Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Chapter 12, Problem 15P

A natural-circulation pillow-block bearing with l/d = 1 has a journal diameter D of 2.500 in with a unilateral tolerance of ‒0.001 in. The bushing bore diameter B is 2.504 in with a unilateral tolerance of 0.004 in. The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 300 lbf in shaft-stirred air at 70°F with α = 1. The lateral area of the pillow-block housing is 60 in2. Perform a design assessment using minimum radial clearance for a load of 600 lbf and 300 lbf. Use Trumpler’s criteria.

Expert Solution & Answer
Check Mark
To determine

The design for load 600lbf.

The design for load 300lbf.

Answer to Problem 15P

The design has been successful for load 600lbf.

The design has been successful for load 300lbf.

Explanation of Solution

Write expression for minimum thickness.

    c=bd2                                                                                                       (I)

Here, bore diameter is b, and journal diameter is d

Write expression for journal radius.

    r=d2                                                                                                            (II)

Write expression for radial clearance ratio.

    cr=rc                                                                                                         (III)

Write expression for nominal pressure.

    P=Wld                                                                                                         (IV)

Here, length of bearing is l , load on bearing is W.

Write expression for viscosity.

    μ=μoexp[bT+95]                                                                                    (V)

Here, viscosity is μ, viscosity at standard temperature is μo,constant is b , and temperature is T.

Write expression for Somerfield number.

    S=(rc)2μNP                                                                                             (VI)

Here Somerfield number is S, radial ratio is rc, and rotational speed of shaft is N

Write expression for heat generated due to friction.

    Hgen=25451050(W×N×c)(frc)                                                                   (VII)

Here, Coefficient of friction is f,

Write expression for heat loss.

    Hloss=(hCRAα+1)(TfT)                                                                         (VIII)

Here, overall coefficient of radiation and convection radiation is hCR,α is constant dependent on lubrication scheme and bearing housing geometry, ambient temperature is T ,and film temperature is Tf.

Write expression for heat loss by linear approximation.

    Hloss=mTf+b                                                                                           (IX)

Here, constant is m and constant is b.

Write expression for inlet temperature.

    T1=Tf(ΔTf2)                                                                                           (X)

Here, film temperature is Tf, temperature rise of film ΔTf.

Write expression for maximum temperature of the lubricant.

    Tmax=Ts+ΔTf                                                                                           (XI)

Here, the sump temperature is Ts

Conclusion:

Substitute 2.500in for d,2.504in for b in Equation (I)

    c=2.504in2.500in2=0.0.004in2=0.002in

Substitute 2.500in for d in Equation (II).

    r=2.500in2=1.250in

Substitute 1.250in for r, and 0.002in for c in Equation (III).

    cr=1.25in0.002in=625

Design for 600lbf:

Substitute 600lbf for W ,2.5in for l ,and 2.5in for d in the Equation (IV).

    P=(600lbf)(2.5in)(2.5in)=(600lbf)(6.25in2)=96psi

The nominal pressure P is lower than given design pressure 300psi, so The Trumpler’s design criteria is satisfied.

Refer to figure 12.1 “Viscosity-temperature chart in SI units”, obtain the value of viscosity μo as 0.0136, and constant b as 1271.6°F for oil SAE 20 at film temperature 150°F.

Substitute 0.0136 for μo, 1271.6°F for b, and 150°F for T in Equation (VI).

    μ=(0.0136)exp[(1271.6)(Tf+95°F)]

Substitute 1.25in for r, 0.002in for c, 33.33rev/min for N and 96psi for P in Equation (V).

    S=(1.250in0.002in)2(μ)(1120rev/min)(96psi)=(625)2×(μ)(1120rev/min)(96psi)|1rev/s60rev/min|=(625)2×(μ)(18.66rev/s)(96psi)=0.0760μ

Substitute 600lbf for W, 1.25in for r, 0.002in for c, 18.67rev/s for N and 96psi for P in Equation (VII).

    Hgen=(242)[(600lbf)×(18.67rev/s)×(0.002in)](frc)=(242)[(600lbf)×(18.67rev/s)×(0.002in)](frc)|12ft1in|=[(145200lbf)×(18.67rev/s)×(0.024in)](frc)|12ft1in|=(54.3)(frc)

Substitute 2.7Btu/(hft2°F) for hCR, 60in2 for A, 1 for α in Equation (VIII).

    Hloss=[(2.7Btu/hft2°F)(60in2)1+1](Tf70°F)=[(2.7Btu/hft2°F)(30in2)](Tf70°F)|12ft1in||144ft1in2|=0.5625(Tf70°F)

Calculate the value of Hgen,(frc), Hloss, S,and μ at different film temperature 200°F and 240°F as in below table.

TfμS(frc)HgenHloss
2200.7700.0591.9103.284.4
2400.6050.0461.792.395.6

Substitute 0.545 for m, 84.4 for Hloss, and 223.1 for b in Equation (IX).

    (84.4)=(0.545)Tf+223.184.4223.1=(0.545)Tf138.7=0.545TfTf=254.49°F

The value of Hgen,(frc), Hloss, S, and μ at film temperature of 254°F is tabulated as below.

TfμS(frc)HgenHloss
2540.5170.03931.6392.393.2

Write expression for minimum film thickens.

    hoc=0.21                                                                                                 (XII)

Solve Equation (XII) for ho.

    ho=0.21c                                                                                              (XIII)

Substitute 0.002in for c in Equation (XIII).

    ho=0.21×0.002in=0.0042in

Refer to figure 12-24 “Temperature rise and Somerfield graph”

Write expression for temperature rise.

    ΔT°F=Ppsi9.70[0.349109+6.00940S+0.047467S2]                                 (XIV)

Here, Temperature rise is ΔT°F (in Fahrenheit), The pressure is Ppsi(in psi)

Substitute 96psi for P, and 0.047 for S in Equation (XIV).

