Engineering Materials Design

Lecture. 1

Simple Stresses

By

Dr. Mohammed Ramidh

Introduction to Engineering Materials Design

Engineering materials design is a complex undertaking, requiring many skills.

Extensive relationships need to be subdivided into a series of simple tasks. The

complexity of the subject requires a sequence in which ideas are introduced

and iterated. To design is either to formulate a plan for the satisfaction of a specified need or

to solve a problem. If the plan results in the creation of something having a

physical reality, then the product must be functional, safe, reliable,

competitive, usable, manufacturable, and marketable. The complete design process, from start to finish, is often outlined as in Fig. 1–1.

The process begins with an identification of a need and a decision to do

something about it. After many iterations, the process ends with the presentation

of the plans for satisfying the need. Depending on the nature of the design task,

several design phases may be repeated throughout the life of the product, from

inception to termination.

Figure 1–1

Simple Stresses

Load and stress analysis

Load

It is defined as any external force acting upon a machine part. The following

four types of the load are important from the subject point of view:

1. Dead or steady load. A load is said to be a dead or steady load, when it does not

change in magnitude or direction.

2. Live or variable load. A load is said to be a live or variable load, when it changes

continually.

3. Suddenly applied or shock loads. A load is said to be a suddenly applied or

shock load, when it is suddenly applied or removed.

4. Impact load. A load is said to be an impact load, when it is applied with some

initial velocity.

Note: A machine part resists a dead load more easily than a live load and a live

load more easily than a shock load.

Stress When some external system of forces or loads act on a body, the internal forces

(equal and opposite) are set up at various sections of the body, which resist the

external forces. This internal force per unit area at any section of the body is known

as unit stress or simply a stress. It is denoted by a Greek letter sigma ().

Mathematically,

Stress, = P/A

where P = Force or load acting on a body.

A = Cross-sectional area of the body.

In S.I. units, the stress is usually expressed in Pascal (Pa) such that 1 Pa = 1 N/ .

In actual practice, we use bigger units of stress i.e. megapascal (MPa) and

gigapascal (GPa), such that

1 MPa = 1 × N/ = 1 N/

and 1 GPa = 1 × N/ = 1 kN/

Strain When a system of forces or loads act on a body, it undergoes some deformation.

This deformation per unit length is known as unit strain or simply a strain. It is

denoted by a Greek letter epsilon ().

Mathematically,

Strain, = l / l or l = .l

where l = Change in length of the body, and

l = Original length of the body.

1. Tensile Stress and Strain

When a body is subjected to two equal and opposite axial pulls P (also called

tensile load) as shown in Fig. (a), then the stress induced at any section of the body

is known as tensile stress as shown in Fig. (b). A little consideration will show that

due to the tensile load, there will be a decrease in cross-sectional area and an

increase in length of the body. The ratio of the increase in length to the original

length is known as tensile strain.

Let P = Axial tensile force acting on the body,

A = Cross-sectional area of the body,

l = Original length, and

l = Increase in length.

Tensile stress, t = P/A

and tensile strain, t = l / l

2. Compressive Stress and Strain When a body is subjected to two equal and opposite axial pushes P (also called

compressive load) as shown in Fig. (a), then the stress induced at any section of the

body is known as

compressive stress as shown in Fig. (b). A little consideration will show that due to

the compressive load, there will be an increase in cross-sectional area and a

decrease in length of the body. The ratio of the decrease in length to the original

length is known as compressive strain.

Let P = Axial compressive force acting on the body,

A = Cross-sectional area of the body,

l = Original length, and

l = Decrease in length.

Compressive stress, c = P/A

and compressive strain, c = l/ l

2.1 Young's Modulus or Modulus of Elasticity Hooke's law* states that when a material is loaded within elastic limit,

the stress is directly proportional to strain, i.e.

modulus of elasticity. In S.I. units, it is usually expressed in GPa i.e. GN/ or

kN/ . It may be noted that Hooke's law holds good for where E is a constant of

proportionality known as Young's modulus or tension as well as compression. Example 1. A coil chain of a crane required to carry a maximum load of 50 kN, is

shown in Fig. Find the diameter of the link stock, if the permissible tensile stress in

the link material is not to exceed 75 MPa.

Example 2. A cast iron link, as shown in Fig., is required to transmit a steady

tensile load of 45 kN. Find the tensile stress induced in the link material at sections

A-A and B-B.

Example 3. A hydraulic press exerts a total load of 3.5 MN. This load is carried by

two steel rods, supporting the upper head of the press. If the safe stress is 85 MPa

and E = 210 kN/ ,

find :1.diameter of the rods,and 2.extension in each rod in a length of 2.5 m.

