CHAPTER - 12 Strength of Materials

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CHAPTER - 12 Strength of Materials

Mechanical Properties of Metals: The following properties of metals are important from the point of view of

their use in engineering generally and their use in boilers in particular:

Strength: It is the ability of a material to resist the externally applied forces without breaking or yielding.

Elasticity: Elasticity is the property that enables a metal to return to its original shape when the force that causes the change of shape is removed. Whenever a single force (or a system of forces) acts on a body, it undergoes some deformation and the molecules offer some resistance to the deformation. When the external force of resistance vanishes, the body springs back to its original position. But it is only possible, if the deformation, caused by the external force, is within a certain limit. Such a limit is called Elastic Limit. A body is said to be perfectly elastic, if it returns completely to its original shape and size after the removal of external forces. With Elasticity is related another property namely, Stiffness which is the ability of a material to resist deformation under stress. The modulus of elasticity is the measure of stiffness.

Plasticity: It is property of a material due to which it retains the deformation produced under loads.

Hardness: Hardness refers to the ability of a metal to resist abrasion, penetration, cutting action, or permanent deformation. Structural parts are often formed from metals in their soft state and then heat treated to harden them so that the finished shape will be retained and these parts will resist abrasion, penetration and cutting action of any other objects coming in contact with them. Hardness and Strength are closely associated properties of all metals.

Brittleness: Brittleness is the property of a metal that allows negligible bending or deformation without shattering. In other words, a brittle metal is apt to break or crack without change of shape when subjected to stress causing forces.

Malleability: A metal that can be hammered, rolled, or pressed into various shapes without cracking or breaking or other detrimental effects is said to be malleable. Copper is one example of a malleable metal.

Ductility: Ductility is the property of a metal that permits it to be permanently drawn, bent, or twisted into various shapes without breaking. This property is essential for metals used in making wire and tubing. Ductility is similar to malleability.

Toughness: A material that possesses toughness will withstand tearing or shearing and may be stretched or otherwise deformed without breaking.


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Resilience: It is the property of a material to absorb energy and to resist shock and impact loads. It is measured by the amount of energy absorbed per unit volume within elastic limit.

Creep: When a part is subjected to a constant stress at high temperature for a long period of time, it tends undergo a slow and permanent deformation called creep.

Fatigue: When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue. The failure is caused by means of a progressive crack formation which is usually fine and microscopic size.

Stress and Strain:

Stress: Solid materials subjected to external forces on it resist these external forces and tries to retain its shape and size. This resistance of the material comes from within the material. Stress is defined as the internal force per unit area on the material which resists external forces on the material. It is expressed in pounds per square inch or kg per cm2 [Stress is always expressed by engineers with the notation lb per sq in. or kg per sq cm. to differentiate it from the psi designation of pressure, which is an external force per unit area on the material].

Suppose F is external force acting at a right angle to the cross-sectional area ‘a’ of a bar.

This force produces a normal Stress σ = F/a. Stress can be of the following types:

Tensile Stress:

When a section is subjected to two equal and opposite pulls, as a result of which it

tends to lengthen, the stress induced is called Tensile Stress (σ T). The

corresponding strain is called Tensile Strain.

Compressive Stress:

When a section is subjected to two equal and opposite pushes, as a result of which the body tends to shorten its length, the stress induced is called

compressive stress (σ C). The corresponding strain is called Compressive Strain.

Shear stress:

Shear stress has the same units as normal stress (force / area) but represents a stress that acts parallel to the cross section. This is different from normal stress which acts perpendicular (normal) to the cross section. Torsion is a force that causes shear stress but this is not the only force that can cause shear stress. For example, a beam that supports a shear force also has a shear stress over the section (even without torsion).


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Bending stress: A beam supported on each end and loaded in the middle will develop a bending

stress. The beam, when bending, will actually be under a tension stress on one side and a compression stress on the opposite side. Flat plates and stayed surfaces in boilers are some elements that are subjected to bending stress.

Strain: When a body or material is subjected to external forces, internal forces are developed within the body or material to resist these forces, but there is always some deformation with load. This deformation is called Strain. For example, a steel rod will stretch or get strained when pulled upon by an external force but at the same time the steel rod will develop stress which will resist the external forces by exerting equal and opposite forces so that the further deformation or ‘strain’ is stopped. The total stretch is expressed in a length measurement such as inches or centimeters. Strain is defined as the stretch per unit length, or deformation of a body per unit length, and is always expressed as inches per inch, centimeters per centimeter, etc.

Thus strain ε = ΔL/L or ΔL = ε * L,

Where, ε = strain, [This symbol is called epsilon]

ΔL = change of length of the body

L = original length of the body.

