Chapter-2: Slope Deflection Method By Prof. H.P.Sudarshan ...

Chapter-2: Slope Deflection Method

By Prof. H.P.Sudarshan

Sri Siddhartha Institute Of Tech, Tumkur

Example: Analyze the propped cantilever shown by using slope defection method. Then draw Bending moment and shear force diagram.

Solution: End A is fixed hence A =0 End B is Hinged hence B 0

Assume both ends are fixed and therefore fixed end moments are

FAB

wL2 12

,

FBA

wL2 12

The Slope deflection equations for final moment at each end are

MAB

FAB

2EI L

2A

B

wL2 12

2EI L

B

(1)

MBA

FBA

2EI L

2B

A

wL2 12

4EI L

B

(2)

In the above equations there is only one unknown B .

To solve we have boundary condition at B;

Since B is simply supported, the BM at B is zero

ie. MBA=0.

From equation (2) MBA

wL2 12

4EI L

B

0

EIB

wL3 48

- ve sign indicates the rotation is anticlockwise

Substituting the value of EIB in equation (1) and (2) we have end moments

MAB

wL2 12

2 L

wL3 48

wL2 8

- ve sign

indicates moment

is anticlockwise

MBA

wL2 12

4 L

wL3 48

0

MBA has to be zero, because it is hinged.

Now consider the free body diagram of the beam and find reactions using equations of equilibrium.

MB 0

RA

L

MAB

wL

L 2

wL2 wL L 5 wL

8

28

RA

5 8

wL

V 0

RA RB wL

RB

wL RA

wL

5 8

wL

3 wL 8

Problem can be treated as The bending moment diagram for the given problem is as below

The max BM occurs where SF=0. Consider SF equation at a distance of x

from right support

SX

3 8

wL

wX

0

X 3L 8

Hence the max BM occurs at 3 L from support B 8

Mmax

MX

3 wL 8

3L 8

w 3 L 2 2 8

9 wL2 128

And point of contra flexure occurs where BM=0, Consider BM equation at

a distance of x from right support.

MX

3 8

wLX w

X2 2

0

X 3L 4

For shear force diagram, consider SF equation from B

SX

3 8

wL

wX

SX

0

SB

3 8

wL

SX

L

SA

5 8

wL

Example: Analyze two span continuous beam ABC by slope deflection method. Then draw Bending moment & Shear force diagram. Take EI constant

Solution: Fixed end moments are:

FAB

Wab2 L2

100

4 62

22

44.44KNM

FBA

Wa 2b L2

100 42 2 62

88.89KNM

FBC

wL2 12

20 52 12

41.67KNM

FCB

wL2 12

20 52 12

41.67KNM

Since A is fixed A 0 , B 0, C 0,

Slope deflection equations are:

MAB

FAB

2EI L

2A

B

44.44

2EI 6

B

44.44

1 3

EI

B

MBA

FBA

2EI L

2B

A

88.89 2EI 2B 6

88.89

2 3

EIB

(1) (2)

MBC

FBC

2EI L

2B

C

41.67

2EI 5

2B

C

41.67

4 5

EIB

2 5

EIC

MCB

FCB

2EI L

2C

B

41.67

2EI 5

2C

B

41.67

4EI 5

C

2 5

EIB

(3) (4)

In all the above four equations there are only two unknown B and C . And

accordingly the boundary conditions are

i -MBA-MBC=0

MBA+MBC=0

ii MCB=0 since C is end simply support.

Now

MBA

MBC

88.89

2 3

EIB 41.67

4 5

EIB

2 5

EIC

47.22

22 15

EIB

2 5

EIC

0

(5)

MCB

41.67

2 5

EIB

4 5

EIC

0

(6)

Solving simultaneous equations 5 & 6 we get

EI B = ? 20.83

Rotation anticlockwise.

EI C = ? 41.67

Rotation anticlockwise.

Substituting in the slope definition equations

MAB = ? 44.44 +

1 20.83 51.38 KNM

3

MBA = +

88.89 +

2 20.83 75.00 KNM

3

MBC

=

?

41.67+

4 5

20.83

2 5

41.67

75.00

KNM

MCB

=

+

41.67+

2 5

20.83

4 5

41.67

0

Reactions: Consider the free body diagram of the beam.

Find reactions using equations of equilibrium.

Span AB: Span BC:

MA = 0 V = 0 MC = 0

RB?6 = 100?4+75-51.38 RB = 70.60 KN RA+RB = 100KN RA = 100-70.60=29.40 KN RB?5 = 20?5? 5 +75

2 RB = 65 KN

V=0

RB+RC = 20?5 = 100KN RC = 100-65 = 35 KN

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