Q6 - The fatigue behavior of muscle under alternating stress conditions with zero mean is given by the expression o. Nf cyclic stress, N is the number of cycles to failure and C and a are material constants. Complete the data of the table below. = C, where ơr is the range of %3D O Кра Nr x 103 40 30 25 ? 1 20 500

Q6 - The fatigue behavior of muscle under alternating stress conditions with zero mean is given by the expression o. Nf cyclic stress, N is the number of cycles to failure and C and a are material constants. Complete the data of the table below. = C, where ơr is the range of %3D O Кра Nr x 103 40 30 25 ? 1 20 500

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.2P: 5.2 Evaluate the Prandtl number from the following data: , .

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The S-N curve which represents the fatigue strength, which is measured as a stress amplitude, as a function of number of stress cycles, can be approximated by a straight line in therange of 103cycles to 106cycles when plotted as log(S) versus log(N). Beyond 106cycles,the fatigue strength remains constant. The fatigue strength of a certain alloy at 103cyclesis given as 80 % of the ultimate tensile strength and its endurance limit is 50 % of theultimate tensile strength.(a) The alloy part fails in 15 000 cycles. Calculate the percent change (increase or decrease)of the maximum applied cyclic stress amplitude if the life of the part needs to beextended by 3.5 times. Is this considered under a high or low cycle fatigue failure?State any assumptions made. (12 marks)(b) If the applied cyclic stress amplitude remains unchanged when the life is extended by3.5 times, calculate the percent change (increase or decrease) of the ultimate tensilestrength of a new alloy required. (8 marks)(c) If the…
a . Sketch stress strain curve if the result shown in table represent the force and extension happened in steel, and show Mechanical properties that we get from tensile test on curve? (10p)Note : Lo=80 mm , Do=10 mm , use excel to plot the curve ExtensionLoad(mm) (N)0 0.900.83 4694.341.67 4831.412.50 4781.083.33 4918.834.17 4926.585.00 5257.075.83 5437.016.66 5575.888.33 5775.189.16 5847.5210.83 5965.4111.67 6010.5312.50 6042.5713.33 6072.2614.16 6092.9315.00 6113.2416.67 6140.3617.50 6146.3718.33 6148.1419.16 6149.1725.00 5940.2125.83 5675.3326.67 4725.52b. What is meant by modulus of rigidity? if it increases what does happen to material? (2p)
The stress-strain diagram of a reinforcement steel having a cross-sectional diameter of 12 mmdiameter and 100 mm gage length is determined after its tensile strength test as follows. Based on the stressstrain diagram determine the followings properties of the material (Poisson’s ratio of the material is 0.32) : a) Modulus of elasticityb) Yield strengthc) Toughnessd) Resiliencee) Shear modulusf) Bulk modulusg) Ductility as described bypercent change in length
A high-impulse universal testing machine is being used to determine the Young's Modulus of a rectangular solid specimen (l = 20 cm, t = 1.5 cm, w = 2 cm). The machine produces 35,000 Ns that causes the impact force of compression resulted to a 1 mm deformation in the length of the specimen in 0.05 s. Determine the Young's modulus of the specimen expressed in MPa. Round off your answer to the nearest whole number
Data: Tensile Strength b (mm) L (mm) d (mm) P (N) Tensile Strength (MPa) Average (MPa) 49.74 52.14 10.14 760 49.74 52.14 10.14 980 51.18 49 8.74 800 48.56 48.86 10.96 960 49 51.54 9.72 780
please show a step by step solution, thank you! What is the whal factor of a spring if it has the characteristics average diameter 200 mm, wire spring diameter 30mm, modulus of rigidity is 70 GPa and there is 85 turns in the spring. Answer in 4 decimal places
In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph.
In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph. Answer: breaking stress,true stress at fracture and limit of proportionality.
Sintered silicon nitride parts are expected to experience tensile stresses 50% of the ultimate tensile strength of the material. What will be the maximum allowable size of internal flaws for these parts not to fail in service? Given the following properties of the material: Specific surface energy: 0.3 J/m2 its Youngs Modulus: 304 GPa Ultimate tensile Strength: 570 MPa Select one: a. 1.43 µm b. 0.71 µm c. 15 µm d. 85 µm e. 9.5 µm
A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.
In a forming process the diameter of the specimen is 1 cm and its length changes from 50 mm to 60 mm. Show the maximum force value on the graph of Force vs. True Strain plotted in Excel Program or any plotting program not by hand using the flow stress equation given below. σ= Kεn and σ=F/A Where the parameters are σ : True Stress in MPa K: Strength Coefficient, K=210 in MPa ε : True Strain, ε=ln (l/l0) n: Strain Hardening Exponent, n=0.27 F: Force in 10 A : Area in mm2 Take at least 3 digits after the decimal point. Specify the units. You can use Excel or any plotting program not by hand then paste your graph into the solution section.
Civil engineers often use the straight-line equation, = b 0 + b 1x, to model the relationship between the shear strength y of masonry joints and precompression stress, x. To test this theory, a series of stress tests were performed on solid bricks arranged in triplets and joined with mortar. The precompression stress was varied for each triplet and the ultimate shear load just before failure (called the shear strength) was recorded. The stress results for n = 7 resulted in a Coefficient of Determination of 0.8436. Given that r 2= 0.8436, give a practical interpretation of r 2, the coefficient of determination for the least squares model. a. We expect to predict the shear strength of a triplet test to within about 0.8436 tons of its true value. b. About 84.36% of the total variation in the sample of y-values can be explained by (or attributed to) the linear relationship between shear strength and precompression stress. c. In repeated sampling, approximately 84.36% of…