deflections-lecture-2024.pdfmechanics126
- 1. BEAM DEFLECTIONS ENGR. JEAN C. EBALLE ES 126 Instructor
- 2. 1 Deflection diagrams & the elastic curve Elastic-beam theory The double integration method Moment-area theorems Conjugate-beam method Beam Deflections
- 3. 2 Deflection diagrams & the elastic curve In this topic, only linear elastic material response is considered. It is useful to sketch the shape of the structure when it is loaded in order to visualize the computed results & to partially check the results.
- 4. 3 Deflection diagrams & the elastic curve This deflection diagram rep the elastic curve for the points at the centroids of the cross- sectional areas along each of the members. If the elastic curve seems difficult to establish, it is suggested that the moment diagram be drawn first. From there, the curve can be constructed.
- 5. 4 Deflection diagrams & the elastic curve Within the region of –ve moment, the elastic curve is concave downward. Within the region of +ve moment, the elastic curve is concave upward. There must be an inflection point where the curve changes from concave down to concave up.
- 6. 5 Deflection diagrams & the Elastic curve
- 7. 6 Draw the deflected shape of each of the beams shown in the roller at A allows free rotation with no deflection while the fixed wall at B prevents both rotation and deflection.
- 8. 7 Draw the deflected shape of each of the beams shown in In Fig. (b), no rotation or deflection occur at A & B.
- 9. 9 Elastic Beam theory To derive then , we look at an initially straight beam that is elastically deformed by loads applied perpendicular to beam’s x-axis & lying in x-v plane of symmetry as shown in Fig.1 (a).
- 10. Beam Deflection 10 Differential Equation of Deflection v dx dv ds θ tan dv dx = cos = ds dx sin dv ds = ds d = 1 = ds d Recall that 1/ρ is the curvature of the beam. Slope of the deflection curve
- 11. Beam Deflection 11 Assumptions Assumption 1: θ is small. 1. 2. Assumption 2: Beam is linearly elastic. Thus, the differential equation for the deflection curve is: dx ds dx d ds d = 1 tan dv dx = 2 2 d d v dx dx = 2 2 1 d v dx = EI M = 1 2 2 d v M dx EI =
- 12. Beam Deflection 12 Diff. Equations for M, V, and w Recall from before: So we can write: Deflection curve can be found by integrating Bending moment equation (2 constants of integration) Shear-force equation (3 constants of integration) Load equation (4 constants of integration) Chosen method depends on which is more convenient. w dx dV − = V dx dM = 4 4 d v EI w dx = − 3 3 d v EI V dx = 2 2 d v EI M dx =
- 13. Beam Deflection 13 Boundary Conditions Sometimes a single equation is sufficient for the entire length of the beam, sometimes it must be divided into sections. Since we integrate twice there will be two constants of integration for each section. These can be solved using boundary conditions. Deflections and slopes at supports Known moment and shear conditions 0 0 = = B B M V 0 = A M 0 = B M vA vA vB
- 14. 14 The double integration method ) ( ) ( 2 2 b dx v d dx d a and dx dv = = The value of slope tanϴ = dy/dx, may therefore with only small error
- 15. 15 The double integration method If we consider the variation in ϴ in a differential length ds caused by bending, it is evident that ) (c d ds =
- 16. 16 2 2 1 1 dx v d or dx d ds d = = Because the elastic curve is very flat, ds is practically equivalent to dx, so from eqs. (c ) and (b) we obtain
- 17. 17 ) ( 2 2 d M dx y d EI = From the relation: EI M = 1 Equating the values from equations (d) and the Above relation, we have This is known as the differential equation of the elastic curve of a beam. EI is constant along the beam.
- 18. 18 ) ( 1 f C Mdx dx dv EI + = If the equation (e) is now integrated, we obtain This is the slope equation specifying the slope value od dy/dx at any point.
