1
00:00:22,210 --> 00:00:27,270
Welcome to the 4th lesson on the course Strength
of Materials.
2
00:00:27,270 --> 00:00:28,270
.
3
00:00:28,270 --> 00:00:36,050
Today we will continue our discussion on certain
aspects of Analysis of Stress.
4
00:00:36,050 --> 00:00:37,050
..
5
00:00:37,050 --> 00:00:44,329
In this particular lesson it is expected that
once we complete it,
6
00:00:44,329 --> 00:00:47,690
•
One should be able to evaluate stresses on
7
00:00:47,690 --> 00:00:53,630
any plane through stress transformation
equations
8
00:00:53,630 --> 00:00:55,969
•
Evaluate principal stresses and locate principal
9
00:00:55,969 --> 00:00:59,079
planes for two dimensional (2-D)
problems.
10
00:00:59,079 --> 00:01:04,600
In fact in the previous lesson we discussed
about the evaluation of principal stresses
11
00:01:04,600 --> 00:01:07,450
in
three dimensional planes.
12
00:01:07,450 --> 00:01:14,229
Here we will be discussing evaluation of principal
stresses and location
13
00:01:14,229 --> 00:01:15,689
of principal planes for 2-D problems.
14
00:01:15,689 --> 00:01:21,640
•
We are going to evaluate the maximum shear
15
00:01:21,640 --> 00:01:26,459
stresses at a point on a body for 2-D
problems.
16
00:01:26,459 --> 00:01:31,579
•
We will also look into the concept of Mohr’s
17
00:01:31,579 --> 00:01:39,459
circle for stress and we will demonstrate
how to construct Mohr’s circle for stress.
18
00:01:39,459 --> 00:01:40,459
..
19
00:01:40,459 --> 00:01:50,070
Hence the scope of the particular lesson is
the derivation of transformation equation
20
00:01:50,070 --> 00:01:54,960
for
evaluation of stresses for 2-D problems, evaluation
21
00:01:54,960 --> 00:02:02,360
of principal stresses and maximum shear
stresses; construction of Mohr’s circle.
22
00:02:02,360 --> 00:02:04,659
We will look into aspects of how we are going
to draw
23
00:02:04,659 --> 00:02:10,830
Mohr’s circle for the evaluation of the
stresses at a particular point in the body.
24
00:02:10,830 --> 00:02:16,120
We will solve a
few examples to demonstrate how the stresses
25
00:02:16,120 --> 00:02:23,200
can be evaluated at any particular point.
26
00:02:23,200 --> 00:02:24,200
.
27
00:02:24,200 --> 00:02:31,140
Let us look into the derivations of the transformation
of equations.
28
00:02:31,140 --> 00:02:43,269
We have discussed about the
plane stresses in 2-D. If we consider the
29
00:02:43,269 --> 00:02:50,940
stress body at a particular point this is
our reference xaxis and this is the reference
30
00:02:50,940 --> 00:02:51,940
y-axis.
31
00:02:51,940 --> 00:02:57,900
The stress which is acting in the x-plane
the normal stress
32
00:02:57,900 --> 00:03:23,719
.is σ x . The normal stress in the y-plane
is σ y and the shearing stresses are τ xy
33
00:03:23,719 --> 00:03:30,949
. We are interested
now to evaluate the stresses on a plane the
34
00:03:30,949 --> 00:03:42,880
normal to which is at an angle θ with respect
to the xaxis.
35
00:03:42,880 --> 00:03:48,790
The plane is considered in such a way that
the normal direction normal to the plane
36
00:03:48,790 --> 00:03:58,810
coincides with reference axis which we denote
as x prime and y prime.
37
00:03:58,810 --> 00:04:03,150
Since the normal to this
particular plane coincides or is parallel
38
00:04:03,150 --> 00:04:11,870
to the x ' axis we call this plane as x ' plane.
39
00:04:11,870 --> 00:04:19,650
Now let us look into the state of stress on
this particular plane, if we take out this
40
00:04:19,650 --> 00:04:28,830
particular
wedge and if we designate this as A, B, and
41
00:04:28,830 --> 00:04:34,770
C the stresses which are acting on this particular
part
42
00:04:34,770 --> 00:04:51,060
are σ x normal stresses on this surface,
σ y the shearing stresses τ xy . This being
43
00:04:51,060 --> 00:04:55,419
the x ' plane the
normal stress to this particular plane is
44
00:04:55,419 --> 00:05:03,449
σ x ' , and correspondingly the shear stress
will be tau x
45
00:05:03,449 --> 00:05:07,690
prime y prime.
46
00:05:07,690 --> 00:05:17,090
Considering the unit thickness normal to the
plane of the board if we assume the area on
47
00:05:17,090 --> 00:05:29,229
line AC
as dA which is length AC multiplied by the
48
00:05:29,229 --> 00:05:36,020
unit thickness then considering that this
particular
49
00:05:36,020 --> 00:05:51,280
angle being θ , this particular angle is
also θ the area on line AB can be designated
50
00:05:51,280 --> 00:06:01,590
in terms of the
area dA which is dA cos θ and area on line
51
00:06:01,590 --> 00:06:11,050
BC can be designated in terms of dA sin θ
. Hence
52
00:06:11,050 --> 00:06:16,270
the forces which are acting on these planes
are the stresses multiplied by the corresponding
53
00:06:16,270 --> 00:06:21,219
area
will give us the force.
54
00:06:21,219 --> 00:06:29,340
If we wish to write down the equilibrium equations
in the x ' direction and
55
00:06:29,340 --> 00:06:44,229
y ' direction then the equation looks like
this: summation of forces in the x ' direction
56
00:06:44,229 --> 00:06:46,439
that is
∑F x ' is equal to 0.