3. Shear Stress and Strain

When a body is subjected to two equal and opposite forces acting tangentially

across the resisting section, as a result of which the body tends to shear off the

section, then the stress induced is called shear stress.

The corresponding strain is known as shear strain and it is measured by the angular

deformation accompanying the shear stress. The shear stress and shear strain are

denoted by the Greek letters tau () and Pci () respectively. Mathematically,

the area resisting the shear off the rivet,

and shear stress on the rivet cross-section,

● It may be noted that when the tangential force is resisted by two cross-sections of

the rivet (or when the shearing takes place at two cross-sections of the rivet), then

the rivets are said to be in double shear. In such a case, the area resisting the shear

off the rivet,

…………..(For double shear)

and shear stress on the rivet cross-section,

Notes : When the holes are to be punched or drilled in the metal plates, then the

tools used to perform the operations must overcome the ultimate shearing resistance

of the material to be cut. If a hole of diameter ‘d’ is to be punched in a metal plate

of thickness ‘t’, then the area to be sheared, A = πd × t

and the maximum shear resistance of the tool or the force required to punch a hole,

P = A × u = πd × t × u

where u = Ultimate shear strength of the material of the plate.

3.1 Shear Modulus or Modulus of Rigidity

It has been found experimentally that within the elastic limit, the shear stress is

directly proportional to shear strain. Mathematically

= G. or G= ⁄

where = Shear stress,

= Shear strain, and

G = Constant of proportionality, known as shear modulus or modulus

of rigidity. It is also denoted by N or G.

3.2 Relation Between Young’s Modulus and Modulus of Rigidity

The Young's modulus (E) and modulus of rigidity (G) are related by the following

relation,

3.3 Poisson's Ratio

Consider a circular bar of diameter d and length l, subjected to a tensile force P as

shown in Fig. (a). A little consideration will show that due to tensile force, the

length of the bar increases by an amount l and the diameter decreases by an

amount d, as shown in Fig. (b). Similarly, if the bar is subjected to a compressive

force, the length of bar will decrease which will be followed by increase in

diameter.

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,

This constant is known as Poisson's ratio and is denoted by or Poisson’s

ratio may be computed with sufficient accuracy from the relation

Example 4: A pull of 80 kN is transmitted from a bar X to the bar Y through a pin

as shown in Fig. If the maximum permissible tensile stress in the bars is 100

N/ and the permissible shear stress in the pin is 80 N/, find the diameter

of bars and of the pin.

4. Stresses due to Change in Temperature—Thermal Stresses Whenever there is some increase or decrease in the temperature of a body, it

causes the body to expand or contract. A little consideration will show that if

the body is allowed to expand or contract freely, with the rise or fall of the

temperature, no stresses are induced in the body. But, if the deformation of

the body is prevented, some stresses are induced in the body. Such stresses

are known as thermal stresses.

5. BIAXIAL STRESSES : An element in biaxial tension.

6. Torsion and Bending stresses Introduction Sometimes machine parts or structural member are subjected to pure

torsion or bending or combination of both torsion and bending stresses. We

shall now discuss these stresses in detail.

6.1 Torsional Shear Stress When a machine member is subjected to the action of two equal and opposite

couples acting in parallel planes (or torque or twisting moment), then the

machine member is said to be subjected to torsion. The stress set up by

torsion is known as torsional shear stress. Consider a shaft fixed at one end and subjected to a torque (T) at the other

end as shown in Fig.

Fig. . Torsional shear stress.

…………………………… ( 1)

G= Modulus of rigidity for the shaft material ◦ From equation (1), we know that

The strength of the shaft means the maximum torque transmitted by it. Therefore,

in order to design a shaft for strength, the above equations are used. The power

transmitted by the shaft (in watts) is given by

Example 5.1. A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable

diameter for the shaft, if the maximum torque transmitted exceeds the mean by

25%. Take maximum allowable shear stress as 70 MPa.

Example 6.1 A steel shaft 35 mm in diameter and 1.2 m long held rigidly at one

end has a hand wheel 500 mm in diameter keyed to the other end. The modulus of

rigidity of steel is 80 GPa.

1. What load applied to tangent to the rim of the wheel produce a torsional shear of

60 MPa?

2. How many degrees will the wheel turn when this load is applied?

We know that

→

ہ

6.2 Bending Stress in Straight Beams In engineering practice, the machine parts of structural members may be subjected

to static or dynamic loads which cause bending stress in the sections besides other

types of stresses such as tensile, compressive and shearing stresses. Consider a

straight beam subjected to a bending moment M as shown in Fig.

