Calculation of Stiffness in Structural elements

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:-

• unit displacement = u1=1
• Stiffness coefficientof column translation (2 similar) 2 x 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 (`:'`L=2h)
• unit rotation u3 = 1
• Stiffness coefficient of column rotation at end 6EI/h^2
• Stiffness coefficient of column + beam rotation at right end 4EI/h + 4EI/L 4EI/h + 4EI/2h = 6EI/h

[[24EIc/h^3,6EIc/h^2,6EIc/h^2],[6EIc/h^2,6EIc/h,EIc/h],[6EIc/h^2,EIc/h,6EIc/h]]

fs = k.u

EIc/h^3 [[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,6h^2]] {{u1},{u2},{u3}} = {{fs},{0},{0}}

• By applying static condensation method, we are deducting the 3 unknown expression into 3 unknown equations.

{{u2},{u3}} = -[[6h^2,h^2],[h^2,6h^2]^-1 [[6h,6h]]u1 = -6/7h [[1,1]]u1

fs = (24EIc/h^3 - EIc/h^3 6/7h <6h 6h> [[1],[1]])u1

= 96/7 EIc/h^3 u1

Stiffness is the amount of force required for unit displacement. so interchanging the translation u1 to the left side, we arrive te stiffness coefficient as follow

K = 96/7*EIc/h^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

• 

Ans:-

• position of element can be desribed by one coordinate only. u(t)
• The system has single degree of freedom system structure.
• The column has fixed end at bottom. so the stiffness coefficient will be as follow.

K = 3EI/h^3

• The equation of motion shall be m ddot U + Ku = 0
• Natural frequency formula omega = sqrt 3EI/Mh^3

Question 2:

b)

Ans:-

• position of the element can be described by one coordinate only u(t) because beam with infnite stiffness will induce uniform translation in all columns.
• the system has single degree of freedom system structure.
• Two columns with both ends fixed & one column has pinned - fixed ends.so the stiffness coefficient will be follow.

K1 = 12EI/h^3 , K2= 12EI/h^3, K3 = 3EI/h^3

For parallel tiffness elements Ke = (K1+K2+K3) = 27EI/h^3

the equation of motion will be m ddot U + Ku = 0

natural frequency formula omega = sqrt K/m

omega = sqrt 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.

Ans:-

• Both 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 the point, both elements will be respond simultaneously. it means they are connected in parallel (as depicted in the sketch

Effective stiffness of system Ke=Kb+Kc

Kb = 48EI/lb^3 Kc=3EI/lc^3

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

Natural frequency formula omega = sqrt k/m

omega = sqrt 3EI(1/lc^3 + 16/lb^3)/M)

time period = 2 pi / omega

Question 4

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

Ans:

a)

• The stiffness coefficient of beam element is 3EI/l^3
• The stiffness coefficient of spring is K
• Stiffness elements are in series so,

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

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

Ke = 3EIK/(KL^3 + 3EI

b)

Ans:-

• The stiffness coefficient of the both column elements is 3EI/h^3
• The stiffness coefficient of the spring is K, but is inclined. so the horizontal force (P) = P/cosTheta
• inclined component of deformation u = ucosTheta
• stiffness coefficient of spring

K = P/ucos^2Theta :' (K = (p/cos Theta)/cosTheta)

P/u = Ks = K cos^2 theta

cos theta = L/(sqrt L^2+H^2)

cos^2 theta = 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 pinned-pined 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 deformation caused b y strut at the point of 'm' is half of its right end nodal deformation.

mg/2 = Kea,vert X2Xdelta

mg/delta2 = 4 Kea,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 = 4Kea h^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/(48Keah^2/L^2+h^2)

1/Ke = 1/(48EI/L^3) + 1/(48EAh^2/(L^2+h^2)15)

Ke =( (48EAh^2)/((L^2+h^2)^3/2) (48EI/L^3)) / ( (48EAh^2)/((L^2+h^2)^3/2) + (48EI/L^3)