57
00:06:46,439 --> 00:07:03,389
The forces which are acting in the x ' directions
are σ x ˈinto dA is acting in the x ' direction
58
00:07:03,389 --> 00:07:17,000
minus σ x acting on the area dA cos θ and
the component in the x ' direction is cos
59
00:07:17,000 --> 00:07:25,159
θ ; σ y
which is acting in the opposite direction
60
00:07:25,159 --> 00:07:42,219
of σ x ' is the minus σ y dA sin θ the
force and multiplied
61
00:07:42,219 --> 00:07:52,490
by the component sin θ .
The shearing stresses we have is minus τ
62
00:07:52,490 --> 00:08:00,749
xy acting on BC which is dA sin θ component
along x
63
00:08:00,749 --> 00:08:18,749
direction is cos θ minus τ xy which is acting
in the plane x dA cos θ , and component along
64
00:08:18,749 --> 00:08:25,949
sin θ
is equal to 0.
65
00:08:25,949 --> 00:08:44,600
This gives us the equation as σ x prime is
equal to σ x cos square θ plus σ y sin
66
00:08:44,600 --> 00:08:59,540
square θ plus 2 τ xy sin θ cos θ . Writing
sin square θ and cos square θ in terms of
67
00:08:59,540 --> 00:09:09,180
cos 2 θ we
can write this as σ x into 1 by 2(1plus cos
68
00:09:09,180 --> 00:09:35,070
2 θ ) plus σ y 1 by 2 (1 minus cos 2 θ
) plus τ xy sin 2 θ .
69
00:09:35,070 --> 00:09:45,860
If you write sin θ and cos θ as sin 2 θ
, then plus τ xy sin 2 θ ); this we can
70
00:09:45,860 --> 00:10:09,510
write as ( σ x plus σ y )
by 2 plus ( σ x minus σ y ) by 2 cos 2 θ
71
00:10:09,510 --> 00:10:25,130
plus τ xy sin 2 θ . This is the stress in
the x-direction which
72
00:10:25,130 --> 00:10:33,130
is the normal stress σ x prime which are
written in terms of stresses σ x , σ y and
73
00:10:33,130 --> 00:10:45,250
τ xy . Similarly, if
we take equilibrium along Fy prime; ∑Fy
74
00:10:45,250 --> 00:11:02,740
prime is equal to 0 we get tau x prime y prime
is equal to
75
00:11:02,740 --> 00:11:52,660
minus σ x cos θ sin θ plus σ y sin θ
cos θ plus τ xy sin square θ , τ xy cos
76
00:11:52,660 --> 00:12:11,850
square θ minus τ xy sin
square θ . Hence we can write tau x prime
77
00:12:11,850 --> 00:12:29,740
y prime is equal to [(minus σ x minus σ
y ) by 2] sin 2 θ
78
00:12:29,740 --> 00:12:39,070
plus τ xy cos 2 θ . So these are the equations
σ x prime and tau x prime
79
00:12:39,070 --> 00:12:40,070
y
80
00:12:40,070 --> 00:12:54,290
stresses on the plane which is at an angle
of θ with respect to the x-axis.
81
00:12:54,290 --> 00:12:58,340
prime and they are the
82
00:12:58,340 --> 00:13:04,251
.Similarly, if we want to evaluate the stress
in the y prime direction the normal stress
83
00:13:04,251 --> 00:13:12,570
σ y prime,
the stress σ y prime is at an angle of θ
84
00:13:12,570 --> 00:13:20,480
plus 90 degrees, if we substitute in place
of θ as θ plus
85
00:13:20,480 --> 00:13:39,050
90 then sin(180 plus 2 θ ) is equal to minussin
2 θ ; cos (180 plus 2 θ ) is equal to minus
86
00:13:39,050 --> 00:13:45,800
cos 2 θ
and if we substitute these values in the expression
87
00:13:45,800 --> 00:13:51,210
for σ y prime again we get σ y prime is
equal
88
00:13:51,210 --> 00:14:15,860
to ( σ x plus σ y ) by 2 minus( σ x minus
σ y ) by 2 cos 2 θ minus τ xy sin 2 θ
89
00:14:15,860 --> 00:14:24,450
. Thereby if we add
these two σ x prime and σ y prime this gives
90
00:14:24,450 --> 00:14:38,010
us the value as σ x plus σ y .
The stresses σ x plus σ y plus σ z is equal
91
00:14:38,010 --> 00:14:42,269
to σ x prime plus σ y prime plus σ z prime
which
92
00:14:42,269 --> 00:14:47,800
indicates that irrespective of the reference
axis system the summation of these normal
93
00:14:47,800 --> 00:14:52,310
stresses
are constant which we called as stress invariants.
94
00:14:52,310 --> 00:14:59,100
So here this is to prove again that the normal
stresses with reference axis is x primey prime
95
00:14:59,100 --> 00:15:05,460
σ x prime plus σ y prime is equal to σ
x plus σ y is
96
00:15:05,460 --> 00:15:11,740
equal to constant and so are the other stress
invariants.
97
00:15:11,740 --> 00:15:26,430
Hence we have obtained the stresses in
the direction at an angle of θ as σ x ' is
98
00:15:26,430 --> 00:15:40,160
equal to ( σ x plus σ y ) by 2 plus ( σ
x minus σ y ) by 2 cos
99
00:15:40,160 --> 00:15:52,380
2 θ plus τ xy sin 2 θ .
We have seen tau x prime y prime is equal
100
00:15:52,380 --> 00:16:05,090
to minus ( σ x minus σ y ) by 2 sin 2 θ
plus τ xy cos 2 θ .
101
00:16:05,090 --> 00:16:06,842
These are the transformation equations.
102
00:16:06,842 --> 00:16:11,750
That means we can evaluate stresses at any
plane which
103
00:16:11,750 --> 00:16:21,220
is oriented at an angle θ in terms of the
normal stresses σ x , σ y and τ xy . Please
104
00:16:21,220 --> 00:16:25,860
keep in mind that
the rotation of the angle θ we have taken
105
00:16:25,860 --> 00:16:33,140
as anti-clockwise and this is a positive according
to our
106
00:16:33,140 --> 00:16:34,140
convention.
107
00:16:34,140 --> 00:16:35,140
.
