A STUDY OF DESIdH METHODS

OR EARTHQUAKE-RESISTANT BUILDINGS

By

DemeLer George Fertis
O“

A THESIS

Submitted to the School of Graduate Studies of Michigan
State College of Agriculture and Applied Science
in partial fulfillment of the requirements

for the degree of

MASTER OF SCIENCE

Department of Civil Engineering

195“

I'HESIS

(44;; , 575/

TABLE OF CONTENTS

ACKNOWLEDGMENTS . . . . . . . ..... . . . . .....
INTRODUCTION . . . . . . . . . . . . . . . ........
EXPLANATION OF SYMBOLS ...... . . . . . .......

EXPLANATIONOFTERMS....... ...... .....

FREE TRANSVERSE VIBRATIONS OF A SLENDER BEAM . . . . .

Case 1 - Free-free Beam . . .............

Case 2- Hinged-free Beam . . . ..... . . . .

Case 3 - Sliding-free Beam . .........

Case 1+ - Fixed-free Beam . . . . . . . ........
FORCED TRANSVERSB VIBRATION OF A SLRNDER BEAM . ......

case 1 -' FIXBQ-fl‘ee Beam 0 o o o o o o O o e o o 0

Case 2 - Sliding-free Beam . . . . ..........

Case 3 - Hinged-free Beam . . . . . ........

\

comma-n vs. sum. IN DAMPING m Immzons . . . . .

DISTRIBUTION OF SEISMIC SEARS LED MOMENTS . . . .

EFFECT or L/d ON’MOMENT AND SHEAR DEFLECTIONS . . . . . .

APPLICATIONTOBUILDINGS. . . . . . . . . . . . . . . .

ASIISMIC DESIGN OF ATALL BUILDING . . . . . . . . . . .

Analysisabout theX-Xine . . . . . . . . . .

Analysisabout theY-YAxie . .. . . .. .

ASEISMIC DESIGN 013-; SHORT BUILDING . . . . . . . . . .

LnalyeisabouttheX-Xlxis . .. . . . . . . .

35r1’7’81

Canon—Free—freeBeam

\D-q-dmmm

10
12
13
15
17
19
23
25
33
36

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”FETISH

313L133?

Analysis about the Y - Y Axis .......... . .
APPENDIX A ........................
APPENDIX B . . ............ . . .....

PAGE
5h
62
6h
65

lemonnnmm

The author wishes to express his indebtedness to Dr. G. R. Snell
for his valuable advice and suggestions, and also to acknowledge his

thanks to Dr. C. L. Shermer who read the study.

A STUDY OF DESISH METHODS FOR

EARTHQUAKE-RESISTANT BUILDING-

ABS"RACT

There is probably no part of the world that has not experienced
at sometime some degree of earthquake shock. But, the sections in
which earthquakes have been destructive are less extensive.

An earthquake occurs when the fault blocks of the outer shell of
the earth move with respect to one another owing to forces acting under
certain definite but yet unknown physical laws. If this motion is
abrupt, the surrounding strata are fractured, a series of local move-
ments begin and the elastic earth waves. The surface vibration caused
by these waves in passing, is popularly known as an earthquake.

The purpose of this study is to present a practical dynamic system
of aseismic design. The reading matter has been reduced to a minimum
so that the reader may concentrate on basic principles, and the formulas
are derived so that more precise seismic data can become available.

In the study first the earthquake motion of a particle is analysed
and then a beam which is thrown into free and forced vibration. For
the ones of free transverse vibrations of a slender beam. the beam
vibrates under the action of no external forces except such as are re-
quired to maintain the theoretical and conditions. The end conditions
of the beam in both cases were taken as Free-free; Hinged-free; Sliding-

free, and Fixed-free.

From this analysis a number of formulas were derived from which
the unit deflections In. unit shears It, and unit moments “u were
calculated in tabulated form for a number of values of c and 13/73:)
and then the actual deflection y, moment n, and shear V were estimated.

.These tables and formulas, together with the analysis of the
distribution of seismic shear and moment in a building, were used for
the aseismio design of.a tall and short building. The buildings are
classified as tall or short according to the L/d factor. For tall
buildings the Shearing deflection is negligible in comparison to the
moment deflection but for short buildings it should be taken into
consideration.

A complete investigation of the earthquake resistant design of a
building included the following steps:

I. A seismological study of the site and locality with partic-
ular reference to its earthquake history, dominant periods
of vibration, and proximity of faults and rifts.

II. A geological survey of the site to determine the nature of
" the soil, the depth of the strata, and the elevation of the
ground water level.
III. Selection of the type of foundation to be used.
IV. Decision as to the probable end conditions of the building.
Selection of ”C”.

V. Determination of the foundation amplitude.

VI. Design of the structural frame for the statical loads, dead,
live and wind load, making the floor systems as deep and

rigid as possible.

III.

IIII. '

II.

III

III?

hi1“!

hum

VII. Symetricel distribution of the vertical R/G beams.
VIII. Calculation of the transverse fundamontal periods of the
building about the 3-1 and 1-! use.

1):. Calculation or determination of the ratios 'r/sp. If
1' falls in the dangerous bracket, assume that 2/11, a
0.9 or 1.1.

X. Determine the minus! deflections, moments, and shears by
we of caution (9). (10). (n). (18). (19). (20) and
Figures 1 and 2.

III. For very “portant structures, check these values by tests
of models.
XII. Distribute the seismic shears and moments among the vertical
beans. Check the stress in the vertical beams.
1111. Design, detail. and construct the building in conformity
with the principles outlined previously.
XIV. Check the column sections for the combined stresses due to
the statinl and seismic loading.
1'. Measure the periods of the completed structure and check
on the assumed end conditions. computed periods, ratios
s/rp,
Under the statical methods. the maxim shears and moments in

deflections, mounts. and shears.

buildings occur shore the height 1., is sore. But, from this study the
maximum noment my occur at any point betseen 0.00I. and 0.131., and the
maximum shear between 0.001. and 0.751.. Hence, the maximum earthquake

dosage any center at any plans of weakness between 0.00L and 0.751..

 

 

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INTRODUCTION

Earthquakes are possible in all localities. From a report pub.
lished in 19311 which refers to the earthquake activity within a specific
year, 1929, typical of any year. shows that there is probably no re-
gion within the United States and its possessions which is immune from
the possibility of earthquake damage. Also recent earthquakes in the
Ionian Islands of Greece. and Ihessali, the central part of Greece,
where approximately 150,000 people were left homeless. prove the same.

In the following pages the author presents a system of aseisnic
design. The design of earthquake resistant buildings has been based on
statistical assumptions and the reading matter has been reduced to a
minimmmtso that the reader may concentrate on basic principles.

Under statistical methods. the maximum shears and moments in build-
ings occur where the height, L. is sore. From page 279 of T. Naito'sa:
”The majority of those who have observed the different buildings in
Marunuclfi have drawn the conclusion that the point three-tenths of the
total height is the most vulnerable spot from the standpoint of damage
by earthquake. However, a more accurate survey of the damage will
indicate that the above conclusion is not always true....'

From Figures 1 and 2 of this study it is found that the maximum

moment may occur at any point between 0.00L and O.H3L, and the maximum

 

1s. H. Beck and s. n. Bodle of the U.S. Coast and Geodetic Survey.

2
T. Haito: IEarthquake Resisting Construction“.

V

shear between 0.00L and 0.75L. That means that the maximum earthquake
damage may center at any place of weakness between 0.00L and 0.75L.

In the following pages is also shown that the exterior walls of
buildings can be designed to take most of the seismic shear and moment

and the only additional expense is in designing them to take care of

the earthquake forces.

L

Vuuln'laubau"...pnkhlloldvqwcqlsflumvuufilfnvn’nvutfn a

1
2

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EXPLANATION OF SYMBOLS

Young's modulus

Moment of inertia of the section

Deflection at any point X

Moment at 1

Weight per unit length

Mass per unit length

Acceleration due to gravity

Acceleration at any point X and equals d 22y/dt

Area of an element

numerical coefficient depending on p and n

Steel ratio

Ratio of Young's moduli for steel, E3, and for concrete, I

Proportionate depth of the neutral axis

Period of vibration

Length of beam

Constant depending on the and conditions. (Free vibration)

Constant depending on the end conditions

Amplitude

Constant of forced vibration

Unit deflection

Unit shear

Unit moment

Period of forced vibration

Total horizontal seismic force transmitted by the upper floor

Average unit shearing stress

Unit bond stress

Unit stress in the tensile steel

Unit stress in concrete

Deflection constant

Average value along L

II of section parallel to and above the average floor

E1 of section through the average equivalent floor

31 of section through the foundation

a1:= Lineal extent of (El), (EI)2, (31)}, respectively
Napierian base 8 2. 7182%

r = Radius of gyration

c

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EXPLANATION OF TERMS

Modern theory3 indicates that the earth consists of a central core
of molten nickel-iron 1#350 miles in diameter. The outer shell of rock
is 50 miles thick and it is known as the crust. The crust is divided
by planes of cleavage known as M into immense irregular blocks
called gag}; M. The surface trace of a fault plane is known as
. um-

Normally these fault blocks are in equilibrium, but sometimes,
adjacent blocks move with respect to one another due to forces acting
under certain definite but unknown physical laws. This relative motion
is called £11. It may be horizontal, vertical, or a combination of
both.

A slip may be gradual or abrupt. If it takes place gradually,
there is a similar adjustment in the contiguous strata so that it pra.
duces no immediately observable consequence. If the slip is abrupt

the surrounding strata are fractured and a series of local movements

 

3.1.3. Macelwane, ”The Interior of the Earth", Bull. Seis. Soc. Amer.
June 192%.

HJ'. Reid, Report 93; the State Earthquake Investigation Com. 9};
Californiag Vol. 2.

 

J .R. Freeman, Earthquake we a__n_¢_i_ Earthquake Insurance, New York:
McGraw-Rill Book Company, Inc.

F. Omori and S. Sekiya, "Effect of Depth on Amplitude," Jour. Coll.
Sci., Imp. Univ. Tom, Vol. 1‘.

J.J. Crekoff, mags-3 qt; Egthqueékg Resistant figuggures, New York:
chravn-Hill Book Company, Inc.

C)

'1

“.1.
(In

311330

Dravi
sing]

the t

01‘ I"!

F.’
"I
f.
(D

begins. The elastic earth waves. The surface vibration caused by
these waves in passing, is known as an earthquake.
Earthquakes caused by slip are classified as tectonic.
Earthquakes due to volcanic explosions are called volcanic.
Epicenter is the point of the earth's surface directly over the
origin or focus of an earthquake.

Epicentral zone is the area about the epicenter, where the earth»

 

quake is most destructive.

The !§!g_§gplitude is defined as the maximum distance of travel
of a particle from the center of its path.

The term particle as used here, may be a body of any size or shape,
provided that all the forces acting on it are regarded as applied at a
single point and that the motion of the point determines the motion of
the body.

A particle is thrown into free vibration_by a sudden application
or release of a force.

A particle is thrown into a forced vibration by an application of
a periodic force.

If the period of the periodic force approaches the value of the
free period, the phenomenon is known as gynchronism.

The consensus of scientific opinion seems to be that destructive
earthquakes have associated with them a definite bracket of periods
ranging between 1.0 and 1.5 seconds. This is known as the dangerous

bracket gf_periods.

4

Experiments by Omori and Sekiya indicated.that a particle located
at some depth is displaced less than one near the surface.

Fourier has demonstrated that the most complex vibration of a par-
ticle may, in so far as friction can be neglected, be expressed as the
resultant of a series of simple harmonic vibrations. Each of the
simple harmonic vibrations is called a component or a complex vibration.

