ch1 beams bending
DESCRIPTION
ME2114 Lecture NotesTRANSCRIPT
-
Founded 1905
NATIONAL UNIVERSITY OF SINGAPORE
Department of Mechanical Engineering
MECHANICS OF
MATERIALS II
ME2114
Course Lecturer: A/P CJ TAY
-
1
Founded 1905
SESSION 2014-15
Semester 2
ME2114 Mechanics of Materials II Modular Credits: 3
Part I Lecture Notes
A/P CJ TAY
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2
Recommended Books
Basic Text:
A.C. Ugural, Mechanics of Materials, McGraw-Hill, 1993
(Chapter 4, 12 & 13 for part I)
Supplementary Readings:
1. F. P. Beer and E. R. Johnston, Mechanics of Materials, McGraw-Hill, 3rd Ed., 2003.
2. R. C. Hibbeler, Mechanics of Materials, Prentice Hall, 4th Ed., 2000. 3. J. M. Gere and S. P. Timoshenko, Mechanics of Materials, PWS
Publishing Company, 4th ed., 1997.
4. R. R. Craig, Jr., Mechanics of Materials, McGraw-Hill, 2nd ed., 2000. 5. A.L. Window ed., Strain gauge technology, London : Elsevier
Applied Science , 2nd ed., 1992.
6. J.W.Dally & W.F. Riley, Experimental Stress Analysis, McGraw-Hill, 3rd Ed., 1991.
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3
TABLE OF CONTENTS Chapter 1
ELASTIC-PLASTIC BENDING OF BEAMS AND TORSION
OF CIRCULAR BARS 1.1 BENDING OF BEAMS STRESS CONCENTRATION 1.2 BENDING OF BEAMS INELASTIC BENDING 1.3 TORSION OF CIRCULAR BARS ASSUMPTIONS 1.4 FIRST YIELD TORQUE 1.5 ELASTIC-PLASTIC TORQUE 1.6 FULLY PLASTIC TORQUE 1.7 RESIDUAL SHEAR STRESS DISTRIBUTION
Chapter 2 BUCKLING OF COLUMNS
2.1 EULER BUCKLING OF SLENDER COLUMNS 2.2 SUMMARY 2.3 EFFECTIVE LENGTH 2.4 COLUMNS INITIALLY CURVED 2.5 COLUMNS WITH ECCENTRIC LOADS
Chapter 3 EXPERIMENTAL STRESS ANALYSIS
3.1. THEORY
3.2. TYPE OF BRIDGE CIRCUITS
3.3. APPLICATIONS OF STRAIN GAUGES
3.4. GRAPHICAL SOLUTION
3.5. TRANSDUCERS
3.6. TEMPERATURE EFFECT
3.7. THREE-LEAD-WIRE ARRANGEMENT
3.8. SPECIAL PURPOSES GAUGES
3.9. ADVANTAGES AND DISADVANTAGES OF STRIAN GAUGES
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2
CHAPTER 1
BENDING OF BEAMS
1.1 BENDING OF BEAMS STRESS CONCENTRATION
Bending stress formula I
MCmax is used only for a constant
cross-sectional area. For cross-section that changes suddenly,
the stress-strain distributions become nonlinear.
Examples of changes in cross-sections
-
3
The stress distribution for case (a) is :
The max stress occurs at base of grooves, max stress is given
by:
I
MCkmax
k stress concentration factor The value of k and the stresses through the section are
determined by experiment or theory (sometime).
-
4
For a certain beam geometry, the stress concentration values
can be from the following Figs:
Beams with shoulder fillets (Fig. F1):
-
5
Beams with grooves (Fig. G1):
-
6
Example 1.1-1
Beams with shoulder fillets
A steel bar with shoulder fillets as shown in the following
figure is subjected to a bending moment of 5 kNm, determine
the maximum bending stress developed in the steel. Given
that r = 16 mm, h = 80 mm, w = 120 mm, t = 20 mm.
