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 1, Problem 19P

A round cold-drawn 1045 steel rod has a mean strength S y ¯  = 95 .5 kpsi with a standard deviation of σ ^ S y  = 6 .59 kpsi. The rod is to be subjected to a mean static axial load of P ¯ = 65 kip with a standard deviation of σ ^ P = 5.0 kip. Assuming the strength and load have normal distributions. determine the reliabilities corresponding to the design factors of (a) 1.2. (b) 1.5. Also, determine the diameter corresponding to each case.

(a)

Expert Solution
Check Mark
To determine

The reliability corresponding to 1.2 design factor for cold-drawn 1045 steel rod.

The diameter of cold-drawn 1045 steel rod for 1.2 design factor.

Answer to Problem 19P

The reliability corresponding to 1.2 design factor for cold-drawn 1045 steel rod is 94.7%.

The diameter of cold-drawn 1045 steel rod for 1.2 design factor is 1.020in.

Explanation of Solution

Write the expression for coefficient of variance for strength.

Cs=σ^SxyS¯sy (I)

Here, the coefficient of variance for strength is Cs, the standard deviation for shear yield strength is σ^Sxy and shear yield strength is S¯sy.

Write the expression for coefficient of variance for stress.

Cσ=σ^PP¯ (II)

Here, the coefficient of variance for stress is Cσ, the standard deviation for axial load is σ^P and axial load is P¯.

Write the expression for transform variable.

z=n¯d1n¯2dCS2+Cσ2 (III)

Here, the design factor is n¯d and transform variable is z.

Write the expression for axial stress.

σ¯=P¯(πd24)=4P¯πd2 (IV)

Here, the axial stress is σ¯, the axial load is P¯ and the diameter of steel rod is d.

Calculate the reliability for steel rod.

R=1Φ(z) (V)

Here, the reliability for steel is R and the transform function is Φ(z).

Calculate the diameter of steel rod.

n¯d=S¯yσ¯ (VI)

Substitute 4P¯πd2 for σ¯ in Equation (VI).

n¯d=S¯y4P¯πd2n¯d=πd2S¯y4P¯d=4P¯n¯dπS¯y (VII)

Conclusion:

Substitute 6.59kpsi for σ^Sxy and 95.5kpsi for S¯y in Equation (I).

Cs=(6.59kpsi)(95.5kpsi)=0.06901

Substitute 5kip for σ^P and 65kip for P¯ in Equation (II).

Cσ=(5kip)(65kip)=0.077

Substitute 1.2 for n¯d, 0.06901 for Cs and 0.077 for Cσ in Equation (III).

z=((1.2)1)(1.2)2(0.06901)2+(0.077)2=(0.2)0.01278=(0.2)0.1130481.77

Write the formula of interpolation.

y2=(x2x1)(y3y1)(x3x1)+y1 (VIII)

Here, the variables denote by x and y are the value of z and Φ(z).

Refer to table A-10 “Cumulative distribution function of Normal distribution” and obtain the value given in Table below.

xy
1.610.0537
1.6127y2
1.620.0526

Table (1)

Substitute 1.61 for x1, 1.6127 for x2, 1.62 for x3, 0.0537 for y1 and 0.0526 for y3 in Equation (VIII).

y2=(1.61271.61)(0.05260.0537)(1.621.61)+0.0537=(0.00000297)(0.01)+0.0537=0.053403

Substitute 1.77 for Φ(z) in Equation (V).

R=(10.053403)×100%=(0.9465)×100%94.7%

Thus, the reliability of steel rod is 94.7%.

Substitute 65kip for P¯, 1.2 for n¯d and 95.5kpsi for S¯y in Equation (VII).

d=4(65kip)(1.2)π(95.5kpsi)=(312kip)π(95.5kpsi)=(1.0399in2)1.020in

Thus, the diameter of steel rod is 1.020in.

(b)

Expert Solution
Check Mark
To determine

The reliability corresponding to 1.5 design factor for cold-drawn 1045 steel rod.

The diameter of cold-drawn 1045 steel rod for 1.5 design factor.

Answer to Problem 19P

The reliability corresponding to 1.5 design factor for cold-drawn 1045 steel rod is 99.98%.

The diameter of cold-drawn 1045 steel rod for 1.5 design factor is 1.140in.

Explanation of Solution

Write the expression for coefficient of variance for strength.

Cs=σ^SxyS¯sy (IX)

Write the expression for coefficient of variance for stress.

Cσ=σ^PP¯ (X)

Write the expression for transform variable.

z=n¯d1n¯2dCS2+Cσ2 (XI)

Write the expression for axial stress.

σ¯=P¯(πd24)=4P¯πd2 (XII)

Calculate the reliability for steel rod.

R=(1Φ(z))×100% (XIII)

Calculate the diameter of steel rod.

n¯d=S¯yσ¯ (XIV)

Substitute 4P¯πd2 for σ¯ from Equation (XII) in Equation (XIV).

n¯d=S¯y4P¯πd2n¯d=πd2S¯y4P¯d=4P¯n¯dπS¯y (XV)

Conclusion:

Substitute 6.59kpsi for σ^Sxy and 95.5kpsi for S¯y in Equation (IX)

Cs=6.59kpsi95.5kpsi=0.06901

Substitute 5kip for σ^P and 65kip for P¯ in Equation (X).

Cσ=5kip65kip=0.09231

Substitute 1.5 for n¯d, 0.06901 for Cs and 0.09231 for Cσ in Equation (XI).

z=((1.5)1)(1.5)2(0.06901)2+(0.09231)2=0.50.138693.6051

Write the formula of interpolation.

y2=(x2x1)(y3y1)(x3x1)+y1 (XVI)

Here, the variables denote by x and y are the value of z and Φ(z) respectively.

Refer to table A-10 “Cumulative distribution function of Normal distribution” and obtain the value given in Table below.

xy
3.60.000159
3.605y2
3.70.000108

Table (2)

Substitute 3.6 for x1, 3.605 for x2, 3.7 for x3, 0.000159 for y1 and 0.000108 for y3 in Equation (XVI).

y2=(3.6053.6)(0.0001080.000159)(3.73.6)+0.0.000159=(0.00000255)+0.000159=0.00015645

Substitute 0.00015645 for Φ(z) in Equation (XIII).

R=(1(0.00015645))×100%=(0.99984355)×100%99.98%

Thus, the reliability of steel rod is 99.98%.

Substitute 65kip for P¯, 1.5 for n¯d and 95.5kpsi for S¯y in Equation (XV).

d=4(65kip)(1.5)π(95.5kpsi)=(390)kipπ(95.5)kpsi=1.29990in21.140in

Thus, the diameter of steel rod is 1.140in.

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Chapter 1 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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