CIVL 311 - Conjugate eam 1/5 Conjugate beam method The development of the conjugate beam method has been atributed to several strucutral engineers. any credit Heinrich üller-reslau (1851-195) with the develoment of this method, while others, say the method was developed by Christian Otto ohr (1835-1918). The method is based on the similarity between the relationships for loading and shear, and shear and moment. dv wx ( ) dx d V dx V w( x) dx w( x) dx dx d dx w Heinrich Franz ernhard üller was born in Wroclaw (reslau) on 13 ay 1851. In 1875 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 üller-reslau. He founded the so-called "erlin School" of structural theory. The previous expressions relate the internal shear and moment to the applied load. The slope and deflection of the elastic curve are related to the internal moment by the following expressions d dy dx dx dx y dx dx Christian Otto ohr was an enthusiast for graphical tools and developed the method, for visually representing stress in three dimensions known as ohr's Circle. He also developed methods for truss displacements and for analyzing statically indeterminate structures. He founded the so-called Dresden School" of applied mechanics and has differences of option with üller-reslau their careers. Let s compare expressions for shear, V, and the slope, dv w dx d dx What do you see? If you replace w with the term / the expressions for shear force and slope are identical
CIVL 311 - Conjugate eam /5 Let s compare expressions for bending moment,, and the displacement, y d dy w dx dx What do you see? Just as before, if you replace w with the term / the expressions for bending moment and displacement are identical Therefore, the two theorems related to the conjugate beam method are: Theorem 1: The slope at a point in the real beam is equal to the shear at the corresponding point in the conjugate beam. Theorem : The displacement of a point in the real beam is equal to the moment at the corresponding point in the conjugate beam. We will use this relationship to our advantage by constructing a beam with the same length as the real beam referred to as the conjugate beam. The conjugate beam is loaded with the / diagram, simulating the external load w. Conjugate eam Supports When the conjugate beam is drawn, it is important that the shear and moment developed in the conjugate beam correspond to the slope and displacement conditions in the real beam. Real Support Conjugate Support V Pin or roller Pin or roller = 0 0 = 0 V 0 w Conjugate eam Supports w = w(x) x Real beam with applied loading. Determine the bending moment (draw the bending moment diagram) Real Support Conjugate Support V Conjugate beam where the applied loading is bending moment from the real beam Free end 0 0 Fixed end 0 V 0 x Note the sign of loading w and the / on the conjugate beam. Fixed end =0 =0 Free end = 0 V = 0
CIVL 311 - Conjugate eam 3/5 Conjugate eam Supports Real Support Conjugate Support R V L V R L Interior support =0 L = R 0 Hinge = 0 V L = V R 0 L R V L V R 0 Hinge L and R may have different values Interior roller 0 V L and V R may have different values s a rule, statically determinant real beams have statically determinant conjugate beams and statically indeterminate beams become unstable conjugate beams. However, the / loading may provide the necessary equilibrium to hold the conjugate beam stable. Procedure for analysis 1. Construct the conjugate beam with the / loading. Remember when the / diagram is positive the loading is upward and when the / diagram is negative the loading is downward.. Use the equations of equilibrium to solve for the reactions of the conjugate beam. This may be difficult if the moment diagram is complex. 3. Solve for the shear and moment at the point or points where the slope and displacement are desired. If the values are positive, the slope is counterclockwise and the displacement is upward.
CIVL 311 - Conjugate eam 4/5 following beam. ssume that E = 30,000 ksi and I = 300 in 4. 10 k C Construct the conjugate beam and apply the / diagram as loading 10 Remember positive (+) bending moment is a positive (+) loading on the conjugate beam. following beam. ssume that E = 30,000 ksi and I = 300 in 4. y y 50k ft 150kft 50kft Fy 0 (10 ft) VC V C 0 following beam. ssume that E = 30,000 ksi and I = 300 in 4. following beam. ssume that E = 30,000 ksi and I = 300 in 4. F CE 150kft FCE (0 ft) 10 y y 150kft 0 (0 ft)(10 ft) y(0 ft) 150 0 kft Fy (0 ft) y y 50kft y 50kft y y y 50k ft Therefore, the displacement of the beam at point C is equal to the moment at point C on the conjugate beam and the slope is equal to the shear in the conjugate beam. In this problem. the displacement at point C is -0.3 in and the slope is zero. following beam. ssume that E = 30,000 ksi and I = 300 in 4. Example: Determine the slope and the displacement at point for the following beam. ssume that E = 9,000 ksi and I = 800 in 4. 5 k y 150kft 10ft 50kft C 0 C (10 ft) (10 ft) 3 3 1,666.6kft C y 50k ft 3 3 1,666.6kft 1,78in 0.3in 4 3 (30,000 ksi )(300 in ) ft 15 ft 15 ft
CIVL 311 - Conjugate eam 5/5 Example: Determine the maximum displacement at the mid-span of the following beam. ssume that E = 30,000 ksi and I = 800 in 4. End of Defections Part ny questions? 5 k 9 ft 3 ft Example: Determine the slope at point and the displacement at point E for the following beam. ssume that E = 9,000 ksi and I = I DE = 400 in 4, and I D = 800 in 4. 8 k 10 ft C D 10 ft 8 k E Example: Determine the slope at and the displacement at mid-span. w L re there any disadvantages to the conjugate beam method for uniform or high-order loading functions?