Answer:
ok i will on day
Explanation:
A steel plate of width 120mm and thickness of 20mm is bent into a circular arc radius of 10. You are required to calculate the maximum stress induced and the bending moment which will give the maximum stress. You are given that E=2*10^5
Answer:
Hence the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
Explanation:
Width of steel fleet = 120 mm The thickness of steel fleet = 10 mm Let the circle of radius = 10 mNow,
We know that,
[tex]\frac{M}{I} = \frac{E}{R}[/tex]
Thus, [tex]M =\frac{EI}{R}[/tex]
Here
R = 10000 mm
[tex]I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}[/tex]
[tex]E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}[/tex]
Hence, the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
A machine component is loaded so that stresses at the critical location are σ1 = 20 ksi, σ2 = -15 ksi, and σ3 = 0. The material is ductile, with yield strengths in tension and compression of 60 ksi. What is the safety factor according to (a) the maximum normal-stress theory, (b) the maximum-shear-stress theory, and (c) the maximum distortion-energy theory?
Answer:
a) the safety factor according to the maximum normal-stress theory; n = 3
b) the safety factor according to the maximum-shear-stress theory; n = 1.714
c) the safety factor according to the maximum distortion-energy theory is; n = 1.97278
Explanation:
Given the data in the question;
a) What is the safety factor according to the maximum normal-stress theory;
According to the maximum normal-stress theory
n = S[tex]_y[/tex] / σ[tex]_{max[/tex]
since σ₁ = 20 ksi is greater than σ₂ = -15 ksi
σ[tex]_{max[/tex] = 20 ksi and yield strengths in tension and compression S[tex]_y[/tex] = 60 ksi
we substitute
n = 60 ksi / 20 ksi
n = 3
Therefore, the safety factor according to the maximum normal-stress theory; n = 3
b) What is the safety factor according to the maximum-shear-stress theory.
According to maximum-shear-stress theory;
τ[tex]_{max[/tex] = [(σ₁ - σ₂) / 2]
= S[tex]_y[/tex] / 2n
n = S[tex]_y[/tex] / 2[(σ₁ - σ₂) / 2]
n = S[tex]_y[/tex] / (σ₁ - σ₂)
we substitute
n = 60 ksi / (20 ksi - (-15 ksi))
n = 60 ksi / (20 ksi +15 ksi)
n = 60 ksi / 35 ksi
n = 1.714
Therefore, the safety factor according to the maximum-shear-stress theory; n = 1.714
c) the safety factor according to the maximum distortion-energy theory?
By distortion energy theory
σ₁² + σ₂² - σ₁σ₂ = (S[tex]_y[/tex]/n)²
we substitute
(20)² + (-15)² - ( 20 × -15 ) = ( 60 / n )²
400 + 225 + 300 = 3600 / n²
925 = 3600 / n²
n² = 3600 / 925
n = √( 3600 / 925 )
n = 1.97278
Therefore, the safety factor according to the maximum distortion-energy theory is; n = 1.97278
Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient pressure. Be sure to state clearly which explanation is for above and which is for below ambient pressure. ii. Suggest two reasons each why reactors are run at a.) elevated pressures and/or b.) elevated temperatures. Be sure to state clearly which explanation is for elevated pressure and which is for elevated temperature
Solution :
Methods for selling pressure of a distillation column :
a). Set, [tex]\text{based on the pressure required to condensed}[/tex] the overhead stream using cooling water.
(minimum of approximate 45°C condenser temperature)
b). Set, [tex]\text{based on highest temperature}[/tex] of bottom product that avoids decomposition or reaction.
c). Set, [tex]\text{based on available highest }[/tex] not utility for reboiler.
Running the distillation column above the ambient pressure because :
The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.
Run the reactor at an evaluated temperature because :
a). The rate of reaction is taster. This results in a small reactor or high phase conversion.
b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.
Run the reaction at an evaluated pressure because :
The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.
The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.
Một doanh nghiệp có tư bản đầu tư là 600,000 usd, cấu tạo hữu cơ tư bản 3/1. Xác định tiền công trả cho người lao động
Answer:
450.000 USD
Explanation:
Đây,
Cấu trúc hữu cơ 3/1 có nghĩa là ¾ một phần vốn được chuyển vào chi phí cố định và ¼ một phần đi vào chi phí biến đổi.
Tiền lương của người lao động là chi phí cố định và do đó, mức lương
= ¾ * 600.000 USD
= 450.000 USD
The answer to the question mark the park in a
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. 60 45 30
Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Compute the first four central moments for the following data:
i xi
1 45
2 22
3 53
4 84
5 65
Answer:
Compute the first four central moments for the following data:
i xi
1 45
2 22
3 53
4 84Explanation:
7. The binary addition 1 + 1 + 1 gives
11 [2-bit]
011 [3-bit]
0011 [4-bit]
________
1 + 1 + 1 = 3
________
3 = 2 + 1
2¹ 2⁰
3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)
3 = ..011
Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.
