Wongyeec 1969
Wongyeec 1969
Wongyeec 1969
Yee-Chit Wong
277XL
Title:
Abstract approved:
DESIGN
Castiglione's
Yee-Chit Wong
A THESIS
submitted to
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1970
APPROVED:
4.1
Engineering
In Charge of Major
TABLE OF CONTENTS
Introduction
II.
III.
20
29
38
Illustrated Examples
48
Conclusions
Bibliography
Appendix
64
67
68
I.
IV.
V.
VI.
VII.
VIII.
INTRODUCTION
The
How-
II.
Assumptions
In this analysis and design of horizontally curved
beams, the following assumptions are used:
(1)
(2)
(3)
Notations
M
Its
Torsional moment.
its location.
F
Vertical force.
its location.
Poisson's ratio.
G = E/2 (1 +A4)
EI/GJ.
co
5
A
Shear force.
u-
Cm
Ct
Cf
5
Explanation of m:
EI
m =
GJ
2(1;4)1
2 (1 +f)
3
For rectangular section, Figure la, I = bh /12. Where
b, h are the dimension of the section parallel and perpendi1
3
cular to the radial axis respectively. J=3(h-0.53b)b ,where
of the
h is the long dimension and b is the short dimension
2192(h-0.63b)
3
1
Fig.lb.
value
of
Jwill
be
J=-7-Ehb
I or T sections, the
k,2
Jin
Figure
lc.
For box section as shown
I(ds/t). (3)
In this equation, a is the shear area and equals to the
.
hr
t3
h
1,3
hj
Fig. 1a
Fig. lb
It&
Fig. is
Sign Convention
Bending moment and torsional moment are expressed
by moment vectors.
Torsional
Figure 2 shows
Figure 2.
Sign convention.
III.
Statical analysis
lz
Figure 3a.
Figure 3b.
about
Figure 3c.
Figure 4.
10
(04.:-94.40)
For portion BC
(1)
(2)
(3)
au
au
aTb
amb
DU
au
a7";
= 0
amb
r ma me
EI
EI amo
With m
(To Te =
Tr-j 7711777
aTe
- sin 8
5713
[Y(Mbcose
EI
EI
b 0
+mT13.5 (-cosesinGdO)
o
11
+Florm*osOsinede
- Flomr5s0 inGdO
+Pr10 sin(0-00)cosOde
wo
+Prmjd (1-cos(0-00))sinede = 0
o
Ek(g
EI
pinag
li(cos(20-1))
=
Tb(4(cos20-1)
m! mEknag)
521:(1-m) (cos20_1)
Fbmr(cos0-1)J
(m-1) cos00(-11(cos20-cos200))
'ED
(gig)
[(0-0
pi!
EI
'si3O
sin20-sin200]
2o )
+ mPr (cos0-cos00)
EI
2L
sinOcos0(m-1)]
21E:1 Tb(sin20(m-1))
Mb((m+1)
r2
Fb(sin20(m-1) + 2m(cos0-19
+2E1
= .111[(m-1)00s00(sin20-sin200) + sin00(0-00)
(m+1)
2E1
-
(5)
Similarly
By
aTb
= 0 we obtain
2E1
+
Mbsin20(m-1) + 7ET
gch[2msin0
- 95(m+1)
0(m+1) + sinOcos0(m-1)]
sinOcos0(m-1)]
(sin20-sin20o)
(9U
= 0
(6)
we obtain
c5-b
r2.
TI2msin0-0(m+1)
2m(cos0-1)] +
2E1mi-in20(m-1)
+
2E1
uL"
12
- sin Ocos0(m-1)]
+ r3Fb[0(m+1)
2E1
4msin0 + 2m0
+ sin Ocos0(m-1)
1(11-1)cos00(sin20-sin20)
Let
al = 0(m+1) - sinOcos0(m-1)
bi = sin20(m-1)
01 = sin20(m-1) + 2m(cos0-1)
b2 = 0(m+1) + sinOcos0(m-1)
02 = 2msin0 - 0(m+1) - sinOcos0(m-1)
a3 = c1
b3 = 02
03 = 0(m+1) + sinOcos0(m-1)
co = (m-i)cos00(sin20
4msin0 + 2m0
(8)
13
alMb-blTb+ClFbr = aoPr
-biEb+b2Tb+C2Fbr = boPr
(9)
CiMb+C2Tb+C3Fbr = CoPr
Mb=
rib
al
-b1
Ci
= -b1
b2
C2
C1
C2
C3
1
1.A.1
aoPr
-b1
C1
boPr
b2
C2
CoPr
C2
C3
a1
1
Tb = tki
C1
C1
boPr
C2
CoPr
C3.
