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1965

Analysis of flat slabs with various edge supports Analysis of flat slabs with various edge supports

Wu-Mo Chern

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Recommended Citation Recommended Citation Chern, Wu-Mo, "Analysis of flat slabs with various edge supports" (1965). Masters Theses. 5716. https://scholarsmine.mst.edu/masters_theses/5716

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ANALYSIS OF FLAT SLABS WITH

VARIOUS EDGE SUPPORTS

BY

WU-MO CHERN

A

THESIS

submitted to the faculty of the

UNIVERSITY OF MISSOURI AT ROLLA

in partial fulfillment of the work required for the

Degree of

MASTER OF SCIENCE IN CIVIL ENGINEERING

Rolla, Missouri

1965

Approved by

Gli:;vt~~ til"~~# · ··

~~a-f. k (Advisor) rJ j

t{), fk.~~

ABSTRACT

This study presents the solution of flat slabs with various

edge conditions including T - Beam edge supports. The equations

formulated are based on the finite difference method and solved

using an electronic computer.

The results indicate that the solution of these slabs by the

finite difference approximation is a practical method. In addition,

this method is more flexible and yields moment values that are

considerably more accurate than those obtained using the ACI Code.

11

iii

ACKNOWLEDGEMENT

The author wishes to express his sincere appreciation to

Dr. Joseph Harold Senne, Jr., Chairman, Civil Engineering Depart­

ment, for his assistance in suggestion of this study.

The author also would like to take this time to thank Prof.

Ralph Edward Lee, Director of the Computer Center, for his

permission to use the facilities of the Center.

iv

TABLE OF CONTENTS

ABSTRACT ..... 11

ACKNOWLEDGEMENT iii

TABLE OF CONTENTS. iv

LIST OF FIGURES vi

NOTATIONS .... ix

1. INTRODUCTION 1

2. REVIEW OF LITERATURE 2

3. DIFFERENTIAL EQUILIBRIUM EQUATIONS. 3

3-1. Beam Action .. 3

3-2. Plate Action . 4

4. FINITE DIFFERENCE EXPRESSION OF DIFFERENTIAL

EQUATIONS ................ . 9

4-1. Finite Difference Interpretation 9

4-2. Finite Difference Expressions of Beam Action 12

4-3. Finite Difference Expressions of Plate Action. 12

5. DEVELOPMENT OF OPERATORS FOR PLATES WITH

VARIOUS EDGE CONDITIONS. 15

5-1. Properties of Operator . 15

5-2. Plate Supported by T-Beam 15

5-3. Plate Supported by Balcony Beam 18

5-4. Plate with Two Edges Supported by Balcony Beams and Two Edges Fixed ......... . 19

5-4. Elimination of Supplemental Points 20

v

5-6. Plate with Freely Supported Edge 21

a. Method 1. 21

b. Method 2. 24

5-7. Plate with Two Adjacent Free Edges 25

5-8. Operator at Free Corner ...... . 28

5-9. Plate with a Simply Supported Edge 30

5-10. Plate with One Edge Simply Supported and One Edge Free ... 31

5-11. Plate with Fixed Edge 34

5-12. Plate with One Edge Free and One Edge Fixed. 36

5-13. Plate with Two Edges Free and Two Edges Supported by Beams ...... . 37

5-14. Plate w i th Notched Free Edges 39

6. ILLUSTRATIVE PROBLEMS 41

Problem 1. 42

Problem 2. 43

Problem 3. 44

7. CONCLUSIONS. ..... 45

BIBLIOGRAPHY .. 46

VITA ..... . 47

APPENDICES 48

Computer Program. 48

APPENDIX 1. 49

APPENDIX 2 .. 51

APPENDIX 3. 53

Fig. 3-1

Fig. 3-2

Fig. 3-3

Fig. 4-1

Fig. 4-2

Fig. 4-3

Fig. 5-l

Fig. 5-2

Fig. 5-3

Fig •. 5-4

Fig. 5-5

Fig. 5-6

Fig. 5-7

Fig. 5-8

Fig. 5-9

LIST OF FIGURES

Flexural analysis of beam action

Element of plate with applied loads

Element of plate with applied moments

Equally spaced network for a finite difference

approximation

Illustration of finite difference approximation

Biharmonic operator for deflection

Reinforced concrete slab beam floor system

Operator for the composite action of the plate

supported by T- beam

Operator for the moment of the plate at free edge

Corner operator for the plate supported by balcony

beam

Plate with two edges fixed and two edges supported

by balcony beams

Plate grid showing supplemental points

Operator for the plate supported by beam

Operator at free edge of the plate

Operator at one increment length along one axis

vi

3

5

6

9

10

14

15

17

18

19

20

21

23

24

from the corner at the free-to-free edge of plate 26

Fig. 5-10 Operator at one increment length from free edge

of plate 27

Fig. 5-11

Fig. 5-12

Fig. 5-13

Fig. 5-14

Fig. 5-15

Fig. 5-16

Fig. 5-17

Fig. 5-18

Fig. 5-19

Fig. 5-20

Fig. 5-21

Fig. 5-22

Operator at one increment length along both

axes from free-free edge of the plate

Operator at free corner of the plate

Biharmonic operator at one increment length

from simply supported edge

Biharmonic operator at one increment length

from both simply supported edges

Operator at two increments from simply

supported edge of the free -to- simply supported

plate

Operator at one increment length from simply

vii

28

29

30

31

32

supported edge of free-to-simply supported plate 32

Operator at one increment length from free edge

of free-to-simply supported plate

Operator at one increment length from edges of

free -to - simply supported plate

Biharmonic operator at one increment length from

fixed edge

Biharmonic operator at one increment length of

fixed-to-fixed plate

Operator at one increment length from free edge

of free-to-fixed plate

Operator on free edge of free-to-fixed plate

33

34

35

35

36

36

Fig. 5-23

Fig. 5-24

Fig. 5-25

Fig. 5-26

Plate with two edges free and two edges supported

by beams

Operator at one increment length from beam

supported edge of plate

Plate with two edges fixed and other edges free

Operator for the notched plate with two edges

fixed and other edges free

viii

37

38

39

40

Symbol

D

E

h

I

L

n

q

q'

t

w

NOTATIONS

Definition

Flexural rigidity of plate

Modulus of elasticity

Increment length in the finite difference net

Moment of inertia of the beam

Length of the plate

Bending moments per unit length of the sections

perpendicular to X axis

Twisting moment per unit length of sections

perpendicular to X axis

Bending moments per unit length of sections

perpendicular to Y axis

2EI/Dh

Uniformly distributed load per unit area

Uniformly distributed load directly on beam

Shearing forces parallel to Z axis per unit length of

section of plate perpendicular to X axis

Shearing forces parallel to Z axis per unit length of

section of plate perpendicular to Y axis

Thickness of plate

Displacement indirection of Z axis

lX

X,Y,Z

<rx,<ry

v

Rectangular coordinates

Poisson1 s ratio

Radii of curvature of middle surface of a plate m XZ

and YZ plane respectively

Normal components of stress parallel to X, Y axis

respectively

Forces perpendicular to Y, X axes transmitted from

plate respectively

Shearing force in the beam

Shearing stress component in rectangular coordinate

X

I. INTRODUCTION

Since the exact analysis of a flat plate can be extremely

difficult, particularly in the case of reinforced concrete slabs with

special edge supports, many approximate methods including both

experimental and theoretical analyses have been used in dealing with

this problem. The finite difference method as applied to the plate

problem is a numerical technique involving a division of the slab

into a grid or network, and the setting up of operators for the formu-

lation of linear equations to obtain the grid point displacements.

In analysis, this yields a more rapid convergence than use of a

double trigonometric expansion.

Boundary conditions play a decisive role in the construction

of operators. The operator obtained from the evaluation of the finite

difference in accordance with the boundary conditions is applied to

the structure and from this process linear equations are set up from

point to point in the network. The accuracy of the results increase

as the plate is divide d into an incr easing number of segments.

This investigation presents the analysis of the plate supported

by T- beams with both symmetrical and nonsymmetrical flanges.

Also, the ope rators are constructed in such a manner that thos e grid

points that extend beyond the limits of the plate are eliminated.

Methods for analyzing flat slabs have been available in ACI

1

lA

code books for many years. However, these methods are limited

to certain width span ratio and boundary conditions. Even if these

conditions are met the very mture of the simplified ACI procedure

may introduce considerable error into the analysis. The ACI

method has been popular because it was simple for the

designer to use. Due to the rapidly expanding use of electronic

computers in structural analysis and design it is believed that

more exact methods such as finite difference will be employed

in the near future.

2. REVIEW OF LITERATURE

In recent years, two developments, relaxation methods

and electronic computers, have made possible the solution of

engineering problems of great complexity.

