Analysis of ferro cement in bending

ARCH1EF

ANALYSIS OF FERRO-CEMENT

IN BENDING

%%iv oft 44, THE DEPARTMENT _OF NAVAL ARCHITECTURE

January 1970

Hans Fredrick Muhlert

ND MARINE ENGINEERING E UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING

Lab. v. Sciieepsbouwkunde

Toe4c3hnische Hogeschool

Delft

U43

ANALYSIS OF FERRO-CEMENT IN BENDING

HANS FREDRICK MUHLERT

STUDENT

DEPARTMENT OF NAVAL ARCHITECTURE

AND

MARINE ENGINEERING THE UNIVERSITY OF MICHIGAN

ANN ARBOR, MICHIGAN

EASTERN CANADIAN SECTION

SOCIETY OF NAVAL ARCHITECTS AND MARINE ENGINEERS MONTREAL DECEMBER 2, 1969 Bibliotheek va-n d Afdeling _{cheepvaartkunde} nusche Hog2schoo DOCUMENTATiE _ut DATUM'. -Schee

ABSTRACT

My purpose in undertaking the work summarized in this paper has been to provide the designer with the

basic information and methods with which to design. ferro-' cement structures. The premise throughout the paper

is that the standard methods for analyzing reinforced. concrete are applicable.

The paper begins with a description of the physical behavior of ferro-cement experiencing a bending moment.

The strength analysis

for

working loads is covered, and

a

method,

is

suggested for estimating the ultimate. strength, The. appendices include test data, Strength talculationS, and comparisons of calculations with test. results.

My particular interest is in marine applications, hence it is my hope that this paper will permit the

PREFACE

The questions I hear most often about ferro-cement boats are still: How could it float? Won't it crack? Won't it be too heavy? This appears particularly ironic for although it is common knowledge that ferro-cement

small craft and full sized concrete ships have been con-structed for some time, the naval architect who wishes

to design a ferro-cement structure to a given loading

has very little to go on. Not only is there a shortage

of strength data, but also it is difficult to estimate the strength of different types of ferro-cement

config-urations. The naval architect needs (1) some general

qualitative data on behavior and weights, (2)

quantita-tive strength data, and (3) an analysis method for

estimating the strength of configuration for which there

is no test data.

My principal objective in this paper is to focus on

the third category, or specifically, to find an analysis method for ferro-cement in bending.

Rather than assume that ferro-cement is a homogen-eous material and then attempt to find an allowable stress,

I have reverted to reinforced concrete techniques. I have

used working stress analysis to estimate the stress under normal working loads, and ultimate strength analysis for

predicting the ultimate strength.

The approach for this work has been to first test

the ferro-cement components. That is, to test the mortar, wire mesh, and reinforcing rods separately. Secondly, to

fabricate ferro-cement specimens and test them in bending,

and thirdly, to compare the test results with the calculated

strength. The test results, calculations, and comparisons

are given in the appendices whereas the theory is given in

In general the predictions are conservative, and the data scatter is within the limits normally encountered in

reinforced concrete tests. Hence the methods outlined could

be used for analyzing marine structures.

In particular, one could design to the loads, if they

are known. In the absence of loading information the struc-ture could be designed to the equivalent strength or stiff-ness of a known successful structure of a different material.

For example, the shell of a ferro-cement boat could be

designed to the equivalent strength or stiffness of a similar

boat in fiberglass or laminated wood. In any case, the naval

architect should be able to estimate the strength and weight of a ferro-cement boat, and to furnish quantitative answers to those three simple questions that he will undoubtedly be

asked.

There are a number of people who have given generously

of their time in assisting me with this work. The project

was originally suggested by Robert Allan and was done under

the supervision of Professor Amelio M. D'Arcangelo. I am

greatly indebted for the untiring help of Professors Legg and Legatski of the Civil Engineering Department, and for the constant encouragement of Professor Benford, Chairman of the Department of Naval Architecture and Marine

Engin-eering. Thanks also go to Charles Canby for going through

my calculations and making helpful suggestions for the format, and to Debbie Moore, for doing a fine job of a very difficult

typing task. The photographs 5 and 6 in Appendix E were taken

by Terry Little, and the one on page 3 was taken by Dean Runyan.

Martin Iorns, of Fibersteel Corporation, deserves credit for

his encouragement and support. Finally, a word of thanks goes

to M. Rosenblatt and Son, Inc. for assisting with the travel

CONTENTS

Page

FORCE-DEFLECTION CHARACTERISTICS 1

II. WORKING-STRESS ANALYSIS 5

ULTIMATE STRENGTH ANALYSIS 19

IV. CONCLUSIONS 25

BIBLIOGRAPHY 27

APPENDICES,

A- TEST DATA 29

COMPUTER PROGRAM FOR

WORKING-STRESS ANALYSIS 42

ULTIMATE STRENGTH CALCULATIONS 65

GRAPHS 77

FORCE-DEFLECTION CHARACTERISTICS

The typical force-deflection curve of a ferro-cement

specimen experiencing a gradually increasing bending moment is shown in Figure 1. The curve has three distinct parts. The first part is in the uncracked range. It is approximately linear, and the bending moments are quite low. The curve

then changes slope and becomes linear again in the cracked

range. The curve changes slope a second time and becomes

horizontal.

In the first part of the curve the specimen behaves elastically, as is suggested by the constant slope. The

steel and mortar are stressed below their yield strength

on both the tension and the compression sides of the beam. The tensile strength of the mortar, however, is quite low, approximately 400 lb/in.2. Furthermore, since mortar is a

brittle material, its tensile strength is not clearly

defined. Hence, at a relatively low and somewhat unpredictable

bending moment the mortar will crack on the tension side.

(Mb)

Yielding

Uncracked range

Deflection (D)

Fig. 1. Load deflection curve.

Cracked range

2 _{Force-Deflection Characteristics}

The cracks will propagate from the tension surface toward the middle of the specimen and will stop at the neutral

axis where the stress is zero. Figure 3 illustrates this cracking process. These cracks form and propagate between

the uncracked range and the cracked range. As the specimen

cracks there is a load transfer. The tensile load carried

by the mortar is transferred to the steel in the tension

side, and there is a subsequent shift of the neutral axis.

With increasing load a stable situation will develop, and the beam will again behave elastically, but in the cracked

range. The compressive strength of mortar is high compared

to the tensile strength, so the curve follows Hooke's Law

(i.e. is linear) in the cracked range. Eventually, as the

load is increased, the mortar or the steel, or both, will

reach their yield strength. If the steel begins to fail

first, the specimen is underreinforced; if the mortar fails first, the specimen is overreinforced; and if both fail

simultaneously, the specimen is a balanced beam.

(Mb)

Yielding

Cracked Range

Deflection (D)

Fig. 2. Load deflection curve for a

Force-Deflection Characteristics 3

(a)

Fig. 3. (a) Specimen is unstressed or lightly stressed and

remains uncracked. (b) Specimen is cracked to the neutral axis and is operating in the cracked range. Cracks are

micro-scopic. (c) Specimen is failing. Cracks become visible.

(d) Photograph of actual test specimen after failure at point of load application (right side). Cracks were made visible with the aid of a dye.

4 _{Force-Deflection Characteristics}

A beam which is cracked, perhaps because of some previous loading, would not exhibit the initial stiffness

of the uncracked beam. The force deflection curve would

WORKING-STRESS ANALYSIS

Working-stress analysis is valid in the uncracked and

in the cracked regions. Since most beams are designed to

operate in the cracked region, the analysis method will be

described for this range. With one minor modification, it

is directly applicable in the uncracked range.

The analysis of composite beams in the elastic range is fairly simple if we assume that the strains vary linearly

from the neutral axis and are independent of the material. In other words, at a given distance from the neutral axis,

the strains in the different materials, e.g., mortar and

steel, are equal. The stresses are then

a = EE (2.1)

where

u = stress

= strain

E = Young's modulus

If we choose one of the materials as the reference

material, for example the mortar, we can transform the other materials into equivalent areas of the reference

material. We then analyze the transformed beam as a

homogeneous beam by conventional slender-beam theory.

