Design of a short concrete shell

RICHARD ALLAN FINN

Submitted in Partial Fulfillment

JUN 25 1958

L1BRA9S of the Requirements for the

Degree of Bachelor of Science at the

MASSACHUSETTS INSTITUTE OF TECHNOLOGY May 26, 1958

Signature redacted

Signature of Author., .. .op.0-.. . .. ...0. *... Department of Civil and Sanitary Engineering

May 26, 1958

Signature redacted

Certified by.. . . . Thesis ... p.e...

II. Introduction

A. Brief History . . . . . . . . . . . . . . 2

B. Shell Behavior ... ... .. . . . .

2

C.

Choice of Dimensions . . . . . . . . . .

3

D.

Outline of Analysis Procedure . . .

7

III. Procedure and Calculations

A.

Calculation of Properties . . . . . . . . . . . 9

B. Membrane Analysis. . . 10

C. Line Load Analysis . . . . . . 13

D. Statical Check of Simply Supported Shells . . . 24

E. Effects of Continuity Over Supports' . . . . 29

F. Final Internal Forces . . . . . . .

31

G. Determination of Required Reinforcing Steel Areas32

H.

Conclusions and Evaluiations . . .

33

IV. Appendix Table I I Table II Table III Table IV Table V Table VI V. Illustrati Figure 1 Figure 2 Figure 3 Figure 4 Figure 5a Figure 5b Figure 5c Figure 5d Figure 6 Figure 7 Figure 8 Figure 9 Figure 10 nte She Int She Int She Int Int Int ons rnal Forces of Simply Supported er A-B Fre o Simply S .. . . . . ernal Forces of Simply Supported 11 C-D . . . . ernal Forces of Simply Supported 11 E-F . . 1* * * . . . . . . . . . 4.

ernal Forces of Continuous Shell A-B ernal Forces of Continuous Shell C-D ernal Forces of Continuous Shell E-F 37 39 41 43 45 46 Element of Shell Surface . . . 2.

Isometric of Structure . . . 4

Dimension Plan of Structure . . . 5

Longitudinal Section of Structure . . . .

6

Definitions of Internal Forces and Deflections . . . . . . . . 8

Surface Loads . . . 11

Line Loads . . . 13

Forces Acting on Edge Beam . . .

15

Transverse Force T for Simply Supported Shells A-B, C-D, and E-F . . . .. . . 20

Longitudinal Force T for Simply Supported Shells A-B, C-D, andXE-F. . . 21

Shear Force, S, at Ribs A and B, Ribs C and D, Ribs E and F for Simply Supported Shells A-B, C-D, and E-F. ... 22

Transverse Moment, M for Simply Supported Shells A-B, C-D, and. E-F . . . .. . . 23

The object of this thesis is to analyze and design a structure containing five short cylindrical concrete shells. The structure is illustrated in Figure 2. Only the end and center shells will be completely analyzed with the idea that the internal forces of the two remaining intermediate shells

can be interpolated from forces of the other three. A de-tailed design will not be attempted, but the areas of

rein-forcing steel required at various points will be examined.

The analysis and design procedure will follow the methods as set forth in the American Society of Civil Engineers Manual No. 31, Design of Cylindrical Shell Concrete Roofs, which was published in 1952. The analysis consists of three major steps, and appropriate tables and coefficients in the Manual will be used as necessary.

The first step is a membrane analysis, in which the shell is assumed to be statically determinate, and membrane forces are determined. However, longitudinal edge support require-ments of the membrane state cannot be fulfilled by actual con-ditions, and in the second step line loads are applied to the longitudinal edges to fulfill the edge requirements. The in-ternal forces caused by the line loads are superimposed upon

the membrane forces. In the third step, continuity of the shells at the ribs is investigated, and the effects of this continuity are superimposed upon the internal forces of the first two steps.

The forces determined in steps one and two, in which the shells are considered to be simply supported, result in families of smooth curves, indicating that interpolation of forces in the two remaining shells can be carried out. Be-cause the actual stress intensities are low, as found in the third step of the analysis, any errors in interpolation will not greatly affect the relative magnitudes of stress inten-sities. The low values of stress intensities, and negligible transverse moments indicate that the shells operate quite efficiently in the transfer of loads by direct stresses.

INTRODUCTION

A. Brief History

Thin shell construction is relatively new, although a theoretical mathematical analysis of shell behavior was first presented in the late Nineteenth Century. The first cylin-drical shell was built in Germany by Carl Zeiss in 1924, and

other Germans have contributed much in the simplification and systematization of shell analysis. A notable example of Ger-man work is the Grand Market Hall in Frankfurt, GerGer-many.

Among the most notable contributions of American designers is the Onondaga County War Memorial Building in Syracuse, New York, and a hangar at Limestone, (Maine) Air Force. Base.

The roof of the latter building has a transverse span of 340 feet, and a shell thickness of 5 inches.

B. Shell Behavior

Thin shells are able to cover large unobstructed areas because of their structur-al behavior. This behavior is char-acterized by the transmission of any loading by direct

stresses, with small bending stresses, provided that the supports furnish the required reactions.

Norma( 'T nqential Corn povient

Componernt

/-ExIternaI

Load p Shearing FFore. Transvi nerse Force

Figure 1. Element of Shell Surface

Figure 1 illustrates the property of shell action which enables shells to carry loads by direct stresses. Consider-ing the strip as a free body, the normal components of the transverse forces resist the normal component of the external load p. The tangential component of p is resisted by radial

shearing forces on the transverse sections. This transverse behavior, peculiar to only shells, enables the shells to carry any type of loading by direct stress.

For shells with a longitudinal length (L) greater than five times the radius (r), the behavior in the longitudinal direction is similar to beam behavior, and it may be assumed that there is a straight line stress distribution. For

shells with an L/r less than five, stress distribution given by beam theory is no longer valid. The stresses near the edge of the shell are of greater intensity then would be-given by ordinary beam theory, and the compressive stresses also become larger, as less of the shell participates in beam action. _{This has the effect of lowering the neutral}

axis of the shell.

C. Choice of Dimensions

The structure which is to be investi;ated in this thesis will house a theater-auditorium. A thin shell was chosen for

the design because of its directness and architectural sim-plicity, and because of the minimum amount of material re-quired to cover the unobstructed area. The shell is illus-trated in Figure 2.

The dimensions and form of the shell roof depended on three considerations: sufficient room for audience seating, sufficient space and head room in the stage area, and economy of construction. At the same time, it was thought desirable to narrow the structure in order to emphasize the stage as the center of interest. These considerations were reconciled by the use of a constant radius throughout the roof, and by

the reduction of the height of the shell roof at a constant rate as the structure was narrowed down toward the stare end.

