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Dyadic analysis of space rigid framework
DYADIC ANALYSIS OF SPACE RIGID FRAMEWORK. BY
CHEN National
Central
PEI-PING,
University. I.
Chungking.
China.
SUMMARY.
hy space framework consisting of prismatic beams and rigid joints may be analyzed exactly by solving a set of simultaneous linear vector The iteration method may bc equations with dyadic coefficients. applied equally well. II.
I.
GENERAL
RULE.
Notation. Use A, B, C, . . . for scalars, a, b, c, . . . for vectors, 55, ?Ir, . . . for dyadics, 4, *, 4, *. . for nondimensional
and
dyadic factors,
Use joint convention instead of beam convention. 3. Force, displacement, moment, and rotation are vectors, the sense of the latter two being determined by the right hand rule. 4. Let i, j, and rl: be the three nondimensional unit vectors along the torsion axis and the principal bending axes at the centroid of the beam cross section so that these vectors form a right hand orthogonal system. 5. Other assumptions : 2.
i. That the principle of superposition is applicable. ii. That the external loads are applied at the joints or at the beam ends (otherwise see VI). iii. That the secondary effect may be neglected. iv. That the length and direction of any beam are not affected by the strained condition of this framework. III.
FUNDAMENTAL
VECTOR
EQUATION
FOR
A BEAM.
Assume that the external forces fr, fi and external couples ml, m2 are applied at the ends of a beam (Fig. I). It is required to find the relations between the loads, the displacements dl, dz and the rotations al, a2 at both ends. 32.5
326
[J. F. I.
C1IE.N PEI-I'ING.
Resolve each vector into three components along the three beam axes, e.g., fl = iFli + jPli + kFlk, etc. From statics, it gives F2i = -
FIN, Fzj = -
M2i = - Ml:,
Flj,
Fz~ = - FIN,
Mz_+= - Mlj -
FlkL,
and
Mz~ = - MIX + I’ljL,
or, written in the vector form, f2 = -_fl,
and
(Ia>
?n2 = - ml + (kj - jk) .flL.
A beam
Fig.I Similarly from strength of materials, (a1 -
a2) f
=
W
(2 + $q.I,L 3
it gives + (gii
z
+ $ + $tzl, I
(IC)
and [A -_ d, -
1 (kj - jk) .a2L] F2
where all symbols are the conventional ones used by engineers except C which represents the nondimensional torsion constant and ii, jk, etc., which are dyads. Now two special cases are discussed: Case I. The end (I) is hinged. Assuming that a couple ml is applied at the end (I) so as to produce a local rotation ul. If their relation be given by 921 = @*al, then + may be called the moment
dyadic of this beam.
(2) Its value
depends on the dimensions of the beam and the supporting condition of the other end. For example : end (2) is fixed (n2 = 0, dz = 0). E‘rom Eel. (I), it givc5 I. ‘h
@=
F
ii + 4EIjjj
+ 4EIkkk )/L,
and where
(4)
7 may be called the carry-over dyadic factor. 2. The end (2) is also hinged (m2 = o, d, = 0). it gives +’ = JE(Ijjj
and
+ Ikkk),lL,
From Eq. i Il.
(<$I)
f, = - fi = - (kj - jk) .m,,lL,
(51~)
n2 = - V’Ul.
(SC’)
Case II. The end (I) is slotted (i.e., a fictitious joint which can only produce a resisting moment to prevent any rotation but no resisting shear to prevent any displacement). Assume that a forcefl is applied at the end (I) so as to produce a local displacement dl, while the displacement of the other end is d,. lf their relation be given by .fl = *. (dl - dz),
(0)
then q may be called the force dyadic of this beam. Its value also depends on the dimensions of the beam and the supporting condition of the other end. For example: I. The end (2) is fixed or also slotted (u2 = 0). From Eq. (I), it gives
and
ml = m2 = $(kj - jk) .fIL,
(7:1)
fi = - fl
(7b)
+ = (AL2ii + 12Ikjj + I2ljkk) Fz
(7(‘)
Without sidesway, the approximate values will he
and
ml = m2 = 0,
(&I)
\k = AEii/L.
