5. Pressure Vessels and Axial Loading Applications

5. Pressure Vessels and Axial Loading ApplicationsAxial Loading Applications

5.1 Introduction

• Mechanics of materials approach (analysis)- analyze real structural elements as idealized models subjected simplified

loadings and restraints.

5.2 Deformation of axially loaded members

- Uniform member:

- Multiple loads/sizes:

- Nonuniform deformation:1 1

0

,

n ni i

ii i i i

L x

x

L PLLE EA

PLE A

P dxEA

sd e d

d d

d

= =

= = =

= =

=

å å

ò0xEAò


• Example Problem 5-1

DA = 100 mm, DB_in = 100 mm, DB_out = 150 mm, DC_in = 125 mm, DC_out = 200 mm, EA = 73 GPa, EB = 100 GPa, EC = 210 GPa

- Determine the overall shortening of the member



A homogeneous bar of uniform cross section A hangs vertically

- Elongation of the bar due to its own weight W in terms of W, L, A, and E

- Elongation of the bar if it is also subjected to an axial tensile force P at its lower end- Elongation of the bar if it is also subjected to an axial tensile force P at its lower end

5.3 Deformations in a system of axially loaded bars

( )2 2

2 2 2 2 22 2

In a similar manner,

AB f i B B

AB AB B B B

AB B

BE B

L L L v u L

L L L Lv v uv

u

d

d dd

d

= - = + + -

+ + = + + +® @

@

( ) ( )2 2

2 2 2 2 2 2 2 2

In case of ,

cos sin

2 cos 2 cos sin 2 sinsin cos

BE B

BC

BC B B

BC BC B B B B

BC B B

uL R

R u R v R

R R R Ru u R Rv vv u

d

d q q

d d q q q qd q q d

@=

= - + + -

+ + = - + + + +

® @ - = sin cosAB BEq d q-

For small displacements, the axial deformation in any barmay be assumed equal to the component of thedisplacement of one end of the bar taken in the direction ofthe unstrained orientation of the bar.

5.3 Deformations in a system of axially loaded bars


Cross-sectional areas of tie rod AB and pipe strut BC are650 mm2 and 925 mm2, respectively. E = 200 GPa

- Normal stress in AB and BC- Normal stress in AB and BC

- Lengthening or shortening of AB and BC

- Horizontal and vertical components of the displacementof B

- Angles through which members AB and BC rotate

5.4 Statistically indeterminate axially loaded members

- The number of unknowns should be the same as the number of equations.

- For many mechanical systems, the equations of equilibrium are not sufficient forthe determination of axial forces in the members and reactions at the supports.

- Additional equations involving the geometry of the deformations of the system- Additional equations involving the geometry of the deformations of the systemcan be helpful to solve the problems.

- Hooke’s law and the definition of stress and strain can be used to relatedeformations and forces when all stresses are under the proportional limit of thematerials.


- Calculation of the tension of the cable

( )

2

1) Rigid cable

0 : 0

2) Deformable cable

0 : cos 0 cos

cos

cos

B

B

WaM TR Wa TR

WaM TR W a TR

TL AE WaAE L RR R EA WaL

q q

dd q

d q q q

= - = ® =

= - = ® =

= ® =

= ® =

å

å


- Comparison of tensions with various elastic moduli of the cable



A pier has nine 25-mm-diameter steel reinforcing bars (E = 200 GPa) in theconcrete (E = 30 GPa). P = 650 kN

- Stresses in the concrete and the steel bars

- Shortening of the pier- Shortening of the pier



A 0.5 in.-diameter bolt (E = 30,000 ksi) & a sleeve (E = 15,000 ksi) of 0.375 in2

cross-sectional area are deformed by a nut (0.02 in.).

- Stresses in the bolt and the sleeve

5.5 Thermal Effects

- When the deformation of a bar is prevented the thermal strain is offset by themechanical strain in opposite direction.

0total T T L L TL LEs ssd d d e e a= + = + = D + =

5.5 Thermal Effects


A 10-m section steel rail (E = 200 GPa, α = 11.9(10-6)/˚C) has a cross-sectional areaof 7,500 mm2. Deformations in all directions are restricted. For an increase intemperature of 50 ˚C, determine

- Normal stress in the rail- Normal stress in the rail

- Internal force on a cross section of the rail

5.5 Thermal Effects


The temperature increases 100 ˚C. The thermal coefficients of expansion and themodulus of elasticity are 22(10-6)/˚C and 75 GPa for the rod A, and 12(10-6)/˚C and200 GPa for the rod B. The cross sectional area of A and B are 1000 mm2 and 500mm2, respectively. The rod CD is rigid.