There are certain simple modes in each of which the particle exa
ecutes simple harmonic motion. The mode of largest period is known as
the fundamental_and those of lesser periods are called harmonics.

The period of a free vibration depends on the constitution of the
particle, whereas that of a forced vibration is detenmined by the
periodic force.

The term {Eggrfggg.bggm.means that the vibrating beam is free for
rotation and translation; hingggrfggg.means free for rotation but fixed
for translation; slidiggrfreg_means free for translation but fixed for
rotation; and fixedrfree means fixed for translation and rotation.

The term average equivalent floor means the theoretical depth of
floor slab over the gross area of the plan obtained by considering the
transformed volumes of the actual floor skeleton and slab at the average

floor.

"1

Assume a
circumference
n. it a time
tine t,
he minus I

Period

The Vel

The sec
31’ Substitui

Referr'

conventiona.

“‘1 final:
b

"("er1 a

‘
t‘f:

FREE TRANSVERSE VIBRATIONS OF A SENDER 3m

Assume a particle P moving along the i P
circumference at a constant angular velocity m
:1. At a time t, A6‘P = nt. The displacement at B (Tn F} A
time t, “Lg—4-

X = r cos (nt)
The maximum displacement of PX’ r, is known as the amplitude.

Period T = 2n/n

The velocity of P , V = dx/dt = - nr sin (nt)

2 = nar cos (at)

The acceleration, a = dZX/dt
By substitution, sax/at.2 - - n21.

Referring to beams: For deflections within the elastic limit, the
conventional theory of homogeneous, isotropic beams shows that:

EIy =ffuuxux

The first derivative of y with respect to X gives the slope
s1 dy/dx = fl: ax
The second derivative of y with respect to I gives the moment,
31 day/dxa = s
The total shear is given by the third derivative of y with respect to X
m d3y/ax3 = v
And finally the fourth derivative of y with respect to I gives the
mass-acceleration I at any point, or:
II any/ax“ = 4' directly opposite to V.

According to Newton, 1' = ma 2 w/g a, than F = w/g day/dtz

I a ‘I:
5; 5L

Lettin

31m

The 3:

he I]

Inc B.

he s

by “batitntion. 3.12%: L- 161 D or. $3 £2; y
' E g dt2 31g 0

Letting ml} ‘ 93!. than n2 = mung , and because T = _2_n_, then, n2 = %2
Mg :1

Equating and solving for T, T = 2__!_r_
m2 lIg

Free v_i_bration equations:
any/ax” 8 may
The solution is: y = A cos hon!) + C cos (mX) + D sin (mX)+ Bsinhme)

The slope is: .1_ $3; A sin h (a!) +3 cos h(n1)- C sin (mx) + D cos (mx)
m

The 3.1!. is given by: L A one h(mX) + B sin h (IX) - C cos (IX)

a}:

- D sin (mX)
The maximum shear is given by: i

‘1 921,,Asinh(ml) +Bcosh(mX) +c.1n(nx) - Dcos (max)
I

Taking into consideration the four cases of end-conditions of the beams,
it is found“ that:

Case 1 — Tree-free Beam

1' e 0.28 114’: (fundamental), r1 = 0.10 (first
Hg :15 harmonic)

Case g - Hinged-free Beam

 

T 8 O. 41 (fundamental), '1'1 = 0.1} (first
;Ig 31: harmonic)

 

..

l;

J. J. Creskoff, Mos oils: uake Resistant Structures. New York:
HcGraw-Hill Book Co., Inc., pages nil-Em

gas; 3. - Sliding-free Beam

T = 1.12 3§2_ (fundamental), T1
leg

0.21 13': (first
lIg harmonic)
Case n - Fixedpfree Beam

T = 1.79‘IEEE' (fundamental), T1 = 0.29 !;E_(first
11g 31g harmonic)

Tabulating the numerical coefficients of the fundamental and first

 

harmonic periods in cases 1 to M inclusive, and obtaining the ratios

of the fundamentals to the first harmonics we have:

 

 

- g) (2) i (3)
End Conditions Coefficient of T Coefficient of T1 T/Tl
Fixed-free 1.79 0.29 6.2
Sliding-free 1.12 0.21 5.3
Hingedpfree O.hl 0.13 3.2
Pres-free 0.28 0.10 2.8

 

‘- —W W v w

Column 1 shows that the fundamental period of a fixed-free beam
is 1.6 times as large as that of a similar but sliding-free beam, H.“
times as large as that of a similar but hinged-free beam, and 6.4 times
as large as that of a similar but free—free beam.

Column 2 shows that the first harmonic period of a fixedpfree beam
is 1.“ times as large as that of a similar but sliding—free beam, 2.2
times as large as that of a simlar but hinged-free beam, and 2.9 times
as large as that of a similar but free-free beam.

Consequently, if the fundamental or harmonics of a beam.lies within
the bracket of periods of destructive earthquakes, a simple but effective

method of changing the periods of the beam consists in changing its end

conditions. For example, if the fundamental period of a fixed-free
beam is 1.0 seconds, the dangerous bracket may be avoided by making
the end conditions either sliding-free, hinged-free, or free—free, with

periods of 0.62. 0.23, and 0.16 seconds, respectively.

FORCED TRANSVERSI VIBRATIONS OF A SLENDER BEAU
From previous calculations, (1" g - L. $21
IKE BI: 412
During a forced vibration, y =f (X)f( t) = F cos (pt) (at I 8 0)
where #1) = F, f(t) = cos (pt), 1' = amplitude, p = forced angular

velocity, and t = time.

Then:
Forced period, T], = 2.9. , and T2 = (2:322
a p2
from, d g _ Pay, we have, 6.“ 8 E; y
at? if 316

letting m'u 8 £2; , then, P2 = m'h'llg and by equating and
31g w

olvi forT wehav 1‘ =21! w

But from page 6, T: 2!; w

n2 3‘s
' 2.2.1.
1:2 E1
:- an? m- av”;- Mr;-
P ;I—Z-EI-g- . P p

But ml. is a constant of free vibration which depends on the end condi-
tions of a beam. m'L, the constant of forced vibration, can therefore
be computed in terms of the ratio between the free and forced periods

of vibration.

Forced vibration equations}
Substituting N“ for Paw/31g, we have, qu/dx = Why
The solution is, y--= A cos h (m'X) + B sin h (111'!) + C cos (m'l) (1)
+ D sin (m'X) where y is the mimum deflection

at any point I.

10
Maximum slope at any point x is,
__1___ £11: A sin h (m'X) + B cos h (m'X) — C sin (m'X) + D cos (m'l) (2)
m' (11
Maximum BJI. at any point is,

_l__ (121 = A cos h (m'X) + B sin b (m'X) - C cos (m'X) - D sin(m'X) (3)
m'2 cm?
Maximum shear at any point is,

_L_ d3 = A sin 11 (m'X) + B 008 h (m'X) + 0 sin (m'X) _ D cos(m'X) (u)
“03 61}

Taking into consideration the four cases of end-conditions of the

beams, it is found, that for:

Case 1 - Forced vibrations of a fixed—free beam.

 

The conditions at the fixed end are: x =- o, y = r, l/m'(dy/dX) = o

from X: 0 , and equations (3) and (‘4): y = P = A + C. (5)
1 8 a:
5-;— g- 0 3+1) (6)

The conditions at the free end are:

2 3
x = 1., 1 d g _1__ a g
:5 “2 O. “'3 dx 0

From X = L, and equations (3) and (Q):
2

} flag-=o=lcoshm'L+sunhm'L-ecoum'L-

“ (ea)
Dsinm'L

l :113

l:73-(I-x§‘-"=()=Lsinhm'lwl-Bcoshm'I.-i-Csinm'l...

D cos m'L
By equations (5) and (b) A = F - C, and B = -D, and equation (60,)

wehave: Fcoshm'L-Ccoshm'L-Dsinhm‘L-Ccoshm'L-
Dsinm'LIBO (7)

11
F sinhm'L- C sinhm'l - Dcos hm'L+ 0 sin m'L- Dcosm'L=O (8)
Solving equations (7) and (8) for C, equating the two forms of C, and

solving for D, we have:

I? (sinhm'L cos m'L+coshm'L muggy 3:- p
(l + cos h m'L cos m'ID

N]

c = E. (l-sin h m'L sin m'L + cos h m'L cos m'L)
2 (l + cos h m'L cos m'L)
F (l + sin h m'L sin m'L + cos h m'l. cos m'_)

A 8 (l + cos h m'L cos m'L)

MI

The maximum deflection, moment, or shear at any point x is found by
substituting the values of A, B, C, D, and X into equations (1), (3),
and (it).

From equation (1) the maximum deflection, or amplitude, where X = O,
is: y = A + C = I

By equation (3) the maximum moment, where X = O is:

J... 9.1:
m.2 “2 A-C,

2
but as previously stated, B.M. is 319.1 therefore:

ax?

2
Eli—1%: 21 went - o)

By equation (’4), the maximum shear, where X = 0 is:

,1}
m0} dz}
d3
but as previously stated, shear V is III—g- therefore:
dx
d3
31 = Bllm':5 (B - D)

d!

12

Influence Curves for Case 1:

For free vibration m1. = l. 88, but, m'L= mLVE-i then:

\IT
m'L = 1.88 .F' ; letting: X = cL then m'X = mn'l.
P

Therefore: T
m'X = en'L = c 1.88 5;

Assuming F = 15, the maximum unit deflections Y“, unit shear V“, and

unit moment "u: can be calculated for a number of values of c and T/Tp
by means of equations (1), (3) and (it). See Figure l.

The actual maximum deflection y = 1'!u (9), Yu' obtained from the
correct T/Tp in Figure l, and F being the amplitude at X = O.

The actual maximum moment at any point X is:
931 2
= Eldxz = EIm' Flu (10)

“u being obtained from the correct T/‘l‘p in Figure l.

The actual maximum shear V at any point X is:

v = 313% = 1cm:3 rvu (11)

Vu being obtained from the correct 'l'l'l'p in Figure 1.

Case _2_ - Forced vibration of a sliding-free bem.

The conditions at the sliding end are: X = 0, y = 1‘, L $1 = 0,

m' d!
shear at X = 0 cannot be assumed as O.

2
.L d - 1 d3
The conditions at the free end are: X = L, “.2 d—X-E .. o. .73. —% = O

The end conditions for Case 2 are the same as for Case 1. The

solutions are identical. Figure l and equations (9), (10) and (11)

 

Creskoff, @. 911.. p. 59-

13

Forced Vibration of a Fixed-free Beam (F = l)

 

 

’U

‘14
.5
“‘3 M

s/rp m'X Y Mu v c r/r m'X

 

 

 

 

u u

0.5 0.00 1.00 1.16 .1.00 0.0 0.7 0.00 1.00 2.30 -2.52
0.27 1.03 0.77 -1.33 0.2 0.31 1.10 1.50 -2.19

1: 1.08 0.53 1.13 0.u5 -1.05 0.9 131.05 0.03 1.30 0.9M -l.82
B=-0.80 0.80 1.25 0.23 -0.73 0.0 B=—l.26 0.9h 1.71 0.43 -1.3u
C=-0.08 1.00 1.39 0.07 .0.38 0.8 o=.o.05 1.20 2.09 0.09 -0.85
D=O.8O 1.33 1.53 0.00 0.00 1.0 n=1.201.57 2.50 0.00 0.00
0.9 0.00 1.00 7.89 -0.00 0.0 1.1 0.00 1.00 -7.5u 9.00
0.30 1.95 5.55 -0.20 0.2 0.39 0.u -5.03 5.00

a: u.u2 0.71 2.00 3.u0 -5.99 0.9 A:-3.27 0.79 -0.9 -3.00 9.88
3:.3.30 1.07 n.23 1.73 -u.31 0.0 B= 2.33 1.18 -2.91 -l.83 9.13
0:.3.u2 1.92 0.02 0.50 -2.u5 0.8 c= u.27 1.58 -5.22 -0.52 2.59
:0: 3.30 1.78 7.93 0.00 0.00 1.0 D=—2.33 1.97 -7.59 0.00 0.00
1.5 0.00 1.00 -1.50 0.20 0.0 3.0 0.00 1.00 0.12 -1.21
0.90 0.85 -1.35 0.0M 0.2 0.05 0.97 -0.u7 -0.57

1=-0.28 0.92 0.90 -1.00 0.93 0.9. A: 0.50 1.30 0.78 -0.0u 0.01
s: 0.10 1.38 -0.20 -o.5u 0.97 0.0 ‘3=-0.01 1.95 0.31 -o.ug 0.38
c: 1.28 1.8a -1.ou -0.10 0.00 0.8 0: 0.uu 2.01 -0.30 -0.22 0.37
D-0.10 2.30 4.81: 0.00 0.00 1.0 D=0.01 3.20 -1.oo 0.00 0.00

Figurel

apply to both cases. The free periods for two cases differ. So, for beams
identical except for the end conditions, the deflections, moments and shears

rill differ quantitatively.

