The CPU of Ethereum mining machine is broken
Publish: 2021-05-28 05:33:26
1. If it is concentrated in the middle of the span, it is about 0.95 tons. If it is uniform load (500x300), it is 1.9 tons
2. 1、 The bearing capacity of concrete ground is mainly affected by four factors: foundation bearing capacity, concrete grade, concrete thickness and design form
1, foundation bearing capacity (calculation target value): because the bearing capacity of concrete pavement is mainly analyzed, and the three-dimensional structure (15cm loess cushion and 15cm sand gravel cushion) designed by the design institute generally meets the foundation requirements, the foundation in the calculation is considered as infinite width (rigid) Foundation (solved according to the thickness)
2, concrete grade: the grade in concrete is directly proportional to the stiffness, that is, the greater the grade, the greater the stiffness of concrete. Therefore, the selection of too low grade concrete will lead to the network crack of the whole pavement, while the selection of too high grade concrete will lead to the excessive stiffness of the whole pavement, which is brittle, that is, easy to crack as a whole, Therefore, the correct choice of grade is also an important factor for the long-term good condition of concrete pavement, so the concrete grade in this paper is C30 grade designed by the design institute
3, concrete thickness (generally 18cm-30cm): 25cm, 28cm and 30cm are substituted according to the formula. Taking the 25cm thick C30 concrete as an example, the axial compressive strength of C30 is 20.1mpa = 20.1n/mm2 = 20.1 × 1000000 n / m2, equivalent to 20.1 × 100000 kg (five zeros divided by 10, acceleration of gravity), which is 20.1 × 100 tons, 2010 tons, that is, 2010 tons / m2. Because it is 25cm thick concrete, it needs to be multiplied by 0.25. Therefore, it is estimated that the design compressive capacity of 25cm thick C30 concrete per cubic meter is about 502.5 tons / m3 According to the preliminary calculation, C30, 25cm thick, can only bear 63.245 tons at most)
4, design form: because the above factors all consider the compression of concrete (i.e. vertical to the ground), they all follow the plain concrete scheme provided by the Design Institute, without reinforcement treatment
according to the above analysis, it can be seen that the compressive bearing capacity of plain concrete pavement mainly depends on the concrete thickness, so the ultimate bearing capacity can be calculated by formula according to the known thickness< br />Fcd=0.7· β H · FTD · um · h
FCD -- the maximum concentrated return force of concrete< br /> β H-for the thickness less than 300 mm, take 1; FTD -- axial tensile stress (1.39 MPa for C30); Um -- height conversion ratio = 2 · (a + b) + 4h, a = 20cm, B = 60cm (a, B are the width and length of wheel trace respectively); H -- thickness
the value brought in is the corresponding relationship:
C30 concrete 25cm ultimate vehicle bearing capacity: 63.245 tons; The ultimate vehicle bearing capacity of C30 concrete 28cm: 74.104t; The ultimate vehicle bearing capacity of C30 concrete 30cm: 81.732t< Second, the bearing capacity of reinforced concrete can be divided into: 1) flexural bearing capacity. This is relatively simple, under the huge section: M = afbx (h0-x / 2) - FYAs
2) shear capacity of inclined section
3) shear capacity of twisted section (also divided into tension bending, compression bending, bending torsion shear, bending torsion shear compression capacity)
4) punching shear capacity
5) local compression capacity
about so many
what is your member's stress situation, The bearing capacity of different members is different
for example:
question: for the beam concrete of building, the span is 5 meters, the span is 4 meters, the span is 3 meters, the bottom width is 15 cm, and the height is 30. How much rebar is needed
answer:
the thickness of the slab is 12cm, the live load is 0.002n/m ^, the line load is 0.002 * 5000 = 10N / mm, and the reinforced concrete is 25kn / M and 179;, Transverse load = 25 * (0.15 * 0.3 + 0.12 * 5) = 16.125kn/m, i.e. 16.125n/mm & # 179
basic combination: 1.2 * 16.125 + 1.4 * 10 = 33.35n/m
1, bending capacity
1) bending moment
m = QL ^ 2 / 8 = 33.35 * 5000 ^ 2 / 8 = 104218750 n * mm
1) relative compression height
x = H0 ^ 2-2 * m / (1.0 * 14.3 * 150 = 270 ^ 2-2 * 104218750 / (1.0 * 14.3 * 150) = - 24273.65967, less than 0,
