| Introduction | ||
 Cantilever Beam Used to move Electronic Parts | The boss just reviewed your work for the electronic part assembly system, and he does not think the tip deflection is correct (it was previously calculated using the moment-curvature equation in the previous section and was found to be 0.2120 mm). Recall, a mechanical assembly system moves sensitive electronic parts from one location to another using a cantilever beam. The beam has two sections as shown in the diagram. The electronic parts will only be located on the extended section of the beam. The deflection of the beam tip is critical in the assembly process. What is known:
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| Question | ||
| What is the deflection of the beam tip using a method other than the moment-curvature equation? | ||
| Approach | ||
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| In the previous sections, Integration of the Moment Equation, was shown how to determine the deflection if the moment equation is known. This section will extend the integration method so that with additional boundary conditions, the deflection can be found without first finding the moment equation. | ||||||||
| | Moment-Shear-Load Relationships | |||||||
 Differential Element from Beam | When constructing moment-shear diagrams, it was noticed that there is a relationship between the moment and shear (and between the shear and the loading). That relationship can be derived by applying the basic equations to a typical differential element from a loaded beam (shown at the left). First, summing the forces in the vertical direction gives ΣFy = 0 V - (V + dV) - w(x) dx + 0.5 (dw) dx = 0 Both dw and dx are small, and when multiplied together gives an extremely small term which can be ignored. Assuming (dw)(dx) = 0, and simplifying gives,
ΣMright edge = 0 -M + (M + dM) - V dx + [w(x) dx][0.5 dx] = 0 Again, 2nd order terms such as dx2, are assumed extremely small and can be ignored. This gives
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| | Extending the Deflection Differential Equation | |||||||
 w, M, V, Slope, and y Relationships and Sign Conventions | Recall, the basic deflection differential equation (Moment-Curvature Equation) was derived as  This can be combined with dM/dx = V to give
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| Solving the Load-Deflection Differential Equation | ||||||||
 Each Beam Section Requires its Own Deflection Equation | The differential equation EIv´´´´ = w(x) is not useful by itself but needs to be applied to a beam with specific boundary conditions. It is assumed that EI is constant and w(x) is a function of the beam length. Note, the function, w(x), can be equal to 0. In fact, in most situations it does equal 0. Integrating the equation four times gives,  The integration constants, C1, C2, C3 and C4, are determined from the boundary conditions. For example, a pinned joint at either end of a beam requires the deflection, v, equal 0 and the moment, M, equal 0.. A fixed joint requires both the deflection, v, and slope, v´, equal 0, but moment and shear are unknown. Each beam section must have at least four boundary conditions. Details about boundary conditions are given below. Each beam span must be integrated separately, just like when constructing a moment diagram. 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 | ||||||||
| Determining the boundary conditions is usually the most difficult part of solving the deflection differential equation, especially when integrating four times. In particular, boundary conditions for multiple beam sections can be confusing. The basic types of boundary conditions are shown below. Those conditions that require two sections are sometimes called continuity conditions instead of boundary 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. Also, the shear difference will equal the applied point load at that location, and the moment will be equal in both beam sections at that point, M1 = M2. These conditions are needed to solve for the additional integration constants. | ||||||||
 Typical Boundary Conditions (v, v´, V, M) for Beam Sections Using Fourth Order Load-Deflection Equation | ||||||||
 Beam Loading | The deflection at the tip of the equipment can be determined by integrating the basic distributed load equation four times. There are actually four equations, one for each half since the load and beam structure differ in each section. A total of eight boundary conditions will be needed to solve the four integration constants for both equations. | |
| Free-Body Diagram | ||
 Free-Body Diagram | To help determine beam sections and boundary conditions, a free-body diagram should be constructed. Each change in beam geometry and load requires a new beam section and deflection equation. For this cantilever beam, there will be two sections, one from point A to B and a second section from point B to C. The uniform distributed load on the left part of the beam is w = (0.04 N/mm2)(12 mm) = 0.48 N/mm The actual values of the reactions do not need to be determined which is one of the advantages of this method. | |
| Beam Properties | ||
| | The moment of inertia using the equation 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 the Load-Deflection Equations | ||
| The deflection of any beam can be found by integrating the basic load-deflection differential equation, EIv´´´´ = -w(x) for each beam section. Section 1 (from point A to B) The load function w1(x) is the actual uniform distributed load of 0.48 N/mm EIv1´´´´ = -0.48 N/mm EIv1´´´ = -0.48x + C1 => V1 EIv1´´ = -0.24x2 + C1 x + C2 => M1 EIv1´ = -0.08x3 + C1 x2/2 + C2 x + C3 EIv1 = -0.02x4 + C1 x3/6 + C2 x2/2 + C3 x + C4 Note that EIv´´´ is the shear and EIv´´ is the moment. This will be needed when applying the boundary conditions. Section 2 (from point B to C) Section 2 is similar to section 1 except there is no uniform load. Thus, w2(x) is just 0. EIv2´´´´ = 0 N/mm EIv2´´´ = C5 => V2 EIv2´´ = C5 x + C6 => M2 EIv2´ = C5 x2/2 + C6 x + C7 EIv2 = C5 x3/6 + C6 x2/2 + C7 x + C8 | ||
| Boundary Conditions | ||
 Eight Boundary Conditions | There are eight constants of integrating that need to be defined. This requires eight 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). The next two conditions are due to continuity 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. Conditions 5 and 6 are from the free end which cannot have any shear or moment. And finally, similar to slope and deflection, the shear and moment need to be the same between beam sections 1 and 2. The shear is the the same since there is no point load at the joint. Likewise, the moment is the same since there is no applied point moment at the joint. | |
| Determining Constants | ||
| With the eight boundary conditions defined, eight equations can now be constructed. 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 5) V1 = 0 at x = 0 mm 0 = -0.48(0) + C1 C1 = 0 Boundary Condition 6) M1 = 0 at x = 0 mm 0 = -0.24(0) + C1 (0) + C2 C2 = 0 Boundary Condition 7) V1 = V2 at x = 50 mm -0.48x + C1 = C5 -0.48(50) + 0 = C5 C5 = -24 N Boundary Condition 8) M1 = M2 at x = 50 mm -0.24x2 + C1x + C2 = C5x + C6 -0.24(50)2 + 0 + 0 = 24(50) + C6 -600 = -1,200 + C6 C6 = 600 N-mm Boundary Condition 2) v´2 = 0 at x = 100 mm 33.33×106 v´2 = -24x2/2 + 600x + C7 33.33×106 (0) = -12(100)2 + 600(100) + C7 C7 = 60,000 N-mm2 Boundary Condition 1) v2 = 0 at x = 100 mm 33.33×106 v2 = C5x3/6 + C6x2/2 + C7x + C8 33.33×106 (0) = -4(100)3 + 300(100)2 + 60,000(100) + C8 C8 = -5.0×106 N-mm3 | ||
Boundary Condition 4) v´1 = v´2 at x = 50 mm  C3 = 19,720 N-mm2 Boundary Condition 3) v1 = v2 at x = 50 mm  C4 = -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 - 0.2119 mm v2 = -1.2×10-7x3 + 9.0×10-6x2 + 0.0018x - 0.15 mm The maximum deflection at the tip (x = 0) is vx=0 = -0.2119 mm | |
| Example | ||
 Shelf Supports and Loading | A beam is constructed where one end cannot deflect (pinned joint), but can rotate, and the other end cannot rotate, but can deflect.. Determine the deflection equation by integrating from the loading function. | |
| Solution | ||
Starting with the loading function, the deflection can be found by integrating it four times. However, this will require four unique boundary conditions. Those four conditions are
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 Equivalent beam deflections using superposition principle | The loading is a constant distributed load, or EI v'''' = -w Integration this gives the shear function, V(x) = EIv''' = -wx + C1 The 4th boundary condition can be used to determine the integration constant, C1, V(x=L) = -wL + C1 = 0 C1 = wL Integrating again gives the moment equation, M(x) = EIv'' = -wx2/2 + xwL + C2 The 3rd boundary conditions gives, 0 = -w 02/2 + x 0 L + C2 C2 = 0 Thus, the final moment equation is M(x) = -wx2/2 + xwL Next, this equation can be integrated to give the rotation equation, θ(x) = EIv' = -wx3/6 + wLx2/2 + C3 Using boundary condition 1 gives, 0 = -wL3/6 + wL3/2 + C3 C3 = -wL3/3 The final rotation equation is, θ(x) = EIv' = -wx3/6 + wLx2/2 - wL3/3 The deflection equation is just the integral of the rotation equation, EIv = -wx4/24 + wLx3/6 - wL3x/3 + C4 Applying the last unused boundary conditions, number 2, gives, 0 = - 0 + 0 - 0 + C4 ==> C4 = 0 The final deflection equation is v = -w (x 4/8 - Lx3/2 + xL3) / (3EI) | |