- Normal stress in bars A and B- Normal stress in bars A and B

- Vertical component of the displacement of point D

5.6 Stress concentrations

- The stress is concentrated around discontinuities that interrupt the stresspath.


- Stress concentration factor, K, is defined based on an area at the reducedsection (net area) or on the gross area:

PKA

s =


- Kirsch’s solution: stress distribution around a small circular hole in awide plate under uniform unidirectional tension.

2 2 4

2 2 4

2 4

4 31 1 cos 22 2

3

ra a ar r r

a a

s ss q

s ss q

æ ö æ ö= - - - +ç ÷ ç ÷

è ø è øæ ö æ ö

= + + +

( )

2 4

2 4

2 4

2 4

2 40 0 , 3

31 1 cos 22 2

2 31 sin 22

0 ; 1 2cos 2 ; 0

32 1.0742

r

r r

r a

a ar r

a ar r

at r a

a ar r

q

q

q q

q q

s ss q

st q

s s s q t

ss s s= = =

æ ö æ ö= + + +ç ÷ ç ÷

è ø è øæ ö

= + +ç ÷è ø

= = + = =

æ ö= + + =ç ÷

è øo o


- Saint Venant’s Principle:Localized stress concentration disappears at some distance.→ The difference between the stresses caused by statically equivalent loadsystems is insignificant at distances greater than the largest dimension of thearea over which the loads are acting.



The machine part is 20 mm thick and the maximum allowable stress is 144 MPa.

- The maximum value of P

5.7 Inelastic behavior of axially loaded members

- When the stresses in some members extend into the inelastic range,stress-strain diagrams must be used to relate the loads and the deflectionsand solve the problem.

Steel (elastoplastic) Aluminum alloy(strain hardening)

Magnesium alloy(strain hardening)

5.7 Inelastic behavior of axially loaded members


A is a 0.5-in.-diameter steel rod (elastoplastic) which has a proportional limit of 40ksi and a modulus of elasticity of 30,000 ksi. Pipe B has a cross-sectional area of 2in.2 and shows strain-hardening as below. Load P is 30 kip.

- Normal stress in A and B- Normal stress in A and B

- Displacement of plate C

5.8 Thin-walled pressure vessels

- A pressure vessel is a thin walled container whose wall thickness is sosmall that the normal (axial, meridional) stress on a plane perpendicular tothe surface is uniform throughout the thickness.

1) Spherical pressure vessels: no shear stress (strain)

( ) ( )2 22a aprP p r R rtt

p p s s= ® =


2) Cylindrical pressure vessels:

Hoop (tangential, circumferential) stress vs. axial (meridional) stress

( ) ( )2 2h x hprQ Lt P p rLt

pr

s s= ® =

22a h aprt

s s s= \ =


3) Thin shells of revolution:

Shapes made by rotating plane curves (meridian) about an axis – sphere,hemisphere, torus (doughnut), cylinder, cone, and ellipsoid.

2 sin 2 sin 0m m t tP F d F dq q- - =

( )( ) ( ) ( )2 2 2 2 sin 2 2 sint t m m m t t m t m m t

m tt m

m t

p r d r d tr d d tr d d

pd dr r t

q q s q q s q q

s sq q

= +

» ® + =



A cylinder whose diameter is 1.5 m is constructed by 15-mm-thick steelplate, and is under 1500 kPa of inner pressure.

- Normal and shear pressure at welded edge forming 30˚ from hoop- Normal and shear pressure at welded edge forming 30˚ from hoopdirection



A pressure vessel of ¼ in. steel plate is closed by a flat plate. The meridianline follows a parabola of y = x2/4. P = 250 psi, rm = (1+(dy/dx)2)1.5/(d2y/dx2)

- Meridional and tangential stress σm and σt at a point 16 in. above the- Meridional and tangential stress σm and σt at a point 16 in. above thebottom

5.9 Combined effects – axial and pressure loads

- A case of an axial load (P) combindedly added to an axial pressure (p)

x xP xps s s= +

5.9 Combined effects – axial and pressure loads


A cylindrical pressure tank whose diameter and wall thickness are 4 ft and¾ in., respectively is under 400 psi of pressure and 30,000 lb of axial load.

