-
Problem 7-41
Determine the x, y, z components of internal loading in the rod
at point E.
Units Used:
kN 103N:=Given:
M 3kN m:=
F
7
125
kN:=
a 0.75m:=b 0.4m:=c 0.6m:=d 0.5m:=e 0.5m:=
Solution:
Guesses
Cx 1N:= Cy 1N:= Bx 1N:=
Bz 1N:= Ay 1N:= Az 1N:=
Given
0
Ay
Az
Bx
0
Bz
+
Cx
Cy
0
+ F+ 0=
d eb c+
0
0
Ay
Az
0
b
0
Bx
0
Bz
+
0
0
a
Cx
Cy
0
+
0
b c+0
F+
0
0
M
+ 0=
-
Ay
Az
Bx
Bz
Cx
Cy
Find Ay Az, Bx, Bz, Cx, Cy,( ):=
Ay
Az
Bx
Bz
Cx
Cy
53.687
109
82116
65.6
kN=
Guesses
NEx 1N:= VEy 1N:=VEz 1N:= MEx 1N m:=MEy 1N m:= MEz 1N m:=
Given
0
Ay
Az
NEx
VEy
VEz
+ 0=
e0
0
0
Ay
Az
MEx
MEy
MEz
+ 0=
NEx
VEy
VEz
MEx
MEy
MEz
Find NEx VEy, VEz, MEx, MEy, MEz,( ):=NEx
VEy
VEz
0
53.6
87
kN=
MEx
MEy
MEz
0
43.526.8
kN m=
-
Problem 7-42
Draw the shear and moment diagrams for the shaft in terms of the
parameters shown; There isa thrust bearing at A and a journal
bearing at B.
Units Used:
kN 103N:=Given:
P 9kN:=a 2m:=L 6m:=
Solution:
P L a( ) Ay L 0= Ay PL a
L:=
x1 0 0.01 a, a..:=
Ay V1 x( ) 0= V1 x( )AykN
:=
M1 x( ) Ay x 0= M1 x( )Ay xkN m:=
x2 a 1.01 a, L..:=
Ay P V2 x( ) 0= V2 x( )Ay P
kN:=
M2 x( ) Ay x P x a( )+ 0=
M2 x( )Ay x P x a( )
kN m:=
-
0 1 2 3 4 5 65
0
5
10
Distance(m)
Forc
e(kN
)
V1 x1( )V2 x2( )
x1 x2,
0 1 2 3 4 5 65
0
5
10
15
Distance(m)
Mom
ent(k
N m
)
M1 x1( )M2 x2( )
x1 x2,
-
Problem 7-43
Draw the shear and moment diagramsfor the beam in terms of the
parametersshown.
Given:
P 8kN:= a 2.5m:= L 6m:=Solution:
x1 0 0.01 a, a..:= x2 a 1.01 a, L a..:= x3 L a 1.01 L a( ),
L..:=
V1 x( )P
kN:= V2 x( ) 0:= V3 x( )
PkN
:=
M1 x( )P x
kN m:= M2 x( )P a
kN m:= M3 x( )P L x( )
kN m:=
0 1 2 3 4 5 610
5
0
5
10
Distance (m)
Forc
e (k
N) V1 x1( )
V2 x2( )V3 x3( )
x1m
x2m
,x3m
,
-
0 1 2 3 4 5 60
10
20
30
Distance (m)
Mom
ent (
kN-m
)M1 x1( )M2 x2( )M3 x3( )
x1m
x2m
,x3m
,
-
Problem 7-44
Draw the shear and moment diagrams for the beam (a) in terms of
the parametersshown; (b) set M0 and L as given.
Given:
M0 500N m:=L 8m:=
Solution:
For 0 x L3
+ Fy = 0; V1 0=Mx = 0; M1 0=
For L3
x 2L3
+ Fy =0;
V2 0=
Mx = 0; M2 M0=
For 2L3
x L
+ Fy = 0; V3 0=Mx = 0; M3 0=
b( ) x1 0 0.01 L,L3
..:= x2L3
L3
1.01, 2L3
..:= x32 L
32 L3
1.01, L..:=
V1 x1( ) 0:= V2 x2( ) 0:= V3 x3( ) 0:=M1 x1( ) 0:= M2 x2( ) M0:=
M3 x3( ) 0:=
-
0 1 2 3 4 5 6 7 8 91
0.5
0
0.5
1
Distance in m
Shea
r for
ce in
N V1 x1( )V2 x2( )V3 x3( )
x1 x2, x3,
0 1 2 3 4 5 6 7 8 9
0
200
400
600
Distance in m
Mom
ent i
n N
- m M1 x1( )
M2 x2( )M3 x3( )
x1 x2, x3,
-
Problem 7-45
The beam will fail when the maximum shear force is Vmax or the
maximum bending moment isMmax. Determine the magnitude M0 of the
largest couple moments it will support.
Units Used:
kN 103N:=Given :
L 9 m:=Vmax 5 kN:=Mmax 2 kN m:=
Solution:
The shear force is zero everywhere in the beam.
The moment is zero in the first third and the last third of the
bam.
In the middle section of the beam the moment is M M0=
Thus the beam will fail when M0 Mmax:= M0 2.00 kN m=
-
Problem 7-46
The shaft is supported by a thrust bearing at A and a journal
bearing at B. Draw the shear andmoment diagrams for the shaft in
terms of the parameters shown.
Given:
w 500Nm
:=
L 1m:=
Solution: For 0 x L