judge that your section is too small, and the upper layer needs to be equipped with bearing reinforcement
2) re reinforcement
x = 0.55 * 270 = 148.5, 150 mm
3) the area of upper compression steel bar
A1 = (m-fbx (h0-x / 2)) / (FY * (h0-30) = (104218750-14.3 * 150 * 150 * (270-150 / 2)) / (300 * (150-30) = 1152.152778m ^, the steel bar with 3-22 can be selected
4) the area of lower tension steel bar
A2 = a1 + FBX / FY = 1152 + 14.3 * 150 * 150 / 300 = 2224.5m ^, the secondary steel bar with 6-22 can be selected
conclusion: the section of 150 can't hold 6 and 22 steel bars, and your section is a little small (because I calculate it by simply supported beam, the actual load is not so large, but the earthquake action is not considered, and my algorithm above is a little small)<
the beam section is re determined for you: 200 * 400
live load: 10N / M; Transverse load: transverse load = 25 * (0.2 * 0.4 + 0.12 * 5) = 17kn / M
basic combination: basic combination: 1.2 * 17 + 1.4 * 10 = 34.4n/m
m = QL ^ 2 / 8 = 34.4 * 5000 ^ 2 / 8 = 107500000n * mm
1) relative compression height x = 0.55 * 370 = 203.5mm, Take 210mm
2) the area of upper compression reinforcement
A1 = (m-fbx (h0-x / 2)) / FY = (111718750-14.3 * 200 * 210 * (370-210 / 2)) / (300 * (370-30)) = - 465.1m2
less than 0, no reinforcement can be provided, and the reinforcement a = 0.2% * BH = 0.2% * 200 * 400 = 160m ^. Option 2 φ 12 (2 and 12 bars, area 226)
3) area of lower compression reinforcement
1. Relative compression height
x = H0 - (H0 ^ 2-2 * m / (1.0 * 14.3 * 200)) ^ 0.5 = 370 - (370 ^ 2-2 * 107500000 / (1.0 * 14.3 * 200)) ^ 0.5 = 122mm
A2 = FBX / FY = 14.3 * 200 * 122 / 300 = 1163.06m ^, select 4 φ 20 (4 and 20 bars, area 1256)<
2. Shear force of inclined section
1) shear force
V = 34.4 * 5000 / 2 = 86kn
2) reinforcement calculation (assuming single stirrup)
A / S = v-0.7fbh0 = 86000-0.7 * 14.3 * 200 * 370 = - 654740 & lt; 0
it is enough to arrange the reinforcement according to the structure, bearing L / 4: 8@150 , mid span 8@250
final answer: Section Selection: 200 * 400
upper reinforcement: 2b12 (2 pieces of 12 secondary reinforcement)
lower reinforcement: 4b20 (4 pieces of 20 secondary reinforcement)
stirrup: densified area 8@150 , non encrypted 8@250
1, foundation bearing capacity (calculation target value): because the bearing capacity of concrete pavement is mainly analyzed, and the three-dimensional structure (15cm loess cushion and 15cm sand gravel cushion) designed by the design institute generally meets the foundation requirements, the foundation in the calculation is considered as infinite width (rigid) Foundation (solved according to the thickness)
2, concrete grade: the grade in concrete is directly proportional to the stiffness, that is, the greater the grade, the greater the stiffness of concrete. Therefore, the selection of too low grade concrete will lead to the network crack of the whole pavement, while the selection of too high grade concrete will lead to the excessive stiffness of the whole pavement, which is brittle, that is, easy to crack as a whole, Therefore, the correct choice of grade is also an important factor for the long-term good condition of concrete pavement, so the concrete grade in this paper is C30 grade designed by the design institute
3, concrete thickness (generally 18cm-30cm): 25cm, 28cm and 30cm are substituted according to the formula. Taking the 25cm thick C30 concrete as an example, the axial compressive strength of C30 is 20.1mpa = 20.1n/mm2 = 20.1 × 1000000 n / m2, equivalent to 20.1 × 100000 kg (five zeros divided by 10, acceleration of gravity), which is 20.1 × 100 tons, 2010 tons, that is, 2010 tons / m2. Because it is 25cm thick concrete, it needs to be multiplied by 0.25. Therefore, it is estimated that the design compressive capacity of 25cm thick C30 concrete per cubic meter is about 502.5 tons / m3 According to the preliminary calculation, C30, 25cm thick, can only bear 63.245 tons at most)
4, design form: because the above factors all consider the compression of concrete (i.e. vertical to the ground), they all follow the plain concrete scheme provided by the Design Institute, without reinforcement treatment
according to the above analysis, it can be seen that the compressive bearing capacity of plain concrete pavement mainly depends on the concrete thickness, so the ultimate bearing capacity can be calculated by formula according to the known thickness< br />Fcd=0.7· β H · FTD · um · h