Case 1 - Forced vibration of a hinged-free beam.

1 1?:
The conditions at the hinged end are: X=O, y = F, —- a: o

m'2 012
By equations (1) and (3) and X = O: Y = l' = A + C (12)

2
J-g-lso-c-l (13)

 

 

3
The conditions at the free end are: X = L, J— d = O, J'— L1,: 0,

“.2 dx? “.3 “3
By equations (3), (h) and X = L,
2
iéuzozn coshm'L+B einhm'L- C cosm'L
m‘ 01:2 -D sin m'L (1n)
1 13.x
—— =O=Asinhm'L+Bcoshm'L+Csinm'L
m'3 613 - D cos m'L (15)

By equations (12) and (13), A = C = F/2, substituting F]? for A and C

in equations (1“) and (15), we have:

’5 cos h m'L + B sin h m'L - gcos m'l - D sin m'I. = O (16)

22-0121 h m'L + 3 cos h mm. + i? sin m'L - D cos m'L = o (1?)
Solving equations (16) and (17) for D, equating the two forms of D, and

solving for B, we have:
F Lsin h m'L sin m'L - cos h m'l. cos m'L +1)

3 = 2 (sin h m'L cos m'L - cos h m'L sin m'L

D 1 (sin h m'L sin m'L + cos h mg. cos m'L - l)
2 sin h m'L cos m'L - cos h m'L sin m'L)

The maximum deflection, moment, or shear at any point X is found by

 

substituting the values of A, B, C, D, and X into equations (1). (3)

and (11),
By equation (1), the maximum deflection at X = 0 is, y = A + C = F
2
By equation (3), the maximum moment at X = O is, ---1--— 31.1 = A - c-
m‘2 a2 ’

2 2
but B.|£. is defined as Flu, therefore, 311-!- = 313'2(A - C) .
ex? ex?

13.

By equation (‘4) the maximum shear at X = O is, -%-3- g B - p;
m

\N

(11

15
. d3! d3z 3
but shear V is defined as M , therefore, E1 = EIm' (B - D)
0x3 0x3
Influence curves for Case 1.

m1- = 3-93

. — '1' .. '1'
but. m'I. - 1111. it; . then m'L - 3-93Vfr5: also, similarly to Case 1.

'x Cm'L c '/3 93 T
m = = . --
K 0

Assuming that F = 16, the mimum unit deflections Yu, moments “u.
and shear Vu, can be calculated for a number of values of C and ‘1']‘1'p
by means of equations'(l), (3) and (14). See Figure 2.

The actual maximum deflections y, moments M, and shears V at any

point X are:

y = FY“L (18)
M = Elm'aFMu (19)
v = nxm'3rvu (20)

F being the amplitude at X = O, and Yu' Mn and Va being obtained from

the proper TITp in Figure 2.

Case ‘1 - Forced vibration of a free-free beam.

 

= 0.

$99-

1
The conditions at X = O, are: X = O, y = F. :55:

The shear at X = O unannot be assumed as zero since the assigned

motion is being accomplished by a periodic force at that point.
2 3

. . __ . __ 1 d 1 d =

The c0nd1t1ons at X - L are. X - L, {“72- d—J-é 0, m (”7* 0

The end conditions for Case a are the same as for Case 3. The
Solutions are identical. Equations (18), (19) and (20) and Figure 2

aP1313! to both cases.
P
creskoff, 92. 01:” p. 03.

lb

Forced Vibrations of a Hingedpfree Beam (F ==l)

 

 

T/T m'x In Mu v c T/T m'X In nu V0

 

 

 

0.5 0.00 1.00 0.00 -0. 3 0.0 0-9 0-00 1-00 0-00 -3-21
0.50 0.82 -0.32 -0.33 0.2 0.75 2.32 -1.98 -1.93

0.50 1.11 0.50 -0.38 0.07 0.1 1: 0.50 1.99 2.59 -2.73 -0.03
0.55 1.07 0.20 -0.20 0.30 0.0 B=-O.59 2.29 1.39 -2.11 1.53
0.50 2.22 -0.20 -0.10 0.28 0.8 0= 0.50 2.98 -0.93 -0.77 1.71
0.28 2.78 -0.73 \ 0.00 0. 1.0 'D= 2.02 3.73 -3.71 0.00 0.00
1.1 0.00 1.00 0.00 2.10 0.0 1.5 0.00 1.00 0.00 -0.07
0.82 -1.29 1.85 1.19 0.2 0.90 0.15 0.25 0.47

A: 0.50 1.05 -2.38 2.38 0.30 0.h A; 0.50 1.92 -0.u0 0.00 0.28
s=.0.uu 2.h7 -1.59 2.u3 -1.u3 0.0 3=-0.n9 2.89 -0.50 0.08 -0.21
c= 0.50 3.30 0.7u 0.92 -l.85 0.8 0= 0.50 3.85 0.07 0.29 -0.u1
n=—2.00 11.12 3.83 0.00 0.00 1.0 D=—O.l+2 11.81 0.90 0.00 0.00

Figure 2

The free periods for the two cases differ. Consequently, for
beams identical except for the end conditions, the deflections, moments,

and shears will differ quantitatively.

CONCRETE VERSUS STEEL IN DAT'PIIFJ TREE VIBRATIONS

A rectangular fixed—free, steel, and R/C beam will be compared to
determine the relative values of concrete and steel in damping free
vibrations. The beams assumed were designed to carry equal loads and
were identical in length and width but had different depths.

For a fixed-free beam the fundamental period,

7’11“ 'Lu
T = 1.79\\ 91:; ,or, T1 = 3. 20-— ETg

2
Using 3 and c for steel and concrete, $5,:

Ts 'sEcIc

'cEsIs

For a uniformly loaded, rectangular, fixed-free beam,

3/2
yr = 1152.. . s = Me I_ 003 2bc3. “d 63 = (11.2.)

i—" 12 3 4bs

3.1/2 3/2
Ec ;(:c (t)

Suyehiro showed that the damping value of a fixed-free beam is covered

then,

346?.)
II

by: X = .-l/2kt LA cos (n' t) + Bsin (n't)] where the damping factor

is: e'llzkt.

2 2
Using Creslsoff's7 formulas, k = L232!)- 3. . and Tc 3:113};E
T I? Ecch.

and equating we have:

kc no we”, 8.3/2
i: 5:0) (a

8
Suyehi ro gives, RC = 7270 and R. = 104,050.

 

7Creslcoff, 22- g____it., p. 50.

81(.Suyehiro, "Engineering Seismology,” Proc. Amer. Soc. Civil Eng.
‘iay, 1932; Bull. Imp. Earthquake Research Inst. Japan, Vol. 6, p. 63.

18
For 2,000 pound concrete, Sc = 700 p.s.i. and for structural steel,
Se = 18,000 p.s.i. Since the beams were designed to carry identical

loads, the maximum resisting moments are equal, therefore:

 

 

c c - s s -
and, Co - 5.070.

Assuming b = 1 foot, d. = 1 feet, then depth of concrete is, d0 = 5.07 feet.

wk = #90 lb. per foot, and wb = 5.07 x 150 = 700 lb. per foot.

then, he _ _
i?’ - 7.1, or, kc - 7.1]:s .
s
Then the damping factor for steel is, e-l/ZRSt, and e'l/2(7'1ke)t, for

the R/C beam.

The two factors compared, indicate that R/C is far more effective
than steel in damping free vibrations. The R/C beam will rest much
sooner than the steel beam, and its abnormal amplitude during synchronism

will be less.

DISTRIBUTION OF SEISMIC ShLARS AND DIOMENTS

With end conditions fixed-free but guided, for steel beam:

(concentrated load at the end)

3
2131

:9

 

0

Maximum deflection due to moment. 35 =

H

Maxims. deflection due to shear, 5's = 1%: . (See Appendix A)

3
Total maximum deflection, Y' = ym + ys = £11.--.)- 111

but, A = I/r2, r, being the radius of gyration;

2 2
- PLCL '+‘3br )
he” Y 1231 (1)

For a similar R/C beam:
Turneaune and Maurer9 devised the following semiempirical moment-
deflection formulas for R/C beams:

= 21113:: , K. ginuwa. 1.... 1.1.2121

q“ 3,003E7'

ll
H
N
H

For a fixed-free but guided R/C beam, C== 1/12, W = P, bd3

D. = ”m3
183,1K'.

Assuming a theoretical rectangular steel beam.with the same and condition:

3
. - EL;__ .. 1 _ 'm
A - 121,1 ' °rv $13-- i'I'S'K' °"' ”m ‘ ”£753:-

2
r L + 0r2
Total deflection of BIG beam, 0:: lSESIK' (2)

 

-mcc

9Creskioff, 22, 213., p. 37.
T. Naito, l'Eart'hquake Resisting Construction".

J. Prescott, "Applied Elasticity".

20

By hypothesis, at the same elevation Y' = D.

2 2
Taking, Z = (L.}}.b£ ). for steel, and G = 913%,16‘11-1 we have:
I" g-PSLZ chG

and D = m...
1811‘.