(8h)
328
CHEN
[J.
PEI-PING.
I;. 1.
With sidesway, the approximate value for the reciprocal dyadic will be (9) 2.
The end
(2)
is hinged (& = o, m2 = 0).
From Eq.
\k = (AL2ii + 3Ikjj + 3I&)
$-
IV.
AN
EXTERNALLY
HINGED
RIGID
(I),
it gives
(10)
JOINT.
The framework shown in Fig. 2 consists of an externally hinged rigid joint only, where a couple ml is applied so as to produce a rotation It is required to analyze this framework. al at that joint.
Assume that the moments exerted on the beams, p, p, . . . at the joint (I) are ml*, mlp, . . . and the corresponding moment dyadics are From Eq. (2), it gives RJ, %, . * * respectively. ml, = +P.al,
ml* = (Pu.al, . ..,
ml = Zmr, = (Z@,) -al, and therefore
al = (Z@p)-l-ml.
(II4
Putting c$+
=
ape
(Imp,
c$l*
=
a*.
(Z@,)-1,
.
. .
(I
Ib)
where &,, &, . * * may be called the moment distribution dyadic factors (at that joint) of which the sum is surely an idemfactor, it gives ml, = &.rnl,
ml, = &*ml,
-...
(I rc)
Having obtained the values of al, mlp, mlq, etc., all beams may be easily analyzed by the aid of Eq. (I). V.
AN
EXTERNALLY
SLOTTED
RIGID
JOINT.
Consider the same framework shown in Fig. 2 except that the rigid joint (I) is externally slotted and that all other ends of the beams are
Nov.,
1944.1
DYADTC
ANALYSIS
either fixed or hinged before, it gives
OF
(d =
RIGID
SPACE
Proceeding
0).
3-39
FKAMEWOKK.
in the same manner
as
(r2a)
dl = (z*,)-l.fl* Putting +I, = lc/y’ (S\k,)-1,
$1, = \k,* (B&J-l,
...
(121))
etc., may be called the force distribution dyaclic factors h, h, (at that joint) of which the sum is also surely an idemfactor, it gives
where
fl,, = VI.
fl,
IC/lP._fl,
EQUIVALENT
=
#l,.fl,
LOAD
. . . *
(13‘)
SYSTEM.
If the load (which may be a force, a couple, or a combination of them) is applied at a point not at the ends, it may be replaced by two forces and two couples (Fig. 3) without influencing the stressed and
Fi$C‘3
Eguivalent
load system
The beam itself, to which the strained condition of any other beam. load is applied, needs a correction obtained by assuming both ends fixed. These forces and couples so suggested may be called the equivalent load system, and are given by the following formulae with secondary effect neglected : fl=(~-u>f+(u-Uz)[(1-2U)(jj+Kk).f+6(kj-jk).mlL],
(r3a)
.fz =_f -f1,
(131))
ml=(I--)m+(U-u~)[(I-U)(~~--jk)~fL-~(jj+~~)~m],
(IX)
and
mz=m-ml+(kj-jk).(Uf-fz)L. For distributed
loads the integration VII.
The equivalent be obtained
(1,3(l)
EQUIVALENT
process is used. DYADICS.
moment and force dyadics of a system of beams may
as in the following
illustrative
example:
(J. F. I.
330
Figure 4 shows a system of beams including an externally hinged one, acted on by a couple ml so as to produce a rotation al. From the above equations, it gives
where +‘sr is obtained from eq. (5a). Hence the required equivalent moment dyadic is cf, = [Q. (R + cpf + @,I + R-l. VIII.
(14)
7s + %‘-p.
DIRECT METHOD.
