Page 45 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers

32 • Using ansys for finite element analysis
Assumptions:

∧ ∧
• The bar cannot sustain shear force, that is, f 1 y = 0, f 2 y = 0.
• Any effect of transverse displacement is ignored.
• Hooke’s law applies, that is, s = E e .
x
x
• No intermediate applied loads.
Two nodes: Node 1, Node 2
Local nodal displacements: ∧ , ∧ (inch, m, mm)
d x1 d x2
Local nodal forces: ∧ ∧
f , f 2 x (lbs., Newton)
x 1
Length L (inch, m, mm)
Cross-sectional area A (Sq. inch, Sq. m, Sq. mm)
Modulus of elasticity E (psi, Pa, MPa)

The stiffness equation for a single spring element in a local coordinate
system can be written as:
 ∧   ∧ 
 f x 1  k  11 k   d x 1    ∧    
∧
∧
12
   
  =     ⇒ f   = k d
 ∧   k 21 k 22   ∧      
f
 2 x   d x 
2
Step 2: Select a displacement function
• Degree of freedom (DOF) per node = 1
• Number of nodes per element = 2
• Total (DOF) per element = 2 × 1 = 2
• Number of coefficients = 2
Assume a linear displacement function:

∧
u = a + ax ∧
2
1
Write in matrix form.
a 
∧  ∧  1
u = 1 x  


a
   2 
^ ∧ ∧
Express u as a function of nodal displacements ( d x1 , d x2 )
Apply boundary condition:
∧ ∧ ∧ ∧ ∧ ∧
u
a
0
0
At x = 0 u = d x ∴ () = a + () = d x ∴ a = d x1
1
1
2
1
1