q
Ib/ft
=
.
4001 bf+
L)
β
attdruta
a
f + - 2+ t
8
ƩF
EYA
=
=
A +β
0
,
400 -
-
400
200 - 10
=
8 200
글
.
10
f
-
Pβ
↑
A
-
.
B
640
let
.
A
M
.
360
becomes
maximum
-
Ʃ
400
.
400
β
P3
\
o
a
tada ↓
.
5
pow
↑360
36 . /
aa
200
-
8
JM
ttt
tM
‰
MA
=
-
M
=
Ʃ
MB
+
M
=
V
=
o
o
=
.
용
a
을
..
=
0
,
a
=
5
.
367
.
.
200
-
-
=
200
.
-
명글 룩 5307 M
53
.
+
=
0
1 b ff
.
400
360 10
-
-
8 글 ( +] + 200 글 룩 2 + M
400 t 6030
=∅
.
."
1688 3
0
윙
360 .
.
=
2
360
=
when
20
=
,
3675
.8+ 국)
200
.
'
5 367
2-
호
.
,
a
=
-
1733 33 + 10 B
-
x= a
y
x
를
-
a
.
At
.
δ
TiH
t
5 367
!
no
N
1000
=
o
=
4266 67
=
VL 1 bl
↑
-
.
β
호
IPm
2
-
-
800
54
-
4000
M (Ibf+ ) Λ
16883
\
-
400
-
5 6n
.
미
4vobittf
x ( f+ )
.
2001 b 1 f
서
o
tor
t
t
a
-
A +B
ƩF = 0 =
MA
=
=
0
=
=
-
-
-
.
-
?
=
400
3333 33
-
a
A+ B
1000
-200 ( 60-
a
=
1000
←
) 는( a 씀 )
+
(
333 33 a
tT
+ 100
33 333 a +
.
maximum moment
400 lift
t
t
AT
5
occurs
t
10 B
+← LoB =
.
.
=
l0 β
02 ) + 10 β
3
-
tβ
,
일를
-
-
100
400
β
「
.
글
- ( 00 ( 0 - µ )
a
을
lott
-
200 9
-
0
t
200 - 10
=
400
- 400
=
t to
o
C
↑
A
Ʃ
'
1
373 33 ,
.
when
dr
626 67
=
=
33 333 a
-
.
.
shear force
trod ' nctet
t
A
becomes
ƩF
A
=
o
A
=
=
zero
.
0
-
200
2응
…
동. 일
5
.
"
(2)
"
( 1)
From
and
(i )
2
33 333 a
-
+
a)
,
626 67 a
-
.
2500
=
a=
0
5 745 f +
.
'
.
A
β
으
435 . 2
=
1b
=
200
-
α
을
=
574 . 1 b,
5
B
,
200
.
1000
=
( 10 µ)
-
.
-
A
늘
=
564 81 b
=
.
425 51b
.
V 11 bh
435 2
=
-
.
a
200
5 745
'
-
139 2
되 !
-
*
ㆀ
'니세
ㅣ
.
-
564 8
^
-
.
명
"
435 2
At
x
M
=
5 f+ ,
Ʃ
1850 8
.
My
=
0
400 - 174 . . 5 글 5
죽
1
=
-
ibft
Mllbft )
( 850 8
스
^
.
400
-
10 sxlft)
띠
4vobttf
2001 b 1t
a
A
t
t
tor
.
t to
C
a
t - rott
2
-
= 174 .1
방)
?
π
.
.
=
5
a
tβ
+
M
내
β
33 333
=
.
373 33
a +
.
maximum moment
π
From
hen
occurs
M
=
.
0
=
=
Y=
=
400
Fnn ( 3 ) ,
33 333 a
-
becomes
zero
: 1 b /ff
Ʃ F
"
A
(1)
"
.
.
shear force
626 67 a
-
400
+
.
-
200
.
=
0
A
=
200
-
뿐
10
=
-
33 333 a
.
같
.
.
x
일
(지
"
=
2
100 x
=
.
