Tài liệu shi20396 chương 6 doc

6-11 The material is brittle and exhibits unequal tensile and compressive strengths.. Decision:Use the Modified II-Mohr theory as shown in Fig.. 6-15 Design decisions required:• Material

2+ (−5)2 = −2.615, −13.385 kpsi

2+ 12 = 12.123, 3.877 kpsi

2+ 1002 = 224.5, −44.5 MPa = σ1,σ3

224.5 − (−44.5) = 1.45 Ans.

[1802+ 3(1002)]1/2 = 1.56 Ans.

(c) σ A,σ B = −160



−1602

2+ 1002 = 48.06, −208.06 MPa = σ1,σ3

G J K L

F E D

A

B C

Scale 1" ⫽ 200 MPa

␴B

␴A

B C

1" ⫽ 100 MPa

6-7 S ut = 30 kpsi, Suc = 100 kpsi; σA = 20 kpsi, σB = 6 kpsi

(a) MNS: Eq (6-30a) n = S σ ut

2+ (−5)2 = −2.61, −13.39 kpsi

2+ 82 = 4, −16 kpsi

−16 = 6.25 Ans.

I H G

A B

1" ⫽ 20 kpsi

␴B

␴A

6-8 See Prob 6-7 for plot.

(a) For all methods: n = O B

Use DE theory for analytical solution For σ, use Eq (6-13) or (6-15) for plane stress and

Eq (6-12) or (6-14) for general 3-D

For graphical solution, plot load lines on DE envelope as shown.

2+ 52 = −0.910, −12.09 kpsi

6-10 This heat-treated steel exhibits S yt = 235 kpsi, Syc = 275 kpsi and εf = 0.06 The steel is

ductile (ε f > 0.05) but of unequal yield strengths The Ductile Coulomb-Mohr hypothesis

(DCM) of Fig 6-27 applies — confine its use to first and fourth quadrants

(c)

(a)

(b) (d)

E

C G

H

D

B

A O

F

1 cm ⫽ 10 kpsi

␴B

␴A

(a) σ x = 90 kpsi, σ y = −50 kpsi, σz = 0
σA = 90 kpsi and σB = −50 kpsi For thefourth quadrant, from Eq (6-13)

(c) σ x = −40 kpsi, σy = −90 kpsi, τx y = 50 kpsi σA, σ B = −9.10, −120.9 kpsi.

Although no solution exists for the third quadrant, use

6-11 The material is brittle and exhibits unequal tensile and compressive strengths Decision:

Use the Modified II-Mohr theory as shown in Fig 6-28 which is limited to first and fourthquadrants

S ut = 22 kpsi, Suc = 83 kpsiParabolic failure segment:

6-12 Since ε f < 0.05, the material is brittle Thus, S ut = S uc and we may use M2M which is

basically the same as MNS

6-14 Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the

DE theory to stress elements A and B with S y = 280 MPa

6-15 Design decisions required:

• Material and condition

Decision 3: Use the Distortion Energy static failure theory.

Decision 4: Initially set n d = 1

π(81 000)

1/3

= 0.922 in

Choose preferred size of d = 1.000 in

F = π(1)3(81 000)

32(15) = 530 lbf

n = 530

416 = 1.274 Set design factor to n d = 1.274

6-16 For a thin walled cylinder made of AISI 1018 steel, S y = 54 kpsi, Sut = 64 kpsi.

The state of stress is

σ t = pd

4t = p(8)4(0.05) = 40p, σl =

pd 8t = 20p, σr = −p

These three are all principal stresses Therefore,

σ= √1

2[(σ1− σ2)2+ (σ2− σ3)2+ (σ3− σ1)2]1/2

= √1

2[(40 p − 20p)2+ (20p + p)2+ (−p − 40p)2]

= 35.51p = 54 ⇒ p = 1.52 kpsi (for yield) Ans.

For rupture, 35.51p = 64 ⇒ p = 1.80 kpsi Ans.

6-17 For hot-forged AISI steel w = 0.282 lbf/in3, S y = 30 kpsi and ν = 0.292 Then ρ = w/g =

0.282/386 lbf · s2/in; ri = 3 in; ro = 5 in; r2

Setting the tangential stress equal to the yield stress,



3.292

8

(5− 3)2

So the inner radius governs and n = 13 000 rev/min Ans.