Solutions:
Given r = 16 mm, h = 80 mm, w = 120 mm
We have
r/h = 16/80 = 0.2 , w/h = 120/80 = 1.5
Second moment of area 4308.002.0
12
1mI
Using I
MCkmax
-
7
From Fig. F1, the value of k is given by 1.45
Using M = 5 kNm, C = 0.04 m,
MPa
x
x340
1053.8
04.0545.1
7max
Stress distribution below the fillets :
Stresses away from the fillets are not affected by the stress
concentration and the max stress is given by:
I
MCmax i.e. k = 1
Hence MPax
x234
1053.8
04.057max
Stress distribution away from the fillets :
-
8
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9
Example 1.1-2
A simply supported beam with thickness of 10 mm is loaded
as shown in the Fig. Determine the length L of the center
portion of the beam so that the maximum bending stress at
section A, B, C is the same.
From vertical equilibrium
RD = RE = (350L)/2 = 175L
BM at section A or B
RD RE
D E
-
10
MA = 175L (0.3) = 52.5L
BM at section C
For r = 8 mm, h = 40 mm, w = 60 mm
r/h = 8/40 = 0.2 , w/h = 60/40 = 1.5
MC = 175L (0.3 + L/2)
- 350(L/2)(L/4)
= 52.5L + 43.75L2
350 N/m
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11
From Fig. F1, the value of k is given by 1.45
Maximum bending stress at either section A or B
LxL
I
CMk AAat
7
3max 1085.2
04.001.012
1
02.05.5245.1
Maximum bending stress at section C
266
3
2
max
1029.71075.8
06.001.012
1
03.075.435.52
LL
LL
I
CMCCat
Problem requires that
CatAat maxmax i. e.
mL
LLL
72.2
1029.71075.81085.2 2667
-
12
1.2 BENDING OF BEAMS INELASTIC
BENDING
1.2.1 STRESS-STRAIN CURVE
In general, deformation of material under load can be divided into
four stages:
I: Linear elastic deformation
-small strains and displacements; Hookes law
II: Non-linear deformation
-permanent "set" after unloading
Str
ess
u
yu yl Rupture
Strain
-
13
III: Large deformation
-eg. in metal forming processes
IV: Rupture/Fracture
Up till now, mainly concerned with stage I. This section considers
stage II and transition from stage I to stage II (elastic-plastic
behaviour).
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14
Why the need to study the elastic-plastic behaviour of structures?
It would be unwise if designers knew nothing of what would
happen to components that were grossly overloaded to the point
where marked yielding and plastic deformation had occurred;
One important aspect - in some circumstances, enhanced
performance can be achieved by prior plastic deformation resulting
in favourable residual stresses eg, in thick-walled pressure vessel.
In elastic-plastic analysis, usually the actual - curves for the real
strain-hardening material introduces some complications in
analyses.
Because of this, semi-idealised behaviour is often assumed in which
(a) strain-hardening occurs linearly from initial yield, or (b) strain-
hardening is ignored.
-
15
strain-hardening assumed linear from initial yield
( bi-linear stress-strain curve)
Strain-hardening is ignored
Y
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16
1.2.2 ASSUMPTIONS
1. That any cross-section of the beam will remain plane during bending as in elastic bending.
2. That the fibres are in a condition of simple tension or
compression.
RECTANGULAR SECTION
(a) Limit of total elastic action
In elastic bending of a beam, there is a linear stress distributed over
its cross-section. The extreme fibres reach the yield stress when the
bending moment is
MY MY
Y
Y
d
b
-
17
6
2
12
1I 2
3
bd
d
bd
yM
I
YM
YYYY
Y
Y
(b) Partial elastic-plastic action
When the bending moment is increased further beyond MY, some of
the fibres near the top and bottom surfaces begin to yield and plastic
deformation penetrates deeper into the beam as shown above. The
Y
Y
c
c
b
d
Plastic zone
Mp > My Mp
(a)
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18
strain at outer fibres of beam may increase, but stress will remain
at Y .
Bending moment Mp in beam is given by:
dyybM p
where b is width of beam at distance y from N.A.