Example I. Two 4” steel beams, s and t, each 20 feet long, form a rigid joint (2) at right angle (Fig. 5). * The end (I) of the beam s is hinged and is acted on by a couple m, while the end (3) of the beam t is fixed. Find the rotation al and the equivalent moment dyadics of this framework at (I) by the direct method.
the unit vectors i, j, k of each beam Solution: (First step)-Tabulate in terms of three standard orthogonal unit vectors X, y, and z (Fig. 5). (Second step)-Determine the number of unknown vectors. In this problem, the known vectors are: ml, = m,
dl = o,
ds = o,
a3
=
0.
Nov.,1944.1DYAI)IC ANALYSIS
RIMI)
OF SPACE
331
lT~u~z~wo~~.
The unknown vectors are:
(I‘hird step)-For C = Substituting
I,
the circular steel beams, Ii/2
= Ij = .Tk = I,
G/E = 0.4.
SilY,
the above values in Eq. (I), it gives (16Ll)
j-G = -j-L3 mz, =
(a1 -- a&!!xI/L [-
&
-
(1.5)
= (2x -
-
ml,
+
(x2 - 2x) -f&L)
(16b)
+ (I .25yy + xx + 22) .rnl,,
(r6c)
x2) .&L/2
j-d (x2 - ~x).e2L]~=(xx+zi).~+(x~--BT).~,
(Itid) iI6C)
Jst = -j-2(, mat = (L&I/L
=
dz
(yz
-
-
L
zy) *fzt ;
$ = (yy
yzjq&L,
(zy -
mzt + +
(1.25xX
+ zz) .fzt f
+
+
(IOf)
(digj
yy + 22) .w4,
(zy - yz) .
7.
(16hj
Take the rigid joint (2) as a free body, ( 16;)
J-z,?+ fAt = 0, ?WJ,+ ??zzt= 0.
and
( I C7.i )
=2 total of ten independent
equations is sufficient to solve for the ten unknown vectors. The general proof may also be sought by resolving Thus the number of unknowns are each vector into three components. increased to three times with equal number of linear scalar equations. (Fourth &$--Solve these equations. Substituting Eqs. (16g), (16h) into (r6d) and eliminating fzt, mz,, it gives (16x~-6xy-6yx+4yy+2~22)~~1,L=(6x2-6~y+212~--_I(~~:z)~nz whence .f,* = _f&
= fzt = _ .f:<, = (2=+!!’
From Eq. (16~) al =
_ 9?7!!)
-<
6oxx + 42~~ + 42YX + r95YY 92
(r&)
. f. A
332
CHEN
[J. I;. I.
PM-PING.
Putting L = 20 and solving for the equivalent moment dyadic of this framework at (I), it gives + = (0.09OXX - 0.02OXY - 0.02OYX + 0.028YY + IX.
ITERATION
0.17
jzz)EI.
(161)
METHOD.
Fig. 6 represents a framework with two rigid joints and Example 2. five 4” steel beams. A force 4oo(y - z) is applied at the midpoint of the beam q. It is required to find the displacement and rotation of joint (I) and the reactions at the ends of beams p by iteration method.
Example 2 Solution: (First step)-Tabulate as in Example I. i
Beam
P
0.8+
Q r
X
t
and
Y z z
X
2 Y
X
(Second step)-Find Eq. (13), it gives
k
j
0.62 0.6x + o&z Y
-Y Y
s
the unit vectors i, j, k of the beams
X
z
L,
ft.
2.5 20 IO
20 20
the equivalent load system if necessary.
From
fl = fi = 2002 - 2ooy, ml = rn2 = 1ooo(y + 2).
(Third step)-Determine whether there is sidesway or not. By observation it is seen that rigid joint (2) has sidesway while rigid joint (I) has not. (Fourth step)-Determine the distribution dyadic factors and also the equivalent dyadics if necessary. It gives I. Consider the joint (I) as hinged. % = (0.2%
+ j&
+ &)4~~lL,
= (0.078~~ - 0.061x2 - 0.0612x f
from Eq. (3a), (15)~ 0.11422 f
0.160yy)EI,
Nov.,
r94-1.1
DYADIC ANALYSIS OF SP.~CE KIGII) FRAMEWORK.
