Project 2- ENGRCEE 283
The equations used are below:
2 U x U i+1, j2 U i , j+ U i1, j
=
x2
x2
2 U y U i , j +12 U i , j+U i , j 1
=
y2
y2
From the above equations we can get:
2 U x 2 U y
U 2 U i , j +U i1, j U i , j+12 U i , j +U i , j1
+
=0= i+ 1,

1/2(62.4)(20)2 = 12,480 Ib (55.51 kN), where w = weight of water,
lb/ft3 (N/m3 ), and h = water height, ft (m), thenj = (1 /s)(20) = 6.67 ft
(2.03 m). 3. Compute the moment of the loads with respect to the base
centerline Thus, M= 18,000(8 - 7.5) + 9000(1

alignment chart, determine that Kx= \ .77. 5. Compute the slenderness
ratio with respect to both principal axes, and find the allowable stress
Thus, KJLIrx = 1.77(28)(12)/6.17 = 96.4; KyLlry = 28(12)73.71 = 90.6.
Using the larger value of the slenderness

(VS)WL = (1/8)(45,000)(25)(12) = 1,688,000 in-lb (190,710.2 N-m);/=
MIS = 1,688,000/56.3 = 30,000 lb/in2 (206,850.0 KPa). This stress is
acceptable. DESIGN OF A BEAM WITH REDUCED ALLOWABLE
STRESS The compression flange of the beam in Fig. Ia will be brace

Also, Lc = 7.6 > 5. Hence, the beam is acceptable. 2. Compute the
equivalent load for a member of A242 steel To apply the AISC Manual
tables to choose a member of A242 steel, assume that the shape selected
will be compact. Transform the actual load to an

2(1.5)(2400)/444 = 16.2 in (411.48 mm). However, the Specification
imposes additional restrictions on the weld spacing. To preclude the
possibility of error in fabrication, provide an identical spacing at the top
and bottom. Thus, smax = 21(0.375) = 7.9 i

(298.906 kN). This solution method used in part b is termed the static, or
equilibrium, method. As this solution demonstrates, it is unnecessary to
trace the stress history of the member as it passes through its successive
phases, as was done in part a; t

for the internal force in a particular member, and calculate the relative
displacement A, of the two ends of that member caused solely by this
force. Now remove this member to secure a determinate truss, and
calculate the relative displacement Aa caused s

the mechanism method. Assume, for part a, a plastic hinge at B and C. In
Fig. 25, construct the force diagram and bending-moment diagram for
span A C. The moment diagram may be drawn in the manner shown in
Fig. 25b or c, whichever is preferred. In Fig. 25

ft (2.890 m). 2. Assume several trial load positions Assume that the
maximum moment occurs under the 10-kip (44.5-kN) load. Place the
system in the position shown in Fig. 4Ib9 with the 10-kip (44.5-kN) load
as far from the adjacent support as the resultan

Fb = 22 kips/in2 (151.7 MPa); Cm = 0.793. 2. Obtain the properties of
the section From the Manual for a W12 x 53, A = 15.59 in2 (100.587
cm2 ); Bx = 0.221 per inch (8.70 per meter); ax = 63.5 x 106 in4 (264.31
x 1Q3 dm4 ). Then when KL = 20 ft (6.1 m),Pal

this force to the W shape. Thus/mean = MyII= 215,600(12)(9.31)/1634 =
14,740 lb/in2 (101,632.3 kPa). Then P = ^/mean = 3.75(14,740) =
55,280 Ib (245,885.4 N). Use a 1X4-Ui (6.35-mm) fillet weld, which
satisfies the requirements of the Specification. The c

ENGRCEE 283
UCI
MTH MTHDS ENGR ANLY
Instructor: Prof. Gao
Fall 2011
APPLYING FILTERS TO FOURIER
SPECTRUM OF IMAGES
- Term Project Report -
Fall 2011
The Lowpass filters attenuate high frequencies and leave the low frequencies.
There are 3 main Lowpass fil