For rectangular section beam (b = constant)
(Elastic portion):
Yc
y ; (Plastic portion): = Y
c
o
d
cY
Yp dyybdyyby
cM
2/2
=
2/2
0
3
232
d
c
Y
c
Y yby
c
b
=
343
41
4
22
2
22 cdb
d
cbdY
Y
(2.1)
When stresses at outermost fibres of beam just reach yield i.e. c =
d/2, the bending moment MY is
M b d d bd
YY Y
2 2 2
4 12 6 (2.2)
Same as Eq. (a)
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19
(c) Total plastic action
When stresses throughout the beam section reach yield as in above,
i.e. c = 0, the "ultimate B.M." or fully plastic moment is
Y
Y
Mult Mult
d
b
From Eq. 2.1
The ratio is the "shape factor"; it depends only on the shape of the cross-section of beam.
ultYYPM
db
cdbM
434
222
5.1
6
42
2
d
b
db
M
M
Y
Y
Y
ult
-
20
EI
M
Rand
R
yEE xx
1
From beam bending equations (ME2113)
R is the radius of curvature of the beam neutral axis
When first yield has just occurred (at the extreme fibres) the value of c is d/2. The radius of curvature of the beam at first yield is
1
2R EY
Y
d
1
R EcY
o
R
cEYx
I.e.
For a partially elastic-plastic beam at a distance c from the neutral axis, the stress in the fibres has just reached the value
Y
(note: C = d/2)
-
21
The moment-curvature relationship for the rectangular beam is
thus linear up to a value M = MY and beyond this point the
relationship is non-linear and asymptotic to Mult as shown in the
figure below.
Curvature
Ben
din
g
Mo
men
t
Mult
My
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22
1.2.3 SYMMETRICAL I-SECTION
(a) When yielding is about to occur at extreme fibres (first
yield)
2
12
12
I
3
11
3
Y
d
dbbd
yM
I
YM
xx
YY
Y
Y
b1/2 b1/2
b
d1 d
Y
Y
-
23
(b) When top and bottom flanges have yielded
Plastic portion: = Y
Elastic portion:
2
1
Y y
d
Y
Y
b2
Y
Y
-
24
Recall M b y dy
2
23
4
2 4
2 2
2
2
1
22
1
0
3
2
1
2
1
0
2
2
1
2
2
1
2
1
0
2
2
12
1
d
dY
d
Y
d d
d Y
Y
d d
d Y
Y
yb
yb
d
dyybdyybd
dyybdyybyd
M
Substituting the value of y and simplifying we have
M b b d d
Y
+b d2
1 12
12
6 4
(c) Fully Plastic Condition
Y
Y
Y
Y
-
25
2
1
0
2
2
12Y 2
d d
d YultdyybdyybM =
Ybd b d2 1 1
2
4
Shape factor,
M
M
bd d
bd d
dult
Y
b
b
21 1
2
31 1
3
12
4 2
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26
Example 1.2-1
The steel wide-flange beam has the dimensions as shown
in Fig. 3.1. If it is made of an elastic perfectly plastic
material having a tensile and compressive stress of
Y = 250 MPa. Determine the shape factor of the beam.
Fig. 3.1
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27
a) Determine the maximum elastic moment MY :
I of X-section:
46
233
1044.82
75.1185.1220012
5.122002
12
2255.12
mm
I
A1 A2
N
A1
A2
A
118.75 mm
56.25 mm
mm
-
28
Applying the beam bending formula:
kNm
yM
I
YM
YY
Y
Y
88.164
125
1044.82250
I 6
-
29
b) Determine the fully plastic moment Mp :
kN
areaflangetopCForce Y
625
2005.12250
2
-
30
kN
areawebtopCForce Y
56.351
5.1125.12250
1
kNm
kNmm
webtopandflangetoptodueMoment
94
25.5656.35175.118625
Hence Mp = 2 x 94 = 188 kNm
The "shape factor" 14.188.164
188
Y
p
M
M
Note: the ultimate moment (Mp) is only 14% higher than the
moment at first yield (MY).
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31
Example 1.2-2
The beam x-section as shown in the following figure is
made of an alloy of titanium that has a stress-strain
relationship as shown. If the material behaviour is the same
in both tension and compression, determine the moment that
is applied to the beam to cause a strain of 0.05 at the
extreme top and bottom fibre of the beam.
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32
The material exhibits elastic-plastic behaviour with linear
strain hardening. The strain distribution is given by:
From similar triangles
We have
mmcmy
y
33.0
050.0
010.0
5.1
The stress and force distribution on the x-section are:
-
33
252000 N=
-
34
1.2.4 ASYMMETRICAL SECTION (Y-Y is only plane
of symmetry)
(a) When yielding is about to occur at top fibre
(a) Determine position of N.A. ie. "h" from bottom
h = 70 15 200 7 5 10 170 85 15 100 15 7 5
70 15 10 170 100 15
. .
