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## CONJUGATE BEAM METHOD

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**CONJUGATE BEAM METHOD**By Murtaza zulfiqar**OBJECTIVES**At the end of this unit, students are supposed to; • Understand what is conjugate beam ? • Be able to understand what is conjugate beam method • Will know Advantages of this method over others methods • Be able to derive Conjugate beam method derivation/proof • Be analyse beams and frames using the conjugate beam method • examples**Conjugate beam method**The conjugate-beam method is an engineering method to derive the slope and displacement of a beam. The conjugate-beam method was developed by H. Müller-Breslau in 1865 • Many credit Heinrich Müller-Breslau (1851-1925) with the development of this method, while others, say the method was developed by Christian Otto Mohr (1835-1918).**Theorm of conjugate beam method**• 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 2 : The displacement of a point in the real beam is equal to the moment at the corresponding point in the conjugate beam**The basis for the method comes from the similarity of Eq. 1**and Eq 2 to Eq 3 and Eq 4. To show this similarity, these equations are shown below. Integrated, the equations look like this.**Procedure for analysis of beams through CBM**• Procedure for analysis • 1. Construct the conjugate beam with the M/EI loading. Remember when the M/EI diagram is positive the loading is upward and • when the M/EI diagram is negative the loading is downward. • 2. Use the equations of equilibrium to solve for the • reactions of the conjugate beam. • This may be difficult if the moment diagram is complex.**Procedure**• 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.**The following procedure provides a method that may be used**to determine the displacement and slope at a point on the elastic curve of a beam using the conjugate-beam method. • Conjugate beam • Draw the conjugate beam for the real beam. This beam has the same length as the real beam and has corresponding supports as listed above.**In general, if the real support allows a slope, the**conjugate support must develop shear; and if the real support allows a displacement, the conjugate support must develop a moment. • The conjugate beam is loaded with the real beam's M/EI diagram. This loading is assumed to be distributed over the conjugate beam and is directed upward when M/EI is positive and downward when M/EI is negative. In other words, the loading always acts away from the beam.[**Equilibrium[**• Using the equations of statics, determine the reactions at the conjugate beams supports. • Section the conjugate beam at the point where the slope θ and displacement Δ of the real beam are to be determined. At the section show the unknown shear V' and M' equal to θ and Δ, respectively, for the real beam. In particular, if these values are positive, and slope is counterclockwise and the displacement is upward**EXAMPLE**& SOLUTIONS**Example**Draw the shear and moment diagrams for the beam shown in Fig. The support at B settles 1.5 in. Take E = 29(103) ksi, I = 750 in4. 20 k 1.5 in A C B 12 ft 12 ft 24 ft Actual Beam**Solution**Principle of Superposition • The beam is first degree statically indeterminate. • The centre support B is chosen as redundant, so that the roller at B is removed. 20 k 1.5 in A C Actual Beam B 12 ft 12 ft 24 ft**20 k**1.5 in A C • Byis assumed to act downward on the beam. Actual Beam B = 20 k B A C Primary Structure ΔB + By B Redundant By applied A C Δ’BB=ByfBB**Compatibility Equation**• With reference to point B, using units of ft, we require • Use conjugate beam method to compute ΔB and fBB since the moment diagrams consists straight line segments. • For ΔB 20 k B A C 20 k A C 15 k 5 k 12 ft 36 ft**20 k**A C Compatibility Equation 15 k 5 k 12 ft 36 ft 16 ft 8 ft 24 ft conjugate beam**Compatibility Equation**16 ft 8 ft 24 ft MB’ VB’ 8 ft 16 ft**1 k**B A C Compatibility Equation • For fBB 1 k A C 0.5 k 0.5 k 24 ft 24 ft conjugate beam 24 ft 24 ft**conjugate beam**Compatibility Equation • For fBB 24 ft 24 ft mB’ vB’ 24 ft 8 ft**Compatibility Equation**• Substituting these results into eq. (1), we have • Expressing the units of E and I in terms of k and ft, we have**Equilibrium Equations**• The negative sign indicates that By acts upward on the beam. 20 k A C By=5.56 k Cy Ay 12 ft 12 ft 24 ft**Equilibrium Equations**20 k A C By=5.56 k Cy=2.22 k Ay=12.22 k 12 ft 12 ft 24 ft**20 k**• Using these results, shear and moment diagrams are A C By=5.56 k Cy=2.22 k Ay=12.22 k 12 ft 12 ft 24 ft V (k) 12.22 x (ft) -2.22 -7.78 -20**20 k**• Using these results, shear and moment diagrams are A C By=5.56 k Cy=2.22 k Ay=12.22 k 12 ft 12 ft 24 ft M (k.ft) 146.7 53.3 x (ft)**Advantages of this method over others methods**• Generally, the is a more direct and effective method than the consistency method in determining reactions and deflections of beams.**References:**• Rc hibler , 2009 edition**Be calm Guys , feel free to ask any questions ….**We are here ,to satisfy U ….. But one by one :P