Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
A structure is designed using 4 circular columns. Due to a quirky design, the four columns will all carry different loads of 1800 N, 2100 N, 2275 N, and 2200 N. A factor of safety of 5 is used to design the columns. The diameter of each of the columns is supposed to be 50 cm, at most. Determine the maximum height of the structure (i.e. the column height) so that the structure will not fail. Assume that all columns may be modeled as Euler columns for your analysis. Assume a pinned-pinned boundary condition for your analysis, and assume the elastic modulus of the column material is 10 MPa.
Answer:
5.16 M
Explanation:
Loads ; 1800N, 2100N, 2275N, 2200N
safety factor = 5
diameter of each column = 50 cm = 0.5 m
Elastic modulus = 10 MPa
Calculate the max height of structure
moment of inertia for a circular section ( I ) = πd^4 / 64
lets represent the required maximum height of the column as L
Applying Euler column theory
The bucker load of the column = ( attached below )
attached below is the remaining solution
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what is tracer lathe machine
Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.
Explanation:
Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction
Answer:
Explanation:
From the given information:
weight of fiber [tex]w_f[/tex] = 3.0 g
weight of composite specimen [tex]w_c[/tex] = 4.0 g
specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g
specific gravity of fiber [tex]S_f[/tex] = 2.4
specific gravity of matrix [tex]S_m[/tex] = 1.3
The weight of the matrix = weight of the composite - the weight of fiber
⇒ (4.0 - 3.0) g
= 1.0 g
The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]
[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]
[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]
[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]
The experimental density [tex]\rho _{ce}[/tex] is determined by using the equation:
[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]
[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]
The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]
[tex]= \dfrac{1.980-1.620}{1.980}[/tex]
= 0.1818
You are given a pot of hot tea (350 mL) at a temperature of 85°C and a quantity of ice at -12°C. Determine the absolute minimum amount of ice you can add to the hot tea so that at equilibrium you have iced tea (ie. a very, very small amount of ice and some water). You can assume tea has the same density (1.00 g/mL) and specific heat (4190 J/kgK) as liquid water. The heat of fusion of H2O is 3.33x10^5 J/kg. The specific heat of ice is 2090 J/kgK.
Answer:
348 g
Explanation:
The heat gained by the ice equals the heat lost by the hot tea.
Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.
So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c = specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.
Making m subject of the formula, we have
mc(T - T') + mL = -MC(T - T")
m[c(T - T') + L] = -MC(T - T")
m = -MC(T - T")/[c(T - T') + L]
substituting the values of the variables into the equation, we have
m = -MC(T - T")/[c(T - T') + L]
m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]
m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]
m = 124652.5 J/358080 J/kg
m = 0.3481 kg
m = 348.1 g
m ≅ 348 g
So, the minimum amount of ice to be added is 348 g.
Describe the main component in an electronic control unit (ECU) used in an automobile. Indicate the function of each ingredient.
Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.
what is geo technical
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
The load on a bolt consists of an axial pull of 10 KN together with a transverse shear force of 5 KN. Find the diameter of bolt required according to I. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal strain theory, 4. Maximum strain energy theory, and 5 Maximum distortion energy theory. Take permissible tensile stress at elastic limit = 100 MPa and poisson's ratio = 0.3
Answer:
hey. its a big question. solved from *c hegg
Explanation:
how to solve circuit theory using mesh analysis
Explanation:
Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.
This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.
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Example
Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...
mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)
mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs
mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0
You will note that the matrix of equation coefficients is symmetric.
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In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.
Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m2 per 10-m length of pipe.
Answer:
f = 0.04042
Explanation:
temperature = 0°C = 273k
p = 600 Kpa
d = 40 millemeter
e = 10 m
change in P = 235 N/m²
μ = 2m/s
R = 188.9 Nm/kgk
we solve this using this formula;
P = ρcos*R*T
we put in the values into this equation
600x10³ = ρcos * 188.9 * 273
600000 = ρcos51569.7
ρcos = 600000/51569.7
=11.63
from here we find the head loss due to friction
Δp/pg = feμ²/2D
235/11.63 = f*10*4/2*40x10⁻³
20.21 = 40f/0.08
20.21*0.08 = 40f
1.6168 = 40f
divide through by 40
f = 0.04042
phân tích phương pháp gia công plasma
Answer:
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