= 90.2 mm
Fibre furthest away from N.A. will yield first.
ymax = 200 - 90.2 = 109.8 mm (i.e. at top fibre)
First yield moment, My
YY NA
I
max
d = 200
h
Y
N A
10
b1/2=30 b1/2=30
di=170
b=100
-
35
47
2
3
2
3
2
3
10556.2
7.821510012
151008.917010
12
170103.1021570
12
1570
mm
INA
Y
Y
YM
57
10328.28.109
102.556
-
36
(b) Bending moment is increased until bottom fibre yields:
Assume Y is the same in tension and compression.
Let y = depth of plastic flow.
From the condition that nett axial force on section is zero:
F and F b dy 0
0152
110015
2
200
210
2
200
2
11015101570
1
1
Y
YYY
y
yy
Y
Y
-
37
2
100
285
2
200
152
200
2
200
152
200
1
1
y
y
y
y
y
y
YY
Y
Only unknown is y , so can be solved.
0151
2y100
2y85
2100
2y85
2y100
2y85
210
2
y200
2
11015y101570
Y
Y
YYY
y y2 245 6975 0 y = 32.9 mm
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38
Note: The position of zero stress is now (200 - y )/2 = 83.6 mm
from the bottom (compared to 90.2 mm for elastic case)
In elastic-plastic loading mode, the N.A. moves to such a position
that the total force over the x-sectional area subjected to the tensile
stress system is equal to that experiencing compressive stresses and
the state of equilibrium remains undisturbed.
The corresponding B.M. is given by M b y dy
-
39
Bending Moment calculation
152
100153
2
215100
5.72
10015100
152
1003
215
2100
210
2100
3
2
2100
210
2100
2
151510
5.72
1001570
1
1
1
y
y
yy
yy
yyy
yy
M
Y
Y
Y
Y
Y
Y- l
-
40
(c) Fully Plastic Case
Since there can be no longitudinal resultant force in the beam,
A Y Y1 2 A
where A1 and A2 are the areas of the c/s above and below N.A.
respectively.
That is A1 = A2 = 1
2A where A is the total area of the cross-section.
Thus for the fully plastic state, the neutral axis divides the cross-
section into two equal areas and the stress diagram is shown above.
n
-
41
Thus
7015 + 10n = 10(170 - n) + 10015
n = 107.5 mm
Note: The position of zero stress is now:
(170-n) + 15 = (170-107.5) + 15 = 77.5 mm from the bottom
(compare to 90.2 mm for elastic case).
M n nn
nn
ult Y Y
Y Y
70 15 7 5 102
10170
2100 15 170 7 5
2
.
.
Hence "shape factor M
Mult
Y
1.3
n line
-
42
1.2.5 RESIDUAL STRESSES
When a beam that has undergone plastic deformation is unloaded,
the unloading is assumed to be linearly elastic.
If applied moment (causing plastic deformation) is Mp and the
unloading moment is Me then for equilibrium,
Mp - Me = 0 (5.1)
(Me is negative as it is applied in the opposite direction)
Mp + Me = 0 (5.1)
-
43
Note: Compressive where it was tensile (and vice versa).
Y e
Mp Mp
e Y
+
+
-
-
e - Y
Me Me
-
44
e is calculated from the equilibrium condition ie.
Mp -Me = 0 (5.1)
For rectangular section, b x d
From Eq. 2.1
3
4
22 cdbM Yp
6 by given is moment recovery elastic The
12
12 Using
2
3
bdMM
bd
dM
eee
e
e
Hence Eq. 5.1 becomes
)2.5(2
2
3
3
4
6
0 6
- 3
4
2
2
222
222
d
c
cdb
bd
bdcdb
Y
e
Ye
eY
Hence we can find e Note: The maximum possible value of e is
when beam is fully plastic ie. c = 0. Then 2
3
Y
e
-
45
Example 1.2-3
The steel wide-flange beam shown in the following figure is
subjected to a fully plastic moment of Mp. If this moment is
removed, determine the residual stress distribution in the beam. The
material is elastic perfectly plastic and has a yield stress of y = 250
MPa.
