Solid Mechanics Course No. ME213

Double-Integration Method/ Macaulay's method Introduction A technique used in structural analysis to determine the deflection of Euler-Bernoulli beams The first English language description of the method was by Macaulay. The actual approach appears to have been developed by Clebsch in 1862 The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to directly work on the equation of the elastic curve. ME213 Solid Mechanics

Double-Integration Method The deflection curve of the bent beam is In order to obtain y, above equation needs to be integrated twice. y Radius of curvature x An expression for the curvature at any point along the curve representing the deformed beam is readily available from differential calculus. The exact formula for the curvature is ME213 Solid Mechanics

Double-Integration Method Integrating once yields to slope dy/dx at any point in the beam. Integrating twice yields to deflection y for any value of x. The bending moment M must be expressed as a function of the coordinate x before the integration Differential equation is 2nd order, the solution must contain two constants of integration. They must be evaluated at known deflection and slope points (i.e. at a simple support deflection is zero, at a built in support both slope and deflection are zero) ME213 Solid Mechanics

Double-Integration Method Integrating once yields to slope dy/dx at any point in the beam. Integrating twice yields to deflection y for any value of x. The bending moment M must be expressed as a function of the coordinate x before the integration Differential equation is 2nd order, the solution must contain two constants of integration. They must be evaluated at known deflection and slope points (i.e. at a simple support deflection is zero, at a built in support both slope and deflection are zero) ME213 Solid Mechanics

Double-Integration Method Sign Convention Positive Bending Negative Bending Assumptions and Limitations Deflections caused by shearing action negligibly small compared to bending Deflections are small compared to the cross-sectional dimensions of the beam All portions of the beam are acting in the elastic range Beam is straight prior to the application of loads ME213 Solid Mechanics

Double-Integration Method M = f(x), successive integration of basic equation will yield the beam’s slope −𝜃≈ tan 𝜃 = 𝑑𝑦 𝑑𝑥 = 𝑀 𝐸𝐼 𝑑𝑥 Equation of elastic curve −𝑦=𝑓(𝑥)= 𝑀 𝐸𝐼 𝑑𝑥 The internal moment in regions AB, BC & CD must be written in terms of x1, x2 and x3 ME213 Solid Mechanics

Double-Integration Method Once these functions are integrated & the constants determined, the functions will give the slope & deflection for each region of the beam. It is important to use the proper sign for M as established by the sign convention used in derivation. +ve v is upward, hence, the +ve slope angle,  will be measured counterclockwise from the x-axis ME213 Solid Mechanics

Double-Integration Method Once these functions are integrated & the constants determined, the functions will give the slope & deflection for each region of the beam. It is important to use the proper sign for M as established by the sign convention used in derivation. +ve v is upward, hence, the +ve slope angle,  will be measured counterclockwise from the x-axis ME213 Solid Mechanics

Double-Integration Method The constants of integration are determined by evaluating the functions for slope or displacement at a particular point on the beam where the value of the function is known. These values are called boundary conditions. Here x1 and x2 coordinates are valid within the regions AB & BC. ME213 Solid Mechanics

Double-Integration Method Once the functions for the slope & deflections are obtained, they must give the same values for slope & deflection at point B This is so as for the beam to be physically continuous ME213 Solid Mechanics

Double-Integration Method ME213 Solid Mechanics

Double-Integration Method ME213 Solid Mechanics

Double-Integration Method Example The cantilevered beam is subjected to a couple moment Mo at its end. Determine the equation of the elastic curve. EI is constant. ME213 Solid Mechanics

Double-Integration Method Solution By inspection, the internal moment can be represented throughout the beam using a single x coordinate. From the free-body diagram, with M acting in +ve direction, we have: Integrating twice yields: ME213 Solid Mechanics

Double-Integration Method Solution Using boundary conditions, dy/dx = 0 at x = 0 & y = 0 at x = 0 then C1 = C2 =0. Substituting these values into earlier equations, we get: Max slope & displacement occur at A (x = L) for which The +ve result for A indicates counterclockwise rotation & the +ve result for vA indicates that it is upwards. ME213 Solid Mechanics

Double-Integration Method Solution Consider the beam to made of steel (E = 200GPa)having a length of 3.6m and is acted by a couple moment of 20kNm If this beam were designed w/o a F.O.S by assuming the allowable normal stress = yield stress = 250kNm/2. ME213 Solid Mechanics

Conjugate-Beam method Conjugate beam is defined as the imaginary beam with the same dimensions (length) as that of the original beam but load at any point on the conjugate beam is equal to the bending moment at that point divided by E Here the shear V compares with the slope θ, the moment M compares with the displacement v, and the external load w compares with the M/EI diagram. Below is a shear, moment, and deflection diagram. A M/EI diagram is a moment diagram divided by the beam's Young's modulus and moment of inertia.

Conjugate-Beam method Integrating we have,

Conjugate-Beam method Here the shear V compares with the slope θ , the moment M compares with the displacement y & the external load w compares with the M/EI diagram To make use of this comparison we will now consider a beam having the same length as the real beam but referred to as the “conjugate beam”

Conjugate-Beam method The conjugate beam is loaded with the M/EI diagram derived from the load w on the real beam From the above comparisons, we can state 2 theorems related to the conjugate beam Theorem 1 The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam Theorem 2 The disp. of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam When drawing the conjugate beam, it is important that the shear & moment developed at the supports of the conjugate beam account for the corresponding slope & displacement of the real beam at its support

Conjugate-Beam method Consequently from Theorem 1 & 2, the conjugate beam must be supported by a pin or roller since this support has zero moment but has a shear or end reaction When the real beam is fixed supported, both beam has a free end since at this end there is zero shear & moment

Conjugate-Beam method

Conjugate-Beam method

Conjugate-Beam method Example Determine the max deflection of the steel beam. The reactions have been computed. Take E = 200GPa, I = 60x106 mm4

Conjugate-Beam method The conjugate beam loaded with the M/EI diagram is shown. Since M/EI diagram is +ve, the distributed load acts upward. The external reactions on the conjugate beam are determined first and are indicated on the free-body diagram. Max deflection of the real beam occurs at the point where the slope of the beam is zero. Assuming this point acts within the region 0  x  9m from A’ we can isolate the section.

Conjugate-Beam method Solution Note that the peak of the distributed loading was determined from proportional triangles, Chapter 8: Deflections Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Conjugate-Beam method Using this value for x, the max deflection in the real beam corresponds to the moment M’. Hence, The –ve sign indicates the deflection is downward .