Muller breslau
- 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...