The use of vertical walls permitted more usuable space for a given roof area, and at the same time, sufficient space and head room in the stage area was provided. The choice of a constant radius will permit the builder to reuse form panels, and thus reduce the construction cost.

The distance between the rib -centers was specified at

<?* c-O 26 D

rigu~re

Z.

Iso mefric

Drawing

of Shell Structure

13

L

ARb8Rib 1)

f:?('b E

Ii

I

.0'l F'9ure

3

o'i mens ion Fpar

0 _(V 0 0 N _______________ 11. 5 a 30.O':/5O. of Figure

4-Longitudmyat 5ec# ion P-A Scale: 1 " = 200' GN 1~~) IC) (V

Q4

=6

I

(V

R1

!

b

1 8 )T I Cm4

culated. _{The dimensions of the structure are given in}

Figures 3 and 4. A shell thickness of 3.5 inches (0.292feet) was chosen; it was felt that this was a reasonable thickness,

since small transverse moments were expected. D. Outline of Analysis Procedure

The analysis consists of three major steps: first a membrane analysis, second, an analysis of the effects of

line loads, and finally an evaluation of the effects of con-tinuity across the ribs. The definitions of internal forces and displacements are given in Figure 5. a.

In the membrane analysis, the shell is assumed to be statically determinate. The shell is considered to be supported along the longitudinal and transverse edges in a manner consistent with the membrane condition. In this state

all surface loads are transmitted by direct forces within the plane of the shell, and there is no transverse moment. How-ever, the reactions and displacements along the edges are in-consistent with actual conditions.

In order to correct for these inconsistencies, line loads are placed along the longitudinal edge in the second step. An edge beam is used to stiffen the shell and reduce edge

deflection. Since it is assumed that the beam can resist only vertical loads and horizontal shear forces, the final line loads applied at the longitudinal edge must be equal and opposite to these beam loads. The vertical displacement of the edge must equal the vertical displacement of the beam, and the longitudinal strains of shell and beam must also be equal in order for continuity to exist at the edge boundary. The line loads create both direct forces and transverse

moments in the shell. These direct forces are superimposed upon the membrane forces previously determined.

It is interesting to note the effects of line loads applied to a longitudinal edge on shells of different physi-cal dimensions. Shells with a small r/L ratio and a small

WS

'Ftr 5a, Definitions of internal forces and displacements T = the direct force component in the transverse direction,

considered positive when tensile;

T = the direct force component in the lonqitudinal direction, X considered positive when tensile;

S = the tangential shearing force, considered positive when it creates tension in the direction of increasing

values of x and

$;

V = the vertical displacement, considered positive when it is in the downward direction;

L = length of shell between supports; r = centerline radius of shell;

t = thickness of shell;

x = longitudinal distance measured from the left support; = angle measured from the right edge of shell;

Ok = angle subtended by the edge of shell measured from the centerline axis;

E = modulus of elasticity of concrete;

Pu = intensity of uniform load on unit area; p = intensity of dead load on unit area.

M = bending moment on the radial face, considered positive when it produces tension in the inner fibers;

v = the tangential displacement of the shell, considered positive in the direction of increasing values of 0; w = the radial displacement of the shell, considered

effects of line loads are not damped out, but carry across the shell to the opposite edge; such shells are known as long shells. _{Shells with larger r/L values and larger ok} angles are not so sensitive to line loads, and the effects of line loads are soon damped out; such shells are known as

short shells. _{The change in behavior is a gradual one, but} the division between long shell and short shell behavior takes place at an r/L of about 0.6 for Ok equal to 30*. In this structure, the shells have an r/L of 2.035, and the smallest Oi is about 37*,

A necessary, though not conclusive statical check of the simply _{supported shells is made at this point. The} horizon-tal forces of each shell and beam are checked by Simpson's Rule, from which the compressive T force is found and

equated to the tensile T force and the force within the edge beam. The sum of the moments is also checked by Simpson's Rule; the sum of the internal moments.caused by the T forces taken about the bottom edge is equated to the sum of the ex-ternal moments also taken about the bottom edge. The sum of the internal shear forces is found by Simpson's Rule and equated to the sum of the external surface loads. In all cases, ten equal arc lengths are used, and values are deter-mined from large scale plots of the forces.

The third step in the analysis is the correction for continuity over the supporting ribs. In the exterior shells A-B and E-F, Tx, S, and M are affected. In the interior

shell C-D, only T and M are affected. These corrections are superimposed upon the forces and moments of the simply supported shells, and the final internal forces and trans-verse moments are determined.

PROCEDURE AND CALCULATIONS' A. Calculation of Properties

The shell radius r is determined from Rib A and chord A. The shell height h in each case is specified; thus the chord length c, and the angle k can be calculated. The

longitudi-nal span length L is 30 feet. Dimensional properties are de-termined at each rib and at the midspan of each shell. A

sample calculation of properties is given: Rib A, h = 50.0', c = 120:0' 4 _(hA)2 _{+ (cA)' 4x(50 - + (120)2} rA 8h -50 . , rA = 61.0 feet Rib B, h = 42.5', r = 61.0' sing

_

__ h 0.h48

2

2r

122

Bin '0.590;

2-

36*10, 72*20' Ok 2h also, tan

k

0.731, c - = 116.2 ft. B. Membrane Analysis

For a thickness of 3.5 inches (0.292 ft.), the dead load is 44 psf, plus 6 psf dead load for a hung ceiling or a total of 50 psf. Live load used will be a 25 psf snow load.

Although these loads are of uniform intensity, an approx-imate loading of sinusoidal shape is used. This is because the corrective line loads applied in the second part of the analysis are of sinusoidal shape, and consistency of loading is required. A comparison of uniform-loading and sinusoidal loading of a Fourier aei'jde,,es, ahows good agreement for moment and deflection if only the first term of a Fourier Series is used for shells of these dimensions. Therefore, pu the

uniform live load is assumed to be uniforni in the transverse direction and varies as L sin n in the longitudinal direc-tion; likewise Pd, the dead load, varies as the weight of the shell, and varies as - sin rr - in the longitudinal direction.

iT

These are illustrated in Figure 5 b.

Data necessary for tho membrane analysis is:

Uniform transverse load, pu = 25 psf Dead load, Pd = 50 psf

r = 61.0 ft. L/r = 0.492 (L/r)2 = 0.242

The forces To and Tx vary as sin in which is measured in radians. When x = L/2, sin

=

1, and To and

LIII7IDI

4EE~

adrfTVm

Figure 5 b

T are a maximum. When x

=

0 or x

=

L, sin l= 0, and T

x Lx

and T also equal 0. The shear force varies as cos T , and LL

this function equals +1 at x = 0, 0 at x = L, and -1 at x = L. In the membrane analysis, T and T are determined at midspan, and S is determined at a supoiort.

Coefficients and equations of Table 1B in Manual No. 31 are used to determine To, T , and S in the membrane analysis.

