Solid Mechanics - eduwavepool.unizwa.edu.om

Solid Mechanics

Dr. Imran LatifDepartment of Civil and Environmental Engineering

College of EngineeringUniversity of Nizwa (UoN)

1

Chapter 1: Tension, Compression and Shear

Why do we study Mechanics of Solids? 2

Anyone

concerned

with

the

strength

and

physical

performance

of

natural/man‐made structures should study Mechanics of Solids

Introduction to Mechanics of Solids 3

Definition: Mechanics of solids is a branch of applied mechanics

that deals with the behaviour of solid bodies subjected to various

types of loading

Compression

Tension

Bending

Torsion

Shearing(streteched)

(twisted)


Fundamental concepts

• stress and strain

• deformation and displacement

• elasticity and inelasticity

• load‐carrying capacity

Design and analysis of mechanical and structural systems


•

Examination of stresses and

strains inside real bodies of

finite dimensions that deform

under loads

•

In order to determine stresses

and strains we use:

–

Physical properties of

materials

–

Theoretical laws and

concepts

Problem solving 6

• Draw the free‐body diagram

• Check your diagram

• Calculate the unknowns

• Check your working

• Compute the problem


• Write the solution


Free Body Diagram 7

Free Body Diagram

The unknowns: RA

, RB

, RC

Free Body Diagram 8

Free Body Diagram

The unknowns: Ax

, Ay

, CB

Normal stress and strain•

Most fundamental concepts in Mechanics of Materials are stress and strain

•

Prismatic bar: Straight structural member with the

same

cross‐section

throughout its length

•

Axial force: Load directed along the axis of the member

•

Axial force can be tensile

or compressive

•

Axial loads: Tension (+) and compression (−)

•

Type of loading for landing gear strut and for tow bar?

9

Structural members subjected to axial loads

Normal stress and strain 10

A truss bridge is a type of beam bridge with a skeletal structure. The

forces of tension, or pulling, are represented by red lines and the

forces of compression, or squeezing, are represented by green lines.

Normal stress 11

Continuously distributed stresses

acting over the entire cross‐section.

Axial force P is the resultant of those

stresses

Stress (σ)

has units of force per unit

area

If stresses acting on cross‐section

are uniformly distributed then:

Units of stress in USCS: pounds per square inch (psi) or kilopounds per square inch (ksi)SI units: newtons per square meter (N/m2) which is equal to Pa

FBD of a segment of the bar

Segment of the bar before loading

Segment of bar after loading

Normal stresses in the bar

Limitations 12

The loads P are transmitted to the bar by pins that pass through

the holes

High localized stresses are produced

around the holes !!

Stress concentrations

Steel eyebar subjected to tensile loads P

Normal strain 13

A prismatic bar will change in length when under a uniaxial tensile

force, and obviously it will become longer

Definition of elongation per unit length

or strain (ε)

If bar is in tension, strain is tensile

and if in compression the strain is

compressive

Strain is a dimensionless quantity

(i.e. no units)

FBD of a segment of the bar

Segment of the bar before loading

Segment of bar after loading

Normal stresses in the bar

Line of action of the axial forces for a uniform stress distribution

14

It can be demonstrated that in

order to have uniform tension or

compression in a prismatic bar,

the axial force must act through

the centroid of the cross‐

sectional area.

Example

A short post constructed from a

hollow circular tube of aluminum

supports a compressive load of 26

kips (26000 lb). The inner and outer

diameters of the tube are d1

= 4 in.

and d2

= 4.5 in. respectively, and its

length is 16 in. The shortening of the

post due to the load is measured as

0.012 in.

Determine:

(a) Compressive stress in the

post.

(b) Compressive strain in the

post.

15

16

Problem 17

Problem

A circular aluminum tube of length L = 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal

direction.

(a) If the measured strain is 550 x 10-6

, what is the shortening of the bar?

(b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

18

Problem

Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle α

= 34°

to the horizontal and wire BC is at an angle β

= 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.)

Determine the tensile stresses AB and BC in the two wires.