108
00:16:35,140 --> 00:16:42,490
.Hence these are the stresses which we have
derived; σ x prime is equal to ( σ x plus
109
00:16:42,490 --> 00:16:47,830
σ y ) by 2 plus
( σ x minus σ y ) by 2 into cos 2 θ plus
110
00:16:47,830 --> 00:16:56,160
τ xy sin 2 θ . tau x prime y prime is equal
to ( σ x minus σ y ) by
111
00:16:56,160 --> 00:17:06,470
2 minus sin 2 θ plus τ xy cos 2 θ . We
have also seen σ y prime like this and if
112
00:17:06,470 --> 00:17:11,210
we add σ x prime
plus σ y prime we will get σ x plus σ y
113
00:17:11,210 --> 00:17:14,850
.
.
114
00:17:14,850 --> 00:17:27,970
Now let us look into the position of the planes
where the normal stresses are at maximum.
115
00:17:27,970 --> 00:17:32,860
We
have obtained that the normal stresses on
116
00:17:32,860 --> 00:17:38,010
a plane σ x prime which is at an angle θ
is equal to
117
00:17:38,010 --> 00:17:55,610
( σ x plus σ y ) by 2 plus ( σ x minus
σ y ) by 2 cos 2 θ plus τ xy sin 2 θ . If
118
00:17:55,610 --> 00:18:01,450
we take the derivative of
the normal stress with respect to θ is ∂ σ
119
00:18:01,450 --> 00:18:13,730
x prime by ∂ θ is equal to minus 2 (σ
x minus σ y ) by 2
120
00:18:13,730 --> 00:18:34,009
sin 2 θ plus 2 τ xy cos 2 θ . If we set
this as equal to zero and take this on the
121
00:18:34,009 --> 00:18:44,710
other side then we
get tan 2 θ is equal to τ xy by ( σ x minus
122
00:18:44,710 --> 00:18:49,999
σ y by 2).
123
00:18:49,999 --> 00:19:04,350
Now this particular equation has two values
of θ . One is θ P with reference to the
124
00:19:04,350 --> 00:19:11,131
axis system we have the plane which is in
the angle of θ P .
125
00:19:11,131 --> 00:19:22,769
Also, we will get another angle which is at
an angle of 180 as tan 180 plus θ is equal
126
00:19:22,769 --> 00:19:36,149
to tan θ .
Hence we have one angle as 2 θ P and another
127
00:19:36,149 --> 00:19:40,190
at angle of 180 plus 2 θ P which will us
two
128
00:19:40,190 --> 00:19:45,549
values.
129
00:19:45,549 --> 00:19:46,549
..
130
00:19:46,549 --> 00:19:56,159
Therefore this is the derivative of the normal
stress and this is the derivatives of tan
131
00:19:56,159 --> 00:20:04,820
θ . We will
get two values of this root 2 θ and 180 plus
132
00:20:04,820 --> 00:20:11,929
2 θ . We have designated these as θ P and
180 plus 2
133
00:20:11,929 --> 00:20:21,399
θ P . In effect when we transform from this
into the stress part we have angle θ P and
134
00:20:21,399 --> 00:20:28,809
90 plus θ P
and that indicates that we have two planes
135
00:20:28,809 --> 00:20:34,940
which is at an angle of θ P and normal to
this is the
136
00:20:34,940 --> 00:20:41,490
plane on which the normal stress is maximum,
and then we have another plane which is at
137
00:20:41,490 --> 00:20:45,789
an
angle of 90 degrees with reference to this
138
00:20:45,789 --> 00:20:48,909
particular plane because the other plane is
at 90
139
00:20:48,909 --> 00:20:56,830
plus θ P . These are the two normal directions
where one will be the maximum and the other
140
00:20:56,830 --> 00:20:57,830
will
be the minimum.
141
00:20:57,830 --> 00:21:04,879
These are the two normal stresses.
142
00:21:04,879 --> 00:21:05,879
.
143
00:21:05,879 --> 00:21:17,000
.Interestingly if we look into the expression
tau x prime y prime is equal to ( σ x minus
144
00:21:17,000 --> 00:21:24,140
σ y ) by 2 sin
2 θ plus τ xy cos 2 θ . If we say tau x
145
00:21:24,140 --> 00:21:30,720
prime y prime is equal to 0 then we get tan
2 θ is equal to 2
146
00:21:30,720 --> 00:21:38,929
tau xy by ( σ x minus σ y ) which is similar
to the expression which we have obtained for
147
00:21:38,929 --> 00:21:43,650
tan 2 θ
setting the derivative of the normal stress
148
00:21:43,650 --> 00:21:45,740
to 0.
149
00:21:45,740 --> 00:21:50,409
And since these two angles match this shows
the
150
00:21:50,409 --> 00:21:58,999
planes where we have obtained the maximum
and minimum principal stresses they coincide
151
00:21:58,999 --> 00:22:03,790
with
the planes where the shearing stress is 0.
152
00:22:03,790 --> 00:22:08,190
And as we have defined before that the planes
on which
153
00:22:08,190 --> 00:22:14,110
shearing stress is 0 the normal stress is
designated as principal stress.
154
00:22:14,110 --> 00:22:18,879
Hence the maximum
normal stresses which we have obtained are
155
00:22:18,879 --> 00:22:21,870
nothing but the principal stresses where the
shear
156
00:22:21,870 --> 00:22:36,530
stresses are 0 and their angles are defined
by θ P where we have evaluated θ P 180 degrees
157
00:22:36,530 --> 00:22:47,960
2 θ P .
.
158
00:22:47,960 --> 00:22:53,029
Let us evaluate the maximum values of the
normal stresses and the principal stresses.
159
00:22:53,029 --> 00:23:03,799
We have
seen that this angle 2 θ P where tan of 2
160
00:23:03,799 --> 00:23:12,149
θ P is equal to τ xy by ( σ x minus σ
y ) by 2.