Southwell's relaxation methods (5), deal with systems of

finite freedom with equations in one independent variable, and

apply this idea to partial differential equations, involving two

independent variables, which satisfy the boundary conditions.

Forsythe and Wasow (8) discussed the finite difference

method of solution to boundary problems by using an equally

spaced network for setting up the difference approximations.

2

3. DIFFERENTIAL EQUILIBRIUM EQUATIONS

3-1. Beam Action

As in the usual analysis, a small element is taken as a free

body from the beam under transverse loading conditions. In Fig.

3-1, moment, shear due to an uniformly distributed load may be

obtained by static equilibrium (7).

,....---~"'·

v v

" n, t-it~

(I.)

,..., '{+JM Vtt/V)

Fig. 3-1. Flexural analysis of beam action

In Fig. 3-1 (a) there is no external load on the small element, the

3

shearing forces on section mn and m 1n1 are equal, and from ~M = 0,

we get:

dM V=­

dx

When a uniform load q is applied to the element, and since

~F y = 0,

then

also

dV q = - dx '

dM = Vdx - qdx·dx/2

4

V = dM/ dx, (neglecting high order differential)

Where

2 M = -EI4

dx

3 V = -EI d w3 and,

dx

d4w q = EI--

dx4

V is the shearing force on the beam

M is the bending moment acting on the beam

q is the uniform load acting on the beam

E is modulus of elasticity

I is the moment inertia of the beam

3-2. Plate Action

Eq. (3-1)

Eq. (3-2)

Eq. (3-3)

Because the fundamentals of slab behavior are well known,

only a brief derivation will be considered. (1)

5

.----~-----.,.---... X

Fig. 3-2. Element of plate with applied loads

When an external load is applied to the differential element

dx•dy, as in figure 3-2, forces summed in the z direction (LF z = 0),

we get the following equation;

aov = --L -q Eq. (3 -4)

ax oy

6

Fig. 3-3. Element of plate with applied moments

Positive resisting moments are shown in Fig. 3-3 (a). For

convenience in analysis, the right hand rule is used and the moment

vector for positive direction is shown. Summing vectors in the X, Y,

and Z direction yield the following equations respectively:

aMY aMxl = Qy, Eq. (3- 5) ay ax

aMx + aMlx = Qx, and Eq. (3 -6) ax ay

aQx + aQY = Eq. (3 -7) ax ay

-q.

The following equations are obtained by applying Hooke's Law

and the equations of equilibrium to the free body of the differential

element:

Twisting Moments

Bending Moments

Shears

Deformation

2 M = - M = D ( 1-IJ. ) a w

xy YX ax• ay 1 Eq. {3-8)

- 1 1 a2w a2w Mx- i-D(Px +IJ. Py)= -D(ax2 +IJ. ay2) Eq. (3-9)

- (1 + 1 - (a 2w + a 2w My-tD- 1-l-)--D - 2 1-l-) Eq. {3-10) p Y p X ay ax2

=_9._ D

Eq. (3-11)

Eq. {3-12)

Eq. (3-13)

7

8

Stresses r-z Qz r::.Z w

- &;.~ (u W + u ) a-x -- -"""""2 --z ~ 1-~ ax ay2

Eq. {3-14)

where

T xy

Ez

Eq. {3-15)

Eq. {3-16)

z is the distance of the point from the middle surface.

Mxy is the twisting moment per unit length of sections of a

plate perpendicular to X axis.

M , M are the bending moments per unit length of the X y

sections of the plate perpendicular to X and Y axes.

Qx, Oy are the shearing forces parallel to Z axis per unit

length of the section of a plate perpendicular to X and

Y axes.

w is the deflection in direction of Z axis

o-x• o- y are the normal components of stress parallel to X

andY axes.

T xy is the shearing stress component in rectangular coordinate5

D is the flexural rigidity of a plate.

~ is Poisson's ratio.

4. FINITE DIFFERENCE EXPRESSION OF

DIFFERENTIAL EQUATIONS

4-l. Finite Difference Interpretation

For the rudimentary equations only the equally spaced grid

9

of Fig. 4-1 will be discussed in explaining the following difference

quotients. Also, for convenience, only the deflection curve projected

in the xz plane needs to be taken into account ( 1) (3 ).

tt

tJ t ltr

111 1 "" r rr

1.1 " l.r

J,~

Fig. 4-1. Equally spaced network for a finite difference

approximation

10

X

Fig. 4-2. Illustration of finite difference approximation

Let the incremental length in the network in Fig. 4-1 be h. Then to

the first order approximation at center point, m is:

( aw) = Wr- Wm

ax mr h

( aw) = wm- wl '

and

a X ml h

( aw) = wr -wl

ax m 2h Eq. (4-1)

Observing that the curvature is the rate of change of slope , one

can first obtain the slope values at the nearest half-grid points, and

11

then operate on these values by Eq. (4-1 ), in terms of the half -grid

length, yielding:

( a2;) Wr- 2\Vm+ Wl ' = 11:1.

ax m

3 Wrr- 2wr + 2W!- Wll (a w) = , and

ax3 2h3 m

4 1 (a w) = (wrr +wn- 4wr- 4w1 + 6wm). ax4 m h4

Similarly,

(ow) Wt -wb

= ay 2h m

a2w wc2wm+wb ' ( ay2) m = h2

3 wtt _ 2wt + 2wb- wbb (a ~) = and

oy3 m 2h3

4 1 (a w) = -- (wtt + wbb - 4wt- 4wb + 6w ). ay4 m h4 m

In like manner it can be proved that

w tr - wbr + wbl - w tl ,

4h2

= w tr + w tl - wbr - wbl - 2w t + 2wb ,

. 2h3

Eq. (4-2)

Eq. (4 -3 )

Eq. (4-4).

Eq. (4-5)

Eq. (4-6)

Eq. (4-7)

Eq. (4- 8 )

Eq. (4- 9 )

Eq. (4-10)

12

3 (~) = 8x·ay2 m

Wtr + Wbr- Wtl- Wbl- 2Wr + 2wl

2h3 , and Eq. (4-11)

4-2. Finite Difference Expressions of Beam Action

By substituting Eqs. (4-2), (4-3), (4-4) into Eqs. {3-1), {3-2),

and (3 -3) respectively, the finite difference expressions of beam

action are obtained:

-EI v =~ (wrr- 2wr + 2wl- wu)' and 2h

EI q = ""h4 (wrr + wn- 4wr- 4wl + 6wm).

4-3. Finite Difference Expressions of Plate Action

Eq. (4-13)

Eq. (4-14)

Eq. (4-15)

After obtaining the plate deformations, the moments, twists,

and shears can be calculated. By substituting the finite difference

expressions in Eq. (4-1) through (4-12) into corresponding Eqs.

(3-8) to (3-16), the moments,shears, twists, and stresses are

obtained in terms of finite difference forms.

Moments

Shears

D Q =--

X 2h3

Q =- D y ~

Twists

M -­xy

Stresses

(f = -X

Deformation

13

=_.9._ D

Eq. (4-16)

From the above equations, an operator can be formed by

placing the coefficients of the various "w" terms in their respective

grid patterns. For example, if the coefficients for "w" in Eq. (4-16)

are arranged in this manner the well-known biharmonic operator

(Fig. 4-3hesults.

14

Fig. 4-3. Biharmonic operator for deflection

15

5. DEVELOPMENT OF OPERATORS FOR PLATES WITH VARIOUS

EDGE CONDITIONS

5-l. Properties of Operator

There are two properties that should be mentioned in this article

First, the algebraic sum of the coefficients of the deflections

appearing in the operator should be zero as long as a symmetrical free

edge condition exists. The symmetrical free edge condition is defined

as the condition symmetrical to the center point, m, of the pattern in

Fig. 4-l.

Second, the axes of the operator may be rotated in the free

edge without changing the deflections.

These two properties are necessary but not sufficient conditions.

5-2. Plate Supported by T -Beam

This type of construction is common in reinforced concrete

slab beam floor systems as shown in Fig. 5-1.

Fig. 5-1. Reinforced concrete slab beam floor system

16

Thomson and Trait (2) found that the force transmitted from the

plate to the beam is:

aMxy 2 2 R = Q - = -D _a_[ a ; + (2 -1-1-) a w] , and X X

ay ax ax ay2

aMxy a 2 a 2w R = Q - = -D- [~ + (2-!-L) -] y y ax ay ay2 ax2

When the plate and beam resist the external force simultaneously

as a rigid body, i. e. when there is composite action and the deflection

of plate is the same as that of the beam, the applied load q' directly

on the beam is equal to the total intensity of load q on the beam and

the force transmitted from the slab.

where

q' = Rx + q

q = EI a4w ay4

Therefore the first boundary condition is:

EI a4w - q' = D _!_ ( a2w + (2-!-L) a2w] ay4 ax ax2 ay2

Eq. (5-l)

The second boundary condition is that the moment on the beam sup-

ported edge is zero.