Stress-strain diagram:

If we look at a metal rod in simple tension as shown in Fig. 1, we see that there will be an elongation (or deformation) due to the tension. If we then graph the tension (force) verses the deformation we obtain a result as shown in Fig. 2.

Fig. 1

Fig. 2 shows that, if the metal rod is tested by increasing the tension in the

rod, the deformation increases. In other words stress increases as strain increases. In the first region the deformation increases in proportion to the force or in other words, the stress is proportional to strain. That is, if the amount of force is doubled, the amount of deformation is doubled or in other words if strain is doubled the stress also becomes double. This is a called


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Hooke's Law and could be written this way: F = k Deformation, where ‘k’ is a constant depending on the material (and is sometimes called the spring constant). After enough force has been applied the material enters the plastic region - where the force and the deformation are not proportional, but rather a small amount of increase in force produces a large amount of deformation. In this region, the rod often begins to 'neck down', that is, the diameter becomes smaller as the rod is about to fail. Finally the rod actually breaks.

Fig. 2

The point at which the Elastic Region ends is called the Elastic Limit, or the Proportional Limit. Strictly speaking, these two points are not quite the same. The Elastic Limit is the point at which permanent deformation occurs, that is, after the elastic limit, if the force is taken off the sample, it will not return to its original size and shape, permanent deformation has occurred. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). Although these two points are slightly different, they will be treated to be identical in this course. It is observed that if the material is stressed beyond elastic limit, in the plastic stage on removal of load the deformation is permanent. Thus the material yields before the applied forces or load. The point at which this occurs is called yield point. The stress corresponding to the yield point is known as Yield Point Stress. The largest stress obtained by dividing the largest value of load reached in a test to the original cross-section area is called Ultimate Stress. After reaching the ultimate stress neck is formed in the sample being tested, resulting into less amount of force required to break the specimen. The stress at the point of failure is called breaking stress.


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Fig. 3

Now let us consider a stress-strain diagram. As we see from Fig. 3, the Stress verses Strain graph has the same shape and regions as the force verses deformation graph in Fig. 2. In the elastic (linear) region, since stress is directly proportional to strain, the ratio of stress/strain will be a constant (and actually equal to the slope of the linear portion of the graph). This constant is known as Young's Modulus, and is usually symbolized by an E or Y. We will use E for Young's modulus. We may now write Young's Modulus = Stress/Strain, or: E = σ/ ε. (This is another form of Hooke's Law). The value of Young's modulus - which is a measure of the amount of force needed to produce a unit deformation - depends on the material. Young's Modulus for Steel is 30 x 106 lb/in2, for Aluminum E = 10 x 106 lb/in2, and for Brass E = 15 x 106 lb/in2.

Hooke’s Law:

When a material is loaded within its elastic limit, the stress is proportional to the strain.

Mathematically, Stress/ Strain = E = a constant This law holds good equally for tension as well as compression.

Modules of Elasticity (Young’s Modulus):

Whenever a material is loaded within its elastic limit, the stress is proportional to strain.

σ = E * ε or E = σ / ε

where, σ = stress;

ε = strain and

E = a constant of proportionality known as modules of elasticity or young’s modulus. Numerically, it is that value of tensile stress, which when applied to a uniform bar will increase its length to double the original length, if the material of the bar could remain perfectly elastic throughout such an excessive strain.


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Poisson’s Ratio

It has been found experimentally that when a body is stressed within elastic limit, the lateral strain bears a constant ratio to the linear strain.

Mathematically, Lateral Strain/ Linear Strain = Constant

This constant is known as Poisson’s ratio and is denoted by 1/m or μ.

Volumetric Strain

When a body is subjected to a system of forces, it undergoes some changes in its dimensions. In other words, the volume of the body is changed. The ratio of the change in volume to the original volume is known as volumetric strain. Mathematically, volumetric strain. Thus, mathematically, ε = δV/V Bulk Modulus

When a body is subjected to three mutually perpendicular stresses, of equal intensity, then the ratio of the direct stress to the corresponding volumetric strain is known as bulk modulus. It is usually denoted by K.

K = Direct stress / Volumetric Strain = σ/ δV/V

Relation Between Bulk Modulus and Young’s Modulus

K = m.E / 3(m-2) = E / 3(1-2μ)

Relation Between Young’s Modulus and Modulus of Rigidity

G=m.E / 2(m +1) = E / 2(1+μ)

Deformation of a body due to Force acting on it:.

Consider a body subjected to a tensile stress.

Let P = Load or force acting on a body. l = Length of a body A= Cross section area of the body σ = Stress induced in the body E = Modulus of elasticity for the material of the body ε = Strain and Δl = Deformation of the body.

Now, σ = P / A Strain, ε = σ / E = P / AE (As σ = P / A)

and

Deformation, Δl = ε * l = (P * l) / A * E (As σ = P / AE)

The above formulae holds good for compressive stress also.


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Stresses due to change in temperature:

Temperature above designed limits has the immediate effect of lowering the permissible stress on the material due to physical change. Certain parts of boilers, particularly tubes, tube sheets, furnaces in scotch marine boilers, and cast parts in CI boilers are very susceptible to temperature or overheating damages. A large temperatures increase in a material, with accompanying lower permissible stress levels, is one of the most common causes of boiler damages. Low water, poor circulation, and scale are some causes of overheating of the material beyond safe stress levels. Let us not forget that the firing side of boilers is hot enough to melt steel. And with existing pressure on the water or steam side, it does not require much overheating to cause ruptures, bulges, and other deformation. Thus if the material is stressed well beyond the yield stress at high temperatures, permanent deformation will take place. In severe cases, the ultimate stress of the material is reached at the elevated temperature level, leading to complete rupture of the affected parts of the boiler. If the boiler part is free to expand or contract, no increase in stress occurs, unless the stress is influenced by too high a temperature rise, thus lowering the permissible stress due to physical changes caused by temperature. But expansion or contraction, even if the part is free to move, can be considerable. This can be calculated by the following equation:

Δl = l * * (T2 – T1) e = nl (T2 – T1)

where, Δl = change in length l = original length T1 = original temperature T2 = final temperature

= coefficient of thermal expansion (change in length per unit of length per degree change in temperature)

If the ends of the bar are fixed to rigid supports so that its expansion is prevented, then compressive strain induced in the bar,

ε = Δl / l = (l * * t) / l = * t

Stress = σ = ε E = * t * E

Circumferential & Longitudinal Stress in Thin Cylinder: Thin-walled cylinders, have two stresses, called

(1) longitudinal and (2) circumferential

By thin wall pressure vessel we mean a container whose wall thickness is less than 1/10 of the radius of the container.

Let us now consider a cylindrical pressure vessel as shown in Fig. 4, where we have cut a cross section of the vessel and have shown the forces due to the internal pressure, and the balancing forces due to the longitudinal stress which develops in the vessel walls. (There is also a transverse or circumferential stress which develops, and which we will consider next.)


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Fig. 4

The longitudinal stress may be found by equating the force due to internal gas/fluid pressure with the force due to the longitudinal stress as follows:

P(A) = (A'); or P(3.1416 * R2) = (2 * 3.1416 * R * t), then canceling terms and solving for the longitudinal stress, we have:

= P R / 2 t

where, P = internal pressure in cylinder; R = radius of cylinder and t = wall thickness

To determine the relationship for the transverse stress, often called the hoop stress, we use the same approach, but first cut the cylinder lengthwise as shown in Fig. 5 below.

Fig. 5

We once again equate the force on the cylinder section wall due to the internal pressure with the resistive force which develops in walls and may be

expressed in terms of the hoop stress, . The effective area the internal pressure acts on may be consider to be the flat cross section given by (2*R*L). So we may write:

P(A) = (A’); or P(2*R*L) = (2*t*L), then canceling terms and solving for the hoop stress, we have:

= P R / t ; where


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P = internal pressure in cylinder; R = radius of cylinder and t = wall thickness

Note that the hoop stress is twice the value of the longitudinal stress, and is normally the limiting factor. Also note that equations for both longitudinal and circumferential stresses (due to pressure) are independent of the length of the vessel. But if a vessel is very long, the bending stress will have to be added to the stress due to pressure

Thin spherical shells:

Next we consider the stress in thin wall spherical pressure vessels. Using the approach as in cylindrical vessels, in Fig. 6, we have shown a half section of a spherical vessel.

Fig. 6

If we once again equal the force due to the internal pressure with the resistive force expressed in term of the stress, we have:

P(A) = (A'); or P(3.1416 * R2) = (2 * 3.1416 * R * t), then canceling terms and solving for the stress, we have:

= P R / 2 t ; where,

P = internal pressure in cylinder; R = radius of cylinder and t = wall thickness

Stress concentration:

If a structural material has an abrupt change in a section, the stress distribution is not uniform over the cross-sectional area of the material. Near the abrupt change the stress is much higher than calculated. The affected section is said to have a stress concentration section, or area.

Stress concentration plays an important part in structural members subject to repeated type of loadings, as the stress concentration can lead to cracks and fatigue failures. If the stress concentration is severe enough (even in normal loading), stresses may be induced far above the normal expected stress. Sharp corners in welded joints and other sharply formed shapes must be avoided. Thus openings cut into plates must be reinforced to strengthen the edges around the opening against stress concentration.