Review: Circle the right one from the multiple choices
Matrix
0 with i j
a ij
1) A square matrix with
is
0 with i j
a) a diagonal matrix
b) an upper triangle
c) a lower triangle matrix
2) Fill the right matrix dimensions (subscripts) for matrix multipl

Homework # 3 (Due Thursday 11/17/2016)
1. Greens Theorem F d r F k dA provides a method to compute the
C
R
area A of a region R enclosed by a closed curve C, just by assuming a vectorial field
F , such that F k 1 , and therefore F d r dA A . Note, that an

web plate and two 20 x 3/4 in (508.0 x 19.05 mm) flange plates. The
unbraced length of the compression flange is 18 ft (5.5 m). If Cb = 1,
what bending moment can this member resist? Calculation Procedure: 1.
Compute the properties of the section The tabl

of maximum deflection Set 6 = O and solve for ;c to locate the section of
maximum deflection. Thus L2 - 3x2 = O; x = L/35 . Substituting in the
deflection equation gives ymsai = M,2 /(9/35 ). MOMENT-AREA
METHOD OF DETERMINING BEAM DEFLECTION Use the
momen

DOUBLE-INTEGRATION METHOD OF DETERMINING BEAM
DEFLECTION The simply supported beam in Fig. 36 is subjected to a
counterclockwise moment N applied at the right-hand support.
Determine the slope of the elastic curve at each support and the
maximum deflectio

Ascertain whether the Specification requires a reduction in the flange
area. Therefore gross flange area = 2(6.94) + 7.0 = 20.88 in2 (134.718
cm2 ); area of rivet holes = 2C/2)(1)4(3 /4)(1) = 4.00 in2 (25.808 cm2 );
allowable area of holes = 0.15(20.88) =

has a specific shear but zero moment; i.e., it is simply supported at A. At
C, the given beam has a specific slope and a specific deflection.
Correspondingly, the conjugate beam has both a , u. ,_ ,. . . . . shear and a
bending moment; i.e., it has a b) F

NL/(6EI). 3. Determine the right-hand slope in an analogous manner 4.
Compute the distance to the section where the slope is zero Area MED =
area MBC(x/L)2 = Nx1 J(TEIL); 0E = B1 - area MED = NL/(6EI) - Nx2
/(2EIL) = Q;x = L/3Q5 . 5. Evaluate the maximum

(102,183.9-kPa) compression;/* = 9000/12.56 - 0.84(900Ox 8)76.28 =
8930-lb/in2 (61,572.3-kPa) tension. SOfL PRESSURE UNDER DAM A
concrete gravity dam has the profile shown in Fig. 34. Determine the
soil pressure at the toe and heel of the dam when the wat

refer to the basic mechanisms and the subscript 3 to their composite
mechanism. Then Mpl = C1Ii1-, Mp2 = C2Ii2. When the basic
mechanisms are superposed, the values of WE are additive. If the two
mechanisms do not produce rotations of opposite sign at any

in the front and center rows and 6 ft (1.82 m) in the rear row. The
resultant of vertical loads on the wall is 20,000 Ib/lin ft (291.87 kN/m)
and lies 3 ft 3 in (99.06 cm) from the front row. Determine the pile load
in each row. (b) Plan (c) Pile reaction

calculation procedure, Rds. = 18,040 Ib (80,241.9 N); Rb =
42,440(0.375) - 15,900 Ib (70,723.2 N); s = 15,900/3290 - 4.8 in (121.92
mm), where s = allowable rivet spacing, in (mm). Therefore, use a 4%-in
(120.65-mm) rivet pitch. This satisfies the require

will be compact The section modulus S = MIf= 340.3(12)/24 = 170.2 in3
(2789.58 cm3 ). (c) Bending-moment diagram FIGURE 3 3. Record the
properties of the beam section Refer to the AISC Manual, and record the
following properties for the Wl 8 x 60; d = 18.

the previous calculation procedure, using the approximate method of
determining the shearing stress in a beam. Calculation Procedure: 1.
Assume that the vertical shear is resisted solely by the web Consider the
web as extending the full depth of the secti