The coefficients for each angle are multiplied by a factor to obtain the force at the particular point. The multipliers for the membrane analysis are as follows.

Uniform Transverse Load pu

T

=

(Mxcoeff,) sin ; M = 4ur 4 x256 = 1,940

T = (Mxcoeff.) sin ; M =pur = -L. 4 x25x6lxO.242 = 470

S = (Mxcoeff.) cos ; Dead Load T = (Mxcoeff.) sin ; M ur L 4 x25x6lx0.492 = 950 T d r 3.14 M = Pdr 325880 Tr

T = (Mxcoeff.) sin . M = 4pdr = 940

S = (Mxcoeff.) cos M = Pdr - 1,900

In Table 1B, coefficients are given in transverse inter-vals of 5 degrees for values of Ok-0. For plotting purposes, intervals of 10 degrees are sufficient. However, since the parameter locates values of the internal force due to the line loads, values of T , Ty, and S in the membrane

analy-sis are also found at 0k-0 values corresponding to the various values of I s . k- is evaluated to the nearest 30 seconds from the exact values of 0 found from the solution, of the parameter. For plotting purposes, however, 0 is

taken to the nearest degree.

The following example will illustrate the method of computation: For 8 - = 1.6, s = 1.6 = 161x292x900 l.6x11.25 or s = 18.0 ft. (in degrees) = x 180 = 18.0x180 16050' r Tr

6l.0x3.l4-therefore Ok-0 = 75*56' - 16050' = 590 and 0 = 170

.mIterpolating in Table 1B between k- = 550008 and Ok-0 = 60*00' gives the following coefficients. The result-ing internal forces for this trcnsverse point are:

T = 1,940(- 0.2658) + 3,880(- 0.5147) or To = - 520 - 1,990 = - 2,510 lb/ft.

T = 470(+ 0.1426) + 940(- 0.1043)

or T

=

+ 70- 100 =-30 lb/ft.

SA = 950(- 0.3851) +

l,900(-

0.5657)

or SA = - 370 - 1,080 = - 1,450 lb/ft.

C. Line Load Analysis

Because in actuality the boundary requirements of the membrane analysis are unfulfilled, the transverse force and

the shear force at the longitudinal edge are unbalanced. Also, the actual edge displacement is not compatible with the mem-brane edge displacement. In order to remedy this situation, corrective line loads are applied at the longitudinal edge. In a simply supported single shell with free edges, the line loads are equal to the unbalanced membrane forces at the edges. However, since the edges of these shells are re-strained by edge. beams, the final line loads are equal to the algebraic sum of the line loads contributed by the edge beam and the line loads required for a free edge condition. . The edge beam used is 18 inches deep by 6 inches wide. A sample calculation of the final line loads of Shell A-B will be made, and the results of the calculation of the final line loads of Shells C-D and E-F will be giben. Line Loads are illustrated in Figure 5-c, wherein TL is the tan-gential line load, RL is the radial line load, and SL is the

shear line load.

In the first step of the determination of final line loads, the shell is detached from the beam and considered separately. The force T created at the edge of Shell A-B is:

T = (+ 80 + 1,060x8.964 + 1,380x4.410) sin

x L

or T = 15,680 sin lbs. per ft.

x L

In the equation above, +80 is the longitudinal force at the edge due to membrane forces, while the two products are effects of line loads which counteract the tangential and shear reac-tions at the edge resulting from the membrane analysis. The

coefficients are taken from Table 3A. In Shell C-D, Tx = 27,350 sin r lbs. per ft., and in Shell E-F, Ty = 40,000

sin

F

lbs. per ft.

The vertical displacement of the edge in the free edge condition is computed by the addition of the vertical com-ponents of the tangential and radial displacement as given

in Table 3B.

= ~t

[(-

1,060x10.13 - 1,380x2.853) 0.9700

+ 103.0(- l,060x0.92o4 - 1,380x0.2029) 0.2431 sin

L TTx -4.7x1 6 Trx

or 4V

=

45,660ft sin r- E sin L

In the equation above, the first term is the vertical com-ponent of the tangential displacement v, and the second term is the vertical component of the radial displacement w, due to the application of tangential and shear line loads. Also, L/t = 103, sin Ok = 0.9700, and cos Ok = 0.2431. If E is assumed equal to 3,000,000 psi, dead load is taken at 2/3 of

the total live load and long time deflection assumed 3.8 times the instantaneous deflection, then:

2x4.7x106x3.8 + 1x4.7x106 _{0.0316 ft.}

V 66

3x3x10 x144 3x3x10 x144 or A _V, 0 .38 in. I ₈ in,

For Shell C-D,Av = 147,890 sin _{,/o A ; in.} L ETX L .92 i

and for Shell E-F, A

=

- 318,400 L sin 1,-o x 2 in

These deflections given in inches are probably 4 to 5 times greater than actual shell deflections. The factor 3.8 is a maximum for beam deflection, and shell creep is less than

the creep of ordinary beams.

The next step is to reunite the common edges of shell and beam and satisfy the requirements of continuity. Only the requirements that vertical displacement and longitudinal strains are coincident need be satisfied. The other two re-quirements of continuity, that of horizontal displacement' and angular rotation can be neglected since horizontal and rotational stiffnesses of the beam are slight as compared with those of the shell.

Continuity requires that line loads acting on the edge of the shell must be equal and opposite to those carried by the edge beam. Since the edge beam is assumed to resist only vertical loads and horizontal shearing loads, forces acting on the shell must correspond to these two loads. If forces acting on the beam are defined as Vb and Sb, the shell line loads are from Figure 5'd:

TL =Vb sin k= - 0.970OVb (la) RL Vb cos _k= 0.2 4_31Vb (lb) S. b (1c) rb V₆ Figure 5 d

the condition that the horizontal edge force equals zero. Longitudinal stress and vertical displacement at the edge are determined by superimposing effects of these line loads created by the beam on those previously determined for the f.7ee edge condition. ThusTx and A V are:

T

x- (15,680 + 8 .96 4TL - x.6374RL + 4 .410SL) sin (2a)

16 = L 45,660 +(- 101.3 TL + 103x0.9204RL - 2.853L) 0.9700

+ 103(-0.9204TL + 103xO.1174RL - 0.2029SL)0.2431 sin T(2b) Substituting into equations (2a) and (2b) the values of TL9

RL, and SL as expressed in equations (1) gives:

T T

T (15,680 - 24.6V + 4.410'sb sin (3a)

(- 45,660 + 12 7 .9Vb ~-9S) sin (3b)

Equations (3a) and (3b) express the longitudinal stress and vertical displacement of the shell's edge in terms of the in-ternal forces acting on the beam. Similar expressions are written for the beam based on its elastic behavior. In order

to maintain compatibility between TL and Vb and between SL and Sb, the vertie&I loading Vb and shearing forces Sb must vary as sin and as cos T respectively.

