IL 32 Influence Lines - Muller-reslau Principle /5 In 886, Heinrich Müller-reslau develop a method for rapidly constructing the shape of an influence line. Heinrich Franz ernhard Müller was born in Wroclaw (reslau) on 3 May 85. In 875 he opened a civil engineer's office in erlin. round this time he decided to add the name of his hometown to his surname, becoming known as Müller-reslau. In 886, Heinrich Müller-reslau develop a method for rapidly constructing the shape of an influence line. In 883 Müller-reslau became a lecturer and in 885 a professor in civil engineering at the Technische Hochschule in Hanover. The Muller-reslau principle states: The influence line for a function (reaction, shear, moment) is to the same scale as the deflected shape of the beam when the beam is acted on by the function. To draw the deflected shape properly, the ability of the beam to resist the applied function must be removed. For eample, consider the following simply supported beam. for the vertical reaction at. Remove the ability to resist movement in the vertical direction at by using a guided roller onsider the following simply supported beam. y y for the shear at the mid-point (point ).
IL 32 Influence Lines - Muller-reslau Principle 2/5 Remove the ability to resist shear at point onsider the following simply supported beam. shear is equal to for the moment at the mid-point (point ). Remove the ability to resist moment at by using a hinge reaction at point M reaction at point reaction at point y y
IL 32 Influence Lines - Muller-reslau Principle 3/5 reaction at point reaction at point y y reaction at point shear at point y y shear at point moment at point shear is equal to
IL 32 Influence Lines - Muller-reslau Principle 4/5 moment at point moment at point M moment at point shear at point M shear at point 0 ft 0 ft 5 ft 5 ft shear is equal to
IL 32 Influence Lines - Muller-reslau Principle 5/5 0 ft 0 ft 5 ft 5 ft y 0 ft 0 ft 5 ft 5 ft y 35 ft 5 ft 0 ft 0 ft 5 ft 5 ft shear is equal to shear at point is equal to The influence lines can be determined by similar triangles the values of 0ft -0.5 0.5 0.5(5 ft ) 0.75 0ft 5ft 0ft 0.5 0 ft 5 ft 5 ft -0.75 etermine the maimum positive moment that can be developed at point in the beam shown below due to a concentrated live load of 4 k, a uniform live load of 300 lb/ft, and a dead load of 200 lb/ft. nd of Influence Lines Part 2 ny questions? 5 ft 5 ft 5 ft 0 ft