The influence of blast and earth pressure loadings …Calhoun: The NPS Institutional Archive Theses...

Calhoun: The NPS Institutional Archive

Theses and Dissertations Thesis Collection

1960

The influence of blast and earth pressure loadings

on the dynamic response of flexible underground

two-hinged arches.

Walls, Worthen Allen

University of Illinois

http://hdl.handle.net/10945/12331

DUDLEY KNOX LIBRARY

NAVAL POSTGRADUATE SCHOOI

MONTEREY CA 93943-5101

THE INFLUENCE OF BLAST AND EARTH PRESSURELOADINGS ON THE DYNAMIC RESPONSE OFFLEXIBLE UNDERGROUND TWO-HINGED

ARCHES

BY

WORTHEN ALLEN WALLSB.S., University of Arkansas, 1948M.S., University of Illinois, 1954

THESISSUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS

FOR THE DEGREE OF DOCTOR OF PHILOSOPHY IN CIVIL ENGINEERING

IN THE GRADUATE COLLEGE OF THE

UNIVERSITY OF ILLINOIS, 1960

URBANA, ILLINOIS

DUDLEY KNOX LIBRARY

NAVAL POSTGRADUATE SCHOOl

MONTEREY CA 93943-5101

ACKNOWLEDGMENTS

This thesis was undertaken under a postgraduate educational

program sponsored by the United States Navy for Civil Engineer Corps

Officers. The author expresses his appreciation to the United States

Navy and to the Bureau of Yards and Docks, United States Navy, for the

opportunity of participating in this program.

Appreciation for timely personal interest and encouragement

is expressed to Dr. N. M. Nevmark, Head, Department of Civil Engineering,

University of Illinois, under whose immediate supervision this thesis

was prepared.

The author is deeply indebted to Dr. J. D. Haltiwanger,

Professor of Civil Engineering, for constructive criticism throughout

the preparation of the thesis, and to Dr. J. L. Merritt, Assistant

Professor of Civil Engineering, for frequent exchanges of ideas.

iv

TABLE OF CONTENTS

Page

CHAPTER I. INTRODUCTION 1

1.1 General Statement of the Problem 1

1.2 Purpose of this Thesis 1

1*3 Method of Approach 4

CHAPTER II. GENERAL DISCUSSION OF THE VARIABLES 7

2.1 Blast Pressure Loading 72.2 Passive Ecjrth Pressure Loading 8

2.3 Structures Considered 9

CHAPTER III. GENERAL RELATIONS COMMON TO EACH LOADING FUNCTION ... 11

3.1 Stress Transformation with Direction 12

3.2 Distributed Load Functions 13

3.3 Basis for the Determination of Analogous Loads 153.3 •! General Equations for the Right Reactions of

Circular-Segment Arches 16

3.3»2 Moment Equations for 180 Degree Arches 18

3.3»3 Moment Equations for 120 Degree Arches 213.3*^ Procedure in Determining Analogous Loads 23

CHAPTER IV. BLAST PRESSURE LOADING 33

4.1 Pressure-Time Relations at a Point 334.2 Harmonic Analysis of Radial Blast Load with Time a Constant 374.3 Blast Loading Approximations 40

4.3.1 180 Degree Arches 434.3.2 120 Degree Arches 48

4.3.3 Summary 52

CHAPTER V. PASSIVE EARTH PRESSURE LOADING 6l

5.1 Unit Passive Earth Pressure 6l5.1.1 180 Degree Arches 625.1.2 120 Degree Arches 63

5.2 Distributed Earth Pressure Load 64

5.3 Analogous Earth Pressure Load 66

CHAPTER VI. ANALYSIS OF THE RESPONSE OF THE ARCH TO THEDYNAMIC LOADING 70

6.1 Method of Analysis 706.2 Sample Calculation 736.3 Discussion of Results 75

CHAPTER VII. QUALITATIVE DESIGN METHOD FOR ARCH RIBS 82

TABLE OF CONTENTS (Continued)

Page

CHAPTER VIII. CONCLUSIONS AND RECOMMENDATIONS 85

8.1 Conclusions 858.2 Recommendations 87

APPENDIX A. BIBLIOGRAPHY 08

APPENDIX B. NUMERICAL CALCULATION RESULTS 89

VITA 125

CHAPTER I

INTRODUCTION

1.1 General Statement of the Problem

When a nuclear weapon explodes, a high pressure wave develops

and moves away^from the point of the explosion, traveling over the ground

surface and through the ground. A shallov-buried arch subjected to this

pressure responds in various ways, depending on the orientation of the

structure relative to the direction of travel of the blast wave, the parame-

ters of the blast loading, and the properties of the soil surrounding the

arch. If the arch is oriented with its longitudinal axis perpendicular to

the direction of travel of the blast wave, the windward side of the arch

is subjected to the pressure a few milliseconds before the leeward side.

This unsymmetrical sequence of loading produces flexural stresses and de-

flection in the arch. The deflection of the arch into the surrounding soil

gives rise to an unsymmetrical passive earth pressure loading.

An understanding of how these loadings occur and their effects on

the arch is necessary for the proper consideration of the design of the

arch rib.

1.2 Purpose of this Thesis

When the arch rib is subjected to the above described loads it

responds in a configuration which is a function of an infinite number of

shapes or modes of vibration, since the arch rib has distributed mass and

elasticity. An infinitely large number of coordinates would be required

to specify the position of the deflected arch. The participation of each

mode depends on the spatial distribution of the loads along the arch rib

2

as veil as the time variation of the dynamic load in relation to the natural

frequencies of vibration of each mode.

The distortion of the arch rib is a function of the stiffness of

the rib, the passive earth pressure loading resulting from the stiffness of

the soil surrounding the arch, the distribution of the soil and dynamic

loads, the time variation of the dynamic load, and the support condition of

the abutments. It is convenient to consider the earth pressure loading due

to the stiffness of the soil as an integral part of the resistance of the

arch. For flexible arches, the stiffness of the soil is greatly in excess

of the stiffness of the arch rib, consequently the properties of the arch

rib may be neglected. Henceforth, unless specifically designated otherwise,

the resistance of the arch rib means the resistance given to the arch by

the soil.

Design of structures for dynamic loading is usually based on the

principle of conservation of energy. When the dynamic load acts on the

structural element, the element deflects. Work is done on the structure by

the load at any point as that point deflects. Due to the speed of the load-

ing, kinetic energy of motion is given to the mass of the element, and, as

the element deflects, strain energy is stored. The strain energy may involve

both elastic and plastic action. A general statement of the principle of

conservation of energy is:

work done * kinetic energy + strain energy

To evaluate this equation the displacement of any point on the

structure under the load must be known. The strain energy is the sum of

all the strain energy in t he element, and may be caused by compressive,

flexural, torsional and shear stresses in the element. The kinetic energy

is the energy of translation or rotation of all the masses in the element.

The evaluation of these quantities would be exceedingly difficult

if no more than the arch rib were involved. Evaluation of the energy-

stored in the soil is a hopeless task, impossible for practical purposes.

Fortunately, the principal modes of response of the arch rib are a sym-

metrical mode corresponding to the compression stresses and an anti symmetrical

mode corresponding to the flexural stresses. This allows a considerable

simplification to be made in the treatment of the problem.

A single-degree-of-freedom system consists of a lumped mass

restrained by a veightless spring. When this system is subjected to a load,

only one coordinate is required to fully define the position of the mass.

By assuming that the antisymmetrical mode of response of the arch rib is

defined by the radial deflection of a point at the quarterpoint of the arch,

which is the point of maximum deflection of the arch, the entire deflected

shape of the arch may be defined with one coordinate, and the arch may be

considered a single-degree-of-freedom system. In essence this neglects any

tangential movements of the arch rib associated with the radial deflection.

After representing the action of the arch rib as that of a single-

degree-of-freedom system, dynamic equivalence relations should be established

between the actual and idealized masses, spring stiffnesses, and loads acting

on the two systems. However, for the purposes of this study, this refinement

is omitted. Instead, the following assumptions are made:

(1) The mass of the single-degree-of-freedom system is equal to

the unit mass of the arch, which is obtained by uniformly distributing along

the arch rib the weight of the soil which creates the passive earth pressure.

(2) The resistance function of the single-degree-of-freedom

system is equal to the unit resistance given the arch by the earth pressure.

(5) The forcing function on the single-degree-of-freedom system

is equal to the unit pressure of the forcing function acting on the arch rib.

Once the mass, resistance function, and forcing function acting

on the single-degree-of-freedom system have been defined, the response of

the system may be determined by the vell-knovn differential equation of

motion of such *a system:

mx + kx = p(t)

vhere k* = the second derivative of the displacement vith respectto time

kx = the resistance of the system as a function of thedisplacement

m = mass of the system

p(t) = the externally applied force as a function of time

The main purpose of this thesis is to investigate the response of

various arch ribs under a limited range of variables -when acted upon by blast

pressure. In order to do this, it is necessary to relate the parameters of

the single-degree-of-freedom system to those of the actual structure under

investigation

.

1.5 Method of Approach

The blast pressure acting on the surface of the ground causes

vertical and horizontal pressures in the soil. These pressures are a

function of the maximum surface pressure, and are time-dependent upon a

pressure wave vhich passes through the soil. Pressure-time relations of

these pressures are formulated.

These pressures are assumed to be principal stresses, the resultant

of vhich acts radially on the arch. The tangential component is omitted.

The total radial load vhich results from these pressures is designated as

a load profile p (<P, t), which is then represented as a distributed load by

the series ) p (t) sin ~-

where pn(t) = iL

f9pr

(cp,t) sin H|-2 Rd 9

lo£ central angle of the arch

R = arch radius

An analogous, uniformly-distributed, antisymmetrical, time-

dependent load, p (t), is then determined. This load has the same generala

r1

effect on the entire haunch of the arch as the load / p (t) sin _ .

The pressure-time diagram for p (t) and the numerical relation between p (t)

and the dynamic load \ p (t) sin are established by equating their

effects at various times as though they both acted statically. The analogous

antisymmetrical load po(t) defines the forcing function acting on the arch

and also on the single-degree-of-freedom system.

The unit, horizontal, passive earth pressure acting on the arch

is determined by Rankine's theory for passive earth pressure. The unit

vertical pressure acting on the arch is found by statics. These two pressures

are assumed to be principal stresses, and their radial resultant is assumed

to resist the movement of the arch. The tangential component is omitted.

The total radial soil load is designated as a load profile p (<p), and is

represented as a distributed load by the series \ p^ sin '

vhere pn= TrJ pr^ 9)

sin ^T2 Rd 9

6

o

An analogous, uniformly-distributed, antisymmetrical load which

has the same general effect on the entire haunch of the arch as the load

Ip sin is determined by statics. The passive earth pressure is

assumed to reach its maximum value when a point located at d/k on the arch

rib deflects radially an amount equal to 0.001B. In addition, the resistance

diagram of this pressure is assumed to be elasto -plastic . The analogous

antisymmetrical load defines the unit resistance of the arch rib and also

the resistance function for the single-degree-of-freedom system.

The weight of the soil that causes the passive earth pressure is

determined by Rankine's theory and by statics. This weight is assumed to

be uniformly distributed along the arch rib and defines the unit mass of the

arch rib and also the mass of the single-degree-of-freedom system.

The response of the equivalent single-degree-of-freedom system is

determined by a method developed by Dr. N. M. Newmark for blast resistant

design, Reference ( 8)

.

CHAPTER II

GENERAL DISCUSSION OF THE VARIABLES

The variables of the problem may be divided into three general

groups corresponding to the blast loading, the earth pressure loading, and

the configuration of the arch rib.

2.1 Blast Pressure Loading

The blast forces acting on the arch rib are conceived to act as a

pressure vave passing through the soil, Figure 2.1. The direction and

velocity of the propagation of this wave in the soil is assumed to be related

to the shock velocity of the surface pressure wave, U, and to the seismic

velocity of the soil surrounding the arch, C . The magnitude of C iss s

assumed to be less than U.

