Optimization of Beams

Abstract

Beam optimization involves both the design of the cross section and the distribution of material along its length. For a relatively solid beam, the necessary cross-sectional area under a given bending moment at any section is determined largely by its permitted height and width. At lower load, when the cross section is relatively thin, a beam becomes liable to buckling under compressive and shear stress. A spreadsheet program is made for an I-section beam loaded in bending and shear under stress, buckling and stiffness constraints. For a geometrically similar family of beams, the cross-sectional area can be related to the bending moment by a non-dimensional coefficient depending only on the chosen shape ratios of the cross section. Geometric similarity is adopted in a spreadsheet program for the optimum distribution of material along the length of a beam, with a finite element analysis for the bending moment distribution. Provided some degree of yielding is permitted, the capacity of a beam is not exhausted when the elastic limit is reached at some point. With increasing load, yield hinges are formed at points along the span, leading to the ultimate collapse of the beam. This is the classic problem of limit design.

Exercises

• 6.1 Calculate the coefficient $n_{h}$ in Eq. (6.2) for an I-section beam with a maximum height and width of 100 mm, under a bending moment of $50 \times 10^{6}$ Nmm. The allowable stress of the material is 500 N/mm². The beam has equal thickness web and flanges.

  Calculate the required thickness so that the maximum stress is equal to the allowable stress. Calculate the cross-sectional area and substitute A, M, $\upsigma_{0}$ and $h_{\hbox{max} }$ in Eq. (6.2). Compare the value of $n_{h}$ with that of a solid rectangular section.

• 6.2 Calculate the coefficients C and $n_{g}$ for a solid circular section bar, and for a hollow circular section with inner diameter equal to one-half of the outer diameter.

  To calculate coefficients C and $n_{g}$ for the solid and hollow circular bars, first choose an arbitrary diameter and calculate A, I and coefficient C. Then for some chosen allowable stress $\upsigma_{0}$, calculate the corresponding maximum bending moment M and coefficient $n_{g}$. Compare the values of $n_{g}$ with those in Table 6.1.

• 6.3 Derive a formula for the minimum volume V of a beam, simply supported at each end, with geometrically similar cross section under loads $P/2$ at one-third span and two-thirds span.

  Calculate the bending moment distribution, and use Eq. (6.5) for the minimum cross-sectional area at any section. Compare the volume with that of a beam loaded by a load P at mid-span in Sect. 6.2.1.

• 6.4 Derive a formula for the yield moment $M_{\text{y}}$ of a hollow, square-section beam of side a and thickness $a/4$, with yield stress $\upsigma_{\text{y}}$. Take $\upvarepsilon_{\hbox{max} } /\upvarepsilon_{\text{e}} = 3$.

  Follow the method in Sect. 6.3.1 . There are now three integrals to evaluate for $M_{\text{y}}$ . Compare the result with the maximum elastic moment for the same beam.

• 6.5 Repeat Example 6.2 for a beam clamped at both ends, under the same loading.

  The two possible collapse modes now have three yield hinges. Modify the formulae in Example 6.2 for the additional yield hinge. With these as constraints, and variables the yield moments $M_{1} ,M_{2} ,M_{3}$ , set up a spreadsheet to use Solver to minimize the volume of the beam. Compare the volume with that of a uniform beam, clamped at both ends.

• 6.6 Use the spreadsheet ‘I-section Beam’ to show the effect of a required minimum bending stiffness $EI_{\hbox{min} }$ on the minimum cross-sectional area of the beam, with minimum flange widths $b_{1\hbox{min} } = b_{2\hbox{min} } = 25\;{\text{mm}}$, under the loading and with the material properties given on the spreadsheet.

  Take a range of $EI_{\min}$ from $20 \times 10^9\,to\,500 \times 10^9\;{\text{Nmm}}^2$ . Use the values on the spreadsheet for other maximum and minimum dimensions. Observe the margins of safety and the optimized dimensions of the cross section at different values of $EI_{\min}$.

• 6.7 Use the spreadsheet ‘I-section Beam’ to show the effect of a required minimum flange width on the minimum cross-sectional area of the beam, under the loading and with the material properties given on the spreadsheet.

  Take a range of minimum flange widths $b_{1\min} = b_{2\min}$ from 20 to 50 mm. Set ${EI}_{\min} = 0$ (to avoid influencing the results). Use the values on the spreadsheet for other maximum and minimum dimensions. Plot the cross-sectional dimensions against minimum flange width, indicating on the plot the ranges over which different constraints are critical.

• 6.8 Use the spreadsheet ‘Beam under Lateral Load’ to find the optimum stiffness distribution for a beam of length 1200 mm, clamped at both ends, with:

  • (a) a load of 4800 N applied at mid-span,

  • (b) a load of 4800 N applied at quarter span,

  • (c) loads of 3600 and 1200 N applied at quarter span and three-quarter span.

    Compare the stiffness distribution and the minimum weight of the beam in each case.

    Use the parameters initially present in the spreadsheet. Modify the constraints for a beam clamped at both ends. Place the load(s) at the appropriate node(s) in each case. Choose an initial second moment of area I = 10,000 mm⁴ for all elements. Use Solver to minimize the weight W. Compare the weight of the stepped beam with that of the tapered beam after recalculation. Plot the bending stiffness to locate points at which it reduces to zero. Note how the beam is reduced to a statically determinate one in each case.

• 6.9 Use the spreadsheet ‘Beam under Lateral Load’ to optimize a beam of length 1200 mm, simply supported at each end and at a third support at mid-span, carrying a load of 4800 N uniformly distributed along the span.

  Apply loads of 200 N at all unsupported nodes to represent the uniformly distributed load. Modify the constraints as necessary to represent the proper conditions of support. Choose an initial second moment of area I = 2000 mm⁴ for all elements. Use Solver to minimize the weight W.

• 6.10 Repeat exercise 6.9 by modelling only one-half of the symmetric beam. Examine any difference in the two results.

  Reduce the length of the beam to 600 mm for the half-beam, with loads reduced to 100 N at each node to represent the same uniformly distributed load. Change the support at the node on the plane of symmetry to displacement v and rotation θ both constrained, to represent the condition of the symmetric beam at mid-span. Choose an initial second moment of area I = 2000 mm⁴ for all elements. The weight of the whole beam is now twice that given in the spreadsheet. Compare the results with those in exercise 6.9, now with in effect twice as many elements for the whole beam, to indicate the accuracy of the finite element model (note that we are using a uniform-stiffness beam element here to model the tapered beam).