IWSD -Module 2-2_1 Static Equilibrium

7/22/2019 IWSD -Module 2-2_1 Static Equilibrium

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Objective:

To be acquainted with basic concepts of static equilibrium from simplestructures and how these are applied in the analysis of simple structural

elements.

Module 2.1: Static equilibrium

1

Scope:

Equilibrium of forces , Equilibrium of moments , Action and reaction forces, Free

body diagram, Shear and moment diagrams, Tension, compression, shear,bending and torsion force components

Expected result:

Review basic principals of statics and dynamics that are relevant to

understanding the behaviour of structures. Explain different forces components

that can act on a structure and how these are defined. Compute of a staticallydeterminate beam or other simple solid. Identify statically determinate and

statically indeterminate structures. Present shear moment diagrams for beam

and frame type structures.

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The behavior of objects withstanding stressesand strains

Various methods of calculating stresses in structural members, such

as beams, columns and shafts.

Predict the response of a structure under loading and its

susceptibility to various failure modes

Take into account of different material properties; yield strength,

ultimate strength, stiffness, Young's modulus, buckling, etc.

In order to obtain the stresses and strains in structural members the

internal forces and moments need to be determined which aid of

static equilibrium

Strength of materials

2IWSD M2.1
http://en.wikipedia.org/wiki/Stress_(physics)http://en.wikipedia.org/wiki/Strain_(physics)http://en.wikipedia.org/wiki/Strain_(physics)http://en.wikipedia.org/wiki/Stress_(physics)


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Where and why Strength of Materials?

3

You name it, Mechanics of Materials is in it!

Main Objective: Predict behavior to prevent failure!

BiomechanicsSource: Royal Inst. of Tech., Sweden

Oil & Gas IndustrySource: Dr. I. Barsoum, PI

Automotive IndusrtySource: Dr. I. Barsoum, Sweden

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4

Medical SurgerySource: VASCOPS Vascular Diagnostics, Sweden

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Where and why Strength of Materials?

You name it, Mechanics of Materials is in it!

Main Objective: Predict behavior to prevent failure!


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Equilibrium of forces

5

Conditions for equilibrium:

Equilibrium

At rest remians at rest

In motion with constant velocity

Static Equilibrium most often an object at rest

To maintain equilibrium, satisfy Newtons 1st law

=

Neccesary and sufficient condition for equilibrium

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Equilibrium of forces

6

Translational Equilibrium

If every part of a system moves in a straight line at a constant speed, wesay it is in translational equilibrium. This includes being at rest.

For a body to be in translational equilibrium, the resultant forces in anytwo perpendicular directions must be zero

This means that using the graphical method of vector addition for the

forces acting on the body, always produces a closed loop:

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Equilibrium of forces, example:

7

E.g. 1

A skier moving at constant speed down a slope:

i. Situation Diagram: ii. FBFD for skier:

Normal contact force from

ground on skier

Gravitational forcefrom skier on ground

Frictional force from

ground on skier

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Equilibrium of forces, example:

9

A 50kg mass on a slope. Friction prevents it from moving.

Determine the frictional force.

Equilibrium parallel to slope:mg sin30 F = 0

0.5 x 50 x 10 F = 0

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Free Body Diagram (FBD)

10

Free Body Diagrams are one of the most important things for you to know

how to draw and use.

What ? - It is a drawing that shows all external forces acting on the particle.

Why ? - It is the key to being able to write the equations of equilibriumwhich

are used to solve for the unknowns (usually forces or moments).

Must account for all known and unknown forces acting on the particle

Isolate the body from itssurroundings

Account for contact forces

Account for forces over a distance (gravity; magnetic; electric)

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Free Body Diagram (FBD)

11

Imagine the particle to be isolated or cut free from its surroundings.

Show all the forces that act on the particle.

Active forces: They want to move the particle.

Reactive forces: They tend to resist the motion.

Identify each force and show all known magnitudes and directions. Show allunknown magnitudes and / or directions as variables


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Free Body Diagram (FBD), example:

12

Since particle A is in equilibrium, the net force at A is zero. The vector sum is zero.

In general, for a particle in equilibrium,

Or, written in a scalar form,

These are two scalar equations of equilibrium (E-of-E)

They can be used to solve for up to twounknowns

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Free Body Diagram (FBD), example:

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Equilibrium of moments

14

The necessary and sufficient condition for the static equilibrium of a

body are that the resultant force and couple from all external forces

form a system equivalent to zero,

00 FrMF O

000

000

zyx

zyx

MMM

FFF

Resolving each force and moment into its rectangular components

leads to 6 scalar equations which also express the conditions forstatic equilibrium,

For a rigid body in static equilibrium, the external forces and

moments are balanced and will impart no translational orrotational motion to the body.

