This presentation summarizes the Muller-Breslau principle for constructing influence lines. The principle states that the deflected shape of a structure under a unit internal load or reaction corresponds to the influence line for that load or reaction. The presentation provides the history of Muller-Breslau, explains the principle using virtual work, and outlines the general procedure for constructing influence lines using conjugate beam analysis and deflected shapes. Examples of influence lines for simply supported and continuous beams are presented.

1. PRESENTATION ON:
MULLER-BRESLAU PRINCIPLE.

2.  SHAIKH MOHD. AMIR N. 150860106062
 PRAJAPATI RIYA P. 150860106054
 BHANDARI NEENAD H. 150860106006
 RATHOD NEHA M. 150860106055

3.  HISTORY
 MULLER-BRESLAU PRINCIPLE
 EXPLANATION
 GENERAL PROCEDURE
 DEFLECTED SHAPES

4.  Heinrich Franz Bernhard Müller was born in
Wroclaw (Breslau) on 13 May1851.
 In 1875 he opened a civil engineer‘s office in
Berlin. Around this time he decided to add the
name of his hometown to his surname, becoming
known as Muller-Breslau.
 In 1883 Muller-Breslau became a lecturer and in
1885 a professor in civil engineering at the
Technische Hochschule in Hanover.
 In 1886, Heinrich Müller-Breslau develop a
method for rapidly constructing the shape of an
influence line.

5.  “IF AN INTERNAL STRESS COMPONENT OR A
RECTION COMPONENT IS CONSIDERED TO
ACT THROUGH SOME SMALL DISTANCE AND
THERE BY TO DEFLECT OR DISPLACE A
STRUCTURE, THE CURVE OF THE DEFLECTED
OR DISPLACED STRUCTURE WILL BE, TO SOME
SCALE, THE INFLUENCE LINE FOR THE STRESS
OR REACTION COMPONENT”.

6.  The Muller-Breslau principle uses Betti's law of
virtual work to construct influence lines. To
illustrate the method let us consider a structure
AB (Figure a).
 Let us apply a unit downward force at a distance
x from A , at point C .
 Let us assume that it creates the vertical
reactions RA and RB at supports A and B ,
respectively (Figure b). Let us call this condition
“System 1.”
 In “System 2” (figure c), we have the same
structure with a unit deflection applied in the
direction of RA . Here Δ is the deflection at point
C .

7. Figure,(a) GIVEN SYSTEM AB,
(b) SYSTEM1,STRUCTURE UNDER A UNIT LOAD
(c) SYSTEM2,STRUCTURE WITH A UNIT DEFLECTION CORRESPONDING TO RA

8.  According to Betti's law, the virtual work done by
the forces in System 1 going through the
Corresponding displacements in System 2 should
be equal to the virtual work done by the forces in
System 2 going through the corresponding
displacements in System 1. For these two
systems, we can write:
(RA)(1) + (1)(- Δ) =0
 The right side of this equation is zero, because in
System 2 forces can exist only at the supports,
corresponding to which the displacements in
System 1 (at supports A and B ) are zero. The
negative sign before Δ accounts for the fact that
it acts against the unit load in System 1.
 Solving this equation we get:
RA= Δ.

9.  In other words, the reaction at support A due
to a unit load at point C is equal to the
displacement at point C when the structure is
subjected to a unit displacement
corresponding to the positive direction of
support reaction at A .
 Similarly, we can place the unit load at any
other point and obtain the support reaction
due to that from System 2.
 Thus the deflection pattern in System 2
represents the influence line for RA .

10.  STEP-1: TO DRAW ILD FOR ANY SUPPORT
REMOVE THAT SUPPORT.
 STEP-2: APPLY UNIT LOAD AT THAT
SUPPORT.
 STEP-3: DRAW BENDING MOMENT DIAGRAM
FOR THAT SUPPORT.
 STEP-4: CONSTRUCT CONJUGATE BEAM.
 STEP-5: FIND DEFLECTION AT SOME
SPECIFIED INTERVALS ( WE KNOW THAT FOR A
CONJUGATE BEAM, DEFLECTION AT ANY
POINT = BM AT THAT POINT).

11.  STEP-6: DIVIDE EACH DEFLECTION BY
DEFLECTION CORRESPONDING TO THE POINT OF
APPLICATION OF UNIT LOAD.
 STEP-7: WE OBTAIN THE ORDINATES FOR THE
INFLUENCE FOR THAT PARTICULAR SUPPORT.
 STEP-8: FOR OTHER SUPPORT REPEAT THE SAME
PROCEDURE.
 STEP-9: FOR ILD OF BENDING MOMENT,
CONSTRUCT A STATIC EQUATIONS FROM THE
BEAM AND SUBSTITUTE VALUES AT DIFFERENT
INTERVAL AND YOU WILL GET ORDINATES OF ILD
FOR BMD.
 STEP-10: FOR SHEAR FORCE DIAGRAM
CONSTRUCT STATIC EQUATIONS AND SOLVE.

12.  ILD FOR SIMPLY
SUPPORTED BEAM:

13.  ILD FOR TWO SPAN
CONTINOUS BEAM:

14.  ILD FOR
THREE SPAN
CONTINOUS BEAM:

15. THANK YOU...