-b1
b2 boPr
C1
C2 CoPr
Solving
= PrCt
-b1 aoPr
al
Fbr =1/71
aoPr
= PrCm
= PrC f
(10)
00, m, and r.
14.
Figure 5 and 6.
To
= PrC
The values of C
Figure 7.
aTe=
a 9
-mbcos0 - TbsinG + Fbrsine =.0
Tang = -0,,br_Tb)
(12)
(04 4 00)
Fbr-Tb-cos00Pr
Mb-Prsin00
fd
\p0-
d,
(13)
- PI
Deflection
re
InJo
,0
Ter(1-coskf-9))ds
(14)
15
A0/2
El
1 5/2
r2
=-==
Ei
Mgr sin(i-0)d0 +
H
mr2
+====--
2c/2
-do
r (1-cos(-0))d0
Te
91 s ire)]
l-cos0) + (T,--F r)( 4
u b
EI
b
- cos-0 ( -r(1-cos0) + (Tb-Flor)(0+sin0)/4
2
0
0 -Nbfd
+ Fb rsin- )- sin-(-r-tp-s1n0) +
2 4
2
Tb-Fbr
(1-cosh)
0
- F br(cosi - 1) )]
Pr3r,
(15)
EI
as shown in Figure 8.
16
0
1P0
165
Mb = PrC m
150
Straight Bear
135
12C
105
90
75
6o
45
30
15
'.05
.1
.15
.20
..25
.30
.35
..40 .45
.50
Cm
Figure 5.
17
00
180
Tb = PrCt
165
m=1
150
m=4
m=8
135
120
105
90
75
6
45
30
.02
Figure 6.
.04
.06 .08
18
oo
180
165
Nmax
150
PrCmm
135
m=8
Straight
Beam
120
105
90
75
6o
45
30
15
0
.03
.06
.09
.12
.15
.18
.21
.24
.27
.30
C,
Figure 7.
19
180
165
150
135
Straight beam
120
105
p 3
EI
90
75
bo
45
30
.04
.08
.12
.16
.20
.24
.28
.32
.36
.40
Cd
Figure 8.
20
IV.
476
Figure 9
Fb
w r
r sin z
6/2
21
BC = re
At any section C
(16)
+ wr20 (1-cose)
wr2 (6-sine)
(17)
2E1
aMe
cos0
8TO - (-sine)
aMb
m =
g aEe
= sinb
alb
cos()
s2Z.2.
alb
au
=
=
..51
Mbcose + TbsinG
wr2 sing
2
Jot
wr2(1-coseicosede
= 0
which gives
IMID[0(m + 1)
aTb
zr Jo
+INJO
(rn -1)
Osin20(m-1)
- sing cos0(m-I)1
= 0
(18)
.440, = 0
[Mbcose + TbsinO 2
rsine + wr2(1-cosGlsin9d9
(I-cosi:3)
wr2(0-sinG)]cosGdG
22
which gives
- 124bsin20(m-1) +
T [0(m+1) + sinOcos0(m-11
2 b
- 1r21102(m+1) + lOsinOcos0(m-1)
2
2
L2
sin20(m-1)
(19)
b = sin20(m-1)
c = sinOcos0(m-1)
d = O(m+1) + mO (1 +cos0) -
sin20(m-1) - 2s1n0(m+1)
-sink cos0(m-1)
- sin20 (m-1)
(20)
bd + e
b2 + c2
-a
11
(21)
23
(22)
TO = wr2 Ct8
The
ELY
sin
g
g
+ wr2(1-cosg)= wr2 [Cmcos! + Ctsin2 - 2sin2
+
(1-cosq= wr2
(24)
mm
0f
- Tbsin9 - Mbcose +
(-Tb +
WT
Tan = (Nb-wr2)
COSR
- lib )
(wr2
co8
= (CM-1)
(Z.
Ct)
(25)
24
Deflection
Mers in
EI
4-6 )
1
ds +
ir
(ds )r (1-cos ( i - 6 )
Under uniform load the maximum deflection occurs at the middle of the span,i.e.,when 6=0/2.
Substituting equations 16
and 17 into the above equation the maximum deflection will be:
max
writ
0
(
EI sing
(0+sin0)(0m-1)/4
(21(
+ (Ct-)(
-cost
Cm(1-cos0)(1+m)/4
0
(1-cos!)(1+m) ) + m ( (cos --1)(C
-1) +
2
2
)
sinL( 1-m+m(Cm-1)(0-sin0)/4 )1
2
"L
EI
cd
(26)
25
Oa
180
165
Mb
= wr2C m
b
Straight Beam
150
135
m=8
120
105
90
75
6o
45
30
15
0.1
0.2
Figure 10.