-5-6 Working-Stress Analysis

where

Fs = total force on the steel bar

As = cross-sectional area of the steel bar

as = stress in the steel bar

We want the load carried by the equivalent concrete

area of the transformed beam to be the same: Fs = A u

SS

1

Fig. 4. (a) Actual reinforced concrete beam.

(b) Transformed beam. An equivalent area of

concrete has been added to replace the steel

reinforcing bar.

A study of the reinforced beam in Figure 4 will

illustrate such transformations. We choose the mortar

as the reference material and want to replace the steel

reinforcing bar with a suitable area of concrete so that the resulting transformed beam behaves like the original

beam.

The load carried by the steel bar is

(2.2)

We assumed that the strain at any distance from the

neutral axis would be the same in both materials. Since

we are placing the equivalent area of concrete at the

location of the steel,

(2.4)

where

es = strain in the steel

cc = strain in the concrete

Since and as

=EE

SS

AE =AE

_cc

S S Working-Stress Analysis 7 (2.3) (2.5) ac = c E

_cc

(2.6) equation (2.3) becomes A (E E ) = A Cc E ) (2.7) C

cc

S S S Since we get

(2.8)

Fc = A

_cc

A s s

8 Working-Stress Analysis

The equivalent concrete area will then be

AE

s s

Ac

-Ec

The ratio of the Young's moduli is usually referred

to as the "modular ratio" (n),

Es ff

= n ;

Having determined this, we now replace the

reinforce-ment bar with the equivalent concrete area,

Ac = nAs (2.10)

at exactly the same distance from the neutral axis. This

is sketched symbolically in Figure 4(b).

With this imaginary transformed beam, we can now

det-ermine the stresses as a function of the bending moment,

b 2a T a Neutral axis

r

d

Fig. 5. Dimensions of the cross

section of the beam.

(2.9)

Or

the bending moment as a function of the stresses6 In

order to do either, We Must find the neutral axis of the. beam.

The neutral axis

will

lie at the centroid of the

transformed beam. Neglecting the tension concrete, but not the transformed steel area, we calculate the first moment of the area about an assumed neutral axis and set

it. equal to zero. Using the dimensions of Figure 5i

aA_-c(nA ) = 0

c s

a(2a(b))-c(nA ) = 0

s

-We have the additional information that

2a+c = d Working-Stress Analysis 2a2b-dnA5 + 2anAs 0

(2b)a2+(2nAda- dnA = 0

(2nAs ±j(2nA5)2-4(2b)(-nA3) A, it" 2 (2b)

i.e.r

we have

4

quadratic equation to solve in order to

(2.15)i

Hence

c = d-2a (2.13)

and equation (2.12) bedomes

a(2a(b))-(d-2a)(nAs) = 0 (2.14)

The only unknown

in

this equation is "a." Hence we can solve for it

=

10 Working-Stress Analysis

locate the neutral axis.

The bending moment and the stress are related by the

equation

where

ax = stress in the x-direction (along the

beam) at a location "y" from the neutral

axis

Mb = bending moment

Iyy = transverse moment of inertia of trans-formed beam about the neutral axis

For the particular beam we are studying,

where

Gx

I = jrzy2dy

= nAsc24-(2a03

3

If we know the bending moment we can calculate the

stress at any location:

axc

ax_s = na_xc

axc = concrete stress

G_xs = steel stress (2.16) (2.17) (2.18) (2.19) (2.20) -Mby x =

Working-Stress Analysis 11

Conversely, for any given stress we can determine the

bending moment.

For ferro-cement the procedure is the same, only there are more layers of steel and hence the moment

equations have more terms.

Sample Working-Stress Calculation for a Ferro-Cement Specimen

The steps in calculating a ferro-cement specimen are:

determining the dimensions and properties

finding the neutral axis

calculating Iyy

-My

solving a

-C

YY

(1) Dimensions and properties

Properties

Wire mesh area _{( Aw)1} .04787 in.2

Rod area (

Ar ) .0829 in.2

acu mortar 5,930 lb/in.2

a wire mesh 91,800 lb/in.2

u wire mesh 107,000 lb/in.2

Specific weight of mortar ( W ) 145 lb/ft3

1 A is the total area of the corresponding layer of reinforcement.

12 Working-Stress Analysis

Young's modulus of mortar

E 2 = (W)"5(33)(G_cu)0.5

= (145)1-6(33) (5930)0.5

= 57,400(77.007)

= 4.42 X 106 lb/in.2

23"

Fig. 6. Beam configuration

2

Ferguson, P. M. Reinforced Concrete Fundamentals. New

L.3 layers of 1/2" X 19 gage galvanized hardware cloth

Fig. 7. Section through beam

Modular ratio Working-Stress Analysis 13 3/16" Compression side .100" 1.00" .156" Rod _.375" .156" Tension side _.213" Transformed areas nAw = 6.561 x .04787 = .3140 in.2 nAr = 6.561 x .0829 Es

n=

Ec 29 x 106 4.42

x 106

= 6.561

14 Working-Stress Analysis

Ar(n)____

W

AdF111

az mi. Apr

Fig. 8. Section through transformed beam.

Dimensions From Figure 8, a = unknown = .5439 in.2 (n-1)Aw = 5.561 x .04787 = .2662 in.2 Mortar Area

Effective area is in the compression side:

Ac = 2a(6) (for a 6" wide beam)

= 12a

Aw(n)-1//

* For explanation of (n-1) see page 18.

1781" 3593" 17195" .2907" b = .5374-2a c = .70935-2a d = 2a-.1781

(2) Neutral axis

Hence,

The neutral axis is at the centroid of the trans-formed beam. Hence

EM

=0

n.a.

Substituting from Figure 8,

Emn.a. = 12a(a)+(n-1)Aw(d)-nAr(b)-nAw(c) = 12(a2)+(.2662)(2a-.1781) -(.5439)(.5374-2a)-(.3140)(.70925-2a) = 12a2+.5324a-.04741-.2923+1.088a -.2227+.628a = 12a2+(.5324+1.088+.628)a -(.04741+.2923+.2227) = 12a2+2.2484a-.56241

=0

-2.2484±V(2.2484)2-4(12)(-.56241) a -24 = .1424 Working-Stress Analysis 15 a

16 Working-Stress Analysis

(a)

Fig. 9. (a) Neutral axis location. (b) Strain distribution

through member. Moment of inertia I = xy2dy YY = (n-1)Aw(d)2+nAr(b)2+nAw(c)2 + (6) (2a) = .2662(.1067)2+.5439(.2526)2+ .3140(.42445)2 6(.2848)3 3 = .13787 in.4

Stresses and bending moments

Suppose we impose a bending moment of 2,000 in.-lb. 2a=.2848

(b)

Compression

y = -.4026

rg

i

as = n YY -2,000(-.4026) = 6.561 .13787 = 38,318 lb

Suppose we want to know what the bending moment will

be if we load the mortar to its ultimate stress.

cc_{u yy} -M -b (5930 lb/in.2)(.13787 in.) .2848 in. in. 2 in. 2 Working-Stress Analysis 17

Mortar compressive stress

y = .2848 -Mby ac Iyy -2,000 in.-1b(.2848 in.) .13787 in.4 = 4,131 lb

Steel tensile stress

= 2871 in.-lb

18 _{Working-Stress Analysis}

Compression Steel

As noted in the sample calculation for the 1.00-in.

ferro-cement specimen, compression steel is converted to

an equivalent mortar area by the ratio (n-1). The explanation for this follows.3

Consider a column experiencing a compressive load (P).

