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.
