 Beam Deflection, v, and Slope, dv/dx | Previously, beam stresses and strains were investigated and various equations were developed to predict bending and shear stresses. In addition to stresses, deflection and slope are important and need to be calculated. This section (and this chapter) will deal with various methods to calculate beam deflections. | |||||||
| | Deflection Differential Equation | |||||||
 Bending Moment, M, and Radius of Curvature, ρ are related | When a moment acts on a beam, the beam rotates and deflects. The relationship between the radius of curvature, ρ, and the moment, M, at any given point on a beam was developed in the Bending Stress and Strain section as  This relationship was used to develop the bending stress equation but it can also be used to derive the deflection equation. Recall from calculus, the radius of curvature for any point of a function, y = f(x), is  | |||||||
 Geometry of Beam Bending and Deflection | For beams, it is convenient to note the deflection as v (upward is positive) instead of y. Also, for small deflections, the first derivative or slope, dy/dx, is small. If it is squared, then it is very small and can be assumed to be zero. In other words, (dv/dx)2 ≈ 0. This simplifies the radius of curvature equation to  Combining this with the bending stress equation gives standard moment-curvature equation,
| |||||||
| Solving the Deflection Differential Equation (Moment-Curvature Equation) | ||||||||
 Each Beam Section Requires its Own Deflection Equation | The differential equation EIv´´= M is not useful by itself but needs to be applied to a beam with specific boundary conditions. Generally, EI is constant and M is a function of the beam length. Integrating the equation once gives, Integration again gives,  The integration constants, C1 and C2, are determined from the boundary conditions. For example, a pinned joint requires the deflection, v, equals 0. A fixed joint requires both the deflection, v, and slope, v´, equal 0. Each beam section must have at least two boundary conditions. Each beam span must be integrated separately, just like when constructing a moment diagram. If the moment curve is discontinuous, then a single equation cannot model the deflection. Thus, each new support or load will start a new beam section that must be integrated. Examples of beam sections are shown at the left. | |||||||
 Boundary Conditions for Beam Sections | Boundary Conditions | |||||||
| Determining the boundary conditions is usually the most difficult part of solving the deflection differential equation. In particular, boundary conditions for multiple beam sections can be confusing. For multiple beam sections, many times the boundary between the sections creates a boundary. These type of conditions are also called "Continuity Conditions". For example, a point force on a beam causes the deflections to be split into two equations. However, the beam's deflection and slope will be continuous at the load location requiring v1 = v2 and v´1 = v´2. These conditions are needed to solve for the additional integration constants. The table at the left summarizes most common boundary (and continuity) conditions for beam deflection and slope. | ||||||||
 Beam Loading | The deflection at the beam tip can be determined by first finding the moment equations and then integrating those equations. There are actually two moment equations, one for each half, since the load and beam structure is not continuous. Four boundary conditions will be needed, two for each beam section. | |
| Moment Equation and Diagram | ||
 Moment and Shear Diagrams | One method to determine beam deflections is to integrate the moment equation. This requires that the moment equation is known before starting the integration. This can be done by cutting each beam section and developing a moment equation as a function of the beam location, x (details for this process can be found in the Shear and Moment Diagrams section). For the cantilever beam, there are two sections. The first one is from point A to B. The second is from point B to C. Making a cut in the first sections and solving for the moment and shear at the cut surface gives, V1 = -0.48x N M1 = -0.24x2 N-mm The second cut in section 2 gives V2 = -24.0 N M2 = -24 (x - 25) N-mm = -24x + 600 N-mm The equations are plotted at the left. | |
| Beam Properties | ||
| | The moment of inertia for a rectangular cross section gives, I1 = 12(3)3/12 = 27.0 mm4 I2 = 16(5)3/12 = 166.7 mm4 The material stiffness, E, is given as E = 200 GPa = 200×109 N/m2 (1 m/1000 mm)2 = 200,000 N/mm2 | |
| Integrating Moment Equations | ||
| The deflection of any beam can be found by integrating the basic moment differential equation, EIv´´ = M However, each section must be integrated separately. Integrating section AB twice gives,  Recall, v is the deflection and v´ is the slope of the beam. The constants of integration, C1 and C2, must be determined from the boundary conditions (see below). Integrating the second beam section BC, gives  | ||
| Boundary Conditions | ||
 Four Boundary Conditions | There are four constants of integrating that need to be defined. This requires four boundary conditions. The first two conditions are due to the fixed joint at the right end. This requires both the deflections, v, and the slope, v´, to be zero. These are listed in the table at the left as conditions 1) and 2). Another two conditions can be identified at the joint between beam sections 1 and 2. Since the beam is continuous, the beam deflection and slope on either side of the joint must be equal. This gives the third and fourth condition, as listed in the table. | |
| Determining Constants | ||
| With the four boundary conditions defined, four equations can now be constructed which will allow all four constants to be determined. Generally, boundary conditions can be applied so that only one constant is present in a given equation. However, sometimes two or three equations will need to be solved simultaneously. Boundary Condition 2) v´2 = 0 at x = 100 mm 33.33×106 v´ = -12x2 + 600x + C3 33.33×106 (0) = -12(100)2 + 600(100) + C3 C3 = 60,000 N-mm2 Boundary Condition 1) v2 = 0 at x = 100 mm 33.33×106 v = -4x3 + 300x2 + C3x + C4 33.33×106 (0) = -4(100)3 + 300(100)2 + 60,000(100) + C4 C4 = -5.0×106 N-mm3 Boundary Condition 4) v´1 = v´2 at x = 50 mm  C1 = 19,720 N-mm2 Boundary Condition 3) v1 = v2 at x = 50 mm  C2 = -1,145,000 N-mm3 | ||
| Final Deflection Equations | ||
 Final Deflection Curve | The final deflection equations for both beam sections are v1 = -3.704×10-9x4 + 0.003652x - 0.2120 mm applies for 0 ≤ x ≤ 50 v2 = -1.2×10-7x3 + 9.0×10-6x2 + 0.0018x - 0.15 mm applies for 50 ≤ x ≤ 100 The maximum deflection at the tip (x = 0) is vx=0 = -0.2120 mm | |
| Example | ||
 Partially Loaded Beam | Beams commonly have distributed loads over only sections of the total beam, similar to the beam shown in the diagram. In this case, what is the deflection of the far right end (point C)? Assume the modulus of elasticity, E, is equal to 10 GPa. | |
| Solution | ||
| | There are two basic ways to solve this problem using integration. First, the moment in both beam sections, AB and BC, can be determined, and then integrated using two boundary conditions for each span. A second, and easier method, would be to find the rotation angle at B, and then just extrapolate the deflection from B to C. This is possible since there are no loads on the beam section BC and thus there will be no bending. The section BC will rotate, but it remain a straight line. Of course, the rotation at B still needs to be determined by integrating the moment equation, but it is only one equation instead of two equations needed in method 1. | |
 Beam Support Reactions | Due to symmetry, each of the two supports will carry half the load, giving, Ay = By = 3(2)/2 = 3 kN The moment equation for the first span, AB, is found by cutting the span at distance x from the left, and summing moments. This gives, M1 + 3x(x/2) - 3x = 0 M1 = 3x - 1.5x2 kN-m Now that the moment equation is known for the span, it can be integrated once to find the beam rotation, and a second time for beam deflection,  The deflection, v1, for the span AB is know at x = 0 and x = 2 m. Using these two boundary conditions, gives v1(x=0) = 0.5 (0)3 - 0.125 (0)4 + C1 (0) + C2 ==> C2 = 0 v1(x=2) = 0.5 (2)3 - 0.125 (2)4 + C1 (2) + 0 ==> C1 = -1 This give the beam rotation as EIv' = (1.5x2 - 0.5x3 -1) kN-m2 | |
 Beam Deflection at Point C | The beam moment of inertia, I, is I = 83(4)/12 = 170.7 cm4 = 170.7e-8 m4 and EI = (10e9 N/m2)(170.7e-8 m4 ) = 17.07 kN/m2 The beam rotation at B is v'(x=2) = (6 - 4 -1)/EI = (1 kN-m2)/(17.07 kn-m2) = 0.05858 radians The final deflection at C can be determined by noting that that beam rotation at B is also the slope at B. The final deflection is just the angle (or slope) times the distance, δC = θ d =0.05858 (2 m) = 0.1172 m | |