Let

A = gross area of mortar

As = steel area

Ac = net mortar area

P = A a +A a

_{cc ss}

as ac

= =

Es Ec

Solving equation (2.22) for the steel stress,

Es

as

= a

Ec

Substituting this into equation (2.21),

P = A a +A (na)

cc

s

= a (A +nA) (2.23)

C c s

3 Ferguson, P. M. Reinforced Concrete Fundamentals. New

York: John Wiley and Sons, March 1966. p 42

Working-Stress Analysis 19 Since Ac =

A-A

_g s

P = a(A-A+nA

c g s s) P = ac(Ag+(n-1)(As)) (2.24)

Hence the modular ratio for compression steel is (n-1). Actually it is common practice in concrete design

to use a modular ratio of (2n) for long time loads because,

after a period of time, the steel is carrying a much greater

portion of the load when the structure was new. The in-creased load is due to concrete creep.

Analysis in the Uncracked Range

Analysis in the uncracked range would be the same as that in the cracked range, except that a modular ratio of

(n-1) is used throughout. This ratio is used for the same reason as that in the case of compression steel.

The end of the uncracked range is difficult to predict

because of the brittle nature of the mortar. Concrete has a

tensile strength of approximately 10 percent of its compressive

strength. Yielding, however, will be highly dependent on

stress concentrations which arise from surface roughness and

/MENEM,

_MINN/

=MIMI

EMMY MFIW

IMF

011. MY

r

acu

ULTIMATE STRENGTH ANALYSIS

-Fc

Yielding of Ferro-Cement

Theoretically, a ferrocement specimen will begin

to yield when

the stress in the mortar reaches the Compres-sion strength, or

the stress in the steel reinforcemeht reaches, the yield strength, or

both (1) And (2) occur simultaneously

Is termed an overreinforced beam,

an underreinforced beam, and

(&) a balanced beam.,

In at underreinforced beam the steel, yields first

And there is considerable deflection before failure. The

failure process is shown in Figure 10. When the steel

.85ocu

,-,

(a) (b) (c) (d)

Fig. 10. Failure Of an underreinforced beam. (a) to (c)

shows the actual failure, and (d) shoWS the linearized

model.

20

21 Ultimate Strength Analysis

yields, the neutral axis moves up. This increases the moment arm between the resultant of the mortar force (Fc)

and the steel force (Fs), which slightly increases the

load-carrying capacity. The stress distribution in the

mortar becomes nonlinear, as shown in Figure 10. Eventually the mortar fails in compression and the entire beam fails.

The fact that the beam will undergo considerable deflection

before failing gives warning of the oncoming failure. For

this reason most land structures are underreinforced.

In an overreinforced beam the mortar fails first, as

shown in Figure 11. As the outer fibers fail, the neutral axis moves down, hence decreasing the distance between

Fs

and

Fc and diminishing the load-carrying capacity. As a

result, the overreinforced beam tends to fail rather

suddenly.

The balanced beam is primarily of academic interest

and will not be considered here.

F

.85acu

Fc

N.A.

(a) (b) (c) (d)

Fig. 11. Failure of an overreinforced beam. (a) to (c)

shows the actual failure, and (d) shows the linearized

model.

Estimating Ultimate Strength

strength of a concrete beam is to replace the actual stress

distribution in the mortar by an equivalent stress block,

Fc

a

jd

Fig. 12. (a) Dimensions of beam. (b) Stress block and

resultant forces.

as shown in Figures 10 and 11. Take, for example, the beam in Figure 12

Fc = A

_cc

where

Ac = cross section area of stress block Ultimate Strength Analysis 22

ac = concrete stress in stress block

The American Concrete Institute recommends a value

of .85

acu for cc. Hence,

Fc =

.85 aA

_cuc

(3.1)

(3.2)

23 Ultimate Strength Analysis

For the tension steel,

where

As = steel area

as = steel stress

The yield stress is used for as. Hence,

Fs = As a

_{y S}

since

EFx = 0

Fs

=F

c

The depth of the stress block (a) remains unknown. However, Ac = ab where b = width of beam SO Fc = .85 _CU(oh) = Fs (3.7)

Therefore we can solve for (a):

Fs a - (3.8) .85acub (3.3) (3.4) (3.5) (3.6) F = A =

The moment arm (jd) is then where and where jd = d - a

7

Ultimate Strength Analysis 24

Fw = A a_w

w-Fw = resultant force of wire mesh

A

- wire mesh

area

ow = yield strength of wire mesh

Fr = A a

ry

Fr = resultant force of rods

Ar = rod area

cry = yield strength of rods

(3.9)

(3.11)

(3.12)

and the bending moment is

Mb = jdFs (3.10)

For beams with multiple layers of reinforcing, the

procedure is similar.

Two simplifying assumptions can be made:

The compression steel is neglected.

All the tension steel is assumed to be yielding.

25 _{Ultimate Strength Analysis}

a - Fs

Fc

jd

Fig. 13. (a) Dimensions of beam. (b) Stress block and

resultant forces. Fs

=F

s + Fr .85u cub jd = d -Mb = jdFs = Moment to fail a

It should be noted that for the specimens tested,

this method predicted ultimate strengths which were less than those actually observed.

(3.13) (3.14) (3.15) (3.16) (a) (b)

--w

IV

CONCLUSIONS

The behavior of ferro-cement in bending follows the

normal load-deflection curve for reinforced concrete, as shown in Figure 1.

Ferro-cement cracks at very low tensile stresses.

These cracks are microscopic and are shown in Figure

3 with the aid of a dye.

Reaching the ultimate strength of either the mortar or the steel does not necessarily mean failure of the

member. Loads are transferred from the failed material to the intact material and the behavior may remain

linear at slightly increased loads.

The point at which the ultimate strength of the weak-est material (usually the mortar) is reached is

con-siderably below the failure point of the member.

Working-stress analysis can safely be used to compute

stresses which are due to working loads.

The ultimate-strength analysis outlined can safely be used to estimate the ultimate strength.

Further study of the failure mechanism is clearly

needed, not to mention study of corrosion and fatigue. Shear stresses due to bending moments should be

inves-tigated.

26

-(7)

-BIBLIOGRAPHY

Canby, C. D. "Ferro-Cement (with Particular Reference to Marine Applications)." Ann Arbor: Department

of Naval Architecture and Marine Engineering, The

University of Michigan, March

1969

Coleman, J. F., Jergovich, N., and Muhlert, H. F.

"Ferro-Cement Trawler Design Report." Ann Arbor: Department of Naval Architecture and Marine

Engin-eering, The University of Michigan, June

1968

Ferguson, P. M. Reinforced Concrete Fundamentals. New

York: John Wiley and Sons, March

1966

Hagenbach, T. M. "Ferrocement Boats." Montreal:

Con-ference on Fishing Vessel Construction Materials, October,

1968,

sponsored by The Federal-Provincial Atlantic Fisheries Committee

27

APPENDIX A

TENSILE STRENGTH OF RODS

0491 in.2

30

-1/4" Rod Test 1 - 2,000 lb Yield Force - 3,120 lb Ultimate Force

Test 2 - 1,900 lb Yield Force - 3,030 Ultimate Force

Average Yield Force - 1,950 lb Average Ultimate Force - 3,075 lb

Stresses: a 1,950 lb 0491 in.2 = 39,800 lb in.2 3,075 lb au = = 62,600 lb in.2

3/16" Rod (Rods showed no yield point)

Test 1 - 2,455 lb Ultimate Force

Test 2 - 2,550 lb Ultimate Force

Average Ultimate Force - 2,503 lb Force

Stresses: au .0276 in.2 2,503 lb = 90,000 lb in. 2

-lb a L. 80% a= 72,500

in./

TENSILE STRENGTH OF WIRE MESH.

The wire mesh is lg gage

Diameter = .0410

Area = .0013 in,

,2

AsSUMing that the proportional limit can be taken as c, and throwing out the Inconsistent high. value on the plot, forces to yield are

Test 1/2 Force_ (1b)

1 61.2

2 61.0

4 62.1

5 60.0

Ultimate forces are

Average force = 61.075 x = 122 lb

KY==

122 lb Y _{.00133 in./} = 91,800 lb in./ Test 1/2 Force (lb) 1 70.0 Average force = 71.3 x 2 2 71.8 ₌ 142. 6 lb 4 72.8 5 70.6 - 142.6 lb -u ,00133 in./ = 107,000 lb in. 2 31 -in. - -2

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COMPRESSION TESTS

4" Diameter

Fig. 15. Test cylinder.