Mechanical properties of materials 19

In order to understand the mechanical

behaviour of materials we need to perform

experimental testing in the lab

A tensile test machine is a typical equipment

of a mechanical testing lab

ASTM (American Society for Testing and

Materials)

Stress () –

strain () diagrams 20

–

diagrams 21

Stress‐strain diagram for a typical structural

steel in tension (not to scale)

Nominal stress and strain (in the calculations we use the initial

cross‐sectional area A)

True stress (in the calculations we use the cross‐sectional area A

when failure occurs)

True strain if we use a

strain gauge

Stress‐strain diagrams

contain important

information about

mechanical properties

and behaviour

–

diagrams 22

OA: Initial region which

is linear and

proportional (Slope of

OA is called modulus of

elasticity)

BC: Considerable

elongation occurs with

no noticeable increase

in stress (yielding)

CD: Strain hardening –

changes in crystalline structure (increased

resistance to further deformation)

DE: Further stretching leads to reduction in the applied load and fracture

OABCE’: True stress‐strain curve

23

Tensile coupon test specimens showing elongation

immediately prior to failure (G.J. Davies)

–

diagrams 24

Stress‐strain diagram for a typical

structural steel in tension (drawn to

scale)

The strains from zero to point A are so

small as compared to the strains from

point A to E and can not be seen (it is

a vertical line…)

Metals, such as structural steel, that

undergo permanent large strains

before failure are ductile

Ductile materials

absorb large

amounts of strain energy

Ductile materials: aluminium, copper,

magnesium, lead, molybdenum,

nickel, brass, nylon, teflon

Aluminum alloys 25

Typical stress‐strain diagram for an

aluminum alloy.

Although ductile, aluminium alloys

typically do not have a clearly

definable yield point

However, they have an initial

linear region with a recognizable

proportional limit

Structural alloys have proportional

limits in the range of 70‐410 MPa

and ultimate stresses in the range

of 140‐550 MPa

Offset method 26

When the yield point is not

obvious, like in the previous

case, and undergoes large

strains, an arbitrary yield stress

can be determined by the

offset method (e.g. 0.002 or

0.2%)

The intersection of the offset

line and the stress‐strain curve

(point A) defines the yield

stress

Brittle materials 27

Typical stress‐strain diagram for a brittle

material showing the proportional limit

(point A) and fracture stress (point B)

Brittle materials fail at relatively

low strains and little elongation

after the proportional limit

Brittle materials: concrete, marble,

glass, ceramics and metallic alloys

The reduction in the cross‐

sectional area until fracture (point

B) is insignificant and the fracture

stress (point B) is the same as the

ultimate stress

28

Problem

Three different materials, designated A, B,and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of

2.0 in. (see figure).

At failure, the distances between the gage marks are found to be

2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be0.484, 0.398, and 0.253 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.

29

Problem 1.3-6

A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?

Solution to problem 30

Modulus of elasticity: 2.35 GPa

Proportional limit: 47 MPa

Yield stress: 52.5 MPa

Material is brittle, because the strain after the proportional limit is exceeded is relatively small.

Elasticity 31

Fig. 1. Elastic behavior

Fig. 2. Plastic behavior

What happens when the load is removed (i.e.

the material is unloaded)?

Tensile load is applied from O to A (Fig 1) and

when load is removed the material follows the

same curve back. This property is called

elasticity

If we load the same material from O to B (Fig

2) and then unloading occurs, the material

follows the line BC. Line OC represents the

residual

or permanent

strain. Line CD

represents the elastic recovery of the material.