    ΔT°F=(96psi)9.70[0.349109+6.00940×(0.047)+0.047467×(0.047)2]=(9.89psi)[0.349109+0.2824+0.0001048]=(9.89psi)[0.6316]=6.25°F

Substitute 237°F for Tf, and 6.31°F for ΔTf in Equation (X).

    T1=237°F(6.31°F2)=233.85°F

Thus, the inlet temperature of the lubricant is 233.85°F.

Substitute 6.31°F for ΔTf and 233.8°F for Ts in Equation (XI).

    Tmax=233.8°F+6.31°F=240.1°F

Since the maximum temperature Tmax is less than the given temperature 250°F, Therefore Trumpler’s design criteria satisfied.

Write expression for minimum film thickness using Trumpler’s criteria.

    ho(0.0002+0.00004d)                                                                         (XV)

Substitute 2.5in for d in Equation (XV)

    ho[(0.0002in)+0.00004(2.5in)][(0.0002in)+(0.0010in)]0.00030in

The minimum radial clearance 0.00042in is greater than 0.00030in. So, the Trumpler’s design criteria satisfied.

Write the equation for nd.

    nd2                                                                                                       (XVI)

Substitute 2 for nd in Equation (XIV).

    22

So, the Trumpler’s criteria satisfied.

Thus, the design is successful for load 600lbf.

Design for 300lbf:

Substitute 300lbf for W ,2.5in for l ,and 2.5in for d in the Equation (IV).

    P=(300lbf)(2.5in)(2.5in)=(300lbf)(6.25in2)=48psi

The nominal pressure P is lower than given design pressure 300psi, so the Trumpler’s design criteria is satisfied.

Refer to figure 12.1 “Viscosity-temperature chart in SI units”, obtain the value of viscosity μo as 0.0136, and constant b as 1271.6°F for oil SAE 20 at film temperature 150°F.

Substitute 0.0136 for μo, 1271.6°F for b, and 150°F for T in Equation (V).

    μ=(0.0136)exp[(1271.6)(Tf+95°F)]

Substitute 1.25in for r, 0.002in for c, 33.33rev/min for N and 96psi for P in Equation (VI).

    S=(1.250in0.002in)2(μ)(1120rev/min)(48psi)=(625)2×(μ)(1120rev/min)(48psi)|1rev/s60rev/min|=(625)2×(μ)(18.66rev/s)(48psi)=0.152μ

Substitute 300lbf for W, 1.25in for r, 0.002in for c, 18.67rev/s for N and 96psi for P in Equation (VII).

    Hgen=(242)[(300lbf)×(18.67rev/s)×(0.002in)](frc)=(242)[(300lbf)×(18.67rev/s)×(0.002in)](frc)|12ft1in|=[(145200lbf)×(18.67rev/s)×(0.024in)](frc)|12ft1in|=(27.15)(frc)

Substitute 2.7Btu/(hft2°F) for hCR, 60in2 for A,1 for α in Equation (VIII).

    Hloss=[(2.7Btu/hft2°F)(60in2)1+1](Tf70°F)=[(2.7Btu/hft2°F)(30in2)](Tf70°F)|12ft1in||144ft1in2|=0.5625(Tf70°F)

The value of Hgen,(frc), Hloss, S, and μ at different film temperatures are tabulated as below.

TfμS(frc)HgenHloss
2100.8790.1363.1585.52578.75
2300.680.1032.978.13590

Substitute 0.369 for m, 163 for b and 90 for Hloss in Equation (IX).

    (90)=(0.369)Tf+223.190163=(0.369)Tf73=0.369TfTf=197.8°F

The value of Hgen,(frc), Hloss, S and μ at film temperature 197.8°F are tabulated as in below.

TfμS(frc)HgenHloss
197.80.6520.1122.5182.2382.32

Write expression for minimum film thickens.

    hoc=0.42                                                                                                (XVII)

Solve Equation (XVII) for ho.

    ho=0.42c                                                                                              (XVIII)

Substitute 0.002in for c in Equation (XVIII).

    ho=0.42×0.002in=8.4×105in

Thus, the minimum radial clearance is 8.4×105in.

Write expression for temperature rise.

    ΔT=P9.70[0.349109+6.00940S+0.047467S2]                                   (IXX)

Substitute 48psi for P, and 0.048 for S in Equation (IXX).

    ΔT=(48psi)9.70[0.349109+6.00940×(0.047)+0.047467×(0.047)2]=(4.90psi)[0.349109+0.2824+0.0001048]=(4.90psi)[0.6316]=5.23°F

Substitute 237°F for Tf, and 5.23°F for ΔTf in Equation (X).

    T1=237°F(5.23°F2)=234.385°F

Thus, the inlet temperature of the lubricant is 234.385°F.

Substitute 5.23°F for ΔTf and 234°F for Ts in Equation (XIV).

    Tmax=234°F+5.23°F=239.23°F

Since the maximum temperature Tmax is less than the given temperature 250°F, So the Trumpler’s design criteria is satisfied.

Write expression for minimum film thickness using Trumpler’s criteria.

    ho(0.002+0.00004d)                                                                           (XX)

Substitute 2.5in for d in Equation (XX)

    ho[(0.002in)+0.00004(2.5in)][(0.002in)+(0.0010in)]0.0030in

Since the minimum radial clearance 0.0030in is greater than 8.4×105in, so the Trumpler’s design criteria is satisfied.

Write equation for nd.

    nd2                                                                                                       (XXI)

Substitute 2 for nd in Equation (XXI)

    22

So, the Trumpler’s criteria is satisfied.

Thus, the design is successful for load 300lbf.

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