Advanced Strength of Materials Laboratory ReportStructural Engineering, Laboratory Report Advanced Strength of Materials (DESN40113) Submission Deadline – 24 / 03 / 2011 Martin Smithurst N 0201171, MSc in structural Engineering with Materials 2010 Martin Smithurst N 0201171 1 12 / 03 / 2011 Differential movement – Martin Smithurst CAD Martin Smithurst N 0201171 2 12 / 03 / 2011 . Loads will be applied to the centroid of the section at right angles to the section on a plate (Figure 1). Dial gauges are placed at the top of the plate will read the differential vertical movement Y of the plate to the right and left of the section (Figure 2). Apparatus – Martin Smithurst CAD Figure 2. resulting in the beam being fixed. Figure 1. The point at which this equalisation of torsion (both +V plus -V = 0) occurs is the shear centre within limitations of the 20mm increments. SHEAR CENTRE Method Using different sections with thin wall steel welded at each end to the section frame. torsion on the section will be equalised both left and right. The load will be moved in 20mm increments from one side of the section to the other. at this point the components of the force are in the Y axis (Figure 3). When the central plate is horizontal with the load applied.Advanced Strength of Materials Laboratory Report Aim The Aim of the laboratory testing is to demonstrate the practical application of the theoretical calculations through experiment. Equal angle. Differential movement – Martin Smithurst CAD Three sections are to be tested. Sections – Martin Smithurst CAD Picture 1. Sections being tested in lab. Channel & Semicircle dimensions as below (Figure 4). Figure 4.63mm thick. featuring the channel – Martin Smithurst Martin Smithurst N 0201171 3 12 / 03 / 2011 . The sections are made from consistent cold formed sheet steel 1.Advanced Strength of Materials Laboratory Report Figure 3. The theoretical shear centre should be at the centre at position 6 in Figure 6. Martin Smithurst CAD Figure 6 – Equal Angle section. The shear centre remains unaffected by the orientation of the angle because the components of the force change in equal magnitude between the X and Y axis transferring from one to the other (Figure 5).Advanced Strength of Materials Laboratory Report Equal Angle. Theory The equal angle is symmetrical resulting in the shear centre being within the centroid of the section. Martin Smithurst CAD Martin Smithurst N 0201171 4 12 / 03 / 2011 . D) Figure 5 – Equal Angle section. (SEWARD. 72 200 -3. This demonstrates that the theory follows the practice.65 4 Equal Angle Displacement (mm) 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 Left Right -1 -2 -3 -4 -5 Position Number Chart 1 – Equal Angle Results From the table it can be seen that position 6 is where the load is 0.06 0.46 2.71 1.93 40 2.92 3.74 100 0. The 0.4 160 -2.18 3.68 140 -1.56 -0.68 20 2.Advanced Strength of Materials Laboratory Report Practical Application Results Equal Angle (mm) from Right Left Dial reading 0 3.29 -1. Martin Smithurst N 0201171 5 12 / 03 / 2011 .38 Table 1 – Equal Angle test results Load Position 1 2 3 4 5 6 7 8 9 10 11 Right Dial reading -4.81 -0. if this were an infinite sliding scale there would be a point of true equilibrium.47 80 0.04 120 -0.02mm from the equilibrium from left to right about position 6.06 180 -2.97 -3. When this data is plotted into a chart the intersection of the two results is the shear centre (Chart 1).79 -2.18 60 1.02mm can be explained by the fixed 20mm position and the thickness of the steel section. The eccentricity of this is dimension “e” (Figure 7). ( Equation 1 – Channel ) ( ) e = eccentricity h = height or depth of the section b = breadth of the section This puts the theoretical shear centre distance “e” of 18. This theoretical result should result in a shear centre in the practical test somewhere between positions 6 and 7 (Figure 7). Figure 7 – Channel. Martin Smithurst CAD Martin Smithurst N 0201171 6 12 / 03 / 2011 . Distance “e” is calculated using the following equation (Equation 1).75mm outside the channel section.Advanced Strength of Materials Laboratory Report Channel Theory As this is non-symmetrical section the shear centre will fall out of the centre line of the section. Advanced Strength of Materials Laboratory Report Practical Application Results Channel Left Dial reading 0.02 -0.035 0. When this data is plotted the intersection of the two dial readings is where the load is at its greatest Y = maximum component and X = 0 being the shear centre of the beam.185 Load Position 1 2 3 4 5 6 7 8 9 10 11 (mm) from Right 0 20 40 60 80 100 120 140 160 180 200 Right Dial reading -0.14 -0.23 -0. This demonstrates that the theory follows the practice.15 -0.1 0. Martin Smithurst N 0201171 7 12 / 03 / 2011 .14 0.08 0.035 -0.08 0.075 -0.015 -0.175 Table 2 – Channel test results 0.2 -0.095 -0. In the graph it is seen that the intersection falls below the 0 line as the whole section is deflected by this displacement.2 Channel Displacement (mm) 0.05 0.1 0 1 2 3 4 5 6 7 8 9 10 11 Left Right -0.125 0.19 -0.3 Position Number Chart 2 – Channel Results As predicted the point of equilibrium falls outside of the section between positions 6 and 7.045 -0.115 -0.1 -0.28 -0.005 0. H.G) ( ) ( ) Equation 2 – Semicircle e = eccentricity r = radius This puts the theoretical shear centre distance “e” at 13. Martin Smithurst CAD Martin Smithurst N 0201171 8 12 / 03 / 2011 . This theoretical outcome should result in a shear centre in the practical test somewhere between again positions 6 and 7 (Figure 8) Figure 8 – Semicircle. T. The eccentricity of this is dimension “e” (Figure 8). Distance “e” is calculated using the following equation (Equation 2).Advanced Strength of Materials Laboratory Report Semicircle. (MEGSON.66mm outside the Semicircle section. Theory Again this is non-symmetrical section the shear centre will fall out of the centre line of the section. 615 160 -0.8 Displacement (mm) 0.7 -0.51 60 0.21 100 0.15 0.09 -1.5 200 -0. Martin Smithurst N 0201171 9 12 / 03 / 2011 .2 1 2 3 4 5 6 7 8 9 10 11 -0.035 0. This demonstrates that the theory follows the practice.23 0.26 180 -0.3 Left Right -0. The intersection falls below the 0 line as the whole section is deflected by this displacement. When this data is plotted the intersection of the two dial readings is where the load is at its greatest Y component and X = 0 being the shear centre of the section.Advanced Strength of Materials Laboratory Report Practical Application Results Load Position 1 2 3 4 5 6 7 8 9 10 11 Semicircle (mm) from Right Left Dial reading 0 0.52 -0.37 80 0.41 0.585 0.2 Position Number Chart 3 – Channel Results Again as predicted the point of equilibrium falls outside of the Semicircle section between positions 6 and 7.7 -1.66 40 0.77 Semicircle 0.8 20 0.475 140 -0.33 -0.07 -0.06 120 -0.65 Table 3 – Channel test results Right Dial reading -1. Martin Smithurst CAD. with an average section of 9.” (http://en. 600mm in length.39mm. March 2011) Method The intention is to demonstrate the changing moment of inertia and the states of unsymmetrical bending from a maximum to a minimum by orientation of the section. Therefore.65mm x 19. it encompasses not just how much mass the object has overall. Picture 2. Using a solid steel section bar. Figure 9 – Unsymmetrical bending rig.org/wiki/Moment_of_inertia. but how far each bit of mass is from the axis. and a load of 20N applied (Figure 9).wikipedia. unsymmetrical being tested in lab– Martin Smithurst Martin Smithurst N 0201171 10 12 / 03 / 2011 .Advanced Strength of Materials Laboratory Report UNSYMMETRICAL BENDING This is an investigation into the moment of inertia described as: “The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis. The change in moment of inertia will alter the deflection of the bar given a constant Young's