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Week - 1 - Consistency of Units

AIM To convert the following measurements to the specified system of units, in addition to specifying the conversion factors: MASS: 1 kgLENGTH: 1 mTIME: 1 sFORCE: 1 NSTRESS: 1 PaENERGY: 1 NmDENSITY: 1 kg/m3YOUNG'S MODULUS: 210 GPaVELOCITY: 56.33 kphGRAVITY: 9.8 m/s2 The above values are to be converted to the following…

LS-DYNA

Vaishak Babu
updated on 05 Aug 2021

AIM

To convert the following measurements to the specified system of units, in addition to specifying the conversion factors:

MASS: 1 kg
LENGTH: 1 m
TIME: 1 s
FORCE: 1 N
STRESS: 1 Pa
ENERGY: 1 Nm
DENSITY: 1 kg/m3
YOUNG'S MODULUS: 210 GPa
VELOCITY: 56.33 kph
GRAVITY: 9.8 m/s2

The above values are to be converted to the following system of units:

Kg cm ms

Kg cm s

gm mm ms

kg m s

g cm s

g cm us

g mm s

ton mm s

slug ft s

kg mm s

g cm ms

kg mm ms

Also, detailed calculations for converting from the SI unit system to the g/mm/ms unit system only is to be shown in this challenge

WHY CONVERT TO A DIFFERENT UNIT SYSTEM?

The reason for this exercise is to familiarize ourselves with the fact that LS Dyna has no consistent unit system. LS Dyna looks to the user to define and provide it with a unit system according to which the solver gives results.

The following defines a consistent system of units for LS-DYNA as enunciated on dynasupport:

1 force unit = 1 mass unit * 1 acceleration unit
1 acceleration unit = 1 length unit / (1 time unit)^2
1 density unit = 1 mass unit / (1 length unit)^3

CONVERTING FROM SI UNIT SYSTEM TO g/mm/ms UNIT SYSTEM

Mass

$1 k g = 1000 g$ $1 kg = 1000 g$ or $10^{3} g$ $10^3 g$

Length

$1 m = 1000 m m$ $1 m = 1000 mm$ or $10^{3} m m$ $10^3 mm$

Time

$1 s = 1000 m s$ $1s = 1000ms$ or $10^{3} m s$ $10^3 ms$

Force

$F = m a$ $F = ma$

$m = 10^{3} g$ $m = 10^3 g$

$a = \frac{10^{3} m m}{{(10^{3} m s)}^{2}}$ $a = (10^3mm)/(10^3ms)^2$ = $10^{- 3} \frac{m m}{{(m s)}^{2}}$ $10^-3 (mm)/ (ms)^2$

$F = 10^{3} g \cdot {(10)}^{- 3} \frac{m m}{{(m s)}^{2}} = 1 \frac{g . m m}{m s^{2}}$ $F = 10^3 g * (10)^-3 (mm)/(ms)^2 = 1 (g.mm)/(ms^2)$

Stress

Stress = Force / Area

$= 1 \frac{g . m m}{m s^{2}} \div {(10^{3} m m)}^{2}$ $= 1 (g.mm)/(ms^2) ÷ (10^3mm)^2$ = $10^{- 6} \frac{g}{m m . m s^{2}}$ $10^-6 g/(mm.ms^2)$

Energy

Energy = Force x Distance

= $1 \frac{g . m m}{m s^{2}} \cdot 10^{3} m m$ $1 (g.mm)/(ms^2) * 10^3 mm$ = $10^{3} g . \frac{m m^{2}}{m s^{2}}$ $10^3 g.(mm^2)/(ms^2)$

Density

Density = Mass/Volume

= $\frac{10^{3} g}{{(10^{3} m m)}^{3}}$ $(10^3 g)/(10^3mm)^3$ = $10^{- 6} \frac{g}{m m^{3}}$ $10^-6 g/(mm^3)$

Young's Modulus

Young's Modulus, $E = 210 G P a = 210 \times 10^{9} P a$ $E = 210 GPa = 210 xx 10^9 Pa$ = $210 \times 10^{9} \frac{k g}{m . s^{2}}$ $210 xx 10^9 (kg)/(m.s^2)$

The SI units are converted into this system's units

= $210 \times 10^{9} \frac{(10^{3} g)}{(10^{3} m m . {(10^{3} m s)}^{2})}$ $210 xx 10^9 ((10^<a href="https://skill-lync.com/electronics-engineering-courses/introduction-cellular-communications-2g-3g" target="_blank">3g</a>))/((10^3mm.(10^3ms)^2)$ = $210 \times 10^{9} (10^{- 6} \frac{g}{m m . m s^{2}})$ $210 xx 10^9 (10^-6 g/(mm.ms^2))$

= $210 \times 10^{3} \frac{g}{m m . m s^{2}}$ $210 xx 10^3 g/(mm.ms^2)$

Velocity

Velocity = 56.33 kmph

1 kilometre = $10^{6} m m$ $10^6 mm$ & 1 hour = $60 \times 60 \times 1000 = 3.6 \times 10^{6} m s$ $60 xx 60 xx 1000 = 3.6 xx 10^6 ms$

Therefore,

$56.33 \frac{k m}{h r}$ $56.33 (km)/(hr)$ = $56.33 (\frac{10^{6} m m}{3.6 \times 10^{6} m s})$ $56.33 ((10^6mm)/(3.6 xx 10^6 ms))$

= $15.647 \frac{m m}{m s} = 15.65 \frac{m m}{m s}$ $15.647 (mm)/(ms) = 15.65 (mm)/(ms)$

Acceleration due to gravity

Acceleration of $1 \frac{m}{s^{2}}$ $1 m/s^2$ $= \frac{10^{3} m m}{{(10^{3} m s)}^{2}}$ $= (10^3mm)/(10^3ms)^2$ = $10^{- 3} \frac{m m}{{(m s)}^{2}}$ $10^-3 (mm)/ (ms)^2$

Acceleration due to gravity = $9.8 \frac{m}{s^{2}} = 9.8 \times 10^{- 3} \frac{m m}{{(m s)}^{2}}$ $9.8m/s^2 = 9.8 xx 10^-3 (mm)/ (ms)^2$

TABLE - GIVEN MEASURED QUANTITIES CONVERTED TO SPECIFIED UNIT SYSTEMS

TABLE - CONVERSION FACTORS OF MEASURED QUANTITIES FROM SI UNIT SYSTEM TO SPECIFIED UNIT SYSTEMS

CONCLUSION

The given quantities were converted to the required unit systems and detailed calculations were also attached to his challenge to show the conversion of the quantities from the SI unit system to the g/mm/ms system.