Structural Analysis - SPRIGN-STIFF, Study notes of Structural Analysis

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Stiffness Method

Beam with Spring Supports

Deflected Shape

5 m 5 m

24 kN 12 kN

A

B C

5 m 5 m

EI / 10

EI EI

EI / 25

D = K =

Applying unit displacements on every segment the corresponding

stiffnesses are obtained.

The matrices for a single segment have been formulated below.

K = Stiffness Matrix

D = Matrix of unknown Displacements

6 EI

L^2

12 EI

L^3

-12 EI

L^3

6 EI

L^2

4 EI

L

6 EI

L^2

-6 EI

L^2

2 EI

L

-6 EI

L^2

-12 EI

L^3

12 EI

L^3

-6 EI

L^2

2 EI

L

6 EI

L^2

-6 EI

L^2

4 EI

L

R 1

M 1

R 2

M 2

In the case of springs, deflection caused at a spring node due to

external forces will be reduced due to the force generated by the

spring proportional to its stiffness co-efficient K.

F = K x ∆

K =

6 EI

24 EI

-12 EI

6 EI

4 EI

6 EI

-6 EI

2 EI

-6 EI

-12 EI

12 EI

-6 EI

2 EI

6 EI

-6 EI

4 EI

EI

5 m 5 m

24 kN 12 kN

A

B C

5 m 5 m EI / 10

EI EI

EI / 25

EI

Solving the above in MATLAB gives the following results:

ΔB = -179.04/EI ΔC = -100.586/EI

θB = 1.269/EI θC = -49.903/EI

Calculating final reactions and moments at fixed ends:

EI

EI

RA

MB

RC

MC

MA

RB

The values of moments and reactions at A are obtained as

follows:

RA = -14.163 MA = -40.

Spring Forces:

F 1 = 0.1 x 179.04 F 2 = 0.04 x 100.

= 17.904 kN = 4.023 kN

Shear Force Diagram: