How to calculate the expansion force
Publish: 2021-04-15 23:13:04
1. Taking the establishment of the calculation model for the instantaneous linear thermal expansion coefficient of carbon steel as an example:
when the temperature of the material changes from Tref (reference temperature of the benchmark) to t, the relative change of the material length L is:
(1)
according to the density ρ It is inversely proportional to L3 ε Th and ρ There is the following relationship between the two factors:
(2)
(3)
then the instantaneous linear thermal expansion coefficient is defined as:
(3)
. Therefore, the key to obtain the instantaneous linear thermal expansion coefficient is to determine the density of carbon steel at different temperatures< According to the characteristics of solidification structure ring cooling (see Fig. 1), carbon steels with 〔 C 〕≤ 0.8% were divided into the following four groups according to carbon content:
I. 〔 C 〕 < 0.09%:
L → L+ δ → δ → δ+γ → γ → α+γ → α+ Fe3C
Ⅱ.〔C〕=0.09 %～0.16 %:
L→L+ δ → δ+γ → γ → α+γ → α+ Fe3C
Ⅲ.〔C〕=0.16 %～0.51 %:
L→L+ δ →L+ γ → γ → α+γ → α+ Fe3C
Ⅳ.〔C〕=0.51 %～0.80 %:
L→L+ γ → γ → α+γ → α+ The solidification structure of Fe3C
carbon steel is a multiphase mixed system, and its density is determined according to formula (4) and formula (5), namely:
(4)
F1 + F2 +... + fi = 1 (5)
where FI is the mass fraction of component I in the system, which can be determined by program calculation according to the lever rule and phase diagram. Component I (I is L δ、γ、α Or Fe3C) as a function of temperature and carbon content ρ〔 T,(i)〕= ρ I (T, c), whose value is from reference [6]
when calculating the coefficient of linear thermal expansion, the solis temperature is selected as the reference temperature. The coefficient of thermal expansion decreases linearly from the value at the solis to zero at the zero strength temperature (that is, the temperature corresponding to the solid fraction FS = 0.8). Above the zero strength temperature, the coefficient of thermal expansion remains zero. In this way, the thermal stress in the liquid region can be avoided< 1 Fe-C phase diagram
1.2 brief introction of slab thermal elastic plastic stress model
1.2.1 calculation of slab temperature field
the heat transfer in drawing direction is ignored, and according to the symmetry, 1 / 4 section slab is taken, and its quadrilateral 4-node isoparametric element grid is shown in Figure 2. The governing equations of unsteady two-dimensional heat transfer are as follows:
Fig.2 calculation domain and FEM meshed for analysis
(6)
the initial temperature is casting temperature, the measured value of heat flux on slab surface is q = 2 688-420 T1 / 2 kW / m2, and the central symmetry line is adiabatic boundary. The thermophysical parameters used in the model vary with temperature, and the equivalent specific heat capacity C is used to consider the effect of latent heat. In addition, the convection effect in the liquid region is achieved by appropriately enlarging the thermal conctivity of the liquid region
1.2.2 calculation of slab stress field
in order to make use of the calculation results of temperature field, the slab grid generation method consistent with the temperature field is adopted. In the system, the mold copper plate is a rigid contact boundary, and the mold taper is characterized by controlling its motion trajectory (including motion direction and velocity). If the distance between a node on the slab surface and the copper plate is less than the specified contact criterion, it is considered that the contact occurs here, and the contact constraint is imposed on the node (to avoid the node crossing the copper plate surface), otherwise it is treated as a free boundary
in the calculation, the liquid and solid regions are regarded as a whole, and the mechanical parameters of the material higher than the liquis temperature are specially treated, so that the stress state of the liquid region can maintain a uniform static pressure state, and the static pressure of the liquid steel applied on the outside can be basically transferred to the inside of the solid shell. According to the symmetry, a fixed displacement constraint in the vertical direction should be applied on the central symmetry line, but because only the displacement field of the shell is concerned, and the thickness of the shell is generally not more than 15 mm, the constraint is only applied within 15 mm from the surface. The range beyond 15 mm is basically liquid phase region, and hydrostatic pressure is applied on its outer edge (at the symmetrical line) (the pressure value is proportional to the distance from the meniscus)< The force balance equation of the above system is:
(7)
