| Introduction | ||
 Windmill Structure | When a new wind power generated was being constructed, the generator housing was offset 3 inches from the center line of the vertical support column. The design engineer is worried that the eccentric loading of the column will failure. What is knows:
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| Questions | ||
| 1. Does the actual stress exceed the allowable stress including the factor of safety? 2. What is the maximum weight the column can support assuming the weight is 3 in off center? | ||
| Approach | ||
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| Eccentric Axial Loads | |||||
 Pin-Pin Column with Eccentric Axial Load | Generally, columns are designed so that the axial load is inline with the column. These types of columns were analyzed in the previous sections (Basic Columns and Fixed Columns). However, there are situations that the load will be off center and cause a bending in the column in addition to the compression. This type of loading is called eccentric load and is analyzed differently. For simplification, only pin-pin columns will be analyzed in this section for eccentric loading. Stresses for the other beam types (pin-fixed, fixed-fixed, and fixed-free) with eccentric loading can be determined by using the effective length concept. | ||||
 Column Deflection due to Eccentric Axial Load | When a column is load off center, bending can be sever problem and may be more important than the compression stress or buckling. To better understand this, take an eccentrically loaded column and cut it at a distance x from the bottom pin as shown in the diagram on the left. The horizontal direction is labeled as v which represents the deflection of the column. At the cut surface, there will be both an internal moment, m, and the axial load P. This partial section of the column must still be equilibrium, and moments can be summed at the cut surface, giving, ΣM = 0 m + P (e + v) = 0 Recall, from beams, bending in a structure can be modeled as m = EI d2v/dx2, giving EI d2v/dx2 + Pv = -Pe This is a classical differential equation that can be solved using the general solution, v = C1 sin kx + C2 cos kx - e where k = (P/EI)0.5. The constants C1 and C2 can be determined using the boundary conditions. First, the deflection, v, at x = 0 is 0 giving 0 = C1 0 + C2 1 - e C1 = e The second boundary condition specifies the deflection, v, at X = L is also 0, giving, 0 = e sin kL + C2 cos kL - e  Now that the constants are known, the final deflection equation as a function of x is
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| Maximum Deflection | |||||
 Load-Deflection Diagram for Two Different Eccentric | The maximum deflection occurs at the column center, x = L/2, since both ends are pinned. This gives,  If the eccentricity is 0, then the deflection is 0 until the column buckles as shown in the diagram. For any non-zero value for e, then the deflection is gradual and then quickly grows as the load P increases. | ||||
| Maximum Stress | |||||
 Eccentric Loaded Column Stress | Unlike basic column buckling, eccentric loaded columns bend and must withstand both bending stresses and axial compression stresses. This can be illustrated by looking at both these stresses separately. The axial load P, will produce a compression stress P/A. Since the load P is not at the center, it will cause a bending stress My/I. The maximum compression stress will be total of both these stress, giving  While generally compression stress is noted as negative, the maximum is considered an absolute value and thus positive. The c term is the largest distance from the neutral axis. The maximum moment, Mmax, is at the mid-point of the column (x = L/2), Mmax = P (e + vmax) Combining the above equations gives  But I and A can be related using the radius of gyration, r, as I = Ar2. This gives the final form of the secant formula as
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| There is a lot of terms in the secant formula that may be confusing. The first issue is to understand what is unknown. Generally, the load P is not known for a given column, material type, and maximum allowable stress, σmax. One might ask, if P is the unknown, why is the equation not formulated with P on one side of the equation? The problem is P cannot be isolated on one side of the question since this is a nonlinear equation. Thus P cannot be solved using algebra but a numerical method is needed such as a root finder or even trial and error. | |||||
So, back to the terms. The stress maximum, σmax, is generally the yield stress or allowable stress of the column material, which is known. The geometry of the column, length L, area A, radius of gyration r, and maximum distance from the neutral axis c are also known. The eccentricity, e, and material stiffness, E, are considered known.
The secant formula can be better understood if it is plotted as function of the slenderness ration, L/r and the pure axial compression stress, P/A. The stiffness, E, maximum stress, σmax, and eccentricity ratio, ec/r2, need to be set. In the diagram, the σmax, is set at 36 ksi and the eccentricity ratio is shown for various values. The stiffness can be changed by the user (move the slider).
For the eccentricity ratio of 0, the Euler equation is recovered and is shown on the graph as the red curve
| Generally, columns are loaded at their center so that there is no eccentric forces. However, due to construction problems or design reasons some columns must support loads that are off set from their center, which are called eccentric loads. In this problem, a wind generator system is attached to vertical 6 in pole at its edge. Thus, the weight of the generator, 5 kips, is off center by 3 in. This eccentric load will cause the pole to bend and induce added bending stresses. This problem can be solved using the standard secant formula. | ||
| Total Stress | ||
 Wind Generator Column with Off Center Load  Pole Cross Section | The secant formula provides a method to determined the stress in a column with an eccentric load. The standard form of the formula is  Since the load P is given, 5 kips, the equation is easy to use. The other parameters are, A = (ro2 - ri2)π = (32 - 2.752) π = 4.515 in2 e = 3 in (eccentric distance) c = 3 in (furthest distance to outside edge) I = (ro4 - ri4)π/4 = (32 - 2.752) π/4 = 18.70 in4 r = (I/A)0.5 = (18.70/4.515)0.5 = 2.035 in E = 29,000 ksi The column height, L, needs to be converted to effective length since this is not a pin-pin column. For a fixed-free column, the effective length is Le = 2L = 2(30 ft) = 60(12) in = 720 in Substituting these values into the secant formula gives,  = 1.107 [1 + 2.173 sec(1.0931)] = 1.107 [1 + 2.173 (2.175)] | |
| The final stress which includes both compression and bending stress is σ = 6.339 ksi The failure stress of the material is 36 ksi, but there is a factor of safety of 3. Thus the allowable is only 12 ksi. However, this the actual stress, 6.34 ksi is less than the allowable, so the design is still good. | ||
| Maximum Load | ||
| From the calculations above, a load of 5 kips will only produce a stress of 6.34 ksi which is below the allowable or 12 ksi. So how high can the load go before the stress hits 12 ksi? The secant formula can be used again, but this time the unknown will be P, not σ. Substituting the values in gives,  The variable P cannot be solved using just algebra. A numerical method needs to be used or even trail and error. One simple and effective method to solve any single equation is to graph the error as a function of the variable. This is done by first rearranging the equation as  The solution for P is when R goes to zero. This equation is shown in the diagram on the left. | ||
 Solution Curve for Load P | The variable P cannot be solved using just algebra. A numerical method needs to be used or even trail and error. One simple and effective method to solve any single equation is to graph the error as a function of the variable. This is done by first rearranging the equation as The solution for P is when R goes to zero. This equation is shown in the diagram on the left. The accuracy can be improved by narrowing the equation range. Using this method, the final solution is approximately P = 6.660 kips | |
