Uploaded by Noppagorn Wongnikhom

In Class Practice 65010500

advertisement
H.
6 :
r
5
Erratri
=
Fx
-
ri:
simm
=
10
10
x
2.1655
=
x
-210
M2s
100 mt
=
N
My (10 103) (200
:
rortSmm
187
+
18)/150x183)
2kN.m
=
-1.3kN.m
=
+
2kNiM
Torsion:1:Mo
51018m
3 ro=
Yx,::
(l)_
21.59
MPa
&. 18358188
I=
Transverse shear:A:Ers
for
for
fx
v=
3r!-3r?=29.6840
G:00a;
pipe
10013N
=
Mx,
semicircle
=
tsrelra):12mm: 12-18
-(10x18)/27.6841784
12.09278103/12%8
s
11.02
8
m
MPa
8
=
x
24.37+11.02:65.39
↑xy:
Says
Rs
MPa
35.39 MPa *
6 max:GaugtR:
Emin:
MPa
o
35.39
Saug-:
↑max:R
35.39
=
35.49
MPa
MPaxy
bj Er
=
=
mm:27.684-18m
PointMG, so
Bending:
a
Download