Me16achapter2

ME16A: CHAPTER ONE

STATICALLY DETERMINATE STRESS SYSTEMS

INTRODUCTION 

A problem is said to be statically determinate if the stress within the body can be calculated purely from the conditions of equilibrium of the applied loading and internal forces.

2.1 AXIALLY LOADED BARS, STRUT OR COLUMN 2.1 AXIALLY LOADED BARS, STRUT OR COLUMN The external force applied at the ends of the member is balanced by internal force which is average stress x cross sectional area. F

F

σX If the axially loaded bar is cut perpendicular to the axis into two: F=

F A

σx A i. eσx =

(tensile stress)

If the bar is cut at an angle to the axis, two components of stress will be created: one normal to the plane, parallel to the plane,

τs .

σh τs . τs .

σh

σh and the other

2.1.1. Principle of St. Venant 

It states that the actual distribution of load over the surface of its application will not affect the distribution of stress or strain on sections of the body which are at an appreciable distance (> 3 times its greatest width) away from the load

Principle of St. Venant Contd.     

e.g. a rod in simple tension may have the end load applied. (a) Centrally concentrated (b) Distributed round the circumference of rod (c) Distributed over the end crosssection. All are statically equivalent.

Principle of St. Venant Concluded

F

F

Uniform stress – stress distribution not affected by distribution of load but by its resultant.

Alternatively:

The principle states that the stress distribution at sections far removed

from the point of application of concentrated forces depends on stress resultants and not on the actual distribution of forces.

Example 

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The piston of an engine is 30 cm in diameter and the piston rod is 5 cm in diameter. The steam pressure is 100 N/cm2. Find (a) the stress on the piston rod and (b) the elongation of a length of 80 cm when the piston is in instroke. (c) the reduction in diameter of the piston rod (E = 2 x 107 N/cm2; v = 0.3).

Solution p = 100 N/cm2

σx

F Piston rod (dia. = 5 cm) Piston (a)

For horizontal equilibrium of forces

π x 52 π x (302 − 52 ) 2 σx = 100 N / cm x 4 4 σ x = σ rod (b) Elongation =

(c)

εy =

π / 4(100)(302 − 52 ) = = 3500 N / cm2 2 π /4 x 5 FL σ L 3500 N / cm2 x 80 cm = = = 0.014 cm AE E 2 x 107 N / cm2

Change in dia 1 1 = [σ y − ν σ x ] = [0 − 0.3(3500)] Original dia E 2 x 107

= − 5.25 x 10−5 Change in diameter = 5.25 x 10 -5 x 5 = 0.0002625 cm

30 cm

2.2 THIN-WALLED PRESSURE VESSELS

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Cylindrical and spherical pressure vessels are commonly used for storing gas and liquids under pressure. A thin cylinder is normally defined as one in which the thickness of the metal is less than 1/20 of the diameter of the cylinder.

THIN-WALLED PRESSURE VESSELS CONTD 

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In thin cylinders, it can be assumed that the variation of stress within the metal is negligible, and that the mean diameter, Dm is approximately equal to the internal diameter, D. At mid-length, the walls are subjected to hoop or circumferential stress, and a longitudinal stress, .

Hoop and Longitudinal Stress

2.2.1 Hoop stress in thin cylindrical shell

Hoop stress in thin cylindrical shell Contd. 

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The internal pressure, p tends to increase the diameter of the cylinder and this produces a hoop or circumferential stress (tensile). If the stress becomes excessive, failure in the form of a longitudinal burst would occur.

Hoop stress in thin cylindrical shell Concluded

Consider the half cylinder shown. Force due to internal pressure, p is balanced by the force due to hoop stress,

σ h.

i.e. hoop stress x area = pressure x projected area

σ h x

2Lt = Pxd L

σ h =

(P d) / 2 t

Where: d is the internal diameter of cylinder; t is the thickness of wall of cylinder.

2.2.2. Longitudinal stress in thin cylindrical shell

Longitudinal stress in thin cylindrical shell Contd.

The internal pressure, P also produces a tensile stress in longitudinal direction as shown above. Force by P acting on an area longitudinal stress,

π dt

π d2 4

is balanced by

acting over an approximate area, σ L

(mean diameter should strictly be used). That is:

π d2 σ dt= Px L xπ 4

Pd σ = L 4t

Note 

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1. Since hoop stress is twice longitudinal stress, the cylinder would fail by tearing along a line parallel to the axis, rather than on a section perpendicular to the axis. The equation for hoop stress is therefore used to determine the cylinder thickness. Allowance is made for this by dividing the thickness obtained in hoop stress equation by efficiency (i.e. tearing and shearing efficiency) of the joint.

