Answer:
Tensile stress = 0.1855Kpsi
Total deformation = 0.0012243 in
Unit strain = 1.855 *10^-5 or 18.55μ
Change in the rod diameter = 5.02 * 10^ -6 in
Explanation:
Data given: D= 0.82 in
L = 5.5 ft * 12 = 66 in
load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm
Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²
∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²
Rt = 0.1855 Kpsi
∴ Total deformation = PL / AE = Rt * L/ Eal
= 0.1855 * 10³ * 66 / 10000 * 10³
= 0.0012243 in
∴the unit strains = total deformation / L = 0.0012243/ 66
=0.00001855 = 1.855 *10^-5
= 18.55μ
∴ Change in rod Δd/ d = μ ΔL/L
= (0.33) 1.855 *10^-5 * 0.82
= 5.02 * 10^ -6 in
Draw a sinusoidal signal and illustrate how quantization and sampling is handled by
using relevant grids.
A signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ration of 0.9, calculate the minimum necessary cycle length.
Answer:
[tex]T_o=82.1sec[/tex]
Explanation:
From the question we are told that:
Lost Time [tex]t=12secs[/tex]
Sum of critical flow ratios [tex]X=0.72[/tex]
Generally the Webster Method's equation for Optimum cycle time is is mathematically given by
[tex]T_o=\frac{1.5t+5}{1-x}[/tex]
[tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]
[tex]T_o=82.1sec[/tex]
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.
Answer:
1. Drawing grid.
2. Orthogonal.
3. Geometries.
4. CTRL+N.
5. Thirteen (13).
Explanation:
CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.
Some of the features of an AutoCAD software are;
1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.
2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.
3. Tangent is a point where two geometries meet at just a single point.
4. If you want to create a new drawing, simply press CTRL+N for the short cut key.
5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx
Answer:
Interval: x∈ ( 0, ∞ )
There are no transient terms
Explanation:
x (dy/dx) – y= x^2sinx
Attached below is the detailed solution of the Given problem
There are no transient terms found in the general solution
Interval: x∈ ( 0, ∞ )
A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.
Answer:
a) Vr = - a^2/r cosθ + aß / r
Vθ = 1/r [ -a^2/r * sinθ ]
b) attached below
Explanation:
potential function
Ø= a^2 /r cosØ + aßlnr ----- ( 1 )
a = radius , ß = constant
a) Expressions for Vr and Vθ
Vr = dØ / dr ----- ( 2 )
hence expression : Vr = - a^2/r cosθ + aß / r
Vθ = 1/r dØ / dθ ------ ( 3 )
back to equation 1
dØ / dr = - a^2/r sinθ + 0 --- ( 4 )
Resolving equations 3 and 4
Vθ = 1/r [ -a^2/r * sinθ ]
b) expression for stream function
attached below
Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?
Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]
Therefore, the current consumed is 3.3 A
Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?
Answer:
jsow
hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh
Explanation:
j
grudbDo you know who Candice is
Answer: Can these nuts fit in your mouth?
Explanation:
im just here for the points >:)
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:
Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2