From example 3.1
I = 82.44 x 106 mm4
Mp = 188 kNm
Applying the beam bending formula:
-
46
MPa
I
YM ee
1.285
1044.82
125101886
6
From similar triangles, we have
mmy
y
61.109
1.285
250
125
Superposition of the above stresses give the residual
stress distribution as shown below:
e = 285.1 MPa
e = 285.1 MPa
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47
1.2.6 RESIDUAL CURVATURE
For a beam which has reached yield, unloading will NOT result in
the beam returning to its original (straight) condition.
We have,
Plastic curvature - Unloading curvature = Residual curvature
ie. re
111
p
For an elastic case, Ey
1 where is the radius of the N.A.
-
48
For the elastic-plastic beam, use the extremity of the elastic stresses
ie. p
1 =
Ec
Y
During unloading, stresses behave elastically:
Mp Mp
c
+ Y
Me Me d/2
- e
e
-
49
2/
1
dE
e
e
2/ -
1
dEEc
eY
r
From Eq. (5.2)
2
2
3
2
2
d
cYe
Hence
3
4
c
1
13
2
dd
c
E
Y
r
Note: Amount of elastic (unloading) movement is known as
SPRINGBACK
-
50
Example 1.2-4
A square steel bar (25 x 25 mm) of an elastic-perfectly plastic
material is formed into part of a circle using a round mandrel. What
mandrel diameter would be required so that an elastic zone of 16 x
25 mm is attained? Determine the final curvature after springback
and the residual stress distribution in the bar. Assume Y = 250
MPa, E = 200 x 103 MPa.
(a) On loading: recall for the elastic case EyEI
M
1
For elastic-plastic beam, elastic core of 16mm x 25mm is obtained.
1
3 15625.0
008.0200x10
250=
1
mEc
Y
p
p = 6.4 m
Mandrel
Beam
25
25
16
+ Y
-
51
Hence mandrel diameter = 12.8 m
What is the elastic-plastic moment?
Mp 2 25 45 250 2 25 8. .
8
3
2 250 8 25
2
1 +
= 843.2 Nm
(b) On unloading: stresses are wholly elastic
2
dM p
e
Sectional area of bar
+Y
Sectional area of bar
-
52
= 12
4
3
1012
25
105.122.843
= 323.8 MN/m2
Unloading curvature 0125.010200
8.323
2/
1
3
dE
e
e
= 0.12952 m-1
Residual curvature r
1 = 0.15625 - 0.12952
= 0.02673 m-1
Hence r = 37.4 m
After springback, final diameter of formed curve is 74.8m
(c) Residual stress at y = 12.5mm is
250-323.8 = -73.8 MPa
(stress recovery is in the opposite direction, hence substraction)
e at y = 8mm
-
53
e = 12
4
3
1012
25
1082.843
= -207.2 MPa
(stress recovery is in the opposite direction, hence -ve)
Residual stress at y = 8mm is 250 - 207.2 = 42.8 MPa
(a) Loading (b) Unloading
25
25
16
Beam Mandrel
-73.8
+42.8
+y - e
(c) Residual Stress
+Y
-
54
1.3 TORSION OF CIRCULAR BARS
ASSUMPTIONS An ideal stress-strain relationship for the material as
shown below
A plane cross-section of the shaft remains plane when in
the plastic state;
The radius remains straight after torque is applied.
YY
Y
-
55
Recall that when a torque T is applied on a circular bar, the
torque is given by:
RR
drrrdrrT0
2
022 (3.1)
-
56
The shear stress distribution is given by:
From similar triangles,
R
r
R
rwhen
R
r
Y
YYR
R
,
-
57
If the torque is further increased, the yielded region will
extend inwards resulting in an outer plastic annulus and an
inner elastic core.
If the torque is increased, yielding will first occur at the outer most fibres.
Let TY be the torque at which the bar first reaches
yield: From Eq. (3.1)
1.4 FIRST YIELD TORQUE
3
0
4
0 0
22
2
4222
R
R
rdrr
R
rdrrT
Y
R
YR R Y
Y
(4.1)
1.5 ELASTIC-PLASTIC TORQUE
-
58
The torque in the bar is then :
T = 2 2 2r dr r r dro
c
Yc
R
(Elastic) (Plastic)
T = Y
r
cr dr r dr
o
c
Yc
R
2 2
2 2
1232
3
2
2c
2c
2T
33
334
2
0
3
cR
cRc
drrdrr
Y
YY
R
cYc
Y
(5.1)
-
59
1.6 FULLY PLASTIC TORQUE
If the torque is further increased until the bar is fully plastic
The torque TP is given by:
3
0
2
3
2
2
R
drrT
Y
RYP
(6.1)
Comparing with the first yield torque (Eq. 4.1) we have
i.e. the plastic torque is 33% greater than the first yield torque.