Longitudinal stress in thin cylindrical shell Concluded

Example 

A cylindrical boiler is subjected to an internal pressure, p. If the boiler has a mean radius, r and a wall thickness, t, derive expressions for the hoop and longitudinal stresses in its wall. If Poisson’s ratio for the material is 0.30, find the ratio of the hoop strain to the longitudinal strain and compare it with the ratio of stresses.

Solution Hoop stress will cause expansion on the lateral direction and is equal to σy while the longitudinal stress is Hoop stress, σh

=

Longitudinal stress,

p d p x 2r p r = = 2t 2t t

σL =

σx ieσy

p d p x 2r p r = = i. e.σx 4t 4t 2t

(a) Stress ratio = 2 (b) εx =

1 1 pr pr 0.2 pr [σx −υσy ] = [ −0.3 ] = ( Longitudinal strain) E E 2t t E t 1 E

1 pr pr 0.85 pr −0.3 ] = ( Hoop strain) E t 2t E t Hoop strain 0.85 Ratio of strains = = = 4.25 Longitudinal strain 0.2

εy = [σy −υσx ] = [

2.2.3 Pressure in Spherical Vessels 2.2.3 Pressure in Spherical Vessels Problems dealing with spherical vessels follow similar solutions to that for thin cylinders except that there will be longitudinal stresses in all directions. No hoop or circumferential stresses are produced.

i.e

Pd 4t

σL =

2.3 STRESSES IN THIN ROTATING RINGS 

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If a thin circular ring or cylinder, is rotated about its centre, there will be a natural tendency for the diameter of the ring to be increased. A centripetal force is required to maintain a body in circular motion. In the case of a rotating ring, this force can only arise from the hoop or circumferential stress created in the ring.

STRESSES IN THIN ROTATING RING

STRESSES IN THIN ROTATING RINGS CONTD. Consider a thin ring of mean radius, r, density,

ρand

having a cross-sectional area, A, to be rotating about centre O with an angular velocity, w (rad/s). For an elemental length which sustends an angle dθ at O, as shown in Fig. (a). Circumferential length of element = r A dθ

Volume of element = Mass of element =

r dθ

ρ r A dθ

Centripetal force to maintain circular motion = mass x w2 r = =

ρw2 r2 A dθ

ρr A dθ w2 r

STRESSES IN THIN ROTATING RINGS CONTD. If the hoop stress created in the ring is Force F acting on cross-section =

σh

σh . A (see diagram b)

Radial component of the force, F = 2 ( σh . A) sin dθ2 = 2 ( σh . A) dθ2 ( for small dθ) This radial component of forces, F supplies the required centripetal force to maintain the element in circular motion. Thus: 2 ( σh . A) dθ2 = i.e.

ρw2 r2 A dθ

σh = ρw2 r2

Putting velocity, V = wr ;

σh = ρV2

STRESSES IN THIN ROTATING RINGS CONCLUDED 

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Hence: Hoop stress created in a thin rotating ring, or cylinder is independent of the crosssectional area. For a given peripheral speed, the stress is independent of the radius of the ring.

EXAMPLE 

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A thin steel plate having a tensile strength of 440 MN/m2 and a density of 7.8 Mg/m3 is formed into a circular drum of mean diameter 0.8 m. Determine the greatest speed at which the drum can be rotated if there is to be a safety factor of 8. E = 210 GN/m2.

SOLUTION 440 MN / m2 = 55 MN / m2 Factor of safety (8)

ρ σ ρ ρ σ ρ π

Greatest stress to be applied =

== 78 . Mg / m3 7800 kg / m3

hoop stress,

h

V2 =

=

w2 r2

55 x 106 N / m2 w= = = 209.9 rad / s 2 3 2 2 r 7800 kg / m x 0.4 (m ) h

rad = 1800,

rad = 57.2960

In 3600 (1 rev), we have 360/57.296 = 6.283 rad i.e 209.9 rad /s = 33.407 rev/s = 2004.4 rev/min

2.4 STATICALLY INDETERMINATE STRESS SYSTEMS

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There is the need to assess the geometry of deformation and link stress and strain through modulus and Poisson’s ratio for the material.

2.4.1 Volume Changes 

Example: A pressure cylinder, 0.8 m long is made out of 5 mm thick steel plate which has an elastic modulus of 210 x 103 N/mm2 and a Poisson’s ratio of 0.28. The cylinder has a mean diameter of 0.3 m and is closed at its ends by flat plates. If it is subjected to an internal pressure of 3 N/mm2, calculate its increase in volume.

