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MECH 260 Solution Guidelines to Final Examination
Question 1
(a) Shear strain in the rubber piece2mm
0.120mm
g = =
Shear stress (average) in the rubber piece Gt g=
6 60.5 10 (Pa) 0.1= 0.05 10 Pa= ´ ´ ´
Area of shear 2 6 250 40mm 50 40 10 m A -= ´ = ´ ´
Load 6 60.05 10 50 40 10 N = 100N P At -= ´ = ´ ´ ´ ´
g
(b) Shear load on the bolt (see free-body diagram)
50 N2
P = =
P
P
2 P
2 P
d
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2 P
2 P
P
Bottom Part
of Arm
Bolt diameter 310mm=10 10 md -= ´
Average shear stress on bolt X-section2
2
4
P
d p
=
6
23 2 2 2
50(N) 50 50 410 Pa = MPa
10(10 10 ) m 10
4 4
p p p-
´= = ´
´´ ´ ´
2MPa 0.64MPa
p= =
Question 2
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Question 3
T = 160 Nm L = 0.5 m
tallow = 82 MPa G = 80 GPa
(a):
.T a
J t =
4
.
2
T a
a
tp
= 32
allow
T a
p t=
×
(b):
GJ
LT .=f
4
.
2
T L
G a
tp
=
×
(c):
a = 1.075x 10-2
m = 10.75 mm
d = 2a = 21.50 mm
f = 0.05 Rad
f = 3.18º
7 points
6 points
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. o
hollow
T a
J t =
4 4
.
( )2
o
o i
T a
a a
tp
=
× -
&2
oi
aa =
Sub for ai and solve for ao : 3
2
1(1 )
16
o
allow
T
ap t
=× × -
(d): The conclusion is that the inner core does not carry much load (torque). With a slight
increase in diameter from 21.5 mm to 21. 96 mm (i.e., keeping the size more or less the
same), a core of diameter 10.98 can be removed, with significant savings in material while
providing the same strength.
The weight ratio,
22
2 22 2
) ) ) 4
) ) ) ( )4
solid solid solid solid solid
hollow hollow hollow hollow hollow o io i
d W m g V g A L g A d
RW m g V g A L g A d d
d d
pr r
pr r
× × × × × ×= = = = = = =
× × × × × × --
22
2
)98.10()96.21(
)50.21(
-= R
Question 4
(a) Using a virtual cut, separate a segment of the beam at distance x from the free end.
M
V
Px
L
x
Equivalent external force (transverse) on the segment Px
L=
ao
= 1.098x 10
-2
m = 10.98 mm
d o = 2C o = 21.96 mm 6 points
d i = ½d o = 10.98 mm
R = 1.27 6 points
The solid shaft is 27% heavier than the hollow shaft.
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From force balance, internal shear force at the X-section of virtual cut
PxV
L=
From moment balance (about the point of virtual cut)
02
Px x M L- - ´ =
or
2
2
Px M
L= -
P L
2
PL-
P
V
P
0
L
M
0
2
PL-
Note: The maximum internal shear force P = . It occurs at the clamped end.
The maximum internal bending moment (magnitude)2
PL= . It also occurs at the
clamped end. (b) For the box section, the neutral axis is the horizontal axis of symmetry (i.e., centroidal
axis) of the X-section.
2nd moment of area 4 4 4 41 15 5[(2 ) ]
12 12 4 I a a a a= - = =
Apply My
I s = -
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The maximum bending stress occurs at y a= (i.e., top edge of the X-section)
The bending moment with maximum magnitude occurs at the clamped end, where
2
PL M = -
We have
max 34
2( )
52 5
4
PL a PL
aa
s = - - =
Note: This occurs at the top edge of the beam X-section at the clamped end.
(c) ApplyVQ
It t =
The maximum moment of area occurs about the neutral axis.
Corresponding sectional thickness 2t a a a= - =
Corresponding moment of area37
(2 ) ( )2 2 4 8
a a a aQ a a a= ´ ´ - ´ ´ =
The maximum internal shear force occurs at the clamped end, where
V P =
Hence3
max 24
778
5 10
4
a P P
aa a
t´
= =
´
Note: This occurs at the neutral axis level (i.e., horizontal axis of symmetry) of the X-
section at the clamped end.
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