where [k] is the total stiffness matrix of the system{ δ i} Is the node displacement array{ Rexter} is the equivalent nodal load array caused by external forces (static pressure of molten steel and contact reaction force of copper wall){ R ε 0} is the equivalent nodal load array caused by thermal strain. Considering the effect of peritectic phase transition, the calculation of {r} ε The linear thermal expansion coefficient curve of carbon steel calculated above is used when the temperature is less than 0}
the thermal elastic plastic model is used in the calculation. It is assumed that the slab section is in the generalized plane strain state, obeys Mises yield criterion and isotropic strengthening law, and the hardening curve is piecewise linear
2 calculation results and discussion
three kinds of carbon steel with carbon content of 0.045%, 0.100% and 0.200% were taken as calculation objects, and the same calculation conditions were adopted, that is, the section size of slab was 150 mm × The casting temperature is 1 550 ℃, the length of mould is 700 mm, the taper is 0.8%, and the distance between meniscus and top of mould is 100 mm
2.1 instantaneous thermal expansion coefficient of three kinds of carbon steel
Fig. 3 shows the calculated instantaneous linear thermal expansion coefficient curve of carbon steel. It can be seen that when [C] = 0.045%, the coefficient of thermal expansion changes suddenly below the solis temperature. This is e to the primary corrosion of molten steel after solidification δ Phase → γ The transformation of phase and the change of specific volume make the coefficient of thermal expansion rise sharply; When [C] = 0.100%, the coefficient of thermal expansion changes abruptly from the two-phase region. This is because when the liquid steel solidifies, the liquid phase and δ The peritectic reaction takes place and the phase is transformed into γ Phase, remaining δ One after another γ Phase transition. The change of specific volume in the transformation process also leads to the sharp rise of thermal expansion coefficient
Fig. 3 instantaneous linear thermal expansion coefficient curve of carbon steel
in the three curves, the starting point of non-zero value is the corresponding point of zero strength temperature
A, B and C are the corresponding points of solis temperature
Fig.3 instant linear thermal expansion
coefficient of carbon steel
in addition, [C] = 0.045% δ Phase → γ The phase transition temperature range is narrow and the transition is fast (see Fig. 1), so the abrupt change value of linear thermal expansion coefficient is large. In contrast, the abrupt change of coefficient of thermal expansion with [C] = 0.100% is smaller. However, because of the wide temperature range of the latter, the temperature range of abrupt change of the coefficient of thermal expansion is also wide. It can be inferred that the effect of peritectic phase transformation on the solidification shrinkage of primary shell at [C] = 0.100% is greater than that at [C] = 0.045% δ Phase → γ Phase transition
[C] = 0.200% steel has no abrupt change in coefficient of thermal expansion. The reason is that although peritectic transformation also occurs, it only occurs at a certain temperature level (about 1 495 ℃), so it has little effect on the coefficient of thermal expansion< The results show that the surface shrinkage of 〔 C 〕 = 0.045%, 0.100% and 0.200% of the three steels varies along the drawing direction and cross section direction (the space inclined plane at the bottom is the mold copper plate
Fig. 4 〔 C 〕 = 0.045%) b) 〔C〕=0.100 % ( c) [C] = 0.200%
Fig.4 surface shrinkage of billet
inner wall). It can be seen from the figure that the corner of the slab shrinks and breaks away from the copper plate of the mould at the early stage of solidification, and almost always contacts the copper plate near the middle (only the steel with [C] = 0.100% keeps separation near the outlet). The closer to the corner, the earlier the contraction and detachment, and the larger the contraction
under the static pressure of molten steel, the shrinking shell will be pressed back to the mold copper plate, so that the shell shrinkage fluctuates [the surface of shrinkage surface is dog tooth shape (see Figure 4)]. The shell near the meniscus is thinner and the fluctuation is obvious. In addition, the closer to the corner, the more obvious the fluctuation is. The shrinkage fluctuation of primary shell will lead to stress concentration, which is easy to ince surface defects such as cracks
comparing the surface shrinkage of three kinds of carbon steel billets, it can be seen that: [C] = 0.100% steel has the most significant shrinkage, the largest shrinkage fluctuation (meniscus area), and the most extensive expansion along the cross section direction The shrinkage of 0.200% steel is the smallest
2.3 shrinkage of primary shell at the corner of meniscus region
