Teaching Deflections of Beams: Advantages of Method of Model Formulas versus Those of Conjugate Beam Method

Transcription

1 Teaching Deflections of eams: dvantages of Method of Model Formulas versus Those of Conjugate eam Method Ing-Chang Jong Professor of Mechanical Engineering Universit of rkansas Proceedings of the 2012 SEE nnual Conference & Exposition Mechanics Division 1

2 Objectives: Share the idea of teaching to methods: method of model formulas (MoMF), established in 2009, (I. C. Jong, n lternative pproach to Finding eam Ractions and Deflections: MoMF, IJEE, pp , Univ. of rkansas) and conjugate beam method (CM), established in 1921, (H. M. Westergaard, Deflection of eams b the Conjugate eam Method, JWSE, pp.69-96, Univ. of Illinois). Provide comparisons beteen the ne MoMF and the traditional CM via head-to-head contrasting solutions of same problems. Enrich students stud and sets of skills in analzing beam reactions and deflections. Mechanics Division 2

3 n Effective pproach to Teaching the MoMF: First, teach the traditional method of integration (MoI), including the use of singularit functions. Go over briefl the derivation of the four model formulas in terms of singularit functions. Demonstrate solutions of several beam problems b the MoMF vs. the CM and point out the advantages: Needs no integration. Handles ell concentrated & distributed loads. Uses just a page of excerpt of 4 model formulas. Mechanics Division

4 Prerequisite to Using MoMF: Singularit Functions n n < x a > = ( xa) if xa 0 and n> 0 n < x a> = 1 if xa 0 and n= 0 n < x a> = 0 if x a< 0 or n< 0 x < x a> n dx = < x a> n+ 1 n 1 if 0 x n n 1 x a dx x a + n n + 1 < > = < > if > 0 d n n1 < x a> = n< x a> dx if n> 0 d n n1 < x a> = < x a> dx if n 0 Mechanics Division 4

5 Excerpt from the Method of Model Formulas: Model loads on beam ab & its positive deflections Mechanics Division 5

6 Model Formulas: Eqs. (1) and (2) Va 2 M a P 2 K 1 0 = θ a + x + x < x xp>+ < x x x x K> < > 2EI EI 2EI EI 6EI < x x > + < x u > + < x u > 24 EI x 24 EI x m0 2 m0 2 + < xxm> < xu m> 2EI 2EI ( u ) 6EI ( u ) (1) V M P K = + x+ x + x < x x + < x x > x x 6EI 2EI 6EI 2EI 24EI < x x > + < x u > + < x u > 120 EI ( ) m0 m0 + < x xm > < x um> 6EI 6EI a a a θa P> < K > ( u x) 24EI 120EI u x (2) Mechanics Division 6

7 Model Formulas: Eqs. () and (4) θ b 2 Va L Ma L P 2 0 ( L xp) K ( L xk ) ( L x) = θa + 2EI + EI 2EI + EI 6 EI ( L x ) + ( L u ) + 24EI 24 ( Lu ) m0 2 m0 2 + ( Lxm) ( Lum) 2EI 2EI ( ux) 6EI EI ( ux) () 2 VL a MaL P K b = a + θ al+ + ( L xp) + ( LxK) ( Lx) 6EI 2EI 6EI 2EI 24EI ( L x) + ( L u ) + ( Lu ) 120 EI( u x) 24 EI 120EI( u x) m0 m0 + ( Lxm) ( Lum) 6EI 6EI (4) Mechanics Division 7

8 Guiding Rules in the CM Snthesized from: Journal of the Western societ of Engineers, Vol. XXVI, No. 11, pp.69-96, 1921 Rule 1: Conjugate beam and given beam are of the same length. Rule 2: Load on conjugate beam is M / EI of the given beam. Existing support condition in Corresponding support condition in the given beam: the conjugate beam: Rule : Fixed end Free end Rule 4: Free end Fixed end Rule 5: Simple support at the end Simple support at the end Rule 6: Simple support not at the end Unsupported hinge Rule 7: Unsupported hinge Simple support Rule 8: Conjugate beam is in static equilibrium. Rule 9: Slope of given beam is equal to the shear force in conjugate beam. Rule 10: Deflection of given beam is equal to the moment in conjugate beam. Mechanics Division 8

9 Example 1. Determine for the beam ith EI (a) slopes θ and θd at and D, (b) deflection at. Solution b MoMF: Eq. () : θ Eq. ( 4) : 0 Eq. (2) D / 2 ( P ) L P L PL L = θ + L + L 2EI 2EI EI 2 P L PL L ( 5 / ) P L 2 = θ L + L L 6EI 6EI + 2EI 2 2 We obtain: θ 14 and 17 = PL θ PL D = 81EI 162EI : x= L/ L 5 P/ L 2PL = = θ + = 6EI 486EI We obtain: = 2PL 46 8 EI Mechanics Division 9

10 Example 1. (Continued) Solution b CM: Conjugate beam FD of conjugate beam Equilibrium of the conjugate beam ields: ppling the guiding rules in the CM, e obtain: θ = 14PL 81 2 EI 2 θ = D PL EI = 2PL 486EI Mechanics Division 10

11 Example 2. For the beam ith EI, determine the reactions and slopes (a) and θ at, (b) and θ at. Solution b MoMF: (assuming = and = ) 2 (2 L) 2 /2 ( /2) 4 Eq. (): 0 = θ + (2 LL) (2 L) (2 L) 2 EI 2 EI 6EI 24 EIL ( /2) 4 + (2 L L) + (2 LL) 6 EI 24 EIL (2 L) /2 ( /2) Eq. (4): 0 = θ (2 L) + (2 LL) (2 L) (2 L) 6EI 6EI 24EI 120 EIL 4 ( /2) 5 + (2 L L) + (2 LL) 24EI 10 2 EIL /2 4 ( /2) 5 Eq. (2): 0 = θ L + L L L 6 EI 24 EI 120 EL I 2 /2 ( /2) 4 Eq. (1): θ = x= L= θ + L L L 2 EI 6 EI 24 EI L 4 5 Mechanics Division 11

12 The preceding four equations ield the values: = = 9L 140 1L 56 L θ = 140EI θ = 2L 1680EI We report that the MoMF ields the folloing solutions: = 9L 140 θ = L 140EI = 1L 56 θ = 2L 1680EI Mechanics Division 12

13 Example 2. (Continued) Solution b the CM: Conjugate beam FD of conjugate beam Mechanics Division 1

14 Equilibrium of the conjugate beam ields: = 9L 140 = 1L 56 c = L 140EI We report = 9L 140 = 1L 56 ppling the guiding rules in the CM, e have θ = V = = c c L 140EI θ = V c = 2L 1680EI θ = L 140EI θ = 2L 1680EI Mechanics Division 14

15 Conclusions The method of model formulas requires the use of just a printed page of excerpt of model loads and four model formulas. The four model formulas serve to provide material equations, besides the equations of equilibrium for a beam. The formulas in the MoMF can account for most of the loads encountered in undergraduate Mechanics of Materials. The conjugate beam method require the use of the first 7 guiding rules to construct the conjugate beam and dra the free-bod diagram of the conjugate beam. Then, appl the last guiding rules to compute and report the requested solutions. Generall, the MoMF is a more direct and effective method than the CM in determining reactions and deflections of beams. Mechanics Division 15

16 Questions Mechanics Division 16