the engineer s notebook 2013

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1 the engineer s notebook 013 Consider a straight beam that is comprised of a material that is homogenous, linear- elastic, and isotropic. The beam is subjected to a uniform loading across its entire length (e.g. its own weight), has a constant cross section, and is simply supported at two points spaced symmetrically about its length. The spacing of the two support points can be chosen such that when the beam is deflected, the slope at each end of the beam is zero and the end faces are both vertical. The spacing of these two support points, known as the Airy points, can be found using the relation s = L 3, (1) where s is the spacing of the supports and L is the overall length of the beam. This result has been derived using Euler- Bernoulli beam bending theory, and thus is only valid for cases that adhere to the requisite assumptions. Airy points are named after Sir George Airy, who originally investigated the requirements for supporting precision length standards. These points are commonly used in precision metrology when supporting precision line and end standards. In addition to being applicable in precision metrology, the concept of Airy points may be of interest to the engineer who is developing a design where it is desirable to minimize angularity of the end faces of a beam or structure. In cases of great precision, the engineer will need to be sure to closely respect all assumptions used in derivation of the Airy points when they formulate their design. In non- precision designs, deviation from these assumptions may be acceptable if the behavior of the structure is within required limits. In both design cases, structuring supports to occur at the Airy points has minimal impact to product cost or complexity, and so it is worthwhile for the design engineer to consider the relevance of this concept when developing designs. Derivation of the result presented in equation 1 is provided in the remaining portion of this technical note. 1

2 the engineer s notebook 013 Derivation i. Nomenclature: Figure 1 illustrates the geometric nomenclature that will be used to derive the Airy point spacing. A coordinate system is established with the origin at the intersection of the left edge and the neutral axis of the beam. The x coordinate represents horizontal position, and the v coordinate represents deflection. The overall length of the beam is denoted as L, and the support points are positioned symmetrically at a spacing denoted as s. The modulus of elasticity of the beam and the area moment of inertia of the cross- section are assumed constant and denoted as E and I, respectively. Figure 1. Geometric nomenclature for derivation ii. Reaction Force Solution: A free body diagram is useful to help determine reaction forces at the beam support points. Figure displays two forces, R 1 and R, which have replaced the two supports shown in Figure 1. In addition to reaction forces, the beam weight is shown as a distributed loading of magnitude w (Units of weight/length). Figure. Free body diagram Figure 3 further simplifies the free body diagram by replacing the distributed loading with a point loading equal to the beam weight, wl, at the center of gravity of the beam. Figure 3. Free body diagram further simplified

3 the engineer s notebook 013 With the free body diagram complete, Newton s second law can be used to investigate forces on the beam. Since the beam is static, the sum of all forces must be equal to zero; that is, = 0 F () Equation is expanded based on the free body diagram of Figure 3 resulting in: R 1 + R wl = 0 (3) Equation 3 is then solved for one of the unknown reaction forces, R 1, so that R = wl () 1 R Again, the second law can be used, this time considering moments applied to the beam. Since the beam is static, the sum of all moments must be equal to zero; that is, = 0 M (5) Equation 5 is expanded based on the free body diagram of Figure 3 and the geometric nomenclature in Figure 1 resulting in: wl L R & & 1 ( R ( = 0 ' ' (6) Equation 6 is simplified resulting in: wl R 1 L + R 1 s R L R s = 0 (7) The result in equation is now substituted into equation 7 to eliminate one of the unknown reaction forces. The remaining equation is wl ( wl R )L + ( wl R )s R L R s = 0 (8) Equation 8 is simplified to obtain the solution for one of the reaction forces: R = wl (9) 3

4 the engineer s notebook 013 Using the result for R provided in equation 9, and substituting it back into equation, a solution can be determined for the other unknown reaction force, R 1. This process is illustrated below. R 1 = wl R = wl wl = wl (10) R 1 = R = wl (11) Since the reaction forces are equal in magnitude and direction, the subscripts previously used with the reaction forces will be eliminated. R 1 and R will be notated as R in the remainder of the analyses that follow. iii. Beam Slope Solution: The information presented in sections i and ii will be used in conjunction with the beam bending moment equation, M ( x) = EI d v dx (1) to derive the slope of the beam as a function of position along its length. Because the beam is supported at two intermediate points, the bending moments across the length of the beam cannot be represented by a single continuous function. The discontinuities in bending moment that occur at the support points make it necessary to formulate the bending moment equation for each of the three sections of the beam. Note that the same coordinate system will be utilized for each section. The moment in section 1 of the beam is formulated and substituted into equation 1, resulting in: Equation 13 can be further simplified to:! wx x & = EI d v dx (13) wx = EI d v dx (1) Equation 1 is then integrated to obtain the slope characteristic of the first section of the beam. wx dx = EI d v dx dx (15)