modulus (the stiffness of the material). The components X and Y deflection can be recorded as the angle of rotation is changed (Figure 10).H. Martin Smithurst CAD. T. Theory The moment of inertia I for a square section is calculated by (Equation 3) (MEGSON. this changing orientation of the section will change the moment of inertia I in the X and Y axis.G) Equation 3 – moment of inertia Ix = moment of inertia about axis X Iy = moment of inertia about axis Y b = breadth of section d = depth of the section The components of the load are given by the following (equation 4) (MEGSON. Figure 10 – Unsymmetrical bending rig.G) Martin Smithurst N 0201171 11 12 / 03 / 2011 .H. The bar is rotated though 105º of rotation from 45º to 150º. T.Advanced Strength of Materials Laboratory Report This will demonstrate the various states of unsymmetrical bending through the symmetrical state. 866 Sin Ɵ 0.1422 5802.835553 1.590914 0.866 0.42 5860.5 0. constant (N/mm²) Theoretical Results Theory bd³/12 Angle (°) 45 60 75 90 105 120 135 150 E = 210 x 10E3 210000 210000 210000 210000 210000 210000 210000 210000 L³ = Length³ 216000000 216000000 216000000 216000000 216000000 216000000 216000000 216000000 W= Weight (N) 20 20 20 20 20 20 20 20 Ix 5860.1422 5802.966 0.013285 1.585037 Gauge 2 0.966 1 0.42 5860.259 0 0.42 5860.42 5860.707 0.5 Gauge 1 0.590914 0.42 5860.42 bd³/12 Iy 5802.023464 Table 4 – Theoretical results Martin Smithurst N 0201171 12 12 / 03 / 2011 .306093 0 0.1422 5802.835553 0.013285 0.1422 Cos Ɵ 0.866 0.1422 5802.8272439 0.827243 1.1422 5802.42 5860.1422 5802.130292 1.1422 5802.5 0.259 0.707 0.Advanced Strength of Materials Laboratory Report Equation 4 – Deflection of the beam dx = deflection about axis X (mm) dy = deflection about axis Y (mm) W = Weight applied (N) L = Effective length (mm) = Angle of rotation (degrees) E = Module of elasticity.707 0.130292 1.306093 0.42 5860.707 0.170075 1. 6 0. laboratory results Martin Smithurst N 0201171 13 12 / 03 / 2011 .24 Laboratory results Deflection at free end (mm) 5 4 3 2 1 0 -1 40 60 80 100 120 Angle of rotation 140 160 Gauge 1 Gauge 2 Chart 4 – unsymmetrical bending.21 2.Advanced Strength of Materials Laboratory Report Theoretical results Deflection at free end (mm) 1.58 2.92 2.1 0.25 19.62 Unloaded 22.08 21.82 21.985 3.05 24.1 1.26 22. theory results Gauge 1 Gauge 2 Practical Application Results Actual Angle (°) unloaded 45 60 75 90 105 120 135 150 21.28 19.28 3.22 21.175 3.34 Actual Y 3.72 17.2 0 40 60 80 100 120 140 160 Angle of rotation Chart 4 – unsymmetrical bending.2 1 0.58 BOTTOM (Y) Loaded 26.11 0.8 0.01 20.91 17.705 21.11 21.46 21.04 TOP (X) Loaded 20.7 16.86 25.51 16.42 Actual X 1.73 20.19 21.72 4.11 1.52 20.88 3.22 3.28 22.08 4.63 21.38 21.905 17.78 20.92 21.13 22.29 16.4 1.14 25.4 0. 4 0. Laboratory results Deflection at free end (mm) 1.8 0.Advanced Strength of Materials Laboratory Report The colouration of the graphs denotes a similar trend showing the moment of inertia transforming as the beam is rotated.2 0 40 140 240 340 Angle of rotation Guage 1 Guage 2 Figure 11 – Full rotation of the section Martin Smithurst N 0201171 14 12 / 03 / 2011 .2 1 0.4 1. Taking this to 315° of rotation as shown below.6 0. this demonstrates that with consistent dimensions there will be maximum inertia of the section twice in rotation for a rectangle section. March 2011) The work submitted is mine alone and not the product of plagiarism. Second Edition. Hampshire. T. Palgrave WIKIPEDIA (http://en. Second edition. as defined by the regulations of the University. Understanding Structures.H.org/wiki/Moment_of_inertia. Structural and Stress Analysis. Oxford SEWARD. Basingstoke. collusion or other academic irregularity.G 2005. Martin Smithurst ………………………………………………….19/03/2011 Martin Smithurst N 0201171 15 12 / 03 / 2011 . Butterworth Heinemann.Advanced Strength of Materials Laboratory Report References: MEGSON. D 1998.wikipedia. Publishing.