Chapter 8 Supplement: Deflection in Beams Double Integration Method
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1 Chapter 8 Supplement: Deflection in Beams Double Integration Method 8.5 Beam Deflection Double Integration Method In this supplement, we describe the methods for determining the equation of the deflection curve and for finding deflections at specific points along the axis of the beam. The calculations of deflection are essential for the following reasons. 1. Analysis of statically indeterminate beams. 2. Verify that deflections do not exceed allowable values: large deflections are associated with poor appearance, and with too much flexibility in the structure (which can lead to a sense of discomfort or broken glass panels). For the derivation of this method, we start with the following equations. dx = R dθ 1/R = dθ/dx and 1/R = M/EI where R = radius of curvature M = bending moment E = modulus of elasticity I = moment of inertia of the beam cross section EI = flexural rigidity of the beam For small rotations: 1/R = dθ/dx = d 2 y/dx 2 = M/EI This equation is the basic differential equation of the deflection curve of a beam. Application of Elastic Curve Double Integration The equation for the deflection curve is found by starting with the following equation. y = d 2 y/dx 2 = M/EI Finding the deflection curve consists of successive integrations of this equation, with the resulting constants of integration being evaluated from the boundary conditions of the beam. Page 1 of 9
2 Integrating once gives an expression for the slope θ of the elastic curve at any point x. θ = y = dy/dx = (M/EI) dx Integrating a second time yields an expression for the deflection of the elastic curve at any point x: = y = (M/EI) dx 2 The following sign convention is used for this method. 1. Downward deflection is negative. 2. Upward deflection is positive. Summary of Procedure First, the equations for the bending moment are written, using free body diagrams and the equations of equilibrium. Next, the following integral equation is written. y = d 2 y/dx 2 = M/EI or EI y = EI (d 2 y/dx 2 ) = M Then, the integration is performed and the constants of integration are evaluated from boundary conditions pertaining to and θ at the supports of the beam and from continuity equations. Double Integration Use of Symmetry The double integration method becomes laborious when the constants of integration are difficult to determine. When simply supported beams are symmetrically loaded, the solution becomes simpler. - The maximum deflection occurs at mid-span. - The slope of the elastic curve is zero at mid-span. Summary 1. Write the equations for bending moment. 2. Write the following integral equation. y = d 2 y/dx 2 = M/EI or EI y = d 2 y/dx 2 = M 3. Perform the integrations. 4. Evaluate the constants of integration. Page 2 of 9
3 Examples Deflection by the Double Integration Method Given: The simply supported beam with a uniformly distributed load. Find: Write equations for Δ and θ. Solution Write an equation for moment using the free body diagram at the right. M cut = 0 = - (wl/2) x + w x (x/2) + M M = (wl/2) x wx 2 /2 EIy = M = (wl/2) x - wx 2 /2 EIy = EIθ = (wl/2)(x 2 /2) - (w/2)(x 3 /3) + C 1 EIy = EIΔ = (wl/2)(x 3 /6) - (ω/2)(x 4 /12) + C 1 x + C 2 Solve for the constants of integration. Boundary conditions: At x = 0, Δ = 0 So C 2 = 0 At x = L, Δ = 0 0 = (wl/2)(l 3 /6) - (ω/2)(l 4 /12) + C 1 L + 0 C 1 L = - (wl/2)(l 3 /6) + (ω/2)(l 4 /12) = (ωl 4 /24)( ) = - ωl 4 /24 So C 1 = - ωl 3 /24 Write the equations for Δ and θ. EIy = EIθ = (wl/2)(x 2 /2) - (w/2)(x 3 /3) - ωl 3 /24 = w/24 (6Lx 2-4x 3 - L 3 ) Thus, θ = (w/24ei)(6lx 2-4x 3 - L 3 ) EIy = EIΔ = (wl/2)(x 3 /6) - (ω/2)(x 4 /12) - (ωl 3 /24)x + 0 = wx/24(2lx 2 - x 3 - L 3 ) Thus, Δ = (wx/24ei)(2lx 2 - x 3 - L 3 ) Page 3 of 9
4 Example Given: The cantilever beam with a uniformly distributed load. Find: Write equations for Δ and θ. Solution Write an equation for moment using the free body diagram at the right. M cut = 0 = wx(x/2) + M M = wx 2 /2 EIy = M = - wx 2 /2 EIy = EIθ = - (w/2)(x 3 /3) + C 1 EIy = EIΔ = - (ω/2)(x 4 /12) + C 1 x + C 2 Solve for the constants of integration. Boundary condition: At x = L, θ = 0 0 = - (w/2)(l 3 /3) + C 1 C 1 = + wl 3 /6 Boundary condition: At x = L, Δ = 0 0 = - (ω/2)(l 4 /12) + (wl 3 /6) L + C 2 C 2 = + wl 4 /24 - wl 4 /6 = (wl 4 /24)(1-4) = - 3wL 4 /24 = - wl 4 /8 Write the equations for Δ and θ. EIy = EIθ = - (w/2)(x 3 /3) + wl 3 /6 = w/6 (- x 3 + L 3 ) Thus, θ = (w/6ei) (L 3 x 3 ) EIy = EIΔ = - (ω/2)(x 4 /12) + (wl 3 /6) x - wl 4 /8 = (w/24)(- x 4 + 4xL 3-3L 4 ) Thus, Δ = (w/24ei)(- x 4 + 4xL 3-3L 4 ) Page 4 of 9
5 Example Given: Simply supported beam loaded as shown. Find: Write equations for Δ and θ. Solution Find the reactions at the supports. F x = 0 = A x A x = 0 M C = 0 = - 20 A y + 2 (10) A y = 300 A y = 15.0 k M A = 0 = 20 C y - 2(10)5 20 C y = 100 C y = 5.0 k Write an equation for moment for the left half of the beam using the free body diagram at the right. 0 < x < 10 M cut = 0 = - 15 x + 2x (x/2) + M M = 15x x 2 EIy = M = 15x - x 2 EIy = EIθ = 15 (x 2 /2) - (x 3 /3) + C 1 = 7.5 x 2 - x 3 /3 + C 1 EIy = EIΔ = 7.5(x 3 /3) - (x 4 /12) + C 1 x + C 2 = 2.5 x 3 - x 4 /12 + C 1 x + C 2 Solve for the second constant of integration. Boundary condition: At x = 0, Δ = 0 So C 2 = 0 Page 5 of 9
6 Write an equation for moment for the right half of the beam using the free body diagram at the right. 10 < x < 20 M cut = 0 = - M + 5 (20 x) M = 5 (20 x) = 100 5x EIy = M = 100-5x EIy = EIθ = 100 x 5 (x 2 /2) + C 3 = 100 x x 2 + C 3 EIy = EIΔ = 100 (x 2 /2) (x 3 /3) + C 3 x + C 4 Solve for the constants of integration. Boundary condition: At x = 10, EIθ 1 = EIθ x 2 - x 3 /3 + C 1 = 100 x x 2 + C (10) 2 - (10) 3 /3 + C 1 = 100 (10) (10) 2 + C 3 C 1 C 3 = = Boundary condition: At x = 10, EIΔ 1 = EIΔ x 3 - x 4 /12 + C 1 x + C 2 = 100 (x 2 /2) (x 3 /3) + C 3 x + C 4 2.5(10) 3 - (10) 4 /12 + C 1 (10) + 0 = 100 [(10) 2 /2] [(10) 3 /3] + C 3 (10) + C C 1 = C 3 + C 4 C 4 = C C 3 = 10 C C 3 C 4 = C 1-10 C 3 = (C 1 - C 3 ) = (333.33) C 4 = Boundary condition: At x = 20, Δ = 0 0 = 100 [(20) 2 /2] - [2.5 (20) 3 /3] + C 3 (20) + C 4 0 = 20, C = 20 C , C 3 = - 14, C 3 = C 1 C 3 = C 1 = C = C 1 = Page 6 of 9
7 Write the equations for Δ and θ. 0 < x < 10 EIy = EIθ = 7.5 x 2 - x 3 /3 + C 1 EIθ = 7.5 x 2 - x 3 /3-375 Thus, θ = (1/EI)(7.5 x 2 - x 3 /3-375) EIy = EIΔ = 2.5 x 3 - x 4 /12 + C 1 x + C 2 = 2.5 x 3 - x 4 / x + 0 EIΔ = 2.5 x 3 - x 4 /12-375x Thus, Δ = (1/EI)(2.5 x 3 - x 4 / x) 10 < x < 20 EIy = EIθ = 100 x x 2 + C 3 EIθ = 100 x x Thus, θ = (1/EI)(100 x x ) EIΔ = 100 (x 2 /2) (x 3 /3) + C 3 x + C 4 EIΔ = 100 (x 2 /2) (x 3 /3) x Thus, Δ = (1/EI)[100 (x 2 /2) (x 3 /3) x ] Page 7 of 9
8 Example Given: The simply supported beam with a uniformly distributed load. W10 x 49 (I x = 272 in 4 ) E = 29,000 ksi Find: Write equations for Δ and θ. Compute Δ and θ at mid-span. Solution Write an equation for moment using the free body diagram at the right. M cut = 0 = M 24x + 4x (x/2) 0 = M 24x + 2x 2 M = 24x 2x 2 EIy = M = 24x - 2x 2 EIy = EIθ = 24(x 2 /2) - 2(x 3 /3) + C 1 = 12x 2 (2/3) x 3 + C 1 EIθ = - 2x 3 /3 + 12x 2 + C 1 EIy = EIΔ = 12 (x 3 /3) - (2/3)(x 4 /4) + C 1 x + C 2 = 4x 3 x 4 /6 + C 1 x + C 2 EIΔ = - x 4 /6 + 4x 3 + C 1 x + C 2 Solve for the constants of integration. Boundary conditions: At x = 0, Δ = 0 So C 2 = 0 At x = 12, Δ = 0 0 = - (12 4 /6) + 4 (12 3 ) + C 1 (12) = C 1 12 C 1 = = C 1 = Page 8 of 9
9 Write the equations for Δ and θ. EIy = EIθ = - 2x 3 /3 + 12x 2 + C 1 = - 2x 3 /3 + 12x Thus, θ = (1/EI)(- 2x 3 /3 + 12x 2-288) EIy = EIΔ = - x 4 /6 + 4x 3 + C 1 x + C 2 = - x 4 /6 + 4x 3-288x + 0 Thus, Δ = (1/EI)(- x 4 /6 + 4x 3-288x) Compute Δ and θ at mid-span (i.e. x = 6 ). EIθ = - 2x 3 /3 + 12x = - 2(6) 3 /3 + 12(6) = = 0 Thus, θ = 0 (i.e. horizontal tangent) at mid-span (as expected due to the symmetry of the loading). EIΔ = - x 4 /6 + 4x 3-288x = - (6) 4 /6 + 4(6) 3 288(6) = = (Deflection is down.) Δ = 1080/EI = 1080 (12 / ) 3 /[29,000 (272)] Thus, Δ = For a simply supported beam with a uniformly distributed load, the deflection at mid-span may be determined by the following standard formula. Δ = 5wL 4 /384EI Δ = 5(4)(12) 4 (12 / ) 3 /[384 (29,000) 272] = Page 9 of 9
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