L L

The results of the flexural theory for homogeneous beams for this type of loading , as given in Manual 31, are as

follows. The extreme fiber stress at the top of the beam is:

f = 0.60793x (v + 143) + 1.2732 b sin (4)

bh[0I 79xh b L

in which the sign convention for fiber stress conforms to that used in the shell analysis; i-e. positive indicates

4 weight of the beam (112.5lbs. per ft.) multiplied by T.

The vertical displacement of the beam is: SL3 0.12319L(Vb _{+ 143) + 0.193513b}

sin ₍₅₎

Equating stresses and deflections given by equations (3) to those of equations (4) and (5) and multiplying by t and respectively gives: 15,680 -

2

4 .6Vb + 4_.4_10Sb= Lt [-0.60793 (Vb + 143) + 1.273283 (6a) also, +45,660 - 12 7 .9Vb + 7.9Sb = Lt _{0.12319L(Vb + 143) + 0.193513} (5b) Substituting values of:

L _{x 30 x} 11.67 h b 1.5 x = 11.67x = 233.3 R) b 1.5 and (h!)8 T; 1 2331.3x 1.5 -4,667

into equations (6) and simplifying gives:

llTVb + 19.2Sb = -36,080

-70OVb - 37.3Sb = 36,140

from which a simultaneous solution results in: Vb = +72 lb. per ft.

Substitution of these values in equations (1) gives line loads which reunite the common edges of barrel and beam:

TL = - 0.9700x72 = -70.0 lbs. per ft. RL = 0.2431x72 = 17.5 lbs. per ft. SL = -2,320 lbs. per ft.

Final line loads are the sum of the line loads created by the beam and the line loads required in the free edge con-dition. They are:

TL = - 70 + 1,060 = 990 lb. per ft. RL = 17.5 lb. per ft.

SL = - 2,320 + 1,380 = - 940 lb. per ft.

The contributions of the line loads to the internal forces of the shell are evaluated at various values of the parameter S - , and superimposed upon the membrane forces

/rtL

previously determined at these points. Coefficients are given in Table 3A. For all internal forces, the total line load contribution equals the sum of the contributions of the

individual line loads. At each value of %-s , the

in-/r-tL

dividual line load contribution equals the appropriate co-efficient times a Multiplier.

The multipliers for the different line loads are as follows: For T,, To, and S, the multipliers for the tangen-tial and shear line loads are TL and SL while the multiplier of the radial line load for these forces is RL x L _L t = 17.5 x 103 = 1,805. Multipliers for the tangential, radial, and

shear line loads for the transverse moment Mo are respectively: T x t = 990x0.292 = 289

RL x L = 17.5x30 = 525

SL x t = - 940x0.292 = -274

A sample calculation is made to illustrate the method of

de-termining internal forces. For Shell A-B at S - = 1.6, rtL

T = + 70 - 100 + f(TL) _{+ f(RL) + f(SL} T = + 70 - 100 + 990(-0.263) ₊ 1,805(-0.024) _- 940(-0.178) T = + 70 - 100 - 260 - 80 + 170 or T = -200 lbs. per ft. also M= 289(-0.3689) ₊ 525(-0.03322) _- 274(-0.0825) M= - 110 + 20 + 20 or M= -70 ft.-lb. per ft.

The same calculations are used in determining the in-ternal forces for Shell C-D and Shell E-F. The values of the

internal forces are summarized in Tables I, II, and III in the Appendix. _{The internal forces for the simply supported}

shells are also plotted in Figures 6, 7, 8, and 9.

Several changes occur in the magnitude of internal forces because of the changes in dimensions of the shells. In contrasting Shells A-B and E-F, the deflection of Shell E-F is greater than the deflection of Shell A-B because of the reduced height and smaller angle Ok of Shell E-F. Tx in the free edge condition is also considerably larger, again because of the reduced height of Shell E-F. To in the mem-brane state is larger than in Shell A-B, and consequently, all three of these changes increase the magnitudes of the line loads. And finally, an increase in the values of line loads again increases all of the internal forces of the simply

supported shells.

In deciding whether the other two Shells B-C, and D-E need to be fully analyzed, the curves for the internal forces

should be evaluated. The curves for T are the only ones which do not have the same general shape at each shell. Al-though a sizable error in the interpolation of To for Shell

D-E may result from the irregular shape, the magnitude of T in the region of Shell D-E is about 5,000 lbs. per ft. com-pression. This force is equivalent to a unit concrete stress of 5,00, or about 120 psi compression, which is hardly

Flure

to

Tro ns verse Fr-rce,

To,

a f --L/.2 70-IJ 4. 4. A, 1- 4. 1 J~

A

I

C-D -

I

1

1

1--1 1 4 4. ~-- I I

~1

I

-I 4. 1 I I. 4 I -3.0 _-2.0 -110 300-V z POO a Ed~ 0

T

' 4oo0 /t, o lpt

For SiMply Suppor fed

She//s5 P-0B C-P)

gE-c

40*3O -

20-ia'

C row~r N -7o0 -6 _.0 _-50 -4.0 Sr- Ve ri - o1 -0 Crown -D

70' b0 500 400 300 200 Edje 0* A, I I I ___________________ I I

-3 0 -zo -. 0 0 ₊₁₀ #20 .f3.0 +40 +& 6.0 +Io

L-onitudmnol For-ce, 7 , p; /,000 /bs per Ft.

i'O Figure 7

Crown c- Longit udin al Force.Ty

af X L/Z,F4or- Simp ly Supported She-1le A-f

c-O, and E-F

Crown EF

E-F

-40

Figure

8

Shear Force, 5, ot Ri/bs P

and 8, Rbs C and D Rbs

E qard F AaP- Simply Supported sheils 1 -8, C-0, in4 E-F.

S /.z I t I A' / I / / 0 00 too SFS _4s d- ~ 0 ________________ I___________ -5.0 ///15/r) 0 -1.0

Shea.r Force ei, /,000 lbs. per Ft.

JN~ 70* 601-50 ,

4

3 :6.0 Crow m, RI'b 8 Crown, Rib C Crown, RibD Crown, Ri, E Crown, RIbF

1-a wer Edge

3. 0 2. 0

Fltgure 9 Trculsverse Momevf,tf6 of X a L/z For Sirmply Su por+PJ 6hells P9-8, C- ' anat E- F, +;zoo

Tran 5verse Momant , M4, ('V% Ft-lbs per foot

(I

/5

E-F

c-o

-30* 200 C 100 EdSe 0 -200 -/00 0 +(00 43+300 +400 --C-D

critical. _{Therefore, the values of T for Shells D-E and B-C} can be safely interpolated without any important errors in design consideration occuring. The values of T , S, and M

for Shells B-C and D-E cam also be interpolated from the curves of Shells A-B, C-D, and E-F. The errors in inter-polation for these forces will be reasonable, because of the

smooth transitions of the curves. Again, no error in inter-polation will significantly affect the design consideration because of the relatively low unit stresses in the simply

supported shells.