5 the engineer s notebook 013 The resulting slope equation for the first section of the beam is: 6 + C 1 dx (16) The moment in section of the beam is formulated and substituted into equation 1, resulting in:! wx x (! & R x + * &- = EI d v ), dx (17) Equation 17 can be further simplified to: wx Rx + R ' = EI d v & dx (18) Equation 18 is then integrated to obtain the slope characteristic of the second section of the beam. ( wx Rx + R + * '-dx = EI d v dx ) &, dx (19) Completing integration provides: 6 Rx + R 'x + C & dx (0) Using equation 11, a substitution can be made in equation 0 for R, producing: 6 wlx + wl 'x + C & dx (1) The resulting slope equation for the second section of the beam is: 6 wlx + wl x wlsx + C dx () The moment in section 3 of the beam is formulated and substituted into equation 1, resulting in:! wx x (! & R x + (! * &- R x + * &- = EI d v ), ), dx (3) 5

6 the engineer s notebook 013 Equation 3 can be further simplified to: wx Rx + R ' Rx + R ' = EI d v & & dx () Equation is then integrated to obtain the slope characteristic of the third section of the beam. ( wx Rx + R ' Rx + R + * '-dx = EI d v dx ) & &, dx (5) Completing Integration results in: 6 Rx + R 'x Rx & + R 'x + C 3 & dx (6) Using equation 11, a substitution can be made in equation 6 for R, producing: 6 wlx + wl 'x wlx & + wl 'x + C 3 & dx (7) Further simplification leads to the resulting slope equation for the third section of the beam: 6 wlx + wl x + C 3 dx (8) The general slope equations are now defined. However, it is still necessary to solve for the unknown integration constants. This will be accomplished by using two known conditions. Since the beam is symmetrically supported the deformed shape will be symmetric about the center. This leads to the first condition, which is that the slope at the midpoint of the beam must be zero. Substituting this condition into equation provides the following result: w 6 3! L & wl! L & + wl! L & wls! L &+ C = 0 (9) After simplification of equation 9, a final solution is determined for the integration constant C. C = wl3 1 + wl s 8 (30) 6

7 the engineer s notebook 013 Since the beam has a smooth shape without abrupt change, the slope must be continuous over the length of the beam. This leads to the second known condition, which is that the slope calculated at the support points must be equivalent as determined by both equations that apply for that given support point. Since E and I are equivalent for all sections of the beam, they can be lumped with the slope to simplify reduction of the equations. Using equations 16 and this condition is setup in equation C 1 = wx3 6 wlx + wl x wlsx + C (31) Prior to substituting the coordinate for the first support point in for x, some simplifications are made to make subsequent operations easier. C 1 = wlx Now the coordinate for the first support point is substituted into equation 3. C 1 = wl ' + wl & + wl x wlsx + C (3) ' wls & '+ C & (33) Simplifying equation 33, the final result for integration constant C 1 is: C 1 = wl3 8 + wls 16 (3) The same condition is applied for the second support point, using equations and 8. 6 wlx + wl x wlsx + C = wx3 6 wlx + wl x + C 3 (35) Prior to substituting the coordinate for the second support point in for x, some simplifications are made to make subsequent operations easier. wlx wl x wlsx + C = C 3 (36) Now the coordinate for the second support point is substituted into equation 36. wl! & wl! & wls! &+ C = C 3 (37) 7

8 the engineer s notebook 013 Simplifying equation 37, the final result for integration constant C 3 is: C 3 = 7wL3 8 wls 16 (38) With a solution determined for all three of the integration constants, it is now possible to list the complete slope equations for each section of the beam. The slope equation for section 1 of the beam, from the left end to the first support is: 6 wl3 8 + wls 16 = EI dv dx (39) The slope equation for section of the beam, from the first support to the second support is: 6 wlx + wl x wlsx wl3 1 + wl s dv = EI 8 dx (0) The slope equation for section 3 of the beam, from the second support to the right end of the beam is: 6 wlx + wl x 7wL3 8 wls dv = EI 16 dx (1) iv. Beam Deflection Solution: The beam slope equations derived in the previous section can now be integrated to obtain equations defining deflection of the beam as a function of position along its length. Equation 39 is integrated to determine the deflection characteristic for the first section of the beam. 6 wl3 8 + wls ' dx 16 & dx& ' dx wx wl3 x 8 + wls x + C = EIv 16 () (3) Equation 0 is integrated to determine the deflection characteristic for the second section of the beam. 6 wlx + wl x wlsx wl3 1 + wl s ' dx 8 & dx& ' dx wx wlx3 1 + wl x wlsx wl3 x wl sx + C 5 = EIv 8 () (5) 8

9 the engineer s notebook 013 Equation 1 is integrated to determine the deflection characteristic for the third section of the beam. 6 wlx + wl x 7wL3 8 wls ' dx 16 & dx& ' dx wx wlx3 + wl x 7wL3 x 6 8 wls x + C 6 = EIv 16 (6) (7) After integration is complete it is still necessary to determine the integration constants. This is accomplished by identifying several known conditions. The first condition considered is that deflection at the first support point must be equal to zero. This condition is applied to equation 3. w ' wl3 '+ wls '+ C = 0 & 8 & 16 & (8) Simplifying equation 8, the final result for integration constant C is: C = 3wL 38 3wL s + wls3 6 ws 38 (9) The same condition is now applied to equation 5. w ' & wl 1 3 ' & + wl 8 ' & wls 8 ' wl3 & 1 '+ wl s & 8 '+ C 5 = 0 & (50) Simplifying equation 50, the final result for integration constant C 5 is: C 5 = 7wL 38 wl3 s 3 wl s 6 + wls3 3 ws 38 (51) Similar to the previous condition, deflection at the second support point must also be zero. This condition is applied to equation 5. w! & wl 6! 3 & + wl! & 7wL3! 8 & wls! 16 &+ C 6 = 0 (5) Simplifying equation, the final result for integration constant C 6 is: C 6 = 11wL 38 + wl s 6 + wls3 ws 38 (53) 9

