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
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]
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
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 ]
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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
