Deflection of Beams. Equation of the Elastic Curve. Boundary Conditions

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1 Deflection of Beams Equation of the Elastic Curve The governing second order differential equation for the elastic curve of a beam deflection is EI d d = where EI is the fleural rigidit, is the bending moment, and is the deflection of the beam (+ve upwards). Boundar Conditions Fied at = a: Simpl supported at = a: Deflection is zero ) =a Slope is zero ) d d =a Deflection is zero ) =a fourth order differential equation can also be written as EI d d = w where is w is the distributed load. Here, two more boundar conditions are needed in terms of bending moment and shear force.

2 Boundar Conditions Free at = a: Bending moment is zero ) = EI d d =a Shear force is zero ) = EI d d =a Simpl supported at = a: Bending moment is zero ) = EI d d =a Notes on Integration Z Z (a + b)d = Z ad + bd + C 1 Z a + b + C 1 = a + b + C 1 Z a Z d = d + bd + = a + b + C 1 + C Z C 1 d + C Problem 1. Calculate the tip deflection for the cantilever beam shown below. P Figure 10: Problem 1. Bending moment = P

3 Hence, EI d d = = EI d d = EI = P P + C 1 [integrating with respect to ] P + C 1 + C [integrating again with respect to ] Use boundar condition d/d and at =. d d = P Figure 107: Problem 1: Free-bod diagram. ) C 1 = P = ) P + C 1 + C ) C = P Hence, the equations of the deflection and slope becomes = 1 P EI + P P d d = 1 P EI + P The tip deflection and the rotation =0 = P EI d = P d EI =0 Problem. Calculate the maimum deflection for the beam shown. The support reactions are = B = P/ 0 apple apple /: Bending moment = P

4 P B B Figure 108: Problem. Hence, EI d d = = P EI d d = P + C 1 [integrating with respect to ] EI = P 1 + C 1 + C [integrating again with respect to ] Use boundar condition at. = P/ Figure 109: Problem : For 0 apple apple /. C / apple apple : Bending moment Hence, = P( ) EI d P( ) = = d P EI d d = P EI = P = P P + C [integrating with respect to ] P 1 + C + C [integrating again with respect to ] Use boundar condition at =. Figure 110: Problem : For / apple apple. B = P/ 0 = P C + C = P 1 + C + C P Now, use compatibilit condition that deflections and slopes from both these equations at = / should match.

5 Or, due to the smmetr of the problem slope at = / should be zero, i.e., d/d at = /. From the equation for the first half of the beam EI d d =/ = P 1 + C 1 ) C 1 = P 1 Similarl, from the equation for the second half of the beam EI d = P d =/ P 1 + C ) C = P 1 ) C = P C = P 8 Hence, the equations of the elastic curve 8 < 1 P P EI 1 1 for 0 apple apple / = : 1 P EI 1 + P P 1 + P 8 for / apple apple Hence, maimum deflection at the midspan =/ = P 9EI P EI = P 8EI [using the first equation] ) ma = P 8EI Check: =/ = P 9EI + P 1EI P EI + P 8EI = P 8EI [using the second equation] Slope at the left end d d =0 = P 1EI Slope at the right end d d = = P 1EI Problem. Calculate the maimum deflection for the beam shown.

6 Figure 111: Problem. = 5 KN/m B = 10 m We will convert all units to N and m. So, our will be in m. The vertical support reactions are = B = / = 5 kn. Bending moment at a distance of from left end = (5000) = (5 10 ) = 5000 N = 5 KN/m Hence, EI d d = = EI d d = EI = C 1 [integrating with respect to ] C 1 + C [integrating again] Use boundar conditions at and = = 10 m. = 5 kn / Figure 11: Problem : Free-bod diagram. =0 ) C ) =10 m 500 (10) 1500 (10) + + C 1 1 (10) =0 C Hence, the equations of the elastic curve and the slope of the curve = ( ) EI 1 d d = EI

7 aimum deflection at the midspan =5 m = EI ) ma = EI = 5w 8EI Problem. Calculate the maimum deflection at the tip for the beam shown. We will convert all units to N and m. So, our will be in m. Figure 11: Problem. = 10 KN/m = 5 m Bending moment 1 () = kn = 1000 N w = = KN/m = 1000 = 1000 Hence, EI d d = = EI d d = EI = C 1 [integrating with respect to ] + C 1 + C [integrating again with respect to ] / Figure 11: Problem : Free-bod diagram.