10 the engineer s notebook 013 With all integration constants derived, it is possible to assemble complete deflection equations for each section of the beam. The deflection equation for section 1 of the beam is: wx + wls x wl3 x wL 38 3wL s + wls3 6 ws 38 = EIv (5) The deflection equation for section of the beam is: wx wlx3 1 + wl x wlsx wl3 x wl sx + 7wL 8 38 wl3 s 3 wl s 6 + wls3 3 ws 38 = EIv (55) The deflection equation for section 3 of the beam is: wx wlx3 + wl x 7wL3 x 6 8 wls x + 11wL wl s 6 + wls3 ws 38 = EIv (56) v. Airy point Solution: Notice that the deflection and slope equations derived up to this point are valid for any spacing, s, of the support points. In order to derive a single solution specifically for the Airy points, an additional constraint must be considered. This constraint is to set the slope equal to zero when x is equal to zero. This condition is substituted into equation 30 and the resultant equation is solved to determine the relationship between L and s. w 0 6 wl3 8 + wls 16 = EI 0 (57) wl3 8 + wls 16 = 0 (58) wls 16 = wl3 8 (59) s = L 3 (60) s = L 3 (61) 10

11 the engineer s notebook 013 References and Additional Reading [1] Haque, Serajul. Effect of Supports. Manufacturing Science. By M. I. Khan. New Delhi: Prentice Hall India, Print. [] Hibbeler, R. C. Mechanics of Materials. Upper Saddle River, NJ: Prentice Hall, 000. Print. [3] Jain, K. C. Points of Support. A Textbook of Production Engineering. New Delhi: PHI Learning, Print. [] Moore, Wayne R. Support of Length Standards. Foundations of Mechanical Accuracy. Bridgeport, CT: The Moore Special Tool Company, Print. [5] Smith, S.T., and D.G. Chetwynd. Appendix A. Foundations of Ultraprecision Mechanism Design. Boca Raton, FL: CRC Press, Print. Key words, Structural Analysis, Precision Metrology, Beam, Deflection, Support, Ideal 11

12 the engineer s notebook 013 Scilab Solver Code //Airy Point Calculator //Given input parameters, the calculator will determine Airy point locations and calculate the deflection vs length characteristic clear; //Input Parameters //Beam Length [m] L = 1; //Beam Height [m] h = 0.01; //Beam Thickness [m] b = 0.01; //Beam Modulus [Pa] E = ; //Beam Density [kg/m^3] Beamdensity = 700; //Calculations //Airy Point Spacing [m] s = L/sqrt(3); //Beam Section Second Moment of Area [m^] I = 1/1*b*h^3; //Beam weight [kg] Beammass = Beamdensity*L*h*b; Beamweight = Beammass* ; w = Beamweight/L; //Support Coordinate 1 [m] X1 = (L - s)/; //Support Coordinate [m] X = ()/; //Beam Descritization Parameters resolution = 1000; increment = L/resolution; i = 1; //Calculate Deflection for i = 1:(resolution+1) //Determine current x coordinate x = (i - 1)*increment; //Calculate Deflection for Section 1 if (x <= X1) then deflectionx(i,1) = x; deflectiony(i,1) = - (w*x^/+w*l*s^*x/16- w*l^3*x/8+3*w*l^/38-3*w*l^*s^/6+w*s^3*l/- w*s^/38)/(e*i); end 1

13 the engineer s notebook 013 //Calculate Deflection for Section if ((x < X) & (x > X1)) then deflectionx(i,1) = x; deflectiony(i,1) = - (w*x^/- w*l*x^3/1+w*l^*x^/8- w*l*s*x^/8- w*l^3*x/1+w*l^*s*x/8+7*w*l^/38- w*l^3*s/3- w*l^*s^/6+w*l*s^3/3- w*s^/38)/(e*i); end //Calculate Deflection for Section 3 if ((x>= X) & (x <= L)) then deflectionx(i,1) = x; deflectiony(i,1) = - (w*x^/- w*l*x^3/6+w*l^*x^/- 7*w*L^3*x/8- w*l*s^*x/16+11*w*l^/38+w*l^*s^/6+w*l*s^3/- w*s^/38)/(e*i); end i = i + 1; end //Plot Results plot(deflectionx,deflectiony) 13

BEAM DEFLECTION THE ELASTIC CURVE

BEAM DEFLECTION THE ELASTIC CURVE BEAM DEFLECTION Samantha Ramirez THE ELASTIC CURVE The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of a beam. Supports that apply a moment