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0c
26
0
180
165
150
135
120
105
90
75
6
45
30
15
Figure 11.
27
oo
180
165
150
135
120
105
90
75
6o
45
0.03
Figure 12.
.27
.30cmm
Variation of maximum span moment coefficients with span angle for uniformly
loaded curved beam.
28
0.
180
Straight beam
165
150
m=8
135
m=1
120
wr
max
Cd
EI
105
90
75
60
45
30
15
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
Cd
29
V.
The existance of torsional moment makes the distribution of shearing stress on a section of the curved beam
more complicated than that of a straight beam.
Generally
closed sections.
V = VQ
-Ib
c
30
Figure 14
The maximum shearing stress distributes uniformly along the
width at neutral axis.
t
4
A-A
ELEVATION
V- DIAGPAM
Figure 15.
B-
31
rectangular section
1
Irmax
V
bh
(2)
(o
_bre = Fb - P
(Pfc,
o.400)
9-40)
V
bd.!
or V=
In practical
rectangular beams.
32
N. A
Figure 16
re
1721.7'
(27)
h/b =
ot
...-
1.0
1.5
2.0
2.5
VALUES CF cd,.
3.0
4.0
6.0
10.6
a-
33
The shearing stress at any point P on the longer side
of the section with a distance x from the mid-point A is
calculated by the formula
P
= r.
(1
(2 x)2)
(28)
A and equal to
(TA) = (
TB =
3t)
(29)
Tr, =TB
=h
h
2Y ) )
(3o)
_2_y__ \
34
t1
Figure 17
T = 2 a H
H=
T
2 a
where
3_5
H
t
11%.
= -7711-t
2 b h t3
LB = 2 b h
( 31 )
t2
1111Tilli111
ELEVATION
4Z
4--
4--
6
25
36
0 i .
positive shear flow around the section that is, the shear
will act upward on the inside face of the section and downward on the outside face.
v+
(32)
where
Portion
Sign
outside face
-.-S0 e.501
01, 0
0/24e 4(95
(0 -
0/2
el)
37
can be obtained.
as follows
A7
`' n
6.0
-2----
)(
6-(3
)2 + X82
'
(33)
where
varies with 0.
38
VI.
Figure 19.
Be-
39
2U
2 Mb
('31) =
U
Z Tb
lb =
B U
Fb
-L
(34)
and
U
Mb
3
= (A7b =
u
Tb
U
B Fb
(35)
- Ab
al Pib
2 EI
2 EI
r
- b1 Mb + b2 Tb 4- C2 Fb r
C1 Fib + C2 Tb + C3 Fb r = Ab
2rEI
=0
2rEI
= 0
2rEI
(36)
and
Mb
C1
- b1 Mb + b2 Tb + C2 Fb r
Ci Mb + C2 Tb + c3 Fb r = Ls b
2rEI
2rEI
_2r EI
(37)
C3
b2 c3 -
c1 c3
2
1113
(38)
4o
(101C2+ b2C1)
Fbr =
mb
C3b2-
(39)
Ci(b2C1 + b1C2)
C3b2 - C22
b1(b1C3 - C1C3)
Mb ai
b2C3- C22
2E1
2 (b2C3- C22)
Mb = (a (h c _ C22)
2
1.-2-3
-1 -1-3 4 c-1c 2 )
h (11 c
C 1 (h-2Ci+
'- b1C2)
EI
(4o)
b1C1 + C2C1
T
b
a1C3 - C12
(41)
(42)
2,
2 ( a103- Ci
Tb =
EI
C727370-aic2. r
(43)
Fbrsin0 = Mb(cos0
b1C3+ C1C2
0 +
b2C3- C22 sin
= Mbraba
(b1C2+ b2c1)ropf)
C3b2- C22 -i-'-''
(44)
41
=Tb tb ma
Expressing
Ma
(45)
biC3 + C2C1
a1C3- C12 Ta
Ta to ma
(46)
+ cosh
a1C3_. C12
a1C2 + b1C1
(1-cos0)1 Tb tba
a1C3 - C12
(47)
b1C3 + C1C2
= Ma mata
(48)
b2C3
b2c1)
C3b2 - C22
(1 -cosy
MbMbta
(49)
Sm
-
42
(2)
'2(a1C3 -
2 (a1C3 - C12)
EI
C12) - b1(b1C3+ C1C2) - C2(b1C1 +a1C2)
r (9)
tlml =
t21 =
a1C3
al1C3
b3 -C3
CC 1
- C122
(54)
+ C2C1
a1c3- C12
sing + Go
- a lC2
+
(1-cosg)
a1c3- C12
(55)
mitt
biC3 + C1C2
b2C3-. C22
(56)
C1C2
m2t1 = b1C3
cosg - s1/10 - b2C2+ b29; (1-cosg)
b2C3 - C22
C3b2- C22
(57)
M1 = m2m21
Ni = T2t2m1
141 = Titimi
T1 = T2t21
Ti = Mimiti
T1 = M2m2t1
(58)
43
N2 = N1 m12
142 = T1 t1 ml
M2 = T2 t2 m2
T2 = T1 t12
T2 = 142 m2 t2
T2 =
mi t2
(59)
Figure 20.