Fig. . Bending stress in straight beams.

The stress distribution of a beam is shown in Fig. The bending equation is given by

Where = Bending moment acting at the given section,

= Bending stress, I = Moment of inertia of the cross-section about the neutral axis,

y = Distance from the neutral axis to the extreme fibre,

E = Young’s modulus of the material of the beam, and

R = Radius of curvature of the beam. From the above equation, the bending stress is given by

Since E and R are constant, therefore within elastic limit, the stress at any point is

directly proportional to y, i.e. the distance of the point from the neutral axis.

Also from the above equation, the bending stress,

⁄

The ratio I/y is known as section modulus and is denoted by Z. Example 6.2. A pump lever rocking shaft is shown in Fig. 5.5. The pump lever

exerts forces of 25 kN and 35 kN concentrated at 150 mm and 200 mm from the

left and right hand bearing respectively. Find the diameter of the central portion of

the shaft, if the stress is not to exceed 100 MPa. Solution. Given : = 100 MPa = 100 N/ Let and = Reactions at

A and B respectively. Taking moments about A, we have

Example 6.3. An axle 1 metre long supported in bearings at its ends carries a fly

wheel weighing 30 kN at the centre. If the stress (bending) is not to exceed 60 MPa,

find the diameter of the axle. Solution. Given : L = 1 m = 1000 mm ; W = 30 kN = 30 × N ; = 60 MPa =

60 N/ The axle with a flywheel is shown in Fig. 5.6.

Let d = Diameter of the axle in mm.

7. Principal Stresses and Principal Planes we have discussed about the direct tensile and compressive stress as well as

simple shear. But it has been observed that at any point in a strained material,

there are three planes, mutually perpendicular to each other which carry direct

stresses only and no shear stress. It may be noted that out of these three direct

stresses, one will be maximum and the other will be minimum.

The perpendicular planes are known as principal planes and the direct

stresses along these planes are known as principal stresses. Determination of Principal Stresses for a Member Subjected to Bi-axial

Stress . When a member is subjected to bi-axial stress (i.e. direct stress in

two mutually perpendicular planes accompanied by a simple shear stress),

then the normal and shear stresses are obtained as discussed below: Consider a rectangular body ABCD of uniform cross-sectional area and unit

thickness subjected to normal stresses and as shown in Fig. (a). In

addition to these normal stresses, a shear stress also acts. ( a ) ( b )

Fig. Principal stresses for a member subjected to bi-axial stress. the normal stress across any oblique section such as EF inclined at an angle

θwith the direction of , as shown in Fig. (a), is given by

Since the planes of maximum and minimum normal stress (principal planes) have

no shear stress, therefore the inclination of principal planes is obtained by

equating = 0 in the above equation (ii).

The maximum and minimum principal stresses may now be obtained by

substituting the values of sin2θand cos2θin equation (i).

Maximum principal (or normal) stress,

and minimum principal (or normal) stress, The planes of maximum shear stress are at right angles to each other and are

inclined at 45° to the principal planes. The maximum shear stress is given by one-

half the algebraic difference between the principal stresses.

Notes: 1. When a member is subjected to direct stress in one plane accompanied

by a simple shear stress as shown in Fig. (b), then the principal stresses are

obtained by substituting = 0 in equation (iv), (v) and (vi).

the maximum principal stresses, due to the combination of tensile or compressive

stresses with shear stresses may be obtained.

Example 7.1. A hollow shaft of 40 mm outer diameter and 25 mm inner diameter

is subjected to a twisting moment of 120 N-m, simultaneously, it is subjected to

an axial thrust of 10 kN and a bending moment of 80 N-m. Calculate the

maximum compressive and shear stresses. Solution. Given: do = 40 mm ; di = 25 mm ; T = 120 N-m = 120 × N-mm ; P

