside
piles (and the maximum may occur on any of the outside piles) must be taken bycantilever steel in the bottom of the raft slab as indicated earlier .
Th e Conical Dome .
Another surface of revolution which may be used to support a load at thecrown is the right cone , the section through the axis of which is a triangle .
Here again,the stability of the Shell is attained b ydirect forces acting down
the shell through the apex and circumferentially at all levels , without transverseor bending stresses . The following analysis shows the principles involved andthe method of estimating the stresses . The weight lb per unit of surface is deemedto include the weight of the shell
,covering
,and superload . Differing from the
spherical dome,the angles which the meridional lines make with the axis of
revolution are constant at any level and are half the vertex angle , say O ,in Fig. 1 6 .
30
DES IGN OF DOMES .
Consider an elemental ring at a depth y from the vertex and having a heightdy .
The vert ical component of the total meridional thrust T at this level isequal in magnitude to the weight of the cone ab ove . Expressed algebraicallythis is
w y
2 cos2 O
On the underside of the elemental ring the meridional thrustincreases to (T dT) , and hence the problem is to find theeffect of this small increas e in the thrust which in turn causeshoop compression around circles of latitude .
The value of the hoop compression H is given byHarry . tan O ds Fig 17
The component of T acting horizontally and radially from the axis of revolation is T sin O . This radial component causes a hoop tension equal to T simO
sin2 Otimes the radius of the ring of a depth y ,that is ,
T sinOy tanO Ty95
At the same time the thrusts (T dT) have a similar effect
(T mo dyfinz
jbut in the opposite direction and causing hoop compression ; the difference ofthe two effects determines the hoop force in the elemental ring . For a cone thelatter is gr eater than the former , resulting in hoop compression at all points onthe surface .
The difference expressed algebraically isHy H(y + dy ) dcosO cosO cosO
y2
sin”i
cos O sin 2 O (1
COS 3 O (I;2w Sln
2
O2y
2 cos O10 tan 2O.y
or H w tanzOyAs shown by the negative sign , the cone is in hoop compression at all levels ,
and the supports have to take an inclined thrust giving a horizontal reactionwY sin OT srn O2 cos
2 Oring tension as described for the flat dome . Actual values for any example mayreadily be substituted in the expressions given .
This force may be taken by a band of reinforcement in
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