+‘a = (0.04xx
+ 0.2oyy
a, = (0.08yy
+ 0.40xX + 0.40zz)EI,
333
+ 0.20ZZ)EI,
an d cps,l = (0.09xx
- 0.02xy
- 0.02yx
The reciprocal
of their sum is
(X@)+ = (1.66%~ + 0.07xy Substituting
+ 0.028yy
+ 0.07yx
+ 0.17.5zz)EI, from Eq. (161).
+ 0.12~2 + 0.122~ + 2.14~~ + I.I322)/El.
into Eq. (I Ib), the moment
distribution
factors
are
41P = 0.123~~ + 0.005xy
- 0.061x2 + 0.011yx + 0.342yy - 0.0902x + 0.12222,
&
= 0.066~~ + o.oogxy
+ 0.004~~ + 0.014~~ + 0.428~~ + 0.0232~ + 0.22722.
&
= 0.663~~ + 0.028~~ + 0.047~2 + 0.006~~ + 0.171~~ + 0.0472x + 0.453%
and $Q,~ = 0.148~~
- 0.036~~ + O.OIOXZ - 0.031~~ + 0.059~~ + 0.020~~ + 0.19822,
of which the sum is xx + yy + zz. (Check.) Since it has no sidesway, it is 2. Consider the joint (I) as slotted. From not necessary to distribute the unbalanced force to joint (2). Eq. (8b), it gives \k, = i,i,EA/L, = (0.0256~~
+ 0.0192~2 + 0.01922~ + o.o144zz)EA,
\E, = o.o~jxxEA, and
‘P’, = o. 10yyEA.
The reciprocal (N-1 Substituting
of their sum is = (20xX - 26.7~2 - 26.72~ + 10522 + ~oyy)/EA. into Eq. (12b), the force distribution
factors
are
*lP = zz + 1.33~2, *I* = X.2 and
1.33x.2,
$17 = YY,
of which the sum is xx + yy + zz. (Check.) (Fifth step)-Distribute the unbalanced moments succession until the required precision is obtained.
and
forces
in
I. Consider the joint (I) as externally the corresponding beams : Ill 1{’= -
[_I.1;.I.
Per-Prxc.
crTEN
334
hinged and distribute
ml to
56s + 342~1 + 1225,
ml, = 7X + 4283’ + 2272, ml,. = 75X + 17Iy + 453s. nml
ml,
-
=
2&Y +
5')_+3' -I-rc$+Y.
From Eq. (3b), it gives flli = 12X + 8y - 162, .fl~ = 17y - 32% fl?
and
=
68x
fls = -
112,
-
13X - 4y -
from Eq. (16k).
22,
Hence the total reaction at the “hinged” joint is 67X + 21y - 612. But actually this joint is free and is under a load ZOO(Z- y). The unbalanced force is fl* = - 67X - 221~ + 2612. 2. Consider the joint (I) as esternally the corresponding beams :
.fl,
slotted and distribute fl* to
= 347X + 261%
fl, = - 414X, and
fir
=
-
221y.
All joints are now in equilibrium. (Sixth step)-Superpose the results obtained in I and 2: fiq = -
j-4q
=
-
414X + 17y - 322,
ml, = 7x + 428~ + 227% and
m4, = 71q.rnl, = -
The true values are obtained system : fl,
=
-
7x + 214~ + 1142.
by correcting
414x + 217~ - 2322,
fdq = 414X + 183~ ml,
and
for the equivalent
= 7X - 572Y -
1682,
7732,
m4, = - 7x + 1214~ + 11142.
Also
al = (D&-l. ml = (170x + 2140~ + I 1302) /EI radians,
and
dl = (Z\k)-l.fr*
= (-
8300X - 2040~ + 292002)/EA
feet.
load
CHEN National
Central
PEI-PING,
University. I.