Fig. 5 shows the shrinkage of corner of three kinds of carbon steel near meniscus region. It can be seen that: within 20 mm from the meniscus, the corner of the slab is separated from the mold copper plate, and the steel with 〔 C 〕 = 0.045% is the earliest to separate from the mold copper plate, because the solis temperature of the steel is the highest, and it is the earliest to solidify and form the shell C] = 0.100% steel shrinks strongly after forming primary shell, but it is pressed back by the increased hydrostatic pressure 50 mm away from the meniscus, and then continues to shrink. The shrinkage of the primary shell of this steel is the most significant, and the shrinkage fluctuation is also the largest, so it is most likely to ince the surface defects of the slab The results show that the shrinkage and shrinkage fluctuation of primary shell of 0.045% steel decrease obviously C] = 0.200% steel has the smallest shrinkage and shrinkage fluctuation< 5 shrinkage of initial shell of pellet corner at Meniscus
3 conclusion
(1) for peritectic steel with carbon content around 0.1%, the shrinkage of primary shell at the top of mold and near the corner is very irregular, which is easy to ince surface defects
(2) the irregular shrinkage of the shell is mainly concentrated in the range of 100 mm below the meniscus. Therefore, the taper of the upper part of the mold is not suitable for shell shrinkage. Therefore, the casting speed should be improved by optimizing the mold taper. An important guiding principle is to use a larger taper in the upper part of the mold to make the shell in good contact with the copper plate.
when the temperature of the material changes from Tref (reference temperature of the benchmark) to t, the relative change of the material length L is:
(1)
according to the density ρ It is inversely proportional to L3 ε Th and ρ There is the following relationship between the two factors:
(2)
(3)
then the instantaneous linear thermal expansion coefficient is defined as:
(3)
. Therefore, the key to obtain the instantaneous linear thermal expansion coefficient is to determine the density of carbon steel at different temperatures< According to the characteristics of solidification structure ring cooling (see Fig. 1), carbon steels with 〔 C 〕≤ 0.8% were divided into the following four groups according to carbon content:
I. 〔 C 〕 < 0.09%:
L → L+ δ → δ → δ+γ → γ → α+γ → α+ Fe3C
Ⅱ.〔C〕=0.09 %～0.16 %:
L→L+ δ → δ+γ → γ → α+γ → α+ Fe3C
Ⅲ.〔C〕=0.16 %～0.51 %:
L→L+ δ →L+ γ → γ → α+γ → α+ Fe3C
Ⅳ.〔C〕=0.51 %～0.80 %:
L→L+ γ → γ → α+γ → α+ The solidification structure of Fe3C
carbon steel is a multiphase mixed system, and its density is determined according to formula (4) and formula (5), namely:
(4)
F1 + F2 +... + fi = 1 (5)
where FI is the mass fraction of component I in the system, which can be determined by program calculation according to the lever rule and phase diagram. Component I (I is L δ、γ、α Or Fe3C) as a function of temperature and carbon content ρ〔 T,(i)〕= ρ I (T, c), whose value is from reference [6]
when calculating the coefficient of linear thermal expansion, the solis temperature is selected as the reference temperature. The coefficient of thermal expansion decreases linearly from the value at the solis to zero at the zero strength temperature (that is, the temperature corresponding to the solid fraction FS = 0.8). Above the zero strength temperature, the coefficient of thermal expansion remains zero. In this way, the thermal stress in the liquid region can be avoided< 1 Fe-C phase diagram
1.2 brief introction of slab thermal elastic plastic stress model
1.2.1 calculation of slab temperature field
the heat transfer in drawing direction is ignored, and according to the symmetry, 1 / 4 section slab is taken, and its quadrilateral 4-node isoparametric element grid is shown in Figure 2. The governing equations of unsteady two-dimensional heat transfer are as follows:
Fig.2 calculation domain and FEM meshed for analysis
(6)
the initial temperature is casting temperature, the measured value of heat flux on slab surface is q = 2 688-420 T1 / 2 kW / m2, and the central symmetry line is adiabatic boundary. The thermophysical parameters used in the model vary with temperature, and the equivalent specific heat capacity C is used to consider the effect of latent heat. In addition, the convection effect in the liquid region is achieved by appropriately enlarging the thermal conctivity of the liquid region
1.2.2 calculation of slab stress field
in order to make use of the calculation results of temperature field, the slab grid generation method consistent with the temperature field is adopted. In the system, the mold copper plate is a rigid contact boundary, and the mold taper is characterized by controlling its motion trajectory (including motion direction and velocity). If the distance between a node on the slab surface and the copper plate is less than the specified contact criterion, it is considered that the contact occurs here, and the contact constraint is imposed on the node (to avoid the node crossing the copper plate surface), otherwise it is treated as a free boundary