= M21
t12
= t21
t1 t1
=-t2 m2
mi ti
=m2 t2
44
-4t2 mi
t2
(60)
- m2 ti
ti m2
Figure 21 shows the application of the above carryThe sign before the factor indicates its
over factors.
proper sign.
+ 11 ig
rn2
1.
t,
+ rnz tz
-t, m,
Itarnz
T
tit
Figure 21.
- tv
45
TABLE III.
IrN
1-4
1
2
3
4
5
6
7
8
1
0 4
5
6
7
8
1
1.1-
15k
5
6
7
8
1
2
3
4
5
6
7
8
1
1f)
2
3
ti
4
5
6
7
8
3n2
St
15.175
15.140
15.106
15.073
15.039
15.006
14.973
14.940
3.907
1.996
1.359
7.433
7.368
7.305
7.244
7.184
7.127
7.070
7.015
2.081
1.123
.802
.640
.542
.476
.428
4.789
4.700
4.616
4.537
4.463
4.392
4.324
4.260
1.524
.878
3.426
3.320
3.225
3.138
3.057
2.983
2.914
2.849
1.279
2.580
2.466
2.367
2.279
2.200
2.128
2.062
2.001
1.153
.735
1.041
.849
.722
.63o
.562
.391
.657
.543
.472
.423
.386
.358
.781
.604
.509
.448
.404
.371
.344
.579
.492
.434
.391.
.358
.332
miti
m2t1
.935
.905
.877
.850
.825
.800
.777
.o65
.065
.065
.065
.065
.065
.065
.065
.000
.000
.001
.001
.001
.00l
.001
.001
.467
.855
1.182
1.462
1.706
1.920
2.110
2.281
.877
.777
.695
.626
.567
.517
.473
.435
.131
.130
.130
.129
.129
.128
.128
.127
.002
.004
-0.026
-0.064
-0.107
-0.154
-0.201
-0.249
-0.297
-0.344
.617
1.040
1.353
1.595
1.791
1.953
2.090
2.210
.755
.596
.484
.403
.342
.294
.257
.227
.196
.194
.193
.191
.190
.188
.187
.186
.008
.012
.015
.018
-0.582
-0.603
-0.621
-0.638
-0.652
-0.665
-0.676
-0.687
-0.053
-0.118
-0.186
-0.252
-0.316
-0.378
-0.437
-0.493
.700
1.093
1.352
1.541
1.687
1.806
1.905
.627
.438
.327
.256
.209
.176
.152
1.991. .136
.261
.257
.253
.250
.247
.245
.242
.240
.020
.028
.035
.041
.o46
.051
.056
.060
-0.627
-0.655
-0.678
-0.697
-0.089
-0.180
-0.266
-0.345
-0.418
-0.486
-0.548
-0.606
.729
1.067
1.275
1.422
1.535
1.626
1.702
1.767
.326
.318
.312
.307
.302
.299
.295
.293
.040
.054
.065
.074
.082
.089
.095
.100
t1m1
t21
.254 .967
-0.505
-0.507
-0.508
-0.510
-0.512
-0.513
-0.515
-0.517
t2m1
-0.001
-0.003
-0.007
-0.011
-0.016
-0.021
-0.028
-0.035
.496
.726
.945
1.154
1.354
1.545
1.727
-0.521
-0.527
-0.533
-0.539
-0.545
-0.551
-0.556
-0.561
-0.009
-0.023
-0.042
-0.065
-0.090
-0.116
-0.144
-0.173
-0.546
-0.560
-0.572
-0.583
-0.594
-0.604
-0.613
-0.622
1i21
- 0.71.3
-0.727
-0.739
-0.749
.509
.319
.225
.174
.143
.125
.115
.109
.005
.006
.007
.008
.009
.010
.021
.024
.027
.029
9t7
arm'
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6oe'o6e8'o658-0-
zzz'o65e*0-
2.059817*
941'
eoz'
6zz'
o- 349'
9t5"0t190-29-06I4'0194'0-
t9t'
C173'
LOT'T 5i71'
65T-I 175T303-1 591'
eez*T
4517'
3517-
ett*
e5z*
39389z*
1717t7
C2,3'
L55*
5e5'
zz5*
CIS.