Referring to Figure 3:

v. = Pleza .___ PsbLzb= an = D
123, 13,121,“
= ELGE = L—PbLGb= PanGn =Y'
188. 181318].
Therefore:
2G P 2G
1:32:01 $3.25.: CZ. —§£=——— —-——=C3. £=4=Cho
P31, 2 Pm 2,, Pa. 32. Feb 32;
P 20
An: .._’.‘. - c,5
Pen 32.
and:
P _ P - P _ P _ P
Fab---s-‘~-°Psn" mrpco’iapcb--B-£°Pcn 35-
01 02 C3 01, 05

Letting n1, n2, n3, nu, n5, and nb be equal to the number of similar beams:

P = ans‘ + n2P8b + n3P8x1 + nuPca + nSPcb + ancn

and by substitution:

P= P +3.3. +3.1 +“‘4 +52%
nlsa. clpsa capes, 639580 can“ +“Psa

then: 35‘ =

-- -----'. --".-‘..-

n1+n_2_+:§_+:lt+:§_+:
c1 02 03 on 05

PC a'

 

With Pea known we can determine Psbv P Pcb, and Pen

811'

24" 24" 72"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

72" 24" 2.4"
I l a I”? V1 V1
P2,, IV/[////l W/A IV/A 17411-1217” 338.
Ta: "Ti-FE scram ,
Rb Rb E1
-_R—n _
El
1 Q
‘1 In: a:-
‘1 8- 3%
p
"‘5 PLAN
, 3
1'3; P7:-
in
A 31
321
s» B j
55-;- I ' IV/f/l T/AIW/A 75;:
1
s-
LTlu,t___r my _______ In_ ___'____'I_gr
fill an IRs'HIIEb: 1le”! I I Pea |||
MI PM I'I,I II'I
"I I'm I'llI 'III
“I: . 1H! , Ihll , a “H
Ulric—Ln Willi-Q1 FEE-11H c° LU
ELVATION
F113

DISTR\BUT|ON or Susmc SHEARS4 MOMENTS

22
Ilijiflbuti‘op‘ q}; Money-E:
Since the seismic moment taken by the vertical beams at any floor

is proportional to the seismic shear, we have:

 

u = U
m” In n ~~-n—-~ VIV—“v fl.
n1+£+l+i+§5+25
c1 c2 G3 on 05
= M .. M __ M _ “$8 = M
M“ Ea' I‘m-in“ “as- ”83' Mob-“r“ Mon '33-
1 Ca (33 61+ 05

Shearing, tensional and compressive stresses:10

l'or steel beams: v = :11; . f 2; “sad
bd 21

 

 

 

P P ll 21!
For B C beams: v = W?“ 11 = cn f = on f -.- cn
/ 1 ha 6.211.. 8 Fl bdz c jkbda
10 ‘

A.C.I., ”Reinforced Concrete Design Handbook."
D. Peabody, "The Design of Reinforced Concrete Structures.”

Creskoff, 92. gi_t_. ,

A.I.S.C. Steel Construction Manual.

EFFECT or L/d ON MOMENT AND SHEAR DEFLECTIONS

m3 PL L2 .
y“ = 1.2.1:? , y‘I = 1‘3" . 3:1: 3d This equation shows where L/d

is large, the shear deflection is negligible compared to the moment

deflection, but, where the ratio is small, the shear deflection must i
be considered. That means, that for tall buildings where the ratio

L/d is large the shear deflection is negligible. ‘But, for short 4
buildings where the ratio L/d is smll, the shear deflection should be

considered.

From equation y, = 1% substituting III. for P, for uniform load

 

-3:sz . 11-1: 39.4.3. _ (12
y,- n .slnce - d.and.1 12 . YS‘Ym {ca-17(1)

This formula gives the shear deflection when L/d is small.

The free period of a beam my be expressed as:

T = 232‘- Lavgm- from, T = 22 if; where, c is a constant depending
c m

on the end conditions. Also they can be expressed as: T2 = 11f}: ,
where, T, is the free period of beam whose L/d is large and, whose
static deflection is therefore almost due to moment. Static moment
deflection of a beam may also be given as, ym = 01 g: where, 01 is a
static deflection constant.

2 t
Then: in = 1“?- , and '1' = (kaym)1/2 here the free period of a beam is

proportional to the square root of its maximum moment deflection.

 

Assuming the same for small L/d, since Y = ym + ys, T, = Vk2(ymTys);

2h

where, ya, is the maximum static due to shear, and, Tt’ the true period.

From equation (1) substituting its equivalent for ya, and Ta/ké for Wm!

r“ ' 2"

' d
T=- l+-—-2-
‘ ‘T2( “1”)

APPLICATION TO BUILDINGS

For earthquake resistant structures the following should be
considered:

1. In investigating the site of an important structure we should *-
determine, a) the proximity of rifts and faults, b) the dominant periods
of vibration, c) the soil characteristics, and, d) the character of

substrata.

 

 

2. It is not right, even in the absence of all signs of seismic
activity, to locate a building over a known fault or rift, for it is
highly probable that in the event of a slip of any.magnitude the
building will be destroyed.

3. The closer the periods approach those in the dangerous bracket
of periods, or those of the proposed building, the larger the amplitudes
assumed.

u. The nature of the materials constituting the site, their elasti-
city, moisture content, and uniformity determine the damping value of
the site, its probable amplitude during a destructive earthquake, and
the character of foundation required. For example, if the material is
bedrock whose elastic value is high, the damping value is poor and the
amplitude assigned is low. If the site is composed of dry, compact,
uniform alluvial or diluvial deposits whose elastic value is less than
that of bedrock, the damping value is better and the amplitude assigned
is larger. Ihere the materials contain an appreciable amount of moistura
the elastic value becomes low, the damping is excellent, and the ampli-

tude assumed is high, its value depending on the cohesive strength of

26
the soil and on how closely it approaches the condition of semi-fluidity.
Finally, if the soil is not uniform but consists of large adjoining
deposits of varying physical properties, the damping value is good and

the amplitude assumed is an average value depending on the nature of

 

the component materials. {
5. During an earthquake, the soil forces the foundation of a i

building through a definite amplitude. An effective way to decrease 5

that amplitude is to insulate the foundation from the native soil by Ev”

eliminating any condition of fixity between the foundation and the soil. -
That means by creating discontinuity. A three foot layer of gravel or

broken stone introduced between the foundation and the native soil is

deep enough to create discontinuity. By experiment a three foot bed of

gravel or broken stone will decrease the amplitude transmitted to the

foundation by one half.

b. Locating foundation at a depth greater than is required ordin—
arily, does not materially decrease the amplitude transmitted. An
aseismic foundation is a beampand-slab mat. This cancels the adverse
effect of differential soil movements, also footings are not digging
themselves in and destroying the desired state of discontinuity, and
also we have a rigid and monolithic design.

7. If individual footings are used, the elevations of the bottoms
of the footings should all be identical and the footings should be
interconnected by beams able to maintain them in that same relative
position.

8. Owing to the oscillation of the center of gravity of a building

during vibration, the exterior parts of the foundation should be

27
designed to withstand soil pressures as high as four times the static
soil pressure.

9. If the soil is not uniform, yields excessively under pressure,
or contains considerable moisture, a beampandpslab foundation is
indicated.

10. Where the site consists of wet, loose fill, or approaches a
state of semifluidity, a beambandpslab foundation on piles is required.

11. Where soil values penmit, short piles, which allow the build—
ing to rotate in a vertical plans, should be used rather than long
piles driven to hardpan.

12. The fundamentals of satisfactory earthquake-proof design are:
symmetry of plan, mass, and rigidity. The center of mass should coin-
cide with the center of rigidity. Both mass and rigidity should be as
nearly uniform throughout the building as possible, so that the build—
ing cannot rotate about a vertical axis owing to one part being heavier
or stiffer than another.

1}. Rigid bents should be symmetrical with respect to the floor
plan.

1“. All parts of the building should be so connected to the frame
that the building will vibrate as a unit, the connections being strong
enough to overcome the inertia of the separate parts.

15. In tied columns, the ties should be spaced closely for some
distance above and belew the floors. Diagonal bands of steel should
extend back from the corner column into the floor slab. Provide addi-
tional and restraint for columns by means of extra bars in the floor-

slab and across the columns.

28
lb. The floors act as horizontal girders and should be designed as
such.

17. The floor system should be so braced diagonally that it will
deflect as a unit and will be strong enough to transmit the lateral
forces to all the bents.

18. The full depth of the wall spandrels should be designed as
beams, so as to attain the maximum.possib1e floor rigidity.

19. Floor slabs should be of rock concrete, rigid and monolithic.

20. Horizontal diagonal bars shall be used between floor beams
connecting interior columns.

21. Exterior walls should be of reinforced concrete, of uniform
thickness, and designed as vertical beams. Use ties and diagonal bars.

22. Diagonal bars shall be used across all wall Openings.

23. All construction joints shall be doweled, as this increases
their reliability.

2“. wall bents should be continuous for the full height of the
building.

25. Temperature steel should be placed diagonally.

26. The end conditions at the top of a building will be considered
one free.

27. The and condition at the bottom of a building is determined
'by the nature of the soil, the type of foundation, and the state of
discontinuity. For example,

28. If the footings are fixed into bedrock, or into dry, compact,
alluvial or diluvial deposits, the end condition will be assumed as

fiXOde

 

 

29
29. Where a layer of gravel separated the foundation from the native
soil, and the construction is designed to permit the foundation to de-
flect laterally, the end condition will be considered as Eggygga

30. If the foundation is fixed into wet, compressible deposits of

.3.
L!
I

__.P‘

such nature that rotation in a vertical plane can occur, the end condi-

.
~.
.‘ma -.

tion will be assumed as hinged.
31. Where a layer of gravel separates the foundation from the

soil, and the building is free to move in an horizontal as well as in

7‘4‘-“ _—-:m—

713.:

a vertical plane, the end condition will be considered as approximately
{£23,

32. After all the conditions involved are evaluated, it becomes a
matter of engineering Judgment to decide how closely any of the arbitrary
standards are approached and what values of '0" shall be assumed.

33. From equations of free-vibrated beams we have the values of

"C" (fundamental)

 

 

rr..-f¥35' Hinged-free Sliding-free Fixed:free

0.28 O.N1 1.12 1.79

 

"W‘— —_

3%. Assuming that the vibration of a tall structure, one where the
ratio L/d.= greater than 5, is similar to that of a slender beam - the
period of vibration of tall buildings may be expressed as:

T = C E T = free period, second, C = a constant depending on
E18 the and conditions. w:= weight, pounds per inch of
length. L = height of building, inches. E‘= Young's modulus, p.s.i.
I = moment of inertia of the section perpendicular to the plane of

vibration, inches“,

30
g'= acceleration due to gravity, inches per second per second.
35. For short buildings, where the ratio L/d is less than 5, the

value of T obtained should be modified by equation:

“
, . #F‘- ‘M

 

 

 

{ 2 \ where T is the true period of a short
r =“'T2 1 + ~9-75‘ t *-m~
t K #CIL } building, and 01 is the deflection ; E
J ‘1
constant whose values are given below: — 1
Values of C1 are based on the assumptions of the above values of C. Q 1
Fixed-free Sliding-free Hingedpfree Free-free pg
0.13 0.20 0.55 0.80

 

36. The quantity BI is taken as the average value along L. It
will be calculated by the formula:

a1(EI)1 + .2(r1)2 + a2(EI)3
l .3. e_._

 

E1 =

(II)1 =IEI of section parallel to and above the average floor.
(II)2 = 21 of section through the average equivalent floor.
(11)} = II of section through the foundation.
a1, a2, and a3 = lineal extent of (II)1, (EI)2, and (BI)3, respectively.
37. Average equivalent floor means the theoretical depth of floor
slab over the gross area of the plane obtained by considering the trans-
formed volumes of the actual floor skeleton and slab at the average floor.
38. The gravest stresses will occur as Tp approaches T, as the
period of the earthquake synchronizes with that of the structure. The
worst case which need be considered practically will be covered by as-

suming a ratio of T/Tp of 0.9 or of 1.1 because complete synchronism is

31

highly improbable, but, if should occur, the inherent factor of safety

of the structure will be ample to take care of the temporary increased

stresses provided that the structure has been designed for a T/Tp of 0.9

or of 1.1.

39-

A complete investigation of the earthquake resistant design ef

a building includes the following steps:

I.

II.

III.

IV.

V.

VI.

VII.

VIII.

II.

A seismological study of the site and locality with particular
reference to its earthquake history, dominant periods of vibra-
tion, and proximity to faults and rifts.

A geological survey of the site to determine the nature of the
soil, the depth of the strata, and the elevation of the ground
water level.

Selection of the type of foundation to be used.

Decision as to the probable end conditions of the building.
Selection of "C”.

Determination of the foundation amplitude.

Design of the structural frame for the statical loads, dead,
live and wind load, making the floor systems as deep and rigid
as possible.

Symmetrical distribution ef the vertical R/C beams.

Calculation of the transverse fundamental periods of the buildp
ing about the X - X and Y - Y axes.