in the calculation, the liquid and solid regions are regarded as a whole, and the mechanical parameters of the material higher than the liquis temperature are specially treated, so that the stress state of the liquid region can maintain a uniform static pressure state, and the static pressure of the liquid steel applied on the outside can be basically transferred to the inside of the solid shell. According to the symmetry, a fixed displacement constraint in the vertical direction should be applied on the central symmetry line, but because only the displacement field of the shell is concerned, and the thickness of the shell is generally not more than 15 mm, the constraint is only applied within 15 mm from the surface. The range beyond 15 mm is basically liquid phase region, and hydrostatic pressure is applied on its outer edge (at the symmetrical line) (the pressure value is proportional to the distance from the meniscus)< The force balance equation of the above system is:
(7)
where [k] is the total stiffness matrix of the system{ δ i} Is the node displacement array{ Rexter} is the equivalent nodal load array caused by external forces (static pressure of molten steel and contact reaction force of copper wall){ R ε 0} is the equivalent nodal load array caused by thermal strain. Considering the effect of peritectic phase transition, the calculation of {r} ε The linear thermal expansion coefficient curve of carbon steel calculated above is used when the temperature is less than 0}
the thermal elastic plastic model is used in the calculation. It is assumed that the slab section is in the generalized plane strain state, obeys Mises yield criterion and isotropic strengthening law, and the hardening curve is piecewise linear
2 calculation results and discussion
three kinds of carbon steel with carbon content of 0.045%, 0.100% and 0.200% were taken as calculation objects, and the same calculation conditions were adopted, that is, the section size of slab was 150 mm × The casting temperature is 1 550 ℃, the length of mould is 700 mm, the taper is 0.8%, and the distance between meniscus and top of mould is 100 mm
2.1 instantaneous thermal expansion coefficient of three kinds of carbon steel
Fig. 3 shows the calculated instantaneous linear thermal expansion coefficient curve of carbon steel. It can be seen that when [C] = 0.045%, the coefficient of thermal expansion changes suddenly below the solis temperature. This is e to the primary corrosion of molten steel after solidification δ Phase → γ The transformation of phase and the change of specific volume make the coefficient of thermal expansion rise sharply; When [C] = 0.100%, the coefficient of thermal expansion changes abruptly from the two-phase region. This is because when the liquid steel solidifies, the liquid phase and δ The peritectic reaction takes place and the phase is transformed into γ Phase, remaining δ One after another γ Phase transition. The change of specific volume in the transformation process also leads to the sharp rise of thermal expansion coefficient
Fig. 3 instantaneous linear thermal expansion coefficient curve of carbon steel
in the three curves, the starting point of non-zero value is the corresponding point of zero strength temperature
A, B and C are the corresponding points of solis temperature
Fig.3 instant linear thermal expansion
coefficient of carbon steel
in addition, [C] = 0.045% δ Phase → γ The phase transition temperature range is narrow and the transition is fast (see Fig. 1), so the abrupt change value of linear thermal expansion coefficient is large. In contrast, the abrupt change of coefficient of thermal expansion with [C] = 0.100% is smaller. However, because of the wide temperature range of the latter, the temperature range of abrupt change of the coefficient of thermal expansion is also wide. It can be inferred that the effect of peritectic phase transformation on the solidification shrinkage of primary shell at [C] = 0.100% is greater than that at [C] = 0.045% δ Phase → γ Phase transition
[C] = 0.200% steel has no abrupt change in coefficient of thermal expansion. The reason is that although peritectic transformation also occurs, it only occurs at a certain temperature level (about 1 495 ℃), so it has little effect on the coefficient of thermal expansion< The results show that the surface shrinkage of 〔 C 〕 = 0.045%, 0.100% and 0.200% of the three steels varies along the drawing direction and cross section direction (the space inclined plane at the bottom is the mold copper plate
Fig. 4 〔 C 〕 = 0.045%) b) 〔C〕=0.100 % ( c) [C] = 0.200%
Fig.4 surface shrinkage of billet
inner wall). It can be seen from the figure that the corner of the slab shrinks and breaks away from the copper plate of the mould at the early stage of solidification, and almost always contacts the copper plate near the middle (only the steel with [C] = 0.100% keeps separation near the outlet). The closer to the corner, the earlier the contraction and detachment, and the larger the contraction
under the static pressure of molten steel, the shrinking shell will be pressed back to the mold copper plate, so that the shell shrinkage fluctuates [the surface of shrinkage surface is dog tooth shape (see Figure 4)]. The shell near the meniscus is thinner and the fluctuation is obvious. In addition, the closer to the corner, the more obvious the fluctuation is. The shrinkage fluctuation of primary shell will lead to stress concentration, which is easy to ince surface defects such as cracks
comparing the surface shrinkage of three kinds of carbon steel billets, it can be seen that: [C] = 0.100% steel has the most significant shrinkage, the largest shrinkage fluctuation (meniscus area), and the most extensive expansion along the cross section direction The shrinkage of 0.200% steel is the smallest
2.3 shrinkage of primary shell at the corner of meniscus region
Fig. 5 shows the shrinkage of corner of three kinds of carbon steel near meniscus region. It can be seen that: within 20 mm from the meniscus, the corner of the slab is separated from the mold copper plate, and the steel with 〔 C 〕 = 0.045% is the earliest to separate from the mold copper plate, because the solis temperature of the steel is the highest, and it is the earliest to solidify and form the shell C] = 0.100% steel shrinks strongly after forming primary shell, but it is pressed back by the increased hydrostatic pressure 50 mm away from the meniscus, and then continues to shrink. The shrinkage of the primary shell of this steel is the most significant, and the shrinkage fluctuation is also the largest, so it is most likely to ince the surface defects of the slab The results show that the shrinkage and shrinkage fluctuation of primary shell of 0.045% steel decrease obviously C] = 0.200% steel has the smallest shrinkage and shrinkage fluctuation< 5 shrinkage of initial shell of pellet corner at Meniscus
3 conclusion
(1) for peritectic steel with carbon content around 0.1%, the shrinkage of primary shell at the top of mold and near the corner is very irregular, which is easy to ince surface defects
(2) the irregular shrinkage of the shell is mainly concentrated in the range of 100 mm below the meniscus. Therefore, the taper of the upper part of the mold is not suitable for shell shrinkage. Therefore, the casting speed should be improved by optimizing the mold taper. An important guiding principle is to use a larger taper in the upper part of the mold to make the shell in good contact with the copper plate.
2. The object expands and contracts because of the change of temperature. The change ability is expressed as the volume change caused by the change of unit temperature under constant pressure, i.e. the coefficient of thermal expansion
the coefficient of thermal expansion α=Δ V/(V* Δ T) Where.
where Δ V is the given temperature change Δ Strictly speaking, the above formula is only the difference approximation of the differential definition when the temperature changes in a small range; Define requirements accurately Δ V and Δ T is infinitely small, which also means that the coefficient of thermal expansion is usually not constant in a large temperature range
when the temperature change is not great, α The volume expansion of solid and liquid can be expressed as follows:
VT = V0 (1 + 3) αΔ T) For ideal gas,
VT = V0 (1 + 0.00367) Δ T)
VT and V0 are the volumes of the final state and the initial state respectively
for an object that can be regarded as one-dimensional approximately, length is the decisive factor to measure its volume. At this time, the coefficient of thermal expansion can be simply defined as the ratio of the increase of length per unit temperature change to the original length, which is the coefficient of linear expansion
for three-dimensional anisotropic materials, the linear expansion coefficient and volume expansion coefficient can be divided into two parts. For example, the graphite structure has significant anisotropy, so the coefficient of linear expansion of graphite fiber also shows anisotropy, which shows that the coefficient of thermal expansion parallel to the layer direction is far less than that perpendicular to the layer direction
there are many relationships between the macroscopic thermal expansion coefficient and the axial expansion coefficient, and the mrozowski formula is generally accepted:
the relationship between the macroscopic thermal expansion coefficient and the axial expansion coefficient is more than one α= A α c+(1-A) α a
α c, α A is the thermal expansion rate of a-axis and c-axis respectively, and a is called the "structural end face" parameter
pure hand beating
the coefficient of thermal expansion α=Δ V/(V* Δ T) Where.