33

-Capping Material Series Specimen Curing Time

Force (lb) Average acu (1b/in.2) 1 1 7 days 61,000 59,000 4,760 2 1 to 3 7 days 60,500 62,500 4,885 3 1 to 3 8 days 76,000 73,000 5,930 3 4 28 days 104,000 95,500 7,937 Mortar Mix Cement 16.5 lb Pozzolan 4.5 lb Sand 30.0 lb Water 3,500 cc -. ...

Series 1, Specimen 1 P (lb) DEFLECTION (in.) 100 .0065 200 .013 300 .020 400 .028 500 .036 600 .045 700 .0605 800 .082 900 .100 1,000 .121 1,100 .141 1,200 .162 1,300 .187 1,400 .212 1,500 .240 1,600 .274 Mortar acu = 4760 in.2 Wire Mesh 1/2" x 19 gage galv. hardware cloth. Four layers on each side.

Rods

1/4" diameter hot rolled steel. Long-itudinal rods spaced

2". Transverse rods

spaced 2".

34

-6"

Section through specimen

Compression side Tension side Location of reinforcement .1875 I.250

.500

250 -I--1 . 187 5 23.5" Loading 1.375" lb

-Wire Mesh

1/2' x 19 gage galv. hardware cloth.. Four

layers each aide.

Rods

1/4" diameter hot rolled steel: Long-itudinal rods spaced

2". TransVerse rods spaced

2°4

35,

-Compression side .0935 Tension side Location of reinforcement Series

2,

Specimen 1 P (lb) DEFLECTION

(in.)

R. 100 .0105 200 -022 23.5" 300 .031 Loading

400

.044

500 .0705 600

.080

700

.104

1100

.127

900

.150

1,000

.174

1,100

1,200

.198,

.230

1.185"

1,300 .268 6" 1,400 .315

Mortar

# 4,885

lb

Section through Specimen

in. 2

.250

.500

.250

.0915 -

-Wire Mesh

1/2" x 19 gage galv. hardware cloth. Three

layers each side.

Rods

3/16" cold rolled steel. Longitudinal rods spaced 2". Trans-verse rods spaced 2".

36 -23.5" Loading 6" .375 Location of reinforcement Section through Specimen

Compression side .1563 .1562 .1562

7.1563

Tension side

/

Series 2, Specimen 2 P (lb) DEFLECTION (in.) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1,000 1,050 Mortar .002 .009 .014 .019 .0265 .0375 .0515 .068 .084 .1025 .121 .1405 .161 .185 .209 .233 .266 .292 .362 .408 .480 acu = 4,885 1.00" lb in. 2

Wire Mesh

1/2" x 19 gage galv. hardware cloth. Three layers each side.

Rods 3/16" cold rolled steel. Longitudinal rods spaced 2-1/2". No transverse rods. 37 -Tension sid

Section through Specimen

Compression side 1.125 .1562 f.125 .1875 11-11562 Location of reinforcement Series 2, Specimen 3 (in.) P (lb) DEFLECTION 50 .0075 100 .0215 150 .0625 23.5" 200 .1275 250 .2075 Loading 300 .2905 350 .3755 400 .4545 450 .5565 500 .7075 Mortar 75"----cicu = 4,885 lb in. 2 _6"

-Series 3, Specimen 1 P (lb) DEFLECTION (in.) 50 100 .0735 150 .086 200 .102 250 .124 300 .154 350 .183 400 .212 450 .238 500 .268 550 .298 600 .332 650 .370 700 .410 750 .460 800 .580 850 .700 Mortar ccu = 5,930 Wire Mesh 1/2" x 19 gage galv. hardware cloth. Three layers each side.

Rods 3/16" cold rolled steel. Longitudinal rods spaced 2-1/2". No transverse rods.

-

38 Tension side 23" Loading .onemmww.momonrnleM=.==mosEmn-armam...rmwall. 6"

Section through Specimen

Compression side 1.1562 .1875 .1562 tO .0 Location of reinforcement .75" lb in. 2

3/16" cold rolled

steel. Longitudinal

rods spaced 2". Trans-verse rods spaced 2".

39 -Location of reinforcement Series 3, Specimen 2 P

(lb)

DEFLECTION (in.) SO .094 100 .101 23" 150 .107 200 .113 Loading 250 .127 300 .147 350 .168 400 .189 450 .209 500 .234 550 .262 .75" 600 .287 650 .310 700 .336 6" 750 .365

800 .400 Section through Specimen

850 .439 900 .483 Compression side 950 .575 lb .100

Mortar ac= 5,930

u .1562 in.2 Wire Mesh .375 1/2" x 19 gage galv. hardware cloth. Three

layers each side.

.1562 .2126

Tension side

Compression side

Wire Mesh i.2

1/2" x-19 gage galv..

hardware cloth. Three

layers each. side, Rods

1/4" diameter hot rolled steel. Long-.

itudinal rods spaced

2". Transverse rods. spaced 2", 6 Tension side 23" Loading

Section through Specimen.

Location of reinforcement Series. 3, Specimen 3 (in.) P (lb) DEFLECTION

100

.061

200 .069 300 .077 406 .085 500 .0925 600 .101 706 .109 800 .121 900 .136 1,000 .156 1,100 .179 1.40" 1,200 .203 1,300 .249 1,400 .309 Fl 1,500. .383 lb Mortar acu = 5,930 in. 2 40 -.25 5 .25

7.2

Series 3, Specimen 4 (lb) DEFLECTION (in.) 50 .032 100 .047 150 .063 200 .084 250 .112 300 .144 350 .177 400 .210 450 .243 500 .279 550 .314 600 .351 .75" 650 .394 700 .444 750 .496 800 .562 850 .756 1/2" x 19 gage galv. hardware cloth. Three layers each side.

Rods 3/16" cold rolled steel. Longitudinal rods spaced 2-1/2". No transverse rods. lb Mortar ocu = 7,937 in.2 Wire Mesh 41 -t 23" Loading 6"

Section through Specimen

Compression side Tension side 1.1562 T.1875 ;.1562

70.0

Location of reinforcement

-APPENDIX B

COMPUTER PROGRAM FOR

COMPUTER PROGRAM FOR WORKING-STRESS ANALYSIS

The computer program shown on pages 47 and 48 was used to do the stress analysis of the ferro-cement specimens by

working stress theory. The computer is not necessary for this analysis. If available, however, it saves time,

mini-mites the chance of error, and makes sensitivity studies

easy to perform. In this case the input consisted of the

material properties and the beam geometry, while the output consisted of the stress-bending moment relationship as well

as some intermediate data. The program could probably be streamlined to use computer time more efficiently. As the

program stands, it uses about 10 seconds of CPU time on the

IBM 360 computer.

Language

The computer language is FORTRAN IV.

Theory

The program uses the stress analysis outlined in

Chap-ter II. Some changes in notation were made. For example,

the layers of reinforcement have been numbered starting from

the bottom and working up. Hence we have steel areas A,

A, A, ...

at distances y y y, ... from the bottom

2 3 1 2 3

surface.

Another superficial change is the generalization of the

neutral axis calculation. The explanation follows.

Taking moments about the neutral axis, (see Figure 16)

NA = (T - NA)2 /j7-- + (y3 - NA)A

3

-

43

-+ (y - NA)A

+

(y1

-

NA)A W = width of beam

-Y3

Y2

NA

Fig. 16. Transverse section through a ferro-cement specimen, showing nomenclature.

