CHAPTER 1

STRESS AND STRAIN

STRESS AND STRAIN

Axial loading

Normal stress

Shear stress

Bearing stress

Strain

Stress-strain relations

Safety factor

Deformation of axial members

Statically indeterminate axial loading

Thermal stress in axial members

Stress And Strain

Mechanics of material is a study of the relationship between the external loads applied to a deformable body and the intensity of internal forces acting within the body.

Stress = the intensity of the internal force on a specific plane (area) passing through a point.

Strain = describe the deformation by changes in length of line segments and the changes in the angles between them

Normal Stress and Normal Strain

Normal Stress, the intensity of force, or force per unit area, acting

normal to A

A positive sign will be used to indicate a tensile stress (member in tension)

A negative sign will be used to indicate a compressive stress (member in compression)

= P / A

(a)

(b)

(c)

Stress ( ) = Force (P)

Cross Section (A)

Unit: Nm -²

Example 1.1:

Two solid cylindrical rods AB and BC are welded

together at B and loaded as shown. Knowing that

d1=50mm and d2=30mm, find average normal stress at

the midsection of (a) rod AB, (b) rod BC.

Solution:

2

2AB

ABAB m/MN7.35

4

m05.0

N70000

A

P

2

2/44.42

4

03.0

30000mMN

m

N

A

P

BC

BCBC

Two solid cylindrical rods AB and BC are

welded together at B and loaded as shown.

Determine the average normal stress at the

midsection of (a) rod AB, and (b) rod BC.

Example 1.2

Example 1.3:

The 80 kg lamp is supported by two rods AB and BC as shown in

the figure. If AB has a diameter of 10 mm and BC has a diameter of

8 mm, determine the average normal stress in each rod.

Solution:

Internal loading:

Average normal stress:

Example 1.4:

Member AC as shown in the figure is subjected to a vertical force of

3 kN. Determine the position x of this force so that the average

compressive stress at the smooth support C is equal to the average

tensile stress in the tie rod AB. The rod has a cross-sectional area of

400 mm2 and the contact area at C is 650 mm2.

Solution:

Internal loading:

Average normal stress:

Normal strain, is the elongation or contraction of a line

segment per unit of length

strain can be expressed as a

percentage strain

L

L

lengthoriginal

lengthchange instrain

Normal strain,

%100L

Lstrain

Determine the corresponding strain for a bar of length

L=0.600m and uniform cross section which undergoes

a deformation =15010-6m.

Example 1.5:

66

6

150 10 m250 10 m m

L 0 600m

250 10 250

/.

@

Example 1.6:

A force acting on the grip of the lever arm as shown in the figure

causes the arm to rotate clockwise through an angle of =0.002 rad.

Determine the average normal strain developed in the wire BC.

Solution:

Since =0.002 rad is small, the stretch in the wire CB is:

Example 1.7:

The plate is deformed into the dashed shape as shown in the figure.

If in this deformed shape horizontal lines on the plate remain

horizontal and do not change their length, determine the average

normal strain along the side AB .

Solution:

Line AB, coincident with the y-axis,

becomes line AB’ after deformation. The

length of this line is:-

The average normal strain for AB is

therefore:-

The negative sign indicates the strain

causes a contraction of AB

Example 1.8:

A short post constructed from a hollow circular tube of aluminium supports a compressive load of 240 kN. The inner and outer diameters of the tube are d1=90mm and d2=130mm, respectively, and its length is 1m. The shortening of the post due to the load is measured as 0.55 mm. Determine the compressive stress and strain in the post. (Disregard the weight of the post itself, and assume that the post does not buckle under the load.)

Solution:

232221

22 m1091.6m09.0m13.0

4dd

4A

MPa7.34m91.6

N000,240

A

P

2

55010550mm000,1

mm55.0

L

6

Assuming that the compressive load acts at the center of the hollow

tube, we can use the equation =P/A to calculate the normal stress.

The force P equals to 240kN and the cross-sectional area is:-

Therefore, the compressive stress in the post is:-

The compressive strain is:-

Stress-Strain Relations

Tensile test is a experiment to determine the load-

deformation behavior of the material.

