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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)7/22/2019 IWSD -Module 2-2_1 Static Equilibrium
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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
IWSD M2.1
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
F = 250NIWSD M2.1
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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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Example: review of statics and design
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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Example: review of statics and design
17
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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Example: review of statics and design
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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Example: review of statics and design
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
Reactions equivalent to a
force of unknown direction
and magnitude.
Reactions equivalent to a
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
23
More unknowns
than equations Fewer unknowns than
equations, partially
constrained
Equal number unknowns
and equations but
improperly constrained
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Beams
24
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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Example: Shear and bending moment diagrams
28
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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Example: Shear and bending moment diagrams
29
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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Example: Shear and bending moment diagrams
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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Example: Shear and bending moment diagrams
31
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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Example: Shear and bending moment diagrams
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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Example: Shear and bending moment diagrams
33
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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Example: Shear and bending moment diagrams
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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Example: Shear and bending moment diagrams
35
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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Example: Shear and bending moment diagrams
36
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.
The change in the bending moment between x = 4 m
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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Example: Shear and bending moment diagrams
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.
The change in the bending moment between x = 7 m
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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Example: Shear and bending moment diagrams
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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Example: Shear and bending moment diagrams
39
Determining the shear force and the bending moment equations
0 m
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Example: Shear and bending moment diagrams
40
Determining the shear force and the bending moment equations
4 m
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Example: Shear and bending moment diagrams
41
Determining the shear force and the bending moment equations
7 m
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Example: Shear and bending moment diagrams
42
Determining the shear force and the bending moment equations
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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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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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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Bending example: Lightweight design of a mobile crane
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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Bending example: Lightweight design of a mobile crane
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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Problem 2.1.1
67
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!
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