2 a 2w aw+I-L = 0 ax2 ay 2

Eq. (5-2)

Finite difference expressions of the two boundary conditions

and the corresponding operators Fig. 5-2, 5-3 are shown as

follows:

17

wtt(n)+ ~b (n) + Wt (-4n)+wb (-4n)+wm (6n)+wrr (-1)

+wr (6-2[J.)+wl (2[J.-6) +wu (1 )+wtr ([J.-2)+wbr (tJ.-2)

2 ,·h3 + wtl (2- fJ. )+wbl (2- fJ. )==-q~-

D

where

Eq. (5-3)

Eq. (5-4)

E and I are the modulus of elasticity and moment of inertia,

of the beam beneath the plate respectively, and

n== 2EI

Dh

If there is no beam beneath the plate then the term n vanishes.

+I -/

2.-)1-

n.

Fig. 5-2. Operator for the composite action of the plate supported

by T-Beam

18

)/'

m

I I -2 -2)A I J I I I

)A

Fig. 5-3. Operator for the moment of the plate at free edge

5-3. Plate Supported by Balcony Beam

This type of construction is composed of the section with ha lf

T-beam or the T- beam with unsymmetrical flange, as shown in

Fig. 5-1.

The boundary c onditions on the e dge are

The corresponding operators are the same as that in Art. 5-2.

An additional boundary condition, Mxy = 0, should be considered

at the corner of the beam e dges ; the corner ope rator of which is

shown in Fig. 5-4.

Eq. (5-4)

19

-I I 0 +I I

m.

I 0

+I I

0 -/ I

Fig. 5-4. Corner operator for the plate supported by balcony

beam

5-4. Plate with Two Edges Supported by Balcony Beams and Two

Edges Fixed

The boundary conditions used in Art. 5-3 are still valid.

20

Fig. 5-5. Plate with two edges fixed and two edges supported by

balcony beams

5-5. Elimination of Supplemental Points

Supplemental points are the grid points that e xte nd one to two

increments beyond the boundary of the plate, as shown in Fig. 5-6.

Elimination of the supplemental points is a simplified procedure

in which the number of finite difference equations may be decreas e d

thus greatly simplifying the problem.

The number of equations necessary to construct the new

operator must be one more than the number of supplemental points.

Suppose the number of points extending outside the plate is N, then

the total number of equations composed of the bounda ry conditions

and the deflection equation of the plate must be N + 1. This extra

equation is the basic deflection equation of the operator.

21

In order to avoid any mistake which might occur during the

evaluation of the numerous equations, the property of the operator

in 5-1 is used, which means the algebraic sum of the coefficients of

the deflection in the operator is equal to zero. All operators described

in the articles that follow have the supplemental points removed.

r- .,.- -r- ,-- r- --. 1 I 1 I I I J.--~• I I ___ J 1 I

I ~-- -- -•, 1 I/ SLtr1,/emc"t~ I .,.;,t t--- __ ..J

I ~ ~ _ _ _ _ -: Eti/' 'f pl•f~

I I I I I I L_.J_ -'--.l-..J--1

Fig. 5-6. Plate grid showing supplemental points

5-6. Plate with Freely Supported Edge

a. Method 1.

The operator for the boundary conditions is shown in

Fig. 5-2 and 5-3. From the second boundary condition on points

t, b in Fig. 4-1, we get eq. (5-5) and (5-6). These two equations,

together with the equations in (5-2) and finally the deflection equation

(4-16) for the plate on point m, eliminate the points tr, br, r, and rr.

Eq. (5-5)

Eq. (5-6)

22

+2wl)]=q', Eq. (5-8)

Wtr• Wbr• and Wr can be eliminated from Eq. (5-5), (5-6),

{5-7). Wrr is then obtained by substituting these values into Eq.

(5-8).

using n = 2EI Dh

the final equation (5-9) and the operator, Fig. 5-7 resulting

from the substitution of the four above equations into Eq. (4-16) are

shown below:

wm (6n-6f.l 2 -8f.l+ 16) + wt (4f.l2-4n+4f.l-8) + wb(4f.l2-4n+4f.l-8)

+w1 (4 f.l-12) + wtt (- f.l2+n+ 1) +wbb (- f.l2+n+ 1)+ wbl (4- 2 f.l)

+ w tl ( 4- 2 f.l ) + wn ( 2 )

= ~ h 4 + q I. 2h 3 D D

Eq. (5-9)

23

n-pl-t I

Beam SoAt,porf,J eJae --.......... ~

4- -2p. 4;/- Lf.Y\ ttt,u. -g

rn +l 4)).-12 6rt-6);l- 8pt /6

4-2)1. /f}l· -4n -tlfp-8

rt- p"t I

Fig. 5-7. Operator for the plate supported by beam

For the freely supported edge, with the beam removed merely

substitute

n = 0 and

q' = 0

into Eq. (5-9)

The deflection equation and operator are obtained as follows:

wm(l6-6f.L 2 -8f.L) +wt (4f.L 2+4f.L-8) +wb (4f.L 2+4f.L-8) + w 1 (4f.L-12)

+ w tt ( 1- f.L 2 ) + wb b 0-f.L 2) + wbl ( 4- 2 f.L) + w tl ( 4- 2 f.L ) + wll ( 2)

24

= 3._. h4. Eq. (5-10) D

1-p. a. J

ft'e~ ~J;e ----'+ -2jJ- 4p.,.+4)A-B

1'1'\

+2 J/.}4 -/). 1

16- '~ -I.IA

4-2~ 4fl-t4~-f

I. 1-)A

Fig. 5-8. Operator at free edge of the plate

b. Method 2

This method makes direct use of the boundary conditions

on the free edge. There are three boundary conditions, i.e.

moment Mx = 0, shear Qx = 0 and twisting moment Mxy = 0. These

boundary conditions may be combined into two equations:

25

Eq. (5-11)

Eq. (5 -12)

The first equation shows that there is no twisting moment and

shearing force; the second indicates that there is no moment on the

free edge (6 ).

By rewriting boundary conditions (5-11), (5-12) in the finite

difference form and using Mx = 0 on points t and b, we get Eqs. (3 ),

(4 ), (5 ), (6 ). Substituting these four equations into the biharmonic

operator yields the final equation below:

2 0wm + (wrr +wll +wtt +wbb)- 8 (wr +wl +wt +wb) + 2 (wtr +wt1 +wbr

q 4 +wb1) = D h (7)

wm (16- 61-! 2 - 81J.) +wt (41-! 2 +41-l- 8) +wb (41-! 2 +41-l- 8) +wbl (41-! -12)

+ wtt ( 1- 1-l 2 ) + wbb ( 1 -I-! 2 ) + wn (2) + wtl (4 - 21-!) + wb1 (4 - 21-!) =So.. h 4 (8) D

Eq. (8) coincides with Eq. (5-10) in Method 1.

5-7. Plate with Two Adjacent Free Edges

If the free edges pass through point t, parallel to x andy axes,

the points tr, r, br, tt and rr have to be eliminated.

26

Use Mx = 0 at points t, b, My= 0 at t and boundary conditions

on point m, then substitute these equations into the biharmonic

operator. The following equation and operator are obtained.

Wm ( 15- Sf.! 2- 8f.!) + Wt (2f.! 2+4f.!- 6 )+wb(4f.! 2+4f.!- 8)+w 1 (4f.!- 12)

+wbb (1-f.! 2 )+w 11 (2)+wtl (4- 2f.! )+wbl (4- 2 f.!) = 7J . h4

4-2.# 2Jf-ttj.)1-6

Eq. (5-13)

m. +2

2 l.fp.-12 15-5P -8.P

/.1-- 2)A 4/+4p-B

1-).1-J.

I Fig. S-9. Operator at one increment length along one axis from the

corner at the free-to-free edge of plate

For the edge through point r parallel to the y-axis using

M = 0 at point r and substituting this boundary condition into the X

biharmonic operator yields:

wm ( 19)+wr (2~J.-6)+w 1 ( -8)+wt(- 8)+wb( -8)+w 11 +wbb +wtt

+wtr(2-1J.)+wtl(2)+wbr(2-IJ.)+wbl(2) = £,. h4.

27

Eq. (5 -14)

The operator obtained from the equation is shown in Fig. 5-10.

I

I

2-.P

Fig. 5-10. Operator at one increment length from free edge of

plate

If the center of the pattern is at one increment length from

both edges, the operator is obtained by evaluating Mx= 0 at point r,

and My = 0 at t and substituting these into the biharmonic operator

which yields:

wm{l8)+wr(2J-L-6) +wl(-8)+wt(2J-L-6) +wb(-8) +wll +wbb

+wtr(2-2J-L)+wt1(2-J-L)+wbr(2-J-L)+wbl(2)= S... h4 . D

28

Eq. {5-15)

The operator obtained from the equation is shown in Fig. 5-11.

