SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
10th Edition
ISBN: 9781259489563
Author: BUDYNAS
Publisher: INTER MCG
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Textbook Question
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Chapter 8, Problem 51P

8–51 to

8–54 For the pressure cylinder defined in the problem specified in the table, the gas pressure is cycled between zero and pg. Determine the fatigue factor of safety for the bolts using the following failure criteria:

(a)    Goodman.

(b)    Gerber.

(c)    ASME-elliptic.

Problem Number Originating Problem Number
8–51 8–33

(a)

Expert Solution
Check Mark
To determine

The fatigue factor of safety for the bolts using Goodman criteria.

Answer to Problem 51P

The fatigue factor of safety for the bolts using Goodman criteria is 7.55.

Explanation of Solution

Write the expression of the length of the material squeeze between the bolt face and washer face.

    l=A+B                                 (I)

Here, the length of the material squeeze between the bolt face and washer face is l, and thickness of pressure vessel  inside the bolt is A, the thickness of the cylinder inside the bolt is B.

Write the expression for the length of the bolt.

    Ll+H                                      (II)

Here the length of bolt is L, length of the material squeeze between the bolt face and washer face is l and the thickness of the bolt is H.

Write the expression of the threaded length for hexagonal bolt.

    LT=2d+6                               (III)

Here, the threaded length is LT.

Write the expression of the length of the unthreaded portion in grip.

    ld=LLT                                          (IV)

Here, the length of the unthreaded portion in the grip is ld.

Write the expression of the length of the threaded portion in grip.

    lt=lld                                                 (V)

Here, the length of threaded portion in the grip is lt.

Write the expression of the major area diameter.

    Ad=π4d2                                           (VI)

Here, the nominal diameter of the bolt is d.

Write the expression of the stiffness for the bolt.

    kb=AdAtEsAdlt+Atld                                (VII)

Here, the bolt stiffness is kb, major area diameter of the fastener is Ad, tensile stress area is At, length of threaded portion is lt, length of unthreaded portion is ld and young’s modulus of elasticity of Es.

Write the expression of stiffness for the steel cylinder.

    k1=0.5774πEsdln[(1.155t+Dd1.155t+D+d)(D+dDd)]     (VIII)

Here, the stiffness of the steel cylinder is k1, young’s modulus for steel is Es, effective sealing diameter of the gasket sealing is D, thickness of gasket is t.

Write the expression for the stiffness of the cast iron pressure vessel.

    k2=EcEs×k1                                        (IX)

Here, the stiffness of the cast-iron pressure vessel is k2 and the young’s modulus for pressure vessel is Ec.

Write the expression for the stiffness of the member.

    km=[1k1+1k2]1                                   (X)

Here, the stiffness of the member is km.

Write the expression of joint constant.

    C=kbkb+km                                            (XI)

Here, the joint constant is C.

Write the expression of initial tension in the bolt.

    Fi=0.75AtSp                                           (XII)

Here, the tensile stress area is At and the proof strength of material of the bolt is Sp.

Write the expression of the effective area of the cylinder.

    Ag=π4(C2)                                         (XIII)

Here, the effective area of the cylinder is Ag, diameter of the cylinder is C.

Write the expression for the total force acting on the assembly.

    Ptotal=Agpg                                            (XIV)

Here, the total load acting on the assembly is Ptotal, stress on the cylinder sealing is Sp.

Write the expression for the load acting on each bolt.

    P=PtotalN                                                 (XV)

Here, the number of bolt is N.

Write the expression for the initial stress in the bolt.

    σi=0.75SP                                             (XVI)

Write the expression for the average stress.

    σa=CP2At                                                 (XVII)

Write the expression for the mean stress.

    σm=CP2At+FiAt                                        (XVIII)

Write the expression for factor of safety by Goodman criteria.

    nf=Se(Sutσi)σa(Sut+Se)                                     (XIX)

Here, the ultimate strength is Sut and the endurance strength is Se.

Conclusion:

Substitute 20mm for A and 20mm for B in Equation (I)

    l=20mm+20mm=40mm

Refer to Table A-31 “Dimensions of the hexagonal nut” to obtain 10.8mm for H.

Substitute 40mm for l and 10.8mm for H Equation (II).

    L40mm+10.8mm=50.8mm55mm

Substitute 12mm for d in Equation (III).

    LT=2×12mm+6=24mm+6=30mm

Substitute 55mm for L and 30mm for LT in Equation (IV).

    ld=55mm30mm=(5530)mm=25mm

Substitute 40mm for l and 25mm for ld in Equation (V).

    lt=40mm25mm=(4025)mm=15mm

Substitute 12mm for d in Equation (VI).

    Ad=π4(12mm)2=(0.785398)(144)mm2=113.09mm2=113.1mm2

Refer to Table 8.1 “Diameter and Area of Unified Screw Threads UNC and UNF” obtain 84.3mm2 for At.