D. Statical Check of Simply Supported Shells

As a precaution against numerical errors, a statical check is made of each simply supported shell. The sum of the horizontal T forces and horizontal beam forces at a

transverse section must equal zero. The sum of the vertical components of the shearing force at each rib must equal one half the total load on the shell. The total internal lon-gitudinal moment at midspan due to the T forces must equal the external moment.

The procedure employed for the statical check will be illustrated by the statical check of Shell A-B. In the check of the horizontal T forces, and beam forces at

mid-span, Simpson's Rule will be used to evaluate (+) Tx and

(-)

T forces. Simpson's Rule, as applied to this check is 6A (y + 4y, + 2ya ₊ 4ya.... _{+ y}

where the y values are equally spaced ordinates of the curve of the T forces, in Figure 7, n is an even number, and4A is the distance between ordinates. The distance from the edge to the crown is divided into 10 equal arc lengths.

A _ rfk(radians) _ 6=x75.93x3.14 8.08 ft.

10 10x180

also,AO equals 75 ₁₀3 = 7.590

The internal stress at the top of the beam at midspan is given by:

-0.60793x (72 + 143) + 1.27-2(-0

t

5x11.5

[

0 1.5 l.73(9320~

or, f = - 104,500 - 118,000 = 13,500 psf The fiber stress at the bottom at midspan is:

f = +

104,500

- 1189000 = 45,500 Psf

The forces at the top and bottom of the beam : T (top) = 13,500x0.5 = 6,750 lb. per ft. T (bottom) = 45,500x0.5 = 27,750 lb. per ft.

The assumed straight line variation of the forces results in a total horizontal beam force (F) of :

F = (6,750 + 27,750) x 1.5 = 25,850 lb. 2 Theref'er@fH = 8 1+ 3,680 + 4(-1,600)+ 2(-350) + 4(-75) + 2(-100) + 4(-200) + 2(-250) + 4(-275) + 2(-500) + 4(-327,) - 330J + 25,850 .H 8.08 ( 12,320 + 3,680) + 25,850 H = - 23,300 + 25,850 = +2,550 The rroris 2,550xl10 --5,2 The error is 23,300 + 25,850 = 5.2

For the check of the vertical sheear at a support, Simp-son's Rule takes the form of:

V 4A sin(k-) + 4yisin( ) + 2yasin(

... + yn sin(fk~n) x 2

Where.VI is the internal shear, and the values of y sin(Ok-) give the vertical components for the ordinates of the shear

curve in Figure 8. Since the shell is symetrical about the crown, the VI is multiplied by 2 to obtain the total Vie

The external shear (VE) is:

L 44

IVE = ~T Pu x + p dx20AAx + Ib + Vb

Where pu = 25 psf, c is the chord length, pd = 50 psf, 20AA is the arc length of the shell at the rib, Wb is the dead weight of the beam, and Vb is the vertical beam load as determined in the line load analysis.

For the shear at Rib A: Ok = 79.7*, n = 10 equal arc lengthsA 0 = 7.970, and AA = 8.5'

The following example will illustrate the computation: At point 3, 0 = 23.91*, Ok-O = 55.79*, sin(kk-0) = 0.827,

y3 from Figure 8 = -1,350, and 4ya sin($k-O) = 3,970

.Vy

=-5

x 2F- 2,260 - 9,130 - 2,150 - 3,970 - 2,000 - 1,060

- 1,30C -

50

-

170

-

07

or'..V- -144,000 lb.

=.E _{1 4}f 25xl2Oxl.273 + 50x2x8.5x1.27 + 21 or 7.VE = -142,000 lb.

In which case VI is 1.4% greater than VE. The same check is carried out for Rib B of Shell A-B, and it is found that V is 8.4% greater than V1.

The internal moment (14) produced by horizontal forces at any section is always the same regardless of where the axis is chosen. Here the moment is computed about the bottom

edge of the shell, and Simpson's Rule takes the form

Where the y values are the ordinates to the T curves of Figure 7 _{at the same locations used in the horizontal forces} check, and h is the companion moment arm about the bottom edge for each y value. _{The moment arm. equals r Ccos($k~0)} -cosokJ. _{The summation is multiplied by 2 to include moment} at either side of the crown.

The final M is the sum of the moment of the Tx forces about the edge and the moment (MF) of the force in the edge beam about the lower edge of the shell. Assuming a straight line 'distribution of forces with the beam, the force diagram of Figure 10 results. The diagram is divided into 2 triangles, and the moment of each is taken about the lower edge; the

moment arm of the beam force (F) is then found.

TX t Ty.b _ _

k

h

Ficure 10 Txt + Txb MF= -F(k) = (h) 7- F or

or

M7 (h)2 x + .T₃

The external moment (Mr) is:

Z u

[x c x + Pd + 10 x 2 x4A +b j

Where the numerical values are the same as those used in the vertical shear check of the shell, except that c is the chord

at midspan, and the arc length is also taken at midspan. The following example of Shell A-B illustrates the method of computation of the moment check:

At point 3, ya = - 350; Cos 0k = 0.243; a = 15.20; - = 60.730; Cos ( = 0.489; h = 61(0:489 - 0.243) = 19.3 ft. Therefore, 4yahs ₌ -10,500 M = (l~)'~(6*7502 279750~ MF =-1.5)4 6 + ,51 = -23,300 ft.-lbs. ZM= 8 _.08_x 2 0- 49,200 - 10,500 - 5,790 - 5,580 26,650 - 18,900 - 45,500 - 26,400 - 59,100 - 15,200 -23,300 or, YM₁ = - 1,418,000 - 23,300 = 1,441,300 ft.-lb. also, YME = - 25x118.3x1.273 + 50x2x10x8.08xl.273 + 143) or _M = -1,295,000 ft.-lb.

In which case MI is 11.3% greater than M E

The results of statical checks on Shells C-D and E-F are as follows. In Shell C-D (+)T is 2.3% greater than

(-)T ; MI is less than 1% greater than ME; at Rib 0, V, is 1.7% greater than VE, at Rib D, VE is less than 1% greater than V₁. In Shell E-F, (-)T is 8.6% greater than (+)T ; ME is 3.3% greater than MI; at Rib E, V, is 4.5% greater than

and at Rib F, VE is 8.9% greater than VI.