During unloading the material is partially

elastic

Plasticity 32

“

Plasticity

is the characteristic of a material which undergoes inelastic

strains beyond the strain at the elastic limit ”

When large deformations occur in a ductile material loaded in the

plastic region, the material is undergoing plastic flow

33

If the material is in the elastic range, it can

be loaded, unloaded and loaded again

without significantly changing the behaviour

When loaded in the plastic range, the

internal structure of the material is altered

and the properties change

If the material is reloaded (fig 1‐19), CB is a

linearly elastic region with the same slope as

the slope of the tangent to the original

loading curve at origin OBy stretching steel or aluminium into the

plastic range, the properties of the material

are changed

Reloading of a material

Creep 34

When loaded for periods of time, some

materials develop additional strains and

are said to creep

Even though the load P remains constant

after

time t0, the bar gradually lengthens

Relaxation is a process at which, after time t0

,

the stress in the wire gradually diminishes and

eventually is reaching a constant value

Creep is more important at high temperatures

and has to be considered in the design of

engines and furnaces

Hooke’s law 35

Many structural materials such as metals, wood, plastics and ceramics

behave both elastically and linearly when first loaded and their

stress‐

strain curve begin with a straight line passing through origin (line OA)

Linear elastic materials are useful for designing structures and

machines

when permanent deformations, due to yielding, must be avoided

Hooke’s law 36

Robert Hooke

(1635‐1703)

The linear relationship between stress and strain for a bar in

simple tension or compression is expressed by:

σ

= E εσ

is axial stress

ε

is axial strain

E is modulus of elasticity

Hooke’s law

The above equation is a limited version of Hooke’s Law relating only the

longitudinal stresses and strains that are developed during the

uniaxial

loading of a prismatic bar

Robert Hooke

was an English inventor, microscopist, physicist, surveyor,

astronomer, biologist and artist, who played an important role in the

scientific revolution, through both theoretical and experimental

work.

Modulus of Elasticity 37

E is called modulus of elasticity or Young’s modulus and

is a constant

It is the slope of the stress – strain curve in the linearly

elastic region

Units of E are the same as the units of stress (i.e. psi or

Pa)

Thomas Young

was an English polymath, contributing to the scientific

understanding of vision, light, solid mechanics, energy, physiology, and

Egyptology.

When a prismatic bar is loaded in tension the

axial elongation is accompanied by lateral

contraction

The lateral strain ε’

at any point in a bar is

proportional to the axial strain ε

at the same

point if the material is linearly elastic

The ratio of the above two strains is known as Poisson’s ratio

(ν)

Poisson’s ratio 38

lateral

contractionlongitudinal

extension

ν

= ‐

(lateral strain / axial strain = ‐

(ε’

/ ε

)

Poisson’s ratio 39

Simeon Denis Poisson(1781‐1840)

Siméon‐Denis Poisson

was a French mathematician, geometer, and

physicist.

The minus sign in the equation is because the lateral

strain is negative (width of the bar decreases) and the

axial tensile strain is positive. Therefore, the Poisson’s

ratio will have a positive value.

When using the Poisson’s ratio equation we need to

know that it applies only to a prismatic bar in uni‐axial

stress

Poisson’s value of concrete = 0.1 – 0.2

Poisson’s value of rubber = 0.5

Limitations 40

Poisson’s ratio is constant in the

linearly elastic range

Material must be homogeneous

(same composition at every

point)

Materials having the same

properties in all directions are

called isotropic

If the properties differ in

various directions the materials

called anisotropic

Shear stress and shear strain 41

tangentialperpendicular

Shear stress

acts tangential to the surface

of the material and not perpendicular

Consider the bolted connection of Figure

where A is a flat bar, C a clevis and B a bolt

When load P is applied, the bar and clevis

will press against the bolt and bearing

stresses

will be developed

The bar and clevis tend to shear

the bolt


If we have a closer look from the side

view (fig b) and draw a FBD (fig c)

bearing area

total bearing force

Bearing stresses exerted by the clevis against the

bolt appear on the left‐hand side (1 and 3)

Stresses from the bar are on the right‐hand side (2)

Based on the assumption of uniform stress

distribution we can calculate an average bearing

stress σb


The

bearing

area

Ab

is

defined

as

the

projected

area

of

the

curved

bearing

surface.