[T2 -

2T (NA)

+ (NA)2]

-Ay -ANA

₃₃

₃

Ay -ANA

₂₂

2

Ay -ANA

₁₁

1 T2 -2

- 2T-2 (NA) +

(NA) 2

Ay -ANA

₃₃

3

Ay -ANA

₂₂

2

Ay -ANA

₁₁

= -2 (NA) 2 + (NA)

- A

3

- A

2

- A)

1

+ (T2-

+ Ay3 + Ay2 +

44

-y3 - NA

T - NA

- NA

NA

- NA

= A

(2T

-

NA-2 7

Let b = (-TN - A - A - A )

3 2 1

andc=

(T17--

+Ay +Ay +Ay)

3 3 2 2 1 1

b + j b2 - 4

H

C

2

b

J b2

-

2WC

Take the plus sign.

It should also be noted that in using this program

we assume an initial neutral axis location. When

calcu-lating the transformed areas, the steel area below the

assumed neutral axis will be multiplied by the factor (n), while the steel above will be multiplied by (n-1). If the

calculated neutral axis is found to be between layers of reinforcement different from the assumed one, then we rerun

the program with the calculated neutral axis location as the new assumed location.

Data Input

Card 1

Punch length (in.) width (in.)

thickness (in.)

mortar compressive strength (1b/in.2) density of mortar (1b/ft3)

E of steel (1b/in.2)

assumed N.A. location from bottom (in.) yield strength of steel (1b/in.2)

45

-Format (8F10.4)

Card 2

Punch number of layers of steel kode - 0 for cracked analysis

- 1 for uncracked analysis

Format (212)

Card 3

Punch area of each steel layer (in.2)

Format (8F10.4)

Card 4

Punch distance of each layer of steel from bottom of beam, in same order as on Card 3 (in.)

Format (8F10.4)

Results

The object of the stress analysis was to determine the stresses as a function of the material, the location

(y), and the bending moment. The stresses in any of the component materials will vary linearly with the bending moment as well as with (y). Let us study the stress as a functional bending moment. We know that b = 0 at Mb =

0. Hence, if we know one more point on the curve, the

curve is defined. The computer program provides one more point for both the steel and the mortar stress curves. We can therefore plot stress as a function of bending

moment. Note that the curves for the steel stresses shown on the following pages apply at the extreme fibers on the

tension side, and those for the mortar stress apply at the extreme fibers on the compression side. Stresses at other

(y) distances can be found by changing the values in statements 25 and 26 in the program.

46

-$CCMFILE

DIMENSICN A(10),ATR(10)0(10)

2 REAL L,NAA,NA,I,N

3 1 REAL(5,200)L,W,T,SCU,WC,ES,NA4,SSY

KOLE = 0 , CRACKEL ANALYSIS

KCDE = 1 , UNCRACKEC ANALYSIS

4 REAC(5,210)NC,KCDE 5 REAC(5,200)(A(J),J=1,NC) 6 REAC(5,200)(Y(J),J=1,NC) E CF MCRTAP EC=(INC**1.5)*33*SCR1(SCIA MODULAR PATIO 8 N=ES/EC TRANSFORMEE AREAS 9 DC 11 J=1,NC 10 IF(KOCE.EC.1)G0 TO 10 11 IF(Y(J).GT.NAA)G0 TC 10 12 ATR(.1)=N*A(J) 13 GO TC II 14 10 ATR(J)=(N-1)*A(J) 15 11 CONTINUE

COEFFICIENTS FOR CUACRATIC FCRPULA

16

17 C=T*I*W/2

18 DO 13 J=1,NC

19 B=B-ATR(J)

20 13 C=C+ATR(J)*Y(J)

SOLVING OUACRATIC FCRMULA IC FINE NEUTRAL AXIS

21 NA=(-8-SORT(8*8-2*W*G))/W MOMENT OF INERTIA

22 I=((T-NA)**3)*W/3

23 DO 14 4=1,NC

24 14 I=I+(Y(J)-NA)*(Y(J)-NA)*ATR(J) CALCULATING BENCING MCMENTS

25 BMSCUC=SCU*I/(T-NA)

26 BMS5Y=SSY*I/((Y(1)-NA)*N)

27 IF(KCIDE.E0.0)GO IC 15

28 EMSCUT=0.1*SCU*I/(-NA) PRINTING OLT INPUT CATA

29 15 WRITE(6,206)7 30 WRITE(6,201) 31 WRITE(6,202)L,W,T,WC,SCUIS5Y,ESINAAINOIKCCE 32 WRITE(6,229) 33 CO 16 J=1,NC 34 16 WRITE(6,23C)J,Y(J),A(J),ATR(J)

PRINTING OUT RESULTS

35 WRITE(6,203)EC,N,NA,1,13MSSY,BP5CUC

36 IF(KOCE.E0.0)G0 TO 1

1 37 WRITE(6,205)BYSCUT 38 GC TO 1 FORMATS 39 200 FCRNAT(8F10.4) 40 210 FCFPAT(242)

_41 206 FERMAT( 11RESULTS FOR ',F8.4,41. INCH SPECIPEN')

42 201 FOFMAT( /,'INPUT CATA',//)

2C2 FORMAT( ILENETF=',F10.3,/,:__

l'WICTF =

1' CEPTF = 'F10.5,/,

PUENSITY CF OCRIAR =

l'ULTIMATE STRENGTH CF MORTAR =_

',F10.0,/,.

PYIELD STRENGTH OF STEEL =

',F10.0,/,

l'YCLNGS MODULUS CF STEEL = ',F1C.0,/,

l'ASSUMED NEUTR41.7-A-ki-S-= ',F1C.4,/,

,UNLMBER OF. STEEL LAYERS =

"PKECE * *

-44 229 HORNAT( /,'STEEL REINFERCENENT

EATA 4,//,

AREA

l'LAYER NUMBER CISTANCE(Y)

1 AREA')

45 230 FORMAT( 117.3F20.5)'

46 203 FCRMAT( //,'RESULTS',//,

TRANSFORMED

l'YCLNGS MODULUS CF PERTAR = 1, F10.0,/,

I'MEDULAR RATIO (N) =

't

F1E.51/v

205

l'NELTRAL AXIS = 1, F10.5,/,

l'MCNENT CF INERTIA = ',F10.5,/,

liBENDING PCNENT AT YEE-LC STRESS -CO STEEL

11BENDI_NG_M:EMENT AT ULTIMATE MORTAR COMPRESSIVE_ STRENGTH = 4,

1.1FIC.01

FERMAT( 'BENDING MOMENT AT TENSILE STRESS CF FICRTAF = 1.,F10.2t_

ENE

_

1 _ I

.

-.

---

_ i

,-4 4 ',F10.2,/, = 48 -43

IRE LE VNI DE UI YI YC AS NU ,

SULTS, FOR 1.375C INCH SPECIMEN Series 1

. -PVT GALA . _ \GTH 23.500

DTH =6.000

---?PH = 1.375CC \SITY CF MORTAR =

145.00_-476

TIMATE STRENGTH CF PCRTAF =

--

C.

ELD STRENGTH OF STEEL =_

_{sleo_c._______}

LNGS MODULUS CF STEEL. = 2-9CCOOCO.,

SUMEG NEUTRAL AXIS = 0.7000

_ A

vBER CF STEEL LAYERS = 3

DE = C _{____}

EEL REINFORCEMENT sCATA_ _{-____}

ER NUMBER CISTANCEIY1 AREA TRANSFORMED AR

1 C.31250 0.C638C 2 _0.56250 C.147CC 0.46542 1.07237 _ 3 1..06250 0.C638'0 0.40162' SUITS NGS MODULUS Cf MCRTAR = 3975294. DULAR RATIO (N) = 7.29506 -JTRAL AXIS = C.,92261 VENT OF INERTIA = - 0.50534

'DING MOMENT AT YIELC STRESS OP STEEL = -10423.

vDING MOMENT AT ULTIMATE MORTAR CCMPRESSIVE STRENGTH = 531.7.

_

---,---

-_

---

-' ---

---

,

-

--

---.

