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Calculation of Stiffness in Structural elements

Question 1: AIM - 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 Elb = Elc ANS - ASSUMING THE THREE DEGREE OF FREEDOM SYESTEM . CASE - 1 UNIT DISPLACEMENT u1=1 STIFFNESS…

Harshal virkhare
updated on 18 Jul 2022

Question 1:

AIM - 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

ANS - ASSUMING THE THREE DEGREE OF FREEDOM SYESTEM .

CASE - 1

UNIT DISPLACEMENT u1=1

STIFFNESS COEFFICEANT OF COLUMN TRANSLATION (2 SIMMILER)

= 2x 12EI/h

STIFFNESS COEFFICIENT OF BEAM ROTATION (2 END)

= EI/h^2

CASE - 2

UNIT ROTATION U2 - 1

STIFFNES COEFFICIENT OF COLUMN ROTATION AT RIGHT END

= 6EI/h^2

STIFFNES COEFFICIENT OF COLUMN + BEAM ROTATION AT LEFT END

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

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

CASE -3

UNIT ROTATION u3 =1

STIFFNESS COEFICIENT OF COLUMN ROTATION AT LEFT END

= 6EI/h^2

STIFFNESS COEFICIENT OF COLUMN + BEAM ROTATION AT RIGHT END

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

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

[ 6EIc/h^2 6EIc/h 6EIc/h ]

Fs = k.u

EIc/h^3 = [ 24 6h 6h ]

[ 6h 6h^2 h^2 ]

[ 6h h^2 6h^2 ]

by applying static condensation method , and deducing the three unknown expression in to 2 unknown equiation.

{u2}

{u3} = - [ 6h^2 h^2 ] ^-1 [6h] [1]

[h^2 6h^2 ] = [6h] = u1 = -(6/7H) [1] u1

fs ( 24ELc/h^3 - ELc/h^3 * 6/7h (6h 6h) [1] ) u1

[1]

= 96/7 ELc/h^3 u1

stiffnes is the ammounrt of the force required for unit displacement so interchanging the translation u1 to the left side, we are arrive the stiffnes coefficient as follow.

k= 96/7 ELc/Gh^3

ans .

AIM -

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 THE ELEMENT CAN BE DESCRIBE BY ONE COORDINATE ONLY.

1. number of degrees-of-freedom for dynamic analysis - the syestem has single degree of freedom structure.

2. Establish the equation of motion -

the column has fixed end at bottom, so the stiffnes coefficient will be as follow.

k = 3EI/H^3

THE EQUATION OF THE MOTION SHALL BE - MU + KU = 0

3. Calculate their natural frequencies - natural frequency formula ω = √ k/m

ω = √ 3EI/Mh^3

POSITION OF THE ELEMENT CAN BE DESCRIBED BY 1 CORDINATE ONLY u(t).

BECAUSE BEAM WITH INFINITE STIFFNES WILL INDUCE UNIFORM TRANSLATION IN ALL COLUMN

1. number of degrees-of-freedom for dynamic analysis - the syestem has single degree of freedom structure.

TWO COLUMN WITH BOTH END FIXED AND ONE COLUMN HAS PINNED FIXED END. SO THE STIFFNESS COFFICIENT WILL BE AS FOLLOW.

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

for parallel stiffnes elements Ke = (k1+k2+k3) = 27ei/*h^3

THE EQUATION OF THE MOTION SHALL BE - MU + KU = 0

NATURAL FREQUENCY FORMULA -

ω = √ k/m

ω = √ 27EI/Mh^3

AIM -

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 - 1. EFFECTIVE STIFFNES OF THE SYESTEM - Ke = Kb+Kc

KB = 48EI/lb^3

Kc = 3EI/lc^3

ke = 3EI (1/Lc^3 + 16/Lb^3)

NATURAL FREQUENCY FORMULA -

ω = √ k/m

ω = √3EI(1/Lc^3 + 16/Lb^3)/M

TIME PERIOD - T = 2π/ω $2^{3}$ $2^3$ ans .

AIM -

Question 4

Determine an expression for the effective stiffness of the following systems

ANS - A.

DETERMINNING THE EXPRESSION FOR THE EFFECTIVE STIFFNES FOR THREE ELEMENT WHICH IS SHOWN IN BELEOW.

1. THE STIFFNESS COEFFICIENT FOR THE BEAM ELEMENT IS = 3EL/ $L^{3}$ $L^3$

2. THE STIFFNESS COEFFICIENT FOR THE SPRING IS 'K'

3.THE STIFFNESS ELEMENT ARE IN SEREIS SO.