8 Use boundar conditions d/d and at = = 5 m. d d =5 m 50 (5) ) + C 1 ) C 1 = ) =5 m 50 (5)5 + C 1 (5)+C C Hence, the equations of the elastic curve and the slope of the curve = ( ) EI d d = EI aimum deflection at the tip = =0 ) ma = EI EI = 0EI Problem 5. Estimate the deflection curve for the beam shown. Figure 115: Problem 5. / B B

9 Using the equations for equilibrium 0 apple apple /: Bending moment  F + B = 1 =  B = = 1 ) B = 1 ) = 1 = = + = 1 w0 = w = Hence, EI d d = = EI d d = 1 EI = 1 + C 1 [integrating with respect to ] C 1 + C [integrating again] = / Figure 11: Problem 5: For 0 apple apple /. / apple apple : Bending moment = = 1 ( ) 1 1 Hence, EI d d = = 1 EI d d = 1 EI = 1 + C [integrating with respect to ] + C + C 7 [integrating again] Figure 117: Problem 5: For / apple apple. B = P/

10 Use boundar conditions at and =. =0 ) C = ) C + C = Net, use the compatibilit condition that at = / deflection and slope from both of these epressions should match ) 8 =/ epression 1 = =/ C 1 = ) C 1 = ) C 1 = ) C 1 + C = 17 0 d ) 1 d =/ C epression 1 ) + C 1 = + C ) C 1 C = C 0 = d d =/ 1 ( 1 )+C 1 = 1 Solving for C 1 and C gives epression C + C [C = 1 epression + C C ] C 1 = C = 10 ) C = 80 Hence, the equations for the elastic curve 8 < 1 w0 5 1 EI = : 1 w0 EI for 0 apple apple / for / apple apple

11 ethod of Superposition ethod of superposition can be used if ou have two or more loads acting on the beam. Problem. Estimate the deflection of the beam as shown. P Figure 118: Problem. Using method of superposition combine results from the following two cases: P Figure 119: Problem : ethod of superposition. / For the first case: Figure 10: Problem : Case I

12 We will use the fourth order governing differential equation. EI d d = w = ) EI d d = + C 1 ) EI d d = + C 1 + C ) EI d d = + C 1 + C + C ) EI = + C 1 + C + C + C Use boundar conditions d/d and at = and bending moment = EI d d at and shear force = EI d d at. = EI d d =0 ) C 1 = EI d d =0 ) C d d = ) C = = ) + C + C ) C = = 8 Hence, for the first case = 1 w0 EI + 8 d d = 1 w0 + EI

13 For the second case: P / Figure 11: Problem : Case II 0 apple apple /: Bending moment. Hence, / apple apple : Bending moment Hence, EI d d = = EI d d = EI = EI d d = EI d d = C 1 [integrating with respect to ] EI = C 1 + C [integrating again] = P( /) P P + P + P + C [integrating with respect to ] P + P + C + C [integrating again] Use boundar conditions d/d and at =. d d = ) P + P + C ) C = ) P + P + C ) C = P 1 Net, use the compatibilit condition that at = / the slope and Figure 1: Problem, Case II: For 0 apple apple /. / Figure 1: Problem, Case II: For / apple apple. P

14 the deflection should match. 0 1 d d =/ ) C 1 = P 8 + P 0 1 =/ epression 1 = P 8 epression 1 0 = d 0 = ) C 1 + C = P 8 + P 1 + C ) C = 5P 8 d =/ =/ 1 1 epression epression Hence, for the second case, the equations for the elastic curve 8 < 1 P 5P EI 8 8 for 0 apple apple / = : 1 P EI + P P 1 for / apple apple Combining case I and II, the elastic curves for the original beam 8 < 1 w0 EI P 5P 8 8 for 0 apple apple / = : 1 w0 EI + P 8 + P P 1 for / apple apple The tip deflection and rotation =0 = 8EI d = + P d EI 8EI =0 5P 8 Staticall Indeterminate Beams Clever use of superposition can be utilized here. Problem 7. Calculate the support reaction here. ethod I: et us use the method of superposition and divide the problem into the following two cases. From Problem 5, the deflection for the first case = 1 EI w0 + 8

15 Figure 1: Problem 7. B Figure 15: Problem 7: ethod of superposition. Hence, the tip deflection in this case =0 = 8EI For the second case, use the result from Problem 1. The tip deflection in this case = EI =0 However, due to the presence of the roller support at, the deflection at should be zero. This leads to EI 8EI = 8 ethod II: ssume the unknown reaction at as and calculate the bending moment as follows: Bending moment at a distance of from left end = ( ) + / Hence, = EI d d = = + + EI d d = + + C 1 [integrating with respect to ] EI = + + C 1 + C [integrating again] Figure 1: Problem 7: Free-bod diagram.

16 Use boundar conditions at and d/d, at =. =0 C d d = ) + + C 1 ) C 1 = = ) + + C 1 + C ) +! + ) = 8 Problem 8. Calculate the support reaction at. Figure 17: Problem 8. = 10 KN/m B = 5 m Using method of superposition this problem can be divided into two cases. From Problem, the tip deflection for case I =0 = 0EI = EI

17 Figure 18: Problem 8: ethod of superposition. From Problem 1, the tip deflection for case II =0 = EI = 1.7 EI However, due to the presence of the roller support at, the deflection at should be zero. This leads to EI EI = 5000 N = 5 kn

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