= 10 kN = 10 × N ; M = 80 N-m = 80 × N-mm

We know that cross-sectional area of the shaft,

Example 7.2. A shaft, as shown in Fig. is subjected to a bending load of 3 kN, pure

torque of 1000 N-m and an axial pulling force of 15 kN. Calculate the stresses at A

and B. Solution. Given : W = 3 kN = 3000 N ; T = 1000 N-m = 1 × N-mm ; P = 15

kN = 15 × N ; d = 50 mm; x = 250 mm

We know that cross-sectional area of the shaft,

Tensile stress due to axial pulling at points A and B,

Bending moment at points A and B, Section modulus for the shaft,

Bending stress at points A and B,

This bending stress is tensile at point A and compressive at point B. Resultant tensile stress at point A,

and resultant compressive stress at point B,

We know that the shear stress at points A and B due to the torque transmitted,

Stresses at point A

We know that maximum principal (or normal) stress at point A,

Minimum principal (or normal) stress at point A,

and maximum shear stress at point A,

Stresses at point B

We know that maximum principal (or normal) stress at point B,

Minimum principal (or normal) stress at point B,

and maximum shear stress at point B,

8. Mohr’s Circle for Plane Stress

Suppose the dx dy dz element of Fig. (a) is cut by an oblique plane with a normal n

at an arbitrary angle φ counterclockwise from the x axis as shown in Fig. (b). This

section is concerned with the stresses σand τthat act upon this oblique plane. the

stresses σand τare found to be

(a) (b)

………….. (1)

………… (2) …………… (3)

Equation (1) and (2) are called the plan-stress transformation equations.

Differentiating Eq. (1) with respect to and setting the result to zero gives, Eq.(3).

the maximum normal stress σ1 and the other, the minimum normal stress σ2.

These two stresses are called the principal stresses, and their corresponding

directions, the principal directions. The angle between the principal directions is

90°. It is important to note that Eq. (3) can be written in the form

In a similar manner, we differentiate Eq. (2), set the result equal to zero, and obtain ………… (4)

The angle between the surfaces containing the maximum shear stresses

is 90°. Equation (4) can also be written as Substituting this into Eq. (1) yields ………………. (5) Formulas for the two principal stresses can be obtained by substituting the angle

from Eq. (3) in Eq. (1). The result is

……………. (6) In a similar manner the two extreme-value shear stresses are found to be …………….. (7)

The relationship between σ and τ is that of a circle plotted in the σ, τ plane,

where the center of the circle is located at C = (σ, τ )=[(σx + σy)/2, 0]

and has a radius of R =√ ( ) .

Mohr’s Circle Shear Convention

This convention is followed in drawing Mohr’s circle:

• Shear stresses tending to rotate the element clockwise (cw) are plotted above the

σ axis.

• Shear stresses tending to rotate the element counterclockwise (ccw) are plotted

below the σ axis

Fig. C

In Fig C. we create a coordinate system with normal stresses plotted along the

abscissa and shear stresses plotted as the ordinates. On the abscissa, tensile

(positive) normal stresses are plotted to the right of the origin O and compressive

(negative) normal stresses to the left. On the ordinate, clockwise (cw) shear stresses

are plotted up; counterclockwise (ccw) shear stresses are plotted down.

The right face is called the x face where φ=0◦.If σx is positive and the shear

stress is ccw as shown in Fig. a, we can establish point A with coordinates

(σx , ) in Fig. C. Next, we look at the top y face, where φ= 90◦,which

contains σy , and repeat the process to obtain point B with coordinates

(σy , ) as shown in Fig. C.

Points A and B are the same vertical distance from the σ axis. Thus, AB must be

on the diameter of the circle, and the center of the circle C is where AB intersects

the σ axis. With points A and B on the circle, and center C, the complete

circle can then be drawn.

EXAMPLE 8-1: A stress element has σx = 80 MPa and τxy = 50 MPa cw, as

shown in Fig. a.

(a) Using Mohr’s circle, find the principal stresses and directions, and show these

on a stress element correctly aligned with respect to the xy coordinates. Draw

another stress element to show τ1 and τ2, find the corresponding normal stresses,

and label the drawing completely.

(b) Repeat part a using the transformation equations only.

Solution (a) In the semigraphical approach used here, we first make an

approximate freehand sketch of Mohr’s circle and then use the geometry of the

figure to obtain the desired information. Draw the σand τaxes first (Fig. b) and from the x face locate σx = 80 MPa, we

see that the shear stress is 50 MPa in the cw direction, Thus, point A (80, )

MPa. the y face, the stress is σ= 0 and τ= 50 MPa in the ccw direction. point B

(0, ) MPa. The line AB forms the diameter of the circle, which can now be

drawn. The intersection of the circle with the σ axis defines σ1 and σ2 as

shown. the triangle ACD,the legs AD and CD as 50 and 40 MPa,. The length of the

hypotenuse AC is

the principal stresses are now found to be

The angle 2φ from the x axis cw to σ1 is

To draw the principal stress element (Fig. c), sketch the x and y axes parallel

to the original axes ,from x measure 25.7°(half of 51.3°) clockwise to locate theσ1

axis. The σ2 axis is 90° from the σ1 axis .