$\frac{1}{K E}$ $1 /( KE$ = $\frac{L^{3}}{3 E I}$ $L^3/(3EI)$ + $\frac{1}{K}$ $1/K$

$\frac{1}{K E}$ $1 /( KE$ = $K L^{3} + 3 E \frac{I}{3 E I K}$ $KL^3+3 EI /( 3EIK$

Ke = $3 E I \frac{K}{K L^{3} + 3 E I}$ $3EIK/(KL^3+3EI$

ELEMENT (B.) ANS

`. THE STIFFNESS COEFICIENT OF THE BOTH THE COLUMN ELEMENT IS $3 E \frac{I}{h^{3}}$ $3EI/h^3$

2. THE STIFFNESS COEFFICIENT FOR THE SPRING IS 'K' BUT IS INCLINED SO THE HORIZONTAL FORCE (P) AND DEFORMATION (U) HAS TO BE RESOLVED IN TO INCLINED COMPONENTS TO ARRIVE ACTUAL COEFFICIENT OF THE SPRING.

INCLINED COMPONENET OF LOAD 'P' = $\frac{P}{C O S \emptyset}$ $P/(COS emptyset$

inclined componenet of deformation'u' = u $C O S \emptyset$ $COS emptyset$ `

stiffness coefficient of spring $k = \frac{p}{u} {cos}^{2} \emptyset$ $k=p/ucos^2emptyset$

$\frac{p}{u} = k s = k {cos}^{2} \emptyset$ $p/u=ks=kcos^2emptyset$

$cos \emptyset = \frac{l}{{\sqrt{l}}^{2}} + h^{2}$ $cosemptyset=l/sqrtl^2+h^2$

${cos}^{2} \emptyset = \frac{l^{2}}{l^{2}} + h^{2}$ $cos^2emptyset=l^2/l^2+h^2$

Ks = K $\frac{l^{2}}{l^{2}} + h^{2}$ $l^2/l^2+h^2$ `

THE EFFECTIVE STIFFNESS OF THE SYESTEM = KE= $6 E \frac{I}{h^{3}} + k \frac{l^{2}}{l^{2}} + h^{2}$ $6EI/h^3+kl^2/l^2+h^2$ ANS.

ELEMENT (C) ANS -

1. THE STIFFNESS COEFICIENT OF THE PINNED BEAM IS K1 = 48EI/L^3

2. THE EFFECTIVE STIFNESS COEFICIENT FOR THE SYESTEM HAS TO BE FORMED BY ADDING THE RELATIVE AXIAL STIFFNESS PROVIDED BY THE INCLINED STRUT CONNECTED AT THE RIGHT END.

3. 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 NODEL DEFORMATION.

$m \frac{g}{2} = k e a . v e r t \cdot 2 \cdot △ 2$ $mg/2 =kea.vert *2*triangle2$

$m \frac{g}{2} = 4 . k e a . v e r t$ $mg/(2 )= 4.kea.vert$

$k 2$ $k2$ =4 kea.vert`

we assume vertical axial stiffness in our calculation.but we have in inclined axial stiffness in the given syestem so we have to resolve the (kea.vert) into (kea).

$k 2 = 4 k e a \frac{h^{2}}{h^{2}} + l^{2}$ $k2=4keah^2/h^2+l^2$

the effective stiffness shall be arive as . $\frac{1}{k} e = \frac{1}{k} 1 + \frac{1}{k} 2$ $1/ke=1/k1+1/k2$

` $\frac{1}{k} e = \frac{1}{48 E \frac{I}{L^{3}}} + \frac{1}{4 K e a \frac{h^{2}}{l^{2}} + h^{2}}$ $1/ke=1/(48EI/L^3)+1/(4Keah^2/l^2+h^2)$

$\frac{1}{k} e = \frac{1}{48 E \frac{I}{l^{3}}} + 1 - (4 E A \frac{h^{2}}{{(l^{2} + h^{2})}^{15}})$ $1/ke=1/(48EI/l^3)+1-(4EAh^2/(l^2+h^2)^15)$

$k e = (4 E A \frac{h^{2}}{{(l^{2} + h^{2})}^{3}} / 2) 48 E \frac{I}{l^{3}} / 4 E A h^{2} \frac{{(l^{2} + h^{2})}^{3}}{2} + 48 E \frac{I}{l^{3}}$ $ke=(4EAh^2/(l^2+h^2)^3/2)48EI/l^3/4EAh^2(l^2+h^2)^3/2+48EI/l^3$