161
00:23:12,149 --> 00:23:21,169
Hence the
value of this hypotenuse R is equal to square
162
00:23:21,169 --> 00:23:31,720
root of (( σ x minus σ y ) by 2) whole square
plus τ xy
163
00:23:31,720 --> 00:23:33,210
square.
164
00:23:33,210 --> 00:23:53,139
Hence value of cos 2 θ P is equal to ( σ
x minus σ y ) by 2R sin2 θ p is equal to
165
00:23:53,139 --> 00:24:02,490
τ xy by R. If
we substitute the values of cos2 θ P and
166
00:24:02,490 --> 00:24:06,070
sin2 θ P , in the expression of the normal
which we have
167
00:24:06,070 --> 00:24:21,600
evaluated σ x prime is equal to ( σ x plus
σ y ) by 2 plus ( σ x minus σ y ) by 2
168
00:24:21,600 --> 00:24:31,529
cos2 θ plus τ xy sin2 θ .
Now if we substitute for cos2 θ and sin2
169
00:24:31,529 --> 00:24:37,580
θ in terms of θ P this we get as the maximum
stresses,
170
00:24:37,580 --> 00:24:48,830
so σ x prime, maximum or minimum
which are nothing but the principal stresses
171
00:24:48,830 --> 00:25:03,669
σ 1 and σ 2 is
equal to ( σ x plus σ y ) by 2 plus (( σ
172
00:25:03,669 --> 00:25:24,009
x minus σ y ) by 2) whole square 1 by R plus
τ xy square by R.
173
00:25:24,009 --> 00:25:30,070
This is equals to at this particular part,
(( σ x minus σ y ) by 2) whole square by
174
00:25:30,070 --> 00:25:35,269
Rand τ xy square
by R and as we have denoted the R as root
175
00:25:35,269 --> 00:25:37,929
of this, so the R square is the top part.
176
00:25:37,929 --> 00:25:39,899
So σ 1 2 can be
177
00:25:39,899 --> 00:25:51,080
.written as ( σ x plus σ y ) by 2 plus R
square by R, these get cancel so, this eventually
178
00:25:51,080 --> 00:25:59,870
gives us
( σ x plus σ y ) by 2 plus square root of
179
00:25:59,870 --> 00:26:10,090
(( σ x minus σ y ) by 2 ) whole square plus
τ xy square.
180
00:26:10,090 --> 00:26:20,520
So
this is the value of maximum stress or one
181
00:26:20,520 --> 00:26:24,020
of the stresses we get.
182
00:26:24,020 --> 00:26:36,730
Now we have obtained that, σ maximum or minimum,
let us call it as σ 1 is equal to ( σ x
183
00:26:36,730 --> 00:26:43,779
plus σ y )
by 2 plus square root of (( σ x minus σ
184
00:26:43,779 --> 00:26:51,120
y ) by 2 ) whole square plus τ xy square.
185
00:26:51,120 --> 00:26:57,039
We have seen that
the normal stresses are the constants summation
186
00:26:57,039 --> 00:27:03,159
of σ x plus σ y is equal to σ x plus σ
y is equal to
187
00:27:03,159 --> 00:27:06,989
σ 1 plus σ 2 ; because these are the two
normal stresses at perpendicular plane.
188
00:27:06,989 --> 00:27:12,960
We can write σ 1 plus σ 2 which are two
normal stresses at perpendicular plane as
189
00:27:12,960 --> 00:27:24,070
equals to σ x
plus σ y , σ 2 from here is σ x plus σ
190
00:27:24,070 --> 00:27:33,350
y minus σ 1 ; σ 1 is given by this, which
can be write this is
191
00:27:33,350 --> 00:27:46,620
equals to ( σ x plus σ y ) by 2 minus square
root of (( σ x minus σ y ) by 2) whole square
192
00:27:46,620 --> 00:27:53,200
plus τ xy
square.
193
00:27:53,200 --> 00:28:06,320
Hence the stresses σ 1 or σ 2 is given as
( σ x plus σ y ) by 2 plus or minus square
194
00:28:06,320 --> 00:28:12,710
root of
(( σ x minus σ y ) by 2) whole square plus
195
00:28:12,710 --> 00:28:14,210
τ xy square.
196
00:28:14,210 --> 00:28:23,669
Hence these are the values of principal
stresses, maximum and minimum principal stresses.
197
00:28:23,669 --> 00:28:24,669
.
198
00:28:24,669 --> 00:28:29,779
Now these are the values of principal stresses
maximum and minimum values are ( σ x plus
199
00:28:29,779 --> 00:28:34,480
σ y )
by 2 plus square root of (( σ x minus σ
200
00:28:34,480 --> 00:28:51,370
y ) by 2) whole square plus τ xy square 0.
201
00:28:51,370 --> 00:28:52,370
..
202
00:28:52,370 --> 00:29:03,119
Maximum shear stress:
We have seen shear stress on any plane, tau
203
00:29:03,119 --> 00:29:13,700
xˈyˈ is equal to minus ( σ x minus σ y
) by 2 sin 2 θ
204
00:29:13,700 --> 00:29:28,480
plus τ xy cos2 θ . If we take derivative
of this with respect to the θ , ∂tau xˈyˈ
205
00:29:28,480 --> 00:29:39,879
by ∂ θ is equal to
minus 2( σ x minus σ y ) by 2 cos 2 θ minus
206
00:29:39,879 --> 00:29:53,639
τ xy 2 sin2 θ . So τ xy sin 2 is equal
to minus
207
00:29:53,639 --> 00:30:10,120
( σ x minus σ y ) by 2 cos 2 θ ; or tan
2 θ is equal to ( σ x minus σ y ) by 2
208
00:30:10,120 --> 00:30:19,789
by τ xy . Here also as we
have noticed earlier two values of θ defining
209
00:30:19,789 --> 00:30:22,850
two perpendicular planes on which the shear
stress
210
00:30:22,850 --> 00:30:32,899
will be maximum and those angles being, 2
θ s and 180 degrees plus 2 θ s . So in the
211
00:30:32,899 --> 00:30:41,039
stress body it
will be θ s plus 90, or θ s and θ s plus
212
00:30:41,039 --> 00:30:44,629
90 perpendicular plane on which shear stress
will be
213
00:30:44,629 --> 00:30:46,840
maximum.