Within the stressed-soil zone, the vertical pressure p (t) is

assumed to be equal to the surface overpressure P (t), but to have a dif-s

ferent pressure-time curve to allow for the retardation effects on the shock

transmitted through the soil. The horizontal pressure P^t) is assumed to

be proportional to the vertical pressure.

The time of arrival t of the pressure wave at any point in the

soil, or on the arch, is determined by the propagation velocity of the

wave. The time of rise t of the vertical pressure to its maximum value Pr m

is assumed to be a function of the vertical transit time t, of the pressure

wave. The rate of decay of the earth pressure wave is assumed to be the

same as that for the surface pressure. These relations are shown

qualitatively in Figure 2.2.

Study of Figure 2.1 indicates that as the shock velocity of the

blast wave increases relative to the seismic velocity of the soil, the

8

unsyrametrical loading decreases, since the pressure wave approaches the

arch more nearly vertically. For the usual range of these variables the

earth pressure wave qualitatively approaches the arch as shown in Figure 2.1.

It should be noted that this pressure wave is considered to have

been induced solely by the surface pressure. The direct ground shock wave

associated with" a nuclear weapon explosion is not considered in this thesis.

Reference (l) states that for pressures of less than 300 psi and for depths

of less than fifty feet the direct ground shock may be ignored.

The arch is assumed to be acted upon by the radial resultant of

the vertical and horizontal pressures. If the ratio of the horizontal

pressure to the vertical pressure is as great as 1:1, the radial pressure

is equal to the vertical pressure. Moreover, the total radial force in-

creases as the ratio of the horizontal to the vertical pressure increases.

The effect of the rise time may be seen in Figure 2.2. The

pressure wave arrives at some point cp, on the arch, Figure 2.1, and the

pressure increases to its maximum value in some time t . At some adjacent

point cp the pressure is at a different level, depending on the relative

arrival and rise times. Reference (2) recommends that the rise time be

taken equal to one-half the vertical transit time.

The following range of variables will be studied in this paper:

P = 100 and 200 psi with U = 3000 and 1*000 fps

C - 1000 and 2000 fpss

t » t . and 1/2 t.r t x>

Ph/py- 0.25, 0.50, 0.75 and 1.00

2.2. Passive Earth Pressure Loading

The unsymmetrical blast load on the arch causes the arch to deflect

into the soil on the leeward side. This deflection brings into action the

9

passive earth pressure of the soil as well as the inertia of the mass of

soil associated with this pressure, since the blast load acts dynamically.

This pressure contributes materially to the resistance of the arch and must

be considered.

While precise evaluation of the passive earth pressure under static

conditions is complex, evaluation under dynamic conditions is equally, or

even more, complex. The passive earth pressure can vary in the length of

the arch due either to the properties of the soil or to the construction

technique used in placing the backfill. Rankine's theory has been chosen as

the simplest method for evaluating this passive earth pressure.

In the application of Rankine's formulas, Reference (3)> the unit

weight of the soil is assumed to be 120 pounds per cubic foot and the angle

of internal friction is assumed to be thirty degrees.

2.3 Structures Considered

The blast loading on arches with central angles of 180 and 120

degrees and radii of 150 and 300 inches is investigated to determine the

effect of the soil variables, arch geometry, and relative rise time on the

arch loading, and to determine a conventionalized dynamic loading function.

Subsequently investigated is the response of arches with central angles

of 180 and 120 degrees and radii of 200 to 600 inches to this conventionalized

dynamic loading function.

10

Air Blast

Wave Front

Air InducedGround Pressure Wave

Ground Surface

Figure 2.1 Underground Arch Subjected to Blast Pressure

For $

For d>

Time

Figure 2.2 Vertical Pressure-Time Diagram for 9

11

CHAPTER III

GENERAL RELATIONS COMMON TO EACH LOADING FUNCTION

To predict the effect of the loads on the axch rib, the loads must

be defined in a convenient manner.

Certain general relations may be derived for defining either the

blast load or the earth pressure load. These general relations concern the

transformation of stress with direction, the method used to determine the

value of the coefficients p (t) and p for the distributed radial load

functions \ p (t) sin _ and \ p sin • which represent the actual

load on the arch, and the basis for assuming analogous loads to represent

these distributed loads.

For purposes of dynamic analysis, the arch is idealized to a

single-degree-of-freedom system. The time-dependent load \ p (t) sinfl

is replaced by a uniformly-distributed, antisymmetrical, time-dependent load

p (t), and the distributed earth pressure load \ p sinn ™

is replaced

by a uniformly-distributed antisymmetrical load p . These antisymmetricalct

loads act outwardly on one half of the arch, and they act inwardly on the

other half.

The direction of the analogous earth load is opposite to that of

the analogous blast load. These anti symmetrical loads define the effects

of the actual loads on the arch haunch.

The values of these analogous loads are found by the application

of the principles of statics. The value of the analogous load for the soil

pressure may be found by equating the moment produced by the distributed load

Ip sin —r— at a point 6/k to those of the analogous antisymmetrical loadn v

p at the same point. The replacement of the dynamic load \ p (t) sinn ™

with an analogous load requires that a basic assumption be made. It is

12

assumed that if two generally similar static loads produce equal static

effects on a structure at a specified point, then the same two loads applied

dynamically with similar pressure-time diagrams of equal duration would

cause equal dynamic effects at the same point. Of course the dynamic

effects of the loads are not defined. The point considered is 0/k.

To determine the analogous loads, the moments caused by the anti-

symmetrical loads p (t) and p are equated with the moments caused by thea a

Z/\ ntf cp V nfl cp

p (t) sin' and \ p sin respectively. In

addition, the effect of a uniform compressive load on the central third of

the arch coupled with a uniform outward load on the outer thirds of the arch

is determined. This load provides an analogous load for study of the crown

of the arch.

5«1 Stress Transformation with Direction

Let it be assumed that a horizontal unit stress p, and a vertical

unit stress p are acting on a segment Rdcp of a semicircular arch rib located

at the angle cp measured from the right abutment. Assuming that p and p.

are principal stresses, the radial pressure p (cp) at cp may be found by the

well-known method for determining the relations between stresses at a point

on different planes passing through the point, Reference (k) .

Rdcp may be assumed to be a plane surface for an infinitesimal

segment. The vertical projection of the plane Rdcp is Rdcp cos cp; the hori-

zontal projection is Rdcp sin cp. Equating the sum of the radial forces to

zero yields the following:

2 2p (cp) Rdcp = u Rdcp cos cp + p Rdcp sin cp

Using the trigonometric identities

pcos cp = 2(1 + cos 2cp)

psin cp m ^(l - cos 2cp)

13

and cancelling Rdcp from both sides of Equation (a),

Pr ((P) - iU + cos 29) 1^ + i(l - cos 2<p) py

- id^ + Pv ) + 2(1^ - Pv ) cos 2t) when the pressures

considered arise from the blast pressure acting on the surface of the ground.

Let PyXt) = Cp (t). Then Equation (l) becomes:

Pr(<P,t) = C py

(t) + py(t) (1 - C) sin

2<p (2)

This equation is evaluated for various values of C as follows:

For C * 0.25, p„(<P,t) m fc.25 + 0.75 8in2cp) p (t) (3)

For C = 0.50, p (cp,t) =(0.50 + 0.50 sin2cp) p (t) (h)

For C - 0.75, Pr(<P,t) =(0.75 + 0.25 sin

2cp) Py(t) (5)

For C = 1.00, pr(cp,t) = py

(t) (6)

Equations (3) through (6) express the time-dependent unit radial

pressure acting on the arch rib at any point cp for the various assumed

ratios of the horizontal pressure Pj^t) to the vertical pressure p (t).

Evaluation of these equations is set forth in Table I.

3.2 Distributed Load F unctions

Any radial load profile p (cp), such as that shown in Figure J.l,

when acting upon a segment of a circular arch rib with a central angle 6,

may be represented as a distributed load by a trigonometric series in the

following manner:

Pr((P) = Px

sin -|- + p2sin—- + + Pq sin — +

The coefficient p is obtained by multiplying both sides of

Ik

(7)

/ \ nic 9Equation (7) by sin and integrating over the length of the arch rib.

J pr(<p) eln^Rdcp =J px

sin 2 8in £2 Rd9o w

/.6

fP2

sin -2 sin £2 Rd9o

.2 nncp+ + / p sin —-2- Rdcp + --X pn~ a

ou °

(8)

The value of the integral / sin —^nsin —— Rd9 is zero provided

that m and n are integers and m does not equal n. Equation (8) then yields:

2n«9

= i v R0n

Pn=

$R

o0

J pr(cp) sin £2 Rdcp (9)

If p (cp) can be expressed analytically, the value of p may be

found by integration. If p (9) is an irregular load profile for which

there is no convenient analytical expression, the integral may be evaluated

by numerical methods since the integral / p (9) sin £2- Rdq> represents the

nwparea under the load profile multiplied by sin —~ .

The sign convention adopted throughout this thesis is that a

positive load acts inwardly and a negative load acts outwardly. The

15

xiupharmonic components of loading p 6in —~ shown in Figure ).l are in

n u

qualitative agreement with the original load p (<p); therefore the sign of

the coefficient p2 is definitely negative. The sign of the coefficient p_

has been taken as negative, although no positive statement can be made that

this is the case for the original load p (<p) in Figure 3.1.

Included in Figure 3*1 are two cases of uniform radial loading.

p sin —

r

2- on the crown and haunchn 6

of the arch may be represented by such uniformly distributed loads.

If the load profile p (cp) is time-dependent, it is expressed as

P (t). The procedure described above may be used to determine the co-

efficient p (t), with the qualification that p (t) must be evaluated at a

specified time.

Reference (5) contains a full development of the methods of

harmonic analysis as applied to structural engineering problems.

3.3. Basis for the Determination of Analogous Loads

As stated at the beginning of this chapter, the analogous loads

p (t) and p are derived by equating their moments to the sum of the moments& EL

of the distributed loads \ p (t) sin —S- and \ p sin —2J., respectively.

The time dependency of p (t) is not considered in this section since thecL

load \ p (t) sin -— is assumed to be acting statically at the instant

the analogous load p (t) is evaluated.

Moment equations for all load profiles in Figure 3»1 &ce found

in this section after general equations for the right reaction of any two-

hinged semicircular arch are established. Since the blast load and soil

Zn«Pp sin —~ , one set

of moment equations suffices for both analogous loads.

16

5.3«1» General Equations for the Right Reactions of C ircular-Segment Arches

Since the circular-segment two-hinged arch, Figure 3.2, is one

degree indeterminate, one equation based on Castigliano's theorem,

Reference (k), is used for the determination of the reactions. The horizontal

thrust at the right reaction is chosen as the redundant. It is assumed that

the arch rib may* be treated as a thin ring, making it unnecessary to consider

the theory of curved beams. The effects of rib shortening also are not

considered

.

The following geometrical relations are derived in accordance with

Figure 3*2:

Figure 3.2a BC = 2R sinJ9

AB = BC cos^ d - <p) = 2R singCp cog^< e - 9)

AC = BC sin^{0 - 9) = 2R sini9 sin^ 9 - 9)

BD = 2R sinê

Figure 3.2b DF = 2R sinj(e - 0)

EF = 2R sini(e - p) coeê - 0)

= moment arm of P about D

Figure 3.2c FG = R sin (9 - P)

a moment arm of P about G

Figure 3. 2d AB = moment arm of V about C

AC = moment arm of H about C

DB = moment arm of V about D

In Figure 3»2d, the total strain energy U in the arch rib due to

the moment M caused by the radial load P acting at a point on the rib

located by p may be expressed as

17

°"/ 2©^=/ 2EIRdCP

Since the abutments are assumed not to move, the contribution of the right

reaction H to the strain energy is zero and may be expressed as

ai-J ^ Rdcp =< a >

Two equations are required to express the moment in the arch rib;

one equation for angle <p less than or equal to £, and one for cp equal to or

greater than p . For < cp 5- 0,

1^ = V(AB) - H(AC)

» 2VR sin ^p cosi(e - cp) - 2HR sin^cp sinj(0 - cp)

(b)

^g- -2R sin^cp sin^S - cp) (c)

For B < cp < 6,

Mg = V(AB) - H(AC) - P(FG) (d)

* 2VR sin^qp cosi( - 9) - 2HR sin^cp eln^ ©-cp) - PR sin(9"-£)

^g- = -2R sin^cp sin^<0 - 9) (e)

Substituting Equations (b) through (e) into Equation (a),

°-j 1S. OH"

RdcP + JM2 5T Rd<P

= / [ 2VR sin^cp cosj( 0-cp) - 2HR sin^cp sini< 0-cp) ]

J o

[ -2R sin^cp sinj(0-cp)] Rdcp +

r e

/ [ 2VR sin^cp cos^( 0-cp) - 2HR sinjcp sini( 0-cp)

- PR sin(cp-B)][-2R sin^cp sin5<0-cp)] Rdcp (f)

Solving Equation (f) for H yields

18

e

o4/ VR5 sin

2 §sin(±2) cos(^)d*

-2 / PR5 sin(cp-3) sin | sin(-^)d<p

H % («)

i*/6E5 Bin

2| si„

2(±2) av

Referring to Figure 3.2, the sum of the moments equated about

the left reaction gives

VcZiM=2PR sin(ffi cos(ffi p^^)

1)52R sin| 2 sin |

Substituting Equation (10) into Equation (g),

P {"SUfl [l . * sine - i cose] + [cosI

-I cos(f - (3)

- £cos(|2-0) +(6^) sin(f -e)J|H =

e - 2/ e% ~ . e e^ + e cos (75) - 3 sin ^ cos £

(11)

Equations (10) and (ll) define the right reaction of any circular-segment

two-hinged arch acted upon by a radial load P located at any angle £.

3.3*2. Moment Equations for l8o Degree Arches

For a semicircular arch, the central angle 9 is * , and

Equation (ll) reduces to

H =I [sin + (p - «) cos p] (12)

Using Equation (12), the horizontal right reaction may now be

DSRPfound for the harmonic load component p sin —~ .

19

For P » p, sin — Rdcp = p sincp Rdcp, since Q = x,

if* P1R

H «= — / p, sin <p[sin

-a f -p2

sin2cp[sincp + (cp-x) coscp] Rdcp = - PgR

For P = -p., sin 39 Rdcp,

if* 5H = - I -p, sin39[sincp + (cp-*) coscp] Rdcp « 4 p R

nxcpThe loading configurations p sin —~ are shovn in Figure 3»1«

n o

It is also necessary to investigate the effect of the uniform

antisymmetrical load p and the uniform symmetrical load p shovn in

Figure 3.1. For the uniform antisymmetrical load,

n

H = - / -P tsincp + (cp-x) coscp] Rdcp + - / p[sincp + (cp-x)coscp]Rdcp = pR

2

For the uniform symmetrical load,

x 2x

H = j I ^-p[sincp + (cp-x) coscp] Rdcp + i / ^ p[sincp + (cp-a)coscp] Rdcp

3

+ j \2% p [sincp + (cp-x) coscp] Rdcp = | f$ pR

TEquation (10) is used to find the right vertical reaction.

With 6 = x, Equation (10) reduces to V = £ P sincp.

20

f n

For P = pxsin -2 Rd9, V = £ / px

sin 9 Rd9 = ^KR

For P = -p2

sin ^2 Rd9, V = - £ / pg

sin2<p sin9 Rd9 =

p, sin39 sincp Rdcp =

o 5

For the uniform antisymmetrical load,

it

V = £ /2

-P sincp Rdcp + £ / p sincp Rdcp =

2

For the uniform symmetrical load

* 2*

V = i I^ -P sin Rd + £ /

5p sincp Rdcp

5

2j2« p sincp Rdcp =

The value of zero for the last four loads was to be expected from

the symmetry of the arch and the loading configurations. It is interesting

to note that the two-hinged semicircular arch is statically determinate for

the last four loadings.

All the above values for the right reactions for a 180 arch are

tabulated in Table II.

The moment at any point on the arch in Figure 3-1 located by 9

due to a distributed load p, sin -3- Rd£ acting at (3 isi &

dM = -p sin£[R sin (9-0)1 Rdp

The total moment at 9 is

21

M - -H(AC) + V(AB) - / P1sinp [R sin(cp-p) ] Rdp

. - !£ [2R six, | sin^] + !|f

[2R sin | cos(i2)]

+ p^2

£ sinq> sin2p - cos(p ( | - -^^ )

p.R x - ( *-2q>) cos9 - 3 sin<p1u

A similar procedure is used to find the moment at 9 due to the

other loading configurations in Figure 3«1« The results are tabulated in

Table III.

3. 3.3« Moment Equations for 120 Degree Arches

When the central angle of the arch is 120 degrees, Equation (11)

reduces to

sin(^ - p)

sin .

3

p-¥+ cos - -

I cos(| - p) - j cos (n-p) +g

? sin(| - p) |»

™« 2« 2 « I ! * *:= + -=- cos =• - 3 sin rr cos =3 3 3 3 3

P Ucosp + 2j- sinp)( | -3J *) + i -(g + -jr) cosp

+(f

- ^) sinp + -£ p cos p - £ p sinp

= P I.83979 - 2.70578 cosp + O.16665 sinp

+ 1.59330 p cos p - O.9199O p sinp (13)

22

Equation (13) may be used to find the right horizontal reaction

of an arch with a 120 degree central angle acted upon by a concentrated

radial load P located at any angle p. The reaction for p, sin -^ Rdcp is

found as follows:

H = / T Pxsin

Icp 1.83979 - 2.70578 coscp + O.16665 sincp

+ 1*59330 9coscp - 0.91990 cpsincp Rdcp = O.53076p R

The horizontal reactions for the remaining harmonic loads and the

uniform loads are found in the same manner, and are set forth in Table II.

Equation (10) is used to find the vertical reaction.

jtcp

For p1sin ~ Rdcp ,

v = /5

I^—5 Pi sin I x

3

The remaining vertical reactions have been found, and are stated

in Table II.

For a segment Rd£ at £ loaded with p1 sin -& Rd £ , the moment

at cp in Figure 3.1 is

dM = -p sin | £[R sin(cp-P) ] Rd£

and the total moment at cp is

M = -0.53076 pxRlAC] + 0.6p

1RtAB]

- / Pxsin

I plRsin(<p-e) 1 Rd?

P- cp

= -0.53076 pxR J2R sin I

sin(-2_I)j

2*9

+ 0.6 p^R 2R sin I cos(-2—g)

-/ px sin ^ p [R sin(cp-p)] Rdp

= (0.78^99 - 1.35965 sin9 - 0.781*99 coscp + 0.8 sin ^) ^R2

23

The moment equations for the remaining load configurations shown

in Figure 3*1 for an arch with a 120 degree central angle have been found

and are contained in Table III.

3 .3 • ^ • Procedure in Determining Analogous Loads

The moment equations tabulated in Table III have been evaluated

for every ten degrees. The results are tabulated in Table IV and shown

in Figures 3*3 &a& 3«^« These values are for static loads. They do not

indicate the dynamic moment which may exist at any time in the arch.

Inspection of Figures 3«3 and 3*^ shows that the effect of the

analogous antisymmetrical load is almost identical to that of the load

2»Pp? sin —-r1 • If Pp bas a value of one, then the value of the antisymmetrical

load must be slightly less than one if the effect of each is to be the same.

However, the antisymmetrical load must also include the effect of the loads

p, sin ~- and p, sin ^^- . Therefore, the moment that \ p sin "— causes10 3 / n 6

at a point e/h is first determined, and subsequently a value of the anti-

symmetrical load is found which will cause the same moment at e/h. The

point 6/k is selected since the unsymmetrical blast load produces the maximum

effect at this point. The time must remain constant for each numerical

evaluation of p (t), as described below,a

The actual blast load on the arch is p (<p, t) . At any given time

the load profile p {y,t) can be determined and analyzed to give a distributed

load \ Pn(t) sin —^- . By assuming that \ p (t) sin —~ acts statically,

its effects may be found at 6/k. An analogous load can be found which

causes the same effect. By repeating this operation at various times, the

pressure-time diagram of p (t) can be determined. This load must then be

applied dynamically to determine its actual effect on the arch.

2k

In determining the static load of the earth, the analogous load

is found by equating the moments at Q/k of the distributed static load

\ p sin ^~- with the moment of the anti symmetrical load. Since the earth

pressure is not a dynamic load, only one evaluation is needed.

Further inspection of Figures 3-3 and J.k shovs that the effect of

Znfltpp sin —^ on the central third of the arch may be

represented by the symmetrical load which acts in compression on the central

third of the arch and in tension on the outer thirds. This load is

designated p . If the distributed load is a time-dependent load, the same6

procedure is used in finding the pressure-time diagram for p (t) as that

used for p (t)

.

25

TABLE I

COEFFICIENTS FOR RESOLVING VERTICAL AND HORIZONTAL PRESSURE INTORADIAL PRESSURE

Angle Radial pressure p (9>t) in terms of vertical pressure

10

20

30

ko

50

60

70

80

90

100

110

120

130

140

150

160

170

180

Assumed ration P^/p

0.50 0.75

0.250 0.500 0.750 1.00

0.273 0.515 0.758 1.00

0.338 0.559 0.779 1.00

0A38 0.625 0.813 1.00

0.560 0.707 0.853 1.00

0.690 0.79^ 0.897 1.00

0.815 0.875 0.938 1.00

0.913 0.9U2 0.971 1.00

0.978 0.985 0.993 1.00

1.000 1.000 1.000 1.00

0.978 0.985 0.993 1.00

0.913 0.91*2 0.971 1.00

0.813 0.875 0.938 1.00

0.690 0.794 0.897 1.00

0.560 0.707 0.853 1.00

0A38 0.625 0.813 1.00

0.338 0.559 0.779 1.00

0.273 0.515 0.758 1.00

0.250 0.500 0.750 1.00

26

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29

(a) Radial Load Profile p (<p) and Harmonic Load Components

* n * q>

p sin —t-1^n

(b) Uniform Symmetrical and Antisymmetrical Loadings

Figure 3.1 Various Load Configurations

50

L CBE =-£ $

L CBA =-+-(#-<£

4-^-/3)

DE // To FO

E F is a Tangent Line

Figure 3.2 Geometry of a Circular Arch Segment

51

+ 0.5

180 120 90 60

Angle<f> , degrees

Figure 5.3 Moment vs Angle cp for 180 Degree Central AngleTwo-Hinged Arch

52

1

/- Uniform Antisymmetricol

/ • 2t</>

t r p2s,n-

a

g «^^l

/—Uniform S

/"S/--p3

sir

ymmetrical

3tt<£

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v>•ocoQ.

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c0)

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120 90 60 30-0.20

Angle 4> , degrees

Figure 3.U Moment vs Angle cp for 120 Degree Central Angle Tvo-Hinged Arch

33

CHAPTER IV

BLAST PRESSURE LCADIHG

k.l Pressure-Time Relations at a Soint

As stated in the general discussion of the variables, Section 2.1,

the blast pressure acting on the surface of the ground above the point of

interest causes a vertical pressure at that point. Accompanying this

vertical pressure is a horizontal pressure of varying magnitude, depending

on the soil. The horizontal pressure may be as low as fifteen percent in

loose, dry soil, fifty percent in moist soil, and one hundred percent in

saturated soil, Reference (l). One of the purposes of this thesis is to

study the effects of this variation.i

The vertical pressure is propagated downward with a velocity

equal to the seismic velocity of the soil. At the point of interest, the

time for the pressure to increase to a maximum is a function of the vertical

transit line. The effect of varying the rise time will be studied.

It has been shown in Section J.l that the horizontal pressure

may be expressed in terms of the vertical pressure. Vertical pressure-time

relations at a point on the arch in terms of surface pressure are there-

fore required to define the total radial load profile p (t) acting on

the arch. The equations below are derived from the geometry of Figure 4.1.