905'
To5'
86t*
56t*
85z'
063'
6oe'
c917-0-
Ile-
tez'
w6'0-
COO'T
1796*
z
C
806'
CC9'
694*
914*
699*
839*
e65*
t9*
505*
948'0668-0316*0Tz6*o8z6'0z6'09e6'06C600-
893'0OTt'0oT5'o-
085' 6173'
552.' 991'
098' 951*
o- L85'
tC6*
9t9'0869*06eL'o-
066'
tIC*055t'o-
Tot'
L600056'0o- 456'
196'0-
517e'
1796'0-
50C'
L96'0696'0046'0-
9
4
8
z
C
5
9
808*
zzL*
59'
465*
055*
015'
Te6*
92,t
1743'
5tt
613*
819*
zet'
63I*
LCE'
ttI'
05T'
90t0Z6t0995'0639'0599'0eCL'o-
364*
99C*
5ze*
362*
493'
160'
Lot'
6I1*
5I1'
17e-
nt*
68e'
LLC'
99C'
19e'
55e*
15C'
/MC'
tte
TqZm
040*
ettc-
IgTm
05t'
zee'o06e-0968-0-
t
-cSk
oz4'
o- 173'
o- 854'
oevo-
am'
Two-
395*
Isk
1SX.
ImIg
COL'I
639'T
395*1
zo5'T
Ltt'T
5
9
090'I
804'
ItaZq
Coo*T
TLL'I
666'1
tee*T
s,
tTL'o6e460-
Ws
III (pqraT41100))
0- 1744'
o- 6175'
819'0049'0314'091Leo544*0-
54r
teo'T 88T'
040*T
001'T ZIZ'
035*
549'
894'
Ce8'
388'
616'
056'
546*
tez'
64T*
zeT*
961'
313'
L33'
otz*
z5z*
40t*
ZOt'
86C'
t6e*
865'
8235*
33C
zee*
8CC*
tte'
817C*
z9e'
68e*
tot'
995'
655*
55'
055'
Lt5'
oCt'
5175'
C17'
tW
13t*
9317'
47
TABLE III (Continued)
Sm
.655
.576
.514
.465
.424
.390
2
3
4
5
6
7
8
1
c0
2
3
4
5
6
7
8
St
.894
M21
t2m1
.335
.589
.456
.378
.324
.285
.255
.231
-0.982
-0.986
-0.988
-0.989
-0.990
-0.991
-0.991
-0.992
-0.358
-0.494
-0.579
-0.639
-0.683
-0.718
-0.746
-0.768
timi
.459
.599
.682
.739
.782
.814
.840
.861
.535
.462
.406
.362
.327
.298
.274
.253
.854
.558
.430
.354
.302
.265
.236
.213
-1.000
-1.000
-1.000
-1.000
-1.000
-1.000
-1.000
-1.000
-0.399
-0.526
-0.601
-0.652
-0.689
-0.716
-0.738
-0.756
.399
.526
.601
.652
.689
.716
.738
.756
.361
t21
.237
.204
.217
.236
.253
.269
.282
.293
.254
.240
.259
.280
.298
.313
.326
.336
.598
.595
.594
.592
m2t1
.489
.505
.514
.519
.522
.525
.527
.528
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
.637
mitt
.626
.612
.605
.601
48
VII.
Example 1.
ILLUSTRATED EXAMPLES
given as follows.
Figure 22.
Given:
= 1
= 2 I1 = 2
12
EI
= GJ
Regtd:
mAB = mCD = 1
IBBC
= 2
EAB
= EBC = ECD
Solution:
49
Span BC
Span AB, CD
m =
_60 m =
01
30
sm
7.433
sm
= 3.320
st
2.081
st
= 0.781
M21
0.521
M21
= 0.603
t2 mi
0.009
t2 mi
= 0.118
t2 m2
0.467
t2 m2
= 1.093
t21
= -0.877
t21
= -0.438
m2 t2
0.131
m2 t2
= 0.257
m2 t1
= -0.002
m2 tl
= -0.028
02
Sm
I/r
D. F.
AB
7.433
1/20 = 0.37165
0.527
BC
3.320
2/20 = 0.3320
0.473
Total = 0.70365
1.000
St
I/r
D. F.
AB
2.081
1/20 = 0.10405
0.571
BC
0.781
2/20 = 0.0781
0.429
Total = 0.18215
1.000
50
fixed end torsional moments from Figure 11.
r2 Ct = 1.5 x 202 x 0.0005 = 0.3 k - ft.
TAB = TBA =
0.521
+o.52I
-13.9
+54.1
+39.5
+15.4
+ 25.6
t 4.4
+
+
+13.9
1 0.473
- 116.5
+48.5
+13.8
- t9.2
-
8.3
2.4
+ 3.9
1.3
o.4
0.7
+ 0.3
1.2
4-
0.2
+ 0.8
+75.8
+ 14.7
- 158.3
1.1
0.2
+67.8
- 9o.5
+90.5
0. 603
+.--.