For design purposes, close agreement is not necessary, and errors up to 12 or 15% are not considered excessive

be-cause of the small stress intensities within the shells. The statical check shows up only large and significant errors in the computation of internal forces. There are errors inherent in the statical check itself. Among these errors due to plotting and scaling of ordinates (although

in the actual computation, plots to a larger scale than those illustrated in this presentation are used), errors due to the approximate integration given by Simpson's Rule, errors due to forces not considered in the statical check, and errors due to the fact that the height of the shells

and the angle

are not constant along the shell. E. Effects of Continuity Over Supports

The effect of continuity over the supports on stresses in shells is similar to the effect of continuity on stresses in ordinary beams. However, because longitudinal behavior of short shells is not similar to beam behavior, direct sol-ution of the effects of continuity in short shells by approx-imating similar effects in beams is inapplicable. Instead relationships ffbav the rigorous satisfaction of boundary con-ditions are used to compute the effects of continuity over

supports. The relationships which apply to this structure are given in Table 5, page 47, of Manual 31, under the head-ing Case 2- Three Equal Spans.

In the subsequent discussion , the following symbols will be used: Cou, correction for continuity of uniform

transverse load; Ced, correction for continuity of dead load; C L, correction for continuity of line loads; Cc, the correc-tion for continuity of any internal force or moment; the

sub-script s denotes the internal force or moment in a simply supported shell; and the superscript ' indicates that the value is to be taken at x = 0 or x = L/2, whichever gives the maximum value.

The corrections for continuity of exterior Shells A-B and E-F are handled exactly alike. The distance x is

measured from the interior rib in each case or Rib B, and Rib E. The Cc for each internal force is the sum of Ccu'

cd, and CcLI

The following relationships hold for the shear force S

.Cc , 12

Cd b s 12

10-k8o. -k

80-CcL S + A(e -e Wherein:

k8 ,

$

is radians, and

A C cu-2k8 d cL for C' Cd, and CL taken at # =0 e

For Longitudinal Force T,:

C =-T 12 Ocu = Ixs _{T [5 + 3} 0 ,T 12

o

=

-cd xs n +2 Tr [,- 12( -kr)k8 0_{cL 12 T' + A}

f

_(e 8 _{+ e} )

Wherein A and k8 are the same numerical values used for the

shear force computation. For Transverse Moment M

0 -MI 12 rr

c Os 5

The following example will illustrate the computation of a Cc in exterior Shells A-B and C-D.

For Shell A-B at 0 = 170, considering s:

k8 = 3.15, A = -280 C = -420 x 0.07 = -30 cu C d = -1,000 x 0.022 = -20 CcL 280x0.244 + (-280) (0.3816) = 60 - 110 = -40 Therefore, C0 = -90 lbs. per ft.

Considering T :

Ccu = - (70 x 0.22) = -20 C d = -(-100 x 0.07) = +10

CcL = - (-170.765) + (-280) _{(1.5) (0.383) =} 130 + 170 = 300

Therefore, C = +290 lbs. per ft.

In interior Shell C-D, the Cc for T is computed exactly the same as the C for T in the exterior spans. There is no 0 for the shear force. A is evaluated from the values of Pu' Pd, and L.L. at

$

= 0 using the same multipliers as used

in Shells A-B and E-F for these loads. Fo

-MC

--

M'

12TTr x

For M _{sc 10 Xr}

where x can be evaluated from either support.

F. Final Internal Forces

Final internal forces for Shells A-B, C-D, and E-F are summarized in Tables IV, V, and VI of the Appendix. Values are taken at the supports and quarterpoints, for various values of 0 and s. The values at 0 = 260, s = 2-:Q ft.,

B = 2.4 are taken from the curves of the simply /tL4

supported shells.

For exterior Shells A-B and E-F, the final forces are:

T = T' sin

T =T' sin LE+ Cce-X

s=S' Trx + C

MO M' _{sin T + C c}

- _{+ I}

For interior ShellC-D, the final forces are: T$= T' sin

= T' sin + C

s S' dos x

8 OL

M1= M + C C (_

The following example will illustrate the computation

of final internal forces. For Shell A-B, = 170, s = 18.0 ft.

S' = -1,140 lbs. per ft., C = -90 lbs. per ft. at x = 0, S = +1.000(-1,140) + (-90) =.-1,230 lbs. per ft, at x = L/4, S = o0.707(-1,140) + (-90) = -950 at B = 1/2, S = 0(-1,14o) + (-90) = -90 at x = 3L/4, S = -0.707(-l,140) + (-90) = +720 at x = L, S = -1,000(-1,140) + (-90) = +1,050

The values of M for Shell C-D are not summarized, be-cause as a result of the Cc values for MO, no final internal M is greater than 10 ft.-lbs. per ft.

G. Determination of Required Reinforcing Steel Areas:

A detailed design is not attempted in this section, but only examples for the steel required to resist the internal forces are discussed. A working stress value (f ) of

20,000 psi is assumed for the steel in tension; a 0 psi value is assumed for concrete in tension; for an ultimate compressive strength value of 3,000 psi for concrete; a

working stress value F'c of 1,350 psi compression is assumed. A compressive stress of 1,700 psi is equal to compressive force of (1,350x3.5x12) lb. per foot or 56,800 lbs. per foot. Since there is no compressive force which is that large any-where within the shells, no compressive reinforcing steel is required. However, nominal reinforcement is required in the compressive zones for shrinkage and temperature requirements.

The transverse moment M may be neglected because even at the maximum values, the stress contribution is extremely

small.

The reinforcing used is two layers of welded wire fabric which serves as nominal compressive reinforcement, and is able to resist, all tensile and shear forces except in a few areas of the shell structure. In these areas steel rein-forcing bars or additional fabric is added.

If two layers of 4"x4" fabric of no. 4 gauge is used, the steel area per foot is 0.240 in., or approximately 0.6 per cent. This steel will resist 4,800 lb. per foot tension.

The largest tensile force in the shell structure equals 4,500 lb. in Shell E-F. Therefore, no horizontal bars are

re-quired to resist the horizontal force. The concrete can re-sist a shear stress up to 0.3 f'c or 90 psi, which is equal to 3,800 lbs. per foot. Where the shear force is greater than 3,800 lbs. per foot, steel bars are placed at 45* to resist the excess. For example in Shell E-F at x = 0

(at Rib E) the shear force equals 7,140 lbs. per foot at s = 2.25; the steel area required is 20,000

=

0.17 sq. in. per foot. Where the wire fabric is lightly stressed by its primary load, it may be used to resist diagonal shear.

How-ever, as pointed out in Manual 31, the effectiveness in the 45* direction is only half of the normal value. In the

example cited, the fabric substantially loaded, so its con-tribution to resist shear is negligible.

H. Conclusions and Evaluations

As has been stated in Section C, it is possible to interpolate the internal forces of Shells B-C and D-E from the calculated forces of the other three shells. Shells B-C and D-E are then subject to a statical check, and the cor-rections of continuity and calculations of final internal

forces are made as has been illustrated in the shells which were completely analyzed.