Data from tensile test can be plot into stress and

strain diagram.

Example of test specimen

- note the dog-bone geometry

28

Universal Testing Machine - equipment used to

subject a specimen to tension, compression,

bending, etc. loads and measure its response

29

Stress-Strain Relations:

Universal Testing Machine

Stress-Strain Relations: Stress-Strain

Diagrams

30

A number of important mechanical properties of materials that be

deduced from stress-strain diagram are illustrated in figure above.

Point O-A = linear relationship between stress and strain

Point A = proportional limit (PL)

The ratio of stress to strain in this linear region of stress-strain

diagram is called Young Modulus or the Modulus of Elasticity.

• At point A-B, specimen begins yielding.

Point B = yield point

Point B-C = specimen continues to elongate without any increase in stress. Its refer as perfectly plastic zone

Point C = stress begins to increase

Point C-D = refer as the zone of strain hardening

Point D = ultimate stress/strength ; specimen begins to neck-down

Point E = fracture stress

< PL

Unit: MPa

31

Stress-Strain Relations: Stress-Strain

Diagrams (cont.)

Point O to A

Point C to D

Point D to E

At point E

Normal or engineering stress can be determine by dividing the applied load by the specimen original cross sectional area.

True stress is calculated using the actual cross sectional area at the instant the load is measured.

31

Stress-Strain Relations: Stress-Strain

Diagrams (cont.)

• Some of the materials like aluminum (ductile), does not have clearly yield point such as structural steel. Therefore, stress value called the offset yield stress, YL is used in lieu of a yield point stress.

As illustrated, the offset yield stress is determine by;

Drawing a straight line that best fits the data in initial (linear) portion of the stress-strain diagram

Second line is then drawn parallel to the original line but offset by specified amount of strain

The intersection of this second line with the stress-strain curve determine the offset yield stress.

Commonly used offset value is 0.002/0.2%32

Stress-Strain Relations: Stress-Strain

Diagrams (cont.)

• Brittle material such as ceramic and glass have low

tensile stress value but high in compressive stress.

Stress-strain diagram for brittle material.

33

Stress-Strain Relations: Stress-Strain

Diagrams (cont.)

Elasticity and Plasticity

Elasticity refers to the property of a material such that it returns

to its original dimensions after unloading

Any material which deforms when subjected to load and

returns to its original dimensions when unloaded is said to be

elastic.

If the stress is proportional to the strain, the material is said to

be linear elastic, otherwise it is non-linear elastic.

Beyond the elastic limit, some residual strain or permanent

strains will remain in the material upon unloading

The residual elongation corresponding to the permanent strain

is called the permanent set

34

• The amount of strain which is recovered upon unloading is

called the elastic recovery.

35

Poisson's Ratio,

When an elastic, homogenous and isotropic material is subjected

to uniform tension, it stretches axially but contracts laterally along

its entire length.

Similarly, if the material is subjected to axial compression, it

shortens axially but bulges out laterally (sideways).

The ratio of lateral strain to axial strain is a constant known as the

Poisson's ratio,

strainaxial

strainlateralv

36

Axial elongation and lateral contraction of a

bar in tension:

(a) before loading, and (b) after loading

Example 1.9:

A steel pipe of length L=1.2 m, outside diameter d2=150mm, and inside diameter d1=110mm is compressed by an axial force P=620kN. The material has modulus of elasticity E=200 GPa and Poisson’s ratio =0.30.

Determine the following quantities for the pipe:-

(a) the shortening,

(b) the lateral strain, ’

(c) the increase d2 in the outer diameter and the increase d1 in the inner diameter

(d) the increase t in the wall thickness

(e) the increase V in the volume of the material, and

(f) the dilatation e

Solution:

The cross-sectional area A and longitudinal stress are determined as

follows:

The axial strain may be found from Hooke’s Law:-

(a) The Shortening,

MPa9.75m10168.8

N620000

A

P

m10168.8m11.0m15.04

dd4

A

23

232221

22

5.379or105.379

MPa10200

MPa9.75

E,E 6

3

mm455.0mm102.1105.379LL,L

L 36

Solution (cont.):

(b) The lateral strain is obtained from Poisson’s Ratio

The positive sign for ’ indicates an increase in the lateral dimensions, as

expected for compression.

