stress concept

FF

Faculty of Engineering and Materials Science

Prof. Dr. E.I. Imam Morgan Head of Mechatronics Department

Vice Dean of EMS for Academic Affairs

[email protected]

C7.110

Strength of Materials I

(1)

mailto:[email protected]

Office hours: Monday, 2nd slot (10:00 12:00)

Teaching Assistant: Eng. Bola George, Bishoy Emil, Micheal Wahba [C7:116]

Eng. Mona Nader [C7:101], Eng. Sara Elkhamisy [C3:123]

Eng. Andrew Nagy [C3:127]

Text book:

Beer, F.P., and Johnston, E.R. Mechanics of Materials,

fourth Edition (2006), Fifth edition is now available.

McGraw Hill Publishing Co., New York, NY, ISBN 007-124999-0.

and:

J.M. Gere Mechanics of Materials,

sixth Edition (2004), eights edition is available.

Brooks/Cole, Belmont, CA, ISBN 0-534-41793-0.

Course Assessment:

10% 4 Assignments

25% 2 Quizzes (best 2 out of 3)

25% Mid term exam

40% Final exam.

Attendance: 75% of the course must be attended

How to drop:

Dropping of the course not later than 2 weeks after the start by

notice to the admission office. (2)

Chapter 1

Introduction

3

This chapter is devoted to the study of stresses occurring in many of the elements contained in the shown excavator, such as two-force members, axles, bolts, and pins.

(4) Prof. Dr. Imam Morgan

Head of MCTR Department

Strength of Materials

or, (Mechanics of Deformable Bodies)

or, (Mechanics of Materials) ??

The Strength of Materials is the branch of applied mechanics that deals

with the behavior of elastic bodies subjected to various types of loading.

Bars (axial loading)

Shafts (torsion)

Beams (bending)

Columns (compression)-buckling

Objective Is to provide the future engineer with the means of

analyzing and designing the various components of

any structure or any machine such that they must

operate safely.

The bodies under investigations represent the components of a machine or

a structure.

These components may be

1 Introduction

Prof. Dr. Imam Morgan


(5)

Beams (B.M load)

Shafts

(torsion load)

The most important

concepts in Strength of

Materials are:

Stress and Strain

These concepts will be

illustrated (in this chapter)

by considering a prismatic

bar subjected to axial

forces.

Later, these loadings

produce other types of

stresses.

But, there are other

types of loadings such

as torsion (in shafts)

and bending moment

loads (in beams).

Bars (axial load)



The first step in studying an element is to determine the

loads applied on it in order to decide whether it can support these loads or not.

Therefore, methods of statics are used to determine the static

forces in the supports of the structure as well as the forces applied on the component itself. In machines, the applied dynamic forces are calculated using a similar technique

based on the method of inertia forces (Later in theory of machines).

The steps of study (analysis and design) of a certain structure as well as the concept of stress are investigated using the following example.

2 Static Review



(7)

The shown structure is designed

to support a 30 kN load

The structure consists of a boom AB and rod CB joined by pins at the junction B and supports A and C.



(8)

Static Analysis

Perform a static analysis to

determine the internal force in

each structural member and the

reaction forces at the supports.

The boom and rod are 2-force

members, i.e., the members are

subjected to only two forces which

are applied at member ends.

For equilibrium, the forces must be

parallel to an axis between the

force application points, equal in

magnitude, and in opposite directions.

Therefore, in this case we can use the

method of equilibrium of pin at B (or resolve

the 30kN force into 2 directions)that had been

given in the previous semester [Mechanics I].



(9)

Joints must satisfy the conditions for static equilibrium of concurrent forces

which may be expressed in the form of a force polygon (triangle):

kNF

.FF

kNF

.FF

AB

ABx

BC

BCy

40

080500

50

030600

Note : According the obtained positive values we conclude that:

member AB is in compression. member CB is in tension.

kN50

kN40

3

30

54

BC

AB

BCAB

F

F

FF

Or, from force

polygon



Can the structure safely support

the 30 kN load?

A question is waiting

for an answer!!