+I

+I

Fig. 5-11. Operator at one increment length along both axes from

free-free edge of the plate

5-8. Operator at Free Corner

In this case the free edges pass through point m parallel to x

and y axes.

29

Assume

Wbr = wtl'

This assumption makes the supplemental points decrease to

wbr• Wr, wrr and wtr instead of the original seven points.

Eqs. (5-12), (5-11), (5-6) and biharmonic operator are used,

the other equation is obtained by applying Mx = 0 and rotating the

axes 45° clockwise.

The result of the evaluation of these five equations is shown

below and the re suiting operator in Fig. 5-12.

w m ( 4 1--L 3 - 12~--L + 8 ) + wl { 4~--L- 16 ) + wb (- 8 !J- 3 + 2 0 1--L ) + wn ( 3 )

+wb1 (4~-L 2 -8~-L+4)+w11 (3)+wbb(4~-L 3 -4!-L 2 -4!-L+ 1) = .9.....h4 D

m

+3 ifp-lb 1-- ~)-ll)J+8

lf)l-8p -tlf. 2.D;J-I)i

4p-4il-~~op-rl

I Fig. 5-12. Operator at free corner of the plate

Eq. (5-16)

30

5-9. Platew.itha Simply Supported Edge

The moment and deflection on the simply supported edge are

equal to zero and the radius of curvature is infinite. Therefore, the

second derivative with respect to the edge axis becomes zero, from

which Wr is equal to -w1.

In the biharmonic operator, (Fig. 4-3) if the simply supported

edge passes through r parallel to the y axis, or point t parallel to x

axis, the coefficient of the deflection in the middle point of the opera-

tor becomes 19 (Fig. 5-13); and 18 (Fig. 5-14) if the axes pass through

br parallel to x and y axes.

I

2 -8

111

l -8 I~

2 -8

I

Fig. 5-13. Biharmonic Operator at one increment length from

simply supported edge

I

I

2 -8

til.

-8 18

Fig. 5-14. Biharmonic operator at one increment length

from both simply supported edges

5-10. Plate with One Edge Simply Supported and One Edge Free

If the simply supported edge passes through point tt or t

31

in Fig. 5-8, 5-9, 5-10, 5-11 and is parallel to x axis, the deflection

on the simply supported edge is zero and the cor responding operator

is obtained in Fig. 5-15, 5-16, 5-17, 5-18 respectively.

32

/Free

4-l_~A ll.

f.l.j) +14-p- g

m

2 4-p -12 lb-6p.,.-Hp

4 -).}A 4Ji'+4JA-B

l-p"'

Fig. 5-15. Operator at two increments from simply supported edge

of the free-to- simply supported plate

,~Free eJ m.

.2. '+}'« -12. IS-s)l-ap

4-2p 4)l+4)A-8

1-)-'.,.

I Fig. 5-16. Operator at one increment length from simply supported

edge of free-to- simply supported plate

33

s i"'r 1 t d swppor e edJe ..._.

I fr

2 1-- -8 1-- .2j4A

n'\

+I 1-- -3 1- I 'I r--- 1.)"- 6

I 2 ....,__ -8 1- 2-,..M

+J

Fig. 5-17. Operator at one increment length from free edge of

free-to-simply supported plate

34

1 -8

2 2- }/-

+I

Fig. 5-18. Operator at one increment length from edges of free-to­

simply supported plate

5-11. Plate with Fixed Edge

The boundary conditions at the fixed edge are that

w = 0 and slope = 0 i.e., wm = Wrr

The procedures in 5-9 are repeated and the corresponding

operators are obtained as shown in Fig. 5-19, 5-20.

35

I

1 -8

m

I - -8 .2/

2 -K

I

Fig. 5-19. Biharmonic operator at one increment length from fixed

edge

I

2. -8

rn

I -8 21

Fig. 5-20. Biharmonic operator at one increment length of fix-to-fixed

plate

36

5-12. Plate with One Edge Free and One Edge Fixed

If the fixed edge passes through t parallel to x axis in Fig.

5-10, 5-8, the deflection and slope on the fixed edge a re zero, from

which the operators in Fig. 5-21, 5-22 are obtained.

m /Fr•~ .ed3~

--~-....

2

I

Fig. 5-21. Operator at one increment length from free edge of

free-to-fixed plate

F d d . jJ(~ ~ ~~ '1. 1/Fr~Jl.. •

2. 4;t-U. ~ 17-T.JA3.-f).{ m

'1--l)l 1- 4-JA.l.. t-Jfp -t ..

1-AJ.'l.

l ,:

Fig. 5-22. Operator on free edge of free-to-fixed plate

37

5-13. Plate with Two Edges Free and Two Edges Supported by Beams

When AB and AD are supported by interior beams, and if CD,

CB are free edges of Fig. 5-23, the twisting moment on the interior

beam is zero and deflections of both the plate and beam are equal.

Due to symmetrical structure, two equal shearing forces are trans-

mitted from both sides of the plate to the interior beam. This leads

to the boundary condition:

4 2 2 Eia w- q' = 2·D~[a w + (2-1.1.) a wJ

ax4 ay ay2 ax2 Eq. (5-17)