[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]
[ 128 64 32 16 8 4 2 1 ]
Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of 100w/m^2 from the outer surfance of the pipe to the soil determine the exit temperature of the water.
a. 129.1
b. 111.1
c. 82.1
d. 68.1
Complete Question
Complete Question is attached below
Answer:
Option A
Explanation:
From the question we are told that:
inner Diameter of pipe [tex]d_i=100^c[/tex]
Thickness [tex]t=50mm[/tex]
Outer diameter of pipe [tex]d_o=1.1m[/tex]
Length [tex]l=5m[/tex]
Temperature [tex]T_i=130^oC[/tex]
Generally the equation for Heat Balance is mathematically given by
[tex]q*\pi d_oL=mC_p(T_i-T_o)[/tex]
Therefore
[tex]T_o=T_i+\frac{q*\pi d_oL}{mC_p}[/tex]
[tex]T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}[/tex]
[tex]T_o=129.136^oC[/tex]
Therefore the exit temperature of the water.is [tex]T_o=129.136^oC[/tex]
Option A
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Answer:
Explanation:
do you have any other questions besides "tyuuyiopopiouyttrrtrffrlkl,k;;';'l.l"
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The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 49300 0.11 61300 0.21 What true stress is necessary to produce a true plastic strain of 0.25
Answer:
64640.92 psi
Explanation:
True stress ( psi ) True strain
49300 0.11
61300 0.21
Determine the true stress necessary to produce a true plastic strain of 0.25
бT1 = 49300
бT2 = 61300
бT3 = ?
∈T1 = 0.11
∈T2 = 0.21
∈T3 = 0.25
note : бTi = k ∈Ti^h
∴ 49300 = k ( 0.11 )^h ----- ( 1 )
61300 = k ( 0.21)^h ------ ( 2 )
solving equations 1 and 2 simultaneously
49300/61300 = ( 0.11 / 0.21 )^h
0.804 = (0.52 )^h
next step : apply logarithm
log ( 0.804 ) = log(0.52)^h
h = log 0.804 / log (0.52)
= 0.33
back to equation 1
49300 = k ( 0.11 )^0.33
k = 49300 / (0.11)^0.33
= 102138
therefore бT3 = K (0.25)^h
= 102138 ( 0.25 )^ 0.33
= 64640.92
A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameterand is 65 ft long. The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft. Please estimate:
This question is incomplete, the complete question is;
A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.
The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.
Please estimate: The required brake horsepower.
Answer:
The required brake horsepower is 1400.08
Explanation:
Given the data in the question;
Power required to drive the pump can be determined using the formula;
P = r[tex]_w[/tex]QH / η₀(0.745)
given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.
Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec
the head rise across the pump is 320 ft,
H = 320 ft = ( 320 × 0.3048 )m = 97.536 m
the efficiency η₀ = 81% = 0.81
r[tex]_w[/tex] = 9.81 kN/m³
so we substitute our values into the formula
P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)
P = 844.87926528 / 0.60345
P = 1400.08 HP
Therefore, The required brake horsepower is 1400.08
A long power transmission cable is buried at a depth (ground-to-cable-centerline distance) of 2 m. The cable is encased in a thin-walled pipe of 0.1-m diameter, and, to render the cable superconducting (with essentially zero power dissipation), the space between the cable and pipe is filled with liquid nitrogen at 77 K. If the pipe is covered with a super insulator (ki = 0.005 W/mK) of 0.05-m thickness and the surface of the earth (kg = 1.2 W/mK) is at 300 K, what is the cooling load (W/m) that must be maintained by a cryogenic refrigerator per unit pipe length?
Answer:
[tex]q=9.9w/m[/tex]
Explanation:
From the question we are told that:
Ground-to-cable-center line Distance [tex]d_g=2m[/tex]
Diameter of Cable case [tex]d=0.1m[/tex]
Temperature of Nitrogen [tex]T_n=77K[/tex]
Insulator [tex]ki = 0.005 W/mK[/tex]
Thickness [tex]t=0.05[/tex]
Mass [tex]kg = 1.2 W/mK[/tex]
Temperature of earth surface [tex]T_e=300K[/tex]
Generally the equation for Heat rate per unit length is mathematically given by
[tex]q=\frac{T_g-T_n}{R_g+R_e}[/tex]
Where
[tex]R_g=[kg(\frac{2\pi}{(in4d_g/d_x)}]^{_1}[/tex]
[tex]R_g=[(1.2)(\frac{2\pi}{(in4(2)/0.2)}]^{-1}[/tex]
[tex]R_g=0.489[/tex]
And
[tex]R_e=\frac{In(\frac{D_0}{D_1})}{2\pi ki}[/tex]
[tex]R_e=\frac{In(2)}{2*3.142 0.005}[/tex]
[tex]R_e=22.1[/tex]
Therefore
[tex]q=\frac{223}{0.489+22.064}[/tex]
[tex]q=9.9w/m[/tex]
A spring having a stiffness k is compressed a distance δ. The stored energy in the spring is used to drive a machine which requires power P. Determine how long the spring can supply energy at the required rate.
Units Used: kN = 103 N
Given: k = 5 kN/m; δ = 400 mm; P = 90 W
Answer:
30w
Explanation:
When you double the velocity, you _____ the kinetic energy.
Answer:
you quadruple the kinetic energy
