Calculation of Stiffness in Structural elements

1)

Assume L=2h; EIb=EIc;

Case 1;

Unit displacement u1=1

Stiffness coefficient off column translation (2 similiar) = 2*12EI/h^3

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

Case 2;

Unit rotation u2=1

Stiffness coefficient of column at right end = 6EI/h^2

Stiffness coefficient of column + beam rotation at left end = 6EI/h

Case 3;

Unit rotation u3=1

Stiffness coefficient of column at left end = 6EI/h^2

Stiffness coefficient of column+ beam rotation at right end = 6EI/h

A matrix is formed with the three equations,

we know that, fs= k*u

By applying static condensation method 3 unknown expressions are deduced into 2 unknown equation.

Stiffness is the amount of force required for unit displacement so interchanging the translation u1 to the left side, we arrive at stiffness coefficient , k= (96/7*EIc/h^3)

2)

a)

• Position of the elemnt can be described by one coordinate only u(t)
• The system has a Single degree of freedom structure.
• The column has fixed end at bottom, therefore the stiffness coefficient is, k=3EI/h^3
• The equation of motion shal be Mu+ku=0
• Natural frequency w= root of k/M;
• We know that k= 3EI/h^3.
• Therefore Natural frequency w= root of 3EI/Mh^3 .

b)

• Position of element can be described by 1 coordinate only u(t), stiffness of beam is infinite therefore it will induce uniform translation in all columns.
• The system has Single degree of freedom of structure
• Two columns with both ends fixed; One column with end pinned, so the stiffness coefficient is as follows

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

• For parallel stiffness elements ke=(k1+k2+k3)= 27EI/h^3.
• The equation of motion shall be Mu+ku=0
• Therefore Natural frequency w= root of 27EI/Mh^3. 

3)

• Material and section of both elements are the same, therefore the Modulus of elasticity and the Second moment of Inertia are also the same. The lengths of the elements are different.
• When a mass M is applied at the point both elements will respond simultaneously because they are connected in parallel.
• Therefore Effective stiffness of system ke= kb+kc.

kb=48EI/lb^3; kc= 3EI/lc^3; 

therefore, ke= 3EI (1/lc^3+ 16/lb^3)

• Natural frequency w= root of k/M;

w= root of 3EI (1/lc^3+ 16/lb^3)/M

• Time period T= 2pi/w.  

4)

a)

• Stiffness coefficient of beam= 3EI/l^3
• Stiffness coefficient of spring is K
• The stiffness of elements are in series.

So, 

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

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

ke= 3EIK/kL^3 +3EI

b)

• Stiffness coefficient of both column elements is 3EI/h^3
• Stiffness coefficient of spring is K and it is inclined, so the horizontal force (P) and deformation (u) has to be resolved into inclined components to arrive at actual coefficient of spring.
• Inclined component of load P= p/cos0.
• Inclined component of deformation u=ucos0
• Stiffness coefficient of spring k=P/u(cos^2)0 

p/u= ks =K(cos^2)0

• cos0= L/root of L^2+h^2
• cos^2o= L^2/L^2+h^2
• ks=K (L^2/L^2+h^2) 
• The effective stiffness of system, Ke= (6EI/h^3) + (KL^2/L^2+h^2)

• Stiffness coefficient of 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 defromation caused by the strut at the point of m is half of its right end nodal deformation.

mg/2=kea,vert*2*delta2

mg/delta2=4kea,vert

k2=4kea,vert

• Vertical axial stifness is assumed in the calculation, but inclined axial stiffness is present in the given system.
• Therefore (kea,vert) must be resolved in (kea)

k2=4kea*h^2/(L^2+h^2)

• The effective stiffness is, 1/ke=1/k1 + 1/k2 because they are connected is series.
• Therefore ke= (4AEh^2/(L^2+h^2)^3/2)*48EI/L^3/ (4AEh^2/(L^2+h^2)^3/2)+48EI/L^3