SOLUTION Hoop stress, σ h = (P d) / 2 t =

3 N / mm2 x 300 mm = 90 N / mm2 2 x 5 mm Longitudinal stress, σ L = (P d) / 4 t = 45 N/mm2 Longitudinal strain,

εL =

1 1 [σ L − υ σ h ] = [45 − 0.28 x 90] = 0.00009429 3 2 E 210 x 10 N / mm

SOLUTION CONCLUDED Hoop strain,

1 1 ε h = [σ h − υ σ L ] = [90 − 0.28 x 45] = 0.0003686 3 2 E 210 x 10 N / mm Volumetric strain =

ε L + 2 ε h = 0.00083134 (See Section 1.4)

Original volume of cylinder is equal to :

π x 3002 x 800 = 56.5487 x 10−6 mm3 4 Increase in volume = 56.5487 x 10−6 x 0.00083134 = 47009 mm3

Example 

The dimensions of an oil storage tank with hemispherical ends are shown in the Figure. The tank is filled with oil and the volume of oil increases by 0.1% for each degree rise in temperature of 10C. If the coefficient of linear expansion of the tank material is 12 x 10-6 per 0C, how much oil will be lost if the temperature rises by 100C.

SOLUTION For 100C rise in temperature: Volumetric strain of oil = 0.001 x 10 = 0.01 Volumetric strain of tank =

3 α T

= 3 x 12 x 10-6 x 10 = 0.00036 Difference in volumetric strain = 0.01 - 0.00036 = 0.00964 Volume of tank = π x 102 x 100 + 4/3 x π x 103 = 10000 π + 1333.33 π = 11333.33 π m3 Volume of oil lost = strain difference x volume of tank = 0.00964 x 11333.33 π m3 = 343.2 m3.

2.4.2 IMPACT LOADS

L

W h x = dl

IMPACT LOADS CONTD. Consider a weight, W falling through a height, h on to a collar attached to one end of a uniform bar. The other end of the bar is fixed. Let dl be the maximum extension caused and

be the stress set up. σ

Let P be the equivalent static or gradually applied load which would cause the same extension, dl Strain energy in the bar at this instant = 1/2 P. dl Neglecting loss of energy at impact:

L

W h x = dl

IMPACT LOADS CONTD. Loss of potential energy of weight, W on impact = Gain of strain energy of bar i.e. W (h + dl) = 1/2 P dl Recall that : i.e

PL dl = AE

PL 1 P2 L W (h + ) = ( ) AE 2 AE

PL 1 P 2 L Wh + W = AE 2 AE 1 P 2 L WPL i. e. − − Wh = 0 2 AE AE AE Multiplying by L P2 WhAE − WP − = 0 2 L

L

W h x = dl

IMPACT LOADS CONTD. Recall the quadratic equation formula:

−+ b b 2 −4ac P= 2a i. e. P =W + W 2 +4 x

1 WhAE 2 L

L

W

Using only the positive root:

h

P =W + W +2WhAE / L 2

2hAE =W[1 + 1 + ] WL

From which

PL dl = AE

x = dl

and

P σ= A

can be obtained.

IMPACT LOADS CONTD.

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Note: 1. For a suddenly applied load , h = 0 and P = 2 W i.e the stress produced by a suddenly applied load is twice the static stress. 2. If there is no deformation, ‘ x’ of the bar, W will oscillate about, and come to rest in the normal equilibrium position.

IMPACT LOAD CONCLUDED

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3. The above analysis assumes that the whole of the rod attains the same value of maximum stress at the same instant. In actual practice, a wave of stress is set up by the impact and is propagated along the rod. This approximate analysis, however, gives results on the “safe” side.

EXAMPLE

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A mass of 100 kg falls 4 cm on to a collar attached to a bar of steel, 2 cm diameter, 3 m long. Find the maximum stress set up. E = 205,000 N/mm2.

SOLUTION CONCLUDED π x 202 Area of bar = = 314.2 mm2 4 W = 100 x 9.81 = 981 N

2 x 40 x 314.2 x 205000 P = 981[1 + 1 + ] = 42029.65 N 981 x 3000 P 42029.65 N 2 Stress = = = 133 . 77 = 134 N / mm A 314.2 mm2

ALTERNATIVE SOLUTION Using Alternative Equation:

σ ' =σ +(σ 2 +

σ =

2σ Eh 1/ 2 ) L

By Benham and Crawford (1987)

100 x 9.81 981 = = 3122 . N / mm2 ; σ 2 = 9.748 2 20 314.2 πx 4

σ ' = 3122 . + 9.748 +

2 x 3122 . x 205000 x 40 3000

= 133.80 = 134 N/mm2.