'(

~~~u~--------------------,c

Fig. 5-23. Plate with two edges free and two edges supported by beams

At the intersection of free and beam supported edges, there is

no moment or shear on the beam.

M =0 X Eq. (5-18)

Eq. (5-19)

..

38

Eqs. (5-18) and (5-19) are expressed in finite difference form

1n Eqs. (4-2) and (4-3).

For the equal stiffness of plate and beam, the direct intensity

of load on the beam q' is equal to the product of the total intensity and

increment length, from which the shearing force transmitted from the

plate is equal to zero. For %f = 0, the expression may be obtained

one increment length inward from the edge beam.

wtt + (2[J.-6)wt + (2-[J.)wtr + (2-fJ.)wtl + (6-2[J.) b + (loL-Z)wbr

+ ([J.-2)wbl - wm = 0 Eq. (5- 20)

The operator is shown in Fig. 5-24 .

..,..,

-~+Z. -6+2).( -~-rz.

Wl

-I

-l+.)). -2p.-t-&

;ITJ..--2-t-)1- ~

Fig. 5-24. Operator at one increment length from beam supported

edge of plate

39

5-14. Plate with Notched Free Edges

The operator at point m in Fig. 5-25 may be obtained by

using the boundary conditions (V x)m = 0, (Mx>r = 0, and (Mx)b = 0

as well as deflection equation of the plate.

The equations of the points rr, br, and bb are in terms of

the points inside the plate. The final equation (5-21) and the

opera tor 5-26 are obtained by substituting the above boundary con-

ditions into the general deflection equation .

• rr

.bb

Fig. 5-25. Plate with two edges fixed and other edges free

Eq. (5- 21)

40

r

Fig. 5-26. Operator for the notched plate with two edges fixed and other

edges free

41

6. ILLUSTRATIVE PROBLEMS

To illustrate the finite difference method of solution for plate

deflection, uniformly loaded plates with three different edge support

conditions were considered.

The procedure for solution of plate deflections is given by the

following steps:

Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

Divide the plate into an equally spaced network depending on the

accuracy desired. If symmetrical conditions exist, only

a quadrant or half of the plate need be considered.

Calculate the stiffness of the plate, any attached beams

and the loading term ..S... • h 4 D

Set up deflection equations in accordance with the operator

for the various edge conditions.

Solve these linear simultaneous equations for the plate

deflections using, preferably; an electronic computer.

Once the deflection has been determined, shear and moment

may be obtained by using the expressions in Art. 4-3.

The complete solutions of these problems are given in the

Appendices.

42

Problem 1.

Given plate as shown in the figure, AB and CD are simply

supported edges, AC and BD are free edges.

Given data:

1. Plate Properties: E = 30 x 10 6 psi t = 4 in. 1.1 = 0. 3 L = 72"

2. Loading Condition: Uniformly Distributed Load q = 300 psi

Find the deflections of the plate.

·. ·~ 11 J ' ~ I ,, 114- liD ' J

,., I'" Ill 17 s

11 116 Ill- ~ l.f

D

Answers

w1 = . 2706 in. W8 = . 6506 in . w15 = . 5668 in.

w2 = . 4960 in. w9 = . 2429 in. W16 = . 6121 in .

w3 = . 6441 in. w10 = . 4455 in. w17 = . 2365 in .

w4 = . 6957 in. wll = . 5787 in. w 18 = . 4336 in .

ws = . 2528 in. w12 = . 6250 in. w19 =. 5631 in .

w6 = . 4636 in. w13 = . 2380 in. w 20 = . 6082 in .

w7 = . 6023 in. w14 = . 4364 in .

(M 20 ) = 190,656 in-1b (020 ) = o y y

6 (cr 20 ) = - 30 X 10 x 2 [(.5631-2x.6082+.5631)+.3x(.6121-2x.6082+.6121)]

y 81 X (1 .. 0. 09)

=711 518psi

43

Problem 2.

Given plate as shown in the figure, AB and AD are fixed edges,

CD and CB are free edges.

Given data:

1. 6

Plate Properties: E = 4 x 10 psi t = 8 in. f.l. = 0. 3 L = 5 ft.

2. Loading Condition: Uniformly Distributed Load q = 300 psi

Find the deflections of the plate.

D ~ V' ~

c

/7 .. .. , ~ " ... /

~ / IZ. .f ~ "' '

3

~ / ,,

13 'I ~ " ~ ,/ If 17 ,,. 11 ~ ~

5

.:v ll ~~~ ,, 15" II

., .,,,,, .,,, , ' :r ., .,, .,, ., ., '/; A B Answers

wl ::: 1. 2608 in. w8 = . 6607 in. w15 = . 0644 in. W2= 1. 0432 in. W9 = . 4558 in. W16 = . 2394 in. W3::: . 8089 in. w10 = . 2544 in . w17 = . 132 7 in. W4::: . 5590 in. Wll = . 0854 in . W18 = . 0437 in.

W5 ::: . 3113 in. Wl2 = .5076in . W19::: . 0733 in.

W6 = . 1032 in. wu = . 3491 in . w2o::: . 0240 in.

W7 = . 8562 in. W14 = . 194 0 in. wz1 ::: . 0076 in .

-MB = 110000 in. -lb.

-MB = 138000 in. -lb.* differ ence = 20 o/o

>:<ACI code method 3 case 4

44

Problem 3.

Given plate as shown in the figure. CB, CD are fixed edges,

AB, AD supported by balcony beam.

Given data:

Plate and Beam Properties: E = 4 x 1 o6 psi t = 8.0 in. 1.1 = 0. 3

L = 5 ft. Area of beam= 9. 5 in xl5 in

WI =

w2=

w3 =

w4 =

W5 = W6 =

W7 = W8 =

Loading Condition: Uniform Load q = 300 psi

Find the deflections of the plate.

D

A .

:n

. 0065 in.

. 0160 in.

. 0249 in.

. 0326 in.

. 0378 in.

. 0403 in.

. 0409 in.

. 0645 in.

///. /... ...... .../ .... .,.,, ......... .... L.

v /I

~ / v ,,

.1.1 ~0

, ./

/7 / ,,. g

/ / '" 13 If

/7 14 /O

,g 15 II

Answers

w9 = . 0858 in. w10= . 0991 in. wn = . 1 03 0 in . w 12 =. 1020 in. Wl3 =. 1379 in . w14= .1592 in . Wl5 =. 1642 in . Wl6 =. 1816 in .

l ~

3 / /

If ~ ~

5 ~ ~ v

6 '0

W17 = WI8 = w19 = W20 = W21 = W22 =

. 2077 in.

. 2143 in.

. 2366 in.

. 2442 in.

. 2521 in.

. 2366 in.

-MB = 141400 in. -lb. 6

(!TB) = _ 4 x 10 x 4 [(.0403-Zx0+.0403)+0.3x(0-2xO+O)] x 100x(l-0.09)

= - 14, 176 psi

44A

Problem 3. (continued)

Stress calculation for beam at point B.

(o-B) == Me = -c X I I

= -24, 180 psi

= -15x4 x 106 (.0403-2x0+.0403) 2x100

It should be noted that for all examples a uniform load of 300

psi was used. The deflections, moments and stresses for other values

of uniform load can be found by proportion. For example, if the loading

is changed to 3 psi then all values shown should be multiplied by 0. 01.

7. CONCLUSIONS

The stiffness ratio of a supporting beam and slab system

affects to a considerable extent the deflection of the slab.