The two maximum shear stresses occur at points E and F in Fig. 3–11b. The two

normal stresses corresponding to these shear stresses are each 40 MPa, as indicated.

Point E is 38.7° ccw from point A on Mohr’s circle. Therefore, in Fig. d, draw a

stress element oriented 19.3° (half of 38.7°) ccw from x. The element with

magnitudes and directions as shown.

(b) The transformation equations are programmable. From Eq. (3),

From Eq. (1), for the first angle

= −25.7◦,

The shear on this surface is obtained from Eq. (2) as

which confirms that 104.03 MPa is a principal stress. From Eq.(1), for φp =64.3◦,

the principal stresses are calculated they can be ordered such that σ1 ≥ σ2.

Thus, σ1 = 104.03 MPa and σ2 = −24.03 MPa.

To determine τ1 and τ2, we first use Eq. (4) to calculate φs :

For φs = 19.3◦, Eqs. (1) and (2) yield,

For φs = 109.3◦, Eqs. (1) and (2) give σ = 40.0 MPa and τ = +64.0 MPa.

Using the same logic for the coordinate transformation we find that results again

agree with Fig. d.

Homework

1. For each of the plane stress states listed below, draw a Mohr’s circle

diagram properly labeled, find the principal normal and shear stresses, and

determine the angle from the x axis to σ1.Draw stress elements as in Fig. c

and d and label all details.

(a) σx = 12, σy = 6, τx y = 4 cw

(b) σx = 16, σy = 9, τx y = 5 ccw

(c) σx = 10, σy = 24, τx y = 6 ccw

(d) σx = 9, σy = 19, τx y = 8 cw 2. For each of the stress states listed below, find all three principal normal

and shear stresses. Draw a complete Mohr’s three-circle diagram and label all

points of interest.

(a) σx = 10, σy = −4

(b) σx = 10, τx y = 4 ccw

(c) σx = −2, σy = −8, τx y = 4 cw

(d) σx = 10, σy = −30, τx y = 10 ccw

3. A spindle as shown in Fig. 5.34, is a part of an industrial brake and is

loaded as shown. Each load P is equal to 4 kN and is applied at the mid point

of its bearing. Find the diameter of the spindle, if the maximum bending stress

is 120 MPa. [Ans. 22 mm]

4. Two circular rods of 50 mm diameter are connected by a knuckle joint, as

shown in Fig. by a pin of 40 mm in diameter. If a pull of 120 kN acts at each

end, find the tensile stress in the rod and shear stress in the pin. [Ans. 61 N/mm2;

48 N/mm2]

5. Find the minimum size of a hole that can be punched in a 20 mm thick mild

steel plate having an ultimate shear strength of 300 N/mm2. The maximum

permissible compressive stress in the punch material is 1200 N/mm2. [Ans. 20 mm]

6. What is the difference between modulus of elasticity and modulus of rigidity? 7. What useful informations are obtained from the tensile test of a ductile

material? 8. What do you mean by factor of safety? 9. Explain the difference between linear and lateral strain. 10. Hooke’s law holds good upto

(a) yield point (b) elastic limit (c) plastic limit (d) breaking point 11. The ratio of linear stress to linear strain is called

(a) Modulus of elasticity (b) Modulus of rigidity (c) Bulk modulus

(d) Poisson's ratio

12. When a hole of diameter ‘d' is punched in a metal of thickness `t', then the

force required to punch a hole is equal to.

13. In a body, a thermal stress is one which arises because of the existence of

(a) latent heat (b) temperature gradient (c) total heat (d) specific heat 14. An aluminium member is designed based on

(a) yield stress (b) elastic limit stress (c) proof stress (d) ultimate stress 15. A steel shaft 50 mm diameter and 500 mm long is subjected to a twisting

moment of 1100 N-m, the total angle of twist being 0.6°. Find the maximum

shearing stress developed in the shaft and modulus of rigidity.

[Ans. 44.8 MPa; 85.6 kN/m2] 17. Design a suitable diameter for a circular shaft required to transmit 90 kW at

180 r.p.m. The shear stress in the shaft is not to exceed 70 MPa and the

maximum torque exceeds the mean by 40%. Also find the angle of twist in a

length of 2 metres. Take G = 90 GPa. [Ans. 80 mm; 2.116°] 18. Distinguish clearly between direct stress and bending stress. 19. If di and do are the inner and outer diameters of a hollow shaft, then its polar

moment of inertia is 20. The neutral axis of a beam is subjected to

(a) zero stress (b) maximum tensile stress (c) maximum compressive stress

(d) maximum shear stress