Application of Second Order Differential Equations …

1 Chapter 4 Application of Second Order Differential Equations in Mechanical Engineering AnalysisTai-Ran Hsu, ProfessorDepartment of Mechanical and Aerospace EngineeringSan Jose State UniversitySan Jose, California, USAME 130 Applied Engineering AnalysisChapter Outlines Review solution method of Second Order , homogeneous ordinary Differential Equations applications in free vibration analysis- Simple mass-spring system- Damped mass-spring system Review solution method of Second Order , non-homogeneous ordinary Differential Equations - applications in forced vibration analysis- Resonant vibration analysis- Near resonant vibration analysis Modal analysis Part 1 Review Solution Method of Second Order , Homogeneous Ordinary Differential EquationsTypical form0)()()(22=++xbudxxduadxxud( )where a and b in Equation ( ) are constantsThe solution of Equation ( ) u(x) may be obtained by ASSUMING.

2 U(x) = emx( )in which m= constantto be determinedIf the assumed solution u(x) in Equation ( ) is valid solution, it must SATISFY theDE in Equation ( ).That is:()()()022=++mxmxmxebdxedadxedBecause: ()mxmxemdxed222=()mxmxmedxed=and(a)Upon substitution of the above into Equation (a) leading to:()()02=++mxmxmxebemaemBecause emxin the expression cannot be zero (why?), we thus have:m2 + am + b = 0( )Equation ( ) is a quadratic equation, and its solution for m are:m2+ am + b = 0 The quadratic equation:The TWO roots of the above quadratic equation have the forms:baamandbaam421242122221 = + =( )This leads to two possible solutions for the function u(x) in Equation ( ).

3 ()xmxmececxu2121+=( )where c1and c2are the TWOarbitrary constants to be determined by TWO specified conditions, and m1and m2are expressed in Equation ( )Because the constant coefficients a and b in Equation ( ) are fixed with the DE, the relative magnitudes of the a, bwill result in significant forms in the solution in Equation ( ) due to the square root parts in the expression of m1and m2in Equation ( ). Square root of a negative number will lead to a complex number in the solution of the DE, which requires a special way of expressing thus need to look into 3 possible casesinvolving relative magnitudes a and 4b > 0:In such case, we realize that both m1and m2 are real numbers.

4 The solution of the Equation ( ) is:( )Case 4b < 0:As described earlier, both these roots become complex numbers involving imaginary parts. The substitution of the m1and m2into Equation ( ) will lead to the following:( )in which . The complex form of the solution in Equation ( ) is not always easily comprehended and manipulative in engineering analyses, a more commonly used form involving trigonometric functions are used:( )where A and B are arbitrary expression in Equation ( ) may be derived from Equation ( ) using the Biot relationthat has the form.

5 += 2/422/41222)(xbaxbaaxececexu += 224224212)(abixabixaxececexu1 =i + = xabCosBxabSinAexuax222421421)( SiniCosei = Case = 0:Recall Equation ( ):baamandbaam421242122221 = + =The condition a2 4b = 0 will lead to a situation: m1= m2= a/2 Substituting the m1 and m2 into Equation ( ) will result in: xaxacexuoreccxu21221)()()( =+=with only ONEterm with a constant in the solution, which is not complete for a 2ndorder , we will have to find the missing termin the solution u(x).

6 Realizing the fact that the assumed solution u(x) = emxin Equation ( ) results in onemissing term, we need to find another assumed solution. Let us try the following:u2(x) = V(x) emx( )where V(x)is a function of x, and it needs to be determinedWe may follow the same procedure before in determining function V(x), that is theassumed Second solution of Equation ( ) must satisfy Equation ( )(b)0)()()(22=++xbudxxduadxxudThe DE:The assumed Second solution: u2(x) = V(x) emxWe must have:()[]()[]()[]022=++mxmxmxexVbdxexVda dxexVdOne would find.

7 ()[]()()dxxdVeexmVdxexVdmxmxmx+=and()[]( )()()()2222dxxVdedxxdVmedxxdVeexmVmdxexV dmxmxmxmxmx++ +=(c)After substituting the above expressions into Equation (c), we get:()0)()()2()(222=+++++xVbammdxxdVamdx xVd( )Since m2 + am + b = 0 in Equation ( ), and m = m1= m2= a/2 in Equation (b),0)(22=dxxVdboth 2ndand 3rdterm in Equation ( ) drop out. We thus only have the first term toconsider in the following special form of 2ndorder DE:( )The solution of DE in Equation ( ) is:V(x) = xThe solution V(x) = x leads to the Second solution of the DE0)()()(22=++xbudxxduadxxudin Case 3 with a2 4b = 0as:()22)(axmxmxxexeexVxu ===The complete solution of the DE in this case thus becomes:u(x) = u1(x) + u2(x)or()2212221)(axaxaxexccexcecxu +=+=( )Summary on Solutions of 2ndOrder HomogeneousDEsThe equation.

8 0)()()(22=++xbudxxduadxxud( )with TWOgiven conditionsCase 1: a2 4b > 0: += 2/422/41222)(xbaxbaaxececexu( )Case 2: a2- 4b < 0: + = xabCosBxabSinAexuax222421421)(( )Case 3: a2 - 4b = 0:()2212221)(axaxaxexccexcecxu +=+=( )The solutionswhere c1, c2, Aand Bare arbitrary constants to be determined by given conditionsA special caseExample the following Differential equation ( ):(a)Solution:We have a = 5 and b = 6, by comparing Equation (a) with the typical DE in Equation ( ).

9 This will lead to:a2 4b = 52-4x6 = 25 24 = 1 > 0 - a Case 1 situation with 0)(6)(5)(22=++xudxxdudxxudConsequently, we may use the standard solution in Equation ( ) to be: += 2/422/41222)(xbaxbaaxececexuor1142== ba()()xxxxxececececexu32212/22/12/5 +=+=where c1and c2are arbitrary constants to be determined by given conditionsExample the following equation with given conditions (p. 84):0)(9)(6)(22=++xudxxdudxxud(a)with given conditions:u(0) = 2 (b)and0)(0==xdxxdu(c)Solution:Again by comparing Equation (a) with the typical DE in Equation ( ), we have: a = 6 and b = examining a2 4b = 62 4x9 = 36 36 = 0, leading to special Case 3 in Equation ( ) forthe solution:()2212221)(axaxaxexccexcecxu +=+=or ()()xxexccexccxu3212621)( +=+=( )(d)Use Equation (b) for Equation (d) will yield c1= 2, leading to.

10 ()()xexcxu322 +=Differentiating Equation (e) with condition in Equation will lead to the following manipulation:(e)()() ( )[]06232023230= =+ == =cxcecedxxduxxxxSo, we solve for c2 = 6 Hence the complete solution of Equation (a) is:()xexxu3312)( += Application of 2nd Order Homogeneous DEs for Mechanical Vibration AnalysisPart 2 Mechanical vibration is a form of oscillatory motionof a solid or solid structure of a Sources of Mechanical Vibrations:(1) Time-varying Mechanical force or pressure.