Once the plate operators are derived it is a simple matter

to formulate the linear equations for deflection and solve them

using a computer. This makes it a practical method.

A comparison of the method with the ACI code indicates

that finite difference is considerably more accurate and versatile

than handbook methods.

45

46

BIBLIOGRAPHY

l. S. F. Borg and Joseph J. Gennaro, ADVANCED STRUCTURAL

ANALYSIS, D. Van Nostrand Company, Inc. , 19 59

2. S. Timoshenko, THEORY OF PLATES AND SHELLS, p. 90,

McGraw-Hill Book Company, Inc. , 1940

3. C. T. Wang, APPLIED ELASTICITY, McGraw-Hill Book Company,

Inc. , 19 53

4. D. N. DE G. Allen, RELAXATION METHODS, McGraw-Hill Book

Company, Inc., 1954

5. R. V. Southwell, RELAXATION METHODS IN ENGINEERING

SCIENCE, Oxford University Press, 1940

6. PROCEEDINGS OF SYMPOSIA IN APPLIED MECHANICS Vol.

McGraw-Hill Book Company, Inc., 1950

7. S. Timoshenko and G. H. MacGullough, ELEMENT OF STRENGTH

OF MATERIALS, McGraw-Hill Book Company, Inc., 1951

8. Forsythe and Wasow, FINITE-DIFFERENCE METHODS FOR

PARTIAL DIFFERENTIAL EQUATIONS, John Wiley and Sons,

Inc., New York., 1960

9. F. B. Hildebrand, INTRODUCTION TO NUMERICAL ANALYSIS

McGraw-Hill Book Company, Inc., 1956

10. Building Code Requirements for Reinforced Concrete (ACI 318-63}

VITA

Wu-Mo Chern, son of Mr. and Mrs. Mou-Lin Chen was born

on November 6, 1940 at Yin-Ko, Taipei Shen, Taiwan, China.

He entered the Provincial Cheng-Kung High School after

graduation from Yin-Ko Elementary School, where he received his

secondary education from 1953 to 1959.

47

Subsequently he entered the Provincial Cheng-Kung University

on the recommendation of the High School Authorities. He received

the degree of Bachelor of Science in Civil Engineering in 1963.

He served one year in the Chinese Air Force after graduation

from the University.

In September 1964, he entered the University of Missouri at

Rolla to continue his graduate study.

48

APPENDICES

Computer Program

The numerical solution of problems 1, 2 and 3 were program­

med in Fortran II.

The Gauss-Jordan method was used in solving a system of

linear simultaneous equations A· X = B. The Pivot method was also

used to reduce the round-off errors. (9)

The input data are the coefficients of the deflections, and the

output data are the deflections of the numbered points.

The run time of 22 simultaneous linear equations does not

exceed 2 minutes.

The IBM 1620, a medium sized digital computer, was used

for the solution of these equations.

49

APPENDIX I. PLATE WITH TWO EDGES SIMPLY SUPPORTED AND

TWO EDGES FREE

1. The plate is divided into 8x8 network as shown in Frob. 1

2. D =

.9.... h4 = 0. 0112 D

= 17 5. 7 X 106

3. Set up deflection equations.

(1) Using biharmonic operator and Fig. 5-13 to point 9 up to

point 20~

w 1 - 8w 5 + 2w 6 + 1 9w 9 - 8w 1 0 + w 11 - 8w 13 + 2w 14 + w 1 7 = 0. 0 11 2

w 2 + 2w 5 - 8w 6 + 2w 7 - 8w 9 + 2 Ow 1 0 - 8w 11 + w 1 2 + 2w 13 - 8w 14

+ 2w 1 5 + w 18 = 0. 0 11 2

w 3 + 2w 6 - 8w 7 + 2w 8 + w 9 - 8w 1 0 + 2 1 w 11 - 8w 12 + 2w 14 - 8w 15

+ 2w l6 + w 19 = 0. 0112

w 4 + 4w 7 - 8w 8 + 2w 1 0 - 1 6w 11 + 2 Ow 12 + 4 w 15 - 8w 16 + w 2 0 = 0. 0 11 2

w 5 - 8w 9 + 2w 1 0 + 2 Ow 13 - 8w 14 + w 1 5 - 8w 1 7 + 2w 18 = 0. 0 11 2

w 6 + 2w 9 - 8w 1 0 + 2w 11 - 8w 13 + 21 w 14 - 8w 15 + w l6 + 2w 1 7

- 8w 18 + 2w 1 9 = 0. 0 11 2

- 8w 1 9 + 2w 2 0 = 0. 0 112

ws+4w ll - 8w 12 + 2w 14 - 16w 15 + 21 w 16 + 4w 19 - 8w20 = 0. 0112

2w9- 16w 13 +4w 14 + 19w 17 - 8w 13 +w 19 = 0. 0112

2wl0+4wu -16w14+4w 15 - 8w17+2ow 18 - 8w19+w20 = 0.0112

2w 1 1 + 4w 14 - 16w 15 + 4w 16 + w 1 7 - 8w 18 + 2 1 w 1 9 - 8w2 0 = 0. 0 112

2w 12 +sw 15 -16w16 +2w 18 -16w 19 +2ow 20 = 0.0112

50

(2) Using Fig. 5-18 to point 5, Fig. 5-17 to point 6, Fig. 5-10 to

points 7, 8, and Fig. 5-16 to point 1, Fig. 5-15 to point 2, Fig. 5-8

to points 3, 4.

-5. 4w1 + 1. 7w2 + 18w5 - 8w6 +w7- 8w9+ 2wlO+w 13 = 0. 0112

1. 7w1-5• 4w2+1. 1w3-8ws+19w6-8w7+w8+2w9-8w 10+2w11+w 14 = 0. 0112

1. 7w 2-s. 4w3+ 1. 7w4+w 5 -8w6+2ow7 -Bw8+2w 10 -Bw 11 +2w 12+w 15 = o. 0112

3. 4w3 -5. 4w4+ 2w6-16w7+ 19wg_+4w 11 -Bw 12+w 16 = 0. 0112

12.1Sw 1-6.44w2+0.91w3 -10.8ws+3.4w6+2w9 = 0.0112

-6. 44w 1+13. 06w2 -6. 44w3+o. 91w4+3. 4w5-1o. 8w6+3. 4w7+2w 10 = 0. 0112

0. 91w 1 - 6. 44w 2 + 13. 97w3- 6. 44w4 + 3. 4w6 - 10. 8w7 + 3. 4w8

+ 2w 11 = 0. 0112

1. 82w2 - 12. 88w3 + 13. 06w4 + 6. 80w7 - 10. 8w8 + 2w 12 = 0. 0112

The total number of linear equations that must be solved is twenty.

51

APPENDIX 2. PLATE WITH TWO EDGES FIXED AND TWO EDGES

1.

2.

FREE

Divide the plate into 6 x 6 network as shown in Prob. 2

D = 1 7,5 . 4 X 1 0 6

.9.... h4 = 0. 0171 D

3. Set up deflection equations

(1) Using biharmonic operator Fig. 4-3, 5-19, 5-20 at points

12 up to 2 I.

2w 3 + 2w 7 - 16w 8 + 4w 9 + 2 Ow 12 - 16w 13 + 2w 14 + 2w 16 = 0. 0 1 7 1

w4 + 3wg- 8w9 + 2w10- 8w 12 + 22w 13 - 8w 14 +w15- 8w 16 + 3w 17 = 0.0171

w 5 + 2w 9 - 8w 1 0 + 2w11 + w 12 - 8w 13 + 2 Ow 14 - 8w 15 + 2w 16 - 8w 1 7

+ 2w 18 + w 19 = 0. 0171

2w9 + 2w 12 - 16w 13 + 4w 14 + 20w 16- 16w17 + 2w 18 + 2w 19 = 0. 0171

w1o+ 3w 13 - 8w14 + 2w15- Bw 16 + 22w 17 - 8w18- 8w 19 + 3w20 = 0. 0171

w11 +2w 14 - 8w1 5 +w 16- 8w17+21w18 +2w 19 - 8w20 +w21 = o. 0171

w 15 + 3w 17 - Bw 18 + 23w20 - Bw21 - Bw 19 = o. 0171

2w 18 + 2w 19 - 16w20 + 22w21 = o. 0171

(2) Using Fig. 5-10, 5-11, 5-21 to point 7 up to point 11.

2. 7w2- 5. 4w3 + 1. 7w4 - 8w7 + 21w8 - 8w9 +w1o- 8w 12 + 3w 13 = 0.0171

1. 7w5 - 5. 4w6 +w9 - Bw 10 + 20w 11 + 2w 14 - Bw 15 +w 18 = 0. 0171

I. 4w1- 10. 8w2 + 3. 4w3 + 18w7 - 16w8 + 2w9 + 2w 12 = 0. 0171

1. 7w 3 - 5. 4w 4 + 1. 7w 5 + w 7 - 8w 8 + 1 9w 9 - 8w 1 0 + w 11 + 2w 12

- 8w 13 + 2w 14 +w 16 = 0. 0171

1. 7w 4 - 5. 4w5 + 1. 7w6 + ws - 8w9 + 19w 1 o- 8w 11 + 2w 13 - 8w !4

+ 2w 15 + w 1 7 = 0. 01 7 1

(3) Using Fig. 5-8, 5-9,5-12, 5-22to points on the free edge.