Substitute 113.1mm2 for Ad, 84.3mm2 for At, 15mm for lt, 25mm for ld and 207GPa for Es in Equation (VII).

    kb=(113.1mm2)(84.3mm2)(207GPa)(113.1mm2)(15mm)+(84.3mm2)(25mm)=(113.1mm2)(84.3mm2)(207MPa)(103N/mm21MPa)(113.1mm2)(15mm)+(84.3mm2)(25mm)=(1973606.31×1033804N/mm)(103N/m1N/mm)(1MN/m106N/m)=518.8MN/m

Substitute 207GPa for Es, 20mm for t, 12mm for d and 18mm for D in Equation (VIII).

    k1=0.5774π(207GPa)(12mm)ln[1.155(20mm)+18mm12mm](18mm+12mm)[1.155(20mm)+18mm+12mm](18mm12mm)=(4505.8657(mm)(GPa))(103N/mm21GPa)ln2.740=(4470.108×103N/mm)(103N/m1N/mm)(1MN/m106N/m)4470MN/m

Substitute 4470MN/m for k1, 207GPa for Es, and 100GPa for Ec in Equation (IX).

    k2=100GPa207GPa×(4470MN/m)=2159.4MN/m2159MN/m

Substitute 4470MN/m for k1 and 2159MN/m for k2 in Equation (X).

    km=[14470MN/m+12159MN/m]1=[(2159MN/m)+(4470MN/m)(4470MN/m)(4470MN/m)]1=[6.86891×104MN/m]1MN/m1456MN/m

Substitute 518.8MN/m for kb and 1456MN/m for km in Equation (XI).

    C=518.8MN/m518.8MN/m+1456MN/m=518.81947.8=0.26270.263

Refer to Table 8-11 “Metric Mechanical Property Classes for Steel Bolts, Screws and Studs” to obtain 650MPa for Sp.

Substitute 650MPa for Sp, 84.3mm2 for At in the Equation (XII).

    Fi=0.75(84.3mm2)(650MPa)=(41096.25N)(1kN1000N)=41.09kN41.1kN

Substitute 100mm for C in Equation (XIII).

    Ag=π4(100mm)2=(0.785398)(10000)mm2=7853.98mm27854mm2

Substitute 7854mm2 for Ag and 6MPa for pg in Equation (XIV).

    Ptotal=(7854mm2)(6MPa)=(47124N)(1kN1000N)=47.124kN

Substitute 47.124kN for Ptotal and 10 for N in Equation (XV).

    P=47.124kN10=4.7124kN/bolt

Refer to Table 8.17 “Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Threads” to obtain 140MPa for Se with respect to ISO10.9.

Refer to Table 8.11 “Metric Mechanical-Property Classes for Steel Bolts, Screws, and Studs” to obtain 900MPa for Sut with respect to medium carbon.

Substitute 650MPa for Sp in Equation (XVI).

    σi=0.75(650MPa)=487.5Mpa

Substitute 4.7124kN/bolt for P, 0.263 for C and 84.3mm2 for At in the Equation (XVII).

    σa=(0.263)(4.7124kN/bolt)2×84.3mm2=1.5599×103mm2×4.7124kN/bolt(103N1kN)=7.350MPa

Substitute 4.7124kN/bolt for P, 0.263 for C, 84.3mm2 for At and 41.1kN for Fi in Equation (XVIII).

    σm=0.263(4.7124kN/bolt)2×84.3mm2+41.1kN84.3mm2=1.5599×103mm2×4.7124kN/bolt(103N1kN)+41.1kN84.3mm2(103N1kN)=7.350MPa+0.4875MPa7.8375MPa

Substitute 7.350MPa for σa, 487.5Mpa for σi, 900MPa for Sut and 140MPa for Se in Equation (XIX).

    nf=140MPa(900MPa487.5Mpa)7.350MPa(900MPa+140MPa)=140MPa(412.5Mpa)7.350MPa(1040MPa)=5775076447.55

Thus, the fatigue factor of safety for the bolts using Goodman criteria is 7.55.

(b)

Expert Solution
Check Mark
To determine

The fatigue factor of safety for the bolts using Gerber criteria.

Answer to Problem 51P

The fatigue factor of safety for the bolts using Gerber criteria is 11.4.

Explanation of Solution

Write the expression for the factor of safety using Gerber criteria.

    nf=12σaSe[SutSut2+4Se(Se+σi)Sut22σiSe]     (XX).

Conclusion:

Substitute 7.350MPa for σa, 487.5MPa for σi, 487.5MPa for σi, 900MPa for Sut and 140MPa for Se in Equation (XX).

    nf=12(7.350MPa)(140MPa)[(900MPa)[(900MPa)2+4×140MPa(140MPa+487.5Mpa)](900MPa)22(487.5MPa)(140MPa)]=12058(MPa)2[969914.4292(MPa)2(900MPa)2136500(MPa)2]=11.37711.4

Thus, the fatigue factor of safety for the bolts using Gerber criteria is 11.4.

