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Questions and Answers of Mechanical Engineering

The worm shaft shown in part a of the figure transmits 1.35 hp at 600 rev/min. A static force analysis gave the results shown in part b of the figure. Bearing A is to be an angular-contact ball
In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an L10 life of 115 h and an L80 life of 600 h. The basic load rating of this bearing is 39.6 kN.
A 16-tooth pinion drives the double-reduction spur-gear train in the figure. All gears have 25◦ pressure angles. The pinion rotates ccw at 1200 rev/min and transmits power to the gear train.
Different bearing metallurgy affects bearing life. A manufacturer reports that a particular heat treatment increases bearing life at least threefold. A bearing identical to that of Prob. 11–15
Estimate the remaining life in revolutions of an 02-30 mm angular-contact ball bearing already subjected to 200 000 revolutions with a radial load of 18 kN, if it is now to be subjected to a change
The same 02-30 angular-contact ball bearing as in Prob. 11–18 is to be subjected to a two-step loading cycle of 4 min with a loading of 18 kN, and one of 6 min with a loading of 30 kN. This cycle
The expression Fa L = constant can be written using x = L/L10, and it can be expressed as Fa x = K or log F = (1/a) log K − (1/a) log x. This is a straight line on a log-log plot, and it is the
A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20◦ pressure angle. The pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face
A steel spur pinion has a diametral pitch of 12 teeth/in, 16 teeth cut full-depth with a 20◦ pressure angle, and a face width of ¾ in. This pinion is expected to transmit 1.5 hp at a speed of
A steel spur pinion has a module of 1.25 mm, 18 teeth cut on the 20◦ full-depth system, and a face width of 12 mm. At a speed of 1800 rev/min, this pinion is expected to carry a steady load of
A steel spur pinion has 15 teeth cut on the 20◦ full-depth system with a module of 5 mm and a face width of 60 mm. The pinion rotates at 200 rev/min and transmits 5 kW to the mating steel gear.
A steel spur pinion has a module of 1 mm and 16 teeth cut on the 20◦ full-depth system and is to carry 0.15 kW at 400 rev/min. Determine a suitable face width based on an allowable bending
A 20◦ full-depth steel spur pinion has 17 teeth and a module of 1.5 mm and is to transmit 0.25 kW at a speed of 400 rev/min. Find an appropriate face width if the bending stress is not to
A 20◦ full-depth steel spur pinion has a diametral pitch of 5 teeth/in and 24 teeth and transmits 6 hp at a speed of 50 rev/min. Find an appropriate face width if the allowable bending stress
A steel spur pinion is to transmit 15 hp at a speed of 600 rev/min. The pinion is cut on the 20◦ full-depth system and has a diametral pitch of 5 teeth/in and 16 teeth. Find a suitable face
A 20◦ full-depth steel spur pinion with 18 teeth is to transmit 2.5 hp at a speed of 600 rev/min. Determine appropriate values for the face width and diametral pitch based on an allowable
A 20◦ full-depth steel spur pinion is to transmit 1.5 kW hp at a speed of 900 rev/min. If the pinion has 18 teeth, determine suitable values for the module and face width. The bending stress
A speed reducer has 20◦ full-depth teeth and consists of a 22-tooth steel spur pinion driving a 60-tooth cast-iron gear. The horsepower transmitted is 15 at a pinion speed of 1200 rev/min. For
A gear drive consists of a 16-tooth 20◦ steel spur pinion and a 48-tooth cast-iron gear having a pitch of 12 teeth/in. For a power input of 1.5 hp at a pinion speed of 700 rev/min, select a
A gearset has a diametral pitch of 5 teeth/in, a 20◦ pressure angle, and a 24-tooth cast-iron spur pinion driving a 48-tooth cast-iron gear. The pinion is to rotate at 50 rev/min. What
A 20◦ 20-tooth cast-iron spur pinion having a module of 4 mm drives a 32-tooth cast-iron gear. Find the contact stress if the pinion speed is 1000 rev/min, the face width is 50 mm, and 10 kW of
A steel spur pinion and gear have a diametral pitch of 12 teeth/in, milled teeth, 17 and 30 teeth, respectively, a 20◦ pressure angle, and a pinion speed of 525 rev/min. The tooth properties
A milled-teeth steel pinion and gear pair have Sut = 113 kpsi, Sy = 86 kpsi and a hardness at the involute surface of 262 Brinell. The diametral pitch is 3 teeth/in, the face width is 2.5 in, and the
A 20◦ full-depth steel spur pinion rotates at 1145 rev/min. It has a module of 6 mm, a face width of 75 mm, and 16 milled teeth. The ultimate tensile strength at the involute is 900 MPa
A steel spur pinion has a pitch of 6 teeth/in, 17 full-depth milled teeth, and a pressure angle of 20◦. The pinion has an ultimate tensile strength at the involute surface of 116 kpsi, a
A commercial enclosed gear drive consists of a 20◦ spur pinion having 16 teeth driving a 48-tooth gear. The pinion speed is 300 rev/min, the face width 2 in, and the diametral pitch 6 teeth/in.