33.13
4
2
3
2
3
3
r
R
T
T
Y
Y
Y
P
-
60
Example 1.6-1
The tubular shaft as shown in the figure is made of an aluminum
alloy that is assumed to have a an elastic plastic stress-strain -
relationship as shown in the figure.
Determine
(a) the torque applied to the shaft when first yield occurs,
(b) the fully plastic torque that can be applied to the shaft,
(c) the shear strain at the outer radius when yield first occurs at
the inner radius.
-
61
First yield plastic torque:
We have
kNmT
T
J
cT
Y
Y
YY
42.3
03.005.02
05.01020
44
6
Fully plastic torque:
From Eq. (6.1)
kNm
r
drr
drrTR
YP
1.4
31066.125
10202
2
05.0
03.0
36
05.0
03.0
26
0
2
-
62
Shear strain at the outer radius:
Since the shear strain remains linearly distributed over the
x-section, the plastic strain at the outer fibres is:
From similar triangles
33 10477.030
5010286.0
30
50
30
50
mm
mm
mm
mm
mm
mm
rR
r
R
Plastic shear stress distribution Plastic shear strain distribution
50 mm 30 mm
Plastic shear strain distribution
-
63
Example 1.6-2
A solid circular shaft has a radius of 20 mm and length of 1.5
m. The material has an elastic plastic stress-strain -
relationship as shown in the figure. Determine the torque
needed to twist the shaft by an angle of 0.6 rad.
Given that:
angle of twist = 0.6, R = 0.02 m, L = 1.5 m
-
64
We have
rad
LR
008.0
5.102.06.0
L
-
65
From similar triangles
mm
m
r
R
r
4
004.0
02.0008.0
0016.0
008.0
0016.0
-
66
Using Eq. (5.1)
kNm
cRT Y
25.1
12
0.004
3
0.02 10752
12
3 2
336
33
c
-
67
1.7 RESIDUAL SHEAR STRESS DISTRIBUTION
Case (a)
Consider a bar subjected to a fully plastic torque TP
Case (b)
If the bar in case (a) is unloaded, the bar unloads elastically.
R
-
68
For elastic unloading we have:
J
RTPR
From Eq. (6.1)
3
3
2RT YP
Y
Y
R
R
RR
3
4
2
3
2
4
3
Unloading from fully plastic torque is equivalent to superposing
case (a) and case (b).
The resultant is:
+ =
R
R- Y
-
69
Example 1.7-1
A solid shaft diameter 50 mm is twisted so that an elastic
core of diameter 20 mm remains. Assuming elastic-
perfectly plastic behaviour, calculate: (a) applied torque,
(b) residual stress distribution on unloading, (c) residual
twist.
Assume y = 150 MN/m2 G = 77x103 MN/m2
(a) Torque TR c
p y
2
3 12
3 3
2 150
0 025
3
0 01
12
3 3
. .
= 4.83 kNm
(b) Elastic unloading gives
205.0
32
05.0104.83
2
D
4
3
e
J
Tp
= 196.8 MN/m2
-
70
r at r = 10mm is 4.83 103
0 01
320 054
.
.
= 78.7 MPa
Residual at r = 25 mm is 150 - 196.8 = -46.8 MPa
Residual at r = 10 mm is 150 - 78.7 = 71.3 MPa
Residual Shear Stress Distribution
At r = 10 mm, = 71.3 MPa
At r = 25 mm, = -46.8 MPa e- Y
-
71
(c) Residual Twist:
Recall
GrLor
Gr
L
GJ
Lr
J
r
JTei
Tr
G
LT
..J
andJ
Initial twist:
01.01077
10150
01.0
9
6
GL
y
= 0.195 rad/m
Elastic unloading:
025.01077
10196.8
025.0
9
6
GL
e
= 0.102 rad/m
Residual twist = Initial twist Twist due to elastic unloading
= 0.195 - 0.102 = 0.093 rad/m