1.
2.
CH. 18
Award: 7.14 out of114- polnts 5W WEW
Penne Pharmaceuticals sold 8 million shares of its $1 par common stock to provide funds for research and
development.
If the issue price is $12 per share, what is thejoumal entry to record the sale of the
NOVA SOUTHEASTERN UNIVERSITY
H. Wayne Huizenga School
of Business and Entrepreneurship
Masters Programs
ACT 5060 - Accounting for Decision Makers (Evening)
PROJECT APPENDIX A 14 PTS LAST DAY OF CLASS
REQUIREMENTS:
1. CHOOSE A PUBLICLY TRADED COMPANY THAT
CHAPTER 18
COST-VOLUME-PROFIT
SUMMARY OF QUESTIONS BY STUDY OBJECTIVES AND BLOOMS TAXONOMY
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35.
sg
36.
sg
37.
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CHAPTER 18
COST-VOLUME-PROFIT
SUMMARY OF QUESTIONS BY STUDY OBJECTIVES AND BLOOMS TAXONOMY
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CHAPTER 13
CORPORATIONS: ORGANIZATION AND CAPITAL STOCK
TRANSACTIONS
SUMMARY OF QUESTIONS BY LEARNING OBJECTIVES AND BLOOMS
TAXONOMY
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109.
110.
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112.
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114.
115.
116.
117.
118.
y
= 240 MPa = 240 N/mm2 ; F.S. = 1.5 ; b = 125 MPa = 125 N/mm2 ; b = 240 mm
Fig. 9.33 Fig. 9.34
Diameter of rivets
Let d = Diameter of rivets.
We know that direct shear load on each rivet,
Ps
=
3
100 10
25 000 N
4
Pn
= =Riveted Joints 335
= 150 000 N
Efficiency of the joint
We know that the strength of the unriveted or solid plate,
P = p t t = 100 20 120
= 240 000 N
Efficiency of the joint
=
Least of , and P P P t s c
P
=
150 000
240 000
= 0.625 or 62.5% Ans.
9.15 Design of Boiler Joints
F F l 16 008 N
l
= = .(Q l5 = l2)
and F
6=
6
11
1
F F l 28 846 N
l
= = .(Q l6 = l1)
By drawing the direct and secondary shear loads on each rivet, we see that the
rivets 1, 2 and 3
are heavily loaded. Let us now find the angles between the direct and se
t d = 23.5
h
2
h
1
X
h
1
h
2
X
All dimensions in mm.
Fig. 9.36338 A Textbook of Machine Design
Tensile stress in plates = 100 MPa ; Shear stress in rivets = 75 MPa; and bearing
stress in rivets and
plates = 150 MPa.
circumferential joint. The longitudinal joint is used to join the
ends of the plate to get the required diameter of a boiler. For
this purpose, a butt joint with two cover plates is used. The
Control lever
Air in
Air out
Cylinder
Piston repeatedly forced
D=
D
A
A
3 300
0.15 N
20 300
=
lP
FP
l
The secondary shear loads on each rivet act at right angles to the lines joining the
centre of the
rivet to the centre of gravity of the rivet system as shown in Fig. 9.30.
Now let us find out the resultant shear loa
1
F 4( ) 2( ) l l
l
+
60 103 200 = 1
90.1
F
[4(90.1)2 + 2(50)2] = 416 F1
or F
1 = 60 103 200 / 416 = 28 846 N
Since the secondary shear loads are proportional to the radial distances from the
centre of
gravity, therefore
F2
=
4
d2=22
4
(25)2 100 = 196 375 N
(iii) Crushing resistance of the rivets
Since the joint is double riveted, therefore the strength of two rivets is taken. We
know that
crushing resistance of the rivets,
P
c
=ndt
c
= 2 25 20 150 = 150 000 N
Strength of the
R
D
R
C
FA
FB
30
30
30
30
* M.I. of four rivet holes about X-X
= M.I. of four rivet holes about their centroidal axis + 2 A(h1)2 + 2 A(h2)2
where A = Area of rivet hole.Riveted Joints 337
=
113322
9. What do you understand by the term efficiency of a riveted joint? According to
I.B.R., what is the
highest efficiency required of a riveted joint?
10. Explain the procedure for designing a longitudinal and circumferential joint for
a boiler.
11. Descri
F
F
A = P 100 3 / 2000 = 3 P/20= 0.15 P N
Since the secondary shear loads are proportional to their radial distances from the
centre of
gravity, therefore
FB
=
B
A
A
3 100
0.05 N
20 300
=
lP
l
1 = l3 = l4 = l6 = (75) (50) 2 2 + = 90.1 mm
and l
2 = l5 = 50 mmRiveted Joints 333
Now equating the turning moment due to eccentricity of the load to the resisting
moments of the
rivets, we have
Pe=1222222
123456
1
F()()()()()()llllll
l
+
=122
12
90
37 500 37 500 N
90
=
l
F
l
Now let us find the resultant shear load on each rivet.
From Fig. 9.34, we find that angle between FA and Ps = A = 150
Angle between FB and Ps = B = 150
Angle between FC and Ps = C = 30
Angle between FD and Ps = D = 30
Resul
Consider the weakest section of the plate (i.e. the section where it receives four
rivet holes of
diameter 23.5 mm and thickness t mm) as shown in Fig. 9.36. We know that
moment of inertia of the
plate about X-X,
I
XX = M.I. of solid plate about X-X *M.I.
4 1.5
=dd
d 2 = 60 455 / 125.7 = 481
or d = 21.9 mm
From Table 9.7, we see that the standard diameter of the rivet
hole (d ) is 23.5 mm and the corresponding diameter of rivet is
22 mm. Ans.
Thickness of the plate
Let t = Thickness of the plate in
mm,
b
F
A = 7500 103 / 200 = 37 500 N
Since the secondary shear loads are proportional to their radial distances from the
centre of
gravity, therefore,
FB
=
B
A
A
30
37 500 12 500 N
90
=
l
F
Ps
=
3
60 10
10 000 N
6
Pn
=
Let F
1, F2, F3, F4, F5 and F6 be the secondary shear load on the rivets 1, 2, 3, 4, 5 and 6
at
distances l
1, l2, l3, l4, l5 and l6 from the centre of gravity (G) of the rivet system. From the
symmetry of
the figure, we find
From Fig. 9.34, we find that
lA
=l
D = 60 + 30 = 90 mm and lB = lC = 30 mm
We know that
P e = AAA B C D A B 2 2 2 2 2 2
AA
F F l l [( ) ( ) ( ) ( ) ] 2( ) 2( ) l l l l l l + + + = +
.(Q lA = lD and lB = lC)
7500 103 = A 2(90) 2(30) 200 2 2 A
90
+=
F
F