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Stress concentration Factor:

It is defined as the ratio of the maximum stress in a member ( at a notch or fillet) to the nominal stress at the same section based on net area.

Stress concentration factor Kt = maximum stress / nominal stress

Torsion:

A turning force is applied to transmit energy by rotation. This turning force is applied either on the rim of a pulley keyed to the shaft, or to any other suitable point at same distance from the axis of the shaft. The product of this turning force, and the distance between the point of application of the force, and the axis of the shaft is known as torque, turning moment or twisting moment. The shaft is said to be subsisted to torsion. Due to this torque every cross-section of the shaft is subjected to a shear stress.

Torsional stress and strain:

Consider a shaft fixed at one end, and subjected to a torque at the other end as shown in Fig. 9. Let, T= Torque in kg-cm; L = Length of the shaft and R = Radius of the shaft

As a result of this torque, every cross section of the shaft will be subjected to shear stresses. Let the line CA on the surface of the shaft be deformed to CA and OA to OA1 s shown in figure. Let LA OA1 = 0 LA OA1= 0 in radians

σs = shear stress induced at the surface.

C = Modulus of rigidity, Also known as torsional rigidity of the shaft material.

Shear Strain = deformation per unit length.

= AA1 / l = tan = [ tan being very small tan = 0 ]

Fig. 9

The arc AA’ = R*

A

A1

θ Φ

L

T

T

O O

R


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= A A’ / L = R / L ------------------- (1)

If σs is the intensity of shear stress on the outer most layer and C the modulus

of rigidity of the shaft, then,

= σs / c ----------------------------- (2)

From equation (1) & (2), we find that

σs / c = R / L

OR σs / R = c / L

If be the intensity of shear stress, on any layer at a distance from the centre of the shaft then.

/ r = σs / R = c / l ---------------------------(3)

Strength of a shaft:

By the strength of a shaft is meant, the maximum torque or horse power, it can transmit.

Consider a shaft subjected to a torque T as shown in Fig. 9

Fig. 10

Take an element of area da of thickness dx at a distance x from the center as shown in figure 10.

da = 2 r * dx -----------------(1)

Shear stress at this section

σx = σs * x / R ----------------(2)

Where σs = maximum shear stress and

R = Radius of the shaft.

Turning force = Stress * Area = σx * da

x

dx

R


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= σs * (x / R) * da [ As from eqn. (2), σx = σs *

(X/R)]

= σs * (X/R) * 2 x * dx [ from eqn. (1) da=2 r * dx ]

= [(2 σs) / R] * x2 * dx

We know that turning moment of this element dT = Turning force x distance of the moment from the axis of the shaft.

= [(2 σs) / R] * x2 * dx * x

= [(2 σs) / R ] * x3 * dx

The total torque, which the shaft can with stand, may be found out by integrating the above equation between 0 and R i.e.

R

T = [(2 σs) / R] * x3 * dx 0 R R

= (2 σs) / R x3 * dx = (2 σs ) / R [ x4/ 4]

0 0

= ( / 2) * σs *R3

= ( / 16) * σs * D3

Where D is the diameter of the shaft and is equal to 2R ( i.e. 2 * radius)

Strength of a hollow shaft: Consider a hollow shaft subjected to a torque T as shown in Fig. 8. Take an element area da at a distance x from the center of the shaft and of thickness dx as shown in fig. 11

Fig. 11

dx

x

R r


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da = 2 x * d x -------------------------------------(1) Shear stress at this section,

σx = σs * (x / r)

Where, σs = maximum shear stress

And R = radius of the shaft

Turning force = Stress * area = σx * da

= σs * ( x/r) * da ( As σx = σs : x/r)

= σs * (x/r) * 2 * dx ( As da = 2 x * dx)

= (2 σs / R) * x2 * dx

Turning moment of this element dT = Turning force x distance of the element From the axis of the shaft.

2 σc 2 σs

= --------- * x2 * dx * x = -------- * x3 * dx R R Total torque, which the shaft can withstand, may be found out by integrating the above equation between r and R. R

T = [( 2 σs) / R] * x3 * dx

r R R

= (2 σs) / R x3 * dx = (2 σs) / R [ x4 / 4]

r r

= (2 σs) / 4 { ( R4 – r

4) / 4}

= ( / 16) σs { ( D4- d4) / D } -----------------------

Where D = 2R External diameter of the shaft

And d = 2r Internal diameter of the shaft

The shear stress developed of a point proportional to its distance from the center of the shaft. It is thus obvious, that in the central portion of a shaft the shear stress induced is very small, in order to utilize the material, the fuller extent hollow shafts are used.