(c) The increase in outer diameter equals the lateral strain times diameter:

(d) The increase in wall thickness is found in the same manner as the increases in

the diameters; thus,

Note that under compression, all three quantities increase.

66 109.113)105.379(30.0,

mm0125.0mm110109.113dd

mm0171.0mm150109.113dd

611

622

mm00228.0mm0125.0mm0171.02

1

2

ddt

,or

mm00228.0mm20109.113tt

12

6

Solution (cont.):

(e) The change in volume of the material is,

The volume change is negative, indicating a decrease in volume, as expected for

compression.

(f) Finally, the dilatation is,

3m10488.160.01105.379m2.1m10168.8

21AL21VV

6623

o

00015.0 0.60-110379.5-2-1e -6

Shear Stress

Shear stress, : The intensity of force, or force per unit

area, acting tangent to A.

Average shear stress distributed over each sectioned area

that develops shear force is defined by:

avg = Average shear stress at the section, which is assumes

to be the same at each point located on the section

V = Internal resultant shear force at the section determined

from the equitions of equilibrium

A = Area at the section

A

Vavg

41

Depending on the type of connection, a connecting element (bolt, rivet, pin) may be subjected to single shear or double shear as shown.

Rivet in Single Shear

4

d

P

A

V

2

42

Shear Stress (cont.)

22

2

)4

(2d

P

d

P

A

V

43

Rivet in Double Shear

Shear Stress (cont.)

Example 1.10:

The inclined member as shown in the figure is subjected to a

compressive force of 3000N. Determine the average shear stress

along the horizontal plane defined by EDB.

Solution:

The average shear stress acting on the

horizontal plane defined by EDB is:

Shear Stress (cont.)

A

F

A

Pave

Single Shear

A

F

A

P

2ave

Double Shear

Shear Strain

The effect of shear stress is to distort the shape of a body

by inducing shear strains

The shear strain, is a measure of the angular distortion of

the body.

L

Vx

L

x

44

Example 1.11:

The plate is deformed into the dashed shape as shown in the figure.

If in this deformed shape horizontal lines on the plate remain

horizontal and do not change their length, determine the average

shear strain in the plate relative to the x and y axis .

Solution:

The angle of BAC, 90 between the sides

of the plate, referenced from the x, y axes

changes to ’ due to the displacement of

B to B’. Since xy =/2-’ is shown in the

figure, thus,

Example 1.12:

A punch for making holes in steel plates is shown in the figure a.

Assume that a punch having a diameter of 20mm is used to

punch a hole in a 6.5mm plate as shown in the cross-sectional

view (figure b). If a force P=125kN is required, what is the

average shear stress in the plate and average compressive

stress in the punch?

Solution:

The average shear stress in the plate is,

The average compressive stress in the plate is,

MPa306

m0065.0m02.0

N10125

dt

P

A

P 3

savg

MPa398

4/m02.0

N10125

4/d

P

A

P

22punch

c

3

Bearing Stress

Bearing stress is also known as a contact stress

Definition: The bearing stress σb is the average compressive

normal stress acting over a contact surface. It is given by the

load divided by the projected contact area.

Bearing stress occurs in joints, connections, pins, etc.

– anywhere you have surfaces in contact.

Bearing stress in rivet and plat;

td

Pb

45

The connection shown in the figure consists of five

steel plates, each 2.5 mm thick, to be joined by a

single bolt. Determine the required diameter of the bolt

if the allowable bearing stress, σb, is 180.0 MPa and

the allowable shear stress, allow, is 45.0 MPa?

Example 1.13:

Solution:

Solution (cont.):

Shear Modulus

It also known as Shear Modulus of Elasticity or the Modulus

of Rigidity.

Value of shear modulus can be occur from the linear region

of shear stress-strain diagram.

The modulus young (E), poisson’s ratio() and the modulus

of rigidity (G) can be related as

G Unit : Pa

)1(2

EG

48

Example 1.14:

A bearing pad of the kind used tosupport machines an bridge girdersconsists of a linearly elastic material(usually an elastomer, such asrubber) capped by a steel plate asshown in the figure a. Assume thatthe thickness of the elastomer is h,the dimension of the plate are ab,and the pad is subjected to ahorizontal shear force V.