214
00:30:46,840 --> 00:30:52,720
Now if we look in to the values of tan 2 θ
s and compare with the values of previously
215
00:30:52,720 --> 00:30:59,679
calculated
values of tan 2 θ P we find that tan2 θ
216
00:30:59,679 --> 00:31:08,960
s is equal to minus 1 by tan2 θ P and tan2
θ P we have
217
00:31:08,960 --> 00:31:14,230
already evaluated earlier as τ xy by ( σ
x minus σ y ) by 2.
218
00:31:14,230 --> 00:31:32,070
So this is equals to minus cot 2 θ P
which we can write as, tan 90 plus 2 θ P
219
00:31:32,070 --> 00:31:47,460
. This indicates that, 2 θ s is
equal to 90 plus 2 θ P . Or θ s
220
00:31:47,460 --> 00:31:57,940
is equal to 45 degrees plus θ P . This indicates
that maximum shear stress occurs in the plane
221
00:31:57,940 --> 00:32:10,100
which is at angle 45 degrees with maximum
or minimum principal shear stresses.
222
00:32:10,100 --> 00:32:11,100
..
223
00:32:11,100 --> 00:32:26,570
This is the value of tan2 θ evaluated, hence
we find that two mutually perpendicular planes
224
00:32:26,570 --> 00:32:31,380
on
which maximum shear stresses exists; and of
225
00:32:31,380 --> 00:32:35,460
maximum and minimum shear stresses form an
angle of 45 degrees with the principal planes
226
00:32:35,460 --> 00:32:38,979
is just seen.
227
00:32:38,979 --> 00:32:53,619
Now let us look in to the value of
principal stress where shear stress is at
228
00:32:53,619 --> 00:32:55,600
maximum.
229
00:32:55,600 --> 00:33:15,039
Now we have calculated that tan 2 θ is equal
to minus ( σ x minus σ y ) by 2 by τ xy
230
00:33:15,039 --> 00:33:21,249
. If we place
this in geometrical form, this we have to
231
00:33:21,249 --> 00:33:34,610
take a 2 θ s , this is τ xy , this is minus
( σ x minus σ y ) by 2.
232
00:33:34,610 --> 00:33:47,929
Hence the value of R again, square root of
(( σ x minus σ y ) by 2) whole square plus
233
00:33:47,929 --> 00:33:52,440
τ xy square.
234
00:33:52,440 --> 00:34:05,440
Likewise then the cos θ , rather cos2 θ
s is equal to τ xy by R and sin2 θ s is
235
00:34:05,440 --> 00:34:15,419
equal to minus ( σ x
minus σ y ) by 2R if we substitute the values
236
00:34:15,419 --> 00:34:18,660
of cos and sin in the values of the normal
which is
237
00:34:18,660 --> 00:34:36,980
σ x ˈ is equal to ( σ x plus σ y ) by
2 plus ( σ x minus σ y ) by 2 cos2 θ plus
238
00:34:36,980 --> 00:34:44,629
τ xy sin2 θ .
Now in this in place of sin2 θ and cos2 θ
239
00:34:44,629 --> 00:34:50,960
if we substitute cos2 θ s and sin2 θ s we
will get the
240
00:34:50,960 --> 00:34:53,609
values of normal stress.
241
00:34:53,609 --> 00:34:58,510
Also we will get the values of shear stresses
explained.
242
00:34:58,510 --> 00:35:05,230
Now if you
substitute these values we get this is equal
243
00:35:05,230 --> 00:35:12,520
to ( σ x plus σ y ) by 2, these two terms
get cancelled
244
00:35:12,520 --> 00:35:18,779
once you substitute the values.
245
00:35:18,779 --> 00:35:19,779
..
246
00:35:19,779 --> 00:35:26,930
Also we have seen the values of shear stress
as, tau x prime y prime is equal to minus
247
00:35:26,930 --> 00:35:39,099
( σ x
minus σ y ) by 2 sin2 θ plus τ xy cos2
248
00:35:39,099 --> 00:36:00,270
θ . If we substitute the values of sin2 θ
and cos2 θ , sin2 θ
249
00:36:00,270 --> 00:36:22,490
we have obtained as ( σ x minus σ y ) by
2, so this is (( σ x minus σ y ) by 2) whole
250
00:36:22,490 --> 00:36:31,740
square 1 by R
plus τ xy square by R. Then R is equal to
251
00:36:31,740 --> 00:36:34,480
square root of (( σ x minus σ y ) by 2)
whole square plus
252
00:36:34,480 --> 00:36:42,280
τ xy square, so this will be going to equal
to (( σ x minus σ y ) by 2) whole square
253
00:36:42,280 --> 00:36:46,190
plus τ xy square.
254
00:36:46,190 --> 00:36:50,430
So this us the tau max.
255
00:36:50,430 --> 00:36:55,789
In fact the minimum stress is the negative
of this.
256
00:36:55,789 --> 00:37:03,310
So tau max or min is
equal to plus or minus square root of (( σ
257
00:37:03,310 --> 00:37:09,240
x minus σ y ) by 2) whole square plus τ
xy square.
258
00:37:09,240 --> 00:37:13,510
These
are the values of maximum stresses and we
259
00:37:13,510 --> 00:37:19,050
observed that the value of the normal stress
on the
260
00:37:19,050 --> 00:37:25,910
plane where shear stress is maximum is equal
to ( σ x plus σ y ) by 2.