Chungking.
China.
SUMMARY.
hy space framework consisting of prismatic beams and rigid joints may be analyzed exactly by solving a set of simultaneous linear vector The iteration method may bc equations with dyadic coefficients. applied equally well. II.
I.
GENERAL
RULE.
Notation. Use A, B, C, . . . for scalars, a, b, c, . . . for vectors, 55, ?Ir, . . . for dyadics, 4, *, 4, *. . for nondimensional
and
dyadic factors,
Use joint convention instead of beam convention. 3. Force, displacement, moment, and rotation are vectors, the sense of the latter two being determined by the right hand rule. 4. Let i, j, and rl: be the three nondimensional unit vectors along the torsion axis and the principal bending axes at the centroid of the beam cross section so that these vectors form a right hand orthogonal system. 5. Other assumptions : 2.
i. That the principle of superposition is applicable. ii. That the external loads are applied at the joints or at the beam ends (otherwise see VI). iii. That the secondary effect may be neglected. iv. That the length and direction of any beam are not affected by the strained condition of this framework. III.
FUNDAMENTAL
VECTOR
EQUATION
FOR
A BEAM.
Assume that the external forces fr, fi and external couples ml, m2 are applied at the ends of a beam (Fig. I). It is required to find the relations between the loads, the displacements dl, dz and the rotations al, a2 at both ends. 32.5
326
[J. F. I.
C1IE.N PEI-I'ING.
Resolve each vector into three components along the three beam axes, e.g., fl = iFli + jPli + kFlk, etc. From statics, it gives F2i = -
FIN, Fzj = -
M2i = - Ml:,
Flj,
Fz~ = - FIN,
Mz_+= - Mlj -
FlkL,
and
Mz~ = - MIX + I’ljL,
or, written in the vector form, f2 = -_fl,
and
(Ia>
?n2 = - ml + (kj - jk) .flL.
A beam
Fig.I Similarly from strength of materials, (a1 -
a2) f
=
W
(2 + $q.I,L 3
it gives + (gii
z
+ $ + $tzl, I
(IC)
and [A -_ d, -
1 (kj - jk) .a2L] F2
where all symbols are the conventional ones used by engineers except C which represents the nondimensional torsion constant and ii, jk, etc., which are dyads. Now two special cases are discussed: Case I. The end (I) is hinged. Assuming that a couple ml is applied at the end (I) so as to produce a local rotation ul. If their relation be given by 921 = @*al, then + may be called the moment
dyadic of this beam.
(2) Its value
depends on the dimensions of the beam and the supporting condition of the other end. For example : end (2) is fixed (n2 = 0, dz = 0). E‘rom Eel. (I), it givc5 I. ‘h
@=
F
ii + 4EIjjj
+ 4EIkkk )/L,
and where
(4)
7 may be called the carry-over dyadic factor. 2. The end (2) is also hinged (m2 = o, d, = 0). it gives +’ = JE(Ijjj
and
+ Ikkk),lL,
From Eq. i Il.
(<$I)
f, = - fi = - (kj - jk) .m,,lL,
(51~)
n2 = - V’Ul.
(SC’)
Case II. The end (I) is slotted (i.e., a fictitious joint which can only produce a resisting moment to prevent any rotation but no resisting shear to prevent any displacement). Assume that a forcefl is applied at the end (I) so as to produce a local displacement dl, while the displacement of the other end is d,. lf their relation be given by .fl = *. (dl - dz),
(0)
then q may be called the force dyadic of this beam. Its value also depends on the dimensions of the beam and the supporting condition of the other end. For example: I. The end (2) is fixed or also slotted (u2 = 0). From Eq. (I), it gives
and
ml = m2 = $(kj - jk) .fIL,
(7:1)
fi = - fl
(7b)
+ = (AL2ii + 12Ikjj + I2ljkk) Fz
(7(‘)
Without sidesway, the approximate values will he
and
ml = m2 = 0,
(&I)
\k = AEii/L.