FCD -- the maximum concentrated return force of concrete< br /> β H-for the thickness less than 300 mm, take 1; FTD -- axial tensile stress (1.39 MPa for C30); Um -- height conversion ratio = 2 · (a + b) + 4h, a = 20cm, B = 60cm (a, B are the width and length of wheel trace respectively); H -- thickness
the value brought in is the corresponding relationship:
C30 concrete 25cm ultimate vehicle bearing capacity: 63.245 tons; The ultimate vehicle bearing capacity of C30 concrete 28cm: 74.104t; The ultimate vehicle bearing capacity of C30 concrete 30cm: 81.732t< Second, the bearing capacity of reinforced concrete can be divided into: 1) flexural bearing capacity. This is relatively simple, under the huge section: M = afbx (h0-x / 2) - FYAs
2) shear capacity of inclined section
3) shear capacity of twisted section (also divided into tension bending, compression bending, bending torsion shear, bending torsion shear compression capacity)
4) punching shear capacity
5) local compression capacity
about so many
what is your member's stress situation, The bearing capacity of different members is different
for example:
question: for the beam concrete of building, the span is 5 meters, the span is 4 meters, the span is 3 meters, the bottom width is 15 cm, and the height is 30. How much rebar is needed
answer:
the thickness of the slab is 12cm, the live load is 0.002n/m ^, the line load is 0.002 * 5000 = 10N / mm, and the reinforced concrete is 25kn / M and 179;, Transverse load = 25 * (0.15 * 0.3 + 0.12 * 5) = 16.125kn/m, i.e. 16.125n/mm & # 179
basic combination: 1.2 * 16.125 + 1.4 * 10 = 33.35n/m
1, bending capacity
1) bending moment
m = QL ^ 2 / 8 = 33.35 * 5000 ^ 2 / 8 = 104218750 n * mm
1) relative compression height
x = H0 ^ 2-2 * m / (1.0 * 14.3 * 150 = 270 ^ 2-2 * 104218750 / (1.0 * 14.3 * 150) = - 24273.65967, less than 0,
judge that your section is too small, and the upper layer needs to be equipped with bearing reinforcement
2) re reinforcement
x = 0.55 * 270 = 148.5, 150 mm
3) the area of upper compression steel bar
A1 = (m-fbx (h0-x / 2)) / (FY * (h0-30) = (104218750-14.3 * 150 * 150 * (270-150 / 2)) / (300 * (150-30) = 1152.152778m ^, the steel bar with 3-22 can be selected
4) the area of lower tension steel bar
A2 = a1 + FBX / FY = 1152 + 14.3 * 150 * 150 / 300 = 2224.5m ^, the secondary steel bar with 6-22 can be selected
conclusion: the section of 150 can't hold 6 and 22 steel bars, and your section is a little small (because I calculate it by simply supported beam, the actual load is not so large, but the earthquake action is not considered, and my algorithm above is a little small)<
the beam section is re determined for you: 200 * 400
live load: 10N / M; Transverse load: transverse load = 25 * (0.2 * 0.4 + 0.12 * 5) = 17kn / M
basic combination: basic combination: 1.2 * 17 + 1.4 * 10 = 34.4n/m
m = QL ^ 2 / 8 = 34.4 * 5000 ^ 2 / 8 = 107500000n * mm
1) relative compression height x = 0.55 * 370 = 203.5mm, Take 210mm
2) the area of upper compression reinforcement
A1 = (m-fbx (h0-x / 2)) / FY = (111718750-14.3 * 200 * 210 * (370-210 / 2)) / (300 * (370-30)) = - 465.1m2
less than 0, no reinforcement can be provided, and the reinforcement a = 0.2% * BH = 0.2% * 200 * 400 = 160m ^. Option 2 φ 12 (2 and 12 bars, area 226)
3) area of lower compression reinforcement
1. Relative compression height
x = H0 - (H0 ^ 2-2 * m / (1.0 * 14.3 * 200)) ^ 0.5 = 370 - (370 ^ 2-2 * 107500000 / (1.0 * 14.3 * 200)) ^ 0.5 = 122mm
A2 = FBX / FY = 14.3 * 200 * 122 / 300 = 1163.06m ^, select 4 φ 20 (4 and 20 bars, area 1256)<
2. Shear force of inclined section
1) shear force
V = 34.4 * 5000 / 2 = 86kn
2) reinforcement calculation (assuming single stirrup)
A / S = v-0.7fbh0 = 86000-0.7 * 14.3 * 200 * 370 = - 654740 & lt; 0
it is enough to arrange the reinforcement according to the structure, bearing L / 4: 8@150 , mid span 8@250
final answer: Section Selection: 200 * 400
upper reinforcement: 2b12 (2 pieces of 12 secondary reinforcement)
lower reinforcement: 4b20 (4 pieces of 20 secondary reinforcement)
stirrup: densified area 8@150 , non encrypted 8@250
3. The calculation principle is determined according to the length width ratio. Then the maximum bearing capacity is calculated according to the section. The soil slab should be calculated according to the rectangular section, and the specific formula is shown in the national code.