- xy, , and x ys s t-

- Normal and shearing stresses on an inclined plane oriented at +30˚ fromthe x-axis

xyx y

5.10 Thick-walled cylindrical pressure vessels

( )( ) 2 sin 02

0 (neglecting higher-order terms)

r r r t

rr t

dd d d dL d dL d dL

dd

qs s r r q s r q s r

sr s sr

+ + - - =

® + - =

- Force equilibrium in radial direction


( )( ) ( ) 11 1 2 from Hooke's lawa a r t r t a aE CE

e s n s s s s s en

= - + ® + = - =

- Obtaining σr and σt

10 2 2r rr t r

d d Cd ds sr s s r sr r

+ - = ® + =

( )

( )

( )

22

1

22 2

1 1 2

2 21 1 12 2

2 2

Integrating 2

, 2

r rr

rr

r t r t

d d

d d Cd d

dC C C

dC CC C C

r r

r s sr rs rr r

r sr r s r

r

s s s sr r

= + =

= ® = +

\ = + = - + =Q


( )

( )

2 22 2

1 22 2 2 2

at at

r i

r o

i oi o

p ap b

a b p pa p b pC Cb a b a

s rs r

= - =

= - =

--® = = -

- -

- Applying boundary conditions

( )( )

( )( )

2 22 2

2 2 2 2 2

2 22 2

2 2 2 2 2

i oi or

i oi ot

a b p pa p b pb a b a

a b p pa p b pb a b a

sr

sr

--\ = -

- -

--= +

- -


( )

( )

2 2

When 01

t r t

r t

a

d pd e prd e r

s

e s ns

= =

=

=

= -

- Displacements

( )

( )

1t t r

r t r

E

E

e s ns

rd s ns

= -

® = -

5.10 Thick-walled cylindrical pressure vessels- Case 1: Internal pressure only (po = 0)

( )( )

( )( )

( ) ( ) ( )

2 22 2 2 2

2 2 2 2 22 2 2

2 22 2 2 2

2 2 2 2 22 2 2

22 2

1

1

1 1

i oi o ir r

i oi o it t

i

a b p pa p b p a p bb a b a ab a

a b p pa p b p a p bb a b a ab a

a p b

s sr

s sr

rd s ns d n r n

- æ ö-= - ® = -ç ÷- -- è ø

- æ ö-= + ® = +ç ÷- -- è ø

é ù= - ® = - + +( ) ( ) ( ) ( )2 22 2

1 1ir t r r

a p bE b a Erd s ns d n r n

ré ù= - ® = - + +ë û-

- Case 2: External pressure only (pi = 0)

( ) ( ) ( )

2 2

2 2 2

2 2

2 2 2

22 2

2 2

1

1

1 1

or

ot

or

b p ab a

b p ab a

b p bb a E

sr

sr

d n r nr

æ ö= - -ç ÷- è ø

æ ö= - +ç ÷- è ø

é ù= - - + +ë û-

-Case 3: External pressure on a solidcircular cylinder (a = 0)

1

r o

t o

r o

pp

pE

ss

nd r

= -

= -

-= -

5.10 Thick-walled cylindrical pressure vessels• Example Problem 5-19

A steel cylinder ( E = 30,000 ksi and n = 0.30) whose inside and outside diameters are 8 in. and 16 in., respectively is subjected to an internal pressure of 15,000 psi. The axial load is zero.

- The maximum tensile stress- The maximum shearing stress- The maximum shearing stress- The increase of the inside diameter- The increase of the outside diameter

5.11 Design- Failure is defined as the state in which a member or structure no longer

functions as intended.

- Failure modes are limited to elastic failure (yielding) here.

- Allowable stress design: Strength ³ Stress

- Factor of safety: Strength ³ Factor of safety ´ Stress

5.11 Design• Example Problem 5-20

An axially loaded circular bar is subjected to a load of 6500 lb with the factor of safety of 1.5. The yield strength of the bar is 36 ksi.

- Minimum diameter of the bar


An axially loaded circular bar has a yielding stress of 250 MPa. The factor of safety is to be 1.8.

- Minimum diameter of the bar

5.11 Design• Example Problem 5-22

A 40-lb light is supported by a rigid wire whose yield stress is 62 ksi. The factor of safety should be 3.

- Minimum diameter of the wire

Problems: 5-1, 13, 28, 42, 59, 71, 89, 105, 124, 136 by : May 14

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5. Pressure Vessels and Axial Loading Applications