!ij
O'
o
i
r-470
Nt
ci
41\
A-
#
1.6.571
-0.877
to.3
-1.5
+ 0.3
+1.44
+9.9
- 0.2
(1)
cr
0
+0.27
+10.2
t1.71
+11.91
Figure 23.
- o.4.38
04291
1.2.73
1.1.01;
-17.4
+ 1.9
- 1.4
- 0.47
-13.19
4 0.20
+1.28
-11.91
51
To illustrate how
102.6
k - ft.
= + 48.5
= + 13.8.
52
from MBC
= (0.571)
(+ 0.47) = + 0.27
TBC1
= (0.429)
(+ 0.47)
= + 0.19.
One cycle
For different
13.9
+39.5
0.009 x (+ 1.71) =
total =
+25.6
0.0
53
TA:
+0.3
54
EXAMPLE 2.
0
Figure 24.
= 3000 psi.
55
Solution
By Working Stress Design method
1.
= bd3 =
2 1.,4
12
J =
(d -
Tb = P r ct
Mmax =PrC
Mb = w r2 Cm
Tb = w r Ct
Ct = 0.018
Cmm = 0.085
Ct = 0.022
Cmm = 0.176
=3o in.
D.L. - 30 x 14 x 150 = 425 lb/ft. = 0.425k/ft.
144
Fixed end bending moment
Mb =
(0.425 +1) x 20
= - (132 + 89.6)
- 221.6 ft - kips.
x 0.233 + 15 x 20 x 0.299
56
Mid-span bending moment
= 48.6 + 52.8
= 101.4 ft-kips.
x 0.018 + 15 x 20 x 0.022
= 10.2 + 6.7
= 16.9 ft-kips.
jkbfe /2
k = 0.383
fo = 1350 psi
j = 0.872
/221600 x 12
= 28.2 in.
d -
236 x 14
bending moments.
(1) at support
221.6 x 12
Ass
fsjd -20x0.872x29
= 5.14
2
2
As = 5.12 in
(2) at mid-soan
101.4
-2.4 in2.
221.6
Use 3 #8 round bars, As = 2.37 in
As = 5.14
57
3.
Shearing stresses
(1) at support
1
(
b-h
h
=
b
14
2.1
= 0.25
16.9 x 12000
0.25
14
x 29
= 217 psi.
V = 0.5P +
wr0
= 0.15 x 15 + 1.425 x 20 x
= 7.5 +22.4 = 29.9 kips.
IT
29900
14 x 29 - 73.7 psi.
bd
v
= 1.1)77r = 60 psi
= 2 x 0.20 = 0.40 in
=
e
58
T
1:0 = 8.6 Te
e=
v' =
ve = 195 - 60 = 135
# 4 stirrup spacing is
0.44 x 20000
s =
- 4.65 in.
135 x 14
Deflection
c
n= 29/3.16 = 9.2
Use n =9
59
Section at center
Section at support
Figure 25.
In figure 25 let
pt_
bt
b[U3 + ( t-U
Is = As( d-U
)3]
+ AL( U-d' )2
60
P =
5.12
= 0.0121
14 x 31
Pi =
(n-1) = 8
d = 29 in.
di = 2 in.
+
0.194
14.5 + 0.87
= 13.8
U =
1 + 0.03 + 0.0968
4
tl
4
.
2.37
14 x 31
1.58
0.0056
Pt
14 x 31
= 0.00376
U =
1 + 0.0436 + 0.03
c
4
x 14 (3220 + 4290) = 3550o in .
= 733 in4.
t2
= 35500 + 8 x 733
4
= 41360 in .
wr4
From equations 15 and 162N =-
-C
EI
dl
Cd2
+
EI
C dl = 0.0253
C d2 = 0.0198
61
203x 12-'
x 15 x 0.0253
3160 x42765)
1.425 x 204 x 124
, x 42765
+ ( 3100
) x 0.0198
= 0.039 + 0.7
= 0.739 in. < L/360. O.K.
62
2#8
/4" ...1
SECTION A
SECTION B
wOTES:
iNJ
STRENGTH REINiropecEmENT TO
Figure 26.
63
VIII.
CONCLUSIONS
term m = GJ
64
moment.
(
and
'max
4.
This analysis
65
BIBLIOGRAPHY
1.
2.
3.
Cutts, C. E. Horizontally curved box beams. Transactions of the American Society of Civil Engineers
118:517-544. 1953. (Paper no. 2555)
4.
5.
6.
7.
