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
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 t
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)7
(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
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 sel
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 spacin
(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
occurs within the interval rather than at a boundary section, Q= 1; ZVr=
16.5(12)71.99 = 99.5;/; = 22,000-0.679(99.5)2 = 15,300 lb/in2
(105,493.5 kPa);/2 = 12,OOO,000/[16.5(12)(4.85)] = 12,500 lb/in2
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 tru
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 sho
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
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 in
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
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
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
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 , suc
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 Proce
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-A
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
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
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. ,_ ,.
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)
(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
so
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
mech
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. Determin
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 rec
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 Co