GENG 360 Eng. Economy
Fall 2012
QUIZ NO. 5
STUDENT NAME: _
STUDENT ID:
_
Problem 1:
A finishing machine has a first cost of $20,000 with a $5,000 salvage value after 5 years. Using
MACRS method (table is given at the back of the question sheet), find:
(a)
Homework 1
GENG 360 Economics - Fall 2012
Class L02
Dr. Anand Kumar
Done by: Hamzah Amen Lotfy
QU ID: 201000568
3/10/2012
Problems
?What is meant by the term time value of money
.The change in the amount of money over a given time period
.1
1
?What is the
Problem 1 For an interest rate of 2% per quarter, determine the nominal interest rate per
(a) semiannual period = 2*2% = 4: (Given 2% per 3 months, semiannual = 6 months = 2 compounding
periods)
(b) year = 4*2% = 8% (1 year = 4 compounding period)
(c) 2 y
GENG 360: Engineering Economy
Homework #4:
Each problem worth 10 points
Due Date: Nov 28, Wednesday, beginning of the class.
1. Swagelok Enterprises is a manufacturer of miniature fittings and valves. Over a 5- year period, the
costs associated with one p
Chapter 9, Problem 3.
Identify each cash flow as a benefit, disbenefit, or cost.
(a) $500,000 annual income from tourism created by a freshwater reservoir
(b) $700,000 per year maintenance by container ship port authority
(c) Expenditure of $45 million fo
GENG 360: Engineering Economy
Homework #3:
Due Date: October 24th @ beginning of class
NOTE: EACH PROBLEM IS WORTH 4POINTS.
Homework and exams should meet the following guidelines:
- Indicate name, date, page #, and course # on cover page.
- Clearly ident
GENG 360: Engineering Economy
Homework #1:
Due Date: October 3rd @ beginning of class
NOTE: EACH PROBLEM IS WORTH 4POINTS.
Homework and exams should meet the following guidelines:
- Indicate name, date, page #, and course # on cover page.
- Clearly identi
4 8 BEAM AND GIRDER DESIGN DES
When Lp<L,SL,
The exural design strength of noncompact I or C rolled shapes bent about the major '
axis is determined by: ,
W. = MM. (MM M.) [if 12% <1)an
In the Load Factor Design Selection Table, in the case
J"
F.
mm~
wwwrw
-a
W
K Moment of inertia
M ,1, , 1m, Flexural strength defined in Equations
SYMBOLS 6 l9
Specified minimum yield stress of the web, MPa (Table B5.l)
Shear modulus of elasticity of steel, MPa (77 200 MPa) (F12)
Horizontal force, N (Cl)
6l7
Symbols
The section number in parentheses after the definition of a symbol refers to the section
where the Symbol is first defined.
A . Cross-sectional area, mm2 (F1.2)
AB Loaded area of concrete, mm2 (12.4)
Ab Nominal body area of a fastener, mm2 (13
4 _ 54 BEAM AND GIRDER DESIGN
Fy = 250 MPa
BEAMS
W Shapes
Maximum factored uniform loads in kN
for beams laterally supported
For beams laterally unsupported, see page 4-113
In -
1.02 1120 1010 915 823 762
1370 1220
Span (m)
cu
m.
E
o
m
(\I
H
u?
Properti
4 - g4 . , BEAM AND GIRDER DESIGN
F = 345 MPa
BEAMS
W Shapes
Maximum factored uniform loads in kN
for beams laterally supported
For beams laterally unsupported, see page 4-137
._
- 3820 3410
5040 4540 4040
W 530
Properties and Reaction
WW
4 _ 32 . BEAM AND GIRDER DESIGN
2. Enter the Factored Uniform Loads Table for E = 345 MPa and
Wu 2 614 kN.
For W460X106: W, = 660 kN > 614 kN; however, Lb = 3 m > L1, =
1.83 m.
For W460X113: W, = 737 kN > 614 kN; however, L, = 3 m > L1, =
2.81 m.
3. Si
4 - 40 BEAM AND GIRDER DESIGN
W 840 BEAMS
W Shapes
Maximum factored uniform loads in kN
for beams laterally supported
For beams laterally unsupported, see page 4-113
we
3 30
Properties and Reaction Values
ZX / 103 mm3
ch kN-m
<
3,:
SYMBOLS
Required flexural strength in member assuming there is no lateral trans
lation of the frame, Nmm (C1)
Plastic bending moment, N-mm (F1 . l)
Moment defined in Equations AH3-5 and AH36, for use in alternate
interaction equations for combined ben
(H
\v \v
\y
DESIGN STRENGTH OF BEAMS 4 _ 9
Table 4-1 .
Values of Cb for Simply Supported Beams
Lateral Bracing
Along Span
1.67 1.67
1.14 T
At load points
7.67 I. 0 7.67
For Lm > L,
Cbn E] G] 4CWM2
L = V y + 1 L
Mp 2 1
SYMBOLS
h
0 Separation ratio for builtup compression members 2 27 (E4)
ib
Translation deflection of the story under consideration, mm (Cl)
A011
7 Depth tapering ratio (Appendix F3). Subscript for tapered members
(Appendix F3)
7 Unit weight of water,
4 76 BEAM AND GIRDER DESIGN
Fy = 345 MPa,
w 920 BEAMS
W Shapes
Maximum factored uniform loads in kN
for beams laterally supported
For beams laterally unsupported, see page 4-137
6700 6330 5920 5630
Fy= 345 MPa
Span (m)
Properties and Rea
Dug;
FACTORED UNIFORM LOAD TABLES
The following expression may be used for calculating the tabulated uniformly distributed
factored load W, on a simply supported beam or girder:
Wu=bVK/L,kN
For compact shapes, the tabulated c
W 1100 BEAMS
, W Shapes
: I Maximum factored uniform loads in kN
for beams laterally supported
For beams laterally unsupported, see page 4-137
'-_
345 MPa
Fy=
Properties and Reaction Values
zx/1o3 mm3 18100
:1)ch kN-m
4)an kN
H1 kN
R2 kN/mm
rR3 kN
rR5 W
6-38
DESIGN REQUIREMENTS [Chap B
TABLE 35.1
Limiting Width-Thickness Ratios for
Compression Elements
Description of Element ' (non compact)
Flanges of I -shaped rolled b /t 0. 83VE / (F 69)
beams and channels' In flexure
, Flanges ofI- -shaped hybrid or b