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Horsepower transmitted by a shaft: Let, N = No. of revolution per minute and T = Average torque in kg-m Then work done per minute. = Force xdistance = Average torque x anopeler displacement

= T * 2 N = 2 N T Kg –m Power = Work done in Kg-m / min ----------------------------- 4500

= (2 N T / 4500 ) HP


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Examples

Q.1 State whether the following statements are true or false and correct them

where necessary.

Longitudinal stress in a thin cylinder is equal to twice the tangential stress. ---------- False

Longitudinal stress is half the circumferential or hoop stress.

Total pressure on the top of a closed cylindrical vessel completely filled with liquid is directly proportional to (radius)2. ------------- False

F = r2 * p

p = F / r2

i.e. p 1 / r2 ( inversely proportional to r2)

Creep is the fast & continuous deformation of metal under unsteady stress which is considerable higher than yield point.--------------- Permanent deformation, which occurs in a material when subject to a long time loading, is called the Creep. The amount of creep depends upon stress and time.

Q.2 Find the working pressure in a fire tube boiler 2.5 m in diameter. The tube thickness is 25 mm. Assume that longitudinal stress is not to exceed 10 kgf/cm2.

Solution:

We know that circumferential stress is double the longitudinal stress. σ1 = p * d/ 2 * t= circumferential stress σ2 = p * d /4 * t= longitudinal stress d = 2.5 x 100 = 250 cm t = 2.5 cm

σ2 = 10 kg/cm2 p = σ2 x 4t /d = 10 x 4 x 2.5 / 250= 0.4 kg / cm2 Ans


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Q.3 A boiler shell of 2 m diameter is made up of mild steel plates of 20 mm thick. The efficiency of the longitudinal and circumferential joints is 70% and 60% respectively. Determine the safe pressure in the boiler if permissible tensile stress in the plate section through the rivets is 1000 kg/cm2. Also determine the circumferential stress in the solid plate section and longitudinal stress through the rivets.

Solution: Here,

(d = 200 cm t = 2.0 cm

l = 70%

c = 60% σ = permissible stress = 1000 kg/cm2) Circumferential stress in solid stress section, σ1 = Maximum permissible stress = 1000 kg/cm2 ------------------------------Ans

Safe pressure P = (σ1 x 2t) x c / d 1000 x 2 x 2 x 0.6 = ------------------ = 12 kg/cm2 ------------------------------Ans 200 & Longitudinal stress through the rivets p*d 20 x 200 σ2 = ----------- = --------------- = 500 kg/cm2 -------------------------------------------Ans 4* t 4 x 2

Q.4 Find out the thickness of C.I. pipe to carry 30 m3/min of compressed air at a

pressure of 7 kg/cm2. The velocity of air in the pipe is limited to 8 meters per second. Assume tensile stress intensity in the pipe material to be 150 kg/cm2.

Solution:

Here, Discharge Volume V = (π/4) * d2 * velocity V = 30 m3/min 30 / 60 = (π /4) * d2 * 8 = (30/ 60) x m3/sec Velocity = 8 meters/sec d2 = 0.0796 meter d = 28.21 cm Now, p = 7 kg/cm2 d = 28.21 cm f = 150 kg/cm2 and t = thickness


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Circumferential tensile stress = p * d / 2 * t 150 = 7 * 28.21 / 2 * t t = 0.6582 cm ------------------------------------------Ans.

Q.5 A cast iron pipe is to deliver water at the rate of 1,50,000 litres per hour with a

velocity 0.6 metre/sec. The Maximum pressure of the fluid is not to exceed 10.5 kg/cm2. Determine the diameter and wall thickness. The permissible stress in cast iron is 140 Kg/cm2. Consider the pipe as thin cylinder

Solution:

Here, Volume flow rate 150 000 litres/hr = 150 m3/hr. = (150/ 3600) m3/sec Velocity of the fluid = 0.6 metre/sec Pressure in pipe = 10.5 kg/cm2 Permissible stress in cast iron = 140 kg/cm2

Thickness of the pipe = t

Diameter Of the pipe = d

Volume flow rate = area of pipe x velocity of fluid. = (π/4) * d2 * 0.6

150/3600 = (π /4) * d2 * 0.6

d2 = 150 x 4 / 3600 x π x 0.6 = 0.0884 d = 0.2974 metre = 29.74 cm ---------------------------------------Ans.

Thickness of pipe We known that circumferential tensile stress

σ = p * d / 2 * t

t = p * d / 2 * σ = 10.5 x 29.74 / 2 x 140 = 1.115 cm ---------------------------------------Ans.