For

example

(for

stresses

labeled

1)

the

projected

area

on

which

the stresses

act is a

rectangle

with

height

equal

to

the

thickness of the clevis and width equal to the diameter of the bolt

The bearing force Fb

(for stresses labeled 1) is equal to P/2

The same area and force apply for stresses labeled 3For

bearing

stresses

labeled

2

the

bearing

area

is

a

rectangle

with

height

equal

to

the

thickness

of

the

flat

bar

and

width

equal

to

the

diameter of the bolt. The force is equal to P


The FBD (fig. c) shows that there is a tendency to shear the bolt along the

cross sections mn and pq

From the FBD (fig. d) of the portion mnqp of the bolt we see that the shear

forces V act over the cut surfaces of the bolt. There are two planes of shear

(plane mn and plane pq). Therefore, the bolt is in double shear

The shear stresses acting on the cross section mn are shown (fig. e)Shear stresses are denoted by τ

Single shear 45

The axial force P in the metal bar is

transmitted to the flange of the

steel column through a bolt

A cross section of the column (fig b)

shows more details

Fig c shows the assumed distribution

of the bearing stresses acting on the

bolt

Cutting through the bolt at section

mn (fig d) we see the shear force V

(equal to load P). V is the resultant

of the shear stresses that act over

the cross‐sectional area of the bolt

Single shear 46

Shear stress and strain 47

Discussing about bolted connections we disregard friction which is

produced by tightening the bolts

Average shear stress on the cross section of a bolt is obtained by dividing

the total shear force V by the area A of the cross section on which it acts:

Shear stresses have the same units as normal stresses

The two previous examples (double and single shear) are examples

of

direct shear

Direct shear arises in the design of bolts, pins, rivets, keys, welds and

glued joints

48

Consider a small rectangular parallelepiped

element

Assume that a shear stress τ1

is uniformly

distributed over the right‐hand side area bc

(τ1bc

)

For equilibrium in the y direction the τ1bc

must

be balanced by an equal and of opposite

direction shear force on the left‐hand side

The forces τ1bc

acting on the right‐hand and

left‐hand side faces form a couple having a

moment about z‐axis equal to τ1

.abc

(counterclockwise direction)

Small element of material

subjected to shear stresses

Equality of shear stresses on perpendicular planes 49

Similarly, in order to have equilibrium of

the element, we have a shear force τ2

.ac

and consequently a clockwise couple of

moment τ2

.abc

It is therefore evident that for moment

equilibrium we have:

τ1

= τ21.

Shear stresses on opposite and parallel faces of an element

are equal in magnitude and opposite in direction

2.

Shear stresses on adjacent and perpendicular faces of an

element are equal in magnitude and have directions such

that both stresses point toward, or both point away from,

the line of intersection of the faces

50

Shear stresses

acting on an element of

material (fig a) are accompanied by shear

strains

The lengths of the sides of the element do

not change but, the shear stresses

produce a change in the shape of the

element

Rectangular parallelepiped becomes

oblique parallelepiped. Front and rear

faces become rhomboids

The angle γ

(fig b) is a measure of

distortion of the element and is called

shear strain

Hooke’s law in shear 51

We can plot shear stress‐strain diagrams

Hooke’s Law in shearτ

= Gγ

shear modulus of elasticity

shear strainshear stress

G has the same units as E (Young’s modulus)

G and E are also related by:

G = E / (2(1+ν)) Poisson’s ratio

Allowable stresses & allowable loads 52

The principal design interest is

strength

Strength is the capacity of the object

to support or transmit loads

The actual strength of a structure

must exceed

the required strength

Factor of safety must be greater than

1 if failure is to be avoided

Factors of safety from slightly above 1

to as much as 10 are used

Factor of safety (n) = Actual strength / Required strength

Allowable stresses & allowable loads 53

Allowable stress = Yield strength / Factor of safety

σallow

= σY

/ n1 τallow

= τY

/ n2

tension shear

Allowable stress = Ultimate strength / Factor of safety

OR…

σallow

= σU

/ n3 τallow

= τU

/ n4

tension shear

Allowable loads 54

Allowable load is also called permissible load or

the safe load

Allowable load = (Allowable stress) (Area)

For bars in tension or compression:Pallow

= σallow

. A

For pins in direct shear:Pallow

= τallow

. A

Permissible load based upon bearing:Pallow

= σb

. A

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Solid Mechanics - eduwavepool.unizwa.edu.om