261
00:37:25,910 --> 00:37:35,301
Now if the normal stresses are the principal
stresses then we get that maximum shear stress
262
00:37:35,301 --> 00:37:42,420
is
equal to (σ 1 minus σ 2 ) by 2 from this
263
00:37:42,420 --> 00:37:47,440
expression, if the σ x is σ 1 , and σ y
is σ 2 , and this is being
264
00:37:47,440 --> 00:37:54,030
the principal stress τ xy is equal to 0,
so tau max gives us the value in the terms
265
00:37:54,030 --> 00:37:58,500
of principal
stresses as ( σ 1 minus σ 2 ) by 2.
266
00:37:58,500 --> 00:38:03,809
This gives the maximum shear stress in the
terms of principal
267
00:38:03,809 --> 00:38:08,269
shear stresses.
268
00:38:08,269 --> 00:38:09,269
..
269
00:38:09,269 --> 00:38:16,320
Hence, we say that planes of maximum shear
stresses are not free from normal stresses.
270
00:38:16,320 --> 00:38:20,280
As we
have seen in case of principal stresses, the
271
00:38:20,280 --> 00:38:24,950
principal stresses acts on the shear stresses
are 0,
272
00:38:24,950 --> 00:38:31,900
whereas the planes on which the shear stresses
are maximum there we do have normal stresses
273
00:38:31,900 --> 00:38:40,130
and the value of normal stress is ( σ x plus
σ y ) by 2.
274
00:38:40,130 --> 00:38:41,130
.
275
00:38:41,130 --> 00:38:45,720
Now let us look into another representation
or the evaluation of the stress from a concept
276
00:38:45,720 --> 00:38:52,099
which
is given by Otto Mohr of Germany in 1895 which
277
00:38:52,099 --> 00:38:57,059
we popularly designate as Mohr’s Circle
of
278
00:38:57,059 --> 00:39:00,780
.Stress.
279
00:39:00,780 --> 00:39:17,140
As we have seen we have a stress body in which
the normal stresses are σ x and σ y
280
00:39:17,140 --> 00:39:28,160
and
we have the corresponding shear stresses,
281
00:39:28,160 --> 00:39:32,390
now we can represent this stress system in
terms of
282
00:39:32,390 --> 00:39:34,590
this circle.
283
00:39:34,590 --> 00:39:35,590
.
284
00:39:35,590 --> 00:39:41,660
Now let us look in to the expression for normal
stress and shearing stress.
285
00:39:41,660 --> 00:39:53,220
σ x prime is equal to
( σ x plus σ y ) by 2 plus ( σ x minus
286
00:39:53,220 --> 00:40:07,010
σ y ) by 2 cos2 θ plus τ xy sin2 θ . Now
tau x prime y prime is equal
287
00:40:07,010 --> 00:40:25,020
to minus ( σ x minus σ y ) by 2 sin2 θ
plus τ xy cos2 θ . From the first of this
288
00:40:25,020 --> 00:40:33,550
equation you can write
this as σ x ˈ minus ( σ x plus σ y ) by
289
00:40:33,550 --> 00:40:50,550
2 is equal to ( σ x minus σ y ) by 2 cos2
θ plus τ xy sin2 θ . Now, if
290
00:40:50,550 --> 00:40:58,329
we square this equation and the second equation
and add them up we get ( σ x ˈ minus
291
00:40:58,329 --> 00:41:12,700
( σ x plus σ y ) by 2) whole square plus
tau x prime y prime square is equal to (( σ
292
00:41:12,700 --> 00:41:26,799
x minus σ y ) by 2)
whole square
293
00:41:26,799 --> 00:41:43,789
and (sin square 2 θ and cos square 2 θ is
1) plus τ xy square; the other terms get
294
00:41:43,789 --> 00:41:45,029
canceled.
295
00:41:45,029 --> 00:41:58,660
This particular equation can be represented
as (x minus a) whole square plus y square
296
00:41:58,660 --> 00:42:03,710
is equal to
b square.
297
00:42:03,710 --> 00:42:10,270
This particular equation is a well known equation
which is that of a circle where the
298
00:42:10,270 --> 00:42:19,730
centre of the circle lies at the coordinates
(plus a, 0) the radius of which is equals
299
00:42:19,730 --> 00:42:30,839
to b and x
represents σ x prime, and y represents tau
300
00:42:30,839 --> 00:42:39,660
x prime y prime . If we draw a circle whose
centre is at (a,
301
00:42:39,660 --> 00:42:46,619
0), where a is equal to ( σ x plus σ y ) by
2 on the σ x prime axis which is representing
302
00:42:46,619 --> 00:42:51,730
x, with
radius of b is equal to square root of (( σ
303
00:42:51,730 --> 00:42:56,430
x minus σ y ) by 2) whole square plus τ
xy square then we
304
00:42:56,430 --> 00:43:08,559
get the circle, and that is what is represented
as here in terms of Mohr’s circle.
305
00:43:08,559 --> 00:43:17,049
The centre of this particular circle is at
a distance of from the origin, consider this
306
00:43:17,049 --> 00:43:25,559
as σ x axis or σ
axis and tau axis then, this at the distance
307
00:43:25,559 --> 00:43:35,940
of ( σ x plus σ y ) by 2 which is the average
stress.
308
00:43:35,940 --> 00:43:36,940
This
309
00:43:36,940 --> 00:43:53,220
.particular point represents the stress which
we have at the particular body which is σ
310
00:43:53,220 --> 00:43:54,220
x , σ y , and
311
00:43:54,220 --> 00:44:00,529
τ xy . This particular point, point on this
particular circle represents the value of
312
00:44:00,529 --> 00:44:04,180
xy which is
nothing but the σ and tau at a particular
313
00:44:04,180 --> 00:44:08,529
orientation which is representing a plane.
314
00:44:08,529 --> 00:44:19,630
Here this particular point we are representing
as σ x and τ xy . The σ x and τ xy on
315
00:44:19,630 --> 00:44:25,890
this circle is
representing this particular plane.
316
00:44:25,890 --> 00:44:31,970
Hence this being σ x and this being ( σ
x plus σ y ) by 2, the
317
00:44:31,970 --> 00:44:43,339
distance here, this particular distance, σ
x minus ( σ x plus σ y ) by 2 is equal to(
318
00:44:43,339 --> 00:44:47,589
σ x minus σ y ) by 2.