Iyse, I.
8.
504 p.
9.
10.
Spyropoulos, P. J.
Circularly curved. beams transProceedings
(Journal) of the American
versely loaded.
Concrete Institute 60:1457-1469. 1963.
11.
12.
APPENDIX
66
The following notations used in the programs represent those used in the text.
Programs
Text
TX
TXZ
00
P,M
CTD, DTXZ
Cd
CMC
Cme
CTC
Cte
67
PROGRAM 1.
PROGRAM WON
REAL r 'i, NNAX
DIMENSION A(3,3),X(3),D(3)
EQUIVALENCE (X,B)
KCOUNT=1
WRITE(10,10)
10 FORY.AT(1H1,////////)
DO 60 111=30,180,30
TX=III*O.01745329
DO 60 MM=1,8
E=MM
DO 70 JJ=15,180,30
TXZ=JJ*0.01745329
IF (TXZ.GT.TX) GO TO 70
CALCULATION OF CONSTANTS
A(1,1)=TX(E+1.)-SIN(TX)*COS(TX)*(M-1.)
A(2,1)=SIN(TX)**2*(Y-1.)
A(3,1)=SIN(TX)**2*(M-1.)+2.*N*(COS(TX)-1.)
A(1,2)=A(2,1)
A(2,2)=TX*(M+1.)+SIN(TX)*COS(TX)*(M-1.)
A(3,2)=2.*M*SIN(TX)-TX*(M+1.)-SIN(IX)*C0S(TX)*(X-1.)
A(1,3)=A(3,1)
A(2,3)=A(3,2)
A(3,3)=TX*(M+1.)+SIN(TX)*COS(TX)*(M-1.)-4.*M*SIN(TX)
1+2.*E*TX
B(1)=(M-1.)*COS(TXZ)*(SIN(TX)**2
2-SIN(TXZ)"2)+SIN(TX2)(TX-TX2)
3*(M+1.)-(SIN(TXZ)*(M-1.)/2.)*(SIN(2.*TX)-SIN(2.*TXZ))
4+2.*M*(COS(TX)-COS(TXZ))
B(2)=-COS(TXZ)*(TX-TXZ)*(E+1.)-COS(TXZ)*(M-1)*
5(SIN(2.*TX)-SIN(2.*TXZ))/2.-(M-1 .)*SIN(TXZ)*(SIN(TX)
6**2-SIN(TXZ)**2)+2.*M*(SIN(TX)-SIN(TXZ))
B(3)=(M-1.)*COS(TXZ)*(SIN(2.*IX)-SIN(2.*TXZ))/2.
7+(TX-TXZ)*(COS(TXZ)*(M+1.)+2.*M)+(M-1.)*SIN(TXZ)*
8(SIN(TX)**2-SIN(TXZ)**2)-2.*P1*(1.+COS(TXZ))*(SIN(TX)
9-SIN(TXZ))+2.*M*SIN(TXZ)*COS(TX)-CCS(TXZ))
CALL SIMQ(A,B,3,0.00001,0)t
HMAX=X(1)*COS(TXZ)+X(2)*SIN(TXZ)-X(3)*SIN(TXZ)
WRITE(10,101)III,JJ,M,(X(I),I=1,3),MMAX
IF(TXZ.EQ.TX/2,) 40,70
40 DTXz=sIi,;(rix/2.MX(1)*(Tx+siN(TX))/4.+(x(2)-X(3))*
1(1.-COS(TX))/4.)-COS(TX/2.)*(X(1)*(1.-COS(TL))/4.
2+(X(2)-X(3))*(TX/4.-SIN(TX)/4.))
3+M*(X(1)*(COS(TX/2.)-1.)+(K(2)-X(3))*SI(TL/2.)+X(3)
4*(TX/2.))-IT.'COS(TV2.)*(-2L(1)*(1.-0O3(M:))/4.+(x(2)5X(3))*(TX+SI(TX))/4.+-i(3)*SIZ(I.V2.))-::*SIL7.(TX/2.)
68
6*(-X(1)-:,(TX-SIN(TX))/4.+(X(2)-X(3))*
7(1.-COS(TX))/4.-X(3)*(COS(TX/2.)-1.))
KCOM=1
101 FORi'AT(I 12, 15.1, 3F7.4, 2F9.4)
70 CONTINUE
60 CONTINUE
END
69
PROGRAM 2.
PROGRAM WON
UNIFORM LOAD
KCOUNT =1
WRITE(10,10)
10 FORMAT(1H,///////)
DO 40 1=15,180,15
TX-I*0.0174533
DO $) J=1,8
P=J
DO 30 K=16,181,15
KK=(K-1)
TXC=KK*0.0174533
IF(TXC.GT.TX) GO TO 40
CM=-((2.(P+1.)*SIN(TX)-P*TX*(1.+COS(TX)))/
1(TX*(P+1.)-SIN(TX)*(P-1.))-1.)