Obtain formulas for the averageshear stresses in the elastomer andthe horizontal displacement d of theplate (figure b).

Solution:

Assume that the shear stresses in elastomer are uniformly distributed

throughout its entire volume. Then the shear stress on any horizontal

plane through the elastomer equals the shear force V divided by the area

of plane (figure a):

ab

Vavg

From the equation of Hooke’s Law in shear;

,Ge

eee

avg

abG

V

G

ab

V

G

In which Ge is the shear modulus of the elastomer material.

eabG

Vtanhtanhd

The horizontal displacement d is equal to h tan ;

In most practical situations the shear strain is a small angle, and in

such cases we may replace tan by and obtain;

eabG

hVhd

Solution (cont.):

Stress on an inclined plane

50

sincos PVPF

• Resolve P into components normal and

tangential to the oblique section,

cossin

cos

sin

cos

cos

cos

00

2

00

A

P

A

P

A

V

A

P

A

P

A

F

• The average normal and shear stresses on

the oblique plane are

Stress on an inclined plane (cont.)

Allowable Stress/ Safety Factor

Applied load that is less than the load the member can fully support. (maximum load)

One method of specifying the allowable load for the designor analysis of a member is use a number called the Factorof Safety (FS).

Allowable-Stress Design

allow

fail

F

FFS FS > 1

FSor

FS

yieldallow

yieldallow

63

Deformations Under Axial Loading

AE

P

EE

• From Hooke’s Law:

• From the definition of strain:

L

• Equating and solving for the deformation,

AE

PL

• With variations in loading, cross-section or

material properties,

i ii

ii

EA

LP

Example 1.15:

Bar aluminium AB, bar tembaga BC, dan bar keluli CD adalah dihubungkan secara sempurna diantara satu sama lain dan diikat tegar pada dinding seperti yang ditunjukkan dalam rajah dibawah. Luas keratan rentas AB, BC dan CD masing-masing 200mm2, 150mm2, dan 100mm2. Tentukan jumlah pemanjangan yang berlaku keatas bar ABCD akibat daripada beban-beban yang dikenakan. Modulus keanjalan bagi aluminium, tembaga, dan keluli adalah masing-masing 70GPa, 110GPa, dan 200GPa.

12 kN10 kN 8 kN

A BC D

0.6m 0.2m0.4m

Solution:

kN12P

0PkN12

0F

CD

CD

x

kN4P

0PkN8kN12

0F

BC

BC

x

10 kN 8 kN

A B C D

0.6m 0.2m0.4m

12kN

12 kNPCD12kN8kNPBC

m102.1

)m/N10200)(m10100(

m2.0N12000

AE

PLL

4

292

CDCD

6

m10697.9

)m/N10110)(m10150(

m4.0N4000

AE

PLL

5

292

BCBC

6

Solution:

kN6P

0PkN10kN8kN12

0F

AB

AB

x

m10571.2

)m/N1070)(m10200(

m6.0N6000

AE

PLL

4

292

ABAB

6

10 kN 8 kN

C D

12kN

B

PAB

m10013.4m102.1m10697.9m10571.2

LLLL

554

CDBCABABCD

4

Example 1.16:

Bar tegar ABC dalam rajah di bawah berada dalam keadaan mengufuk ketika tiada beban dikenakan. Dengan beban sebanyak 10kN dikenakan di C, panjang serta luas keratan rentas rod BD masing-masing 1m dan 80mm2, dan modulus keanjalan rod BD adalah 200GPa, tentukan anjakan yang berlaku keatas titik C.

10kN

0.5m1mA B C

D

Solution:

N15000m0.1

)m5.1(N10000P

0)m0.1(P)m5.1(N10000

0M

rod

rod

A

m10375.9

)m/N10200)(m1080(

m1N15000

AE

PLL

4

2926rod

rod

AX

AY10kN

Prod

0.5m1m

B CA

+

1m 0.5m

LCLrod

m10406.1

10375.95.1

L5.1

5.11

L

3

4

rodC

Crod

Example 1.17

Determine the deformation of

the steel rod shown under the

given loads.

in. 618.0 in. 07.1

psi1029 6

dD

E

SOLUTION:

• Divide the rod into components at

the load application points.