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Example: review of statics and design

15

The structure is designed tosupport a 30 kN load

Perform a static analysis to

determine the internal force in

each structural member and

the reaction forces at the

supports

The structure consists of a

boom and rod joined by pins

(zero moment connections) at

the junctions and supports

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16

Free Body Diagram

Structure is detached from supports

and the loads and reaction forces are

indicated

Ay and Cy can not be determined from

these equations

kN30

0kN300

kN40

0

kN40

m8.0kN30m6.00

yy

yyy

xx

xxx

x

xC

CA

CAF

AC

CAFA

AM

Conditions for static equilibrium:

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In addition to the complete structure, eachcomponent must satisfy the conditions for

static equilibrium

0

m8.00

y

yB

A

AM

Consider a free-body diagram for the boom:

kN30yC

substitute into the structure

equilibrium equation

Results: kN30kN40kN40 yx CCA

Reaction forces are directed along boom

and rod

Component Free Body Diagram

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18

Conclusion: the strength of member BC isadequate

MPa165all

From the material properties for steel, theallowable stress is

From a statics analysis

FAB= 40 kN (compression)

FBC= 50 kN (tension)

Can the structure safely support the 30 kN

load?

dBC= 20 mm MPa159m10314

N1050

26-

3

A

P

BC

At any section through member BC, the

internal force is 50 kN with a force intensity

or stress of

Stress analysis

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19

Design Design of new structures requires selection ofappropriate 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 (sall= 100 MPa). What is an

appropriate choice for the rod diameter?

mm2.25m1052.2

m1050044

4

m10500

Pa10100

N1050

226

2

26

6

3

Ad

dA

PA

A

P

all

all

An aluminum rod 26 mm or more in diameter

is adequate

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Reactions at support and connections

20

Reactions equivalent to a

force with known line of

action.

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Reactions at support and connections

21


force of unknown direction

and magnitude.


force of unknown

direction and magnitude

and a couple.of unknown

magnitude

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Reactions at support and connections, Equilibrium

22

For all forces and moments acting on a two-

dimensional structure,Ozyxz MMMMF 00

Equations of equilibrium become

000 Ayx MFF

where A is any point in the plane of the

structure.

The 3 equations can be solved for no more

than 3 unknowns. The 3 equations can not be augmented with

additional equations, but they can be

replaced

000 BAx MMF

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Statically indetermined reactions

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More unknowns

than equations Fewer unknowns than

equations, partially

constrained

Equal number unknowns

and equations but

improperly constrained

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Beams

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Beams- structural members supporting loadsat various points along the member

Objective - Analysis and design of beams

Transverse loadings of beams are classified as

concentratedloads or distributedloads

Applied loads result in internal forces

consisting of a shear force (from the shear

stress distribution) and a bending couple

(from the normal stress distribution)

Normal stress is often the critical designcriteria

S

M

I

cM

I

Mymx

Requires determination of the location and

magnitude of largest bending moment


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Beams classification of beam supports

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Shear and bending moment diagrams

26

Determination of maximum normal and

shearing stresses requires identification

of maximum internal shear force andbending couple.

Shear force and bending couple at a point

are determined by passing a section

through the beam and applying anequilibrium analysis on the beam portions

on either side of the section.

Sign conventions for shear forces Vand V

and bending couples Mand M

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Example: Shear and bending moment diagrams

27

A beam is loaded and supported. For this beam:

a) Draw complete shear force and bending moment diagrams

b) Determine the equations for the shear force and the bending moment as

functions ofx

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Overall/Global Equilibrium

We start by drawing a free-body diagram of thebeam and determining the support reactions.

Summing moments about the left end A of the beam

gives

Then, summing forces in the vertical direction

gives

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Drawing the Shear force diagram

Sometimes we are not so much interested in the equationsfor the shear forceand bending moment as we are in knowing the maximum and minimum values

or the values at some particular point. In these cases, we want a quick and

efficient methodof generating the shear force and bending moment diagrams

(graphs) so we can easily find the maximum and minimum values. That is the

subject of this first part of the problem.

Concentrated Force @ A

The 30-kN concentrated force (support

reaction) at the left end of the beam causes

the shear force graph to jump up (in the

direction of the force) by 30 kN (the

magnitude of the force) from 0 kN to 30 kN

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30

Distributed Load @ A and B

The downward distributed load causes the shearforce graph to slope downward (in the direction of

the load). Since the distributed load is constant,

the slope of the shear force graph is

constant(dV/dx= w= constant).