(8h)
328
CHEN
[J.
PEI-PING.
I;. 1.
With sidesway, the approximate value for the reciprocal dyadic will be (9) 2.
The end
(2)
is hinged (& = o, m2 = 0).
From Eq.
\k = (AL2ii + 3Ikjj + 3I&)
$-
IV.
AN
EXTERNALLY
HINGED
RIGID
(I),
it gives
(10)
JOINT.
The framework shown in Fig. 2 consists of an externally hinged rigid joint only, where a couple ml is applied so as to produce a rotation It is required to analyze this framework. al at that joint.
Assume that the moments exerted on the beams, p, p, . . . at the joint (I) are ml*, mlp, . . . and the corresponding moment dyadics are From Eq. (2), it gives RJ, %, . * * respectively. ml, = +P.al,
ml* = (Pu.al, . ..,
ml = Zmr, = (Z@,) -al, and therefore
al = (Z@p)-l-ml.
(II4
Putting c$+
=
ape
(Imp,
c$l*
=
a*.
(Z@,)-1,
.
. .
(I
Ib)
where &,, &, . * * may be called the moment distribution dyadic factors (at that joint) of which the sum is surely an idemfactor, it gives ml, = &.rnl,
ml, = &*ml,
-...
(I rc)
Having obtained the values of al, mlp, mlq, etc., all beams may be easily analyzed by the aid of Eq. (I). V.
AN
EXTERNALLY
SLOTTED
RIGID
JOINT.
Consider the same framework shown in Fig. 2 except that the rigid joint (I) is externally slotted and that all other ends of the beams are
Nov.,
1944.1
DYADTC
ANALYSIS
either fixed or hinged before, it gives
OF
(d =
RIGID
SPACE
Proceeding
0).
3-39
FKAMEWOKK.
in the same manner
as
(r2a)
dl = (z*,)-l.fl* Putting +I, = lc/y’ (S\k,)-1,
$1, = \k,* (B&J-l,
...
(121))
etc., may be called the force distribution dyaclic factors h, h, (at that joint) of which the sum is also surely an idemfactor, it gives
where
fl,, = VI.
fl,
IC/lP._fl,
EQUIVALENT
=
#l,.fl,
LOAD
. . . *
(13‘)
SYSTEM.
If the load (which may be a force, a couple, or a combination of them) is applied at a point not at the ends, it may be replaced by two forces and two couples (Fig. 3) without influencing the stressed and
Fi$C‘3
Eguivalent
load system
The beam itself, to which the strained condition of any other beam. load is applied, needs a correction obtained by assuming both ends fixed. These forces and couples so suggested may be called the equivalent load system, and are given by the following formulae with secondary effect neglected : fl=(~-u>f+(u-Uz)[(1-2U)(jj+Kk).f+6(kj-jk).mlL],
(r3a)
.fz =_f -f1,
(131))
ml=(I--)m+(U-u~)[(I-U)(~~--jk)~fL-~(jj+~~)~m],
(IX)
and
mz=m-ml+(kj-jk).(Uf-fz)L. For distributed
loads the integration VII.
The equivalent be obtained
(1,3(l)
EQUIVALENT
process is used. DYADICS.
moment and force dyadics of a system of beams may
as in the following
illustrative
example:
(J. F. I.
330
Figure 4 shows a system of beams including an externally hinged one, acted on by a couple ml so as to produce a rotation al. From the above equations, it gives
where +‘sr is obtained from eq. (5a). Hence the required equivalent moment dyadic is cf, = [Q. (R + cpf + @,I + R-l. VIII.
(14)
7s + %‘-p.
DIRECT METHOD.