The velocity of the wave propagation is:

v = (1*)

where v = velocity of propagation of the earth pressure wave

Cs= seismic velocity of the soil

U = surface blast wave velocity

3*

The time required for the pressure wave to arrive at any point <f>

on the arch rib is:

=R - R sin(cp -

p)a v x '

where <p = angle locating a point on the arch rib

R = radius of the arch

t = time measured after initial contact of the pressurevave with the arch rib

c*

£ = arc tan rr-

The time required for the pressure at 9 to rise to its peak

is:

R - R sin? + Ht = S. for t » t. (16)r C ' r t % '

8

R - R sincp + Htr

«gg

£ , for tr

= i tt

8

where H = depth of cover on the crown

t = rise time of the pressure

t. = vertical transit time of the pressure wave to 9

R - R sin<p + Ht = £t C

8

The vertical pressure-time variation at a point on the arch rib

is shown qualitatively in Figure 2.2. The pressure-time diagram is assumed

to be a straight line in the rise part of the diagram and to have an ex-

ponential decay rate after reaching its peak value P . The total impulse

of the force pulse at any point must be the same as the total impulse of

the surface force pulse. This means that as the rise time increases, the

decay rate must also increase in order to maintain the same impulse in

55

the force pulse. It is assumed in this study that the decay rate at the

shallow points in the soil considered in this thesis is the same as the

decay rate of the surface force pulse. This is a reasonable assumption

since the maximum unsymmetrical loading occurs within the first few milli-

seconds after the blast wave begins to act on the arch.

The decay rate for a side-on overpressure P having a positive

phase duration t, may be obtained from Reference (6) . The decay rates for

the two overpressure levels studied in this thesis are shown in Figure k.2.

For the 100 psi overpressure, t, is assumed to be 1500 milliseconds, while

for 200 psi overpressure, t, is assumed to be l6i*0 milliseconds. The

weapon yield is 1 HT.

The complete vertical pressure-time diagram for any point can be

determined only by numerical evaluation of the pressure on that point at

various times. A sample calculation of a pressure-time diagram follows.

Assume that a 180 degree central angle arch with a radius of

twelve and one-half feet is subjected to a 200 psi overpressure blast wave

originating from a 1 MI weapon. The duration of the positive phase of the

loading is l6kO milliseconds. The arch is buried in soil having a seismic

velocity of 1000 fps, with zero depth of cover on the crown. The time of

rise of the earth pressure at a point in the soil is assumed to equal the

vertical transit time of the earth pressure wave. The ratio of the

horizontal to vertical pressure is assumed to be 0.25

•

Reference (7) give6 the following equation for the shock velocity

U of the surface blast wave:

U= cq(1 + 6p/7P )

2

where c = ambient sound velocity ahead of the shocko

36

p = peak overpressure behind the shock

p = ambient pressure ahead of the shock

.'. U - 1130 (1 + 1200/103) 2 - **023 fps

By Equation (HO, v = S » 1000 x UOZ}= ^ fps

\|C2

+ U2

\l 10002x 1»023

2

By Equation (15), p « arctan (IOOO/U023) = 13°58'

t - 3~2 [1 - sin( seconds

= 12.5 (l - sin<p)> milliseconds

Let 9 = kO°, Then

t = 12.86 [1 - sin(lK>° - 13 58')3 = 7.2 ms81

t = 12.5 (1 - sin 1+0°) = h.5 ms

Therefore at cp = kQ degrees, the earth pressure wave arrives

7.2 ms after the wave initially contacts the arch rib at 9 = 90° + 13°58'.

The vertical pressure rises at a rate of 200/4.5 = ^5 psi per ms. At 8 ms,

the pressure is (8 - 7.2) V? = 36 psi. At 10 ms, the pressure is

(10 - 7.2)^5 = 126 psi. At 11.7 ms, the pressure reaches a maximum value

equal to the surface overpressure, or 200 psi. Thereafter the pressure

decays at a rate assumed to be equal to the decay rate for the surface

overpressure as shown in Figure k,2. Inspection of Figure k.2 shows that

the decay may be represented accurately by a straight line down to a value

of A p/Ap = O.65 and At/t, = ~ 0.02. The pressure decrement is then

\57

0.55(200) = 70 psi, and the time interval is 0.02(161*0) ~ 55ms. So the

pressure decays at a rate of 70/55 = 2 psi per ms. Therefore at 12 ms, the

pressure is 200 - 2(12 - 11. 7) = 199 psi. At Ik ms, the pressure is

(199 - 4) = 195 psi. This procedure is continued until a time greater than

(55 + 11) ms is reached. Then the time interval is expressed as a fraction

of t, and Ap/Ap *is read from Figure k.2. The complete pressure-timed m

relation for 9 = UO degrees is recorded in Appendix B, Table B5.

To obtain the complete vertical pressure-time relations for the

entire arch, the above procedure is accomplished for every ten degrees on

the arch. Table B5, Appendix B, shovs the vertical pressure-time variation

for the entire arch up to 200 milliseconds. The unsymmetrical loading is

readily seen. For example, at ten milliseconds the windward side of the

arch has pressure acting on its entirety, whereas the leeward side is not

fully loaded until after sixteen milliseconds. The unsymmetrical loading

shifts from the windward side to the leeward side during the time the blast

pressure acts. This effect may be seen at thirty milliseconds. The

pressure on the windward side, cp greater than ninety degrees, averages some

eight pounds less than the pressure on the leeward side. The effect of

the decay rate may be seen by comparing the pressure at the crown with

that at the abutments. For example, at twenty-four milliseconds the pressure

at the crown has reached its maximum and decayed to approximately seventy-

five percent of the maximum surface pressure, while the pressure at the

abutment is approximately equal to the maximum surface pressure.

k. 2. Harmonic Analysis of Radial Blast Load with Time a Constant

Based on the vertical pressures tabulated in Table B5, a radial

load profile p (<?>, t) can be determined for any instant of time given in

38

that table. After p (t) has been defined, the distributed load

I p (t) sin ——• can be evaluated. A sample calculation, for t = 10 milli-

seconds, is given in Tables V and VI.

In Table V, the data in column "p (t)" is obtained from Table B5,

while that in column "p coefficient" is derived from Table I for PyVp

0.25. The data in column "p (t)" of Table V is derived by multiplying

p (t) by the p coefficient. The resulting radial pressure load profile

p (t) is shown graphically in Figure U.3. The coefficients p (t) of

Equation (9) are determined by numerical integration. This is done by

tiR nfltpmultiplying the load ordinate at each ten degree interval by r-g sin —~ ,

2adding the results and multiplying the sum by •= . The resulting dis-

tributed radial loading \ p (t) sin —jp , with t = 10 milliseconds, is

shown as a dotted line in Figure ^.3.

Table VI contains the harmonic analysis of the same blast vave

acting on a 150 inch radius arch with a central angle of 120 degrees. The

radial pressure p (9>t) in Table V from 30 degrees through 150 degrees is

used in Table VI from degrees through 120 degrees. The resulting dis-

tributed radial loading \ p (t) sin ^~ is shown in Figure ^.3»

Inspection of the data in column "p (9,t)", Table VI, reveals that

there is a radial pressure of ten or more pounds acting at all points on

the arch rib. Therefore, a uniform all -around pressure of ten pounds may

be subtracted from the radial pressure before the harmonic analysis is made,

since it is desired to resolve harmonically only the radial pressure causing

flexure in the arch rib. The column "p adjusted" contains the remaining

radial pressure.

The coefficients p (t) of Equation (9) are found by numerical

integration. This is done by multiplying the load ordinate at each ten

degree interval by J| sin £|2 , adding the results, and multiplying by JL .

39

The resulting distributed load, p (t) +\ p (t) sin —* , vith t = 10 milli-

seconds, is shovn as a dotted line in Figure 4.5

•

I

It is assumed that the first three terms of the series

nfltpp sin —~ adequately represent the loading on the arch for the purposesn v

of this thesis.

The radial load on the 180 degree arch at a time t = 10 milli-

seconds is:

p (<p,t) = 166 sin9 - 13 sin29 - 26 sin3<p

Since the time is constant, the load may be considered to be a static load.

An analogous load may be determined which will have the same effect on a

given point of the arch. Let the point be located at cp = 1+5 degrees.

p sin -~ all cause

negative moment at this point. Using the coefficients in Table IV, the

moment of the load is

M= [l66( -0.02262) + 13( -0.33333) + 26( -O.O8857) ] R2

= [-3.75 - 4.33 - 2.30] R2

= -10.38R2

A uniformly distributed antisymmetrical load acting in on the

windward half of the arch and out on the leeward half produces a moment of

0.1al+22p R at cp = k$ degrees. Thereforea

2

P * ^ 25 psi for a time t * 10 msa

o;iute2R

However, the actual load on the arch is time-dependent and consequently p

must be a time-dependent load, p (t). To define the pressure-time diagram

for p„(t), the load p (9>t) on the arch is harmonically analyzed at variousa r

times.

The analogous load for the crown may be found in a similar manner.

The moment caused by the above load at the crown is

1<0

M « [166( 0.035^) + 26(0.125)] R2

2= 9.13 R inch, povinds

A symmetrical load p acting in compression on the central thirds

of the arch and in t ension on the outer thirds of the arch produces a

moment of 0.15^7 p R at the crovn

s

2.'. PR(t) = ? ,:L?R

p = 59 pel, t = 10 ms8

0.15^7R

If .3 Blast Loading Approximations

Appendix B contains the results of numerous harmonic analyses of

the blast load on an arch arising from a 1 MT nuclear weapon. The analyses

were made and analogous loads found as described in the preceding section.

These analyses were made in order to determine the effects of the variables

discussed in Chapter II on the loading the arch receives, and to provide a

basis for ideal! zing the blast load on the structure without the necessity

for actually harmonically analyzing the blast load.

The blast load must be defined in relation to the effect being

studied. In this thesis, the effect studied is the deflection of the arch

due to flexural stresses, the area of primary interest being the haunch of

the arch. The effect of the unsymmetrical blast load on the haunch of the

arch is most conveniently related to an analogous uniformly-distributed

load which acts inwardly on one half of the arch and outwardly on the other

half. The effect of the blast load on the crown is conveniently related

to a uniform load acting in compression on the central third of the arch

and in tension on the outer thirds. Values of such loads for various

assumed conditions are shown in Tables Bll and B12.

kl

After the blast load has been related to the specific effect

being studied, a load-time curve must be established to shov the variation

of the load vith time. It will be noted in Tables Bll and B12 that the

analogous loads are for a specified time only, and that they vary appreciably

with time. Appendix B contains a fev typical analogous load-time curves.

Similar plots were made of all analogous loads.

Study of these load-time curves revealed that the peak values of

the analogous loads are some fairly uniform function of the maximum surface

overpressure P for all assumed values of the variables. In addition, the

shapes of these load-time curves are approximately initially peaked tri-

angular load pulses, which decay rapidly to a relatively small value and

then continue to decay at a much slower rate. The duration of the rapid

decay phase of the loading corresponds approximately to the time required

for the pressure to engulf the arch. The duration of the slow decay phase

of the loading is relatively much longer.

The peak values of the analogous loads found in Appendix B are

shown in Table VTI and plotted in Figures k.k and k.J. The data in

Appendix B and Figures k.k and k.5 may be qualitatively summarized as

follows:

(a) For any given overpressure, the peak value of the

analogous load for the haunch increases as the value of

the seismic velocity of the soil increases. This is ex-

plained by the fact that a higher seismic velocity allows

the pressure to penetrate the soil to a greater extent

and thus a larger segment of the arch on the windward side

is loaded before the leeward side is loaded.

k2

(b) For any given overpressure, the peak value of the

analogous load for the crown decreases as the seismic

velocity of the soil increases. This is explained by

considering the effect of the load profile p (9>t) on

m2P (<fct) sin ^2T Rd<P • ê11 c „ has

%. r v s

a relatively high value in comparison with U, a greater

portion of the segment from 9-, to 92 is located on the

windward side of the arch and the values of the sine

function of the angle 9 are both positive and negative.