0. 9lw 1 - 6. 44w 2 + 13. 06w3 - 6. 44w4 + 0. 91w5 + 3. 4w7 - 10. 8w8

+ 3. 4w9 + 2w12 = 0. 0171

0. 91w2 - 6. 44w3 + 13. 06w4 - 6. 44w5 + 0. 91w6 + 3. 4w8 - 10. 8w9

+ 3. 4w 1 0 + 2w 13 = 0. 01 71

0. 91w3- 6. 44w4 + 13. 06w5- 6. 44w6 + 3. 4w9 - 10. 8w1o+ 3. 4w 11

+ 2w 14 = 0. 01 71

4. 508w 1 - 9. 016w2 + 2. 548w3 + 1. 96w7 = 0. 0171

- 4. 6 2w 1 + 15. 5 5w 2 - 6. 44w 3 + 0. 91 w 4 - 1 0. 8w 7 + 5. 4w 8 = 0. 0 1 71

13. 9 7w 6 - 6. 44w 5 + 0. 91 w 4 - 1 0. 8w 11 + 3. 4w 1 0 + 2w 15 = 0. 01 71

52

The total number of linear equations that must be solved is twenty

one.

53

APPENDIX 3. PLATE WITH TWO EDGES FIXED AND TWO EDGES

SUPPORTED BY BALCONY BEAMS

1. Divide the plate into 6 x 6 network as shown in Frob. 3.

2. D=l75.4xi06

.9....h4 = 0. OI7I D

3. Set up deflection equations

(I) Using biharmonic operator to point I up to point I9 except at

points 6, 11, I5, I8.

2 2w I + 2w 3 + 2w 7 - I6w 2 = 0. 0 I 7 I

-8wi + 23w2 - 8w3 +w4 - 8w7 + 3w8 = 0. OI71

2 I w 3 + w I + w 5 + w I 2 + 2w 7 + 2w 9 - 8w 2 - 8w 4 - 8w 8 = 0. 0 I 7 I

w 2 + 2I w 4 + w 6 + w I 3 + 2w I 0 - 8w 3 - 8w 5 - 8w 9 + 2w 8 = 0. 0 1 7 1

22w5 +w3 +w 14 + 2w 9 + 2wii- 8w4 - 8w6 - 8wi 0 = 0. OI71

2 Ow 7 + 2w 9 + 2w 1 + 2w 12 + 4w 3 - I6w 2 - I6w 8 = 0. 0 I 7 I

20w9 +w7 +w 11 + wi6 + 2w3 + 2w5 + 2w 12 + 2wi 4 - 8w4 - 8w8

-8w10 - 8w13 = o. 017I

2 1 w 1 0 + w 8 + w I 7 + 2w 4 + 2w 6 + 2w 13 + 2w 15 - 8w 5 - 8w 9 - 8w 11

- 8w 14 = 0. 0 I 7 I

w 4 + 3w8 - 8w9 + 2wiO- 8wi 2 + 22w 13 - 8wi 4 +wi 5 - 8w 16 + 3w17 = 0.0171

w 5 + 2w 9 - 8w 1 0 + 2w 11 + w I 2 - 8w 13 + 2 I w 14 - 8w IS + 2w I6 - 8w I 7

+ 2w18 +w1 9 = 0. 0171

54

2w9 +2w12-16w 13 +4w 14 +2aw 16 -16w 17 +2w 18 +2w 19 = 0. 0171

w 1 0 + 3w 13 - 8w 14 + 2w 15 - Sw 16 + 2 3w 1 7 - 8w 18 - 8w 1 9

+ 3w20 = o. 0171

2w 14 + 2w 16 - 16w 1 7 + 4w 18 + 2 2w 1 9 - 16w 2 0 + 2w 21 = 0. 01 7 1

(2) Using Eqs. (5-20) and {5-4), the deflection equations on the

beam may be obtained.

w 3 - 5.4w4 -w5 + 5.4w6 + 1. 7w9 - 1. 7w 11 == 0

w 8 - 5. 4w9 - w 10 + 5. 4w 11 + 1. 7w 13 - 1. 7w 15 + 1. 7w4 - 1. 7w6 = 0

1. 7w9- 1. 7w11 +w12- 5. 4w13- w14 + 5. 4w15 + 1. 7w16- 1. 7w. s= 0

2. 7w 13 - 1. 7w 15 - 5. 4w 16 + 0. 7w17 - 1. 7w20 + 5. 4w 18 = o

w 14 +I.7w16 -5.4w17 -w19 +5.4w20 -1.7w21 = 0

w 15 -4.4w20 +3.4w 17 -5.4w18 +5.4w21 = 0

-wl9+w22 = o

The total number of linear equations that must be solved is twenty

two.

TABLE l. Program for Example No. 1

C C***69812CEX019 WU-MO CHERN C EXAMPLE NO. l

DIMENSION A<40,40l oo 4 I=1,20

-------4~- READ 1v,(A(!,J),J=1,2ll 00 5 1=1,20

5 A(I,21l=O.Oll2 CALL GAUJOR (A,20,21,40,40l PRINT 2J,(A(l,21l,I=l,20l

10 FORMAT (12F6.2l· ------20 FORMAT <F18.8l

CALL EXIT

09/18/65 FORTRAN 2 0000 000 000 000 -

•

...

!

\ ,

- ~ I

END : ·-'··--~ ~~ ... -: .:~-::.:::~- -~-~- - " ·- ·- ·-- . ·c:.,....-;e::e:z..: ·· . ---- -~~-71.:.-· ·· ·· ·-· - · --- - - - -··· - --- . . · -a&he-:a:n:;-~~"Z:.~. -~=--~- --=.;;;~:~:;:;;~.-:.~=--1

TABLE 2. Input Data for Problem No. 1 I ]

1 1.00 o.oo o.oo o.oo -a.oo 2.00 o.oo o.oo 19.oo -a.oo 1.oo o.oo

-8.00 z.oo o.oo o.oo 1.oo o.oo o.oo o.oo ------o:-a·a--·--1 • o o o.oo o.oo z.oo -8.00 z.oo o.oo -a.oo zo.oo -a.oo 1. 0 0 z.oo -a.oo z.oo o .• oo o.oo 1.oo o.oo a.oo o.oo 1.00 o.oo o.oo z.oo 1.oo -a.oo 21.oo -a.oo . -----o-;;cro z.oo ~a.oo 2.uo o.oo o.o o.oo o.oo o.oo 1.00 o.oo o.oo o.oo 2.00-16.00 zo.oo o.oo o.oo 4.oo -a.oo o.oo o.oo

. -- ~------,y~- 0 0 o.oo o.oo o.oo 1.00 o.oo o.oo o.oo -a.oo 2.00 o.oo o.oo 20.00 -8.00 1.00 o.oo -8.oo z.oo o.oo o.oo o.oo o •. oo o.oo o.oo o.oo 1.oo o.oo o.oo z.oo -a.oo 2.00 o.oo

---=-g-~-00 2 I •. oo -8.00 1.00 2.oo -8.oo 2.oo o.oo o.oo o.oo o.oo o.oo o.oo o.oo 1.oo o.oo o.oo 2.oo -a.oo 2·00 1.00 -8.00 22.00 -a.oo o.oo 2.oo -a.oo z.oo

------c-;cro~- o • o o o.oo o.oo o.oo o.oo o.oo r.oo o.oo o.oo 4.oo -a.oo 'C .. o.oo 2.oo-16.oo 21.00 o.oo o.oo 4.oo -a.oo

o. oo o.oo o.oo o.oo o.oo o.oo o.oo o.oo z.oo o.oo o.oo o.oo _______ .. IJl l11

1.11 a-

TABLE 3. Output Data for Problem No. 1

w 1 = . 27054875 W6 = .46364262 Wll:. 57868118 w2 = . 49599725 W7 = .60231269 WI2 = . 62502615 w 3 = . 64413417 w8 = .65056633 w 13 = . 23798016 w 4 = . 69565639 w9 = . 24292341 w 14 = . 43635714 w 5 = . 25279056 w 10 = . 44548666 w15 = . 56676187

Therefore:

(M20)y = - ~ [{\vt-2wm+wb) + f.l. (wr-2wm +wl)] h

Wl6 : . 61212899 W17 = . 23648256 WI8 = . 43358814 WI9 = . 56314242 w2o = . 60821096

6 =- 175. 7x10 [ (w19-2w2o+w19) +o. 3(w16-2w2o+w16)]

81

;: 190, 656 in-lb

(020) Y = - _Q (wtt -4wt +4wb +wtr +wtl-Wbr -wbl -wbb) 2h3

D = - --3 (w18-4w19+4wl9+Wl5+Wl5-WI5-Wl5-Wl8)

2h

= 0

\JI -J

I . TABLE 1. Program for Example No. 2

C C**~69810CEX019 WU-MO CHERN 09/18/65 FORTRAN L ooco 000 000 000 -·--- ---------- ------------- .

C EXAMPLE N0.2 DIMENSION Wl30,3Gl

-- -·-· --- -------- -DO 20 !=1,21

20 REAu 3v, (W(l,J),J=1,22l DO 40 1=1,21

-------- 4.:5 --w·(r;TIT,;~-:-o 111 CALL GA0JOR (~,21,22,3C,3Jl

PRII'\T 50,(W(l,22l,I=l,2ll 30 FOR1"1AT -(10F7.3l 50 FOi~iViAT (F18.8l

CALL t:XIT EN 0-- -------- --------- ---------------··--- --- - -- - .. ------ ---- ..

TABLE 2. Input Data for Problem No. 2

o.ooo o.ooo 2.ooo o.~oo o.ooo za.ooj -16.ao·cf --L .c cc c.ooo

(.1.2:00 o.ooc

o.CJC z.c oo

2.C:JC-l6.CCO "' ,... _-. " Ve\..,.v\..J o. oco

4.000 o.ooc

c.ooo c.ooo

o.ooo o.ooo o.ooo 1.ooo o.ooo o.oco o.oco 3.oco -s.ooo z.ooo O. OCYO .:...e ~-O-J"v_2_[.0b-G--:.:.-d;.::;·vo .1.. COO -o. GO 0 :>. Gv 8 0. OC·O- 0 ~-CJC--C. 0.:.: o.ooo o.ooo 2.COO o.uoo o.ooo

-6.000 o.ooo

o.oc.o o.ooo ""' .. . , ~ \Jo\J\)v 1 • C OC· o.ooo o.cco o.coo z.ovo

1.000 - 8 • 0 0 b "2 (; . c C, ~ - -8 ; ·) ) u ·z . a c-~ - ·-=-6-;· ·a-~ c 2 .-8 00 l.OCC

o.o0o o.aoo o.cco c.ooo 1.~oo o.c:o c.oco "' !'"\_ " ....... v • \ .. .:V v '

o. o J<)·----- - -r--~-c ·() --c---~-a-~ · c. ::, ~:- -2--r - -; c-2--,~-- --c-~-~-c-c·---z;-c2-:--- c ·; ·::-~J-c; - -- - ~-.,-~~-o

-3·000 " -..,1"\1""\ v. I..J 1..) v

.:.CJO ·- " ,.....,-""'"'

.L • · ..... ~ ..... -)

