chapter objectives to analyze the stresses in thin-walled pressure vessels. to develop methods to...

Chapter Objectives

To analyze the stresses in thin-walled pressure vessels.

To develop methods to analyze the stresses in members subject to combined loadings (e.g., tension or compression, shear, torsion, bending moments).

Copyright © 2011 Pearson Education South Asia Pte Ltd

1. Thin-Walled Pressure Vessels

2. Review of Stress Analyses

In-class Activities


THIN-WALLED PRESSURE VESSELS


Assumptions: 1. Inner-radius-to-wall-thickness ratio ≥ 102. Stress distribution in thin wall is uniform or constant

• Cylindrical vessels:

2

:stress alLongitudin

:direction Hoop

2

1

t

prt

pr

THIN-WALLED PRESSURE VESSELS (cont)


• Spherical vessels:

22 t

pr

EXAMPLE 1


A cylindrical pressure vessel has an inner diameter of 1.2 m and a thickness of 12 mm. Determine the maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 140 MPa. Under the same conditions, what is the maximum internal pressure that a similar-size spherical vessel can sustain?

EXAMPLE 1 (cont)


•The stress in the longitudinal direction will be

•The maximum stress in the radial direction occurs on the material at the inner wall of the vessel and is

• The maximum stress occurs in the circumferential direction.

Solutions

(Ans) MPa 28 N/mm 8.2

12

600140

2

1

p

pt

pr

MPa 701402

12

MPa 8.2(max)3 p

EXAMPLE 1 (cont)


• The maximum stress occurs in any two perpendicular directions on an element of the vessel is

Solution

(Ans) MPa 5.6 N/mm 6.5

122

600140

2

2

2

p

pt

pr

REVIEW OF STRESS ANALYSES


• Normal force P leads to:

• Shear force V leads to:

• Bending moment M leads to:

A

Pstressnormaluniform ,

It

VQondistributistressshear ,

beam) curved(for or

beam)straight (for ,

yRAe

MyI

Myondistributistressallongitudin

REVIEW OF STRESS ANALYSES (cont)


• Torsional moment T leads to:

• Resultant stresses by superposition:

Once the normal and shear stress components for each loading have been calculated, use the principal of superposition to determine the resultant normal and shear stress components.

tube)walled- thinclosed(for 2

shaft)circular (for ,

tA

TJ

Tondistributistressshear

m

EXAMPLE 2


A force of 15 kN is applied to the edge of the member shown in Fig. 8–3a. Neglect the weight of the member and determine the state of stress at points B and C.

EXAMPLE 2 (cont)


• For equilibrium at the section there must be an axial force of 15 000 N acting through the centroid and a bending moment of 750 000 N•mm about the centroidal or principal axis.

• The maximum stress is

Solution

MPa 75.340100

15000

A

P

MPa 25.1110040

121

50750003

max I

Mc

EXAMPLE 2 (cont)


• The location of the line of zero stress can be determined by proportional triangles

• Elements of material at B and C are subjected only to normal or uniaxial stress.

Solution

mm 3.33

100

1575

x

xx

(Ans) on)(compressi MPa 15

(Ans) (tension) MPa 5.7

C

B

EXAMPLE 3


The tank in Fig. 8–4a has an inner radius of 600 mm and a thickness of 12 mm. It is filled to the top with water having a specific weight of γw = 10 kNm3. If it is made of steel having a specific weight of γst = 78 kNm3, determine the state of stress at point A. The tank is open at the top.

EXAMPLE 3 (cont)


• The weight of the tank is

• The pressure on the tank at level A is

• For circumferential and longitudinal stress, we have

Solution

kN 56.311000

600

1000

61278

22

ststst VW

kPa 10110 zp w

(Ans) kPa 9.7756.3

(Ans) kPa 50010

2

10006002

10006122

100012

1000600

1

st

st

A

W

t

pr

EXAMPLE 4


The member shown in Fig. 8–5a has a rectangular cross section. Determine the state of stress that the loading produced at point C.

EXAMPLE 4 (cont)


• The resultant internal loadings at the section consist of a normal force, a shear force, and a bending moment.

• Solving,

Solution

kN 89.32 kN, 21.93 kN, 45.16 MVN

EXAMPLE 4 (cont)


• The uniform normal-stress distribution acting over the cross section is produced by the normal force.

• At Point C,

• In Fig. 8–5e, the shear stress is zero.

Solution

MPa 32.1

25.005.0

1045.16 3

A

Pc

EXAMPLE 4 (cont)


• Point C is located at y = c = 0.125m from the neutral axis, so the normal stress at C, Fig. 8–5f, is

Solution

MPa 16.63

25.005.0

125.01089.323

21

3

I

Mcc

EXAMPLE 4 (cont)


• The shear stress is zero.

• Adding the normal stresses determined above gives a compressive stress at C having a value of

Solution

MPa 5.6416.6332.1 I

Mcc

EXAMPLE 5


The rectangular block of negligible weight in Fig. 8–6a is subjected to a vertical force of 40 kN, which is applied to its corner. Determine the largest normal stress acting on a section through ABCD.

EXAMPLE 5 (cont)


• For uniform normal-stress distribution the stress is

• For 8 kN, the maximum stress is

• For 16 kN, the maximum stress is

Solution

kPa 1254.08.0

40

A

P

kPa 375

4.08.0

2.083

121max

x

xx

I

cM

kPa 375

8.04.0

4.0163

121max

y

xy

I

cM

EXAMPLE 5 (cont)


• By inspection the normal stress at point C is the largest since each loading creates a compressive stress there

Solution

(Ans) kPa 875375375125 c

chapter objectives to analyze the stresses in thin-walled pressure vessels. to develop methods to...

Documents

maximum stress

state of stress

stress distribution

uniaxial stress

solution slide

shear stress components

longitudinal stress

maximum internal pressure