Calculation or determination of the ratios T/Tp. If T falls

in the dangerous bracket, assume that T/Tb = 0.9 or 1.1.
Determine the maximum deflections, moments, and shears by means

of equations (9), (10), (ll), (18), (19), (20) and the Figures
1 and 2.

XI.

XII.

XIII.

XIV.

XV.

32
For very important structures, check these values by tests of
models.
Distribute the seismic shears and moments among the vertical
beams. Check the stresses in the vertical beams.
Design, detail, and construct the building in conformity with
the principles outlined previously.
Check the column sections for the combined stresses due to the
statical and seismic loadings.
Measure the periods of the completed structure and check on the
assumed end conditions, computed periods. raties T/Tp, deflec-

tions, moments, and shears.

ASEISMIC DESIGN OF A TALL BUILDING

A #5 story office building with two basements is to be erected.
Plan dimensions 75 feet - h inches by 111 feet - four inches.
The story height is 12 feet - 0 inches and the distance between
any floor slab and the bottom of the spandrel beam is 2 feet - 6 inches.
It is estimated that the theoretical floor slab depth is 8 inches
over the gross area of the plan.
Lineal extent of the theoretical floor slab along the height of
the building is:
(l + #5 + l)0.b7 = 31.“ feet.
Assuming a 6 foot thick foundation slab we have total height of build—
ing L = 570 feet.
Lineal extent of vertical beams = 570 - 31.“ - 6 = 532.6 feet.
L/d = 570/1084 = 5.2) 5 and 570/72.7 = 7.8 > 5
building is classed as tall.
Investigating the site it is found:
a) The lecality is subject to earthquake activity of moderate
intensity.
b) There are no faults or rifts in the immediate neighborhood.
An active fault is 10 miles away.
c) Dominant period of vibration of the locality is 0.1 second.
d) There is a disclosed bedrock at an average depth of 32 feet -
9 inches below the top of street curb where the building is
located.

e) No ground water.

3h
f) A neighboring building 84 feet by 200 feet and height 20b feet,
20 feet away and with footings fixed into bedrock has the
following periods:

Along the 8% foot side:

T = 1.“ seconds, T1 0.2 seconds, T/Tl = 7.0.
Along the 20b foot side:
T = 1.1 seconds, T1 = 0.2 seconds, T/T1 = 5.b.
Average T/T1 = 6.3. Consequently the end conditions are
approximately fixed-free.

To reduce the amplitude transmitted to the foundation a three foot
layer of gravel will be deposited between the foundation of the pro-
posed building and the bedrock.

For free vibration the and conditions of the building will be

between fixed-free and sliding—free.

It was decided that the end conditions be expressed by a value of

0.9M x 1/4 inch = 0.2u inch. See Appendix B.
Amplitude F1: 1/2 x O.2h-= 0.1? inch.
Structural frame consists of a steel skeleton with R/C roof floor
and foundation slabs.

- 7 - 6
E8~3x10,Ec—2x10

 

35

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cb ‘ so

5° ‘ .\\\‘-\\\\‘I

—~ V

Co Le—M 6 \

I&‘“—o————4 ‘ §

2.1'-6"-—— be" I

"”1- —- 3. a?“ s

C“ S

5b H .Sc. H 5; 5c, {:1

' V

c. ; 3

‘° . §

55 H 'Sc E3] Sc. .Sc

Ca ' S

. I

‘° 5 s s. i

C C
Y s. —E}—- m» -—e m a Y

Co 15 . §
i5?

IO: . I '26.“ I

. I a": ‘

Ca 5‘ ‘ - S

s» H ‘ It Ease Is 2 xi ”1“

Ca 63 6; 3". '_ §

. :| ‘34.; I

Cg °= l __

L S

& E5] ! "”-

Ca . ‘L “ “ — §

«._w_'l§'_§"_-- a.-- -, . ._z. _ ._ E

5' w

so. H ' -.\\V..&.\V—1\\VEER\V_\\\VN31_k\\m
0. x '3
“W"; i“-

_ _ _ ._ _. ,_ A; '62
so do J 5:: CE": 1:5 cb <1. 52.. i a; 0.
19.4%

 

ELEVATION AT AVERAGE FLOOR
Fm. 4

S cnon ROUGH V nc L s AGE.

36

 

 

 

 

 

 

 

 

Beam Section A Ix! 1y! r11 ryn
ea 14 x 16 x 26MMF 77.63 3526.0 1331.2 0.7h n.1u
sb 1n x 10 x 398MF 116.98 0013.7 2109.7 7.17 n.31
sc 1“ x 16 x 342MF 100.59 M911.5 1806.9 6.99 H.2h
ca R/C 00" x s" 480.0 2500.0 1uh000.0 2.31 17.32
cb R/C a" x 00" 480.0 1uu000.o 2500.0 17.32 2.31

Analjsis About the J_(_ _-__ }_{ A_J_c_i__8_:

 

then:

11: = Ix' + Adxz
In 2: 1+ £35220 + 77.0300 x 12)é:l+ 10 [0013.7+110.98(30 x 121%

+ 11 [6013.7 + 116.98(18 x 12)?) + 2(6013.7)
'1' 10 [14911.5 + 100.59(18 x 12)? + 504911.53) = 31-62 x 105

EBIxs = 10,296 x 1012 I = moment of inertia

2 about I - X building.

lec = 20 [2560 + “80(36 x 12)“]
’ Ix' = moment of inertia

+ 1+ [11415000 + “8002.5 x 12)§ about its own

2 X - x axis.
+ u [11111,000 + uso<21.5 x 12)]
2 (EI)x = EI of average
+ 1: [11411,000 + 1480(1u.5 x 12)] 1 section taken

2 above the floor.
4- 1. [11111.000 + Meow-5 X 12)]

5 Ixs = moment of inertia
26,h38 x 10 of steel section

about the X - X.

Eclxlc = 5233 x 1012
- - 12 - . 1h
(EDI1 - 1.31” + Eclec - 155811 J: 10 - 150 x 10
ngc = 1/12 x 108.67 x 12(72.07 x 12)3 = 721 x 108
1h

(E1)x2 = EI of average

Eclxgc = l##2 x 10
section taken

(El)x2 = Ests +Ec1x2c through the
equivalent floor
(E1): = 151.15 x 10 _
2 1x20 = moment of inertia

of the equivalent
floor slab section
about x-x axis.

37

Average value of (E1)! Will be:

532,6(31)xl 4» 31.4051)” + 6 Eclxac
570

(21),,

 

2M3 x 101k

(11),
Transverse period about the x - X axis.

(FITIWI ," ' 12
Tx = Cuwgl = 1.50%;12000 I 2183 X 10 .._. 2.143 seconds
) t 5x3 I 2113 x 101 x 386

Assuming a destroying earthquake of period Tp = 1.50 seconds

1/2
T 2.u . T A
ngfig=L62 (ii) =1.27
p " p

First harmonic T1 = 275-? = 0.112 second.

Transverse deflection moments, and shears during forced vibration

about Q; X - X axis:

 

 

0 Yu Mu Vn
0.0 1.00 -1.u0 0.09
0.2 0.55 -l.28 0.5u
o.u o.h3 —o.97 0.85
0. 6 -0. 21 -O.5u 0.92
0.8 “0.98 .Oels 0.614
1.0 -1.77 0.00 0.00

 

m 1/2
mu, 2 “(131] = L88 1 1.27 = 2.39
p

I"

2.3 .. -
m' =bifi= 3.5x10 1+ m' =12.25x10
12

8

m'3 = 112.9 x 10"

(El)" ng F = 2113 x .1011"L x 12.25 x 10"8 x 0.12 = 358 x 106

11+

(E1)Xm'3? --= 2113 x 10 x 112.9 x 10"12 x 0.12 = 125 x 103

 

W‘fl‘-.- I. ..

 

 

 

 

 

(1) (2) (3) (1") (5) (b) (7)
c y, in. Y, in. M, 10.11: v, lb. v', 1b. 11', 10.11..
0.00 0.12 0.00 -501x106 11x10? 22x103 125le
0.20 0.10 0.02 ~M58x106 07.51103 1351103 770110”,
0.110 0.05 0.05 -3h8x102 100x103 2121103 1208x1014
0.00 41.03 0.15 -l93x10b 115x103 230x103 13102110h
0.50 -o.12 0,21. - 511110 80x10} 100x103 913110
1.00 -0.21 0.33 o 0 0 0
C = x/L
y'= qu.= the deflection relative to the space axis.
Y = F - y, or the deflection relative to the foundation,
2
s: '
(EI)xm PM“, or the dynamic moment,
V'= (El)xm33FYfi, or the dynamic shear,
V‘ = 2V, or the static equivalent of the dynamic shear,
M' = moment taken by the vertical beams at any floor. The unp

supported length of the vertical beam is 9.5 feet. The
lever arm is 9.5 feet/2'= 57 inches, therefore, M' = 57V'.

Allowable amplitude = 0.0011. = 6.84 inches) 0.33

Maximum shear and moment at 0.60 of height.

Distribution gf_the seismic shear and moment:

 

(0.K.)

Under the assumptions that the mass of a building is concentrated

at the floor levels, that the floors deflect as units, that the ends of

the columnsand walls are fixed, the columns and walls between two con-

secutive floors may be treated as vertical beams, deflecting equally

by reason of a concentrated horizontal load transmitted by the upper

floor and with end conditions fixed-free but guided.

39

The seisMic shear and moment at any floor are taken by the columns

and exterior and interior walls in the

floor below.

 

‘.’

 

 

 

 

 

 

 

 

 

 

 

 

Steel Beams n A r1. rx.2 Ix' Z
sa M 77.63 6.70 u5.h 3526.0 4.15
sb 16 116.98 7.17 51.3 6013.7 2.u7
so 15 100.59 6.99 M8.8 u911.5 3.00
(L 2 + b )2 2 ,
z = b 3_:;__, = 9.5 feet = 11h inches., Lb = 12,996.
Ix'
13.2., 2 17 P__.. Zn_ 3-0
= ‘— = -" C = ~---
01 P“ z. 5.15‘0°b° 2: PTn 28
Assuming P = 0.007, K' = 0.097.
n/c b a 1 110 2 o
Beams ‘ k' ‘1'
ca 2h 00 6 360 1080 105 3 120.7
ch 16 8 58 1161+ 129800 12600 280 1.83’4
2
G _ (Lb + 3brx‘2)
— IXIK'
c3 = .22: 2.6.2 .. ‘2_Tu9'u = 20.1 on = =.:._:b- :73 w = 0 295
P = P —
sa 1_ ='2 0 000 - ZINC pounds.
(31+:E-1- 22+ :11- 107'
Cl 62 33 on
P 2100 P 21140
P = 8‘ = = 3570 pounds, P = 23L: __
sb El 0.60 .6 ca 071 3020 pounds.
P 2190
P = sa ._ ._ P ._ 2140
ca - - 107 pounds, P = Jfi -- __._ = 7270 pounds
C3 2001 Ch cu 0.295

M 3 1.3410010“)
" 107.h

 

= 122,000 inplb.

”as = _______122,000 = 203,000 in—lb.

Cl 006
M“ = 529.: Leg—£99 = 172,000 in-lb.
C2 .71

122 000
M = 3‘ ‘"““"= 0,080 in-lb.

2.sz = ’19.: 1331-999: 1.11.0000 111.112.
cu 0.295

Shearing tensional, and compressive stresses:

Assuming J = 7/8 and k:= 3/8

 

 

P 7270
- .93.- :
- de - 7 8x8x58 18 pounds (0.K.)
M 919 000 ulh 000
f = = ’ ______ = -—-—‘-——= a a O
' ijdz 0.007x778x8x3366 lbb 2500 P s 1
21.;
f = on _ 828,000 = 828 000 g . .8...
° “25m ' 778x373xs13366 ““3860 93 5 P

The unit bond stress v will also be very low.