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Q. 6 A cylindrical shell of 2200 mm internal diameter is constructed of mild steel pipe. The shell is subjected to an internal pressure of 10 kg/cm2 and the thickness of the boiler shell plate is 20 mm. The efficiency of the longitudinal joint may be taken as 78 %. Determine the factor of safety if the ultimate tensile strength of the steel is 4700 kg/cm2

Solution:

Internal Diameter of the shell = 2200 mm = 220 cm Shell internal pressure = 10 kg/cm2

Thickness of the shell plate = 20 mm = 2 cm ‘η' of the longitudinal joint = 78%

Ultimate tensile strength = Circumferential stress σ1 = 4700 kg/cm2

Longitudinal stress σ2 = 4700/2 = 2350 kg/cm2

σ2 * 4 * t * P = --------------------

F.S. * D 2350 x 4 x 2 x 0.78

10 = --------------------------

F.S. x 220

Factor of Safety (F.S.) = 6.665 -----------------------------------------------Ans

Q. 7 A steam boiler had 75 sq. meter of heating surface and the rate of evaporation

is 20 kg / sq.metre/hour of heating surface. The pressure of steam generation is 8 kg/cm2

gauge. The specific volume of steam is 0.24 cu.meter/kg. Determine the diameter & thickness of the steam pipe to carry steam from this boiler with a velocity of steam in the pipe at 25 meter/sec. The permissible tensile stress intensity in the pipe material is 400kg/cm2.

Solution: Given Heating surface of boiler = 75 m2 Rate of evaporation is = 20 kg/m2/hr. = 1500 kg/hr = 1.5 toners/hr Pressure of steam generation = 8 kg/cm2 (gauge) Specific volume of steam = 0.24 m3/kg Velocity of steam in pipe = 25 metre / sec.

σt permissible tensile stress intensity = 400 kg/cm2

Now, mass flow rate through steam pipe = 1500 kg/hr

= 1500 / 3600 kg/sec = 0.417 kg/sec


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Volume flow rate through steam pipe = mass flow rate x sp. Volume of steam

= 0.417 x 0.24 = 0.0999 m3/sec Say = 0.10 m3/sec ------------(1)

But volume flow rate through the steam pipe.

= π/4 d2 x velocity of steam = π /4 d2 x 25 --------------------------- (2)

Equating the (1) & (2), We get, π /4 d2 x 25 = 0.10

d2 = (0.10 * 4) / (π * 25) d = 0.071 metre

= 7.1 cm

Considering steam pipe as a thin cylinder & taking maximum stress, In thin

cylinder circumferential stress σ1 = p * d / 2 * t

Thickness of steel pipe = t = p * d / 2 * σ1

= 8 x 7.1 / 2 x 400 = 0.071 cm Say, 0.8 mm -----------------Ans.

Q.10 A motor car coupler shaft is made up of steel tube, 4 cm outside diameter and

4 mm thick. The engine develops 12 HP at 1200 rpm. What will be the maximum shear stress in the tube? When the power is transmitted through 4:1 Gear ratio

Solution: Out side dia. Of motor car coupler shaft = 4 cm. Thickness of shaft = 4 mm. Inside dia. = 3.6 cm (4-0.4) Engine HP = 12 Engine speed = 1200 rpm

Gear ratio = 41

Speed of coupler shaft = 1200/4 = 300 rpm.

P = 2 N T / 4500

= 2 * 300 * T / 4500

T =12 * 4500 / 2 300

= 90 / Kg-M

= 90* 100 / Kg-CM

Now, T = (/16) * σs * {( D4-d4) / D}


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(90 * 100) / = ( / 16) * σs * [{ (4)4 – (3-6)4}/4]

σs= 662.90 Kg/cm2 ------------------------------------------Ans.

Q.11 Compare the torsional stiffness of a solid shaft 50 mm diameter & 300 mm long

with that of hollow shaft of the same material having dia. 75, 50mm and length 200 mm

Solution: Torsional stiffness or rigidity = Modules of rigidity x polar moment of inertia

= C * J Modulus of rigidity for both shaft will be same as material is same, but polar moment of inertia will be different.

Torsional stiffness of solid shaft = C * Js

Torsional stiffness of hollow shaft = C * Jh

But, Js = ( / 32) / D4 & Jh = ( / 32)( D4 – d4)

Js = (/ 32) (5)4 & Jh = (/32) [(7.5)4 – (5)4]

Js = 61.359 cm4 & Jh = 249.272 cm4

Torsional stiffness of solid shaft = C * Js = 61.359 * C Torsional stiffness of hollow shaft = C * Jh = 249.272 * C

Ratio of torsional rigidities = [Torsional rigidity of hollow shaft] [Torsional rigidity of solid shaft] = 249.272 * C / 61.359 * C = 4.0625 -----------------------------------Ans.