When C has a lower value in comparison with U, thes

load profile p (9>t) is located more nearly in the central

third of the arch, where the value of the sine function

of the angle 9 is negative. This increases the numeri-

cal value of the analogous crown load.

(c) The higher ratios of P^/p increase the peak value

of the analogous load for the haunch. This is explained

by the fact that the area of the total radial load profile

p (9>t) is greater for the higher ratios, and thus the

total unsymmetrical load is greater.

(d) The p./p ratio has little effect on the peak value

of the analogous crown load. The peak value of the crown

load is primarily a function of the location of the load,

i.e., the central third of the arch. In this area, the

value of the coefficient for determining the resultant

of the horizontal and vertical forces is approximately

one for all values of Ph/p • Consequently, changing the

ratio has little effect on the radial load.

*3

(e) The assumed rise time has relatively little effect

on the peak value of the analogous loads. The peak

valuee of both loads are reached very early in the load-

ing of the arch -when the pressure wave has not penetrated

the soil to any extent. For shall ov depths, the rise

time is* practically equal to the rise time of the surface

pressure, -which is assumed to be instantaneous.

(f) Although the higher ratios of PjVp cause a higher

peak value of the analogous haunch load, they actually

cause a less severe dynamic load effect. If the ratio

is high, the pressure, after the pressure vave has en-

gulfed the arch, is more nearly a uniform compression,

causing less flexural action in the arch. On the other

hand, if the ratio is low, the uniformity of the compres-

sion is reduced, increasing the flexural action in the

arch. For the higher ratios of Pl/p it is found that

the analogous loads change direction, thereby creating

a negative phase of loading. The peak value of the

pressure in this negative phase is small, but could pos-

sibly cause an outward movement of the crown of the arch

since it has a relatively long duration.

k.J.l l8o Degree Arches

Since the blast loading may act over all, or over part, of the

arch segment, one approximation to the peak analogous loads may be obtained

by considering the blast loading acting on the arch in such a way that the

maximum value is given to each component of loading in the series

Uk

I p (t) sin ^^ . The ratio of the horizontal pressure to the verticaln 6

pressure which causes the most severe dynamic loading is 0.25, therefore

this value is used in the following derivations. The effect of the rise

time of the pressure and the time of arrival of the pressure wave is not

considered. The vertical pressure at all points on the arch is assumed to

be equal to P . The unit radial pressure at any point is, by Equation (j),m

Pr(<P,t) = (0.25 + 0.75 sin

29) Pm

Using Equation (9),

*nW - h J pr( *' t) 8in H^Rd,

2P n*maximum pn

(t) = -^ / (0.25 + 0.75 sin <p) sincp Rd9 - 0.955 Pi hk j m

2P_ r*

o

2P r *ntfQf 1 "AT™

tnfnrynmm

P2(t) = iffJn (°* 25 + °* 75 8in = -°-398 Pm

2P r¥p5(t) = ir|j«

5(0 -25 + °- 75 8in2cp

) 8in3<P^ -o. 1^ Pm

The maximum effect of the blast loading which may occur, under an assumed

ratio of PjVp 0.25, can never exceed the effect of the sum of maximum

values of the harmonic components found in this manner. The value of the

individual maximum values found, and the effect of the sum, are always

reduced when the seismic velocity of the soil, the shock velocity of the

surface blast wave, and the pressure-time relationships are considered.

If these components acted statically and separately, they would

produce the following moments at the leeward haunch and crown:

*5

M = - 0.023(0.95 5Pm)R

2+ 0.353(0.398Pm

)R2

+ 0.088(0.4oaPm)R

2

= -0.190 P R2

m

M = 0.035(0.955P )R2

+ 0.125(0. 40&T )R2

= O.O85 P R2

c m m ' m

The peak value of the analogous haunch load is

0.190 P R2

p ( t ) = 2_ = o.U6 P0.4l4 r

2

The peak value of the analogous crown load is

0.084 P R2

These values of the analogous loads can never occur, and, for

vant of a better name, they are designated as the maximax analogous loads.

They are shown in Figure 4.4 so that they may be compared with the analogous

loads found by actual harmonic analysis. Such comparisons indicates that

these approximations are conservative. The following reductions are

arbitrarily made:

Maximum p&(t) = 0.37 P

ffl

(17)

Maximum p (t) - 0.45 P (18)

Equations (17) and (18) are plotted in Figure k.k.

To obtain an idea of the magnitude of these analogous loads, they

may be related to a uniformly distributed load acting on a simply supported

beam. Inspection of Figure 3*3 shows that the moment caused by the analogous

load p on the haunch of the arch is very similar to the simple beamCI

moment. The moment at ^5 degrees caused by the analogous load is

M = 0.37(0.41422) P R2

k6

Assume that the simple beam has a length equal to the length of half the

arch, or L = -r . The moment is wL /8. Equating this to the moment in

the arch, it is found that:

8 x 9-37 x O.M^33 P R2

v =

= 0.5 Pm

A similar analogy may be drawn for the analogous load acting on

the crown, with the exception that the length of the simple beam is now

one-third the length of the arch, or -5- . The moment in the arch at

90 degrees due to the analogous load is

M = 0.45(0. 15^7) P R2

0^ 8 x 0.^5 x (O.15V7) PmR2

and w -— .

<f»

- °-5 p„

The above maximax loads occur within approximately two to four

milliseconds after the loading begins to act on the arch. This fact is

evident by inspection of the data in Appendix B. These loads decrease

rapidly since p~(t) becomes practically zero when the pressure wave has

engulfed the arch, and p,(t) also decreases. However, p,(t) and p,(t) are

still acting. It is now assumed that the pressure has completely engulfed

the arch. Since the pressure has engulfed the arch, the all-around uniform

pressure may be subtracted from the expression for p (<P*t). Then

Pr(<P,t)= 0.75 sin

2q>Pm

hi

This radial pressure may be resolved into harmonic components. It is

found that

22m I * Pp^t) = -j-S

/ (0.75 Bin 9) sinqp Rd9 = O.657 Pm** o

2P r * pp (t) = ^ / (0.75 sin 9) sin39 Rd9 = -0.127 Pm

The moment in the haunch at ^5 degrees caused by these two

components is

J^ - - 0.023(0.637Pm)R2

+ 0.088(0.127Pm)R2

= -0.026 PmR2

The haunch analogous load is

0.026P R2

».<*> " o.fa"°'06 p

m («»

The components p,(t) and p,(t) act for the total duration of the

surface blast pressure. Reference (2) states that the duration of the

positive phase of the side-on overpressure may be considered as

t± = o.Uo (-^9-) v ^5(20)

*m

•where W veapon size in MT

t. = the effective duration in seconds of a triangular pulsehaving the same impulse as the actual shock.

The total duration of p,(t) and p,(t) is considered to be t.

.

The time required for p2(t) to become practically zero is found

by using Equations (15) and (16), with the angle 9=0.

kd

This time is defined as:

t! » t + t = duration of initial triangular impulsed a r

mR f B sin p +

Ra ^

C U 2C ' r * t

fu2

The time t! defines t for the load p„(t) = 0.06 pd. am

4.3.2 120 Degree Arches

Approximations of the blast load acting on a 120 degree arch may

be derived in the 6ame manner as that used in finding the approximations

for the 180 degree arch, as described in Section k.3.1. It is necessary to

modify Equation (3) by substituting (cp + 30) for <$, since in Equation (3)

cp is measured from a horizontal radial line joining the arch abutment with

the center of the circular segment. Equation (3) becomes

Pr(<P,t) = [0.25 + 0.75 sin

2(cp + 30)] Pm

= t- + i*r sin cp + * n? sinCp coscp + *r cos cp Pm

Rdcp

Then the maximum possible values of p (t) are

«« 2*2P r— , fex r —~

~

/

Maximum p,(t) = ^-2 / \ T + ir sin + ^-^ sin<p cosqp

+ ^r cos 9 ) sinjcp Rdcp

= -0.505 Pm

2p r sTMaximum px(t) = -r-^ /

"( r + tr sin *P + o sinCP cosCp

5 I*R ^2* 4 10 a^flR ^2** 9

JL cos2cp ) sin 2 q

2

= -0.417 P.m

The harmonic components p (t) sin —~- would produce moment at the

haunch and crown as shown below, if the blast load is considered to act

statically. The moment coefficients are obtained from Table IV, for cp = 50

and 60 degrees.

M^ = - 0.009(l.l03Pm)R

2+ 0.125(0.505Pm

)R2

+ 0.050(0.4l7Pm)R

2

= - 0.094 P R2

m

Mc

= O.Ol4(l.l03Pm)R

2+ 0.051(0 Al7Pm)R

2

= 0.057 P R2

Using the moment coefficient for <P = 30 and 60 degrees in

Table IV, the analogous loads are found to be

.2

Maximum p (t) =0.094 P R

m

0.1547R2

0.61 Pmm

0.057 PmR

Maximum p (t) = *•

80.06556R

= O.56 P

50

m

These values are plotted in Figure 4.5 so that they may be

compared vith the' actual analogous loads found by harmonic analysis, as

given in Appendix B. Since they are conservative, the following arbitrary

reductions are made:

Maximum p (t) = 0.^5 P (21)

Maximum p (t) = 0.4l P (22)a HI

These equations are plotted in Figure k. 5.

These loads may be related to a simple beam load of 0.5 P by the

same method as that described in section U.3.1 for a 180 degree arch.

After the pressure wave has engulfed the arch, there is a uniform

all -around pressure acting which must be subtracted from the radial

pressure in order to determine the loading which causes flexural action.

This uniform pressure is

p = (0.25 + 0.75 sin2

50) P = A- P•ru x y " ^ ' m 16 m

The remaining radial pressure which is to be resolved harmonically is

Pr (<P) - Pu= ( 5 + ^ sin

2cp + 2_1 sinq> cosq> + ^ cos

2cp)P

m- ^ P

m

= ( - -4 + -fr sin 9 + * ^ sincp cos<P + ^ cos <p) P

Then

51

P-^t) » ^ / 3 (_ 3 + 9 sin2cp + ^__2 8ln<p cos<p + ^ cos

2q)) sin|<P Rd<p

3

0.5^6 Pm

2P2*

P5(t) = --£ / ,

5 (- ^ + -^ Bin29 + 2_^2 Sincp cosq) + ^ cos

2<p)sin|<P Rdqp

—XR o

= -0.020 Pm

These loads vould produce the following moment at the haunch and

crown of the arch if they acted statically:

^ = - O.OC^CO^^PjR2 + 0.050(0.020Pm)R

2

= -0.006 P R2

m

M = 0.014(0. 5k6? )R2

+ 0.05l(0.02OP )R2

= 0.009 P R2

The analogous antisymmetrical load for the haunch is

0.006 P R2

mPa

(t) =

0.155R

0.0U Pm (23)

The analogous load for the crown is

0.009P R2

».W- HO.O6556R

O.lU Pm

{2k)

I

52

The time t that defines the analogous loads in Equations (23)

and (2k) is the time required for the pressure wave to engulf the arch and

rise to its maximum value at the abutments. This time is expressed by-

Equations (15) and (l6). The angle 9 in both of these equations is thirty

degrees

.

^.5.3 Summary

The blast load acting on the arch may be defined by numerically

evaluating at specific times the coefficient p (t) for the distributed load

p (t) sin —7~ . The load-time curve for the analogous loads is obtained

by equating the static effects of the distributed load \ p (t) sin -^- to

those of the analogous loads. This is the procedure employed in Appendix B,

The procedure is extremely tedious, and, in fact, unnecessary, since the

blast load is not accurately known. Therefore, a more convenient method

such as that established in Subsections k.J.l and k.3.2 is desirable.

In these subsections, the blast load acting on the arch has been

idealized to two initially peaked triangular force pulses. For investigat-

ing the effect of the blast load on the haunch of a 180 degree arch, the

peak value of the first pulse is 0.37P> and the peak value of the second

pulse is O.06P . For similar investigation of a 120 degree arch the peak

value of the first pulse is 0.1*5 P , whereas the peak value of the second

pulse is 0.04 P . In each instance the duration of the first pulse is the

time required for the pressure wave to engulf the arch, and the duration

of the second pulse is equal to the effective duration of the blast loading.

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56

Figure ^.1 Geometry of Blast Loading on a 180 Degree Central AngleCircular Segment Arch

57

a.

< a.E

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

.

r- 100 f)Si.

^-200 p1

5i.^,,**,~,

0.1 0.2 0.3 0.4 0.5 0.6

Attw

0.7 0.8 0.9 1.0

Figure k,2 Overpressure as a Function of Time

58

isd 'ajnss9Jdl

D !P Dd

o<£

V .c

o o•—

a. <O -e- a>0>

a>

o> o>c a>

< QO00

OCJ o

isd( 9jnssaJd |D;po^

-8

3olf\H

o «)

IO*g

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a>

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TJ0)

oo „ a> a>

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-e-a>

 Oc O 3< CJ^™

^^^ §ao

a> S

59

(0

Q- 200

0>W

0) I

>O

O9Q_

00

0—9 QHHi—f-

Adopted Value =0.37 P.

Maximax Value = 0.46 P„

20 40 60 80 100

Maximum Value Pa(t), psi

wa. 200

#»

a>k_

3</>

(/>

a>k.Q.k.<V 100>Ojc:

o<D

Q_

100

Maximum Value pc (t), psi.9

CODE

c s= iooo C s

=2000 Ph

Pv

V

t 0.25 1.0

1.00 1.0

V f 0.25 0.5

A 1.00 0.5

Figure k.k Peak Values of Analogous Loads: Antisymmetrical p (t) anda

Symmetrical p (t) for a 180 Central Angle Archs

6o

CL

to

a>>O

o

a.

200

100

to

a)

10CO

>O

Da>

0_

200

100

Maximum Value Pa(t), psi.

Adopted Value = 0.41 P.

Maximum Value Ps(t), psi.

CODE

c s= iooo C s

=2000 J&L Vft

• 0.25 1.0

a 1.00 1.0

7 0.25 0.5

A 1.00 0.5

100

Figure 4.5 Peak Values of Analogous Loads: Antisymmetrical P&("t) and

Symmetrical p (t) for a 120 Degree Central Angle Archs

61

CHAPTER V

PASSIVE EARTB PRESSURE LOADING

If there were no soil surrounding the arch, it would deflect into

the dotted configuration shown in Figure 5«1 under the unsymmetrical blast

loading. However, the soil restricts this deflection by passive earth

pressure and the inertia of the soil mass. To evaluate the passive earth

pressure, the segment of the arch on the leeward side from the springing line

to the crown is assumed to act as a retaining wall. The soil volume and

the relative plane of slip are shown in Figure 5«1»

Under actual conditions, the body of soil located on the windward

side of the arch would most certainly be affected by the blast pressure and

would have to move downward and laterally into the arch to load it. The

energy required to accelerate this mass would lower the energy level in the

blast pressure. In addition, after the surface pressure wave passes the

arch, there would be a vertical pressure of varying intensity acting downward

on the soil body on the leeward side of the arch. This would increase the

shear strength of the soil, the passive earth pressure, and the inertia

force. Evaluation of these later effects would be extremely difficult and

is omitted in this study. This results in a conservative approach.

5.1 Unit Passive Earth Pressure

Rankine's formula for the total horizontal passive earth pressure

force P acting on the vertical plane AV with a height h in Figure 5*1 is

P = iwh^

1 + sin cp'

where K = •? 7—^7 , <P' being the angle of internal* friction of the soil

and w = unit weight of the soil.

62

The unit horizontal pressure along the plane AV in Figure 5*1

varies directly as y varies, y being the distance belov the ground surface.

This pressure then may be expressed as

pp

« wyK (25)

5.1.1 l80 Degree Arches

For a semicircular arch with no cover on the crovn and <p' = 30

degrees, since y = R(l - sincp), Equation (25) becomes

p - vR(l - sincp) J*"*" °*jj = 5vR(l - sincp) (26)p J. - u.5

The angle 9 is measured from the horizontal radial line which joins the

abutment with the center of the circular segment.

In addition to the horizontal passive pressure, it is necessary

to consider the weight of the soil above the arch rib. This soil exerts a

unit pressure

py= wR(l - sinq>) (27)

Using Equations (l), (26), and (27), the resultant radial earth pressure is

Pr (cP) = i (Pv + P

p) + i (P

p- Pv ) cos 29

- i [wR (1 - sinqp) + 3wR (l - sin<p) ]

+ 2 b*R (1 - sin9) - wR (1 - sinq>)3 cos 29

= 2wR (1 - sin9) + wR (1 - sin9) cos 29 (28)

In Figure 5.1, the angle 0, which defines the inclination of the

plane of shear AS with the vertical plane AV, is

6 = i(90°+9')> <P' being the angle of internal frictionof the soil

= i (90° + 30°) = 60° for 9' = 30°

65

From the geometry of Figure 5.1, the total weight of the soil

associated vith the passive earth pressure is

.2

Ws

= § R(R tan 60) + R2

- ^p- 1 v = 1.08 wR2

(29)

Equation (28) is for an arch with no cover on the crovn. If

there is a depth of cover H on the crown, a uniformly distributed hori-

zontal pressure 3*H , as derived from Equation (25), and a uniformly dis-

tributed vertical pressure wH , must be added to the radial pressure actingc

on the arch. For an arch with a central angle of 180 degrees, the resulting

radial pressure due to H , as obtained by Equation (l), is

Pr (cP) - 2 (Pv + P

p) + i (P

p- Pv ) cos 29

= i (wH + JwH ) + i (3vH - wH ) cos 29c c c c

= 2 wH + wH cos 2q> (30)c c x

The total radial pressure on a 180 degree arch with a depth of cover HC

is found by adding the pressures derived from Equations (28) and (30)

•

p (9) = 2wR (1 - sincp) + wR (1 - sin<p) cos 29

+ 2wH + wH cos 29c c

5.1.2 120 Degree Arches

Equation (28) is valid only for a semi-circular arch with the

angle 9 measured from the horizontal radial line. For an arch with no

cover on the crown and a central angle of 120 degrees, Equation (28) is

modified by substituting (9 + 30°) for 9. This yields

Pr(9)= 2wR [l - sin (9 + 30) ] + wR [l - sin(9 + 30)] cos 2(9 + 30)

= wR ( 2 - \ ? sin9 - £ cos9 - ^ cos 39

" "f sin 29 + i cos 29) (31)

6h

If there is a depth of cover H on the crown of the 120 degree arch, the

resulting radial pressure due solely to H is

Pr(9)= i (ph

+ Py) + i (ph- Pv) cos 29

= i (5wH + \fi ) + i (Jvfl - vH ) cos 2(9 + 50)c c c c

= 2 wH + i wH cos 29 - -^ wH sin 29 (32)C C c c

The angle 9 is measured from the radial line Joining the abutment vith the

center.

The total unit radial pressure of a 120 degree arch having a

depth of cover H on the crown is found by adding Equations (31) and (32).

The result is

p (9) = wR ( 2 - \ r sin9 - f cos9 - i cos 39

- -g sin 29 + 5 cos 29) + 2 wH

+ £ wH cos 29 - -i- wfl sin 29 (33)

The total weight of the soil associated with this force is

.2

Ws

= [i (R + Hc )

2tan 6Q - ^-

Jw

= 0.3^2 wR2

+ 1.732 wRH + 0.866 wH2

c c

5.2 Distributed Earth Pressure Load

In considering the unsymmetrical blast loading on the arch rib,

it is advantageous to resolve this loading into harmonic components since

Zi \ n«9p (t) sin —z~- are easier to treat

analytically than the actual irregular blast load. This advantage does not

exist for the passive earth pressure loading, since original analytical

65

expressions are available which may be studied in determining the effect

of the passive earth pressure on the arch. However, resolution of the

passive earth pressure into harmonic loadings does facilitate visualization

of the fact that the earth pressure may be thought of as a distributed

Zn«Pp sin -— in the arch rib. In addition, a desirable uni-

formity of approach is provided for the entire problem.

Equation (28) expresses the unit radial pressure acting on a

180 degree arch with zero depth of cover on the crown. The total force

acting on the arch rib may be resolved into harmonic components by

Equation (9)

•

p (9) = 2wR(l - sinqp) + *R(l - sin?) cos 29

*» - -hi *r<« 8in¥ Rd*

n

Pl=ITr /

2^2vR^ 1 - sinCp) + wR(l - sinq>) cos 2q>]sin |2 Rdqp

= 0.9W wR

=

a

2 [2 Pr(<P) sin2*96

Rd9 = 0.509 wR

2 [2* R Jo

PP(«P) sin 2*9e

Rd<p 1.180 wR

Equation (33) expresses the radial pressure acting on a 120 degree

arch having a depth of cover H on the crovn. The total radial force re-

sulting from this pressure may be resolved into harmonic loading components

as follows:

66

p (9) = vR (2 - 2—-2. sin9 - \ coscp - - cos 39

- «^ sin 2Cp + | cos 29)

+ vHc(2 + | cos 29 - -|- sin 29)

*n - FR /^ Sin *? Rd cos 39

- -^ sin 29 + | cos 293 sin |9 Rd9

*r

+ vH /5 (2 + | cos 29 - -^ sin 29) sin |9 Rd9

o

= 0.111 wR + 0.864 wHc

p2= ~-

J3pr(9)

sin 39 Rd9 = 0.l8l vR + O.987 vHc

p_ = ~-/

5 p (9) sin §9 Rd9= 0.198 vR + 0.557 ^-> |*R J o

r d c

3

5.3 Analogous Earth Pressure Load

In order that the effect of the unsymmetrical blast load on the

arch rib may be determined conveniently, the blast load is represented as

an analogous uniformly distributed radial load acting in compression on the

•windward side of the arch and in tension on the leeward side. This greatly

facilitates the study of the dynamic effects of this load since the problem

67

of dealing with a time-dependent unsymmetrical load is reduced to that of

dealing with a time-dependent uniform load. The problem may be simplified

further by letting this uniform load define the parameters of the dynamic

load acting on the single-degree-of-freedom system.

It is logical to extend the concept of analogous loads to the

unsymmetrical passive earth pressure acting on the arch. In the first place,

a desirable uniformity of approach to the entire problem is obtained. In

addition, by defining the earth pressure load as an antisymmetrical radial

load acting in compression on the leeward side of the arch and in tension

on the windward side, the value of this pressure can be considered as inte-

gral part of the resistance of the arch rib, since the direction of the

earth pressure is directly opposite to that of the blast pressure.

Furthermore, this resistance provided to the arch by the soil may be used

to define the resistance of the single-degree-of-freedom system.

p sin -~ defined in Section 5.2 ,

coupled with the moment coefficients from Table IV, are used to define the

analogous loads for ISO and 120 degree arches. The resulting analogous loads

define the maximum, or yield, resistance given the arch rib by the soil,

since the passive earth pressure is evaluated at its peak value. The

resistance is evaluated as a function of the moment in the arch rib at a

point located by Qjk, the same point used to define the analogous load for

the blast pressure.

For a 180 degree arch with no cover on the crown, the resisting

moment due to the soil is

M= -0.023(0.9^8 wR)R2

+ 0.335(0.509 wR)R2

+ 0.088(1.180 wR)R2

» 0.252 wR^

68

The analogous load Is

0.252 wR^

a0.43A R

2

= 0.611 wR (3k)

For a 120 degree arch with a depth of cover H on the crown, the

resisting moment due to the soil pressure is

M - -0.009(0.111wR + 0.861fw5 )R2

+ 0. 125(0. l8lwR + O.987WH )R2

+ 0.050(0. 198wR + 0.557WHJR2

= 0.03^wR5 + O.lMfwH R2

The analogous load is

0.03ivwR5 + O.lWfwH R2

p = g.

0.155R

0.2l9wR + 0.929WH (35)

69

t V S

»

1 Hc

/ // /////

3 a'

r ""*

RAN Pp\

/y /

//

h /

Figure 5.1 Passive Earth Pressure on a 180 Degree Central Angle Arch

70

CHAPTER VI

ANALYSIS OF THE RESPONSE OF THE ARCH TO THE DYNAMIC LOAD

The arch rib is idealized as a single-degree-of-freedom system,

and the flexural resistance and mass of the arch rib per se are ignored. A

method of dynamic analysis developed by Dr. N. M. Newmark, References (8)

and (9)> is used to predict the response of the arch rib to the given

pressure pulse.

For the arches investigated in this thesis, it is found that the

response of each is in the plastic, or yielded, part of the assumed elasto-

plastic resistance diagram for the soil. Had the response been less than

the deflection required to establish the passive earth pressure, the dif-

ferential equation of motion stated in the introductory chapter of this study

would have been used to find the maximum response.

6.1 Method of Analysis

For an initially peaked triangular force pulse acting on a single-

degree-of-freedom system having an elasto-plastic resistance function,

Figure 6.1, Dr. Newmark presented in Reference (8) the following approximate

equation to predict the maximum response of the structure:

^ . _£_ Vi^TT + i-li_ ( 56)% nt

l 1 + 0.7 T~

where Pm** = maximum value of the pulse loading

a yield resistance of the structural element

u = ductility ratio = maximum deflection/yield deflection

= x /xm' e

71

mx

T = natural period of vibration of the structure = 2« \j-

—

t.. = duration of the loading

Another technique developed by Dr. Newmark, Reference (9)> can

be used to define the dynamic behavior of the structure for other pulse

shapes which may be represented by several initially peaked triangular force

pulses, Figure 6.-1. The relationship between the peak pressure and the

maximum response of the structure is defined approximately as

****[%* 8-

*

where F, , F2

= damage pressure levels for initially peakedtriangular force pulses of duration t,/T andtp/T, respectively, acting separately

c., c2

= ratio of the peak pressure of the first andand second replacement triangles to the totalpeak pressure

fl'

f2

= cl^ pn/qe^

and M 1^^' respectively, the

maximum value of the individual force pulses

To apply the above described methods of dynamic analysis to the

situation at hand it is assumed that:

(1) The configuration of the arch rib is defined by the radial

deflection of the haunch at a point located at &/k. The structure is

idealized as a single-degree-of-freedom system.

(2) The blast loading on the arch is considered to be a uniformly

distributed antisymmetrical loading. The unit value of this loading defines

the forcing function acting on the single-degree-of-freedom system.

(3) The shape of the blast loading force pulse consists of twoi

initially peaked triangular force pulses. The duration of the first pulse,

t_, is defined as the time required for the blast wave to engulf the arch

72

rib and rise to its maximum pressure at the leeward abutment, a time equal

to t plus t . The total duration of both pulses is defined as the effective

duration of the blast loading, t..

(k) The resistance provided to the arch rib by the passive earth

pressure is considered to be a uniformly distributed antisymmetrical load.

The unit value of this resistance defines the resistance function of the

single-degree-of-freedom system.

(5) The weight of the soil associated with the passive earth

pressure is uniformly distributed along the arch rib. This unit weight

defines the unit mass of the single-degree-of-freedom system.

(6) The passive earth pressure reaches its maximum value when

the radial deflection of the arch at a point d/k is O.l^R. The resistance

diagram of this pressure is elasto-plastic and defines the resistance func-

tion of the idealized single-degree-of-freedom system.

(7) The period of vibration of the arch is defined as

T = 2* \/mx /q~ , where m results from the unit weight found in (5)> x is

O.l^R, and a is the unit passive earth pressure resistance found in (h) .

The general nature of assumptions (l) through (7) above is shown

in Figure 6.2.

Numerical values of the parameters in Equation (36) are presented

in Table VIII. Using these parameters, the maximum responses of arches

with various radii subjected to 100 and 200 psi overpressures have been

determined and are plotted in Figure 6.3. It i6 assumed that the arch is

buried in soil with a seismic velocity of 2000 fps. The resistance and

mass of the arch rib have not been included, and soil arching effects are

not considered.

73

6.2 Sample Calculation

A sample calculation of the response of an arch under an assumed

set of conditions follows.

Let: C = 2000 fps W = 1 MT P = 100 psi U = 3000 fpss m

» 2 ft/ms = 3 ft/ms

w 120* pcf R 200 in. 180 degrees- 16.67 ft.

tr =it

tH =c

The parameters are:

pmax= °*57 Pm

= ° -57(10)) = 37 psi

p'.

= 0.06 P = 0.06(100) = 6 psi

R - R sin(9 - P) R - R sin?Zl

=C U 2Cs s

H^2^

l6.67[l - sin(0 - arctan |) ] l6 6_

=' 6

+ ~t~

= 20 ms

t . o.k (M)2/5

w 1/5

m

= UOO ms

T - 0.0076 Vi = 0.0076 n/200 = 0.107 sec.

107 ms

x = 0.001 R - 0.001(200) = 0.2 in.

% „ 0.6H vR = °'y 200 - 8.5 psi

7*

The force pulse is shown in Figure 6.2. Equations (36) and (57)

must be evaluated by trial and error. Reference (9), Chart I, may also be

used to evaluate Equation (36).

Assume that u = 12. For the first initially peaked triangular

force pulse having a duration of 20 ms, the damage pressure level necessary

for a maximum deflection of x = 12x ism e

— -xt^^' 1 +1 + 0.7(1/^)

1 m 1

-sS**- 1 +1 + o.7( w/20) - 8-*-'i

For the second initially peaked triangular force pulse having a

duration of 400 ms, the damage pressure level necessary for a m«.yimnm de-

flection of x = 12x ism e

1 m1

-qf=î ViirTT +1 + 0.7(1^/1*00) - ^ ' F

2

The small increase in the peak value of the second pulse due to

the extension of the pressure line from t = 20 ms to t = ms, an increase

Of)of "zrrr (6) = 0.3 psi, is ignored.

The actual peak pressure in the first force pulse is 31 psi; the

actual peak pressure in the second force pulse is 6 psi; and the actual

ratio of the total peak force to the resistance is

^max 37 , „„

% 8.5

Equation (37) may now be evaluated to determine if the assumed

value of u 12 is satisfactory.

75

L F1

F2

5 = 1.0

where c. >Js • 0.84 c~ = T*7 = O' 1^

.,,_„ / 0.84 0.l6 N . „ . n

• • **«37 ( -gjp + j^-) = 1.02 = - 1.0

Therefore the assumed value of 12 for u is satisfactory, and the

maximum deflection is

Xm

s ^ Xe

o 12(0 .2)

s 2.4 inches

6.3 Discussion of Results

Figure 6.3 contains the quantitative results of the analyses of

the response of various sized arches. The results may be qualitatively

summarized as follovs:

(l) For any given overpressure level, the response of

the arch increases as the radius of the arch decreases.

This is partially explained by considering how the load

on the arch and the resistance given to the arch by the

earth pressure are defined. The blast pressure is de-

fined as a function of the peak overpressure only. The

value of the analogous, uniformly-distributed, anti-

symmetrical load is not a function of the radius of the

arch. The earth pressure is defined as a function of

the radius of the arch and varies directly as the radius

varies. Consequently, the response of the arch to any

76

given overpressure varies indirectly as the radius

of the arch varies.

Other factors which probably influence the re-

sponse of the arch have not been studied in this thesis.

One of these is the value of the resistance of the arch

per se. For arches of smaller radii, this resistance

may be relatively large compared to the resistance given

the arch by the earth pressure.

(2) For equal radius arches, the response of the arch

to any given overpressure level increases as the central

angle of the arch decreases.

(5) The effect of the second triangular force pulse is

very important since it has a relatively long duration.

Several analyses of the response of a 120 degree arch to

a 100 psi overpressure -were made, in which the effect of

this second force pulse was not considered. The response

of a 200 inch radius arch was found to be approximately

three inches, compared to a value of sixteen inches when

the effect of the second pulse is included.

(4) Increasing the depth of cover on the crown reduces

the effect of the blast load on the arch. The effect of

the depth of burial of the arch may be related conveniently

to the ratio of the depth of burial of the arch abutment to

one-half of the arch span. When this ratio is one, for

arches having radii of 200 inches or more and central angles

greater than 120 degrees, the flexural action caused by an

overpressure of 100 psi resulting from alHO weapon may

77

be ignored. For a 180 degree arch, this ratio is one

when there is no cover on the crovn, and the deflec-

tion is approximately tvo inches. For a 120 degree

arch, this ratio is approximately 0.9 "when there is

a depth of cover of 0.3R on the crovn, and the deflec-

tion is approximately one inch.

78

TABLE VIII

PARAMETERS FOR DYNAMIC ANALYSIS OF SOIL RESISTANCE

Parameter 180 Degree Arch 120 Degree Arch 120 Degree ArchH =c

H =c

H = 0.5Rc "

pmbeoc

0. 57Pa 0.^5PmmO.to

P' *0.06pm

O.O^Pm

O.C&Pm

*it + ta r

t + ta r *a

+ tr

*2 t.1 \ \

«e0.6llwR 0.219vR 0.h97yiR

Xe

O.OOIR O.OOIR O.OOIR

m 1.08vR2

* Rg0.3^2wR

20.9UOVR

2

1 *Rg

T 0.0076 >/r 0.0087 >/r O.OO96 >/r

79

q(x)

Mass -^ p(t)

'max.

(a) Single-Degree-of-FreedomSystem

(b) Initially PeakedTriangular Force Pulse

q(x)

1 1

'm

(c) Elasto-Plastic BilinearResistance

max

(d) Two Initially PeakedTriangular Force Pulses

Figure 6.1 Idealized Single-Degree-of-Freedom System, Force Pulses andElasto-Plastic Bilinear Resistance

80

Blast Loading Soil Resistance

q(x)

%

0.001 R m

Soil Resistance Diagram

Figure 6.2 Idealized Loading and Resistance Functions of an UndergroundArch Rib Subjected to Blast Pressure

81

40.0

<U|'d' 4.0

200

100 psi. , 9 = 120,

200 psi. , 9 = 180,

200psi., 0=120,

100 psi., 9 - 180,

100 psi., 9 =120,

Hc

=

Hc=0

HC =0.3R~

Hc =—

HC = 0.3R

400 600 800

Arch Radius , inches

1000 1200

Figure 6.3 Haunch Deflection vs Arch Radius

V » 1 MT C = 2000 fps8 *t

P as noted H as notedc

Q as noted

82

CHAPTER VII

QUALITATIVE DESIGN METHOD FOR ARCH RIBS

The problem of the structural design of the arch rib has not been

studied. However, the following qualitative design procedure, based on the

methods used in this thesis, is suggested:

(1) Idealize the arch as a single-degree-of-freedom system, the

configuration of which is defined by the radial deflection of the arch at

e/k.

(2) Define the fundamental period of vibration of the structure

by the mass and resistance of the soil, and the deflection of the arch at

6/k necessary to establish the full passive earth pressure.

(5) Define the forcing function as consisting of two initially-

peaked triangular force pulses, the first pulse having a duration of t -ft,& r

the second having a duration of t . . These forcing functions are uniformly-

distributed, antisymmetrical, radial loads analogous to the actual blast

loading.

(k) Assume a ductility factor u. The end objective is to

design the arch rib, not to determine the response of the soil surrounding

the arch. However, since the arch cannot deflect without the soil ac-

companying it, this ductility factor also defines the response of the soil.

The assumed value of u should be larger than that desired for the actual

arch rib.

(5) Evaluate the damage pressure level equation to determine

the total resistance required of the combined arch rib-soil system.

(6) Subtract the soil resistance from the total required

resistance and design the arch rib with a resistance equal to the remainder.

The usually recommended allowable stresses for dynamic design are to be used.

85

(7) Determine the yield deflection at d/k of the arch sectioni

designed in (6). The yield deflection must be stated in the same numerical

terms as those used in (2) for finding the period of vibration of the

structure

.

(8) Combine the resistance of the soil and arch rib into an

elasto-plastic resistance diagram. This is accomplished by making the area

of the idealized bilinear resistance diagram equal to the area of the

actual trilinear resistance diagram -when both diagrams extend to a numerical-

ly specified deflection. The ratio x /x is assumed; the initial slope of

the elastic part of the actual diagram is maintained; the value of a is

adjusted to make the areas equal. Determine a new period of vibration for

the combined arch-rib soil system, including the resistance and mass of the

arch section designed in ( 6)

.

(9) Using the elasto-plastic resistance function found in (8),

evaluate the damage pressure level equation to determine the maximum

response of the system. The resulting ratio of the nrnlrnnm deflection to

the yield deflection of the system is found and compared with the ratio

x /x used in combining the soil and arch resistances. If these ratios

are equal, the arch section designed in (6) is satisfactory. If they are

not equal, the arch section must be redesigned and (7) through (9) repeated.

The anticipated results of the design procedure outlined above

are discussed below without proof.

There are two principle modes of response of an arch to blast

loading, Reference (10). One of these is symmetrical, corresponding to

the compression in the arch, and the other is antisymmetrical, corresponding

to the flexural deformation of the arch. Consider an underground arch and

assume that the arch rib has been designed for the static load of the soil

cover and the compression load of the blast force. For the arch sizes,

overpressures, and veapon yield studied in this thesis, it is reasonable

to expect that such arches would have a yield deflection on the haunch of

at least one-half inch. For 100 psi overpressure, the maximum deflection

expected of the haunch, considering the resistance of the soil only, is

two inches for 180 degree arches with zero depth of cover on the crown, and

approximately one inch for 120 degree arches with 0.3R cover on the crown.

Therefore the ratio of the maximum deflection on the haunch to the allow-

able elastic deflection of the arch rib does not exceed k:l, even though

the reduction of the maximum deflection due to energy absorbed by elastic

or plastic deformation of the arch rib is disregarded.

For moderately ductile structures, the above ratio, ductility

factor u, ranges from 10 to 30, Reference (8). For designing flexural

members, including reinforced concrete members, the ductility factor usually

is assumed to be 20.

It is therefore concluded that if the arch rib is designed so that

it has a ductility factor of 20 for flexural action, and is covered as de-

scribed above, no increase in the required resistance for static and blast

compression load is necessary to accommodate the antisymmetrical mode of

response for an overpressure of 100 psi.

85

CHAPTER VIII

CONCLUSIONS AND RECOMMENDATIONS

8.1 Conclusions

The following conclusions are drawn from the results obtained

and the discussions thereof in the preceding chapters.

(1) Harmonic analysis provides a convenient means of analyzing

unsymmetrical loads acting on arches.

(2) The peak values of dynamic loads and their load-time

curves may be determined by harmonically analyzing the load at suitable

time intervals.

(3) The effects of the unsymmetrical blast load acting on the

arch may be studied conveniently through the use of analogous loads. When

studying the effect of the blast load on the haunch, the analogous load

employed is a uniformly distributed radial load which acts in compression

on the windward side of the arch and in tension on the leeward side. The

peak value of this load is approximately forty percent of the peak value

of the overpressure. When studying the effect of the blast load on the

crown, the analogous load employed is a uniformly distributed radial load

which acts in compression on the central third of the arch and in tension

on the outer thirds. The peak value of this load is approximately forty

percent of the maximum overpressure.

(k) The effect of the unsymmetrical passive earth pressure load

on the arch also may be studied through the use of an analogous load. This

analogous load is a uniformly distributed radial load acting in compression

on the leeward side of the arch and in tension on the windward side. It

directly opposes the blast pressure load. Its value varies directly with

the radius of the arch.

86

(5) The peak value of the analogous load employed in studying

the blast effects on the haunch of the arch increases as the seismic

velocity of the soil increases. The peak value of the analogous load used

in studying the crown of the arch decreases as the seismic velocity of the

soil increases.

(6) The relative severity of the dynamic load varies inversely

with the ratio of the horizontal to vertical pressure in the soil. It is

to be noted, hovever, that as this ratio increases, there may be a tendency

for the crown of the arch to move outward.

(7) The relative rise time of the blast pressure to its peak

value has little effect on the peak value of the analogous loads.

(8) The passive earth pressure contributes materially to the re-

sistance of the arch. An overpressure of 100 pounds per square inch from a

one megaton weapon causes a maximum response of approximately two inches

in the 180 degree arches studied. Subjected to the same overpressure, the

response of a 120 degree arch with a depth of cover on the crown equal to

three-tenths of the radius is approximately one inch. This response is

considered negligible.

(9) The passive earth pressure developed by the arch decreases as

the central angle of the arch decreases. It is doubtful if arches with

central angles of much less than 120 degrees would develop significant

passive earth pressure under blast loading.

(10) The rigidity of an arch interferes with the deflection of

the arch required to develop the passive earth pressure. This suggests that

flexible arches have certain advantages in blast resistant construction.

(11) The method of approach presented in this thesis may be

extended to consider soils with different angles of internal friction,

87

different unit weights, and varying vater table levels, as veil as to

cohesive soils, provided the cohesion of the soil can be stated in terms

of an equivalent angle of internal friction.

8.2 Recommendstions

It is recommended that the studies made in this thesis be

extended to:

(1) Consider how the blast loading en the arch is affected by

the soil on the windward side of the arch.

(2) Consider how the passive earth pressure loading on the arch

is affected by the increased shear strength of the soil due to the vertical

blast pressure acting on the leeward side of the arch.

(3) Compare the arch behavior predicted in this thesis with

available test data.

(k) Develop a quantitative procedure for design of the arch rib

under impulsive loading which includes the effect of the passive earth

pressure

.

88

APPENDIX A

BIBLIOGRAPHY

1. Anderson, F. A., "Blast Phenomena from a Nuclear Burst, " Journal ofthe Structural Division, Proceedings of the American Society of CivilEngineers, Paper 1856, St 7, November 195$

•

2. "Protective Construction Review Guide, " issued by Office of Secretaryof Defense, January 1959. Prepared under contract SD-52 by N. M.

Nevmark, R. J'. Hansen and others. Classified Secret Security Information.

5» Huntington, W. C, Earth Pressures and Retaining Walls, John Wileyand Sons, Inc., New York, 1957.

h. Seely, F. B., and Smith, J. 0., Advanced Mechanics of Materials,Second Edition, John Wiley and Sons, Inc., New York, October 1957.

5. Hendry, A. W., and Jaeger, L. G., The Analysis of Grid Frameworks andRelated Structures, Chatto and Windus, London, 1958; Prentice-Hall,Inc., New York, 1959.

6. Brode, H. L., "Numerical Solutions of Spherical Blast Waves," Journalof Applied Physics, v. 26, No. 6, pp. 766-775, June, 1955.

7. The Effects of Nuclear Weapons, Samuel Gladstone, Editor, United StatesDepartment of Defense, June, 1957*

8. Nevmark, N. M., "An Engineering Approach to Blast Resistant Design,"Transactions of the American Society of Civil Engineers, v. 121,Paper No. 2786, pp. 1*5-64, 1956.

9. Melin, J. W., and S. Sutcliffe, "Development of Procedures for RapidComputation of Dynamic Structural Response, " Civil Engineering Studies,Structural Research Series No. 171* University of Illinois, Urbana,Illinois, January, 1959*

10. Merritt, J. L. and N. M. Nevmark, "Design of Underground Structures toResist Nuclear Blast, " Civil Engineering Studies, Structural ResearchSeries No. 1**9, University of Illinois, Urbana, Illinois, April, 195^.

89

APPENDIX B

NUMERICAL CALCULATION RESULTS

This Appendix contains the results of the numerical harmonic

analyses made.

In Tables Bl through BIO are tabulated the vertical pressure-time

values of blast pressure on a 180 degree arch with zero depth of cover on

the crown for various combinations of the variables stated in Chapter II.

Since the 120 degree arch has the same radius, this table also provides the

vertical pressure-time values for the blast loading on a 120 degree arch by-

using the values between 30 and 150 degrees.

Tables Bll and B12 contain the values of the coefficients p (t)

for the harmonic components of loading, as found by numerical analysis.

These values were found by using the vertical pressures in Tables Bl through

BIO, applying the coefficients from Table I to convert the vertical pressure

to radial pressure, and then harmonically analyzing these radial pressures.

These tables also contain the values of the analogous loads p (t) and p (t).a s

The following sign convention is applicable to these loads:

(a) p (t) is positive when the load acts inwardly

(b) p (t) is positive when the uniformly distributed loadacts in compression on the windward half of the archand in tension on the leeward side

(c) p (t) is positive when the uniformly distributed loadacts in compression on the central third of the archand in tension on the outer thirds.

Originally it was intended to study only twenty-five foot radius

arches, some with a rise time equal to one-half the transit time and the

remainder with a rise time equal to the transit time. However, after numerous

calculations had been made, an error amounting to a multiple of two was found.

90

In order to vise the previous work it became necessary to decrease the arch

radius to 12.5 feet, and to increase the rise times of some of the arches

to equal the transit time, and of the remainder to twice the transit time.

This accounts for the data shoving t = 2t.. After the error was dis-

covered, only arches with a twenty-five foot radius and t = ^t were

studied. The results are of such a nature that general conclusions may be

drawn without regard to the disparity in the arch sizes.

Figures Bl through B6 are typical graphs of the values of the

coefficient p (t) and the analogous loads p (t) and p (t).XX fit s

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VITA

Lieutenant Commander Worthen Allen Walls, Civil Engineer Corps,

U. S. Navy, vas born near Little Rock, Arkansas, February 17, 1922.

Following his graduation from Stephens High School, Stephens, Arkansas, in

19^*0, he attended Magnolia A and M College, Magnolia, Arkansas, for two

years.

He entered the Naval Air Corps, U. S. Navy, in June of 19^2 and

after flight training vas designated a Naval Aviator and commissioned as an

Ensign, U. S. Naval Reserve, in 19^3 • Wartime assignments vere primarily

with the Pacific Fleet and included duty on the West Coast, the Aleutians,

Hawaii, Saipan, Peleliu, Okinawa, Iwo Jima, and then Japan in the early

days of the occupation.

After being released to inactive duty in 19k6, he entered the

University of Arkansas, Fayetteville, Arkansas, from which he graduated,

with honors, in 19kS with the degree of Bachelor of Science in Civil

Engineering. He was then employed as an engineer by Phillips Petroleum

Company, Okmulgee, Oklahoma, until April, 19**9 •

In April, 19^9> be returned to active duty and was commissioned

as a Lieutenant (junior grade), Civil Engineer Corps, U. S. Navy. His first

tour of duty was as Assistant Public Works Officer and Assistant Officer in

Charge of Construction, Bureau of Yards and Docks Contracts, U. S. Fleet

Activities, Yokosula, Japan. This duty was followed in 1951 by duty in

the Public Works Department, U. S. Naval Air Station, Corpus Christi, Texas,

until 1955.

In May, 1953> he entered the University of Illinois, for graduate

work in Civil Engineering under a program of postgraduate education

126

sponsored by the U. S. Navy and received the degree of Master of Science

in Civil Engineering in 195^.

Thereafter he vas assigned as Assistant Resident Officer in

Charge of Construction, Bureau of Yards and Docks Contracts, Seville,

Spain, for three years in connection with the Spanish Bases Program. From

1957 to 1958 he served as Public Works Officer and Officer in Charge of

Construction, U. S. Naval Ammunition Depot, Concord, California. He

returned to the University of Illinois in 1958 "to continue his advanced

study in Civil Engineering under Navy sponsorship.

He is a member of the American Society of Civil Engineers,

Society of American Military Engineers, and Tau Beta Pi

.

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