• Apply a free-body analysis on each

component to determine the

internal force

• Evaluate the total of the component

deflections.

SOLUTION:

• Divide the rod into three

components:

221

21

in 9.0

in. 12

AA

LL

23

3

in 3.0

in. 16

A

L

• Apply free-body analysis to each

component to determine internal forces,

lb1030

lb1015

lb1060

33

32

31

P

P

P

• Evaluate total deflection,

in.109.75

3.0

161030

9.0

121015

9.0

121060

1029

1

1

3

333

6

3

33

2

22

1

11

A

LP

A

LP

A

LP

EEA

LP

i ii

ii

in. 109.75 3

Example 1.18:

The rigid bar BDE is supported by two

links AB and CD.

Link AB is made of aluminum (E = 70

GPa) and has a cross-sectional area of

500 mm2. Link CD is made of steel (E

= 200 GPa) and has a cross-sectional

area of (600 mm2).

For the 30-kN force shown, determine

the deflection a) of B, b) of D, and c)

of E.

SOLUTION:

• Apply a free-body analysis to

the bar BDE to find the forces

exerted by links AB and DC.

• Evaluate the deformation of

links AB and DC or the

displacements of B and D.

• Work out the geometry to find

the deflection at E given the

deflections at B and D.

Displacement of B:

m10514

Pa1070m10500

m3.0N1060

6

926-

3

AE

PLB

mm 514.0B

Displacement of D:

m10300

Pa10200m10600

m4.0N1090

6

926-

3

AE

PLD

mm 300.0D

Free body: Bar BDE

ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN300

0M

kN90

m2.0m6.0kN300

0

D

SOLUTION:

Displacement of E:

mm 7.73

mm 200

mm 0.300

mm 514.0

x

x

x

HD

BH

DD

BB

mm 928.1E

mm 928.1

mm 7.73

mm7.73400

mm 300.0

E

E

HD

HE

DD

EE

Static Indeterminacy

• Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate.

A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.

Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations

• Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.

0 RL

65

Determine the reactions at A and B for the steel bar and

loading shown, assuming a close fit at both supports

before the loads are applied.

66

Example 1.19

Solution:

BA

BA

y

RR

RR

F

900

0600300

0

PAD

AAD

ADA

y

RP

PR

F

0

0

PDC

300

0300

0

ADC

DCA

y

RP

PR

F

PCK

300

0300

0

ACK

CKA

y

RP

PR

F

Solution:

PKB

900

0600300

0

AKB

KBA

y

RP

PR

F

kNR

RRRR

E

AE

PL

AE

PL

AE

PL

AE

PL

A

AAAA

KBCKDCAD

AB

KBCKDCADAB

323

010400

15.0900

10400

15.0300

10250

15.0300

10250

15.01

0

0

6666

kNRR AB 577323900900

Two cylindrical rods, CD made of steel (E = 200 GPa) and AC made

of aluminum (E = 72 GPa), are joined at C and restrained by rigid

supports at A and D. Determine

(a) The reactions at A and D,

(b) the deflection of point C.

67

Example 1.20

Solution:

AABAAB

x

RPRP

F

,0

0PAB

PBC

31080

080

0

ABC

ABC

x

RP

RP

F

PCD

310140

06080

0

ACD

ACD

x

RP

RP

F

Solution:

Solution:

Thermal Stresses

• A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports.

coef.expansion thermal

AE

PLLT PT

• Treat the additional support as redundant and apply the principle of superposition.

0

0

AE

PLLT

PT

• The thermal deformation and the deformation from

the redundant support must be compatible.

TEA

P

TAEPPT

0

At room temperature (21°C) a 0.5 mm gap exists between the

ends of the rods shown. At a later time when the temperature has

reached 160°C, determine:

(a) The normal stress in the aluminum rod,

(b) the change in length of the aluminum rod.

Example 1.21

Solution:

Solution: