The ratio of stress to bulk modulus for compression of the material. Select the correct response. Dilatation Modular ratio Bulk's ratio Deformation < Previous
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- Table Q4 belowindicates results on a tensile test of a mild steel specimen. 95 kN Proportional Limit load Original diameter Original ga uge length 30 mm 50 mm Yield Point Load 100 kN Extension at proportional limit 0.050 mm Maximum recorded load 160 kN Final length between gauge points 58.8 mm Final minimum diameter 17.9 mm Table Q4 Calculate the following: a. Modulus of Elasticity for the mild steel. b. The proportional limit stress. C. The ultimate tensile strength. d. The percentage elongation. e. The percentage reduction in area.Define the rigid body used in statics. According to this; rigidity of bolt-nut connected elementBased on the value, the bolt diameter is 20 mm and the safe value of the bolt material.Calculate the stiffness values of the bolt and the intermediate element plate. (Note: Bolt's9x10 ^ 2 times as stress value ??? / cm ^ 2take as.). Strain gaugeElongation in bolt length measured with 250 ??, shortening amount in intermediate element platesIt was measured at 125 ??.The load on a helical spring is 1282 lb and the corresponding deflection is to be 3.4 inches. Rigidity modulus is 11 x 106 psi and the maximum intensity of safe torsional stress is 60,000 psi. If the wire diameter and the mean diameter are 0.5 in. and 3 in., respectively. Determine the number of active coils. Don't round your answer to preferred size. Round your answer to 4 significant figures
- diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the fracture 2.00 in. The data is listed in the table. Plot the stress-strain 8-1. A tension test was performed on a steel specimen n original diameter of 0.503 in. and gage length of PROBLEMS *84. origi the f having an for t and stress. Use a scale of 1 in. Dodraw the elastic region, using the same stress scale but a 20 ksi and 1 in. = 0.05 in./in. strain scale of 1 in.= 0.001 in./in. Load (kip) Elongation (in.) 0. 0. 0.0005 0.0015 1.50 4.60 8.00 11.00 0.0025 0.0035 0.0050 11.80 11.80 0.0080 0.0200 12.00 16.60 0.0400 0.1000 0.2800 20.00 21.50 19.50 18.50 0.4000 0.4600 Prob. 8-1of k3 flnd ) the glabal stiffness gmatrix.Mild steel 1 Young;s modulus 1219.5 Yield strain and stress (0.4101,500.08) Failure stress and strain :not able to find because the given data shows the experiment did not reach the failure point. if the material stress and strain does not reach a failure point ,what dose it means , does it means that the material is more stronger?
- Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…The proof stress of a material under tensile loading can be considered as can be considered as equivalent to the ultimate tensile strength for aluminium alloys being only applied to engineering steels during a tensile test the intercept on the stress stain curve for a given strain where a line draw is parallel to the elastic response line. the stress at yield for materials have no defined yield stressAfter recording the applied twisting moment and resulted angle, fill the following table MT (N.m) T. MPa (degree) (rad.) Now, the twisting moment-twisting angle and shear stress-shear strain curves can be plotted, then determine maximum shear stress, shear stress at proportional limit and modulus of rigidity. 4-6 Problem The following torsion test data were obtained for AA6061-T6 aluminum alloy has a round cross section with 30mm outer diameter, 0 inner diameter and 100 length. 0 32 130 Twisted angle (): 0 1 Torque (N.m): 286 347 487 5.5 591 786 910 1105 1163 1235 1222 3.5 6.5 7.5 9. 10.5 13.5 16.5 21.5 25.5 > Plot the torque-twisted angle curve and determine the proportional limit on it. Plot the shear stress-strain curve the determine the shearing strength and modulus of rigidity. Prof. Adnan N. Abood Asst. Lec. Zainah Waheed 25
- When a bar of 22 mm diameter is subjected to an axial pull of 60 kN the extension on the 49 mm gauge length is 0.1 mm and there is a decrease in diameter of 0.013 mm. Calculate the Young's Modulus. Provide your answer in GN/m2 to the nearest whole number.Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. What tension force is required to cause yielding in a bar of the same material but with a diameter of 20 mm? P, kN AL mm 7.22 0.0839 14.34 0.1636 21.06 26.8 31.7 34.1 35.0 0.241 0.308 0.380 0.484 0.614 36.0 0.924 36.5 1.279 36.9 37.2 1.622 1.994b) Determine the maximum gas pressure and the dimensions of cross -section of the connecting rod for a diesel engine with the following data: (i) Cylinder bore 120 mm and the Length of connecting rod = 360mm (ii) The critical buckling load = 407.1504 KN; (iii) The compressive yield stress =330 N/mm? ; Factor of safety = 6 %3D %3D