The building amply strong about the X.- X axis.

Assume now that a layer of three foot gravel separates the founda-
tion from the soil, and the building is free to move in an horizontal
as well as in a vertical plane, the end conditions will be considered
as approximately free.

Assume then a value of, OP— 0.

r": VH2“
12000 x 2189;: l _
Then fundamental period T; = C 31 16 0°m w\[2u3fi x 101‘ x 386 -

0.30 x 1.07 = 0.501 say 0.50.

 

 

#1
Assume a destructive earthquake of periodTp = 0.50, where the
ratio TtlTp = 0.5/0.50 = 0.9 which is believed to be of the worst
because the period of the earthquake synchronizes with that of the

structure.

T

1/2
T
(.5) = 0.95, first harmonic T1: Q—ga- = 0.10.
P

As previ0usly demonstrated, during a forced vibration the end con»

ditions change to free-free. Using T/Tp = 0.9 it is found:

fl

 

 

c In Mu vu

0.0 1.00 0.00 -3.21

0.2 2.32 -l.98 -1.93

0.0 2.59 -2.73 —o.03

0.6 1.39 -2.11 1.53

0.8 ~0.93 -o.77 1.7u

1.0 -3.71 0.00 0.00

mil—$7: ”M31095 W19
Tr

m' = 2830 = 6. 57 x 10 u, m =u3. 2 x 10‘8, m'3= 283. 8 x 10 12
(El)xm'2F = 2U3 x 1011" x #3.? x 10"8 x 1.88 = 197 x 108

(El)xm'3F = 243 x 101“ x 283.8 x 10"12

x 1.88 = 129.8 x 105
Assume soil from soft wet fill with surface amplitude “.00 (See
Appendix B). The amplitude to the bottom of gravel = 0.9“ x 9.00 = 3.76

Amplitude transmitted to the foundation = 1/2 x 3.70 = 1.88 = F.

42

The actual deflections, moments, and shears are:

 

 

 

1 2 3 4 5 6 7

C y, in. Y, in. M, in-lb V, lb. 17', lb. M', in—lb.
0.00 1.88 0.00 0 g “-392 :105 -78M x10; -446 x10;
0.20 n.37 -2.u9 -390x10 -250 1:105 ~500 x10 ~285 x10
0.40 14.88 -3.00 -538x1og .. 3.ux105 - b.8x105 — 3.9x107
0.60 2.62 -0.7u ~410x108 -198.uxiog ~39b.8x105 226 x107
0.80 -1.75 3.63 ~152x10 226 x10 452 x105 258 x107
1.00 -b.98 8.86 0 o o o

 

C = X/L, y = flu, Y = F - y, 1.1 = (Enxm'aFMu, V = (EDIm'zFVu,

V' = 2V, 1" = 57V".

Column 3 shows that the maximum amplitude is at the top of the
building and equals 8.83 inches. This figure is large compared to the
allowable amplitude of 0.0011. = 0.811 inches. So, steps should be taken
to reduce the amplitude to the foundation. A three foot layer of
gravel is not enough. Should be greater.

Columns 0 and 7 show that the maximum shear and moment occurs at
the bottom of the building and equals, -78,L|-00,000 pounds and
-u,uoo,ooo,0oo inch-pounds respectively much greater than those of the
previous case.

In this case a beam-and-slab foundation on piles is required.
Where the soil values permit, short piles should be used, which allow
the building to rotate in a vertical plane.

Assuming a distractive earthquake of period, T

12
EE=ELZ=O.5' (a), = 0.22

P 1.00 Tp

”3

Using, T/Tb = 0.5, it is found that:

 

c In M In
0.0 1.09 0.00 -0.83
0.2 0.82 -0.32 -0.33
0.u 0.56 -O.38 0.07
0.6 0.20 -0.26 0.30
0.8 -0.26 -0.10 0.28
1.0 —o.73 0.00 0.00

 

m'L = mL‘ 3‘]: = 14.73 x 0.22 = 1.04

 

 

 

 

1.0a -u 2 -8 _
m' = 6836-: 1.52 x 10 , m' = 2.32 x 10 , m'3 = 3.53 x 10 12
(EI)xm'2F = 293 x 1011+ x 2.32 x 10.8 x 1.88 = 101.5 x 107
(EI)xm'3F = 203 x 101“ x 3.53 x 10"12 x 1.88 = 161 x 103
The actual deflections, moments and shears are:
Wm" * 'vsmaz ass—13:33:- mm...“ W" .2:
1 2 W 3 ‘1‘ 5 _ j 7
C y, in. Y, in. M, in-lb V, lb. 7', lb. M‘, inplb.
0.20 1.5h 0.39 - 32.5x10 - 53.2x10 -106.ux103 - 60.6:10
0.u0 1.05 0.83 - 38.6x107 u 11.3x103 22.6110; 12.9x105
0.60 0.38 1.50 - 26.u1107 h8.hxlog 96.8xlO 55.2:105
0.80 -o.h9 2.37 - 11.2:107 u5.1110 90.2x103 51.l+x105
1.00 -1.37 3.25 0 0 0 0

c = I/L, y = nu, Y = r. y, u = (snxm'zmu, v = (snxm'h'vu,

M' = 57w, v'= 2V

Column 3 shows that the anximum amplitude is at the tap of the
building‘and equals 3.25 inches. This figure is safe compared to the

allowable amplitude of 0.001L = 6.80 inches.

Columns 6 and 7 show that the maximum shear and moment are in
the bottom of the structure and equal to, 267 x 103 and 152 x 105,
respectively.

Now comparing the finding for 'r/rp = 0.9, and, 17121) = 0.5 it is
understood that the case of T/‘l‘p = 0.9 is the worst. It can also be
understood by looking on the Table (1). So, on very important struc—
tures, for more safety, a ratio of TITfi= 0.9 or T/Tp = 1.1 should be
used in their design.

The moment and shear produced when T/Tp = 0.9, is 293 times as
large as that of T/Tp = 0.5. The maximum amplitude will be 2.73 times

as large as that for TITp = 0.5.

Analysis About the _Y_- 1 Axis:

*

 

From the theory of the moments of inertia of plane areas,

I, = 1),, + “ye
Letting, (ll)yl =‘II of the average section taken above the floor
In = moment of inertia of the steel sections about the Y - Y axis.
1y. = It [1331.2 + 77.63(51+ x 12)? + 6 [2169.7 + ll6.98(5l+ x 12)?)
+ 1: [2169.7 + 116.98(36112)2] + 1. [2169.7 + 116.98(18 x 12)?)
+ 2 (2169.7) + 3(1806.9) + 6 [1806.9 + 100.5906 x 12?]

+ 6 [1806.9 + 100.59(18 x 13:2]: 6750 x 10‘3

_ 12
11,1” - 20,250 x 10

Letting, I = moments of inertia of the R/C sections about the

Y1c
Y - Y axis:

115
r "2"
=,u[1uu 000 + 1.86050. 5 x 12):) + 1+ 9914.000 + 1.5009 5 x 12)_)

110°
+ n [1014, 000 + 1480(32. 5 x 12)?! + IL [11114, 000 + 080(21. 5 x 1273
+ 9 [1121.000 + 1180(1h.5 x 1331+ h [1119, 000 4- 1480(3. 5 x 12) (
+ 161:2560 + 1480(511 x 12H = 348,511+ x 10‘3

seine = 9703 x 1012 then, (81)“ = Rely + sexylc = 300 x 10“

”Dye = II of the average section taken through the equivalent

floor slab. I = moment of inertia of the equivalent floor slab

Vac

section about the Y - Y axis.

I = 1/12 x 72. 67 x 12(108. 67 x 12)}: 1611 x 108

22° )4
__ 1
then, Iclyac- 3222 x 10 7+
. _ 1
and: (RD’a Rely. + Ec1y2c=31425 x 10

The average value of (BI)y will be:

(RI) 532. 6(Bl)y1+ 31. Mm)ya + 62615.3,

 

=532 6 3‘ 300 x 1011‘ + 31 ‘1 x 3325 x 101“ + 6 x 3222 x 101“
570

= 501! x 1011‘

“W_---

 

__

12
\IwL I.i 000 x 2189 x 10 _1.50 1.35 = 1.7%
= 21. 50
T c llyg J 5014x10‘ 1386

During a destructive earthquake of period Tp = 1.50 seconds:

1/2
'1' 1+ T
Tp 1.50 1.16 and(,rp) 1.08

 

From previous computations T/Tl = 5.8 when G = 1.50;

. 1+
First harmonic, T1 = 1.5-18.: 0.30 second.

 

 

146

Transversg deflectioni, moments, and shears during forced vibration

—-‘.--

Interpolating roughly between TIT; = 1.1 and r/rp = 1.5 in

FigurQCl) it is found:

 

 

0 2h Mu “h
0.0 1.00 -6.u5 3.90
0.2 0.55 J4.85 l41.25
0J1 —o.'70 -2.20 1+.ZO
Oe6 -2.40 “1065 3065
0.8 41.115 -0.50 2.20
1.0 -6.55 0

(7,.“

7!
m'L = 1.88\'§%= 1.88 x 1.08 = 2.03,

8 12

, and, m'3 = 26.2 x 10' .
6

m' = &%= 2.97 x 10"“, m‘2 = 8.81 x 10"

It x 8.81 x 10"8 x 0.12 = 5311 x 10

12

(nI)ym'3r = 50h x 101
(31),.m'3! = 5015 x 10114 x 26.2 x 10' x 0.12 = 158 x 103

The actual deflections, moments, and shears are:

 

‘— m v ‘—~_—_ fl VV ‘_~ “

 

 

 

2 3 h 5 6 7
c y, 1n. Y. in. M. ifl‘lb v, lbe v. . 1b. u. . 1&1b0
. 6 3 3 5
. 0.12 0.00 -3uh0x10 616210 1232210 70ux10

00
20 0.07 0.05 -2590x10g 672x103 13uhx103 766x105
no -0.08 0.20 -1172x106 66ux103 1328x103 756x105
.60 -0.29 0.u1 - 8811106 577x103 115hx10 658x105
80 -0.53 0.65 - 267x10 3h8x103 6961103 397x105
00 -0.79 0.91 0 0 0 0

A

C = X/L. y = My, Y = F - y, M = (Inym'al‘ln, V = (EI)ym'3rvu,
v7 = 2v, u' = 57v'.

 

1*7

From the column 3 the maximum amplitude is at the tap of the

building and within the allowable amplitude of 0.001L = 6.8% inches.

By column 6 the maximum.shear and.moment at any set of vertical beams

at approximately 0.2 of height and equals 1,34h,000 pounds and

76,600,000 inch-pounds respectively.

Distribution g£_§gismic shear and moment:‘

 

Investigating the average section it is found:

 

 

 

 

 

 

St 1 I
30:23 W n _ ‘ ry.‘ ry: 1y. ‘2
sa u 77.63 n.1u 17.1 1331.2 10.23
ab 16 116.98 n.31 18.6 2169.7 6.30
so 15 100.59 n.2u 18.0 1806.9 7.57
2

va

6.

c1 = wig-3 = 0.62, 02 = {2353: 0.71;

Assuming p = 0.007, then, K' = 0.097, then:

—;
.: _

 

 

 

 

R c 2
Balm n _".b_ dfi A 1y, IK' ry, 0
ca 214 8 58 1+6!» 129800 "12600 280 1.8311
ch 16 60 6 360 1080 105 3 129.7
(1,1,2 4‘ 361‘ .2) .2& g Ll'6 = , )4
G = Iy'K' " °3 ‘ 32. 0-75 ° “9

cl} = 36.1: 2H9.“ = 8.13
323 30075

 

'4.“ Q I.‘ v .&

I‘h‘x‘.‘ _ -.‘Lh"‘

in

‘1‘. .