A 20◦ spur pinion with 20 teeth and a module of 2.5 mm transmits 120 W to a 36-tooth gear. The pinion speed is 100 rev/min, and the gears are grade 1, 18 mm face width, through-hardened steel
Repeat Prob. 14–19 using helical gears each with a 20◦ normal pitch angle and a helix angle of 30◦ and a normal diametral pitch of 6 teeth/in.
A spur gearset has 17 teeth on the pinion and 51 teeth on the gear. The pressure angle is 20◦ and the overload factor Ko = 1. The diametral pitch is 6 teeth/in and the face width is 2 in. The
In Sec. 14–10, Eq. (a) is given for Ks based on the procedure in Ex. 14–2. Derive this equation. A speed-reducer has 20◦ full-depth teeth, and the single-reduction spur-gear gearset has 22
The speed reducer of Prob. 14–24 is to be used for an application requiring 40 hp at 1145 rev/min. Estimate the stresses of pinion bending, gear bending, pinion wear, and gear wear and the
The gearset of Prob. 14–24 needs improvement of wear capacity. Toward this end the gears are nitrided so that the grade 1 materials have hardnesses as follows: The pinion core is 250 and the pinion
The absolute value of the pitch variation is such that the transmission accuracy level number is 6. The materials are 4340 through-hardened grade 1 steels, heat-treated to 250 Brinell, core and case,
The gearset of Prob. 14–24 has had its gear specification changed to 9310 for carburizing and surface hardening with the result that the pinion Brinell hardnesses are 285 core and 580–600 case,
The gearset of Prob. 14–27 is going to be upgraded in material to a quality of grade 2 9310 steel. Estimate the power rating for the new gearset.
Matters of scale always improve insight and perspective. Reduce the physical size of the gearset in Prob. 14–24 by one-half and note the result on the estimates of transmitted load Wt and power.
AGMA procedures with cast-iron gear pairs differ from those with steels because life predictions are difficult; consequently (YN) P, (YN) G, (ZN) P, and (ZN) G are set to unity. The consequence of
Spur-gear teeth have rolling and slipping contact (often about 8 percent slip). Spur gears tested to wear failure are reported at 108 cycles as Buckingham’s surface fatigue load-stress factor K.
In Ex. 14–5 use nitrided grade 1 steel (4140) which produces Brinell hardnesses of 250 core and 500 at the surface (case). Use the upper fatigue curves on Figs. 14–14 and 14–15. Estimate the
In Ex. 14–5 use carburized and case-hardened gears of grade 1. Carburizing and case-hardening can produce a 550 Brinell case. The core hardnesses are 200 Brinell. Estimate the power capacity of the
In Ex. 14–5, use carburized and case-hardened gears of grade 2 steel. The core hardnesses are 200, and surface hardnesses are 600 Brinell. Use the lower fatigue curves of Figs. 14–14 and 14–15.
At a constant amplitude, completely reversed bending stress level, the cycles-to-failure experience with 69 specimens of 5160H steel from 1.25-in hexagonal bar stock was as follows: Where L is the
Determinations of the ultimate tensile strength Sut of stainless steel sheet (17-7PH, condition TH 1050), in sizes from 0.016 to 0.062 in, in 197 tests combined into seven classes were Where f is
A total of 58 AISI 1018 cold-drawn steel bars were tested to determine the 0.2 percent offset yield strength Sy. The results were Where Sy is the class midpoint and f is the class frequency.
The base 10 logarithm of 55 cycles-to-failure observations on specimens subjected to a constant stress level in fatigue have been classified as follows: Here y is the class midpoint and f is the
A ½ -in nominal diameter round is formed in an automatic screw machine operation that is initially set to produce a 0.5000-in diameter and is reset when tool wear produces diameters in excess of
The only detail drawing of a machine part has a dimension smudged beyond legibility. The round in question was created in an automatic screw machine and 1000 parts are in stock. A random sample of 50
(a) The CDF of the variety x is F (x) = 0.555x − 33, where x is in millimeters. Find the PDF, the mean, the standard deviation, and the range numbers of the distribution. (b) In the expression
A regression model of the form Show that For the data set Find the regression equation and plot the data with the regression model.