319
00:44:47,589 --> 00:44:56,190
This particular distance is τ xy . So eventually
this particular distance is the square root
320
00:44:56,190 --> 00:45:00,549
of
(( σ x minus σ y ) by 2) whole square plus
321
00:45:00,549 --> 00:45:14,110
τ xy square which is that of radius which
is b.
322
00:45:14,110 --> 00:45:19,349
This particular point represents the maximum
normal stress on which this normal stress
323
00:45:19,349 --> 00:45:24,230
is acting;
this is the minimum normal stress and from
324
00:45:24,230 --> 00:45:33,520
this plane we rotate 2 θ P angle, one of
the maximum
325
00:45:33,520 --> 00:45:40,940
normal stress plane, and if rotate by another
180 degrees, another plane representing the
326
00:45:40,940 --> 00:45:43,369
maximum normal stress.
327
00:45:43,369 --> 00:45:48,261
This maximum normal stress we call as maximum
principal stress
328
00:45:48,261 --> 00:45:58,760
which we represented as σ 1 , and this we
represent as minimum principal stress as σ
329
00:45:58,760 --> 00:46:02,859
2 .
This particular point and this point in the
330
00:46:02,859 --> 00:46:06,529
circle represents the maximum value of the
shearing
331
00:46:06,529 --> 00:46:14,270
stress τ xy is equal to radius is equal to
square root of (( σ x minus σ y ) by 2)
332
00:46:14,270 --> 00:46:17,579
whole square plus τ xy
square.
333
00:46:17,579 --> 00:46:23,890
So plus tau and minus tau are the maximum
and minimum shear stresses.
334
00:46:23,890 --> 00:46:28,160
If we look into
the plane this particular plane is representing
335
00:46:28,160 --> 00:46:33,030
principal stress and this particular plane
representing maximum shear stress and the
336
00:46:33,030 --> 00:46:36,609
angle between these two is 90 degrees which
is twice
337
00:46:36,609 --> 00:46:39,440
of that in the body.
338
00:46:39,440 --> 00:46:44,369
As we have seen that the angle between the
maximum principal plane and the
339
00:46:44,369 --> 00:46:49,760
plane on which maximum shear stress acts is
at an angle of 45 degrees which is being
340
00:46:49,760 --> 00:46:59,079
represented here as 2 θ P is equal to 90;
θ is equal to 45 degrees.
341
00:46:59,079 --> 00:47:00,079
.
342
00:47:00,079 --> 00:47:10,180
.Hence from the Mohr’s circle we can observe
that the maximum normal stress is σ 1 which
343
00:47:10,180 --> 00:47:15,220
we
have designated as maximum principal stress.
344
00:47:15,220 --> 00:47:21,880
The minimum shear stress is σ 2 , which is
minimum principal shear stress and at those
345
00:47:21,880 --> 00:47:26,150
two planes we have seen that no shear stress
exists.
346
00:47:26,150 --> 00:47:31,990
Because that being on the σ axis, the value
of shear stresses is 0 and hence they are
347
00:47:31,990 --> 00:47:34,569
the principal
stresses.
348
00:47:34,569 --> 00:47:40,369
Also, the maximum shear stresses is equal
to the radius of the circle which is square
349
00:47:40,369 --> 00:47:48,200
root of (( σ x minus σ y ) by 2) whole square
plus τ xy square and the radius is nothing
350
00:47:48,200 --> 00:47:52,980
but equals to
in terms of σ 1 and σ 2 has ( σ 1 minus
351
00:47:52,980 --> 00:48:02,069
σ 2 ) by 2 which is we have seen through
our transformation
352
00:48:02,069 --> 00:48:06,529
as well.
353
00:48:06,529 --> 00:48:07,529
.
354
00:48:07,529 --> 00:48:15,539
If σ 1 and σ 2 are equal then, Mohr’s
circle reduces to a point and there are no
355
00:48:15,539 --> 00:48:19,900
shear stresses will
be developed in the x, y-plane.
356
00:48:19,900 --> 00:48:25,029
And if σ x plus σ y is equal to 0, then,
as we have seen, centre of
357
00:48:25,029 --> 00:48:33,130
the circle is located at a distance of (plus
a, 0) which is on the axis and plus a is equal
358
00:48:33,130 --> 00:48:36,790
to
( σ x plus σ y ) by 2, the average stress.
359
00:48:36,790 --> 00:48:45,190
If σ x plus σ y is equal to 0, the centre
coincides with origin
360
00:48:45,190 --> 00:48:49,319
as zero point at the σ tau as reference axis.
361
00:48:49,319 --> 00:48:56,500
Hence, at any point on any plane which is
on the
362
00:48:56,500 --> 00:49:03,500
circumference of the circle representing any
plane at the particular orientation, the values
363
00:49:03,500 --> 00:49:07,900
we will
get are the maximum principal stress as the
364
00:49:07,900 --> 00:49:11,250
tau and also maximum shear stress as tau.
365
00:49:11,250 --> 00:49:17,700
This we
call as the state of pure shear.
366
00:49:17,700 --> 00:49:22,609
Maximum and minimum principal stresses are
also equal to the
367
00:49:22,609 --> 00:49:24,420
maximum shear stress.
368
00:49:24,420 --> 00:49:36,229
These are the important observations from
the Mohr’s circle.
369
00:49:36,229 --> 00:49:37,229
..
370
00:49:37,229 --> 00:49:40,420
Now let us look into how we construct a Mohr’s
circle.
371
00:49:40,420 --> 00:49:47,059
Basically we need to do that, if we know
the stresses: σ x , σ y and τ xy at a particular
372
00:49:47,059 --> 00:49:52,220
point, then we should be able to evaluate
the stresses at
373
00:49:52,220 --> 00:49:58,000
any plane, which are at σ x prime, tau x
prime and tau y prime from the orientation
374
00:49:58,000 --> 00:50:02,230
of the plane
with reference to the x-plane.