49 -ST LA RE YC NE MC BE BE

-100,000L10,000

/7-91,800 90,000 _9,000 80,000 Cl) U) CD 0 M _1 30,000 70,000 -7,000 .Q 60,000 6,000 t 50,000 5,000 -1-) MUYtar Stress 4,760 4.) 20,000 -8,000 -3,000 -2,000 10,000 1,00 2,000 4,000 50 -a 6,000 Steel Stress 5,317 10,423 8,000 Bending Moment (Mb) (in./lb)

Graph 1. Working stress as a function of bending

moment for specimen 1, series 1 (1.375 in.) 40,000 -4,000

10,000

RESULTS FOR 1.1850 _INCH, SPECIMEN Series 2L INPUT CATA 'LENGTH= 23.500 WIDTH = 6.000 DEPTH = 1.1850C DENSITY OF MORTAR = 145.00

ULTIMATE STRENGTH CF MCRTAR = 4885. YIELD STRENGTH OF STEEL =

YCUNGS MCDULUS CF STEEL = ASSUMED NEUTRAL AXIS =

NUMBER OF STEEL LAYERS = KCDE =

STEEL REINFORCEMENT CATA

tAYER NUMBER 1 3, RESULTS, EISTANCE(Y) 0.2165C, 0.4665C 0.96650 9480C. 29000000. C.7000

-YCUNGS MCDULUS CF NCRTAR = 4027152.

- MCDULAR RATIO IN) = 7.20112

NEUTRAL AXIS = C.77429

MCMENT OF INERTIA = 0.39640

BENDING MOMENT AT YIELC STRESS OF STEEL = -9060.

BENDING MOMENT AT ULTIMATE MORTAR COMPRESSIVE STRENGTH

51

-4-715.

F

AREA , TRANSFORMED AREA

0.C6380 0.147CC 0.45943 1.05856 0.06380 0.29563 -0 - -= 2

100,000H 10,000 90,000 - 9,000 80,000 - 8,000 ". 70,000 _ 7,000 60,000 _ 6,000 m rtj 4-1 50,000 5,000 (i) -n -I 40,000 40000 m m m m 30,000

mm

0 3,000

10,000 _1,0

0 Mortar stress a CU

-

52 -Steel Stress 2,000 4,000 6,000 Bending Moment (Mb) (in.-lb)

Graph 2. Working stress as a function of bending

moment for specimen 1, series 2 (1.185 in.)

8,000 10,000

RESULTS_ FOR 1.CCCC INCH SPECIMEN SPries_2 INPUT CATA LENGTH= 23.500 = 6.000 DEPTH = 1.CCCCC DENSITY CF MORTAR = 145.00_

ULTIMATE STRENGTH CF MORTAR = 4885.

KODE = 0

STEEL REINFORCEMENT CATA

RESULTS

YOUNGS MODULUS CF NCRTAR = 4027152. MODULAR RATIC (N) = 7.20112

NEUTRAL AXIS = C.68495

MOMENT OF INERTIA = 0.18081

BENDING MOMENT AT YIELC STRESS OF STEEL = BENDING MOMENT AT LLTIMATE MORTAR CCPPRESSIVE

53

-LAYER NUMBER CISTANCE(Y) AREA TRANSFORMEC AREA

1 0.2344C C.C47E7 0.34472

2 0.40625

c.ce2sc

C.59697

3 0.76555 0.04787 0.29685

YIELD STRENGTH OF STEEL = 91800.

YCLKGS MODULUS CF STEEL = 29000CCC. ASSUMED NEUTRAL AXIS = 0.6000 NUMBER CF STEEL LAYERS = 3

-5116.

STRENGTH = 2804.

4

WIDTH

-IQ) 50,000 40,000 20,000 10,000 5,000 4,000 3,000, - 2,000 Steel stress - 54 Cu a_Sy .Mortar stress, 1,000 2,000 3,000 4,000 Bending Moment (Mb) (in.-lb)

Graph 3. Working stress as a function of bending moment for specimen 2, series 2 (1.001 in.)

5,000 100,00' 10,000, 90,000 9,000 80,0001 8,000 70,000 7,000 60,000 6,000 En TO

0 0

4-) Ica CO 30,000 - ". -H --H -P -1,0 -

-5

RESULTS FOR 0.7500 INCE SPECIMEN Series, 2

INPUT DATA

LENGTH= 23.500 hIDTH = 6.000 DEPTH = C.75CCC

DENSITY CF MCRTAR =

145.00

ULTIMATE STRENGTH CF PCRTAR 4885. YIELD STRENGTH OF STEEL =

91800.

YCLNGS MODULUS CF STEEL = 290000CC.

ASSUMED NEUTRAL AXIS = 0.4000

-NUMBER OF STEEL LAYERS i<OCE = 0

STEEL REINFORCEMENT DATA

LAYER NUMBER CISTANCE(Y) AREA TRANSFORMED AREA

1 2 3 RESULTS 0.20310 C. C4787 0.37500

c.ce2sc

C.34472 0.59697 .0.5469C 0.04787 0.2C685 YCUNGS MODULAR NEUTRAL MOMENT BENDING

MCDULUS OF MCRTAR = AC27152.

--RATIO (N) = 7.20112 AXIS = 0.50SC2

OF INERTIA = 0.07140

MOMENT AT YIELC STRESS CF STEEL = -2975.

BENDING MOMENT AT ULTIMATE MORTAR CCPPRESSIVE STRENGTH = 1447.

-- _, _ _ *,...ft. 1

-_

-55-= =

100,000 90,000 130,000 70,000 60,000 -.4 50,000 10-4 40,000 10,000 9,000 8,000 7,000 _.H, 6i000 m 5,000 $4 P 0 4,000 3,000 2,00 10,1000 Gcu Mortar stress 56 Steel stress 4,000 1,000 2,000 3,000 Bending Moment °lb) (in.-1b)

Graph A. Working stress as a function Of bending moment for specimen 3, series 2 (.75 in.)

5,000 CU C.1) -It 30,000 20,000 00 -

-YIELD STRENGTH OF STEEL = YOUNGS MODULUS OF STEEL ASSUPED NEUTRAL AXIS = NUMBER OF STEEL LAYERS =

KCCE = 0

STEEL REINFORCEMENT EATA

RESULTS

9180C.

29CCOCCC. C.480C

3

YOUNGS MODULUS CF NCRTAR = 4437C38. MCCULAP RATIC IN) = 6.53589

NEUTRAL AXIS = C.46245

MOMENT OF INERTIA = 0.11876

BENDING POPENT AT YIELD STRESS OF STEEL = -433S.

BENDING MOMENT AT ULTIMATE MORTAR COMPRESSIVE STRENGTH

57

-2449.

LAYER NUMBER DISTANCE(Y) AREA TRANSFORMEC AREA

1 C. 07800 0.C47E7 0.31287

2 0.2500C C.CE2SC C.54183

3 0.42200 C.C4787 C.31287

RESULTS FOP C.75CC INCH SPECIMEN Series 3

INPUT DATA

LENGTF= 23.000 WIDTH = 6.000

DEPTH = C.75CCC

DENSITY CF MORTAR = 145.00

M M M 0 4-) 4-) m m r-i 0 0 -4-) Cl) 100,000 - 10,000 80,000 i- 8,000 70,000

-N 50,000 ---1 -.,

-.4

-1 40,000 10,000 0 - _7,000 5,000 - 4 , 000 30,000 3,000

20,000 -2,000

58 -1 Mortar stress 1,000 2,000 3,000 4,000 Bending Moment (Mb) (in.-lb)

Graph 5. Working stress as a function of bending

moment for specimen 1, series 3 (.75 in).

Steel stress 5,000 9,000 90,000 60,000 6,000 .C1) II

RESULTS FCR I.00C0 INCH SPECIMEN _{SPries 3} INPUT CATA LENGTH= 23.000 bIDTH = 6.000 DEPTH = 1.CCOCC CENSITY IF MORTAR = 145.00

ULTIMATE STRENCTF CF MCRTAR = 5930.

YIELD STRENGTH CF STEEL = 9180C.

