Assignment
4. 95
E-
200×109 Pa
p
2.4
=
M
=
I
=
M
350 Nm
=
(
0.008
bh
¥
m
}
0.02
¥
=
✗
f.
0163
=
6 8266667
.
b- 0.02m
-
a)
1-1=3-2 My [ I
Plugging
Ty
Ty
b)
yy
=
Yy
[
I
=
-
[I
750×10-2
'
]
I
in
Ty :)
:
=
273.44×106
292×106 Pa
FEI
292×106×2.4
=
=
-
the values
200
Yy
5¥]
-
=3 -¥
m
3
✗
109
3.504×10-3
Thickness
=
2Yy
=
m
7
.
01×10-3
m
✗
IÉ
me
4. 105
a)
In M¥
=
Jma,
-
e
=
C
=
=
1-2×0.5×0.53
=
A-
IA P¥
-
-
1-
_
-15×10-3 psi
=
-
-
-
=
PK
*
5.2083×10-3 int
=
0.25 in
1
in
0.25
o.j-stf.g.GL?-,o-.s=52in-2P- - :- -=-'s;YP"
=
f- ¥
+
=
288lb
b)
1-
A-
-
e
c.
k
1-2×0.5×0.53
=
=
=
=
F-
p=
5.2083×10-3 int
=
0.25 in
1
in
¥3-
in
f- ¥
+
-
=
o.IS
+
¥÷¥
-÷u=-¥÷
205lb
=
71.882
-2
in
.
#
p
4.111
I
A
¥ ( CF
=
Kfc !
=
Imax
4
=
Tmax
h
I
✗
Iz
A
Atc
30
=
My
* I
3¥
*
in
2
if
"
0-375×0.1684
0.455
4×1
3×9.5835×10-3
=
in
68.67
=
21-2
+
1×53
'
in
=
ce
21.17
in
&
•
B
D
•
2
2. SP
Pa
=
in
At
1.25 in
=
Imax
=
18
18
p
=
=
¥
=
+
37¥
+
25,12¥
-18 Ksi
=
g.
-
D:
Imax
B?
-
0.16838s
=
A
16=14 in
-
f "→-2 )
&
:
Mx
At
Ya
=
-
in
3¢
121×5×63
=
=
=
/ 4-25-12
9.5835×10-3
=
4¥
=
=
=
h
IT
=
(0.3754-0.295)
¥
P¥
1T¥
)
?
C,
¥
=
Teen Eric
4
=
+
4. 147
-
CF )
-
-1A
-
¥4
=
-
37¥
_
Kips
2.5M€
3%8%7
0.0714 P
54.8
_
-
Imax
=
Iz
-
011092 P
2.
-
O
.
1476 P
10
=
%
-0.70714 P
+
3_M×
+
68.67
+
G. 1092 p
p= 53.9 Kips
+
2.271¥
0.1476 P
4.165
R
=
h
=
In
e
E-
F-
=
F
In
¥
-
R
400
=
=
=
-
mm
%-
551-290--30.786
-1¥
=
30-7-86
=
4.214mm
I
120×(-24.214×163)
60×90×10-6 4.26×153×-55×10=3
=
-7¥
-
g-
=
°
F-
=
5.22×106 Pa
I
120×15.786×10-3
60×40
✗
=
✗
10-6×4.214×10-3×15×10-27
-12.49×106
Pa