A complete and detailed analysis and design of this structure would t'ke far more time than we are able to give for this thesis. However, from what has been accomplished, it may indeed be concluded that this type of structure is

very remarkable. _{Its ability to cover this area with a} minimum of materials, and with such low values of internal stresses is almost phenomenal in comparison _{with conventional} steel or concrete longspan structural systems.

BIBLIOGRAPHY

Design of Cylindricdl Concrete Shell Roofs

Manual of Engineering7. Practice No. _{31, 1952} American Society of Civil Engineers

Design of Barrel -Shell Roofs- 1954 Portland Cement Assbciation

Time Saver Standards Pg. 40,7-423

= 0.020 0h 0k /rtL' gk = 75056' Pu Pd . L r/L = 2.035 RL SL T at x = L/2 (lb, per ft,) Multiplier 76* 00 66 10O00'O 56* 2000' 46* 30*00' 36* 40*00'

3.2

340 42000' 1.6 170 59000'

0.8

9* 67*30' 0.4 4' 71030' 0.2 20 74*00' 0 0 75056' 1,940 -1,940 -1,880 -1,710 -1,460 -1,130 -1,070 - 520 - 290 - 200 - 150 - 120 3,880 -3,880 -3,820 -3,650 -3,360 -2,980 -2,880 + -1,990 --1,480 --1,230 ₊ -1,070 + + -~ .940 + 990 1,805 50 -260 + 190 440 -810 + 940 + 990 10 20 100 60 20 10 0 + -940 -5,820 -5,700 -5,360 -4,820 -4,110 10 -3,920 60~-2,690 20 -22,08-0 130 -:2,-906D 130 - 520 80 - 0

0-

₇₀

T at x = L/2 (lb, per ft,) x Multiplier 760 00 66* 10000' 56* 20000' 46' 30*00' 360 40000'

3.2

340 42*00' 1.6 17* 59000' 0.8 9* 67*30'

0.4

40 _{7 1}*3 0\t 0.2 2* 74*00' 0.1 1* 75*00' 0 0' 75*56' + + 470 140 -130 -110

-70

-30 -20 -70 -100 -110 -120 -130 -130 -940 190 190 180 170 150 140 + 100 -990 1,8 30 + 260 -70 -2,200 60 - 250 50 +3,040 50 +5,570 50 +8,900 05 .-940 20- ₃₀ -80 + 170 -330 320 290 240 180 140 200 + 300 + 220 -1,620 + 200 - 900 - 9QO.' - 230 -2,160 + 720 - 620 -3,050 +1,980 -1,150 -4,150 +3,680 M at x

=

L/2 iplier 34* 42*00' 17* 59*00' 90 67*30' 4* 71*30' 2* 74*00' 1 75*00'

0

75056' (t't,-The p.e 289 + 20 - 110 + - 70 + + 20 + + 30 + + 30 + 0 Sum Mult 3.2 1.6 0.8-0.4 0.2 0.1 0 525 0 20 + 40 30 -10 -10 -0 -274 0 20 0 10 10 10 0 + + + + 20 70 30 40 30 30 0

TABLE I (CON'T) INTERNAL FORCES OF SIMPLY.SUPPORTED SHELL A-B rt/L2 = 0.020 rL r/L =

2.035

U Pd TL RL S at Rib a (lb. per ft.) Multiplier 80* 00 700 1000' 60* 2000' 50* 30000' 40* 40*00' 3.2 340 46*00' 1.6 17* 63000' 0.8 90 7100' 0.4 4* 75030' 0.2 20 77030' 0.1 1* 78030' 0 00 79042' 950 1,900 0 0 150 - 210 290 - 410 390 - 600 450 - 780 450 - 870 370 -1,080 280 -1,140 220 -1,170 190 -1,180 170 -1,180 160 -1,190 Ok = 79*42' 990 1,805 - 90 + + 410 -- 880 + -1,640 + -1,350 + - _{850 +} -

0

20 90 10 150 150 100 0 + -+ + -940 0 - 360 - 700 - 990 -1,230 10 -1,380 40 -1,170 240 -2,050 140 -2,740 210 -2,780 520 -2,620 940 -2,290 S at Rib B (lb. per ft.) Multiplier 720 00 62* 10*00' 520 2000' 420 30000' 3.2 34* 40030' 1.6 17* 55*30' 0.8 9 640oo' 0.4 4* 68*00' 0.2 2* 70*00' 0.1 1* 71*00'. 0 00 72020' 950 1,900 0 0 150 - 210 290 - 410 390 - 600 450 - 780 420 -1,000 360 -1,090 310 -1,120 290 -1,140 280 -1,150 260 -1,150 ,k = 72*20' 990 1,805 0 0 - 90 + + 410 -- 880 + -1,640 + -1,350 + - ₈₅₀ -0 20 90 10 150 150 100 0 + -+ + -940 0 0 360 - 700 - 990 10 -1,290 40 -1,140 240 -2,080 140 -2,780 210 -2,840 520 -2,700 940 -2,350 Sum

TABLE ;Il INTERNAL FORCES OF SIMPLY SUPPORTED SHELL C-D rt/L 2

=

0.020 s8

/rtL

pu Ok = 60*48' Pd TL r/L = 2.035 RL SL Sum T at x

='L/2

(lb. per ft.) Multiplier 61*0 51* 10*00' 41* 20*00' 3.2 34* 27*00' 1.6 17* 44000t

0.8

90 52*30'

0.4

5* 56*30 0.2 20 59080' 0.1 1* 600001 0 0 60048' 1,940 -1,940 -1,880 -1,710 -1,540 -1,000 - 720 - 590 - 520 - 480 - 460 3,880 -3,880 -3,820 -3,650 -3,460 -2,780 -2,360 -2,200 -1,990 -1,940 -1,890 2,190 + 120' - 570 -

430

+

960

+1,780 +2,070 -2,190 9,370 -2,050 -+ + -+ + 30 110 500 310 110 30 0 T at x = L/2 (lb. per ft.) Multiplier 61* 00 51* 10*00' 410 20*00' 3.2 34* 27000'

1.6

170 44*00' 0.8 9* 52*30'

0.4

50 56*30' 0.2 20 59*00' 0.1 10 60*00' 0 0 60048' + 470 140 130 -110 -80 -

5-

40-60 -70 -70 -70 -940 2,190 190 190 180 170 + 70 140 -

570

120 - 4,870 110 - 550 100 + 6,710 100 +12,320 90 _+19,650 9,370 2,050 + 90 - 400 +1,610 +1, 040 -1,210 -3,210 -5,960 - 330 - 320 - 290 - 60 - 150 + 370 - 745 + 480 -2,860 -1,960 -1,520

