System of Units

Top  Previous  Next

EngiLab Beam.2D has no default system of units. This is not a limitation of the program, but a deliberate choice in order for the program to work globally, no matter what system of units is used. This way there is no limitation in the system of units that can be used. Any consistent system of units can be used.

Before starting to define any model, you need to decide which system of units you will use. All input data must be specified in consistent units. As a result, the analysis results will also comply to that system. The important point about using consistent units is the necessity to stick with units that work correctly together - not to mix units that do not have a correct relationship with each other.

In order to define a consistent system of units, you have to define first the primary (basic) units and then the derived units which are dependent on the primary units. We propose two different approaches for defining a consistent system of units, as described in detail below:

 

A. Consistent system of units based on Force

Define the three primary (basic) units for Force (F), Length (L), Time (T). For example you can choose to use kN, m, s (sec). The derived units are then the following:

 

Derived Unit    

Formula    

Formula explanation    

In our example (kN, m, s)    

Acceleration

L/T2

(1 Length unit) / (1 time unit)2

m/s2

Mass *

F·T2/L

(1 force unit) / (1 acceleration unit)

kN/(m/s2) = Mg = t *

Density

F·T2/L4

(1 mass unit) / (1 length unit)3

t/m3

Stress

F/L2

(1 force unit) / (1 length unit)2

kN/m2

* The mass unit (in our example 1 t) is the mass that accelerates by the acceleration unit rate (in our example 1 m/s2) when the unit force (in our example 1 kN) is exerted on it.

 

B. Consistent system of units based on Mass

Define the three primary (basic) units for Mass (M), Length (L), Time (T). For example you can choose to use kg, m, s (sec). The derived units are then the following:

 

Derived Unit    

Formula    

Formula explanation    

In our example (kg, m, s)    

Acceleration

L/T2

(1 Length unit) / (1 time unit)2

m/s2

Force *

M·L/T2

(1 mass unit) · (1 acceleration unit)

kg·m/s2 = N *

Density

M/L3

(1 mass unit) / (1 length unit)3

kg/m3

Stress

M/L/T2

(1 force unit) / (1 length unit)2

N/m2

* The force unit (in our example 1 N) is the force required to accelerate the unit mass (in our example 1 kg) at the acceleration unit rate (in our example 1 m/s2).

 

Common consistent systems of units

Some common systems of consistent units are shown in the table below.

Quantity    

SI (MKS)    

MTS    

mmNS    

US Unit (ft)    

US Unit (in)    

Length

m

m

mm

ft

in

Force

N

kN

N

lbf

lbf

Mass

kg

t (tonne)

t

slug

lbf·s2/in

Time

s

s

s

s

s

Stress

Pa (N/m2)

kPa (kN/m2)

MPa

lbf/ft2

psi (lbf/in2)

Density

kg/m3

t/m3

t/mm3

slug/ft3

lbf·s2/in4

Acceleration

m/s2

m/s2

mm/s2

ft/s2

in/s2

 

As points of reference, the mass density of steel, the Young's Modulus of steel and the standard earth gravitational acceleration are given in each system in the table below.

Quantity    

SI (MKS)    

MTS    

mmNS    

US Unit (ft)    

US Unit (in)    

Steel density

7850 kg/m3

7.85 t/m3

7.85E-9 t/mm3

15.231514 slug/ft3

0.0088145342 lbf·s2/in4

Steel E

210E9 N/m2

210E6 kN/m2

210E3 MPa

4385941189 lbf/ft2

30457924.92 psi

Earth Gravity

9.80665 m/s2

9.80665 m/s2

9806.65 mm/s2

32.17404856 ft/s2

386.088583 in/s2

 

Notes:

1 t (tonne) = 103 kg = 1 Mg. It is a mass that accelerates by 1 m/s2 when a force of 1 kN is exerted on it.

1 slug = 1 lbf·s2/in. It is a mass that accelerates by 1 ft/s2 when a force of 1 pound-force (lbf) is exerted on it.

1 MPa = 1 MN/m2 = 1 N/mm2

 

Practical example

The user chooses to use the MTS system, a common choice for structural engineering applications:

Length: m

Force: kN

Time: s

 

EngiLab Beam.2D data have to be given as shown below:

 

Quantity

Unit used

Node Coordinates X, Y

m

Material Elastic Modulus E

kPa = kN/m2

Material Density d

t/m3

Section Area Α

m2

Section Moment of Inertia I

m4

Nodal Force F

kN

Nodal Moment Μ

kN·m

Elemental load f

kN/m

Spring Elastic Stiffness KX, KY

kN/m

Spring Elastic constant KZ

kN·m (/RAD) *

Acceleration

m/s2

 

The results will also comply to that system, thus they will be given as:

 

Quantity

Unit used

Node X, Y Displacement

m

Node Z Rotation

RAD *

Axial, Shear Force at Element end i, j

kN

Moment at Element end i, j

kN·m

Support reaction FX, FY

kN

Support reaction MZ

kN·m

* Rotations are ALWAYS given in RADIANS.

 

Note: In the above example, if one wants to apply self-weight to the structure, he can add the standard earth gravitational acceleration at the -y direction: ay = -9.80665