깊 x 일 ]기 + M
"
(3)
,
을
20
째
400+ 200
o=
626 67
,
and ( 2)
,
()
ƩM
=
; *)
「
Mmax
,
200
4o 0 lbft
A
A
when
173 . 333 a 2 626 .67 a
=
,
a
x= a
=
4
.
. ftf
7
a)
-
d
3 f*
P
7501 b
=
1101 b 1 t
Iadddat
A - 8 ft - 5 + T
A +B
되어지
-
3 750
10031 b
=
83
A
10
-
1300
=
.
5 ( 8+ 2 5 )
-
.
-
β
,
-
홱
453
x
i3
ㅱ
-
M
Mpos
⇒
{
1374
s
"
β
.
891
=
IMneg
f9 에
-
297 1 b
=
-
297
-
+
.
1300
=
.
=
β
V
550
0
=
1750 + 110 5
=
Ik ff
1374 1 bft
=
-
=
송 =π
0
,
=
모
O 375
∞
①
0 375
이
"
.
-
t
-
d
×
2 .bs
0
y. 몰
=
더
92
,
Ioar
A,
Ac
이에도
=
몰
2
375 +
=
0
=
2 5 0 375
.
.
=
.
=
1 6875
.
0 9375
=
.
.
.
0
.
375
-
2 625
.
=
0 9844
.
.
.
√
25
A
A.
.
C
=
2
.
=
0
.
.
9558
d
.
,
,
I )
π
I .
)
I
d
0 7683
=
=
=
0
=
2
7317
.
배름 =
25 ⑬
=
음
,
쁨
.375. .
0
2
0
=
Ialc
=
=
Iπ ,
)
5652
.
. di 2
A
+
0 01099 + 0 9375
.
. 01099
0
=
t
0 7683 +
.
I
1
+
.
Ard
0 5652 + 0
2
9844 0 73172
-
.
.
.
.
.
=
1 6566
.
s
=
용또 ,
Mpi
in
4
=
0
모으
1 84t1 k-
.
(
=
γ
=
044
. 131
in .2
=
Mny [ 몰 서높
:? .
:
.
없
maximmm tensile stress
"
=
9"
=i
20345 8 $ i
.
compressovestress = 13193 . 6ps
max ' mum
=
.9
psi
6168
6566
=
앞
M
M
2
㉚
O 375
.
^
②
미
①
"
(
x
가 C
0 375
allounble
호
2
(
=
18000 psi
=
12000 psi
e
σ.
.
√
2 5
.
진
"
ε
=
M
σ
=
=
.
18000
=
높
,
높
12000 ,
'
1 0917
=
.
=s
높스 10917
500 피 ,
아개쁨
,
두행 * 2
.y
(
0 375 / h - . 37
0
L1
=
h 몰 + 0 .975 2 5
) ( . 375 +
.
.
( h
iosp
.
0
n 5( h - . 3 n 5
o
0
+
375 2 5
.
.
.
h lz
-
f
I used
lex)
ans 2 . 5 .
0
)
.
는
니
0 대+
MATLAB to solve
this
xal th
x (2l →
[2
x (4) →
x(3)
→
problem
.
C,
let
inttial
I
4
values
estimation of
as
h
=
3 in
(
_
π
depth of
required
C)
σ
=
e
18000 pi
M
.=
σ
.
=
.
,
.
음
= 31943 in
h
1
12000
psi
.5
.=
.
3
-
ftd
AF
ƩF
=
IMA
0
=
Iadtiad
O+
T
ATB
:
O
/
q ( / b f+)
β
:
-
β
=
5+ T
P+ 5f
3 P + 83
-
5q
.
10 5
.
=
0
1
A
=
=
P + 5%
0 625 p
.
-
-
합 ( 3 β+ 52 5 % 1
.
1 5625 q
.
frm
Also
( a)
Mpos
,
=
3A
,
=
Mneg 31
A
+ 5(
=
% 시도
. 18750
%
A
}
-b
-
8 A +5 P
=
-
=
12 5 q
2
=
81250
.
-
2 03125 %
.
,
.
β= 6 944 q
.
beam
e
=
is
,
al
s
as
e
ma
q
β=
=
. 0442 ,
2
=
Mng
it
=
=
I
1
.6566
.ft
b
:
.