 

519,0. =8580 1b.,

Psa 253.0 P311 =

rec = 0??) = 7220 1b., Po, .-. gin—9T: 1114,7500 lb.

rcb - 8.3 = 657 1b.,

MS, a: %= 303,000 111.16., 11,1, = 39g??- = 1489.000 in-lb.,
-3%L%0_Q= #10. 000 in- -Mlb., a- -0011300 = 2.539.000 1mm

11 ._ 121.929.: 37 300 in-lb.
cb 8.13

Shearing, tensional and compressive stresses:
Assuming, J = 7/8 and k = 3/8, and taking beam ca, most rigid to

the plane of bending, we have:

,,_. Pa: 1614.700

.7132? 778x8x58'

= 110.0 lb. p.s.i.

8 Ma. 2,539,000 3
I! ma: 0.007 X 7 8 X 8 X 582 15,380 lb. p.s.i.

=W— 2 x 25 =
kad fimm 575 1b. p.s.i.

The steel area required is A. = pbd = 0.007 x 8 x 58 = 3.25 sq. in.

Using three 1-inch rods and one l-inch bar, sectional area = 3.26,

an ,_. 1:11.700 8 .
o x 13,113, The u = 33-59—9- 13.153 x 7/8 x 58 65.5 lb. p.s.i.

For a 2000 pound concrete, u = 100 1b. p.s.i., fs = 18,000 lb. p.s.i.

 

rc = 750 1b. p.s.i., and v = 180 lb. p.s.i..

l1‘he building is strong about the Y - Y axis.

[hub—.- ....‘_..__. +1

‘
~ru~a

ASEISMIC DESIGN OF A SHORT BUILDING

A lZ-story earthquake-proof warehouse is to be built with dimen-
sions 75 feet - u inches by 111 feet - h inches.

Story height = 12 feet - 0 inch and 2 feet - 6 inches from tap floor
slab to the bottom of spandrel beam.

The theoretical floor slab depth is 12 inches over the gross area
of the plane. Linea] extent of theoretical floor slab ==(1 + 12)l.0 =
13 feet.

Assume foundation slab thickness =‘6 feet.

Lineal extent of vertical beams = 162 - l3 - 6'= lh3 feet.

Ratio L/d is:

1_._.__6-? - 122.:
$087 1.5<5 and 27 2.2(5

Building is classed as short.
Investigating the site is found:
a) Locality is subject to considerable seismic activity.
b) There are no rifts or faults in the immediate neighborhood.
The nearest fault being 20 miles away.
c) Dominant period of the locality = 0.3 second.
d) Soil from.dry, springy clay-andpsand mixture.
e) No ground water.
f) No adjoining buildings.
A three-foot layer of broken stone will be interposed between the

foundation and the soil. No special steps to enable the building to

slide freely.

50
It is decided that the end conditions being expressed by a value
of C = 1.00.
Assume maximum surface amplitude excluding the case of synchronism
= 0.50 inch. _
Amplitude at the bottom of the broken stone = 0.‘+9 inch$1 F F
Amplitude transmitted to the foundation, F = 1/2 x 0J8 = 0.25 inch.

Frame consists of steel skeleton with B/C roof, floor, and founda.

 

tion slabs.

IP"--‘§-‘ -.-~"’ a“: - -. -

._ 7 _. 6
xii—3:10, lc-2x10

 

 

Beam Section A II. 11' Ix. r1.
sa 1ux16x1u2 IF 111.85 1672.2 660.1 6.32 3.97
ab 1uxi6x176 I]? 51.73 21149.6 837.9 6.115 11.02
so 111x16x158 '1' 116.117 1900.6 715.0 6.110 1.1.00
ca R/O, 60"x8' 1480.0 2560.0 11m,000.0 2.31 17.32
cb a/c, 8"x60" 180.0 11110000 2,560.0 17.32 2.31

 

Analysis About the L ;_ Axis:

 

Ix = 11' + Ad:2 (31)“ = II of the average section taken above
the floor.

Ixs = moment of inertia of the steelsections
about the X - X axis.

In = u [1672.2 + M18506 x 132] + 10 [2119.6 + 51.7306 x 1202]
+ u [2119.6 + 51.73(18 x 12?] + 2(21u9.6)
+ 10 [1900.6 + u6.l+7(18 x 12?] + 5(1900.6) = 1591 x 10‘3

1.1,, = 14773 x 1012

111° = moment of inertia of the R/C section about X - X axis.

 

1‘ISee Appendix B.

51
lec = 2L1 [2560 + 1480(36 x 1212;) + 1+ [1414000 + l48002-5 x 132]
+ 1.1 [11111000 + u80(21,5 x 12)2] “I“ ’4 [11914000 + 1+5>"O(1‘+._‘3 x 12)2]
+ 1+ [whooo + 14800.5 x 12)2]= 26.638 x 105

 

Eclxlc = 5288 x 101‘2 "+7..
_ __ 11: . F .
(81),,1 - 8,1“ + Eclxlc - 101 x 10 i a
Let:
(ll)x2 = EI of the average section taken through the equivalent
floor slab. 5LFF
Ixec é moment of inertia of the equivalent floor slab section about
the X - X axis,
then:
3 8
1x20 = 1/12 x 108.67 x 12(72.67 x 12) = 721 x 10
.. 1h
__ __ 11+
but: (RI)x2 - Ests + ncIIZC - 1N90 x 10

The average value of (RI)x will be:

 

 

 

 

1113011) + 13(EI) + 62 I
(31), = ‘1 ‘2 ° ’2“ = 262 x 1011*
162
Transverse period about t_h__e_ 5 -_-_ _x_ Axis:
E 10
T = c !L—-- = 1.00 £5000 E—lfiea x410 = 0.19 second.
x (Enxg 262 x 101 x 386

Applying correction formula for short buildings:

2
Tn = x2 (l + d 2 = 0.21 second, (true fundamean period
J “01L about X - X axis)

Assume C1 = 0.26, L2 = 378 x 10", d2 = 76 x 10"

 

Assuming that Tp = 1.00 second.,

52
1/2
T T -
.82.: 0.21, 53!. = 0.06
Tp p
Transverse deflections, moments, and shears during forced vibration

about the X - X axis. End condition - Fixed-free.

 

 

0“ v

 

 

0 2h nu vu

0 0 1.00 0.56 -0.6u

0 2 1.01 0.31 -0.53

0 u 1.0% 0.18 -0.u2

0 6 1.10 0.09 -0.29

0 8 1.16 0.03 -0.15

1.0 1.21 0.00 0.00
m'L = 1.88 x 0.16 = 0.86, m' = 3.2%: 14.14 x 10"h
m'2 = 19.6 x 10'8, m'3 = 86.6 x 10"12

8x0.25=128x107

12 x0.25= 57x10u

(E1)xmfiar = 262 x 101““: 19.6 x 10’

11:

(31)xmfl3r = 262 x 10 x 86.6 x 10'

The actual deflections, moments and shears are:

 

r 1
fi 1

 

 

 

C y, in. Y, in. II, in-lb V, 1b. 7', lb I“, in-lb.
0.0 0.25 0.00 589 x 10g» -365 x 103 730 x 103 #16 x 1056'
0.2 0.25 0.00 397 x 106 -302 x 103 60h x 103 3th x 105
0.h 0.26 0.01 231 x 10 -2u0 x 103 #80 x 10 27h x 105
0.6 0.27 0.02 115 x 102 -165 x 103 330 x 103 188 x 105
0.8 0.28 0.03 38 x 10 - 87 x 103 17k 1 103 99 x 105
1.0 0.30 0.05 0 0 0 0,

Maximum Y = 0.05 at the tap of the building, this is the moment
deflection.

Shear deflection:
2
ya = ym (fig—p); y. = 0.01 inch. y. = shear deflection,
2 ym = momfint deflection,
Y=0.05, d =76x10 , 01=0.26

L2 = 378 x 10“

 

 

53
Maximum deflection at tap of building = (ym + ya) = 0.06 inch.

Maximum shear at the bottom = 730,000 pounds.

Maximum moment taken by any set of vertical beams = l$1,600,000 in-lb.

Distribution of seismic shear and moment:

 

 

 

2:23: n L rx. r162 Ix. 2
as 1+ 141.85 6.3; 110.0 1,672.2 3.611
91: .16 51.73 6. 141.5 2,119.6 6.7!:
sc 15 146.147 6.90 111.0 1,900.6 7.61

 

z = (1.1,? + 36rx621
Ix'

01:85-fgfaoqs 02=§fgzlp=o.88

Assuming p = 0.007, K' = 0.097

 

Lb = 9.5 feet = 1141+ inches, Lba = 12.996

 

 

‘-__.

 

' 2

3343s :1 b d ‘ ‘ 9x' 11' r1. G
ca 21+ 60 6 360 1,080 105 3 129.7
66 16 8 58 1161+ 129,800 12, 600 280 1.8314

 

G = LLbZ + 36rx02)

IK'
03=§%?-§-g-=9.6h cu=%§g§=o.|h2

0 000 - -M_6__60 , ,
p83 a %fi-- 6660 lb., P.‘ 0.78 5980 1b

_h660 g 11660-2.
pee =0.88 5300111.,1=°,,=9—-3II 11811111.,

 

= 32,800 16.

98-:
:3;

51:

111600000 , 265000
.._._____._. = 265,000 imlb., “36 = 0.73 = 31+0,000 16.16.

Msa 156.9

26 000 _ _ 26 000 _
Mac = fig— 302,000 111-11)., 14% - "876!!- 27,500 in-lb.

= 265000 =
ch 0.1172 1,868,000 16.16.

Shearing, tensional, and gpmpressive stresses:

Investigating the beam most rigid in the plane of bending, cb,
we have:
Assume J = 7/8, and k = 3/8.

365-7f8x8x58 p.s. °

 

“ch 18 63000

fs”:I‘D-,j—6-(1'2=0.007x7/8x8x33 =11'320 p°8°1'

 

zucb 3736000

1'8 .. ..
c m-78x38x8x33 -—1§25p.s.i.

The building is strong about the X - X axis.

Analysis About The {:1 Axis:

From the theory of the moments of inertia of plane areas:

Iy == Iy' + my?
Letting (Iny. = II of the average section taken above the floor.
Iys = moment of inertia of the steel sections about the
Y - Y axis.

then:
I” = 1: [660.1 + h1.85(51+ x 12)? 4' 5 [837-9 + 51-73(5‘4 I 132]
+ 1: [837.9 4- 51.7356 x 12)-.5] + 11 [837.9 + 51.73(18 x 12)?

2
+ 2(837.9) + 3W5) + 6 7'6 + 116.1706 x 1211
+ 6 [1% + h6.47(18 x 12)2 = 3155 x 105

55

12

and E81 = 9h05 x 10

ys
Letting: Iylc = moments of inertia of the R/C sections about the Y-Y

axis then:

Inc = 1+ 11.19000 + 1480(505 x 12%] + 14131114000 + 48009-5 x 132]

+ Ll» 1'411000 + “80(32-5 x 12)fl + F4 [IF-“$000 ‘7' F480(21.5 X 12)?

+6Emmo+mmm5x1a§+h wwm+um0511afl

+ 16 2560 + uso(5h x 12)2

= “8,51h x 105
12
and: Eclylc = 9703 x 10
- _ 12 12 _ 1n
(1H),,1 - 88in + Berna — 9465 x 10 + 9703 x 10 - 192 x 10

Letting: (E1)ya = E1 of the average section taken through the equivalent

floor slab; I = moment of inertia of the equivalent floor slab section

Vac

about the Y - Y axis.