4. Simply supported slab is a simple beam type structural member, the topic requirement is bearing capacity, not crack resistance requirements
1. First, calculate the standard value of gravity load kn / M & # 178 on the cover plate Separate permanent load and variable load); Assuming a thickness, the standard value of concrete self weight load is calculated
2. Determine the strength grade of concrete and steel bars; The span m of the cover plate is determined
3. With the ultimate bearing capacity calculation, permanent load and variable load are combined to get the calculated value of load
4. According to the model of simply supported beam under uniform load, the maximum bending moment in the middle of the span and the maximum shear force at the support edge are calculated
5. According to the formula specified in Section 6.2 of code for design of concrete structures (GB 50010-2010), the amount of tensile reinforcement at the bottom of slab is calculated; The shear capacity is calculated according to the formula specified in section 6.3; At the same time, it must meet the basic provisions and structure of the plate in section 9.1 of the code. OK
if the assumed plate thickness is not enough, the plate thickness should be adjusted, or the material with higher strength level should be used for recalculation
this manual calculation needs a function calculator, but calculators on mobile phones and computers don't work.
1. First, calculate the standard value of gravity load kn / M & # 178 on the cover plate Separate permanent load and variable load); Assuming a thickness, the standard value of concrete self weight load is calculated
2. Determine the strength grade of concrete and steel bars; The span m of the cover plate is determined
3. With the ultimate bearing capacity calculation, permanent load and variable load are combined to get the calculated value of load
4. According to the model of simply supported beam under uniform load, the maximum bending moment in the middle of the span and the maximum shear force at the support edge are calculated
5. According to the formula specified in Section 6.2 of code for design of concrete structures (GB 50010-2010), the amount of tensile reinforcement at the bottom of slab is calculated; The shear capacity is calculated according to the formula specified in section 6.3; At the same time, it must meet the basic provisions and structure of the plate in section 9.1 of the code. OK
if the assumed plate thickness is not enough, the plate thickness should be adjusted, or the material with higher strength level should be used for recalculation
this manual calculation needs a function calculator, but calculators on mobile phones and computers don't work.
5. Hello, fellow
I will try to answer your question to see if it is useful to you. For your reference, if you don't agree with me, please don't laugh! Oh, please
the design parameters provided by you are as follows:
the design value of axial compressive strength of C20 concrete: FC = 10N / mm2
the main reinforcement is secondary reinforcement, and the design value of tensile strength: ft = 310n / mm2
according to the specification, the concrete cover of slab: a = 15mm
according to your problem, the slab thickness: H = 60mm
according to your problem, Calculation length of slab: l = 1000-2 * 30 = 940mm
according to your question, calculation width of slab: B = 600mm
take structural importance coefficient Rd = 1.2
reinforcement in slab, 5d12 reinforcement, reinforcement area as = 565mm2
effective height of concrete section H0 = 60-15 = 45mm
according to relevant calculation formula of reinforced concrete structure, There are
Rd * m = FC * b * x * (h0-0.5 * x)
x = FY * as / B / FC
where x is the height of the compression zone of the concrete section, in mm
m is the calculated external load, and in n
the relevant data listed by me are taken into consideration, There are:
x = 29.2mm
m = 4438400n-mm
bending section molus of rectangular section w = b * H ^ 2 / 6
W = 600 * 60 ^ 2 / 6 = 360000mm ^ 3
maximum normal stress that the plate can bear σ= M/W
σ= 4438400 / 360000 = 12.33n/mm2
the total external force that the board can bear is:
F = 12.33 * 940 * 600 = 6953493n = 695t
that's about it. Be careful not to mistake the number. The calculation process is like this
ha ha, I'm sorry, everyone!
