sol guide final exam

7/31/2019 Sol Guide Final Exam

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1

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



2

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



3



4

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



5

. 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.



6

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 = -



7

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.

sol guide final exam

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