1. Supplement To Chapter 1 on OVERVIEW OF ENGINEERING ANALYSIS With additional information on Four-Stage of Engineering Analysis by Tai-Ran Hsu, Professor Department of Mechanical Engineering San Jose State University San Jose, California, U.S.A. September 5, 2016 (Bridge design-Ch1 supplement/ME 130)

2. The Four-Stage of Engineering Analysis - Example on the design of a bridge Stage 1:Identification of the problem: ● Application: To design a bridge over a narrow creek for limited traffic ● Distance of crossing: The length of the bridge is 20’ ● The width of the bridge is 12’ (6’ in the printed notes) ● Maximum designed load is 20 tons (or 40,000 lb f ) ● Steel is chosen as the load-carrying structural material with: Maximum tensile strength for steel = 75,000 psi (from handbook) Maximum shear strength for steel = 25,000 psi (from handbook) ● The induced deflection of the structure is not critical in this case

3. Stage 2:Mathematical idealization: ● The design problem neither specifies the GEOMETRY nor the DIMENSIONS of the structure that is required to carry the load of passing traffic up to 20 tons. ● But we realize the principal load that the bridge structure needs to carry is in the form of BENDING. ● The most commonly used structures for bending-load carrier are I-BEAMS, with cross-section of the geometry: ● ASSUME use two I-beams to carry the passing traffic up to 20 tons (for a width 12’) ● The size of the beams remain unknown ● We need to ASSUME an overall size of each of the I-beam by a dimension of H 1, to be: H 1 =12 inches Once H 1 = 12” is assumed, the other dimensions are supplied by steel beam manufacturer to be: H 2 = 10.92”, b 1 = 5” and b 2 = 0.35” following their standards ● We ASSUME that each of these I-beam with 20’ long carries HALF of the total load of the passing traffic, i.e. 10 tons (or 20000 lb f ) ● We further ASSUME that the passing traffic is a small truck with four wheels each wheel carries ¼ of the total load of 20 tons (i.e. 10000 lb f ) ● The bending load will induce a bending stress in the beam. The maximum induced bending stress must be kept below the Maximum tensile strength for steel = 75,000 psi (from handbook) ● Maximum induced shear stress should also be kept below the maximum shear strength of the material. But the induced shear stress is normally much below the bending stress. H1H1 H2H2 b1b1 b2b2 c

4. Stage 3:Mathematical analysis: ● We may thus set up a math model that we may handle the analysis with the available simple beam theory learned from “strength of materials” class with expressions for induced stresses from the textbook or by available hand books. ● The physical situation for math analysis thus become: From the handbook, the maximum bending stress is: where M = maximum bending moment, c = half the depth of beam cross section I = area moment of inertia One may readily calculate M = Pa = 960,000 in-lb; c = 6”; I = 215.4 in 4 ; leading to: The maximum bending stress:  n,max = 26740 psi at the middle span

5. Stage 4:Interpretation of results: ● A critical question to ask from this type of analysis on structure is: “Would the selected I-beams be strong enough to carry the maximum expected load of 20 tons?” ● Alternatively the question is: “Would the bridge with the selected I-beams e SAFE enough for the designed maximum load of 20 tons?” ● It is therefore for the engineer (analyst) to translate whether the maximum bending stress of 26,740 psi obtained from the math analysis can answer the above questions! ● An intuitive way of translating this stress into physical meanings is to compare it’s magnitude with the Maximum tensile strength for steel = 75,000 psi (from handbook). ● In reality, it is NOT PRACTICAL to do so because we have made many UNREALISTIC IDEALIZATIONS in Stage 2 of the analysis (can you identify the obvious ones we did in the analysis?) ● Consequently, we normally use the “ALLOWABLE STRENGTH” instead of “MAXIMUM TENSILE STRENGTH” of materials for this purpose. ● The ALLOWABLE STRENGTH of materials, σ a is defined as: ● The safety factor (SF) is established depending on the nature and application of the structure

6. Stage 4:Interpretation of results – cont’d: with our maximum bending stress = 26740 psi, and UTS of the steel I-beams = 75000 psi, We get the SF = 75000/26740 = 2.8. However, we need to review the IDEALIZATIONS that we have made in Stage 2 to see if they are realistic NOTE: For analyses that are of NON-STRUCTURAL NATURES, Stage 4 is normally based on the satisfaction of the set DESIGN CRITERIA as illustrated in Figure 1.4 in the bonded lecture notes.

7. About Safety Factor (SF) in Engineering Analysis on products involving structures: SF can be viewed as a DISCOUNT on how much material strength engineers can use in his (her) design analysis. The higher the SF, the less material strength that is used in the design. The value of adopted SF is usually associated with the degree of SOPHYSTICATION of the engineering analysis. Sophistication of analysis ties in with how realistic the engineer made in the Stage 2 in his (her) analysis. The following Chapter-end problem illustrates how un-sophisticated analysis we made in the above example on bridge design. In the Stage 2 of the analysis, we assumed that: (1) The steel I-beams are weightless, and (2) The weight of the concrete pavement was not accounted for in the analysis. Are these hypotheses REALISTIC?

8. Problem 1.9 Conduct an engineering analysis on the example of a bridge structure but include the weight of the steel structure plus the weight of the concrete road surface, 20 feet long x 12 feet wide and 6 inches thick. Solution: We will use Simple beam bending theory for the solution again. (1)The induced bending stress in I-beam by its own weight with mass density of 0.28 lb/in 3 : The weight of the beam structures is considered as the UNIFORM DISTRIBUTED LOADS/unit length in the formulation. We will thus have the math model of bending of beams subject to its own weight shown in the diagram below: Where W s = the weight per unit length of the beam obtained by the expression: Total volume: W s = 2213.28x0.28 = 619.72 lb f.Total weight: Weight per unit length of the beam: w s =619.72/(20x12)= 2.5822 ln/in

9. The maximum bending stress induced in the beam by the uniformly distributed load may be obtained by simple beam theory with formula given in Mark’s ME handbook to be: where and C = half depth of the beam = 6” and I = section moment of inertia = 215.4 in 4, resulting in additional maximum bending stress at the mid-span

10. (2) Induced maximum bending stress by the weight of the concrete pavement with a mass density of ρ c =133 lb f /ft 3 from the same handbook: We assume that each I-beam carries half of the total weight of the concrete pavement (a reasonable assumption) Following the same procedure as we di with the case of including the weight of the steel I-beam, We have the following math model for the current problem: Leading to maximum induced bending stress by the weight of the concrete pavement to be:

11. Interpolation of the analytical results: Design Loading to the Bridge: Gravity loads of structures and loading of passing traffic Maximum bending stress by the passing truck load: σ max-L = 26740 psi (previously computed) Maximum bending stress by the weight of the I-beam: Maximum bending stress by the weight of the pavement: The total maximum bending stress in each I-beam is thus equal to: We realize that the two additional gravity loading of the weight of the I-beam and that of the concrete pavement have increased the total maximum bending stress by a whopping 26.78% And the SF is reduced to: 75000/33902 = 2.21 from 2.8. < uts=75,000 psi SAVE But……