I will try to answer your question to see if it is useful to you. For your reference, if you don't agree with me, please don't laugh! Oh, please
the design parameters provided by you are as follows:
the design value of axial compressive strength of C20 concrete: FC = 10N / mm2
the main reinforcement is secondary reinforcement, and the design value of tensile strength: ft = 310n / mm2
according to the specification, the concrete cover of slab: a = 15mm
according to your problem, the slab thickness: H = 60mm
according to your problem, Calculation length of slab: l = 1000-2 * 30 = 940mm
according to your question, calculation width of slab: B = 600mm
take structural importance coefficient Rd = 1.2
reinforcement in slab, 5d12 reinforcement, reinforcement area as = 565mm2
effective height of concrete section H0 = 60-15 = 45mm
according to relevant calculation formula of reinforced concrete structure, There are
Rd * m = FC * b * x * (h0-0.5 * x)
x = FY * as / B / FC
where x is the height of the compression zone of the concrete section, in mm
m is the calculated external load, and in n
the relevant data listed by me are taken into consideration, There are:
x = 29.2mm
m = 4438400n-mm
bending section molus of rectangular section w = b * H ^ 2 / 6
W = 600 * 60 ^ 2 / 6 = 360000mm ^ 3
maximum normal stress that the plate can bear σ= M/W
σ= 4438400 / 360000 = 12.33n/mm2
the total external force that the board can bear is:
F = 12.33 * 940 * 600 = 6953493n = 695t
that's about it. Be careful not to mistake the number. The calculation process is like this
ha ha, I'm sorry, everyone!
6. It's recommended that you have no mail + 9199
1, 2.5D online games
2, the equipment seems to rely on yourself to play
3, less tasks and easy to complete
4, no gangs and families
5, the characters are not Kawaii, but very exquisite, fighting action is also very handsome
you can go and have a look.
1, 2.5D online games
2, the equipment seems to rely on yourself to play
3, less tasks and easy to complete
4, no gangs and families
5, the characters are not Kawaii, but very exquisite, fighting action is also very handsome
you can go and have a look.
7. Dirty by legendary classic
8. The video memory of a video card is composed of a block of video memory chips, and the total bit width of the video memory is also composed of the bit width of the video memory particles. Memory bit width = memory particle bit width & times; The number of video memory particles. The memory number of the relevant manufacturer is on the video memory granule. You can find the number on the Internet to understand its bit width, and then multiply it by the number of video memory granule to get the bit width of the video card
the bit width of the video memory is the number of bits that the video memory can transmit data in a clock cycle. The larger the number of bits, the greater the amount of data that can be transmitted instantly. This is one of the important parameters of the video memory. At present, there are three kinds of video memory bit width in the market, namely 64 bit, 128 bit and 256 bit. People usually call them 64 bit, 128 bit and 256 bit video cards. The higher the bit width of video memory, the better the performance and the higher the price. Therefore, 256 bit video memory is more used in high-end video cards, while 128 bit video memory is basically used in mainstream video cards
as we all know, video memory bandwidth = video memory frequency x video memory bit width / 8, then when the video memory frequency is the same, the video memory bit width will determine the size of the video memory bandwidth.
the bit width of the video memory is the number of bits that the video memory can transmit data in a clock cycle. The larger the number of bits, the greater the amount of data that can be transmitted instantly. This is one of the important parameters of the video memory. At present, there are three kinds of video memory bit width in the market, namely 64 bit, 128 bit and 256 bit. People usually call them 64 bit, 128 bit and 256 bit video cards. The higher the bit width of video memory, the better the performance and the higher the price. Therefore, 256 bit video memory is more used in high-end video cards, while 128 bit video memory is basically used in mainstream video cards
as we all know, video memory bandwidth = video memory frequency x video memory bit width / 8, then when the video memory frequency is the same, the video memory bit width will determine the size of the video memory bandwidth.
9.
Is steam's built-in frame count software
First, download a steam software on your computer, and then log in with your account and password
10. The crystal remains refresh about once an hour. Creation and magic strategy
at the relic point, run around the relic, and you can see a concave place on one side. Here, you will refresh the relic treasure chest regularly, destroy the relic treasure chest, and open it to obtain crystal drawings.
at the relic point, run around the relic, and you can see a concave place on one side. Here, you will refresh the relic treasure chest regularly, destroy the relic treasure chest, and open it to obtain crystal drawings.
Hot content