Pergunta

Question 1- (10 marks) A ceramic ball with a diameter of Smm is pressed into a hemispherical recess of 10mm in diameter in a steel plate. (a) What normal load is necessary to initiate yield in the steel plate; (b) what is the radius of the contact;and (c) at what depth does yield first occur? Assume Herzian contact behavior and the given parameter are: E_(cerumic)=450GPa;E_(nteel)=200GPa; v_(ceramic)=v_(steel)=0.3;H_(ceramic)=20GPa;H_(steel)=5GPa The load for the onset of yield is given by: W= (0.7pi ^3Y^3R^ast 2)/(E^ast 2) assume H-2.8Y Question 2-(10 marks) A mass of m=0.7kg is pulled along a surface using a spring balance. The spring balance is moved with a constant velocity of v=0.1m/s The motion exhibits a clear stick- slip behaviour. The reading of the spring balance varies between 7 N and 14 N. The stiffness of the spring k=1N/mm. a) Calculate the ratio between the static and dynamic coefficient of friction. b) Calculate the fluctuation of stick-slip motion. c) Calculate the frequency of motion. d) Calculate the natural frequency of the mass spring system.
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RicardoMestre · Tutor por 5 anos
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Question 1:<br /><br />(a) To calculate the normal load necessary to initiate yield in the steel plate, we can use the given formula for the load for the onset of yield:<br /><br />$W = \frac{0.7\pi^3Y^3R^{*2}}{E^{*2}}$<br /><br />where:<br />W is the normal load,<br />Y is the yield strength of the steel,<br />R is the radius of the contact, and<br />E is the Young's modulus of the steel.<br /><br />Given that the yield strength of the steel is 5 GPa, the radius of the contact is half of the diameter of the ceramic ball (2.5 mm), and the Young's modulus of the steel is 200 GPa, we can substitute these values into the formula:<br /><br />$W = \frac{0.7\pi^3(5)^3(2.5)^2}{(200)^2} = 0.35 N$<br /><br />Therefore, the normal load necessary to initiate yield in the steel plate is 0.35 N.<br /><br />(b) The radius of the contact can be calculated using the formula:<br /><br />$r = \sqrt{\frac{3W}{2\pi\sigma}}$<br /><br />where:<br />r is the radius of the contact,<br />W is the normal and<br />σ is the yield strength of the steel.<br /><br />Substituting the values of W and σ into the formula, we get:<br /><br />$r = \sqrt{\frac{3(0.35)}{2\pi(5)}} = 0.089 m$<br /><br />Therefore, the radius of the contact is 0.089 m.<br /><br />(c) The depth at which yield first occurs can be calculated using the formula:<br /><br />$d = \frac{W}{2H_{steel}}$<br /><br />where:<br />d is the depth at which yield first occurs,<br />W is the normal load, and<br />H is the hardness of the steel.<br /><br />Substituting the values of W and H into the formula, we get:<br /><br />$d = \frac{0.35}{2(5)} = 0.035 m$<br /><br />Therefore, yield first occurs at a depth of 0.035 m.<br /><br />Question 2:<br /><br />(a) The static coefficient of friction (μs) can be calculated using the maximum static friction force (Fmax) and the normal force (N):<br /><br />$μ_s = \frac{F_{max}}{N}$<br /><br />Given that the maximum static friction force is 7 N and the normal force is 14 N, we can substitute these values into the formula:<br /><br />$μ_s = \frac{7}{14} = 0.5$<br /><br />Therefore, the static coefficient of friction is 0.5.<br /><br />(b) The fluctuation of stick-slip motion can be calculated using the difference between the maximum and minimum static friction forces:<br /><br />$ΔF = F_{max} - F_{min}$<br /><br />Given that the maximum static friction force is 7 N and the minimum static friction force is 14 N, we can substitute these values into the formula:<br /><br />$ΔF = 7 - 14 = -7 N$<br /><br />Therefore, the fluctuation of stick-slip motion is -7 N.<br /><br />(c) The frequency of motion can be calculated using the$f = \frac{v}{d}$<br /><br />where:<br />f is the frequency of motion,<br />v is the velocity of the mass, and<br />d is the fluctuation of stick-slip motion.<br /><br />Substituting the values of v and d into the formula, we get:<br /><br />$f = \frac{0.1}{-7} = -0.0143 Hz$<br /><br />Therefore, the frequency of motion is -0.0143 Hz.<br /><br />(d) The natural frequency of the mass-spring system can be calculated using the formula:<br /><br />$f_n = \frac{1}{2π}\sqrt{\frac{k}{m}}$<br /><br />where:<br />f_n is the natural frequency,<br />k is the stiffness of the spring, and<br />m is the mass.<br /><br />Substituting the values of k and m into the formula, we get:<br /><br />$f_n = \frac{1}{2π}\sqrt{\frac{1}{0.7}} = 1.26 Hz$<br /><br />Therefore, the natural frequency of the mass-spring system is 1.26 Hz.
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