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

Stiffness Method - Beam with Spring Supports

Typology: Study notes

2010/2011

Uploaded on 09/01/2011

balaagriyan
balaagriyan 🇮🇳

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Stiffness Method
Beam with Spring Supports
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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: