Unit problem in finite element analysis of the hot

2022-08-06
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Unit problem in finite element analysis

most finite element calculation programs do not specify the units of physical quantities used. Different units can be used for different problems, as long as the units of each physical quantity in a problem are unified. However, because many physical quantities in different units may be used in practical engineering problems, if the commonly used units are only used according to custom, the units are unified on the surface, but not unified in practice, resulting in incorrect calculation results

for example, the following units are used in structural analysis: length – m; Time – s; Mass – kg; Force - N; Pressure, stress, modulus of elasticity, etc. – PA, and the unit is unified at this time. However, if the pressure unit is changed to MPa and other units remain unchanged, the unit is not unified; Or at the same time, change the length unit to mm and the pressure unit to MPa, and keep the rest of the units unchanged and the units are not unified. It can be seen that for practical engineering problems, we cannot choose the unit of each physical quantity according to the habit of manual calculation, but must follow certain principles

the unit of physical quantity is related to the unit system adopted. All physical quantities can be divided into basic physical quantities and derived physical quantities. The basic physical quantities in structure and thermal calculation include: mass, length, time and temperature. There are many kinds of derived physical quantities, such as area, volume, velocity, acceleration, elastic modulus, pressure, stress, thermal conductivity, specific heat, heat exchange coefficient, energy, heat, work, etc., which have a definite relationship with the basic physical quantities. The basic tensile testing machine is mainly composed of the host, computer, fixture, sensor and other parts. The unit system used is determined, and then the units of each derived physical quantity can be obtained according to the corresponding formula. The specific method is: first determine the dimensions of each physical quantity, and then get the specific units of each physical quantity according to the different basic physical quantity unit systems

basic physical quantities and their dimensions:

· mass m

· length L

· time t

· temperature T

export physical quantities and their dimensions:

· speed: v = l/t

· acceleration: a = l/t2

· area: a = L2

· volume: v = L3

· density: ρ= m/L3;

· force: F = m · a = m · l/t2

· moment, energy, heat, enthalpy, etc.: e = f · L = m · l2/t2

· pressure, stress, elastic modulus, etc.: P = f/a = m/(T2 · L)

· heat flow and power: ψ= e/t = m·L2/t3;

· thermal conductivity: K= ψ/ (L·T) = m·L/(t3·T);

· specific heat: C = e/(m · T) = l2/(T2 · T)

· heat exchange coefficient: CV = e/(L2 · t · T) = m/(T3 · T)

· viscosity coefficient: kV = P · t = m/(t · L)

· entropy: S = e/t = m · l2/(T2 · T)

· Bochuang "bu6800 super large two plate servo energy-saving injection molding machine" is the most energy-saving two plate injection molding machine with the largest combination of clamping force and injection volume produced in Asia at present. Its mass entropy and specific entropy: S = s/m = l2/(T2 · T)

after selecting the units of basic physical quantities, the units of other physical quantities can be derived. There are many unit systems that can be selected. Here are two common examples

1 basic physical quantities adopt the following unit system:

· mass m – kg; (it should be unified only when mg unit is adopted, which can be derived by referring to the unified unit of material warehouse in RC)

· length L – mm

· time – s

· temperature – K (temperature K is equivalent to c)

the units of each derived physical quantity can be derived as follows, and the relationship with the kg-m-s unit system or some common units is also listed:

· speed: v = l/t = mm/s = m/s

· acceleration: a = l/t2 = mm/s2 = m/s2

· area: a = L2 = mm2 = m2

· volume: v = L3 = mm3 = m3

· density: ρ= m/L3 = kg/mm3 = kg/m3 = g/cm3;

· force: F = m · l/t2 = kg · mm/s2 = kg · m/s2 = Mn (n)

· moment, energy, heat, enthalpy, etc.: e = m · l2/t2 = kg · mm2/s2 = kg · m2/s2 = J (joule)

· pressure, stress, elastic modulus, etc.: P = m/(T2 · L) = kg/(S2 · mm) = 103 kg/(S2 · m) = kPa (PA)

· heat flow and power: ψ= M · l2/t3 = kg · mm2/s3 = kg · m2/s3 = w (Watts)

· thermal conductivity: k = m · l/(T3 · T) = kg · mm/(S3 · K) = kg · m/(S3 · K)

· specific heat: C = l2/(T2 · T) = mm2/(S2 · K) = m2/(S2 · K)

· heat exchange coefficient: CV = m/(T3 · T) = kg/(S3 · K)

· viscosity coefficient: kV = m/(t · L) = kg/(s · mm) = 103 kg/(s · mm)

· entropy: S = m · l2/(T2 · T) = kg · mm2/(S2 · K) = kg · m2/(S2 · K)

· mass entropy and specific entropy: s= l2/(T2 · T) = mm2/(S2 · K) = m2/(S2 · K)

basic physical quantities adopt the following unit system:

· mass m – g

· length L – m (106 m)

· time – MS (10 – 3 s)

· temperature – K (k is equivalent to c)

the units of each derived physical quantity can be derived as follows, and the relationship with the kg-m-s unit system or some common units is also listed:

· speed: v = l/t = m/ms = m/s

· acceleration: a = l/t2 = m/ms2 = m/s2

· area: a = L2 = M2 = m2

· volume: v = L3 = m3 = m3

· density: ρ= m/L3 = g/m3 = kg/m3 = g/cm3;

· force: F = m · l/t2 = g · m/ms2 = 10 – 3 kg · m/s2 = Mn (n)

· moment, energy, heat, enthalpy, etc.: e = m · l2/t2 = g · m2/ms2 = 10 -9 kg · m2/s2 = J (joule)

· pressure, stress, elastic modulus, etc.: P = m/(T2 · L) = g/(MS2 · m) = 109kg/(S2 · m) = 109pa (PA) = GPA

· heat flow and power: ψ= M · l2/t3 = g · m2/ms3 = kg · m2/s3 = w (Watts)

· thermal conductivity: k = m · l/(T3 · T) = g · m/(MS3 · K) = kg · m/(S3 · K)

· specific heat: C = l2/(T2 · T) = m2/(MS2 · K) = m2/(S2 · K)

· heat exchange coefficient: CV microcomputer sends beam movement command according to the value set before the experiment universal testing machine = m/(T3 · T) = g/(MS3 · K) = 103 kg/(S3 · K)

· viscosity coefficient: kV = m/(t · L) = g/(MS · m) = 106kg/(s · mm)

· entropy: S = m · l2/(T2 · T) = g · m2/(MS2 · K) = kg · m2/(S2 · K)

· mass entropy and specific entropy: S = l2/(T2 · T) = m2/(MS2 · K) = m2/(S2 · K)

it can be seen that if you master the method of transformation between units, you can choose the appropriate unit system according to your own needs. See Table 1 for more examples. Table 2 shows the conversion factors between several unit systems and the kg-m-s unit system

note: the last three columns show the factors multiplied when converting the values in the kg-m-s unit system to other unit systems (when preparing the input data); If you need to convert values from other units to kg-m-s units (when analyzing the calculation results), you should divide by this factor

experimental calculation shall be conducted once a day (end)

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