where Δ V is the given temperature change Δ Strictly speaking, the above formula is only the difference approximation of the differential definition when the temperature changes in a small range; Define requirements accurately Δ V and Δ T is infinitely small, which also means that the coefficient of thermal expansion is usually not constant in a large temperature range
when the temperature change is not great, α The volume expansion of solid and liquid can be expressed as follows:
VT = V0 (1 + 3) αΔ T) For ideal gas,
VT = V0 (1 + 0.00367) Δ T)
VT and V0 are the volumes of the final state and the initial state respectively
for an object that can be regarded as one-dimensional approximately, length is the decisive factor to measure its volume. At this time, the coefficient of thermal expansion can be simply defined as the ratio of the increase of length per unit temperature change to the original length, which is the coefficient of linear expansion
for three-dimensional anisotropic materials, the linear expansion coefficient and volume expansion coefficient can be divided into two parts. For example, the graphite structure has significant anisotropy, so the coefficient of linear expansion of graphite fiber also shows anisotropy, which shows that the coefficient of thermal expansion parallel to the layer direction is far less than that perpendicular to the layer direction
there are many relationships between the macroscopic thermal expansion coefficient and the axial expansion coefficient, and the mrozowski formula is generally accepted:
the relationship between the macroscopic thermal expansion coefficient and the axial expansion coefficient is more than one α= A α c+(1-A) α a
α c, α A is the thermal expansion rate of a-axis and c-axis respectively, and a is called the "structural end face" parameter
pure hand beating
3. According to the law of ideal gas,
the pressure of a sealed container with constant volume is directly proportional to the gas temperature, and the pressure at 25 ℃ can be set as one atmospheric pressure.
after converting the centigrade temperature scale into thermodynamic temperature scale, the pressure at 500 ℃ should be 1 * (500 + 273) / (25 + 273) = 2.6 atmospheric pressure
the pressure of a sealed container with constant volume is directly proportional to the gas temperature, and the pressure at 25 ℃ can be set as one atmospheric pressure.
after converting the centigrade temperature scale into thermodynamic temperature scale, the pressure at 500 ℃ should be 1 * (500 + 273) / (25 + 273) = 2.6 atmospheric pressure
4. Expansion work (also known as volume work) - the work done by gas e to volume change in the thermodynamic process Therefore, on the P-V diagram, W12 is the area enclosed by the process line and the horizontal axis. Regulations: the work done by the thermal system to the outside world is positive, and the work done by the outside world to the thermal system is negative. from δ W = PDV: DV & gt; 0, expansion, δ W> 0, the system does work to the outside world; dV< 0, compression, δ W< 0, external work on the system; dV=0， δ W = 0, reactive power transfer between the system and the outside world.
5. Theoretically, under the most ideal condition, the stress of stainless steel expansion screw is M6, M8, M10 and M12, which are 100120200500kg respectively.
6. Theoretically, this is an empirical value, just like dry friction (dynamic friction) = positive pressure × Dynamic friction factor
7. Sorry, yesterday's statement is wrong. The expansion force you said has something to do with the layout of the pipeline. The simple assumption is that there is a steam pipe between two fixed supports, which expands when heated. Then the horizontal force borne by the fixed support is the thermal expansion force, which is very large, Therefore, we generally do not fix the two ends of a straight pipe section directly when we arrange the pipeline. Generally, we need to add a compensation device to rece the thermal expansion force, which is the reason
as for the specific calculation of thermal expansion force, it is relatively complex, which is related to the pipe material (elastic molus, bending section molus, linear expansion coefficient, etc.), thermal displacement caused by pipeline layout, etc. you can see the introction in American power pipeline standard b31.1. If there is no such specification, you can leave your email and I will send it to you
as far as my understanding is concerned, if the two ends of the pipe are not constrained ring thermal expansion, there will be no so-called thermal expansion force
well, it's true that there is no stress in the free expansion within the allowable temperature range of the material, but if it exceeds the service temperature, the material may be damaged, and the service life will be greatly reced in engineering
the specification has been sent to you, please check it.
as for the specific calculation of thermal expansion force, it is relatively complex, which is related to the pipe material (elastic molus, bending section molus, linear expansion coefficient, etc.), thermal displacement caused by pipeline layout, etc. you can see the introction in American power pipeline standard b31.1. If there is no such specification, you can leave your email and I will send it to you
as far as my understanding is concerned, if the two ends of the pipe are not constrained ring thermal expansion, there will be no so-called thermal expansion force
well, it's true that there is no stress in the free expansion within the allowable temperature range of the material, but if it exceeds the service temperature, the material may be damaged, and the service life will be greatly reced in engineering
the specification has been sent to you, please check it.
