During a tensile test on a specimen the following results were obtained: Load (kN) 15 30 40 50 55 60 65 Extension (mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 Load (kN) 70 75 80 82 80 70 Extension (mm) 5.08 7.62 12.7 16.0 19.05 22.9 Diameter of gauge length - 19 mm Gauge length = 100 mm Diameter at fracture = 16.49 mm Gauge length at fracture = 121 mm Plot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine: (a) the modulus of elasticity; (d) the nominal stress at fracture; (b) the percentage elongation; (e) the actual stress at fracture; (c) the percentage reduction in area; (f) the tensile strength. !116 GN/m2; 21%; 24.7%; 247 MN/m2; 328 MN/m2; 289 MN/m2.] 1.10 Figure 1.24 shows a special spanner used to tighten screwed components. A torque is applied at the tommybar and is transmitted to the pins which engage into holes located into the end of a screwed component. (a) Using the data given in Fig. 1.24 calculate: (i) the diameter D of the shank if the shear stress is not to exceed 50 N/mm 2, (ii) the stress due to bending in the tommy-bar, (iii) the shear stress in the pins. (b) Why is the tommy-bar a preferred method of applying the torque? [C.G.] [9.14mm; 254.6 MN/m2; 39.8 MN/m2.]   Can you please do a full solution to this question please.   If you use * canyou please explain what you mean by it.