R. W. Landgraf reported the following axial (push pull) endurance strengths for steels of differing ultimate strengths: (a) Plot the data with Se as ordinate and Su as abscissa. (b) Using the y =
In fatigue studies a parabola of the Gerber type Is useful (see Sec. 6-12). Solved for a the preceding equation becomes This implies a regression model of the form y = a0 + a2x2. Show that the
Consider the following data collected on a single helical coil extension spring with an initial extension Fi and a spring rate k suspected of being related by the equation F = Fi + kx where x is the
In the expression for uniaxial strain the elongation is specified as (0.0015, 0.000 092) in and the length as l ~ (2.0000, 0.0081) in. What are the mean, the standard deviation, and the
In Hooke’s law for uniaxial stress, the strain is given as (0.0005, 0.000 034) and Young’s modulus as E ~ (29.5, 0.885) Mpsi. Find the mean, the standard deviation, and the coefficient of
The stretch of a uniform rod in tension is given by the formula = Fl / AE. Suppose the terms in this equation are random variables and have parameters as follows Estimate the mean, the standard
The maximum bending stress in a round bar in flexure occurs in the outer surface and is given by the equation If the moment is specified as M ~ (15 000,1350) lbf ?in and the diameter is d ~
When a production process is wider than the tolerance interval, inspection rejects a low-end scrap fraction α with x x2. The surviving population has a new density function g(x) related to the
An automatic screw machine produces a run of parts with a uniform distribution d =U[0.748, 0.751] in because it was not reset when the diameters reached 0.750 in. The square brackets contain range
A spring maker is supplying helical coil springs meeting the requirement for a spring rate k of 10 ± 1 lbf/in. The test program of the spring maker shows that the distribution of spring rate is well
The lives of parts are often expressed as the number of cycles of operation that a specified percentage of a population will exceed before experiencing failure. The symbol L is used to designate this
Fit a normal distribution to the histogram of Prob. 20–1. Superpose the probability density function on the f / (Nw) histographic plot
For Prob. 20–2, plot the histogram with f / (Nw) as ordinate and superpose a normal distribution density function on the histographic plot.
For Prob. 20–3, plot the histogram with f / (Nw) as ordinate and superpose a normal distribution probability density function on the histographic plot.
A 1018 cold-drawn steel has a 0.2 percent tensile yield strength Sy = N (78.4, 5.90) kpsi. A round rod in tension is subjected to a load P = N (40, 8.5) kip. If rod diameter d is 1.000 in, what is
A hot-rolled 1035 steel has a 0.2 percent tensile yield strength Sy = LN (49.6, 3.81) kpsi. A round rod in tension is subjected to a load P = LN (30, 5.1) kip. If the rod diameter d is 1.000 in, what
The tensile 0.2 percent offset yield strength of AISI 1137 cold-drawn steel rounds up to 1 inch in diameter from 2 mills and 25 heats is reported histographically as follows: Where Sy is the class
Repeat Prob. 20–25, presuming the distribution is lognormal. What is the yield strength exceeded by 99 percent of the population? Compare the normal fit of Prob. 20–25 with the lognormal fit by
A 1046 steel, water-quenched and tempered for 2 h at 1210°F, has a mean tensile strength of 105 kpsi and a yield mean strength of 82 kpsi. Test data from endurance strength testing at 104-cycle life
An ASTM grade 40 cast iron has the following result from testing for ultimate tensile strength: Sut = W [27.7, 46.2, 4.38] kpsi. Find the mean and standard deviation of Sut, and estimate the chance
A 1038 heat-treated steel bolt in finished form provided the material from which a tensile test specimen was made. The testing of many such bolts led to the description Sut = W [122.3, 134.6, 3.64]
A cold-drawn 301SS stainless steel has an ultimate tensile strength given by Sut = W [151.9, 193.6, 8.00] kpsi. Find the mean and standard deviation.
A 100-70-04 nodular iron has tensile and yield strengths described by Sut = W [47.6, 125.6, 11.84] kpsi Sy = W [64.1, 81.0, 3.77] kpsi What is the chance that Sut is less than 100 kpsi?