CT=-((2.*(P+1.)*(1.-COS(TX))-P*TX*SIN(TX))/
2(TX*(P-1-1.)-SIN(TX)*(P-1.))-TX/1.)
CMC=CM*COS(TXC)+CT*SIN(TXC)-TX*SIN(TXC)/2.+(1.-COS(TXC))
OMM=CM*COS(TX/2.)+Ol*SIN(TX/2.)-(TX/2.)*SIN(TX/2.)
34-(1.-COS(TX/2.))
CTC=CT*COS(TXC)-CM*SIN(TXO)+TX*(1.-COS(TXC))/2.
4-(TXC-SIN(TXC))
CTD=SI.ii(TX/2.)*((TX+SII'T(TX))*(CM-1.)/4.+(CT-TX/2.)*
1((1.-COS(TX))/4.+P-P*(1.-COS(TX))/4.)+SI::(TX/2.)*
2(1.+P)+P*(C::-1.)(TX-SII;(TX))/4.-TX.-:-P/2.)-COS(IX/2.)
3*(C.,;*(1.-COS(TX))*(1.-7)/4.-(1.-0O3(TX))*(1.-P)/4.4.
4(CT-TX/2.)*(TL-SIN(TX)+P*TA+P*SII,,(T_K))/4.+(1.5cos(Tx/2.))*(1.+P))+13*((cos(Tx/2.)-1.)*(ciJ_-1.)
6+Tx**2/8.
wRITE(10,20) I,P,KR,CM,CT,CMC,CTC,CTD
KCOUKT=KCOUNT+1
IF(KCOUT.EQ.51) 50,30
50 WRITE(10,10)
KCOU.f:T=1
20 FOHHAT(15,F5.1,I5,6F9.4)
30 CONTINUE
40 CONTINUE
STOP
END
70
PROGRAM 3.
PROGRAM WON
REAL M21,M1T1,M2T2
MOMENT DISTRIBUTION FACTORS
KCOUNT=1
WRITE(10,10)
10 FORMAT(1H1,////////)
DO 30 1=15,180,15
TX=I*0.0174533
DO 30 J=1,8
P=J
A1=TX*(P+1.)-SIN(TX)*COS(TX)*(P-1.)
B1=SIN(TX)**2*(P-1.)
C1=SIN(TX)**2*(P-1.)+2.*P*(COS(TX)-1.)
A2=B1
B2=TX*(P+1.)+SIN(TX)*COS(TX)*P-1.)
C2=2.*P*SIN(TX)-TX*(P+1.)-SIN(TX)*COS(TX)*(_-1.)
A3=C1
B3=C2
C3=TX*(P+1.)+SIN(TX)*COS(TX)*(P-1.)-4.*P*SIN(TX)+
2.*P*TX
SM=2.*(B2*C3-C2**2)/(A1*(B2*C3-B3*C2)-B1*(B1 *C3+C1 *C2)
1-C1*(B2*C14-B1*C2))
ST=2.*(Al*C3-C1**2)/(B2*(A1 *C3-C1**2)
2-B1*(Bl*C3+Cl*C2)-C2*(Bl*C1+Al*C2))
M21=COS(TX)+(B1 *C3+C1 *C2)*SIN(TX)/(B2*C3 C2**2)
3+(B1*C2+B2*C1)*SIN(TX)/(C3*32-C2**2)
T2M1=SIN(TX)+(B1*C3+C2*C1)*COS(TX)/(A1 *C3-Ca**2)
4+(B1*C1+A1 *C2)*SIN(TX)/(A1*C3-C1**2)
T1M1=(B1*C3+C2*C1)/(Al*C3-C1**2)
T21.COS(TX)-(B1*C3+C2*C1)*SIN(TX)/(A1 *C3-C1**2)
5-(Al*C2+Bl*C1)*(1.-COS(TX))/(Al*C3-C1**2)
m1 TIABl*C3+C1 *C2)/(B2*C3-C2**2)
M2T1.-SIN(TX)+(B1*C3+C1 *C2)*COS(TX)/(B2*C3-C2**2)
6-(B1*C2+B2*C1)*(1.-COS(TX))/(C3*B2-C2**2)
WRITE(10,20) I,P,SM,ST,M21,T2M1, T1M1, T21,M1T1,M2T1
KCOUNT=KCOUNT+1
IF (KCOUNT.EQ.46) 50,30
50 WRITE(10,10)
KCOUNT=1
20 FORMAT(19,F5,1,8F6.3)
30 CONTINUE
STOP
END