The total change in the shear forcegraph between points Aand B is 40 kN (equal to

the area under the distributed load between

pointsAand B) from +30 kN to -10 kN.

We also need to know where the shear force

becomes zero. The full 4 m of the distributed loadcauses a change in the shear force of 40 kN. Since

the distributed load is uniform, the area (change

in shear force) is just 10 b= 30, which gives b= 3

m. That is, the shear force graph becomes zero at

x = 3 m

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Concentrated Force @ B

The 16-kN concentrated force at Bcauses the shear force graph to jump

down (in the direction of the force) by

16 kN (the magnitude of the force) from

-10 kN to -26 kN.

No Loads @ B and C

Since there are no loads between

points B and C, the shear force graph is

constant (the slope dV/dx = w = 0) at -26

kN.

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32

Concentrated Force @ C

The 45-kN concentrated force (supportreaction) at C causes the shear force

graph to jump up (in the direction of the

force) by 45 kN (the magnitude of the

force) from -26 kN to +19 kN

No Loads @ C and D

Since there are no loads between

points C and D, the shear force

graph is constant (the slope

dV/dx= w= 0) at +19 kN.

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Concentrated Force @ D

The 19-kN concentrated forceat D causes

the shear force graph to jump down (in

the direction of the force) by 19 kN (the

magnitude of the force) from +19 kN to 0

kN.

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34

Drawing the Bending moment diagram

Since there are no concentrated momentsacting on this beam, the bending moment

diagram (graph) will be continuous(no jumps) and it will start and end at zero

Decreasing Shear Force

The bending moment graph starts out at

zero and with a large positive slope(since the

shear force starts out with a large positive

value and dM/dx = V). As the shear force

decreases, so does the slope of the bending

moment graph. At x = 3 m the shear force

becomes zero and the bending moment is at a

local maximum (dM/dx= V= 0)For values of x

greater than 3 m (3 < x< 4 m) the shear forceis negative and the bending moment decreases

(dM/dx= V< 0).

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Decreasing Shear Force

The shear force graph is linear (1st

order function ofx),so the bending moment graph is a parabola

(2ndorder function ofx).

The change in the bending moment between x = 0 m

and x = 3 m is equal to the area under the shear

graph between those two points. The area of thetriangle is

So the value of the bending moment atx= 3 m is M= 0

+ 45 = 45 kNm. The change in the bending moment

between x = 3 and x = 4 m is also equal to the areaunder the shear graph

So the value of the bending moment atx= 4 m is M= 45

- 5 = 40 kNm

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Constant Shear Force

Although the bending moment graph is continuousatx= 4 m, the jump in the shear force atx= 4 m causes

the slope of the bending moment to change suddenly

from dM/dx= V= -10 kNm/m to dM/dx= -26 kNm/m.

Since the shear force graph is constantbetweenx= 4 m

andx= 7 m, the bending moment graph has a constantslope between x = 4 m and x = 7 m (dM/dx = V = -26

kNm/m). That is, the bending moment graph is a

straight line.


and x = 7 m is equal to the area under the sheargraph between those two points. The area of the

rectangle is just M= (-26 3) = -78 kNm. So the value of

the bending moment at x = 7 m is M = 40 - 78 = -38

kNm.

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37

Constant Shear Force

Again the bending moment graph is continuous at x= 7m. The jump in the shear force at x = 7 m causes the

slope of the bending moment to change suddenly from

dM/dx= V= -26 kNm/m to dM/dx= +19 kNm/m.

Since the shear force graph is constant betweenx= 7 m

andx= 9 m, the bending moment graph has a constantslope between x = 7 m and x= 9 m (dM/dx = V= +19

kNm/m). That is, the bending moment graph is a

straight line.


and x = 9 m is equal to the area under the sheargraph between those two points. The area of the

rectangle is just M= (+19 2) = +38 kNm. So the value

of the bending moment at x= 7 m is M= -38 + 38 = 0

kNm.

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38

Determining the shear force and the bending moment equations

Sometimes we are not so much interested in the graphsof the shear force and bendingmoment as we are in knowing the equations. In particular, we need to integrate the

equation for the bending moment to determine the shape of beam and how much the

beam will bend as a result of the loads.

The easiest way to get the equations for the shear force and bending moment as

functions of the positionxis to use equilibrium.

