Or in 20
n Y
Problemt
3
lv
2.4
m
Point 1 Both shearand bending
compressive
y
Vn
i
Mn
v
On
ox
E
Linear
y
a
em
f
sx
Quadratic
x
at
f
20
or
ya
I
X
Y
5ketch internalreactions
y
a
z
Ans
TEXT
is
the same
t
Ixy
b
Anse
I
or 20
Point 3
No shear
µIfT
y
01 0
but has bending tensile
3
µ
Tf
Point
nY
ar
for the section justright of A
d
Point 2 shear but no bending
z
I
i
2
as
n
M
z
X
I
G
or 20
I
fI
Are
Gx
3
052t F Lx 051
of Fa Ex
pg
a
Fac x 052 FL x
o
FL x
Fac x OJ t F Lx
o
FL x a
t
f
Q
Fa Lx o
t
f
O E x Ca
Q o
Fa Lx
o
0
0 Ex La
y
x
o
t
o
tf
Cz
6
t F
2
then for
o
Cz O
FL
6
o2
0 Ex
Leg
La
f
J
A
or
Fa_
Etf fFALI
YAB
Cz O
zfFaLzx
x
Cz
for
Cz
y
y
CL
ELI Cz
Since y
For
Csx
Since 0 101 0 then
t
z
a3
2tF2xg
ELI f
FLx
F
2
FLxg a
x
yC
a
CI
y ELI Ea
Thus
FL
FLxz
D
1
ay
f
gFI
x
to B
2
FFI
3
FLx
3
F 7550N
Problem 2
b
l
y
375
4
I
O
I
I
D
II
EI
Need deflection before the
use
V
0
YBC
F
a
e
y
y
375N
so
y
Im
0.5M
2
a2
Zex
EI
Use YAB
2
62 l
5.3
y
EI
IT
fy
5.3712
can use
1
Fg
0.25M
Fa
6 EID
Im
0.5in
Adding all
Need deflection at
I
III
Needdeflection aftertheforce
so use
0.25M
b
e
x
O
YAB
F 7375N
D
1375
I
0.5M
X't 62 e
r6x_6EIe
force so
Sso
t
Im
17.45834y
v
O
0.5 m
I
yma
the force
either yarsory.sc
22E
D
for simplicity
Tggd
a circular
for
cross section
2,0052
23
22
20052
E
IDI
69
C69
l E.FIoaI
4g
22.200516
0002
f 207
109111
I'd
O 0323mor32.3
mn
a
x 05h 375Lx
of 650
II 650 x OJ 375 x
IJM 650 Lx
2
0.55 3752 0.755 t 650 Ex 15
o2550 5SOSx O S
375 Ex 0.7570 650 x i
375 Ex 0.255 5SOSx 0.57
o
05 371 Lx
2
6502
EI O
O2551 SSO x
iz
Lx
0.257
OS
z
375 Ex o 7SJ t 650 X
372
Lx 0.755 6502
2
D
172
CL
0 05
Boundaryconditions
For O S E
D
ELI
0.75 m
x
6520
0
2 072
3725
0.255
2
5250
0.572 CL
2
or
te f
plug
D
0
ELI
e
3721 x
0.255
SEL
x oS
CL
O Sm
t
Cs
Cs
69.5313
Ex L O ZSm
6520
2
0
6 9.53 3
D o
fOCOJ
XZ
69.5313
ELI
For
X
6
2
t
CLI ELI
6521
2
69.5313
4
207 109
Flogged
2 tt 0
3rador
0
O7degree
b) In this case assuming simple supports led to a very small slope. This is an upper bound, as any
rotational constraint would decrease its value. The calculation tells us that we are so close to zero
rotation that fixed or simple supports lead to similar results, so our assumption is warranted. With fixed
supports, we would find reaction moments that get us to zero rotation, but they would be almost zero
i.e. a simple support. All this happens because the shaft is much longer compared to its insertion
length in the bearing, and relatively thick, so it does not deflect much.