C.OCC O.QG~ 0.000 O.CO~ 0.~00 0.000 O.GCC C·COO 2.GCO c.coo

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TABLE 2. Input Data for Problem No. 2 (continued)

-8.000 o.ooo 2.ooo -s.ooo z.oco o.ooo 1·000 o.ooo 0·000 o.ooo 0. t.}OO 0.910 -6.440 13.060 -6.440 0.910 o.ooo 3.400-10.800 3.400 o.ooo

I ----o-:{)··oo 2.000 o.ooo 0.()()0 o.ooo o.ooo o.ooo o.ooo o.ooo o.ooo l o.ooo I o.ooo 0.91U -6.440 l:1.C60 _:-_6_.4~_0 0.910 o.ooo 3.400-10.800 3.400 ~--- ~:~~-~ ··-c-;·oco z-~-ooo --o.coo o.ooo o.ooo o.ooo o-; ooo o.ooo 6-. a·ac1

c.ooo o.ooo 3.400-1o.soo l o.ooo o.oco 0.910 -6.440 13.060 -6.440 I-- n ·-3-;-4o-cr--o • o-c-o-o • 0 0 0 2.ooc o.doo o.coo o.ooo o.ooo o.ooo o.oo_o ____

o.ooo 4o508 -9.016 2.548 voOCC OoOOQ OoOOO 1o960 OoOOO OoOOO OoOOO o. ooo c·~ o·c.G ___ cr;co-o-cr; oo-c--o·;ocrcY-o-. uGo ·a. ooo o. cro·a-·~ooo-o;;ooo ________ _ O.JOO

-4.620 1S.S5J -6.440 r-·-·-··a··;o·acr- o-;-o-60 o.ooo

Oo910 G.COCJ

o.occ e;.ooo

o.ooo-1c.sco a·. \; o o o • o o o

5.400 o.ooo o.ooo o. oco o. ooo o;-oo_o _______ _ o.ooo o.ooo

-=ro.eoo o.ooo

OoOOJ OoOOO Oo910 -6.440 13o970 OoOCO 0.000 OoOOO 3o400 o-~o-oa··--o:-oao·-- o. o oo --z;:so o----o ~-co··o -o. ooo--o-~co-a··--o.ooo ____ o. o a·cf ···-···

t---· -·- -----··- ---- ·------- -----

1 .......... ··-·-- ·-··---·--1

I -------------~---~--------- ----------- -

-- ~------------ ------- ____ .__ ______________________ --------------·

"' 0

TABLE 3. Output Data for Problem No. 2

w 1 = l. 26078160 w6 =. 10317859 W}} = , 08538586 w2 = l. 043 22990 w7 = .85623912 w 12 = • 50 7 6 24 7 6

w3 = • 80888288 w8 = • 660655 52 W}3 = , 34907511

w4 = • 55904167 w 9 = .45581353 w 14 =. 19401277

w5 = • 31126953 WIO = . 25440455 w 1 5 = • 0 64 3 6 7 0 2

Therefore:

M B = - D 2 ( ( w r - 2 w m + w 1) + 0. 3 ( w t - 2w m + wb)] h

6 = -I75.4xiO [(o-o+o)+0.3(.I032-o+.I032)] 100

=- 110,000 in.-lb.

ACI Code Method:

MB = - ~ (0. 032) (300 X 144) (5) 2 X 12

= - 138,000 in. -lb.

Difference = 20 Yo

W}6 :: • 23938574 W}7 : , 13265307 W18 = .• 04372289 w19 = • 07333096 w2o = . 02399587 W21 = . 00758755

0' ,_.

TABLE 1. Program for Example No. 3

C C***69811CEX019 WU-MO CHERN 09/18/65 FORTRAN 2 0000 000 oco 000 -

C EXAMPLE N0.3 DIMENSION W(3Q,301 ----- -----uo --Io--r = 1, 2 2

10 READ t:v,(W!Id>,J=l,23) DO 30 1=1,15

30--~Ttl-,--T:IT = 0. o 17 1 CALL GAUJOR !W,22,23,30,30l

, P R I N T 4 v , ( ~'i ( I , 2 3 l , I = 1 , 2 2 l !- - --------------· .... - - - ·--------- - -----------.,, 20 FORfv1AT !14F5.1l . 4J FORi•1AT !F18.8l . CALL EXIT - ----------END--

TABLE 2. Input Data for Problem No. 3

-----·· -- -----·-

-----

---------·------ ---------- --

22.0-16.0 2.0 o.o o.o o.o ·a ~ -o----v-·;v---o~-00-;v-o~o z.o o.n -- o.o o.o o.o o.o o.o o.o

o.o -8.0 23.0 -8.0 1.0 o.o o.o o.o o.o v.O o.o o.o o.o r~o---=-;-3 ~-o--z1. o-=-a;o 1.0 o.o o.o o.o o.o o.o " " o.o VoV

o.o 1.0 -8.0 21.0 -s.o 1.0 -o ~ o --c5 :o·-- - ~--;o--o :o- o ~-o--cs·~-o­

o.o o.o 1. 0 -a.o 22.0 -a.o 0. ·J o.o

-----0.0 OnJ

-s.o 3.0 o.o o.o 2.0 -s.o o.o o.o o.o 2.0

---------- ·--o.o o.c o .-o o.c ~ " vev VeV o.o o.o o.o o.o

2 • o ..:·-:u;-.-o ---4. o--~cY 0.0 o.o~o.c-16.o o.o o.o o.o c.o o.o 3.0 -8.0 2.0

.~ " V•V

o.o " " v.v o.o

o.o o.c -8.0 .:::.::.c

- --- - - -----------

-------

o.o o.o o.o o.o o.o o.o

2.0 o.o o.o 1.0 c.o o.o

-s.o 2.o o.o o.c 1.0 o.o ---------~--------- -----

z.o -s.o 2.0 o~o o.o 1.0

2.0 o.o 0 • 0 C. • 0 0 • 0 ---6-~ -0- --- -------

-s.o 1.o o.o -s.o 3.o o.o ------ -------

0' N

tABLE 2. Input Data for Problem No. 3 {continued)

o. o· o.o c.c o.o " " VoV o.c o.o o.o o.o o.o 2.0 -a.o ~.o o.o 1.a -s.o 20.0 -8.0 l.O .:::.o -a.o 2.0 o.c 1.0 " ~ u.u o.o o.c c.o c.c o.o

--c;-;o-··a-;o--u.o---2. a·-=-a-;0 2. o o. a··--T;D---=-8; -c; 21.0 -8.0 o.o 2.o -a.o 2•0 OeC leO 0.0 0.0 G.O CoO OoO o.o o.o 1.0 o.o o.o o.a -6.0 o.o-12.0

:~- ~~6-.cr .::c. o~ ·c. o -o·:·a-··-·o-;o·--cf. o -~o----o;·o-- ~------o.o o.o 20.0 o.o 2.0

o.o 0.0 0.0 1o0 OoO 0.0 CeO 3.0 -8.0 2.0 o.o -8.0 22.0 -8.0 1o0 -8.0 3e0 OoO 0.0 0.0 0.0 OoJ

-- ·-- ··a ;a·--o·.--o---v-;u--cr. o 1. o o.o u. o """""o"'.--=o----,2-.--.o,.,.---....,o..,...-...,o,____,2,....__,_o--.-1-• ...,o -s. o .-:-. 1. o -s.o z.o -s.o 2.0 1.0 o.o o.o " " UoV

o.o o.c o.o o.o o.o c.o o.o o.o z.o o.o o.o o. :j ---To ~a= I r;~-u--·z-;·o-·2. ;-o·---v-~ --0~0-~o ---- - ----------- ------------o.o o.o 0.0 o.o o.o 0.0 o.o o.o o.o 1.0 o.o o.o 3.0 -B.o

--------------

2.0 -8.0 23.G -8.0 -8.0 3.0 0.0 o.o ·- - ----- 0 • t/ -- ·o·~-6---·-c • 0 - ··a . 0-G • o - G :-o ·-,a~. --;co:-----;c·-.-u""''----,-0-.-o'~__,o,.--. --::0:----;::o:-.-o;:---o::o--.-oo::-----;o::--. -;::o--z-;o--------------------------

o.o 2.0-16.0 4.a 22.0-16.0 2.0 a.o o.o O.G 1.0 -5.4 -1.0 5.4 0.0 o. o o. a·-- o~ 0-----o ;·a·--·c;o-c-.o --o~o--

o.o o.o o.o 1.7 o.a -1.7 o.o -1.7 o.o o.o o.a o.a o.o o.o

0 ~ rj . - -a· :-u-c:o 0 ~ 0 0 .-0-0 ;o-o.·o 5.4 1.7 o.o -1.7 o.o o.o o.c o.o o.o o.o o.o o.o o.o o.o

-~ 1 -. 1 - s .-z;-- --~~--7---~4--o • -o--=T;'r-- o • o.o o.o 0.0 o.o o.o o.o o.o o.o 1·7 -5.4 o.o -1.0 5.4 -1.7 ·o ~·a ···-·a~ o··-·o-. 0-- o. o o. o -o~ o ·-a~-o-

o.o o.o

o.o o.o

3.4 -5.4 o.o -4.4 5.4 o.o o.o o.o o.o o.o 0.0 o.o o.o

o.o 1.7 o.o -1.7 o.o 1.0 -5.4 -1.0 5.4 c.c o.o o.o o.o

1.7

c.o

r, " v•u -l.7

o.o o.o -·--------

o.o o.o o.o o.o'----u.G o.o o.o 1.0

o.o

o.o o.o

o.o o.o

o.o

o.o o.o o.o

o.o 1.7 o.o

1 • o - s • 4 =-~-a··------------

o.o 2.7 o.o

o.o o.o l.o

o.o o.o o.o

0' \./.)

TABLE 3. Output Data for P :r ob1em No. 3

wl = • 00646396 w2 =. 01604331 w3 = • 02487651 W4 :: • 03254927 w5 = • 03778035 W6 :: • 04032766

Therefore:

W7 :: • 04091630 W8 = . 06453634 w9 = • 08584046 WIQ :: • 09908608 wu = • 10295783 W}2:: • 101 9 6652

W13 :: • 13792817 w14:: • 15916321 w 15 = • 16419 9 21 w 16 = • 18 164 1 7 5 W}7:: • 20768189

- D ( MB - - hL (w6 - 2wB + w6) + fJ. + (0 - 0 + 0)]

=- 141,400in.-1b.

w 1 8 = • 2 14 3 2 84 3 w19 = .23655111 w2o = • 24420768 W21 :: • 25214256 W22 :: • 23655114

"' *""