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39


0 m


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40


4 m


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41


7 m


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42


It is easily verified that these equations have the appropriate character to match theshear force and bending moment diagrams developed in the first part of this

problem. It is also easily verified that these equations match the previous graphs at

the pointsx= 0 m,x= 3 m,x= 4 m,x= 7 m, andx= 9 m.

Assignment: Do it! 10 minutes on your own!

Finally, note that these equations satisfy the load-shear force-bending moment

relationships

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Strength of materials

43

P

P

P

P

P

Mv

Tensile loading

Compressive loading

Shear loading

Bending

Torsional loading

Mv

Multiaxial laoding

P

Different types of loading

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Some basic concepts and defintions

44

Internal force Examples: axial and shear forces and bending and torsinal moments (interior of the

material)

External force Surface and mass forces

Rigid body motion Displacement of the points of a body which do not change the distances

between the points inside the body

Deformation Variation of the distance between any two points inside the solid body

Stress Physical entity which allows the defintion of internal forces independent of the dimensions

and geometry of a solid body (internal force intensity)

Strain Physical entity which allows the defintion of deformation independent of the dimensions

and geometry of a solid body

Meachnics of materials:

aims to find relations between the four main physical entities defined above (external and

internal forces, displacements and deformations)

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Some basic concepts and defintions

45

1) Independent of the properties of the material the body is made of (onlycontinuum hypothesis)

2) Constitutive law; rheological behavior of the material, establish stress-train

relations

3) Kinematic compatibility conditions

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Bending

46

Pure Bending:

Prismatic (non-circularcross section)

members subjected to

equal and opposite

couples acting in the

same longitudinal

plane

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Pure bending - Crane

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48

Eccentric Loading: Axial loading which does

not pass through section centroid producesinternal forces equivalent to an axial force

and a couple

Transverse Loading: Concentrated or

distributed transverse load producesinternal forces equivalent to a shear force

and a couple

Principle of Superposition: The normal

stress due to pure bending may becombined with the normal stress due to

axial loading and shear stress due to shear

loading to find the complete state of stress.

Bending other loading types

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49

Symmetric member in pure bending

From statics, a couple M consists of two equal and

opposite forces.

The sum of the components of the forces in any

direction is zero.

The moment is the same about any axis

perpendicular to the plane of the couple and zero

about any axis contained in the plane.

Internal forces in any cross section are equivalent to

a couple. The moment of the couple is the section

bending moment.

MdAyM

dAzM

dAF

xz

xy

xx

0

0

These requirements may be applied to the sums of

the components and moments of the statically

indeterminate elementary internal forces.

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Bending deformations

50

bends uniformly to form a circular arc

cross-sectional plane passes through arc center and

remains planar

length of top decreases and length of bottom increases

linear elastic material (Hookeslaw)

a neutral surfacemust exist that is parallel to the upper

and lower surfaces and for which the length does not

change

stresses and strains are negative (compressive) above

the neutral plane and positive (tension) below it

Beam with a plane of symmetry in pure

bending:

member remains symmetric

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Strain due to bending

51

Consider a beam segment of length L.

After deformation, the length of the neutral surface

remains L. At other sections,

mx

mm

x

c

y

c

c

yy

L

yyLL

yL

or

linearly)ries(strain va

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St d t b di


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Stress due to bending

52

For a linearly elastic material,

For static equilibrium,

dAyc

dAcydAF

m

mxx

0

0

First moment with respect to neutral

plane is zero. Therefore, the neutralsurface must pass through the section

centroid.

For static equilibrium,

I

My

c

y

SM

IMc

c

IdAy

cM

dAc

yydAyM

x

mx

m

mm

mx

ngSubstituti

2

linearly)varies(stressm

mxx

c

y

Ec

y

E

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B ti ti


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Beam section properties

53

The maximum normal stress due to bending,

modulussection

inertiaofmomentsection

c

IS

I

S

M

I

Mc

m

A beam section with a larger section modulus will

have a lower maximum stress

Consider a rectangular beam cross section,

Ahbhh

bh

c

IS

6

13

6

1

3

12

1

2

Between two beams with the same cross sectionalarea, the beam with the greater depth will be

more effective in resisting bending.

Structural steel beams are designed to have a large

section modulus.

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54

Properties of American Standard Shapes

Beam section properties

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Bending example: Lightweight design of a mobile crane

55

w

h

t Governing equations from beam theory

L

P

max

EI

LP

3

3

I

eLPmax

max

deflection

bending stress

bending stiffnessIE

26

2 whhtI moment of inertia

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56

w

h

t

262 wh

htI

w x h x t60x100x10

Weight = 1

Stiffness = 1Deflection = 1

60x100x6

Weight = 0.61

Stiffness = 0.67

Deflection = 1.49

40x140x5

Weight = 0.61

Stiffness = 1.11

Deflection = 0.9

Reducing weight Increasing stiffness Reducing deformation

stifnessbendingIE

Lightweight Structure

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57

What do we need to design against?

ratiosslendernesplate

t

h

Design against Failure

Plasticity

Yield stress (yield)

Material dependent

Plastic collapse

Elastic instability

Buckilng stress (crit< yield)

Material independent (??)

BucklingPost bucklingcollapse

Lightweight structure

L

PIncreasing plate slenderness ratio

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Torsional load

58

Chassi in a truck

A shaft in a transmission gearbox

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Torsional loads in circular shafts

59

Interested in stresses and strains of

circular shafts subjected to twisting

couples or torques

Generator creates an equal and opposite

torque T

Shaft transmits the torque to the

generator

Turbine exerts torque Ton the shaft

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Net torque due to internal stresses

60

dAdFT

Net of the internal shearing stresses is an

internal torque, equal and opposite to the

applied torque,

Although the net torque due to the shearing

stresses is known, the distribution of thestresses is not.

Unlike the normal stress due to axial loads, the

distribution of shearing stresses due to torsional

loads can not be assumed uniform.

Distribution of shearing stresses is statically

indeterminate must consider shaft

deformations.

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Torsion axial shear components

61

Torque applied to shaft produces

shearing stresses on the facesperpendicular to the axis.

Conditions of equilibrium require the

existence of equal stresses on the faces of

the two planes containing the axis of theshaft.

The slats slide with respect to each other

when equal and opposite torques are applied

to the ends of the shaft.

The existence of the axial shear components

is demonstrated by considering a shaft made

up of axial slats.

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Torsion shaft deformations

62

From observation, the angle of twist of theshaft is proportional to the applied torque and

to the shaft length.

L

T

When subjected to torsion, every cross-sectionof a circular shaft remains plane and

undistorted.

Cross-sections for hollow and solid circular

shafts remain plain and undistorted because a

circular shaft is axisymmetric.

Cross-sections of noncircular (non-

axisymmetric) shafts are distorted when

subjected to torsion.

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Torsion shearing strain

63

Consider an interior section of the shaft. As a

torsional load is applied, an element on theinterior cylinder deforms into a rhombus.

Shear strain is proportional to twist and radius

maxmax and

cL

c

LL

or

It follows that

Since the ends of the element remain planar, the

shear strain is equal to angle of twist.

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Torsion stress in elastic range

64

Jc

dAc

dAT max2max

Recall that the sum of the moments from

the internal stress distribution is equal to

the torque on the shaft at the section,

andmaxJ

T

J

Tc

The results are known as the elastic torsion

formulas,

Multiplying the previous equation by the

shear modulus,

max

Gc

G

max

c

From HookesLaw, G

The shearing stress varies linearly with the

radial position in the section.4

2

1

cJ

41

422

1ccJ

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Torsion Normal stresses

65

Note that all stresses for elements aand chave

the same magnitude

Element c is subjected to a tensile stress ontwo faces and compressive stress on the

other two.

Elements with faces parallel and

perpendicular to the shaft axis are subjected

to shear stresses only. Normal stresses,shearing stresses or a combination of both

may be found for other orientations.

max

0

0max

45

0max0max

2

2

245cos2

o

A

A

A

F

AAF

Consider an element at 45oto the shaft axis,

Element ais in pure shear.

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Torsion Failure modes

66

Ductile materials generally fail in

shear. Brittle materials are weaker

in tension than shear.

When subjected to torsion, a ductile

specimen breaks along a plane ofmaximum shear, i.e., a plane

perpendicular to the shaft axis.

When subjected to torsion, a brittle

specimen breaks along planesperpendicular to the direction in

which tension is a maximum, i.e.,

along surfaces at 45o to the shaft

axis.

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bl


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Problem 2.1.1

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The beamABCD shown in the figure has overhangs at each endand carries a uniform load of intensity q. For what ratio b/L will

the bending moment at the midpoint of the beam be zero?

15 min on your own!

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Problem 2.1.2

The simple beam AB shown in the figure is subjected to aconcentrated load P and a clockwise couple M1 PL/4 acting at

the third points. Draw the shear-force and bending-moment

diagrams for this beam.

15 min on your own!

Documents

IWSD -Module 2-2_1 Static Equilibrium