After fracture, the total length was 53.086 mm and the diameter was 9.982 mm. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity;
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- The (G-E) diagram obtained in the tensile test performed on a metal sample with a diameter of 16 mm is as follows. The loads at points A, B and C and the elongation measured on l. 16 cm gauge length were determined as follows: B A B C Load (kgf) 4800 8400 7200 Elongation (mm) 0.192 28.8 38.4 c) Calculate the fracture work and the maximum elastic energy the metal rod can store. d) Find the cross-sectional area of a 6 m long rod made of this metal such that it can carry 12 tons of load with 2 times the safety of yield strength. How long does the rod extend under this load?The following data were collected from a 0.4-in.-diameter test specimen of polyvinyl chloride (l0 5 2.0 in.): Load (lb) 0 Δl (in.) 0.00000 300 600 900 1200 1500 1660 1600 0.00746 0.01496 0.02374 0.032 0.046 0.070 (maximum load) 0.094 1420 0.12 (fracture) After fracture, the total length was 2.09 in. and the diameter was 0.393 in. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; and (g) the modulus of resilience.During a tensile test on a specimen the following results were obtainedload(KN) 15 30 40 50 55 60 65 70 75 80 82 80 70extension(mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 5.08 7.62 12.7 16 19.05 22.9diameter of gauge length=19mmdiameter at fracture=16.49mmgauge length=100mmgauge lengthat fracture=121mmplot the complete load extension graph and the straight line portion to an enlarged scale .hence determinethe modulus of elasticitythe percentage elongationthe percentage reduction in areathe nominal stress at fracturethe actual stress at fracturethe tensile strength
- A Brinell hardness tester with a diameter of 10 mm and a load of 500 kg gives an impression with a diameter of 1.62 mm in a steel alloy.1-Calculate the Brinell hardness (HB) in the steel alloy. 2-What will be the diameter of the impression in the steel alloy when this has a Brinell hardness of 450 with a load of 500kg.I. Problem Solving: Answer what is being required. Show you complete solution. 1. Compute the compressive stress for the respective ages of the specimen. The diameter of the specimens is 150mm and the height is 300mm. Load (N) 176,000 Stress (MPa) Age 7days 14 days 28 days 252,000 340,000 2. Compute the tensile stress from the value acquired from the experiment if the breaking load is 105KN. The diameter of the specimen is 150mm and the height is 300mm. 3. Compute the flexural stress of the sample specimen if its dimension is 150mm x 150mm x 500mm in a 3 point loading. The effective length is 400mm. The specimen failed at 44,300N in the middle third. 4. Compression test was applied to a certain type of wood. Compute the stress based on the following dimensions. The grains are parallel with the length. а. Dimension (mm) Load (N) Compressive stress (МРа) Compression TestQuestion: A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa sqrt(m) (90 ksi sqrt(in.)) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
- A1 Copy the sketch of the tensile stress-strain test (Figure A1) results for a waisted metallic specimen and indicate the locations representing maximum stress (A), ultimate strain to failure (B), and 0.2% proof (yield) stress (C), and calculate the Young's modulus of the material. Figure A1 300 250 200 150 100 50 0 Stress (MPa) 0.2 0.4 0.6 8.0 Strain (%) 10.0Calculate 6 Calculate the tensile stress when applied force of 120OON on specimen of * diameter of 7mm 111 O 211 O 311 O 411 OAssuming you have been offered a job position as the lead Engineer to supervise the restoration of a steel bridge that collapsed due to fracture. The laboratory results of the wreckage (a truss) given to you show an internal crack whose length is 0.026 mm and radius of curvature is 0.00028mm. If the material theoretical fracture strength is 2000 MPa, determine the allowable applied stress. O a. 145.10 MPa O b. 135.65 MPa OC. 128.17 MPa Od. 156.04 MPa O e. 146.76 MPa
- a. Calculate the ductility of the sample as a percentage (%) b. Calculate the stress required to create a neck in the sample, in MPa.The thin square plate shown is uniformly deformed such that €, = +1445 pE, E, = -674 uE, and y,y = +1260 urad. Determine the normal strain e, in the plate. 60 mm O -157 µE O -309 µE O -198 HE O -244 µE O -256 HEIf the gauge is made of a material having a yield stress equal 200 MPa, determine the minimumrequired diameter d. Using a factor of safety (F.S. = 2)