1

Mohr´s method – for determining a deflection or an angle of rotation in given place

Analogy of

• differential condition: M´´= -q

• Differential equation of elastic curve: EIw´´ = -M

Procedure:1. Moment diagram of the given real beam

2. Fictive (dual) beam to given beam

3. Fictive beam loaded by the moment area of the real beam

4. Deflection:

5. Angle of rotation:

EI

Mw

_

=

EI

Vw

−

== ϕ´−−

VaM- Bending moment and shear force of

the fictive beam from the load of themoment diagram of the real beam

Fictive beam: has to fulfil the same

deformation conditions against M a V as the real beam fullfills against deflection

and angle of rotation.

Real beam Fictive beam

advantage – M function doesn ´t has to be continuousdisadvantage – complicated for continuous load, suitable for force and moment load in points

Example 1Determine by Mohr´s method:

- deflection w in the middle of the

beam;

- angle of rotation ϕ in the leftsupport.

w=0 w=0

a b

c

F

l/2 l/2

Reactions:2

ba

FRR ==

4c

FlM =

Ra Rb

4222ac

FllFlRM =⋅=⋅=

Ra Rb

~~Ra

~

4Fl

]kNm[162

1

24

~~ 2

2

ba

FllFlRR =⋅⋅==

]kNm[4862

~

6

1~

2

~~ 3

3

11ac

FlllQQ

lRM =

−=−⋅=

1

~Q

l/6

]kNm[~q

]kNm[M

M=0 M=0

16242

1~~2

21

FllFlQQ =⋅⋅==

a

c

ba

c

Fictive beam

EI

V

EI

Mw

~~

== ϕMohr´s method:

EI

V

EI

Mw

~~

== ϕ

Mohr´s method:

Ra Rb

~~Ra

~

4Fl1

~Q

l/6

EI

Fl

EI

V2

a16

1~

==ϕ

0c

=ϕ

EI

Fl

EI

V2

b16

1~

−==ϕ

EI

Fl

EI

Mw

3

c48

1~

==

0ba

== ww

Ra

~

– Rb

~

]kNm[~ 2V

]kNm[~ 3

M

+

–

+

ba c

aϕ

bϕ

cw

Elastic curve

Ra

~

EI

Fl

EI

V2

a16

1~

==ϕ

0c

=ϕ

EI

Fl

EI

V2

b16

1~

−==ϕ

EI

Fl

EI

V2

a16

1~

==ϕ

0c

=ϕ

0ba

== ww

EI

Fl

EI

V2

b16

1~

−==ϕ

EI

Fl

EI

V2

a16

1~

==ϕ

0c

=ϕ

EI

Fl

EI

Mw

3

c48

1~

==

0ba

== ww

EI

Fl

EI

V2

b16

1~

−==ϕ

EI

Fl

EI

V2

a16

1~

==ϕ

0c

=ϕ2°

3°

Make the assessment after serviceable limit state: ,

F=10kN

l=1m

E=210GPa

Square section a=100mm

wall= l/1000

....1119,048

3

satisfmmmmEI

Flwc p==

4610338 m.,I y

−−−−====

2

ExamplesEI

V

EI

Mw b

bb

b

~~

== ϕBy Mohr´s method wb, φb :

3