The Conventional Design Process

Abstract

The characteristics of the conventional design process, implying repeated analysis of a structure and resizing of its members until a satisfactory design is obtained, is illustrated by means of some simple truss structures. In this process, it is implicitly assumed that by satisfying as closely as possible all requirements placed on the design this will lead to the ‘best’ design. In terms of the maximum stress in the members, this is the well-known principle of the fully stressed design. Effective as this method often is, common situations are identified where this does not lead to an optimum, minimum weight design. Furthermore, the process may be very slowly convergent, in addition to which it offers no help when conditions other than a simple maximum stress apply or, for example, with the optimum shape of a structure. Minimum weight implies economy of material as well as operational savings directly related to reduction in weight. All this provides justification for the formal optimization methods in the remaining chapters of this book. While this chapter is concerned only with truss structures, conclusions reached can, in principle, be taken to apply more widely to the optimization of many other types of structure. A spreadsheet program for the numerical optimization of a simple seven-bar truss provides a first introduction to use of the Solver optimization tool in Microsoft Excel.

Exercises

• 1.1 Verify the formula in Eq. (1.11) for the minimum volume of the truss structure in Fig. 1.5, and the optimum angle θ.

  Derive formulae for the forces in the members, the corresponding minimum cross-sectional areas of the members and the volume of the truss in terms of the angle θ. Try different values of θ to search for the minimum volume.

• 1.2 A truss structure has to carry two equal loads P/2 over a span L, the loads being placed at 1/3 and 2/3 of the span, respectively. The material of the truss has an allowable stress $\upsigma_{0}$. Try different layouts to find a suitable layout of truss for this loading. Express the result in terms of the coefficient n in Eq. (1.6), where P is the total load on the truss. Compare the value of n with that for the truss in Exercise 1.1 with a load P at mid-span.

  Try a few simple layouts of statically determinate truss, with members at suitably chosen angles. Compare the volume of each layout.

• 1.3 Verify the formula in Eq. (1.15) for the minimum volume of the long truss structure shown in Fig. 1.13. If its height is limited to 1.0 m, calculate the maximum possible span when loaded only under its own weight. The truss is made of steel with an allowable stress of 1000 N/mm² and density 7850 kg/m³.

  In deriving Eq. (1.15), shear force and bending moment distributions are obtained by treating the truss as a continuous beam. Equations (1.13) and (1.14) can then be used to obtain formulae for the volume of the bracing members and horizontal members, respectively, by integration over the span of the truss. These together give the formula for minimum volume of the truss. Treating the weight of the truss as uniformly distributed over its span, use Eq. (1.15) with Eq. (1.10) to calculate the maximum possible span when the height is limited to 1.0 m.

• 1.4 Derive a formula for the minimum volume of a truss similar to the Michell truss in Fig. 1.12a, with only five radial ‘spokes’ at an angle of 45 °C to each other. Compare the result with Eqs. (1.11) and (1.12).

  The forces in the bars can be solved by equilibrium at the nodes. Replace the circular arc by straight bars between the nodes. Note that by equilibrium the force in all five of these ‘circumferential’ bars is the same.

• 1.5 A steel cable is stretched between two towers each of height 100 m above the ground. It may be assumed that the cable forms a shallow parabolic curve, loaded only under its own weight. The tensile strength of the cable is 1000 N/mm² and its density is 7850 kg/m³. What is the maximum possible distance between the towers if the cable is not permitted to touch the ground?

  Derive a formula for the parabolic curve of the cable in terms of the height of the towers and the distance L between them, and from this the angle the cable makes at the towers. The vertical component of the force in the cable at each tower has to be equal to half the weight of the cable. Assume L to be much greater than 100 m, so that the actual length of the cable is approximately equal to L.

• 1.6 A hollow steel tube of diameter 1.0 m is mounted vertically as a tower, fixed rigidly at its lower end and free at its upper end. It is loaded only by its own weight. Under these conditions, the tube can be treated as a column with an effective length $L_{\text{eff}} = 1.122\,L$ in Euler’s formula:

  $$ P = \frac{{\uppi^{2} EI}}{{L_{\text{eff}}^{2} }}, $$

  where L is the actual length of the tube, P is the load at its lower end and I is the second moment of area of the cross section of the tube. The elastic modulus E of the steel tube is 200 GN/m² and its density is 7850 kg/m³. What is the maximum possible length of the tube if it is not to buckle under its own weight?

  The load P at which buckling occurs is equal to the weight of the tube. For a thin circular tube $I = \uppi R^{3} t$, where R is the radius of the tube and t its thickness. Note that its maximum length is independent of the thickness of the tube.

• 1.7 Run Solver in the spreadsheet ‘Seven-bar Truss’ with the parameters and variables already entered in the spreadsheet to optimize the truss. Compare the coefficient n with that given by Eq. (1.11). Try a few different initial values of D and H to test convergence of the optimization.

  Refer to Sect. 1.4 and the Appendix for information on the use of Solver.

• 1.8 In the spreadsheet ‘Seven-bar Truss’, add a constraint $D = 0$ to the Solver dialog box to create the truss in Fig. 1.10b, and run Solver to verify the result for this truss given in Eq. (1.11). Then, delete the constraint just added (to restore the original truss), add a new constraint to restrict the area of one (or more) of the bars to a suitable minimum value and run Solver again to observe the effect of this on the shape of the truss.

  Refer to the Appendix for information on adding constraints in Solver. Ensure that the minimum area of the chosen bar is larger than its optimum value found in Exercise 1.7. Note that the spreadsheet maintains the symmetric shape of the truss.

• 1.9 Use Solver to find the minimum of the function:

  $$ f(x, y) = x + 2y $$

  subject to constraints:

  $$ x + y - 4 \ge 0, $$
  $$ 2x + y - 4 \ge 0, $$
  $$ - 3x + y + 4 \ge 0\,\text{and} $$
  $$ x - 1 \ge 0. $$

  Which constraints are active at the minimum?

  Use the spreadsheet ‘Seven-bar Truss’ as a guide to setting up the spreadsheet and making the appropriate entries in the Solver dialog box. Examine the values of the four constraints after running Solver to see which are active, i.e. equal to zero.

• 1.10 Make a spreadsheet to optimize the three-bar truss under alternative loading in Sect. 1.1.2. Take load P = 100 kN, span L = 1000 mm and allowable stress (both in tension and compression) $\upsigma_{0} = 300$ N/mm². Compare the result with the minimum volume given in Sect. 1.1.2.

  The stresses in the bars are given by Eqs. (1.3)–(1.5). Define constraints in the Solver dialog box to limit each of these to not greater than the allowable stress. Use Solver with variables $A_{1}$ and $A_{2}$ to minimize the volume of the truss.