From a statics analysis

FAB = 40 kN (compression)

FBC = 50 kN (tension) dBC = 20 mm

Consider, for example the rod BC. Its ability to withstand the internal

tensile force FBC depends on: the internal force in the rod, FBC the cross section area of the rod, A the material of which the rod is made

Compare

and

decide

3 Concept of Stress

Combined in

one parameter, called Stress

Each material has its

own mechanical

properties (parameters) Prof. Dr. Imam Morgan


(11)

In general, the force per unit area, or intensity

of the force distributed over a given section is

called the stress (sigma).

As shown in figure:

A

P Pa2 ,orm/N

we have:

1 kPa = 103 Pa

1 MPa = 106 Pa commonly used

1 GPa = 109 Pa Animation



../Animation/Normal_and_shear_stresses.swf

Back to the example:

Analysis From a statics analysis FAB = 40 kN (compression)

FBC = 50 kN (tension)

MPa159m10314

N105026-

3

A

PBC

At any section through member BC, the internal force is 50 kN with a force intensity

or stress of

Conclusion: the strength of

member BC is adequate (safe).

MPa 165all

From the material properties for steel, the allowable stress is

allBC

dBC = 20 mm

Now assume that the rod BC is

made of steel.



(13)

Design of new structures requires selection

of appropriate materials and component

dimensions to meet performance

requirements

For reasons based on cost, weight, availability,

etc., the choice is made to construct the rod from

aluminum all= 100 MPa). What is an

appropriate choice for the rod diameter?

mm2.25m1052.2

m1050044

4

m10500Pa10100

N1050

226

2

26

6

3

Ad

dA

PA

A

P

allall

An aluminum rod 26 mm or

more in diameter is adequate

Design

Choose

material

d = ?

for safety



Normal Stress

Shear Stress

Bearing Stress

Types of Stresses



(15)

The resultant of the internal forces for an axially

loaded member is normal to a section cut perpendicular to the member axis.

A

P

A

Fave

A

0lim

The force intensity (force per unit area) on that

section is defined as the normal stress.

The normal stress at a particular point may not be

equal to the average stress but the resultant of the

stress distribution must satisfy

A

ave dAdFAP

The detailed distribution of stress is statically

indeterminate, i.e., can not be found from statics

alone.

Animation

4 Normal Stress



../Animation/Saint_Venants_principle.swf

If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial

force and a moment.

The stress distributions in eccentrically loaded

members cannot be uniform or symmetric.

This case will be studied in Ch.4.

A uniform distribution of stress in a section implies that the line of action for the resultant of the internal forces passes through the centroid of the section.

A uniform distribution of stress is only

possible if the concentrated loads on the end

sections of two-force members are applied at

the section centroids. This is referred to as

centric loading.

Normal Stress in case of Centric and Eccentric Loadings

(17)

Example (1) P = 115 kN

L

A short post constructed from a hollow

circular tube of aluminum supports a

compressive load of 115 kN. The outer and

inner diameters of the tube are do= 115 mm

and di= 100 mm, respectively. Determine the compressive stress in the post

(average stress).

Solution

MPa

mmddA

NP

A

P

io

4.4591.2532

10115

91.253210011544

10115

3

22222

3

Note that: Substituting P in N and the area A in

mm2

will give, directly, the

stress in MPa.



(18)

Example (2)

Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the

average stress at the midsection of:

(a) rod AB, (b) rod BC.

-80 kN

+

+160 kN

Solution

At first determine the axial force in each

rod. So, draw the Normal Force Diagram.

a- rod AB (rod 1)

N. F. D.

MPa

mmA

NP

A

P

1.1886.4417

1080

86.4417754

1080

3

2

22

2

3

2

2

22

b- rod BC (rod 2)

MPa

mmA

NP

A

P

49.815.1963

10160

5.1963504

10160

3

1

22

1

3

1

1

11

+ tensile stress

Compressive

stress



(19)

Forces P and P are applied transversely to the member AB.

A

Pave

The corresponding average shear stress is,

The resultant of the internal shear force

distribution is defined as the shear of the section

and is equal to the load P.

Corresponding internal forces act in the plane of section C and are called shearing forces.

Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value (Ch.6).

The shear stress distribution cannot be assumed uniform.

Our interest in this chapter

5 Shearing Stress



(20)

P

y

z

x

2

1 1

2

Shearing stress distribution due

to transverse load (Ch. 6).

Note that P acts in plane of the section.



(21)

A

F

A

Pave

Single Shear

Examples of Shearing stress

A

F

A

P

2ave

Double Shear

F/2

F/2

2/FP



(22)

Bolts, rivets, and pins create

stresses on the points of contact

or bearing surfaces of the members they connect.

dt

P

A

Pb

Corresponding average force

intensity is called the bearing stress.

The resultant of the force

distribution on the surface is

equal and opposite to the force

exerted on the pin.

The bearing area is defined as the projected area of the

curved bearing surface

6 Bearing Stress



(23)

We Would like to determine

the stresses in the members

and connections of the

structure shown.

Must consider :

maximum normal

stresses in AB and BC, shearing stress at pins,

bearing stress, on

members at each pinned

connection

From a statics analysis:

FAB = 40 kN (compression)

FBC = 50 kN (tension)

Example (3) Application to Analysis of Simple Structure



(24)

The rod is in tension with an axial force of 50 kN.

The boom is in compression with an axial force of 40

kN and average normal stress of 26.7 MPa.

The minimum area sections at the boom ends are unstressed since the boom is in compression.

MPaA

P

mmA

end,BC 167300

1050

300254020

3

2

At the flattened rod ends, the smallest cross-sectional

area occurs at the pin centerline,

At the rod center, the average normal stress in the

circular cross-section (A = d2/4 = 314.15 mm2) is

BC = +159 MPa.

A- Normal Stresses: (rod BC and boom AB)

Rod BC

Boom AB

FBC



(25)

The cross-sectional area for pins at

A, B, and C,

222

4912544

mmd

A

The pin at A is in double shear with a total force equal to the force exerted by

the boom AB,

MPa.10

A

P 3

ave,A 740491

20

B- Shear Stresses: (pin supports)

MPaA

Pave,C 102

491

1050 3

The force on the pin at C is equal to the force exerted by the rod BC. The pin has single shear.

Pin C

Pin A



(26)

Divide the pin at B into sections to determine the section with the largest shear force,

(largest) kN25

kN15

G

E

P

P

MPa.10

A

P 3Gave,B 950

491

25

Evaluate the corresponding average shearing

stress,

kN50BCF

Pin B

Note that we can consider that the pin is

subjected to double shear with P=50/2 kN



(27)

To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

MPa.

10

td

P 3

b 3532530

40

To determine the bearing stress at A in the bracket, we have t = 2(25) = 50 mm and d = 25 mm,

MPa.

10

td

P 3

b 0322550

40

C- Bearing Stresses: (at supports)

Support at A

Bracket Boom

The bearing stresses at B in member AB, at B and C

in member BC, and the bracket at C are found in a similar way.



(28)

2000 N

250 mm 125 mm

Example (4)

The shown system is used to support 2000N. The upper

portion of link ABC is 10 mm thick and the lower portions

are each 6 mm thick. Epoxy resin is used to bond the two

portions at B. The diameters of pins at A and C are 10 mm

and 6 mm, respectively. Determine:

a- the shearing stress in pin A,

b- the shearing stress in pin C,

c- the largest normal stress in link ABC,

d- the average shearing stress on the bonded surface at B,

e- the bearing stress in the link at C

Solution:

Draw the F. B. D. for the entire

system. Link ABC is two force member.

tensionNF

FM

AC

ACD

3000

02503752000:0

2000 N 250 mm

125 mm

175 mm

45 mm

150 mm

30 mm



(29)

(a) Shear stress at pin A: (A)

pin A is single shear:

MPa.

A

F

A

ACA

238

104

3000

2

(b) Shear stress at pin C: (C)

double shear.

MPa.

A

F

C

ACC

153

64

2

3000

2 2

(c) Largest Normal stress in link ABC: The largest stress occurs at cross section at A

(smallest area)

The area at A is: Anet= 10 x (30-10) = 200 mm2

MPa

A

F

A

net

ACA

15

200

3000

3000 N



(30)

2000 N 250 mm

125 mm

175 mm

45 mm

150 mm

(d) Average shear stress at B: (B)

MPa.

mmA

N/F:sideoneFor

A

F

B

B

BB

111350

1500

13504530

150023000

2

1

1

(e) Bearing stress in link ABC at C : For each portion:

F1 = 1500 N bearing area = 6x6 = 36 mm2

MPa.

A

F

b

b

741

36

15001



(31)

2000 N 250 mm

125 mm

175 mm

45 mm

150 mm

3000 N

Example (5)

The steel bar is to be designed to support a

tension force of P = 120 kN when bolted

between double brackets at A and B. The bar will

be fabricated from 20 mm thick steel plate. The

maximum allowable stresses for this steel are

= 175 MPa, = 100 MPa, and b = 350 MPa.

a- Determine the diameter d of the bolt.

b- Determine the dimension b at each end of the

bar.

c- Determine the dimension h of the bar.

Solution:

Given: Required:

P = 120 kN d = ? (for bolt) t = 20 mm b = ? (at end of the bar) allowable stresses: h = ? (for the bar)

= 175 MPa = 100 MPa b = 350 MPa

20 mm

h



(32)

(a) Diameter of the bolt:

at A or B we have double shear

mmd:usemm.d

d

N/PF:whereA

Fall

28627

4

1060100

10602

2

3

3

Check (for bearing stress on bolt or bar):

all,bb

b MPaMPatd

P

350214

2820

10120 3

OK

Given: Required:

P = 120 kN d = ? (for bolt)

t = 20 mm b = ? (at end of the bar)

allowable stresses: h = ? (for the bar)

= 175 MPa

= 100 MPa

b = 350 MPa



(33)

(b) Dimension b:

mm.b

.adb

mm.a

adbt

A

P

netall

362

14172282

1417

220

1012010120175

33

(c) Dimension h:

mmh:usemm.h

h

th

P

35334

20

10120175

3

Given: Required:

P = 120 kN d = ? (for bolt)

t = 20 mm b = ? (at end of the bar)

allowable stresses: h = ? (for the bar)

= 175 MPa

= 100 MPa

b = 350 MPa



(34)

However, we shall show that either axial or transverse forces may

produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.

Axial forces on a two force

member result in only normal

stresses on a plane cut

perpendicular to the member axis.

Transverse forces on bolts and

pins result in only shear stresses

on the plane perpendicular to bolt or pin axis.

7 Stresses on an Oblique Plane Under Axial Loading



(35)

Pass a section through the member forming

an angle with the normal plane. Consider the left part. Note that +ve is in CCW direction from the vertical.

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:

sincos PVPF

Resolve P into components normal and tangential to the oblique section,

From equilibrium conditions, the distributed

forces (stresses) on the plane must be

equivalent to the force P.

Animation Prof. Dr. Imam Morgan


(36)

Animation/Normal_and_shear_stresses.swf

cossinA

Pcos

A

P

0

2

0

Normal and shearing stresses on

an oblique plane:

The maximum normal

stress occurs when the

reference plane is

perpendicular to the

member axis,

00

m A

P

The maximum shear

stress occurs for a plane

at + 45o with respect to

the axis, ,

00 245cos45sin

A

P

A

Pm

Maximum Stresses



(37)

= ? for which

A member subjected to a general

combination of loads is cut into two

segments by a plane passing through

an internal point Q. Assume that the

plane is parallel to yz plane.

For equilibrium, an equal and

opposite internal force and stress

distribution must be exerted on

the other segment of the member.

A

V

A

V

A

F

xz

Axz

xy

Axy

x

Ax

limlim

lim

00

0

The distribution of internal stress

components may be defined as,

Stress Components

Q

8 Stress Under General Loading Conditions



(38)

Stress components are defined for the planes

cut perpendicular to the x, y and z axes. For equilibrium, equal and opposite stresses are

exerted on the hidden planes.