Example I. Two 4” steel beams, s and t, each 20 feet long, form a rigid joint (2) at right angle (Fig. 5). * The end (I) of the beam s is hinged and is acted on by a couple m, while the end (3) of the beam t is fixed. Find the rotation al and the equivalent moment dyadics of this framework at (I) by the direct method.
the unit vectors i, j, k of each beam Solution: (First step)-Tabulate in terms of three standard orthogonal unit vectors X, y, and z (Fig. 5). (Second step)-Determine the number of unknown vectors. In this problem, the known vectors are: ml, = m,
dl = o,
ds = o,
a3
=
0.
Nov.,1944.1DYAI)IC ANALYSIS
RIMI)
OF SPACE
331
lT~u~z~wo~~.
The unknown vectors are:
(I‘hird step)-For C = Substituting
I,
the circular steel beams, Ii/2
= Ij = .Tk = I,
G/E = 0.4.
SilY,
the above values in Eq. (I), it gives (16Ll)
j-G = -j-L3 mz, =
(a1 -- a&!!xI/L [-
&
-
(1.5)
= (2x -
-
ml,
+
(x2 - 2x) -f&L)
(16b)
+ (I .25yy + xx + 22) .rnl,,
(r6c)
x2) .&L/2
j-d (x2 - ~x).e2L]~=(xx+zi).~+(x~--BT).~,
(Itid) iI6C)
Jst = -j-2(, mat = (L&I/L
=
dz
(yz
-
-
L
zy) *fzt ;
$ = (yy
yzjq&L,
(zy -
mzt + +
(1.25xX
+ zz) .fzt f
+
+
(IOf)
(digj
yy + 22) .w4,
(zy - yz) .
7.
(16hj
Take the rigid joint (2) as a free body, ( 16;)
J-z,?+ fAt = 0, ?WJ,+ ??zzt= 0.
and
( I C7.i )
=2 total of ten independent
equations is sufficient to solve for the ten unknown vectors. The general proof may also be sought by resolving Thus the number of unknowns are each vector into three components. increased to three times with equal number of linear scalar equations. (Fourth &$--Solve these equations. Substituting Eqs. (16g), (16h) into (r6d) and eliminating fzt, mz,, it gives (16x~-6xy-6yx+4yy+2~22)~~1,L=(6x2-6~y+212~--_I(~~:z)~nz whence .f,* = _f&
= fzt = _ .f:<, = (2=+!!’
From Eq. (16~) al =
_ 9?7!!)
-<
6oxx + 42~~ + 42YX + r95YY 92
(r&)
. f. A
332
CHEN
[J. I;. I.
PM-PING.
Putting L = 20 and solving for the equivalent moment dyadic of this framework at (I), it gives + = (0.09OXX - 0.02OXY - 0.02OYX + 0.028YY + IX.
ITERATION
0.17
jzz)EI.
(161)
METHOD.
Fig. 6 represents a framework with two rigid joints and Example 2. five 4” steel beams. A force 4oo(y - z) is applied at the midpoint of the beam q. It is required to find the displacement and rotation of joint (I) and the reactions at the ends of beams p by iteration method.
Example 2 Solution: (First step)-Tabulate as in Example I. i
Beam
P
0.8+
Q r
X
t
and
Y z z
X
2 Y
X
(Second step)-Find Eq. (13), it gives
k
j
0.62 0.6x + o&z Y
-Y Y
s
the unit vectors i, j, k of the beams
X
z
L,
ft.
2.5 20 IO
20 20
the equivalent load system if necessary.
From
fl = fi = 2002 - 2ooy, ml = rn2 = 1ooo(y + 2).
(Third step)-Determine whether there is sidesway or not. By observation it is seen that rigid joint (2) has sidesway while rigid joint (I) has not. (Fourth step)-Determine the distribution dyadic factors and also the equivalent dyadics if necessary. It gives I. Consider the joint (I) as hinged. % = (0.2%
+ j&
+ &)4~~lL,
= (0.078~~ - 0.061x2 - 0.0612x f
from Eq. (3a), (15)~ 0.11422 f
0.160yy)EI,
Nov.,
r94-1.1
DYADIC ANALYSIS OF SP.~CE KIGII) FRAMEWORK.
