Calculation of Stiffness in Structural elements

1.

Answer

• Unit displacement u1 = 1
• Stiffness coefficient of column translation (2 similar) = 2 * 12EI/h^3
• Stiffness coefficient of beam rotation (2 ends) = EI/h^2

 

• Unit rotation u2 = 1
• Stiffness coefficient of column rotation at right end = 6EI/h^2
• Stiffness coefficient of column + beam rotation at left end = 4EI//h + 4EI/L = 4EI//h + 4EI/2h = 6EI/h

 

• Unit rotation u3 = 1
• Stiffness coefficient of column rotation at left end = 6EI/h^2
• Stiffness coefficient of column + beam rotation at right end = 4EI//h + 4EI/L = 4EI//h + 4EI/2h = 6EI/h

 

 

 

2.

a.

 

Answer:- 

 

• Possition of the element can be described by one coordinate only u(t)
• The system has single degree of freedom structure
• The column has fixed end at the bottom. So the stiffness coefficient will be as follows

                                    k = 3EI / h^3

• The equation of the motion shall be Mü+ku = 0
• Natural frequency formula ὠ = (k/m)^1/2  => (3EI/Mh^3)^1/2

 

 

b.

• Possition of the elements can be described by 1 coordinate only u(t) because beam with infinite stiffness will induse uniform translation in all columns 
• The system has single degree of freedom structures 
• Two columns are both end fixed and One column has pinned-fixed end. So the stiffess coefficient will be as follows

         k1 = 12EI / h^3                   k2 = 12EI / h^3            k3 = 3EI / h^3

For parallel stiiffness element ke = (k1+k2+k3) = 27EI/h^3

• The equation of the motion shall be Mü+ku = 0
• Natural frequency formula ὠ = (k/m)^1/2 => (27EI/Mh^3)^1/2

 

3.

• Both the elements have same material and section. It means they have same modulus of elasticity and second moment of inertia. But the lengths are different 
• When a mass M is applied at a point, both elements will respond simultaneously. It means they are connected in parallel

Effetive stiffness of the system ke = kb+kc

         kb = 48EI / lb^3                   kc = 3EI / lc^3

ke = 3EI(1/lc^3 + 16/lb^3) 

Natural frequency formula ὠ = (k/m)^1/2 =>  (3EI(1/lc^3 + 16/lb^3) /M)^1/2

Time period  T= 2π/ὠ 

 

4.

a.

• The stiffness coefficient of the beam element is 3EI/L^3
• The stiffness coefficient of the spring is K
• The stiffness elements are in series

1/ke = L^3/3EI +1/K

1/ke = KL^3+3EI / 3EIK

ke = 3EIK/(kL^3+3EI)

 

b.

• The stiffness coefficient of both the column elements is 3EI/h^3
• The stiffness coefficient of the spring is K, but is inclined. So the horizontal force P and deformation u has to be resolved into inclinedd components to arrive actual coefficient of spring 

 Inclined component of load P = P/cosϴ

Inclined component deformation u = ucosϴ

Stiffness coefficient of spring,

K = P/u

K = P/ucos^2ϴ

p/u = ks = KCos^2ϴ

Cosϴ = L/(L^2+h^2)^1/2

Squaring on both side then

Cos^2ϴ = L^2/(L^2+h^2)

ks = K*L^2/(L^2+h^2)

The effective stiffness of the system, Ke = 6EI/h^3 + KL^2/L^2+h^2

 

c.

• The stiffness coefficient of the pinned to pinned beam is k1 = 48EI/L^3
• 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 the deformation caused by strut at the point of m is half of it's right end nodal deformation 

mg/2 = kea,vert * 2 *Δ2

mg/Δ2 = 4 * kea,vert

k2 =  4 * kea,vert

We assume vertical axial stiffness is our calculation. But we have inclined axial stiffness in the given system, So we have to resolve the (kea,vert) into  (kea)

k2 = 4keah^2/L^2+h^2

The effective stiffness shall be arrived as 1/ke = 1/k1 + 1/k2

1/kea = 1/(48EI/L^3) + 1/ (4kea*h^2/L^2+h^2)

1/kea = 1/(48EI/L^3) + 1/ (4EA*h^2/(L^2+h^2)^15)

ke = 4EA*h^2/(L^2+h^2)^3/2)*48EI/L^3 / 4EA*h^2/(L^2+h^2)^3/2)+48EI/L^3