Calculation of Stiffness in Structural elements

CHALLENGE WEEK 2

ETABS

QUESTION 1: 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

Answer:

• Let us start this computation by substituting 1 as value for each degree of free in different cases.

Case 1:

• Unit displacement u1 = 1
• Stiffness coefficient of column translation (including both)

2 x (12EI/h^3)

• Stiffness coefficient of beam rotation (at both ends)

EI/h^2

Case 2:

• Unit rotation u2 = 1
• Stiffness coefficint of column + Beam roation at left end

(4EI/h) + (4EI/L)

(4EI/h) + (4EI/2h) = (6EI/h)

Case 3:

• Unit rotation u3 = 1
• Stiffness coefficient of column rotation at left end

(6EI/h2)

• Stiffness coefficient of column + Beam rotation at right end

(4EI/h) + (4EI/L)

(4EI/h) + (4EI/2h) = (6EI/h)

• Now the stiffness coefficient from all the cases are made to be computed using matrix calculation.

fs = k.u

• And by this matrix calxulation we obtain the equation

fs = ((24EIc/h^3) - (6EIc/7h^4)(36h^2)[1/1]) = (96EIc/7h^3)u1

• Stiffness is the amount of force required for unit displacement. So interchanging the translation u1 to the left side,
• We arrive the stiffness coefficient as follow.

k= (96EIc/7h^3)

QUESTION 2:For the following structures:

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

Answer:

A)

• Position of teh 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 teh stiffness coefficient will be as follow

k = (3EI/h^3)

• The equation of motion shall be Mü+ku=0
• Natural frequency formula ω = √(k/m)
• Which is

ω = √(3EI/Mh^3)

B)

• Position of the element can be described by 1 coordinate only u(t) because beam with infinite stiffness will include uniform translation in all columns.
• The system has single degree of freedom structure.
• The column with both end fixed & one column has pinned - fixed ends. So the stifness coefficient will be as follow

K1 = (12EI/h^3) K2 = (12EI/h^3) K3 = (3EI/h^3)

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

• The equation of motion shall be Mü+ku=0
• Natural frequency formula

ω = √(k/m)

Which is ,

ω = √(27EI/Mh^3)

QUESTION 3: 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.

Answer:

• Here in this case, both elements have same material and section. It means they have same modulus of elasticity and second moment of inertia.
• But the length of the elements are different.
• When a mass M is applied at the point, both elements will respond simultaneously. It means they are connected in parallel(as depicted in the sketch below).
• Effective stiffness of the system ke = kb + kc

Where ,

kb = (48EI/lb^3)

kc = (3EI/lc^3)

Hence,

ke = 3EI((1/lc^3)+(16/lb^3)

• Natural frequency formula

ω = √(k/m)

after substituting the values of k

ω = √(3EI((1/lc^3)+(16/lb^3)/M)

• Time period T = (2π/ω)

QUESTION 3: Determine an expression for the effective stiffness of the following systems:

Answer:

Case 1:

• The stiffness coefficient of the beam elements is (3EI/L^3)
• The stiffness coefficient of the spring is K
• the stiffness elemts are in series in this case. so,

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

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

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

Case 2:

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

Inclined coomponents of load 'P' = (p/cosθ)

Inclined components of deformation 'u' = ucosθ

Stiffness coeffficient of spring.

K = (p/ucos^2θ)

p/u = ks= Kcos^2θ

cosθ= 1/ √(L^2 + h^2)

cos^2θ = L^2/(L^2 + h^2)

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

• The efficient stiffness of the system , ke = (6EI/h^3)+(KL^2/(L^2+h^2)

Case 3:

• Stiffness coefficient of teh 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 actibg 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 = kea,vert x 2 x Δ2

(mg/Δ2) = 4Kea,vert

k2 = 4Kea, 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 (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/ke) = (1/((48EI/L^3)) + (1/((4Keah^2/(L^2+h^2))

(1/ke) = (1/((48EI/L^3)) + (1/((4Keah^2/(L^2+h^2)^3/2)

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