-4,700

+ 570 -6,640 +2,440 -9,050 +4,620 M at x =L/2 Multiplier 3.2 34*27*00' 1.6 17* 44*00' 0.8 99 52*30' 0.4 4* 56030' 0.2 2* 5900' 0.1 1* 60*00' 0 0 60*48' (TilosPer/ ft. ) 638 2,730 + 40 - 10 + - 240 + 90 + - 160 + 190 + 40 + 130 -+ 80 + 70 -+ 60 + 40 -0 -0 + -5,820 -5,700 -5,360 -4,940 -4,100 -4,050 -1,800 - 800 - 150 - 160 30 140 40 280 280 180 0 -598 10 + 50 -0 + 30 + 20 + 10 + 0, 20 100 30 140 130 90 0

.TABLE 1I (CON'T) INTERNAL FORCES OF SIMPLY SUPPORTED SHELL C-D rt/La = 0.020 k-0 r/L = 2.035 TL RL Pu 8 at Rib C (lb. per ft.) Multiplier 650 0* 55* 1o00'

450

20*00'

3.2 340 3100'

1.6

17* 4800'

0.8

90 56000' 0.4 4* 60*30' 0.2 20 63000' 0.1 1* 64o00' 0 0 64042' 950 1,900 0 0 150 - 210 290 - 410 400 - 620 450 - 900 420 -1,000 390 -1,050 370 -1,080 360 -1,090 350 -1 ,100 , ok = 64*42' 2,190 9,370 -2,050 - 90 + 910 -1,950 -3,620 -2,980 -1,870 0 + -+ + + Sum 0 -- 360 - 700 100 + 20- 990 490 - 100

-1,030

60 + 530 -2,780 770 + 300 -3,990 ,790 - 460 -4,100 530 -1,130 -3,920 0 -2,050 -3,500 S at Rib D (lb. per ft.) Multiplier 570 Q0 470 10*00' 370 20*00' 3.2 34* 23*00' 1.6 17* 40000'

0.8

90 48*00'

0.4

4* 52030' 0.2 2* 55000' 0.1 10 5600' 0 0 560441 950 1,900 0 0. 150 - 210 290 - 410 330 - 470 450 - 780 450 - 900 440 - 960 430 - 990 420 -1,000 420 -1,010 , k = 56*44' 2,190 9,370 -2,050 - 90 + 910 -1,950 -3,620 -2,980 -1,870 0 + -+ + + 100 + 20 490 - 100 60 + 530 770 + 300 790 - 460 530 -1,130 0 -2,050 0 - 360 - 700 + 770 - 910 -2,710 -3,950 -4, 070 -3,990 -3,480 s t SL

TABLE III INTERNAL FORCES OF SIMPLY SUPPORTED SHELL E-F rt/La = 0.020 Ok = 42*50' r/L = 2.035 Pu Pd TL RL Sum T at x

=

L/2 (lb. per ft.) Multiplier 43* 00

3.2

340 90001 1.6 17* 26*00' 0.8 90 34*30' 0.4 _{4* 38*30'}

0.2

2* 41*00' O.1 1* 42*00' 0 0

42 *50'

1,940 3,880 -1,940 -3,880 -1,910 -3,730 -1,570 -3,480 -1,320 -3,190 -1,190 -3,040 -1,100 -2,920 -1 070- 880 -l 40-"840

3,700

20,000 -3,100 + 200 --

960

+ - 720 -+1,630 + +3,010 + +3 500 + +39700 60 240 1,070 660 250 70 0 + -5,820 50 -5,500 210 -5,560 70 -6,370 430 -2,370 420 -1,180 280 - 660 0 - 180 T at x = L/2 (1b. per ft.) Multiplier 43* 0,

-3.2

34* 9*00'

-1.6

170 26*00'

-0.8

9* 34*30'

-0.4

4* 38*30' -0.2 2* 4100' -0.1 1* 42*00' -0 0 42*50' -M at x

=

L/2 Multiplier 3.2 34* 9*00' 1.6 170 26*00' 0.8 9* 34*30'

0.4

4* 38*30' 0.2 20 410o' 0.1 10 42*00' 0 0 42050' 470 140 -130

-90

50

-30 -20 -20 -10 -940 3,700 20,000 190 190 + 120 + 180 170 - 970 - 850 160 - 8,240 + 3,440 150 - 930 + 2,230 140 +11,350 -- 2,590 140 +20,820 ' 6,850 140 +33,200 -12,750 (ft.-1b. per ft.)

1,080

5,850 -+ 60 - 30 -- 400 + 200 + - 270 + 410 + + 70 + 280 -+ 130 -+ 160 -+ 90 + 80 -0 0 -23100 - 330 a 100-- 120 + 550 -1,530 + 720 -4,290 -

2,970

-1,850 - 7,110 +1,490 -10,050 +3,760 -13,680 +6,630 905 10 + 70 -0 + 40 + 30 + 20 + 0 20 130 140 310 260 150 0 8$ _{S L}

TABLE I (CON'T) INTERNAL FORCES OF SIMPLY SUPPORTED SHELL E-F rt/La = 0.020 s k 0 /rtLr r/L = 2.035 TL _RL Pu Pd S at Rib E (lb. per ft.) Multiplier 48* 0 38* 10000' 3.2 340 1400' 1.6 17* 31*00'

0.8

90 39*30' 0.4 _{40 43*30'} 0.2 2* 46o*0' 0.1 1 47*00' 0 0* 47*46' 950 1,900 0 0 150 - 210 210 - 280 400 - 620 450 - 770 450 - 830 450 - 870 450 - 880 890 - 450 -S

$k

= 47046' 3,700 20,000 -3,100 - 330 + 210 +1,540 - 1,050 -3,280 + 120

-6,110

+

1,640

-5,040-+ 1,680 -3,160 + 1,140 0 0 sum 0 - 360 + 30 - 580 - 150 - 680 + 800 -3,580 + 450 -5,300 - 700 -5,380 -1,700 -4,510 -3,100 -4,440

S at Rib F (lb. per ft.) , Ok = 37*2O' Multiplier 37* 0* 3.2 34* 400' 1.6 17o 21000'

0.8

90 2900'

0.4

40 33*00' 0.2 2* 35*00' 0.1 1* 36*00' 0 0! 37*20' 950 1,900 3,700 20,000 -3,100 0 0 60- 80 - 330 + 210 + 3Q 300 - 430 +1,540 - 1,050 - 150 380 - 590 -3,280 + 120 + 800 410 - 660 -6,110 + 1,640 + 450 430 - 690 -5,040 + 1,680 - 700 430 - 710 -3,160 +

1,140

-1,700 440 - 730 0 0 -3,100 0 - 230

-

390

-3,330 -5,090 -5,180 -4,320 -4,270