Tentamen 7 November 2016, antwoorden
Mechanics (NAMECH05E)
Rijksuniversiteit Groningen
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Problem 1
:
y
p
p
AI
t 1
1
as
.
1 O
k N
1
o
KN
20 KN
b)
A
a
= 0
, Ay
Gy
=
(
w + 20 +
io
)
KN
For eoint
A
.
Gt
Ta
MA to
Gy
.
12 m
- 10 KN
. 8m
20
KN
. Gm
- / o
KN.
4m=O
Gy
= 20
KN
{
Ay
= 20 KN
C)
Method
of
joints
.
.
If
the
entire
truss is at
equilibrium
then each
joint
must
he in
equilibrium
Method
of
section
.
.
If
the
entire
truss is at
equilibrium
then each member
must
he in
equilibrium
d) For
joint
A
TT
tFy=O
;
20 -
Fay
sin
450=
FAL
=
KN
→ I7x=
5
FAB
28 Cos 450=
Faa
=
20 KN
Y
KIM
A
g-
FAB
,
ZOKN
For the
joint
L
\y+
ZFr=
FLC =o
fuc
Y
; Fe
nullTry
=o
FLK
=O
\*Ee
FLK
= 28 KN
FLAE
KN
2C
For the
joint
K
y
↳
t Its
-
;
kμ•#g
FKT
1°sin45°
7k
,
cos
(450-26)=
¢§o
null°
FKD =
KN (
compression )
FKL
28
μ p
2C
10kW
KD
A
IFy=
;
locos 450 + 7 (
450-26.
)
Fky
=O
Fks
=
kW
Plastic
deformation
:
Py
=
6ypr2=
. q
|°'2zI
)
=
17
. ,
MN
Stable
structures for
all members
,
no
bucketing
& Hastie
Deformation
Problem
2
y
A)
ooh
My
x
y
the
oee↳⇒tkaEee*
.
g
Z
KIEIFEEEHI
'm
asm
E
Z
ten
.
b)
front
section
M
,
back
section
M
M=5KN
.
- =
15km m
c
)
Fy
=o =
Vy
- 5
(
¥
) = 0
Vy
= 3kW
Fz
= o = Vz
- 5
(
¥
) =
o Vz
=
- 4kW
2-
My
=
0
My
5(4z)
to.
=o
My
=
KN. in
I
Mz = 0
MZ
5( ¥
) 10=0 Mz =
KN . m
Iy
=
Iz
=
####### EN
=
¥
( o
4
=
2 ×
co
m
o =
the
Mtfzy
Bending
cause
only
she
normal stress
forrpoinf A
,
Y=
0
,
-2=
g=
μZY_
- 0
Iz
=
0×03 N. m
to
.o4m
)
0×(10-6) a-
my
= -
17
'9MPa
=
17
for
point
Is
Y=
0
,
2=0.
on
Myz
J
=
0 +
Ly
=
1×
N. M
(
0 ' 04h
)
=
23
d)
,
- ×
Cio
- Ti
m
•
a
-27g
'
- 0
Try
'=
FM
I
Mz•=O
Mz
'
- F
N.
o .5m= 0
M
¥¥↳
,
Mz
"
=
Fm
N .
in
p
.
MZY
Myz
'
r
=
t -
Iz
'
Iy
'
Iz
'=Iy'
=I£=Iy
=
qtr
4
TM =
Mz
'
y
"
null0 =
350
MPA
,
Mz
'
isthemaxmium
of
bending
moment
Problem 3
a)
→
W
B.
••c
p
p
→
→
RA
RB
b)
RAT RB=
W
W
= Zoo
kg
.
9%
=
i. KN
Dmeto
symmetric
structure
RA = RB
=
o/kN
c)
→
,
yeti
'÷
:*
§ }
Np
Ne (
String
)
(
Vessel
)
Cy
Cy
b/c•t→c×
a
,
fd
t
← •o
N
't
@
ight
leg
)
(
Left
leg ) ,
I
RB
Rap
C
→
T
To - the
vessel
→
-27×-0 Nb
cos
45°
N
⇐
cos
450 = 0
No =
Nz
- P
IFY
=o
N
,
sin 450
Nesin
450
W
= 0
fz Np
= W
Np
= NE
=
IZW
=
EZ
.
i.
KN
=
387
KN
choose one side
of
The stand
=
Say
BC
Cy
No
/c•$→c×
heft
leg
PRB
T
←
→
IF
=
o
Nbsim45° + Cx
T
=
0 @
- T
Try
= o
Nd
cos 450 +
Cy
RB =o
@
Me =o
choose
point
C as
origin
RB
. In + T
'
In
ND
.
- m=O
3
f)
TL 2090.
N
.
= - 2nd
=
EA
2° x.
09dg
.
T,
.co#mr=6
=
66μm
0
=
IA
= 6.
Mpa
< Goo
Mpa
,
So
, safe
structure
,
no plastic
deformation expected
Considering
the
following
ratio
foo Mpa
:
Wmax Goo
Wmax
⇒ -
- 655
Mpa
: W
=
KN
Wmax
= 176,
KN
corresponding
to
18031kg
Tentamen 7 November 2016, antwoorden
Vak: Mechanics (NAMECH05E)
Universiteit: Rijksuniversiteit Groningen
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