YOUNGS NCCULUS CF STEEL = 2900000C. ASSUMED NELTRAL AXIS = 0.7000 NUMBER CF STEEL LAYERS =

KCCE = 0

STEEL REINFORCEMENT CATA

RESULTS

YCUNOS MCDULUS CF MCRTAR = 4437C38.

MCCULAR RATIC (N) =

6.53589

NEUTRAL AXIS = C.7I587

MCMENT CF INERTIA = 0.14019

BENDING MOMENT AT YIELD STRESS CF STEEL = -4631.

BENDING MOMENT AT ULTIMATE MORTAR CCMPRESSIVE STRENGTH = 2926.

59

-LAYER NUMBER CISTANCE(Y) AREA TRANSFORMED AREA

1 C.2907C C.C4787 0.31287 2 C.46255 0.CE2CC

C.54183

3

0.82185

C.04787

_0.2650C

3

-100,000 90,000 80,000 7(2,000 . m 50,000 1-5,000

0

-1. $.4 -.., I 4.) tr) tD .-I (11 ,En 40,000 , 4,000, 11) a) Q)

0

-1-) ,ca 30,000 10,000 9,000 8,000 60,000 1 6,000 7,000 Steel Stress 3,000. 0 ! g 1,000 2,000. 3,000 4,000 Bending Moment- (Mb) (in.-lb)

Graph 6. Working stress as a"function of bending moment for specimen 2, series 3 (1.00. in,)

Mortar Stress 60 ocu g*Sy 5,000 20,000 10,000 2,000 1,0 ---

-UITS, FOR 1.400C INCH SPECImEN _{Series_3____} 1 , U1 CA1A . ,. . GTH= 23.CCC TH = 6.000 _ .., ,a-,,,..--,nr."-, , TH = 1.4CCCC SITY CF MCRTAR = 145.00_

IMATE STRENGTH CF NCRTAF = -59-30.

LC STRENGTH OF STEEL = Sl8CC.. NGS MODULUS CF STEEL = 290G0OCC. UMED NEUTRAL AXIS = 0.9CO0L

1

BER OF STEEL LAYERS = 3

E=

0

- ---__

FL REINFORCEMENT CATA

ER NUMBER CISTANCE(Y) AREA TRANSFORMED AREA,

I. G.32500 0.C638C 2 _{C.5750C C.147CC} 0.41699 _- C.9E:0Y7E 1 3 1.07500 0.C638C 0.35319 ULTS _ ?'GS MODUL1JS_CF_1FRTAR = 4437C38._ ' ULAR RATIO (N) = 6.53589 TRAL AXIS = C.S5697 --_.-. - - -- ---r. _ ENT Of INERTIA = 0.48555

DING MCMENT AT YIELD STRESS OF STEEL = -10791.

_ - --"--,,-

,---DING MOMENT AT ULTIMATE MORTAR COMPRESSIVE STRENGTH = 6499.

,

-*-

_

,.

1 -- .-.--_ _ _ _ - _ _ - -, , - - -, ' ' -. 1 - ' . - or ?- - -_ - -

-

61 -RE NP LEN WIC CEP CEN L,L1 YIE YOU .ASS NUM KOD S1E LAY IRES YOU POO NEU PCP BEN BEN -

-100,000 90,000 80,000 70,000 60,000 -I 50,000 ,-I m m m m 40,000 w s-1 4J4-) milm -10,000 -9,000 -8,000 a5 4,000 o H w 30,000 3,000 20,000 2,000 10,000 1,00 7,000 a cu 6,000 :LI 5,000 4c1i)

-

62 -Steel stress Mortar Stress 2,000 4,000 6,000 Bending Moment (Mb) (in.-lb)

Graph 7. Working stress as a function of bending

moment for specimen 3, series 3 (1.4 in.).

8,000 10,000

(i)

b

RESULTS 0,7500 INCH SPECIMEN' _{Series 3} INPUT CATA LENGTH= 23.,000

IIOTH=

_6.000

DEPTH = C.75C0C DENSITY OF MCRTAR = 14

ULTIMATE STRENGTH CF MCRTA

YIELD STRENGTH OF STEEL = YOUNGS MODULUS CF STEEL = ASSUMED NEUTRAL AXIS =

RESULTS 5.00_ = 7937. S180C. -2130-00000.:-0.4800

YOUNGS MODULUS CF MORTAR = 513326E. MODULAR RATIO (N) = 5.64942

NEUTRAL AXIS = C.47493

MOMENT OF INERTIA = 0.10869

,BIENDING MOMENT AT YIELD STRESS OF STEEL = -4449.

iBIENDING MOMENT AT ULTIMATE MORTAR COMPRESSIVE STRENGTH =

3136.

1 2Vrre .... .. ":...p ..4. ...,,... -7 _ +Wt. ..-.._ .._-. ..,._ _4,

--

---r -4.-... r -=---,r,--I "r-i..,--, _-rt -r. --sr _-_-r _ --, .... ... _ . t---_______ . re ..4---,- :Ff.g --....

-, I ..

63

.-- .., --- ---... 1...: 4, ... .... _ . I

'LAYER NUMBER CISTANCE(Y) AREA . TRANSFORMED AREA

1 0.07800 0.047E7 0.27044

2

0.25000

C.CE29C

0.46834

3 0.42200 C.04787 0.27044'

NUMBER OF STEEL LAYERS = 3

KODE = 0

STEEL REINFORCEMENT CATA

rtrrIlerr .

FOR

-100,000 90,000 80,000" 70,000 60,000

50,000

ci) rn (1) 40,000 -P 4-) cn 0 30,000 cf) 20,000 - 64 Mortar stress Steel stress

1000

2,000 3,00G 4,000 Bending Moment (Mb) (in.-1b)

Graph 8. Working stress as a function of bending

moment for specimen 4, series 3 .75 in.),

5,000 _3,000 10,000 10,000

9,000

-8,000

_7,000 6,000 _5,000 0 _ 4 , 0 0 0 - -2,000

APPENDIX C

Prediction for Specimen 1, Series 1 (1.375")

=a A

w wy =- 91,800 x .0638 # 5,857 lb,

F =a A

_r

_ry-r

= 39,800 x

5,851 lb. F =

+F

W r = 5,857 + 5,851 11,708 lb, Fs a.

±

.85a cub 11,708 # A823 in..

.85 x

4760 X 6

= 6"

(a) -Aw = .0638 in.2 Ar = .147 in.2 acu 4,760 lb y = .3125 in.

y2 =

.5625

in.

66 -in. 2 (WI Fig. 17.(a) Dimensions of beam. (b) Stress block and

resultant forces. .147 Fw = -jd = = a N. A y w 2 Fr

EM_{Fr -}= ZMFr IL

Fs

Evaluating:

EMFr = 5,857 x (.5625 - .3125)

= 1,464

in..-lb amFt _1,464 Fs -

11,708

Taking moments about Fr,

172

- Y1)

Fr(0)

Mb = jdFs

= .6963 x 11,708

= 0,152 in -lb

= P 4Mb 4

x 8,152

- 23.5

P =1,388 lb

67

it

-= .125 in.

jd = (1.375 - y

) 2

(1.375 - .5625) + .125 -

4823

₂

.9375 - .2412

= .6963 in.

