| Introduction | ||
 Cantilever Beam Used to Move Electronic Parts | Once again, the boss just reviewed your work for the electronic part assembly system, and he still does not think the tip deflection is correct (it was previously calculated using the moment-curvature equation and the load-deflection). The integration with boundary conditions is too complicated and he wants something easier to understand so he can check the numbers. 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 without integration? | ||
| Approach | ||
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| Finding beam deflections using the moment equation or the load-deflection equation can be tedious and lengthy. An alternative method is to use superposition to find the deflection. Basically, a complex beam with its loading is simplified to a series of basic beams (one span) and with only one load. Then the solution to all the simplified beams are added together to give a final solution. | ||
 Superposition for Beam Deflection  Some of the Basic Beams Shown in the Beam Equations Appendix | ||
| Superposition for Beam Deflection | ||
| While superposition can be used for many problems in engineering, they are particularly useful for beam deflections. Most beam configuration and loading can be split into simpler beams and loading. For example, the beam at the left with a distributed load and a point moment load can be split into two beams. One with the distributed load and one with the moment load. After the beams are simplified, the deflections of the simplified beams are still needed. To assist engineers, the deflection of simple beams are commonly listed in handbooks. This eBook has many basic beams listed in the Beam Equations appendix. The solutions for these simple beams can be derived by integrating the moment equation or load-deflection equation. The deflection equation of the complex beam is the addition of the two simpler beam equations, or δ(x) = δ1(x) + δ2(x) There are a number of assumptions when using superposition. It is assumed that the beam undergoes linear deflection, all deflections are small, elastic material properties, no shear deflection (i.e. no short thick beams, and normal boundary conditions. | ||
| Superposition Example | ||
 Superposition for Beam Deflection | One of the best way to understand the principle of superposition for beam deflection is an example. The same beam as above will be used but with numerical values for the deflection at the beam center. The beam is split into two simpler beams; beam 1) with a distributed load and beam 2) with a point moment load. The deflection for beam 1) with a distributed load over part of the beam is given in the appendix as  For the center deflections, x = 5 ft, L = 10 ft, and a = 6 ft, giving,  The deflection for beam 2) at the beam's center with a point moment load is given as  Summing the two solutions give v = v1 + v2 = -51,325/EI - 25,000/EI = -76,325/EI While this example only has two sub-beam sections, there could be many more, but the process will be the same | |
 Beam Loading | Previously, deflection at the tip of the equipment was determined by integrating both the moment equation and load-deflection equation. Both involved finding numerous integration constants. Using the method of superposition and basic beam deflection equations from the appendix, the tip deflection can be found without the integration. | |
| Reducing Beam into Simpler Beams | ||
 Splitting Beam into Two Simpler Beams | The main concept with superposition is to reduce a complex problem to simpler, smaller problems and then adding those solutions together. The current beam is a basic cantilever beam, but the changing beam cross section and loading makes it complex. First, checking with the appendix, it is noted that there is no solution for this type of beam. Thus it must be simplified. One possible method is to split the beam into two parts at point B, similar with the integration methods previously used. | |
 Equivalent Beam Deflections Using Superposition Principle | It is important that each of the simplified beam sections (sub-beams) are in the appendix (or handbook). This beam has been split into three sub-beams: 1) cantilever with a point load, 2) cantilever with a moment load, and 3) cantilever with a distributed load. (The beam could be split into other sub-beam combinations.) Notice that the beam loading has also been distributed to the sub-beams. The equivalent loading for the uniform distributed load on the right sub-beam is P = [(0.04 N/mm2)(12 mm)] (50 mm) = (0.048 N/mm)(50 mm) = 24 N M = F d = [(0.04 N/mm2)(12 mm)(50 mm)] (25 mm) = [24 N] (25 mm) = 600 N-mm These loads are also the internal shear and moment at the joint between the sub-beams. Sub-beam 3) is modeled as a cantilever beam since its deflection will be in additional to the deflection of the right beam section. The total deflection will be vtotal = vP + vM + vw + θP L + θM L The δ terms are straight forward and simply added together. The right beam also rotates (θP + θM) at its tip due to the point load,P, and moment load, M. These rotations will cause added deflection for the left sub-beam. | |
| Properties | ||
| | Before using the deflection equations from the appendix, the moment of inertia needs to be calculated. 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 | |
| Deflection and Rotation Equations | ||
 Basic Cantilever Beam Deflection and Rotation at Tip | The deflection and rotation equations are listed in the Beam Equation appendix. They are summarized in the table at the left. Only the tip location is required, so the full deflection equation is not needed. Notice the cantilever beams shown at the left are reversed from the appendix listing, which is common. Substituting these relationships gives vtotal = vP + vM + vw + θP L + θM L  vtotal = 0.2119 mm As expected, this is the same deflection as previously calculated using the integration method. | |
| Example | ||
 Shelf Supports and Loading | A student has decided to design a book shelf with a center hanger. The wood shelf, AB, is 10 ft in length and has a cross section of 12 in × 4 in (nominal dimension). The wood elasticity modulus, Ew, is 1.5×106 psi. The hanger, CD, is a 3 foot long steel rod with a yield strength of 36 ksi and an elasticity modulus, Es, of 30×106 psi. The hanger just fits between the shelf and ceiling before books are placed on it. The books exert a uniform load of 400 lb/ft on the shelf. Assuming that the beam will not fail due to bending what should be the diameter of the hanger? Use a factor of safety of 3. | |
| Solution | ||
| The problem can be solved using the method of superposition. This is done by reducing the problem to simpler problems and then adding the solutions together to get the final result. | ||
 Equivalent beam deflections using superposition principle | The shelf AB can be considered as a simply supported beam. This beam can be simplified by splitting it into two sub-beams, 1) a simply supported beam with uniform distributed load and 2) a simply supported beam with a point load acting at the center. The point load is actually the unknown force in the rod. Since the rod will also extend due to the load F, the total deflection at point B will be a combination of the two simplified beams and the rod. This can be summarized as δs = δ1 + δ2 The deflections are shown in the diagram on the left. The moment of inertia for the wood beam is Iw = 11.25(3.5)3/12 in4 = 40.20 in4 Remember, the actual size of a 4x12 wood member is 11.25 in by 3.5 in. The deflection equation of the beam and the hanger can be found in the Beams Equation appendix. δ1 = 5woLw4/(384EwIw) = 5 (400/12) [10(12)]4/ [(384) (40.20) (1.5×106)] = 1.493 in δ2 = -FLw3/(48EwIw) = -F [10(12)]3/ [(48) (40.20) (1.5×106)] = -5.970×10-4 F δs = FLs/AsEs = F (3) (12)/ [30×106 As] = 1.2×10-6 F / As These deflections can now be substituted in the deflection compatibility equation to give 1.2×10-6 F/ As = 1.493 - 5.970×10-4 F F / As (1.2×10-6 + 5.970×10-4 As) = 1.493 F/ As = 1.493 / (1.2×10-6 + 5.970×10-4 As) Using factor of safety of 3, the allowable stress is reduced as Allowable stress = Strength / Factor of safety = 36 / 3 ksi = 12 ksi = 12×103 psi Equating the allowable stress with developed stress, F/As, gives 12×103 = 1.493 / (1.2×10-6 + 5.970×10-4 As) 5.970×10-4 As = 1.244×10-4 - 1.2×10-6 As = 0.2064 in2 The cross-sectional area of steel hanger can be calculated as As = πD2 / 4 0.2064 = πD2 / 4 D2 = 0.2628 D = 0.5126 in | |