□
negM
,
0c
18000
=
so 2
'
97 247 1 b / in
.
6 944 %
=
675 28
.
1b
.
σ
14500
-
=
xx
σ
yy
σ
-
=
=
2530
3500
-
ay
C
R
=
=
=
뿐
=
볼
-
6933 3
.
p
+ㆍ
851도
y
x
j(
.
30 320 (
.
'
-
15448 3
.
,r
'
==
" unp
2θ
30
=
p
n
320
.
20
382
=
=
760
y
x
+
=
π
-
C
=
(
. s 45 .680
-
8515
-
6933 3 6945 680
.
.
,
=
σ,
y
=
-
13359 psi
ε +
=
-
RCos 45 680
=
.
-
8515 + 6933 3
.
c0 s 45
.
3670 9 psi
.
lxly
=
Rsim
=
-
45 . 680
=
.
"
p1
R
0 504795
. 3 sin 45 680
69 3 3
.
4640 4 psi
.
13359
songpai 1380
대
pst
680
py
all
TMurcam
TE
Tlng
circunterential
σ
longitudinal ta
t
=
σ
=
.
MPaq
= = 2 .8
in
out
-
plane
of
-
plane
:
:
y
m
.
max =
max
'
46 667 Mpa
=
τ
τ
=3 333 MPa
맡
y
( b)
σ
y
=
=
=
max
몰
-
r
23 333 MPa
=
.
( r Jy
,
=
0
,
)
46 667 MPa
.
circumferential Ey =
띠
ooaPax
93 333
=
6 667
0
-
.
longitadinal
4 (6 n
=
0 3
C
R
=
=
뿐
=
-
9 333
-
.
.
.
1δ
5
70 MPa
+
(
23 333
=
==
.
"
184
a
.
볼P y
높
tunon
-
.
.
o
d
=
vCat
밑 [ VCrytJ
=
GPa
93
-
3 967
.
Ex
t-
20
p
0
,
2
op
.
counter clockwise
0
=
MPa
.
7o
,
,
_
"
x
"
46667
>
"
+
,
%
int
y
' 93 333
.
.
To the weld ,
perpendicular Jcs
parallel Jy
shear stress (
xg
,
,
=
=
=
C
-
Rcsno
=
0
70 - 23 .333 6 .51700
C + Rosn 00 = 70 + 23 33360 , 700
62
=
=
27 . 317 sin
77 98 MPa
.
'
yat .
ins
"
I
700
=
21
02 MPa
77 98 14 Pa
=
.
.
Rsinyo 0
.
.
926 Mpa
1700
↑
.
조
0.
.
900
o
900
Vx - z d.
Mx 은벌
ㆍM
Af that
point
되니
Mx +
사시
.
=
to thepontloated before A
3
.
0
.
,
=
-
t
Mx
.
,
distance from
as
1x
62
52
~
ef
↑
=
=
900
-
8350
7
700
( )( 1 )
(a
.
-
가
.
블
3 6)
+
β
+
.
옳
( x - .6
3
)
400 이일 ( x
-
450 ( ) ) +
Ʃ( - .6
를)
x
)
3
=
450 ( 기
국
0
-
를)
.
EIV
"
,
=
=
6 2
.
0
,
U
=
π= 6 2
.
0
EIV
EIV
(
=
!
(= 6 .
EI .
θ
=
,
.(
1
=
-
=
=
"
3
6
+
26658 648
.
0
,
(
3
,
=
.
,
. 022216 rad
-3 . ) 후 37 . ( x- 1)
+
C
.
-
136264 058
62
=
.
0
.
를) β+ <
833
8492
02221554
,
1 27290
+
75 (1
-
릎)
3
2
C2
+
=
C 2136264 058
0
.
,
.
.
=
,
= 134922 . 72
=
0
V.
112 436 mM
.
deflectlon
rotation
δ
3
1
θ
=
=
B
(12 436 mm
.
1 27290
.
1
.
C
-
-
.
4
5
=
=
6
28492 833 x
225 ()
+
0
.
ㆍI =
SB
x- . )4 (
1=
=
EIVK
-
x- 1)
50 (
= 28492 . 833
2
EIV
+
8.
'
.
112436 m