It follows that only 6 components of stress are

required to define the complete state of stress

The combination of forces generated by the

stresses must satisfy the conditions for

equilibrium:

0

0

zyx

zyx

MMM

FFF

yxxy

yxxyz aAaAM

0

zyyzzyyz andsimilarly,

Consider the moments about the z axis:



(39)

Finally: The General State of Stress at a point Q is completely

defined by 6 independent stress components, namely:

x , y , and z Normal stresses x .. Stress along x-axis Similar definitions for y , and z.

xy ,yz , and zx Shear stresses xy . Stress perpendicular to x-axis and directed along y-axis. Similar definitions for yz and zx.



(40)

Structural members or

machines must be designed

such that the working stresses

(allowable stresses) are less than the ultimate strength of the material.

Factor of safety considerations:

uncertainty in material properties

uncertainty of loadings

uncertainty of analyses

number of loading cycles

types of failure

maintenance requirements and

deterioration effects

importance of member to integrity of

whole structure

risk to life and property

influence on machine function

stressallowable

stress ultimateFS

safetyof FactorFS

all

u

load allowable

load ultimate

P

PFS

def.ealternativan or,

all

u

Ultimate Stress: is the stress at which the specimen will break or begins to

carry less load. (see next chapter).

9 Factor of Safety



(41)

Example (6)

150 mm

200 mm

dB = dD = 10 mm

dC = 12 mm

Diameter of rod AB is

dA = 11 mm

The rigid beam BCD is attached by bolts to a control rod at B, to a hydraulic cylinder at C, and to a fixed support at D. Each bolt acts in double shear and is made from steel with U = 280 MPa. The control rod

is made of steel with U = 420 MPa. If the minimum

factor of safety is to be 3 for the entire unit, determine the largest upward force which may be

applied by the hydraulic cylinder at C.

Solution:

200 mm 150 mm

F.S. must be 3 in each of the three bolts and in

the control rod.

Statics: Find the force C in terms of the force B

and the force D.

233.20350150:0

175.10200350:0

DCDCM

BCCBM

B

D



(42)

150 mm

200 mm

dB = dD = 10 mm

dC = 12 mm


dA = 11 mm

Calculation of C on the bases of:

Control Rod: For F.S = 3 we have,

MPaallall

U 14034203

kNCSoCFrom

kNNd

B Ball

28.23,3.1375.1:1

3.136.133044

11140

4

22

Bolt at B : For F.S = 3 we have,

kNCCFrom

kNN

d

areashearedBsheardoubleFor

MPa

Ball

all

Uall

66.2566.1475.11

66.144.14663

4

10233.93

42

33.9332803

22

233.2

175.1

DC

BC



(43)

150 mm

200 mm

dB = dD = 10 mm

dC = 12 mm


dA = 11 mm

Bolt at D : For F.S. = 3 we have,

kNCCFrom

kNN

areashearedBDsheardoubleFor

MPaAgain

all

Uall

16.3466.1433.22

66.144.14663

33.9332803,

Bolt at C : For F.S. = 3 we have,

kNCNC

d

areashearedCsheardoubleFor

MPa

Call

all

all

11.217.21110

4

12233.93

42

33.93

22

Conclusion: We have calculated, separately, four maximum allowable

values of C. Therefore we choose the smallest value:

C = 21.11 kN

233.2

175.1

DC

BC



(44)

Revision Problem

Given:

8x36 mm for the four links joining

BD and CE.

16-mm diameter pins at B, D, C, E thickness of lever ABC is 10 mm

Required:

a) maximum normal stress in links

connecting: points B and D points C and E b) normal stress at mid points of the four links.

c) Average shear stress at each pin.

d) At B, find the bearing stress on links and on lever.

Answer:

a) +101.56 and -21.7 MPa

b) +56.42 (BD) & -21.7 (CE) MPa

c) 80.82 (B,D) & 31.08 (C,E) MPa

d) 126.95 MPa on links BD

and 203.13 MPa on lever



(45)

stress concept

Documents

gere mechanics of materials

materials science

axial forces

shown structure

static forces

certain structure

load shafts torsion

branch of applied mechanics