(c)

Expert Solution
Check Mark
To determine

The fatigue factor of safety for the bolts using ASME-elliptic criteria.

Answer to Problem 51P

The fatigue factor of safety for the bolts using ASME-elliptic criteria is 9.73.

Explanation of Solution

Write the expression for the factor of safety using ASME-elliptic criteria.

    nf=Seσa(SP2+Se2)[SPSP2+Se2σi2σiSe]   (XXI)

Conclusion:

Substitute 7.350MPa for σa, 487.5Mpa for σi, 900MPa for Sut and 140MPa for Se in Equation (XX).

    nf=140MPa(7.350MPa)((650MPa)2+(140MPa)2)[(650MPa)(650MPa)2+(140MPa)2(487.5Mpa)2(487.5Mpa)(140MPa)]=140MPa3249435(MPa)3[225650.4668(MPa)2]9.73

Thus, the fatigue factor of safety for the bolts using ASME-elliptic criteria is 9.73.

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Chapter 8 Solutions

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I

Ch. 8 - Prob. 11PCh. 8 - An M14 2 hex-head bolt with a nut is used to...Ch. 8 - Prob. 13PCh. 8 - A 2-in steel plate and a 1-in cast-iron plate are...Ch. 8 - Repeat Prob. 8-14 with the addition of one 12 N...Ch. 8 - A 2-in steel plate and a 1-in cast-iron plate are...Ch. 8 - Two identical aluminum plates are each 2 in thick,...Ch. 8 - Prob. 18PCh. 8 - A 30-mm thick AISI 1020 steel plate is sandwiched...Ch. 8 - Prob. 20PCh. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - A 2-in steel plate and a 1-in cast-iron plate are...Ch. 8 - An aluminum bracket with a 12-in thick flange is...Ch. 8 - An M14 2 hex-head bolt with a nut is used to...Ch. 8 - A 34 in-16 UNF series SAE grade 5 bolt has a 34-in...Ch. 8 - From your experience with Prob. 8-26, generalize...Ch. 8 - Prob. 28PCh. 8 - Prob. 29PCh. 8 - Prob. 30PCh. 8 - For a bolted assembly with eight bolts, the...Ch. 8 - Prob. 32PCh. 8 - 8-33 to 8-36 The figure illustrates the...Ch. 8 - 8-33 to 8-36 The figure illustrates the...Ch. 8 - 8-33 to 8-36 The figure illustrates the...Ch. 8 - 8-33 to 8-36 The figure illustrates the...Ch. 8 - Prob. 37PCh. 8 - Prob. 38PCh. 8 - 837 to 840 Repeat the requirements for the problem...Ch. 8 - Prob. 40PCh. 8 - 841 to 844 For the pressure vessel defined in the...Ch. 8 - Prob. 42PCh. 8 - Prob. 43PCh. 8 - Prob. 44PCh. 8 - Bolts distributed about a bolt circle are often...Ch. 8 - The figure shows a cast-iron bearing block that is...Ch. 8 - Prob. 47PCh. 8 - Prob. 48PCh. 8 - Prob. 49PCh. 8 - Prob. 50PCh. 8 - 851 to 854 For the pressure cylinder defined in...Ch. 8 - Prob. 52PCh. 8 - 851 to 854 For the pressure cylinder defined in...Ch. 8 - 851 to 854 For the pressure cylinder defined in...Ch. 8 - 855 to 858 For the pressure cylinder defined in...Ch. 8 - 855 to 858 For the pressure cylinder defined in...Ch. 8 - 855 to 858 For the pressure cylinder defined in...Ch. 8 - For the pressure cylinder defined in the problem...Ch. 8 - A 1-in-diameter hot-rolled AISI 1144 steel rod is...Ch. 8 - The section of the sealed joint shown in the...Ch. 8 - Prob. 61PCh. 8 - Prob. 62PCh. 8 - Prob. 63PCh. 8 - Prob. 64PCh. 8 - Using the Goodman fatigue criterion, repeat Prob....Ch. 8 - The figure shows a bolted lap joint that uses SAE...Ch. 8 - Prob. 67PCh. 8 - A bolted lap joint using ISO class 5.8 bolts and...Ch. 8 - Prob. 69PCh. 8 - The figure shows a connection that employs three...Ch. 8 - A beam is made up by bolting together two cold...Ch. 8 - Prob. 72PCh. 8 - Prob. 73PCh. 8 - Prob. 74PCh. 8 - A vertical channel 152 76 (see Table A7) has a...Ch. 8 - The cantilever bracket is bolted to a column with...Ch. 8 - Prob. 77PCh. 8 - The figure shows a welded fitting which has been...Ch. 8 - Prob. 79PCh. 8 - Prob. 80PCh. 8 - Prob. 81P
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