Calculation of Stiffness in Structural elements

1. There are 3 degree of freedom ( 2 rotation, 1 transverse)

assume unit displacement u1=1

stiffness coefficient of column translation( 2 similar) = 2* 12 EI/h3

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

unit displacement u2=1

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

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

unit displacement u3=1

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

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

( 24EI/h3 6EI/h2 6EI/h2 )

( 6EI/h2 6EI/h EI/h )

( 6EI/h2 EI/h 6EI/h)

fs=k.u

EI/h3( 24 6h 6h ) ( u1) fs

6h 6h h2 u2 = ( 0 )

6h h2 6h2 u3 0

by applying static condenstion method , we dedudce the 3 unknown expression into 2 unknown equation

( u2) 6h2 h2 6h 1

u3 = - ( h2 6h2 ) ( 6h ) = - 6/7h ( 1 ) u1

fs= ( 24EI/h3 - EI/h3 6/7h (6h 6h ) (1 ) ) u1 = 96/7 EI/h3 u1

1

stiffness is the amount of force required for unit displacement . so interchanging the translation u1 to left side , we arrive the stiffness coefficient as k= 96/7 EI/h3

2. (a) Position of the element can be described by 1 coordinate only u(t)

The system has single degree of freedom structure

The column as fixed end at the bottom. So stiffness coefficient is k= 3EI/h3

The equation of motion Mu..+ ku=0

Natural frequency formula omega=root k/m

(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 & one end hinged . so stiffness coefficient is k1= 12EI/h3 , k2 = 12EI/h3, k3= 3EI/h3

For parallel stiffness elements ke=(k1+k2+k3)= 27EI/h3

The equation of motion Mu..+ ku=0

Natural frequency formula omega=root k/m = root 27EI/Mh3

3. Both elements have same material and section. It means has 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 .

Effective stiffness of system ke = kb+kc

Kb= 48EI/lb3 , kc= 3EI/lc3

Ke= 3EI(1/lc3+16/lb3)

Natural frequency formula omega =root k/m = root 3EI(1/lc3+16/lb3)/M

Time period T= 2pi/omega

4.(a) the stiffness coefficient of the beam element is 3EI/L3

the stiffness coefficient of the spring is K

the stiffness elements are in series. So,

1/ke= L3/3EI+1/K , ke = 3EIK/KL3+3EI

(b) the stiffness coefficient of the both column element is 3EI/h3

the stiffness coefficient of the spring is K , but is inclined. So horizontal force (P) and deformation (u) has been resolved into inclined components to arrive actual coefficient of the spring

inclined component of load P= p/cos teta , inclined component of deformation u= u cos teta

stiffness coefficient of spring K= P/ u cos2teta

cos teta= L/root L2+h2 , cos2teta= L2/L2+h2

ks= k L2/L2+h2

Effective stiffness of system ke= 6EI/h3+ KL2/L2+h2

(c) the stiffness coefficient of pin- pin beam k1=48EI/L3

The effective stiffness 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 * 2*delta2 , k2 =4kea, vert

We assumed vertical axial 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 kea h2/ L2+h2

The effective stiffness will be 1/ke= 1/k1+ 1/k2

1/ ke= 1/ (48EI/L3) + 1/ (4 kea h2/ L2+h2)