Then:
3 8
Iyzc = 1/12 x 72.67 x 12(108.67 x 12) = 1611 x 10
and° E I = 3222 x 101“
' 0 yes 7 ' .
. _ _ 12 16._
and. (81)” — 1131" + 3015,26 - 91465 x 10 + 3222 x 10 -
3317 x 101“

The average value of (BI)y will be:

56

=y.1l+3(EI) + 13(81)ye + 6801yac

(EI)y
162

1”’+ 13 x 3317 x 101“ + 6 x 3222 x 101“

162 “‘A

1M3 x 192 x 10

v“ 2“

1k

556 x 10

 

*‘I‘.‘-.

3 . 10~
'3 - 25 x 10 x 1h28 _x 10 _
= Tyév[;;I;—— 1.0 0J4 556 x 101M x 386 — 0.13 second.

Applying the correction formula for short buildings, we have:

[Mun .u-a‘

 

 

 

.... J ~----.o -'

F 2 \d - 2
\7 d _ 2 (108.67 x 12)
Tyt ‘5’ TV (1 “011:2. ) ( 1+ x 0.26 x 378 x 101;

= 0.16 second.

 

T is the true fundamental free period about the Y - Y axis.

Yt’
Assuming T = 1.00 second- :ll = 93—1-5- = 0.16
I p s 0.1 9

P

 

‘l imv "We: ‘1]?
.F

V-
f

57

Transveggg deflections, moments, and shears during forced vibration

 

about _t_h_e_ Y - Y axis.

Interpolating for a ratio T/Tp = 0.16 in Figure 1 it is found:

 

 

 

 

c {22: u v
0.0 1.00 0.35 -0.u8
0 2 1.01 0.23 -0.ho
0 u 1.0M 0.19 -0.32
0 6 1.08 0.07 -0.22
0 8 1.12 0.02 -0.11
1.0 1.16 0 0
I -— o - ‘1‘
mt=1881mm-oqmm ==zc>.fiE-3.9x1o
19
m"2 = 15.2 x 10'8, m'3 = 59.3 x 10.|2
-8

(11),.nap.= 556 x 101" x 15.2 x 10 x 0.25 = 212 x 107

(:1),m:3r = 556 x 101“ x 59.3 x 10 3

x 0.25 = 825 x 10

The actual deflections, moments, and shears are:

 

 

 

 

:f 1'? 3 ‘ R ‘3 6 7
0 y, in. Y, in. M, in—lb. V, 1b. V', lb 1", in-lb.
0.0 0.25 0.00 783 x 10%» -396 x 103 792 x 103 h52 x 105
0.2 0.25 0.00 h88 x 10 -330 x 103 660 x 103 376 x 10
0.6 0.26 0.01 227 x 10? .26M x 103 528 x 103 301 x 10
0.6 0.27 0.02 1 8 x 106 -182 x 10 36h x 103 208 x 105
0.8 0.28 0.0 1:2 x 10 - 91 x 103 182 x 103 10h x 105
1.0 0.29 0. o 0 0 0

 

The formulas used are the same as for the analysis about the X - x
axis. The maximum Y = 0.011 inch. This is the moment deflection.

The corresponding shear deflection is:

.9.“ 11 x 10 , g: .

 

58
The total maximum deflection = ym + y. = 0.034 + 0.02 = 0.06 inch.
The maximum shear occurs at the bottom.and.= 792,000 pounds. The
maximum.moment taken by any set of vertical beams = ”5,200,000 inch-

Pounds.

Distribution_gf the seismic shear and moment:

 

Investigating the average section as previously, it follows:

 

 

513001 2

 

 

 

 

 

 

Beams n A ry, ry, 1y' 2
sa 6 u1.85 3.97 15.7 660.1 21.3
sb 16 51.73 n.02 16.2 837.9 16.3
sc 15 h6.u7 6.00 16.0 795.0 18.3

-(Lb2+3br'2) ,_ -1681-
- -—-—---I-~y~r'-—-L-—- . Lb - 1299b . cl - 21.3 - 0°77.
02= ;-1-= 0.86; assuming that p = 0. 007, K' =0.097, and:
R/G 2
Bean. n 6 d A 1y. EE:_ 2y, 0
ca 2% 8 58 u6u 129,800 12,600 280 1.83u
66 16 60 6 360 1,080 105 3 12u.7

 

G = (Lb? + 361332)
IK'

 

.. - __‘fis.’i
03-33%—=0.057u, Cu" b3:.9=391

2 000
and: Pea = b . = 1705 pounds, P“ =%l%;”= 2210 pounds,

1 O _ — 1 =
2,, = 668% — 1990 pounds, 26, - 0 057 29.800 pounds.

59

3131-91:
Pcb 3.91 M37 pounds.

Also: 11 — W = 97,500 in—lb., u = Qg%$-= 126,500 in—lb.,

sa ’ 46M.h sb
“so - 2%?53. 113,500 in—lb., Mc‘ 3;357E7* 1,700,000 in-lb.

-915.O_Q.=.
and “ob - 3.91 24,900 inplb.

Shearing, tensional, and compressive stresses:
Investigating the beam most rigid in the plane of bending, ca, is
found:

Assume: j = 7/8, and k = 378, then:

P 2 8
v = --°-§-= fl-gL-g-Qg-x—ggr- 73.5 lb. p.s.i.

- u - 170mm = b. . .1.
f! - Sfizgh' 0.007 x 7/3 x 8 x 582 10'350 1 p s

Ls-

 

31.100000 1

c m=7/8x3/8 xgrx‘S—sa 33386113. ps.eio

The building is strong about the Y - Y axis.

APPENDIX

5 APPENDIX A

 

 

Dr: _D d. c S
I l 7 By Hooke's law the shear strain
I I
II { 4) "’ n. (1) 0| g i: (2)
|¢ l
_"1 l 0 = shear strain.

I I

54 B S = shear stress.

A
J Cube ‘m shear
s

E. = shear modulus.

For all practical purposes the shear modulus Es = 2/5 E (3)

E = Young‘s modulus.
Deflection Due to Shear

Beam subjected to a shear V. Let:

S. = unit shearing stress at distance v

 

d I = moment of inertia of the section
0.5

dd = area of an element.

 

 

 

 

0.5d From the theory of beams:

 

 

 

 

s. ”Ea-fl“ (4) (M = bdv limits
Beam in shear
d/2

d/2 and V
V
S. = Ij vdv or

V

V
s.=a<¥—u3)(m

By equation (3) and that U'i, the average internal work of shear
per unit, equals one-half of the product of the unit stress 3. by the

unit strain we have: S 2

 

, .
p
'97)"! .vy
—.r
I

63
and from equations (u) and (5) We have: v2(dy;8d?v2«+1bvu)
w' =
1 128 12‘. .

 

Multiplying by an element of volume bdvdx, and integrating with respect

to v, with.upper and lower limits d/Z and —d/2, the total internal

work of shear. W1 1:: w = $de ’
1 2110121.“,

f3
F

but for the section: I = bd3/12 and v2 = p2 so:

= 31:21.

w —-—— .
i 5de. 3,.

 

W , the external work done by P in deflecting the beam through the

maximum shear deflection y. is:

H.551:
2
P 2
but. I, ='.. :0. 1.2.: 3.11.11—
2 5des

- .. -111
but, bd—A, andE.—2/53, so, y.- 113°

APPENDIX B

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Loo [ 1 i T .
<19 - E- _1_-__ l i __ l 7 __ a” -
a i . ‘ I
'0 4 ‘
o I 1 1
E u 008,—? “I“ _- ___ ~ _ ,1 _u_+__ f Jr
‘5 ‘° ‘ ‘ : i
a
O z; i l
3 :~ .7 -9 .1- - .82....
.2 ° I ,
5 3 06 ~ ‘
w .9. “My“
' s. I .
‘3 ,3 7 l
m 0.5 , . :2
u l 1‘ .
>‘ L I 1 I . .
O4 __ a - _-_- ' *_--.L__..___+_._.-.. _ __ -‘.- +._..__.l_
I i 1 ‘ \
<13 1 I ' l k L
0 100 200 300 400 500

EFFECT OF DEPTH ON AMPLITUDE12

Excluding the case of synchronism, the maximum.smplitude of a sur-
face particle during a destructive earthquake is seldom greater than:
0.25 inch for a particle in bedrock, 1.00 inch in dry, cohesive soil,

and n.00 inches in marsh or soft, wet fill.

 

12 L.S. Jacobsen, "Motion of a Soil Subjected to a Simple Harmonic Ground
Vibration," Bull. Seis. Soc. Amer., Sept. 1930.

J. Milne, "Effect of Depth 0n Amplitude," Trans. Seis. Soc. Japan, Vol. 10.

F. Omori and S. Sekiya, "Effect of Depth on Amplitude! Jour. Coll.
§21_., Imp. Univ. Tolq'o, v61. 14.

Cresknff, 22: 913,

h. .u- -a._-..

£4 .
"-'l'

BIBLIOGRAH-IY

A.C.I., “Reinforced Concrete Design Handbook.”
A.I.S.C., Steel Constructipn, New York: 19148.

Creskoff, J. J., mnamics of Earthguake Resistant Structures, New York:
McGraw-Hill Book Co., Inc.

Fleming, B., Wind Stresses EBuildigg s, London: Chapman and Hall;
John Wiley and Sons, Inc.

Freeman, J. R., Earthquake Damage and Earthguake Insurance, New York
and London: McGraw-Hill Book 00., Inc.

Grinter, L. 3., Theo‘gzgiuodern Steel; Structures, New York: Macmillan
Company.

Jacobsen, L. 8., ”Motion of a Soil Subjected to a Simple Harmonic
Ground. VibfltiOh.” Mlle SQiSe SOCe mere. Sept. 19300

Macelwane, J. B., "The Interior of the Earth,” Bull. Seis. Soc. Amer.,
June, 192”.

Milne, J., "Effect of Depth on Amplitude,“ Trans. Seis. Soc. 1%
Vol. 10.

 

Naito, T., Earthluake Resistirfi Constructi_o.r_1_.

Omori, F. and S. Sekiya, "Effect of Depth on Amplitude," Jour. Coll.
soy, Imp. Univ. Tokyo, Vol. 1:.

Omori, F., "Imperial Earthquake Investigation Comittee,” Publication
20, p. 73; Jour. Inst. £9211.” Architects, Vol. ’41, p. 1:98.

P.C.A., Analysis 21; Small Reinforced Concrete Buildings £93.; Earthguake
Fozces.

P.C.A., Continuity in Concrete Building Frames.
Peabody, D., The Design 21; Reinforced Concrete Structures.

Porush, l, ”Earthquake Stresses in Rigid Building Frames,“ Bull.
Seis. 5253, Amer., January, 1933.

Prescott, J, Applied Elasticity.

Reid, H. B., Report at; £133 State Earthquake Investigation Com. _i_‘
California, V01. 2.

4- 9": i

‘2

‘ ( :n.——u.-~.- v-. v
E.
n.1,,

66

Suyehiro, K., "Engineering Seismology," Proc. gar. Soc. Civi_1_ £35.,
May 1932; Bull. _I_r_13p_. Earthguake Research Inst. Japan, V01. , p. 63

Tersaghi, x. and n. s. Peck, SoilMechanics in Engineering Practice,
London: John liley and Sons.

Timoshenko, 8., Applied. Elasticity.
Timoshenko, 8., Vibration Problems i_n_ Emineerigg.

———‘—‘fl‘"‘_'_

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nuis-

 

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4" ”A:

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