August 29, 2013Page 1

MECH 321 -Solid Mechanics II

Week 11, Lecture 3

Statically Indeterminate Beams

Method of Superposition

August 29, 2013Page 2

The Method of Superposition has been used earlier in this

course for the analysis of combined loadings.

In order to apply this method to statically indeterminate beams

and shafts we must again first determine the redundant support

reactions.

Then, by removing them from the beam we obtain what is called

the “primary beam”, which is statically determinate and stable.

If we add to the primary beam a succession of similarly

supported beams, each loaded with a separate redundant

reaction, then by the principal of superposition, we can obtain

the actual loaded beam.

Statically Indeterminate Beams

Method of Superposition

August 29, 2013Page 3

To solve for the redundant reactions we again use the

conditions of compatibility that exist at the supports where each

of the redundant reactions act.

Once the redundant reactions are obtained, the other reactions

acting on the beam can be determined from the three equations

of equilibrium.

Statically Indeterminate Beams

Method of Superposition

August 29, 2013Page 4

Statically Indeterminate Beams

Method of Superposition

In the example shown, the redundant

reaction, B

y, is ignored in the first

step.

In order to maintain compatibility the

redundant reaction, By, is then

applied at B where the displacements

vB

and v’

B

are equal and opposite

(and therefore cancel).

Compatibility equation.

BB

vv

′

+

=0

August 29, 2013Page 5

Statically Indeterminate Beams

Method of Superposition

From App. C in Hibbeler.

EI

PL

vB

48

5

3

−=

EI

LB

v

y

B

3

3

=

′

Sub and solve.

EI

LB

EI

PL

y

348

5

0

3

3

+−=

PB

y

16

5

=

August 29, 2013Page 6

Statically Indeterminate Beams

Method of Superposition

Now that we know B

y

, the

reactions at the wall can be

determined from the equilibrium

equations.

The results are…

0=

x

A

PAy

16

11

=

PLM

PLPLM

LP

L

PM

A

A

A

16

3

16

5

16

8

16

5

2

=

−=

⎟

⎠

⎞

⎜

⎝

⎛

−=

August 29, 2013Page 7

As stated previously, the choice of the redundant reactions is

arbitrary (provided that the beam remains stable).

As another example, consider the same beam and loading, but

choose the moment at A, as the redundant reaction.

Statically Indeterminate Beams

Method of Superposition

August 29, 2013Page 8

Statically Indeterminate Beams

Method of Superposition

In this example the capacity of the

beam to resist M

A

is removed, hence

the support at A becomes a pin

support.

There is now a slope at A (θ°) caused

by load P.

In order to maintain compatibility the

redundant reaction, M

A, is then applied

at A where the displacements θ

A

and

θ'A

are equal and opposite (and

cancel).

August 29, 2013Page 9

Statically Indeterminate Beams

Method of Superposition

Compatibility equation.

AA

θ

θ

′

+

=

0

From App. C in Hibbeler.

EI

PL

A

16

2

=

θ

EI

LM

A

A

3

=

′

θ

Sub and solve.

EI

LM

EI

PL

A

316

0

2

+=

PLM

A

16

3

−=

August 29, 2013Page 10

Statically Indeterminate Beams

Method of Superposition

This is the same result (bending

moment at A) as was computed

previously.

Here the negative sign for M

A

means

that the moment acts in the opposite

sense from the direction shown in the

figure.

August 29, 2013Page 11

Statically Indeterminate Beams

Method of Superposition

Here we choose the reaction forces

at the roller supports B and C as

redundant.

When the redundant reactions are

removed, the statically determinate

beam deforms as shown here.

Each redundant force deforms the

beam as shown.

In this case the beam is indeterminate

to the second degree and therefore two

compatibility equations are needed for

the solution.

August 29, 2013Page 12

Statically Indeterminate Beams

Method of Superposition

By superposition, the compatibility

equations for the displacements at

B and C are.

BBB

vvv

′

′

+

′

+

=

0

When the displacements (in terms

of By

and Cy

have been determined,

they can be used with the

equilibrium equations to solve for

the remaining unknowns.

CCC

vvv

′

′

+

′

+

=

0

August 29, 2013Page 13

Statically Indeterminate Beams

Method of Superposition

Procedure for Analysis

The Elastic Curve.

1.Specify the unknown redundant forces or moments that must be

removed from the beam in order to make it statically determinate and

stable.

2.Using the principal of superposition, draw the statically indeterminate

beam and show it equal to a sequence of corresponding statically

determinate beams.

3.The first of these beams supports the same external loads as the

statically indeterminate beam. Each of the other beams that are

“added” to the first beam, show the beam loaded with separate

redundant forces or moments.

4.Sketch the deflection curve for each beam and indicate symbolically

the displacement or slope at the point of each redundant force or

moment.

August 29, 2013Page 14

Statically Indeterminate Beams

Method of Superposition

Procedure for Analysis

Compatibility Equations.

1. Write a compatibility equation for the displacement or slope at each

point where there is a redundant force or moment.

2.Determine all the displacements or slopes using an appropriate

method.

3.Substitute the results into the compatibility equations and solve for

the unknown redundant forces or moments.

4.If a numerical value for a redundant force or moment is positive, it

has the same sense of direction as originally assumed. A negative

numerical value indicates that the redundant force or moment acts

opposite to the assumed sense of direction.

August 29, 2013Page 15

Statically Indeterminate Beams

Method of Superposition

Procedure for Analysis

Equilibrium Equations.

Once the redundant forces and/or moments have been

determined, the remaining unknown reactions can be

found from the equations of equilibrium applied to the

loadings shown on the beam’s free-body diagram.

August 29, 2013Page 16

Example 12.22

Determine the reactions on the beam. Due to the loading and poor construction, the

roller support at B settles 12 mm. Take E = 200GPa and I = 80(10

6) mm4

=

+

August 29, 2013Page 17

Next Time

Columns and Critical Loads

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