Calculation of Stiffness in Structural elements

Q1. Compute lateral stiffness of the one story frame with an intermediate realistic stiffness of the beam. The system has 3 DOFs as shown. Assume L = 2h and El_b = El_c

Ans: The step by step calculation is as follows;

Step 1: Case 1-Unit displacement, u1 = 1

Stiffness coefficient of column translation (2 columns) = 2x(12 EI/h³) = 24 EI/h³

Stiffness coefficient of beam rotation (2 ends) = EI/h²

Step 2: Case 2-Unit displacement, u2 = 1

Stiffness coefficient of column rotation at right end = 6 EI/h²

Stiffness coefficient of column and beam rotation at left end = 4 EI/h + 4 EI/L = 4 EI/h + 4 EI/2h = 6 EI/h

Step 3: Case 3-Unit displacement, u3 = 1

Stiffness coefficient of column rotation at left end = 6 EI/h²

Stiffness coefficient of column and beam rotation at right end = 4 EI/h + 4 EI/L = 4 EI/h + 4 EI/2h = 6 EI/h

Step 4: El_b = El_c

k x u = f_s

Step 5: Static Condensation Method

The values of degrees u₂ and u₃ terms of u₁ can be identified as follow

which will result in

Step 6: Apply the above equation in the first equation of motion

where, k = 96/7 x El_c/h³

Therefore, the lateral stiffness coefficient k is obtained.

Q2. For the following structures:

• Determine the number of degrees-of-freedom for dynamic analysis
• Establish the equation of motion
• Calculate their natural frequencies

Ans:

a)

• Position of the element can be described by 1 coordinate only u(t)
• The system has single degree of freedom structure
• The column has fixed end at bottom. So the stiffness coefficient will be as follow
  • k = 3EI/h³
• The equation of motion shall be
  • Mu + ku = 0
• Natural frequency formula
  • ω = √(k/m) = √(3EI/Mh³)

b)

• Position of the element can be described by 1 coordinate only u(t), because beam with infinite stiffness will induce uniform translation in all columns.
• The system has single degree of freedom structure
• Two columns with both end fixed and one column has pinned-fixed ends. So the stiffness coefficient will be as follow
  • k₁ = 12EI/h³
  • k₂ = 12EI/h³
  • k₃ = 3EI/h³
• For parallel stiffness elements, k_e = (k₁+k₂+k₃) = 27EI/h³
• The equation of motion shall be
  • Mu + ku = 0
• Natural frequency formula
  • ω = √(k/m) = √(27EI/Mh³)

Q3. Consider the propped cantilever shown in the figure below. The beams are made from the same steel section and have lengths as shown on the diagram. Determine the natural period of this system if a large mass, M, is placed at the intersection of the beams at point A. The weight of the beams in comparison with the mass M is very small.

Ans:

• Both the elements have same material and section. It means they have same modulus of elesticity and second moment of inertia. But the lengths are different.
• When a mass 'M' is applied at the point, both elements will respond simultaneously. It means they are conected in parallel.
• The effective stiffness of the system, k_e = k_b + k_c
  • k_b = 48EI/l_b³
  • k_c = 3EI/l_c³
  • k_e = 3EI ( (1/l_c³)+(16/l_b³) )
• Natural frequency formula
  • ω = √(k/m) = √( 3EI ( (1/l_c³)+(16/l_b³) )/M )
• Time period, T = 2π/ω = 2π/√( 3EI ( (1/l_c³)+(16/l_b³) )/M )

Q4. Determine an expression for the effective stiffness of the following systems:

Ans:

a)

• The stiffness coefficient of the beam element = 3EI/L³
• The stiffness coefficient of the spring = K
• The stiffness elements are in series.
  • So, 1/k_e = (L³/3EI) + (1/K) = (KL³+3EI)/3EIK
  • Therefore, k_e = 3EIK/(KL³+3EI)

b)

• The stiffness coefficient of the both the column elements = 3EI/h
• The stiffness coefficient of the spring is K, but is inclined. So the horizontal force (P) and deformation (u) has to be resolved into inclined components to arrive actual coefficient of the spring.
  • Inclined component of load 'P' = P/cosθ
  • Inclined component of deformation 'u' = u cosθ
  • Stiffness coefficient of spring, K = P/u cos²θ
  • k_s = P/u = K cos²θ
  • cosθ = L/√(L²+h²)
  • cos²θ = L²/(L²+h²)
  • Therefore, k_s = K x ( L²/(L²+h²) )
• The effective stiffness of the system, k_e = (6EI/h³)+( KL²/(L²+h²) )

c)

• The stiffness coefficient of the pinned-pinned beam, k₁ = 48EI/L³
• The effective stiffness coefficient for the system has to be formed by adding the relative axial stiffness provided by the inclined strut connected at the right end.
• Force acting at the right end is 'mg/2' and deformation caused by the strut at the point of 'm' is half of its right end nodal deformation.
  • mg/2 = k_ea,vert x 2 x Δ₂
  • mg/Δ₂ = 4 k_ea,vert
  • k₂ = 4 k_ea,vert
• We assumed vertical axial stiffness in our calculation. But we have an inclined axial stiffness in the given system. So we have to resolve the (k_ea,vert) into (k_ea).
  • k₂ = 4 k_ea h²/(L²+h²)
• The effective stiffness, 1/k_e = (1/k₁)+(1/k₂) = ( 1/(48EI/L³) ) + ( 1/( 4 k_ea h²/(L²+h²) ) ) = ( 1/(48EI/L³) ) + ( 1/( 4EAh²/(L²+h²)^1.5 ) )
• Therefore, k_e = (48EI/L³) + ( 4EAh²/(L²+h²)^1.5 )