+‘a = (0.04xx
+ 0.2oyy
a, = (0.08yy
+ 0.40xX + 0.40zz)EI,
333
+ 0.20ZZ)EI,
an d cps,l = (0.09xx
- 0.02xy
- 0.02yx
The reciprocal
of their sum is
(X@)+ = (1.66%~ + 0.07xy Substituting
+ 0.028yy
+ 0.07yx
+ 0.17.5zz)EI, from Eq. (161).
+ 0.12~2 + 0.122~ + 2.14~~ + I.I322)/El.
into Eq. (I Ib), the moment
distribution
factors
are
41P = 0.123~~ + 0.005xy
- 0.061x2 + 0.011yx + 0.342yy - 0.0902x + 0.12222,
&
= 0.066~~ + o.oogxy
+ 0.004~~ + 0.014~~ + 0.428~~ + 0.0232~ + 0.22722.
&
= 0.663~~ + 0.028~~ + 0.047~2 + 0.006~~ + 0.171~~ + 0.0472x + 0.453%
and $Q,~ = 0.148~~
- 0.036~~ + O.OIOXZ - 0.031~~ + 0.059~~ + 0.020~~ + 0.19822,
of which the sum is xx + yy + zz. (Check.) Since it has no sidesway, it is 2. Consider the joint (I) as slotted. From not necessary to distribute the unbalanced force to joint (2). Eq. (8b), it gives \k, = i,i,EA/L, = (0.0256~~
+ 0.0192~2 + 0.01922~ + o.o144zz)EA,
\E, = o.o~jxxEA, and
‘P’, = o. 10yyEA.
The reciprocal (N-1 Substituting
of their sum is = (20xX - 26.7~2 - 26.72~ + 10522 + ~oyy)/EA. into Eq. (12b), the force distribution
factors
are
*lP = zz + 1.33~2, *I* = X.2 and
1.33x.2,
$17 = YY,
of which the sum is xx + yy + zz. (Check.) (Fifth step)-Distribute the unbalanced moments succession until the required precision is obtained.
and
forces
in
I. Consider the joint (I) as externally the corresponding beams : Ill 1{’= -
[_I.1;.I.
Per-Prxc.
crTEN
334
hinged and distribute
ml to
56s + 342~1 + 1225,
ml, = 7X + 4283’ + 2272, ml,. = 75X + 17Iy + 453s. nml
ml,
-
=
2&Y +
5')_+3' -I-rc$+Y.
From Eq. (3b), it gives flli = 12X + 8y - 162, .fl~ = 17y - 32% fl?
and
=
68x
fls = -
112,
-
13X - 4y -
from Eq. (16k).
22,
Hence the total reaction at the “hinged” joint is 67X + 21y - 612. But actually this joint is free and is under a load ZOO(Z- y). The unbalanced force is fl* = - 67X - 221~ + 2612. 2. Consider the joint (I) as esternally the corresponding beams :
.fl,
slotted and distribute fl* to
= 347X + 261%
fl, = - 414X, and
fir
=
-
221y.
All joints are now in equilibrium. (Sixth step)-Superpose the results obtained in I and 2: fiq = -
j-4q
=
-
414X + 17y - 322,
ml, = 7x + 428~ + 227% and
m4, = 71q.rnl, = -
The true values are obtained system : fl,
=
-
7x + 214~ + 1142.
by correcting
414x + 217~ - 2322,
fdq = 414X + 183~ ml,
and
for the equivalent
= 7X - 572Y -
1682,
7732,
m4, = - 7x + 1214~ + 11142.
Also
al = (D&-l. ml = (170x + 2140~ + I 1302) /EI radians,
and
dl = (Z\k)-l.fr*
= (-
8300X - 2040~ + 292002)/EA
feet.
load