- 19. 19 + + = 2 1 C C dx Mdx dx dv EI We now integrate equation (f) to obtain This is the required deflection equation of the elastic curve specifying the value of v for any value of x. the C’s there are the constant of integration that must be evaluated from the given conditions of the beams and its loadings.
- 20. 20 2 1 2 1 2 2 2 C x C x M v EI C x M dx dv EI M dx v d EI o o o + + = + = = The double integration method
- 22. 22 Moment-Area theorems To develop the theorems, reference is made to Fig. 8.13(a). If we draw the moment diagram for the beam & then divide it by the flexural rigidity, EI, the “M/EI diagram” shown in Fig. 8.13(b) results. By equation dx EI M d =
- 23. 23 Moment-Area theorems d on either side of the element dx = the lighter shade area under the M/EI diagram. Integrating from point A on the elastic curve to point B, Fig 8.13(c), we have This equation forms the basis for the first moment-area theorem. / dx EI M B A A B =
- 24. 24 Moment-Area theorems Theorem 1 The change in slope between any 2 points on the elastic curve equals the area of the M/EI diagram between the 2 points. AB / ) ( 1 Area EI A B =
- 25. 25 Moment-Area theorems Theorem 2 The vertical deviation of the tangent at a point (A) on the elastic curve with the tangent extended from another point (B) equals the “moment” of the area under the M/EI diagram between the 2 points (A & B). A B A x Area EI t . ) ( 1 AB / =
- 26. 26 Moment-Area theorems It is important to realize that the moment- area theorems can only be used to determine the angles and deviations between 2 tangents on the beam’s elastic curve. In general, they do not give a direct solution for the slope or displacement at a point.
- 27. 27 Moment Diagram by Parts The resultant bending moment at any section caused by any load system is the algebraic sum of the bending moments at that section caused by each load acting separately. This statement is expressed algebraically by 2 Basic Principles R L M M M ) ( ) ( = =
- 28. 28 Moment Diagram by Parts Where the indicates the sum of the moments caused by all the forces to the left of the section, and is the sum of the moments caused by all the forces to the right of the section. R L M M M ) ( ) ( = = L M ) ( R M ) (
- 29. 29 Conjugate-Beam method w dx dV = w dx M d = 2 2 EI M dx d = EI M dx v d = 2 2 wdx V = dx wdx M = dx EI M = dx dx EI M v =
- 30. 30 Conjugate-Beam method Here the shear V compares with the slope , the moment M compares with the displacement v & 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” as shown in Fig. 8.22.
- 31. 31 Conjugate-Beam method Fig 8.22
- 32. 32 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.
- 33. 33 Conjugate-Beam method Theorem 2 The displacement 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 and displacement of the real beam at its supports.
- 34. 34 Conjugate-Beam method For example, as shown in Table 8.2, a pin or roller support at the end of the real beam provides zero displacement but the beam has a non-zero slope.
- 35. 35 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.
- 36. 36 Conjugate-Beam method When the real beam is fixed supported, both beam has a free end since at this end there is zero shear & moment.
- 37. 37 Conjugate-Beam method Corresponding real & conjugate beam supports for other cases are listed in the table
- 38. 38 Conjugate-Beam method Corresponding real & conjugate beam supports for other cases are listed in the table
- 39. 39 Conjugate-Beam method Shear and Moment at Supports of Conjugate Beam should account for the corresponding slope and deflection of real beam at its supports
- 40. Double Integration Method 40 Problem 1: A simply supported beam 10 meters long carries a uniform load of 24kN/m. Using E = 200GPa and 𝐼 = 240𝑥 106 𝑚𝑚4. Determine the rotation in degree form of the beam at a point 4 meters from the left support using double integration method.
- 41. Double Integration Method 41 Problem 2: A simply supported beam 10 meters long carries a uniform load of 24kN/m. Using E = 200GPa and 𝐼 = 240𝑥 106 𝑚𝑚4. Determine the rotation in degree form of the beam at a point 4 meters from the left support using double integration method.