Q.12 A steel shaft transmit 150 HP at 115 rpm. The maximum moment during each

revolution exceed the mean by 30%. Suggest suitable dia, for solid shaft is the shear stress is not to exceed 650 kg/cm2

Solution:

Given Power P = 150 HP Speed of the shaft, N = 115 rpm

Max sheared stress = σs= 650 kg/cm2 = 650 × 104 kg/ m2

Tmax = Tmean * 1.30 Using the equation,


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P = (2 π N T ) / 4500

T = (P * 4500) / 2 π N = (150 * 4500) / 2 π * 115 = 934.644 kg-m Now, Tmax. = Tmean * 1.3 = 934.644 * 1.3 = 1215.04 kg-m Using the relation,

Tmax. = ( / 16) * σs * D3

D3 = (Tmax. * 16) / ( * σs) = 9.52 × 10-4

D = 0.098 metre --------------------- Ans.

Q.14 A thin steel tyre is shrunk on to a locomotive wheel if 1,2 m diameter. Find

the internal diameter of the tyre, if after shrinking on the hoop stress in the tyre is 100 MPa. Assume E = 200 KN/mm2.Find also the least temperature to which the tyre must be heated above that of the wheel before it could be slipped on. The coefficient of linear expansion for the tyre is 6.5 * 10 -6 per o c.

Solution: D=1.2m=1200mm; σ=100 Mpa = 100 N/mm2 E=200 kN/mm2

α = 6.5 * 10 -6 per o C

Internal diameter of the tyre Let d = internal diameter of the tyre. We know that hoop stress σ 100 = E(D-d) / d = 200 * 1000 (D-d) /d D-d /d = 100 / 200000 = 1/2000 D/d = 1 + 1/2000 = 1.0005 d = D/1.0005 = 1200/ 1.0005 = 1199.4 mm = 1.1994 m Ans. Least temperature to which the tyre must be heated

Let t = Least temperature to which the tyre must be heated. We know that πD = πd + πd.α.t = πd(1+ α.t) α.t = πD / πd – 1 = D – d / d = 1/1999


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t= 1 / α * 1999 = 1 / 6.5 * 10 -6 * 1999 = 77 o C Ans. Q.15 A central steel rod 18 mm diameter passes through a copper tube 24 mm inside

and 40 mm outside diameter. It is provided with nuts and washers at each end. The nuts are tightened until a stress of 10 MPa is set up in the steel. The whole assembly is then placed in a lathe and turned along half the length of the tube removing the copper to a depth of 1.5 mm. Calculate the stress now existing in the steel.

Solution:

Given : ds = 18mm ;dc1 = 24mm ; dc2=40mm;σs = 10 Mpa Cross-sectional area of the steel rod, As = π/4*(ds)

2 = π/4 (18)2 = 254.5 mm2 And the cross-sectional area of copper tube Ac = π/4[(dc2)

2 – (dc1)2] = π/4[40*40 – 24*24] = 804.25 mm2.

When the nuts are tightened on the tube, the steel rod will be under tension and copper tube in compression.

Let σc = Stress in the copper tube. Since the tensile load on the steel rod is equal to the compressive load on the

copper tube, therefore σc * As = σc * Ac When the copper tube is reduced into the area for half of its length, then

outside diameter of copper tube =40 – 2 * 1.5 = 37mm Cross-sectional area of the half length of copper tube, Ac1 = π/4(37*37 – 24*24) = 623 mm2 The cross – sectional area of the other half remains same. If ac2 be the area of

the remainder, then Ac2 = Ac = 804.25 mm2 σc1 = Compressive stresses in the reduced section, σc2 = Compressive stress in the remander, and σc1 = Stress in the fod after turning. Since the load on the copper tube is equal to the load on the steel rod,

therefore Ac1 * σc1 = Ac2 * σc2 = As * σs1 σc1 = As/Ac1 * σs1 = 254.5/623 * σs1 = 0.41 σs1 σc2 = As/Ac2 * σs1 = 254.5/804.25*σs1 = 0.32 σs1 Let δl = Change in length of the steel rod before and after turning. l = Length of the steel rod and copper tube between nuts, δl1= Change in length of the reduced section before and after turning,

and δl2 = Change in length of the remainder section before and after

turning.


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δl = δl1 + δl2 σs - σs1/ Es * l = (σc1 – σc/ Ec * l/2 ) + (σc2 – σc/Ec * l/2) Considering Es = 2Ec or 10 – σs1/2Ec = 0.41 σs1 – 3.16/2Ec + 0.32σs1 – 3.16/2Ec

σs1 = 9.43 N/mm2 = 9.43 Mpa..........Ans.