8. One example:
it is known that the basement retaining wall is 240 thick, 3M high, with vertical and smooth back and horizontal filling surface. The physical and mechanical indexes of the fill are as follows: r = 18kn / m3.
calculation process:
the earth pressure is: q = k0rh
K0 = 0.5, r = 18kn / m3, H = 1.5m, so q = 0.5 × eighteen × 1.5 = 13.5 (KN / M)
when the upper and lower parts are restrained by ring beams and the wall is considered as a fixed end, the bending moment Ma = RG? Under the action of triangular lateral earth pressure? 1/20qL2=1.2 × 1/20 × thirteen point five × 1.52 = 1.82 (KN? M)
shear VA = RG? 7/20qL=1.2 × 7/20 × thirteen point five × 1.5 = 8.5 (KN)
calculation of bending and shear bearing capacity:
mu10 solid coal gangue brick, m10cement mortar, 240 wall thickness
m ≤ ftmw, V ≤ fvbz
design value of bending tensile strength of masonry along tooth joint FTM = 0.8 × 33 = 0.264 (MPA)
shear strength design value FV = 0.8 × 17 = 0.136 (MPA)
take 1m wide wall calculation unit and calculate according to rectangular section,
section resistance moment w = BH2 / 6 = 1000 × 2402/6=9.6 × 106 (mm3)
internal force arm z = 2H / 3 = 2 × 240 / 3 = 160 (mm)
masonry flexural capacity ftmw = 0.264 × nine point six × 106=2.53kN?m> Ma = 1.82 kn? M
masonry shear capacity fvbz = 0.136 × one thousand × 160=21.76kN> VA = 8.5 kn
so the bending and shear capacity of 240 thick retaining wall meet the requirements.
it is known that the basement retaining wall is 240 thick, 3M high, with vertical and smooth back and horizontal filling surface. The physical and mechanical indexes of the fill are as follows: r = 18kn / m3.
calculation process:
the earth pressure is: q = k0rh
K0 = 0.5, r = 18kn / m3, H = 1.5m, so q = 0.5 × eighteen × 1.5 = 13.5 (KN / M)
when the upper and lower parts are restrained by ring beams and the wall is considered as a fixed end, the bending moment Ma = RG? Under the action of triangular lateral earth pressure? 1/20qL2=1.2 × 1/20 × thirteen point five × 1.52 = 1.82 (KN? M)
shear VA = RG? 7/20qL=1.2 × 7/20 × thirteen point five × 1.5 = 8.5 (KN)
calculation of bending and shear bearing capacity:
mu10 solid coal gangue brick, m10cement mortar, 240 wall thickness
m ≤ ftmw, V ≤ fvbz
design value of bending tensile strength of masonry along tooth joint FTM = 0.8 × 33 = 0.264 (MPA)
shear strength design value FV = 0.8 × 17 = 0.136 (MPA)
take 1m wide wall calculation unit and calculate according to rectangular section,
section resistance moment w = BH2 / 6 = 1000 × 2402/6=9.6 × 106 (mm3)
internal force arm z = 2H / 3 = 2 × 240 / 3 = 160 (mm)
masonry flexural capacity ftmw = 0.264 × nine point six × 106=2.53kN?m> Ma = 1.82 kn? M
masonry shear capacity fvbz = 0.136 × one thousand × 160=21.76kN> VA = 8.5 kn
so the bending and shear capacity of 240 thick retaining wall meet the requirements.
9. The pressure, temperature and volume formula of gas is also applicable when the volume of solid is constant, the pressure increases with the increase of temperature, the temperature can not be infinite, the high pressure can not be infinite, so the weight of the object on top of it is limited, and the bucket will break first in the end, which is irresistible
you mean I don't want to use the expansion to measure the pressure
you mean I don't want to use the expansion to measure the pressure
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