+ 1 - -= h - + h -= -= -L =

Prediction for Specimen 1, Series 2 (1.1851!) Fs = 11,708 lb 4

P -

x 7,124 23.5 P = 1,213 lb 68 -N.A. (from a -= h = page Fs 66) Fc jd .850-cub 11,708 .85 x 4,885 x 6 .4699 in .125 (same as page 67) y2 1 jd = (1.185 - y) + h

-

Fig. 18. Stress resultant block and forces. = (1.185 - .4665) + .125 .4699 2 = .8435 - .4699 2 = .8435 - .2350 = .6085 Mb = jdFs = .6085 x 11,708 = 7,124 in.-lb

--

69

-Prediction for Specimen 3, Series 3 (1.4")

F 11,708 lb (from page 66) a - Fs .85acub Fc 11,708 a 6111 .85 x 5,930 x 6 N.A. = h = .3871 in .125 in. (same as page 67) jd r Fw

yl

2 = (1.4 -

y2) +

h

-= (1.4 - .575) + .125 - .3871 Fig. 19. Stress block and resultant forces. 2 = .950

-

.3871 2 = .950 - .1936 = Mb = .7564 jdFs = .7564 x 11,708 = 8,856 in.-lb P - 4 x 8,856 23.0 P = 1,540 = --(1..4

-a=

h -.85a cub 10,405 .85 x 4,885 x 6 = .4176 in. ZNIFr = Fw x .17185

= 4,395 x .17185

= 755.3 in.-lb EMFr Fs 70 -2

Prediction for Specimen 2, Series 2 (1")

N.A. F = w = =

Fr

= = = ow A y w 91,800 4,395

Cr

A y r 72,500 6,010 x lb x lb .04787 .0829 Fc jd a F

r

Fw 755.3 10405 = .0726 a 3d = (1.000 -

y2) +

h

-7

= (1.000 - .40625) + .0726 .4176 2

Fs = Fr + Fw Fig. 20. Stress block and

= 6,010 + 4,395 resultant forces. = 10,405 lb Fs -w

-4

P - x 4,760 23.5

P = 810 lb

71

-Prediction for Specimen 2, Series 2 (1") (Cont'd.)

= .6663 .4176 2 - .6663 - .2088 = .4575 in. Mb = jdFs = .4575 x 10,405 = 4,760 in.-lb

Prediction for Specimen 2, Series 3 (1") F 10,405 (from page 70) F F S c .85acub 10,405 . a5- x 5,930 =4. .34404. in. (from page 70) jd =(.1.000 - y2)

h

_

1

4 P - x 4,563 23.0 P =.794 lb -

72

-jd N.A. :gar,' Fs

= (1.000 - .46255) Fig. 21. Stress block and

resultant forces. .0726 .34404 2 = ,61005 .34404 2 = ,61005 - .17202 .= .43803 Mb = jdFs

=

.4385 x 10,405 = 4,563 in.-lb - -=

a=

a x 6 h = + -Fw 1 +

-Prediction for Specimen 3, Series 2 (3/4") Fs = 10,405 (from page 70) a - Fs .85c b cu 10,405 .85 x 4,885 x 6 = .41764 Impossible situation

(see Fig. 22) i.e.

overreinforced. Suppose the assump-tion is made that the beam will fail

when the stress

block reaches the

Fs

=F

c

Fw = 4,395 lb

Fr = Fs - 4,395

= 4,948 lb

73

-Fig. 22. Stress block and resultant

forces. Note that the stress block

required to balance the steel forces

encroaches on the tension steel. Since

there cannot be both tension and

com-pression at the same (y) location, this is an impossible situation.

= = .85 x 9,343 4,885 x lb 2 6 x .375

Fig. 23. The assumption

the beam will fail

block reaches the

is made that when the stress tension steel.

metal rods (see

y. Fig. 23). a = .75 - y 2 _Fc = .75 - .375 jd = .375 11.1 Fc = .85acuba -Fc -F

Prediction for Specimen 3, Series 2 (3/4") (Cont'd.) ) F EMFr Fw (Y2 - Y1 _r (0) = 4,395 (.375 - .2031) # 755,5

in. -lb

EMFr Fs 755.5 9343 = .08086 4 h -

7

.375 + .08086 2 26836 Mb _jdFs = 26836 x 9 ,34.3 = 2,507 2,507 x 4 p 23.5 P = 427 lb, 74 -= h jd + - -= = in.-lb

Prediction for Specimen 1, Series 3 (3/4") Fs = 10,405 (from page 70) Fs a -85acub 10,405 85 x 5,930 X 6 jd = .4006 X 10,405 - 4,169 4 P - X 4,169 23.0 P = 725 lb 75 -F = .34405 EMFr = Fw (y2 - 171) = 4,395 (.1720) F r + Fr (0)

= 755.9 Fig. 24. Stress block and

h -

ZMFr resultant forces. Fs 755.9 y = .0780 10,405 = .07265

y

2 = .7500 jd = (:75 - y) + h - a

7

2 = .50 + .07265 .34405 2 = .4006 Mb = jdFs a N.A. -

-Prediction for Specimen 4, Series 3 Fs = 10,405 (from page 70) a - Fs 85a cub 10,405 85 x 7,937 x 6 Fc --(---(from page 75) y2) + h -

1

.2571 2 = .4441 x 10,405 = 4,621 - 4 x 4,621 P 23.0 P = 804 lb jd

i

Fig. 25. Stress block and

resultant forces. y = .0780 1 y = .250 2 = .5 + .07265 = .4441 Mb = jdFs = .2571 Ii = .07265 jd = (.75 -Y N.A. 76

-APPENDIX D

GRAPHS

ot-800 -Q, 1,400 1,200 1,000 400, 200 Calculated ultimate strength Deflection,

an.)

Graph 1. Force *plotted as a function of deflection for large specimens.

78 = Specimen 1, series 2 (1.185") Calculated ultimate strength Specimen series 1 (1.375') cL, 6,00 .2 .3 .4 w 0 1, 0 .1

-1,400 1,200 1,000 800 600 400 200 0 .1 79 -.2 .3 Deflection (in.)

NOTE: While testing this

specimen, one of the supports slipped at about 500 lb.

The load was released, the

support adjusted, and the load reapplied. Hence the initial stiffness is not

exhibited in this curve.

.4

Calculated ultimate

strength

Graph 2. Force plotted as a function of deflection for specimen 3, series 3 (1.4").

0

-1,400 1,200 1,000 800

3

oI 600 400 200 Calculated ultimate strength 80 Specimen 2, series 2. Calculated ultimate strength Specimen 2, series 3. .1 .2 .4 Deflection (in.)

Graph 3. Force plotted as a function of deflection

for 1.0" specimens.

-

-0

400

4

f7-41 300 200 100 Specimen 3, series

2.

Calculated ultimate strength Specimen 1, series 3 Specimen 4, series 3. alculated ultimate strength 0 .2 .4 .6 .8 Deflection (in.

Graph 4. Force plotted as a function of deflection for 3/4' specimens. -800 700 600 500' 0 0 1 1 1.0 81

-APPENDIX E

Photo. 1. Compression test of mortar cylinder.

Photo. 2. Tensile test of wire mesh segments.

83

-La

-Photo. 3. The components of a ferro-cement

specimen.

Photo. 4. A ferro-cement specimen ready for

mortar.

84 -I

1. . 3r , 14 u .,61,2IL 4. r- ., 11,I, 0 , 11 II 11." 0 w1 oo ' .1 r.S.,.. 1 w . .1" 1.1 1., 1 '4,-,L ..', ''' _-. 1,14C" *Or I'II1II 1

r1 an04.1100 00) -0 uI71t, AA, 11,1I -,,

,

P Ill111111

0I0k I, RI ,I , In, ,,,1,,,

IRI II IILILI ilIa, a .11 a 11 , 11, I-0 A

, lu II II II,,,,IlmIll11 ,lir: a 0 ,,, 1,,, , a34 a 11[ , ' r11, c .:. W1.1... , .4% r 1,,.... 0 . ,51 yi 1 II, U. III IlIrtil. u,4441.4Hkuple II p 0 0 A a.

4;...".

A ffl ,FX4uNit 11 0 * m I II III 111, nit , cat. 'It 111,'

Photo. 6. _{A ferro-cement specimen ready fOr}

curing.

85

-wuki, 1,"111161, I , " '4 ii2Eptir ,

Photo. 5. _{Working cement mortar into the}

rein-forcing.

0

1.10

UU

-A

-.4a

0,1

Photo. 8. A, ferro-cement specimen set up for

testing.

is

=

Photo .9- _{A ferro-cement specimen tested to}

failure

-

86