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MP2001/AE2008 MECHANICS OF MATERIALS

Dr Chou Siaw Meng

Associate Professor

Tel: 67904958 Office: N3.2-02-71

Email: msmchou@ntu.edu.sg

Nanyang Technological University

Textbook: MECHANICS OF MATERIALS (5th edition, SI units, 2009)

u ors: eer, o ns on, e o an azure

Publisher: McGraw-Hill

Chapter 1- Introduction-Concept of Stress1

The Straits Times4 August 2007

Chapter 1- Introduction-Concept of Stress2

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The Straits Times: Thu 16 Aug 2007

Chapter 1- Introduction-Concept of Stress3

SOME ANSWERS AREINTENTIONALLY LEFT OUT

Chapter 1- Introduction-Concept of Stress4

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REFERENCE

1. En ineerin Mechanics of Solids E.P. Po ov

Prentice Hall.

2. Mechanics of Materials, R.C. Hibbler, Pearson-Prentice Hall.

Chapter 1- Introduction-Concept of Stress5

n ro uc on

The main objective of the study of mechanics ofmaterials is to provide the future engineer with the

means of analyzingand designingvariousmachines and load bearing structures.

involve the determination of stresses,

particular material.

Chapter 1- Introduction-Concept of Stress6

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Conditions for Equilibriumin a free-body diagram (FBD)

0Fi forces in one direction = forces in the opposite

direction(in any direction)

sum of clockwise moments = sum of anticlockwise

i

i

moments(about any point)

Chapter 1- Introduction-Concept of Stress7

M = Fd

M = (10N)(2m)

M = 20 Nmaxisd = 2m F = 10N

M = Fd the roduct of force and the

perpendicular distance from the forces line

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Revision

m = 50 Nd = 3 cm

L = 35 cm ?

mg

d mg

L L

0M L.d.F mg

Chapter 1- Introduction-Concept of Stress9

N583F

1. Body weight2. External loads

.

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RevisionRy

s n53

Rx Tcos53R

2m

4m?

)4)(200()2)(600()8)((Tsin53

o

R

2mN313T

Weight of man = 600 N

Chapter 1- Introduction-Concept of Stress11

Weight of beam = 200 N

Chapter 1

Introduction-Conce t of Stress

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Examples: strut (two-force member)

s aceframes

Chapter 1- Introduction-Concept of Stress15

Introductiontrut two- orce mem er

Chapter 1- Introduction-Concept of Stress16

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Introductiontrut two- orce mem er

Chapter 1- Introduction-Concept of Stress17

Review of Statics struts

a boom and rod joined by

connections) at the

force in each structural

forces at the supports

Chapter 1- Introduction-Concept of Stress18

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Structure Free-Body Diagram - Method I

:0

Conditions for static equilibrium:

m8.0kN30m6.0

xA

e g t ess wrt oa

kN40

xAFBD

kN40 x

xxx

C

:0 yF

yy

Ayand Cycannot be determined30kN

Chapter 1- Introduction-Concept of Stress19

from these equations

Component Free-Body Diagram

Consider a free-body diagram for the

0m8.0:0 yB AMFBD

0 yAsubstitute into the structure e uilibrium

equation (previous slide)

Results:

y

kN40A

x an y representforces from rod BC

kN40xC

Chapter 1- Introduction-Concept of Stress20

y

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CCF2

y

2

xBC

3040 22

BC

50 kN

30 kN

C

'

40 kN

BC

40 kN50 kN

ABF ABF'

AB

Chapter 1- Introduction-Concept of Stress21

Method of Joints - Method II

Direction of FBC

un nowndirection

and FAB can bedeterminethrough the forcepolygonunknown

0BF

knownrec on

Need to assumetwo-force

Chapter 1- Introduction-Concept of Stress22

member

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Method of Joints Joints must satisfy the conditions for

static equilibrium which may be

BCF'

expresse n e orm o a orce r ang e:

0F

600

mm

kN30 BCAB

FF

ABF ABF'

kN50kN40

354

BCAB FF800 mm

A x

Chapter 1- Introduction-Concept of Stress23

from geometry of structure

for an axially loaded member is

perpendicular to the member axis.

The force intensity on that sectionis defined as the normal stress.

PF

AAave

A

0

Chapter 1- Introduction-Concept of Stress24

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may not be equal to the averageassumed distribution must satisfy

constant)

ave dAdFAP

A

Chapter 1- Introduction-Concept of Stress25

Stress Analysisd= 20 mmC Can the structure support 30 kN load?

MPa165: all Given

A B F = 40 kN com ression

From previous analysis

FBC= 50 kN (tension)

A = 314 mm2

dBC= 20 mm At any section through member

BC, the internal force is 50 kN with

BCFABCF

a force intensity or stressof

N1050 3Pa

mm314 2

ABC

or N/mm2

Chapter 1- Introduction-Concept of Stress26

= norma stress to sur ace

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Stress Anal sisd= 20 mmC

600 mm The allowable stress of steel is

all yield

A B

800 mm30kN Conclusion: the strength of

u ma e

member BCis adequateBCF

ABCF

Note: AB than BC

Chapter 1- Introduction-Concept of Stress27

.

d= ? mm

Design of new structures

600 mm

appropriate materialsandFBC= 50 kN

meet performance

= 100 MPamm

30kN

Chapter 1- Introduction-Concept of Stress28

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Desi n Based on several reasons, the choice is made to

all .

What is the rod diameter?A

all

mm500MPa100

N1050 2

PA

all

mm2.25

mm50044therefore,

22

A

d

d

A (all , A , d)

Chapter 1- Introduction-Concept of Stress29

A uniform distribution of stress in asection infers that the line of action forthe resultant of the internal forces

passes through the centroid of thesection.

A uniform distribution of stress is onlycentroid

end sections of two-force members are

.

referred to as centric (concentric)

Chapter 1- Introduction-Concept of Stress30

.

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Eccentric Loadin If a two-force member is

,

resultant of the stress distribution

force and a moment.

e stress str ut ons neccentrically loaded members

canno e un orm or symme r c.

Direction of M and P can befound through equilibrium.

Chapter 1- Introduction-Concept of Stress31

Forces Pand Pare appliedtransversely to the memberAB.

force distribution is defined as the

the load P.

e correspon ng average s ear ngstress is, P

A

ave

Chapter 1- Introduction-Concept of Stress32

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Single Shear Double Shear

Chapter 1- Introduction-Concept of Stress33

Single Shear Double Shear

F = 2P or P = F/2

FP ave FPave

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Bearing Stress in Connections

Bolts, rivets, and pinscreate stresses on the

points of contact or bearingsurfacesof the membersthey connect.

Corresponding average

the bearing stress,

dtAb

(Projected Area)

Chapter 1- Introduction-Concept of Stress35

Stress Analysis & Design ExampleDetermine the stressesin the members andconnections of thestructure

From analysis:F = 40 kN com ression

FBC= 50 kN (tension)

Consider maximumnormal stresses inAB and

BC, and the shearing andbearing stress at each

Chapter 1- Introduction-Concept of Stress36

pinned connection

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Chapter 1- Introduction-Concept of Stress37

2 areas

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Different critical areas for tension and compression

Critical areas(higher stresses)

compression (higher stresses)

Chapter 1- Introduction-Concept of Stress39

Rod & Boom Normal Stresses (on BC)

20 mmCritical

25 mm

40 mmareas

A =

MPa159 50 kN

mm

The smallest cross- mm300mm25mm40mm20 2Asec ona area occurs

at the pin centerline, MPa16710502

3

, NPendBC

Chapter 1- Introduction-Concept of Stress40

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Rod & Boom Normal Stresses (on AB)

The boom is in compression with

an axial force of 40 kN and40kN

average normal stress is

1040 3a.

5030

The minimum area sections at theoom en s an are

unstressed since the boom is in

Chapter 1- Introduction-Concept of Stress41

compress on.

Pin Shearing Stresses (at C and A)

kN50

The cross-sectional area for pins atA, B,and C,

2

2

2 mm25

d=25mm2

kN50bF

force exerted by the rod BC,

3

kN40MPa102

mm491 2,

AaveC

e p n a s n ou e s earw atotal force equal to the force exerted by

kN40bF

p

d=25mm e oom ,

MPa7.40kN)2/40(

P

aveA

Chapter 1- Introduction-Concept of Stress42

kN40bF

p mm

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Pin Shearing Stresses (at B)BC

F

50 kN

30side viewBCF'50

40 50ABF ABF'

40 kN 50

F BC,

00050

2A,

MPa9.50

2(491)

Chapter 1- Introduction-Concept of Stress43

Pin Bearing Stresses (at A on boom and bracket)

To determine the bearing stress atAin the boomAB, we have t= 30 mmand d= 25 mm,

MPa3.53N40000

P

b

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

MPa0.32

mm25mm50N40000

tdPb

Chapter 1- Introduction-Concept of Stress44

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Stresses in Members x a orces on a wo- orce mem er

result in only normal stresses on ap ane cu perpen cu ar o e

member axis.Transverse forces on bolts andpins result in only shear stresseson the plane perpendicular to boltor pin axis.

Either axial or transverse forces may

produce both normal and shearstresses with respect to a plane otherthan one cut perpendicular to the

Chapter 1- Introduction-Concept of Stress45

member axis.

Stress on an Oblique Plane

A o

sincos PVPF

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stresses on the oblique plane are

coscos 2P

A

PF

cos

cossin00 AAA

Chapter 1- Introduction-Concept of Stress47

P' P Maximum Stresseson an oblique plane

2 PP

0APM 00 AA

The maximum normal stress

02' AP

occurs when the reference planeis perpendicular to the member

axis, 00

m A

P

0m

2AP

The maximum shear stressoccurs for a plane at + 45o with

m

'

respect to the axis,45cos45sin m

PP

Chapter 1- Introduction-Concept of Stress48

0 22 00 AA

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initial

final

Chapter 1- Introduction-Concept of Stress49

Factor of Safety

Structural members oractor o sa ety cons erat ons:

uncertainty in material

designed such that theproper es

uncertainty of loadings

less than the ultimate uncer a n y o ana yses number of loading cycles

material. ypes o a ure maintenance requirements and

stressultimate

safetyofFactor

u

FS e er ora on e ec s importance of member to

stressallowableall

s ruc ures n egr y risk to life and property

Chapter 1- Introduction-Concept of Stress50

n uence on mac ne unc on

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Chapter 1- Summary For pin-ended structure (with no transverse load), only

axial force two force members exist

Axial stress due to axial force

A (use N and mm2)

sigma

Shear force produces shearing stress

A

P

tau

Bolted joint - Shearing stress (single shear)(double shear)

Chapter 1- Introduction-Concept of Stress51

Bearing stress

Chapter 1- Summary Axial force produces

Axial/normal stress (on surfaces to force)

Factors of safety

all

u

FS