Structural Analysis - Assignment Example

STRUCTURAL ANALYSIS By Structural Analysis Design Questions a. Structural steel beam is simply supported and has a span length of 6m; as shown in Figure 1. The beam is under perpendicular loadings and also carries a uniformly distributed load of 0.5kN/m as shown below. Calculate and analyse, using appropriate methods, the structural state of this beam and find out its weakest points. Use annotations and justifications throughout. For the member shown above, we first calculate the unknown forces, RA and RB. Since the member is at equilibrium, the forces must balance, i.e., algebraic summation of forces must add up to zero. The Sum of upward forces equals the sum of downward forces. We express the summation of forces as follows: ∑F =0 Therefore, RA +RB – 25-15-10-10-50-0.5x2.5 =0 RA +RB – 111.25 =0 RA +RB – 111.25 =0 RA +RB = 111.25 We then moment about the point of action of RB. Again, since the member is at equilibrium, the algebraic summation of moments at any point is zero. ∑MB =0 The uniformly distributed force, 0.5kN, is equivalent to a force of 0.5 x 2.5kN acting at a distance of 1.75m [(1.25+0.5) m] from point B. 4x RA -23x 3 - 0.5 x 2.5 x 1.75- 15 x 2 -10 x 0.5 + 10 x 0.5 + 50 x 1.5 = 0 4 RA = 32. 1875kN RA ≈ 8kN, RB = 111.25 – 8 = 103.25kN We can, therefore, calculate the Share forces at different points of the member. We start from the far most left side of the member. At x = 0 m => Shear force = 8kN At x = 1 m => Shear force = 8- 25 = - 1.75kN At x = 2 m => Shear force = 8-25- 0.5 x 1 = - 17.5kN At x = 2 m => Shear force = 8 -25 -0.5x 1 – 15 = -32. 5kN At x = 3.5 m => Shear force = 8-25-15 -0.5 x 2.5 = -33.25kN At x = 3.5 m => Shear force = - 33.25 -10 = - 43.25kN At x = 4 m => Shear force = -43.25 + 103.25 = 60kN At x = 4.5 m => Shear force = 60 -10 = 50kN At x = 5.5 m => Shear force = 50-50 = 0kN We therefore calculate bending moments, At x = 0 m => Bending Moment= 0 At x = 1m => Bending Moment= 8 x 1 = 8kNM At x = 2m => Bending Moment= 8 x 2 -25x 1- 0.5 x1 x0.5 = -9.25kNM At x = 3.5m => Bending Moment= 8 x3.5- 25x2.5–0.5 x2.5x 1.25 – 15 x1.5 = -58.5625kNM At x = 4m => Bending Moment=8 x4 -25 x 3-15x2 -10 x 0.5-0.5 x 2.5x1.75 = -80. 1875kNM At x = 4.5 m => Bending Moment= 8 x 4.5-25x3.5-15x2.5-10x1+10x1-+103.25x0.5-0.5x2.5x2.25 = -50.1875kNM At x = 5.5m => Bending Moment= 8x5.5 -25 x4.5 – 15 x3.5 – 10 x 2 -0.5 x 2.5 x 3.25 + 103.25 x 1.5 – 10 x 1 = - 0.1875kNM At x = 6m => Bending Moment= 0 From the shear force and bending moment diagram shown in the next page, it can be seen that the maximum bending moment experienced by the beam is -80.1875KNM. The weakest point of the beam is at x-4m. The beam is likely to break at this point. Shear Force and Bending Moment diagram. b. Calculate the bending stress using the “I” cross section provided as shown in Figure 2, annotate your method and explain the stress distribution with a diagram. Solution We first calculate the section’s inertia. The section can be divided into three parts as shown in the next page. XX shows the section’s centroid. Area A = 100 x 15 = 1500mm2 Area B = 20 x 120 = 2400mm2 Area C = 50 x 15 = 750mm2 IA = 5 mm4 IA = 880000mm4 IC = mm4 Area x A.x A. x2 I I+ A. x2 Area A 1500 142.5 213750 30459375 28125 30487500 Area B 2400 75 180000 13500000 2880000 16380000 Area C 750 7.5 5625 42187.5 14062.5 56250 Total 4650 399375 46923750 Centroid = Centroid = = 85.89mm I = 46923750 – 4650 x 85.892 = 12620271.74mm4 = 12620271.74 x 10-12m4 = 1.26 x 10-5 m4 We can now calculate Tension stress and compressive stress. σ ( Compression) = = = 408MPa σ ( Tension) = = = 546.6MPa The diagram below shows stress distribution on the T- section Compression = 408MPa Stress = 0 Tension = 546.6MPa c. Results Analysis There are several beam section which can be chosen for engineering use. However, the choice should be made carefully by considering the beam strength and the total costs. In engineering construction, material’s strength and the cost are the main parameters to be considered when selecting a particular material. Another important aspect in structural engineering the design. Structures should be designed in such a way that no failure occurs in future. For instance, for the I- Section given, there are two possible stresses. Compressive stress and tensional stress. In designing, the least possible stress should be used for safety. From the stress distribution diagram, it can be seen that stress is zero at the beam axis. This therefore, makes an I-Section the best beam for structural engineering. When the beam bends, the top most fiber is in compression while the bottom most fiber is in tension. The stresses are greatest at the top and the bottom fibers. I-section provides the stiffest beam with the least amount of material. This is because it only requires more materials at the bottom and top flanges. The diagram below shows a cross section through an I-section. This shape is used when the load is parallel with the flange. As you can see, the shape is not so good with lateral forces unless you turn it sideways. When the load will come from two directions, a square tube is used. The section is normally designed so as to minimize materials on. This type of section is better than other sections due to the fact the maximum beam stiffness is achieved with minimum material. To ensure safety of the structure, a safety factor of between 4 and 5 is used. Factor of Safety, F.O. S = We therefore choose the steel strength based on this formula For this case, a factor of safety is taken to be 5. Therefore, for Tension, 5 = Maximum Tension Strength = 5 x 546.6 = 2733MPa For Compression, 5 = Maximum Compression strength = 5 x 408 = 2040 MPa References List Nash, W. (1998). Schaums outline of theory and problems of strength of materials. New York: McGraw-Hill. Timoshenko, S. (1955). Strength of materials. New York: Van Nostrand. Timoshenko, S. and Young, D. (1965). Theory of structures. New York: McGraw-Hill. Read More