A 5160H steel was tested in fatigue and the distribution of cycles to failure at constant stress level was found to be n = W [36.9,133.6, 2.66] in 103 cycles. Plot the PDF of n and the PDF of the
A material was tested at steady fully reversed loading to determine the number of cycles to failure using 100 specimens. The results were Where L is the life in cycles and f is the number in each
The ultimate tensile strength of an AISI 1117 cold-drawn steel is Weibullian, with Su = W [70.3, 84.4, 2.01]. What are the mean, the standard deviation, and the coefficient of variation?
A 60-45-15 nodular iron has a 0.2 percent yield strength Sy with a mean of 49.0 kpsi, a standard deviation of 4.2 kpsi, and a guaranteed yield strength of 33.8 kpsi. What are the Weibull parameters
A 35018 malleable iron has a 0.2 percent offset yield strength given by the Weibull distribution Sy = W [34.7, 39.0, 2.93] kpsi. What are the mean, the standard deviation, and the coefficient of
The histographic results of steady load tests on 237 rolling-contact bearings are: Where L is the life in millions of revolutions and f is the number of failures. Fit a lognormal distribution to
Highway tunnel traffic (two parallel lanes in the same direction) experience indicates the average spacing between vehicles increases with speed. Data from a New York tunnel show that between 15 and
The engineering designer must create (invent) the concept and connectivity of the elements that constitute a design, and not lose sight of the need to develop ideas with optimality in mind. A useful
When one knows the true values x1 and x2 and has approximations X1 and X2 at hand, one can see where errors may arise. By viewing error as something to be added to an approximation to attain a true
Use the true values x1 = √5 and x2 = √6 (a) Demonstrate the correctness of the error equation from Prob. 17 for addition if three correct digits are used for X1 and X2. (b) Demonstrate
Convert the following to appropriate SI units: (a) A stress of 20 000 psi. (b) A force of 350 lbf. (c) A moment of 1200 lbf _ in. (d) An area of 2.4 in2. (e) A second moment of area of 17.4
Convert the following to appropriate ips units: (a) A length of 1.5 m. (b) A stress of 600 MPa. (c) A pressure of 160 kPa. (d) A section modulus of 1.84 (105) mm3. (e) A unit weight of 38.1
Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display. In addition, prefixes should be selected so as to limit number strings
Repeat Prob. 1–11 for:(a) τ = F/A, where A = πd2/4, F = 120 kN, and d = 20 mm.(b) σ = 32 Fa/πd3,
Repeat Prob. 1–11 for: (a) τ = F/A, where A = πd2/4, F = 120 kN, and d = 20 mm. (b) σ = 32 Fa/πd3, where F = 800 N, a = 800 mm, and d = 32 mm. (c) Z = (π/32d)(d4 −
Determine the minimum tensile and yield strengths for SAE 1020 cold-drawn steel.
Determine the minimum tensile and yield strengths for UNS G10500 hot-rolled steel.
For the materials in Probs. 2–1 and 2–2, compare the following properties: minimum tensile and yield strengths, ductility, and stiffness
Assuming you were specifying an AISI 1040 steel for an application where you desired to maximize the yield strength, how would you specify it?
Assuming you were specifying an AISI 1040 steel for an application where you desired to maximize the ductility, how would you specify it?
Determine the yield strength-to-weight density ratios (called specific strength) in units of inches for UNS G10350 hot-rolled steel, 2024-T4 aluminum, Ti-6A1-4V titanium alloy, and ASTM No. 30 gray
Determine the stiffness-to-weight density ratios (called specific modulus) in units of inches for UNS G10350 hot-rolled steel, 2024-T4 aluminum, Ti-6A1-4V titanium alloy, and ASTM No. 30 gray cast
Poisson’s ratio ν is a material property and is the ratio of the lateral strain and the longitudinal strain for a member in tension. For a homogeneous, isotropic material, the modulus of
A specimen of medium-carbon steel having an initial diameter of 0.503 in was tested in tension using a gauge length of 2 in. The following data were obtained for the elastic and plastic states: Note
Compute the true stress and the logarithmic strain using the data of Prob. 2–9 and plot the results on log-log paper. Then find the plastic strength coefficient σ0 and the strain-strengthening
The stress-strain data from a tensile test on a cast-iron specimen are Plot the stress-strain locus and find the 0.1 percent offset yield strength, and the tangent modulus of elasticity at zero
A straight bar of arbitrary cross section and thickness h is cold-formed to an inner radius R about an anvil as shown in the figure. Some surface at distance N having an original length L AB will
A hot-rolled AISI 1212 steel is given 20 percent cold work. Determine the new values of the yield and ultimate strengths

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