6-18 For a thin-walled pressure vessel,

4(0.065) = 6481 psi

σ r = −pi = −500 psi

These are all principal stresses, thus,

6-19 Table A-20 gives S y as 320 MPa The maximum significant stress condition occurs at r i

where σ1 = σr = 0, σ2 = 0, and σ3 = σt From Eq (4-50) for r = r i

6-20 S ut = 30 kpsi, w = 0.260 lbf/in3,ν = 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633 At the inner

radius, from Prob 6-17

Since σ r is of the same sign, we use M2M failure criteria in the first quadrant From Table

A-24, S ut = 31 kpsi, thus,

In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in.

In the x z plane, M B = 848 lbf · in and MC = 1686 lbf · in The resultants are

σ B = 10.06 − 11.27

d3 = −1.21

d3 kpsiFor this state of stress, use the Brittle-Coulomb-Mohr theory for illustration Here we use

S ut(min)= 25 kpsi, Suc(min)= 97 kpsi, and Eq (6-31b) to arrive at

21.33 25d3 − −1.21

97d3 = 1

2.8 Solving gives d = 1.34 in So use d = 1 3/8 in Ans.

Note that this has been solved as a statics problem Fatigue will be considered in the nextchapter

6-22 As in Prob 6-21, we will assume this to be statics problem Since the proportions are

un-changed, the bearing reactions will be the same as in Prob 6-21 Thus

97d3 = 1

2.8 Solving gives d = 1 1/8 in Now compare to Modified II-Mohr theory Ans.

6-23 ( F A)t = 300 cos 20 = 281.9 lbf, (FA)r = 300 sin 20 = 102.6 lbf

T = 281.9(12) = 3383 lbf · in, (F C)t = 3383

5 = 676.6 lbf ( F C)r = 676.6 tan 20 = 246.3 lbf

676.6 lbf

107.0 N

174.4 N 252.6 N

6-27 For the loading scheme shown in Figure (c),



= 4.4

2 (6+ 4.5)

= 23.1 N · m For a stress element at A:

σ x = 32M

πd3 = 32(23.1)(103)

π(12)3 = 136.2 MPa The shear at C is

12

A B

V

M

C

Based on Figure (d) and using Eq (6-15) and the solution of Prob 6-27,

The moment about the center caused by force F

is Fr e where r e is the effective radius This is balanced by the moment about the center caused by the tangential (hoop) stress

"

r

Outer radius is plane stress

σ x = 11.18 kpsi, σy = 19.58 kpsi, τx y = 7.45 kpsi

Inner radius, 3D state of stress

From Eq (6-14) with τ yz = τzx = 0

–2.997 kpsi

9.78 kpsi

22.58 kpsi 6.52 kpsi

z

2 ±

sin2 θ

The stochastic nature of the dimensions affects the δ = |R i| − |Ro| relation in

Eq (4-60) but not the others Set R = (1/2)(Ri + Ro) = 0.3755 From Eq (4-60)

σot = N(48 760, 3445) psi Ans.

6-42 From Prob 6-41, at the fit surface σ ot = N(48.8, 3.45) kpsi The radial stress is the fit

pressure which was found to be

p= 18.70(106)δ

¯p = 18.70(106)(0.001) = 18.7(103) psi

ˆσp = Cδ ¯p = 0.0707(18.70)(103)

= 1322 psiand so

p= N(18.7, 1.32) kpsi

and

σ or = −N(18.7, 1.32) kpsi

These represent the principal stresses The von Mises stress is next assessed

¯σA = 48.8 kpsi, ¯σB = −18.7 kpsi

k = ¯σB / ¯σ A = −18.7/48.8 = −0.383

¯σ= ¯σA(1 − k + k2)1/2

= 48.8[1 − (−0.383) + (−0.383)2]1/2

= 60.4 kpsi ˆσσ = Cp ¯σ= 0.0707(60.4) = 4.27 kpsi

Using the interference equation

These three stresses are principal stresses whose variability is due to the loading From

Eq (6-12), we find the von Mises stress to be

σ =

(18− 9)2+ [9 − (−6)]2+ (−6 − 18)2