Q.16 A copper bar 40 mm in diameter is placed within a steel tube 75 mm external diameter and 50 mm internal diameter of exactly the same length. The two pieces are rigidly fixed together ;by two pins 18 mm in diameter, one at each end passing through the bar and tube. Calculate the stress induced in the copper bar, steel tube ad pins if the temperature of the combination is raised by 50 o C. Take Es = 210 GN/m2 and Ec = 105 GN/m2; αs = 11.5 * 10 -6 / o C and α c = 17 * 10 -8 / o C.

Solution: Given dc = 40 mm; dse = 75 mm ; dsi = 50 mm; dp = 18m = .018 m; t=50 o C

The cross-sectional are of the copper bar, Ac = π/4(ds)

2 = 1257 mm2. And cross-sectional are of the steel tube, As = π[(dse)2 – (dsi)2] = 2455 mm2 Let l = Length of the copper bar and steel tube. Free expansion of copper bar = αc.l.t = 17 * 10 -6 * l *50 = 850 * 10-6 l Difference in free expansion = 850 * 10 -6 l – 575 * 10 -6l Since the free expansion of the copper bar is more than the free expansion of the steel tube, therefore the copper bar is subjected to a compressive stress, while steel tube is subjected to a tensile stress. Let a compressive force P Newton on the copper bar opposes the extra expansion of the copper bar and an equal tensile force P on the steel tube pulls the steel tube pulls the steel tube so that the net effect of reduction in length of copper bar and the increase in length of steel tube equalizes the difference in free expansion of the two. Reduction in length of copper bar due to force P =P.l/Ac.Ec = P.l / 1257 * 10 -6 * 105 * 10 9 =P.l / 131.985 * 10 6 m and increase in length of steel bar due to force P =P.l / As.Es = P.l / 2455 * 10 -6 * 210 *10 9 = P.l/ 515.55 * 10 6 m Net effect in length = P.l / 515.555 * 106 + P.l/ 131.985 * 106


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= 9.52 * 109 P.l Equating this net effect in length to the difference in free expansion, we have 9.52 * 10 -9 P.l = 275 * 10 -6 l P = 28887 N.

Stress induced in the copper bar, steel tube and pins, We know that stress induced in the copper bar, αc = P/Ac = 28887/ 1257 * 10-6

= 22.98 * 106 N n/m2 Stress induced in the steel tube αs = P/As = 28887 / 2455 * 10-6

= 11.77* 106 N/m2 And the shear stress induced in the pins, Τp = P/2 Ap = 28887/ 2 * π/4(0.018)2 =56.76 * 106 N/m2. = 56.76 MPa.


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Examples for practice Example 3: A thin cylindrical pressure vessel of 1.2 m diameter generates steam at

pressure of 1.75 N/mm2. Find the minimum wall thickness, if (a) the longitudinal stress does not exceed 28 MPa ; and (b) the circumferential stress does not exceed 42 MPa.

Example 4: A thin cylindrical pressure vessel of 500 mm diameter is subjected to an

internal pressure of 2 N/mm2. If the thickness of the vessel is 20 mm, find the hoop stress. Longitudinal stress and maximum shear stress.

Example 5: Find the diameter of a solid steel shaft to transmit 20 kw at 200 rpm. The

ultimate shear stress for the steel may be taken as 360 MPa and a factor of safety as 8. If a hollow shaft is to be need i place of the solid shaft, find the inside and outside diameter when the ration of inside to outside diameters is 0.5.

Example 6: A shaft made of mild steel is required to transmit 100 kw at 300 rpm. The

supported length of the shaft is 3 m. \It carries two pulleys each weighing 1500 N supported at a distance of 1 m from the ends respectively. Assuming the safe value of stress, determine the diameter of the shaft.

Example 7: A bar 3 m long is made of two bars, one of copper having E = 105 GN/ m2

and the other of steel having E = 210 GN/ m2 . Each bar is 25 mm broad and 12.5 mm thick. This compound bar is stretched by a load of 50 kN. Find the increase in length of the compound bar and the stress produced in the steel and copper. The length of copper as well as of steel bar is 3 m each.

Example 8:

a) Calculate the bursting pressure of a cold drawn seamless steel tubing of 60 mm inside diameter and 2 mm wall thickness. The ultimate strength of steel is 380 MN/m². b) Calculate the thickness of the metal required for a Cast Iron main 800 mm in diameter for water at a pressure head of 100m, if the maximum permissible tensile is 20 MN/m² and weight of water is 10 KN/m³.

Example10:

a) Derive formula for circumferential or hoop stress and longitudinal stress in

pressure vessels.

b) A steam boiler 1.2 meter in diameter, generates steam at a gauge pressure

of 0.7 N/ mm2. Assuming the efficiency of the riveted joint as 75%, find the

thickness of the shell. Given that ultimate tensile stress = 285 N/ mm2 and

factor of safety = 5

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CHAPTER - 12 Strength of Materials