375
00:50:02,230 --> 00:50:05,799
Now those stresses can be evaluated, either
from the
376
00:50:05,799 --> 00:50:12,480
transformation of the equations as we have
just seen or we can evaluate the stresses
377
00:50:12,480 --> 00:50:14,400
from the
Mohr’s circle as well.
378
00:50:14,400 --> 00:50:19,829
Now let us look at how to construct a Mohr’s
circle based on the given
379
00:50:19,829 --> 00:50:23,359
state of stress.
380
00:50:23,359 --> 00:50:24,359
.
381
00:50:24,359 --> 00:50:49,980
.Now let us say that at a stress body, we
know the stresses: σ x , σ y , and τ xy
382
00:50:49,980 --> 00:50:54,920
. As we have seen the
centre of the circle is located at a distance
383
00:50:54,920 --> 00:51:07,410
of ( σ x plus σ y ) by 2.
384
00:51:07,410 --> 00:51:15,210
Now in this there are two ways
of constructing a circle, one is either we
385
00:51:15,210 --> 00:51:19,109
can represent σ in this direction and tau
in the positive
386
00:51:19,109 --> 00:51:27,170
direction as we have noted earlier in that
case, the angle will be in a clock wise direction,
387
00:51:27,170 --> 00:51:30,730
θ
angle in the Mohr plane will be in clockwise
388
00:51:30,730 --> 00:51:36,549
direction which will be opposite to the convention
which we have assumed while deriving this
389
00:51:36,549 --> 00:51:39,030
transformation equation, where we take θ
in
390
00:51:39,030 --> 00:51:41,619
anticlockwise direction.
391
00:51:41,619 --> 00:51:50,510
Another way to construct is we take σ in
the positive x-direction and tau in the opposite
392
00:51:50,510 --> 00:51:56,100
direction, in that case the representation
of angle in the Mohr’s plane, is in anticlockwise
393
00:51:56,100 --> 00:52:00,349
direction which matches with our transformation
of the equation.
394
00:52:00,349 --> 00:52:07,670
So let us represent the σ in the
positive x-direction and the tau in the lower
395
00:52:07,670 --> 00:52:10,170
direction as positive.
396
00:52:10,170 --> 00:52:14,079
Thereby the angle will be
represented in the anticlockwise direction
397
00:52:14,079 --> 00:52:17,289
which is considered as positive and matches
with our
398
00:52:17,289 --> 00:52:20,430
physical stress system.
399
00:52:20,430 --> 00:52:30,740
Now if we represent this stress σ and tau
on this particular plane, thereby this particular
400
00:52:30,740 --> 00:52:36,720
point on
the circle represents this plane, plane-A,
401
00:52:36,720 --> 00:52:41,960
where the normal stress is σ and shearing
stress is tau.
402
00:52:41,960 --> 00:52:50,769
Likewise in the perpendicular plane we have
stress as σ y and shearing stress as tau
403
00:52:50,769 --> 00:52:53,599
which is at
an angle of 90 degrees, with respect to this
404
00:52:53,599 --> 00:52:54,599
A-plane.
405
00:52:54,599 --> 00:53:03,190
In Mohr’s circle if we go 180 degrees with
respect to this, we get another point, which
406
00:53:03,190 --> 00:53:17,059
is plane-B were we have σ y and τ xy .
Now this if we take the centre, as this O,
407
00:53:17,059 --> 00:53:21,809
and radius as OA and OB, and plot a circle,
this gives
408
00:53:21,809 --> 00:53:38,250
us the Mohr’s circle where in this point
represents the maximum normal stress σ 1
409
00:53:38,250 --> 00:53:40,890
. This
represents the minimum principal stress which
410
00:53:40,890 --> 00:53:46,539
is σ 2 . Now if we are interested to find
out stress
411
00:53:46,539 --> 00:53:54,789
on any plane, whose normal to this particular
plane is oriented at angle of θ , now this
412
00:53:54,789 --> 00:53:58,109
particular
plane which is represented by this point A,
413
00:53:58,109 --> 00:54:05,400
from here it moves at angle of 2 θ , and
joins from the
414
00:54:05,400 --> 00:54:12,190
centre of this the point which we get on the
circle represents the plane normal to which
415
00:54:12,190 --> 00:54:16,369
is at
angle θ from the x-plane.
416
00:54:16,369 --> 00:54:21,059
Thereby the stresses that act over here together
is the normal stress
417
00:54:21,059 --> 00:54:26,580
and the shearing stress on this particular
inclined plane and that is how we compute
418
00:54:26,580 --> 00:54:33,480
the stresses
in the Mohr’s circle.
419
00:54:33,480 --> 00:54:34,480
..
420
00:54:34,480 --> 00:54:44,210
Now having known about the principal stresses
from the transformation equations and through
421
00:54:44,210 --> 00:54:59,280
the Mohr’s circle let us look into some
of the problems.
422
00:54:59,280 --> 00:55:12,130
Let us understand how to evaluate the
mass which is safely supported by the wires.
423
00:55:12,130 --> 00:55:20,079
Now if we take the free body diagram of this
particular part, then here there is the force
424
00:55:20,079 --> 00:55:23,140
F 1 to balance it, the reactive forces.
425
00:55:23,140 --> 00:55:27,170
Here there is force
F 2 to balance it.
426
00:55:27,170 --> 00:55:38,700
If we draw the free body diagram of this whole
part then we have F 1 in this particular direction
427
00:55:38,700 --> 00:55:45,079
and F 2 in this particular direction and the
mass which is hanging from this particular
428
00:55:45,079 --> 00:55:46,079
point.
429
00:55:46,079 --> 00:55:52,349
Let us
call it this force as Mg.
430
00:55:52,349 --> 00:56:00,039
Now let us call this angle as θ 1 , and this
as θ 2 . From the given data, we
431
00:56:00,039 --> 00:56:14,970
can compute cos θ 1 is equal to 0.6 and sin
θ 1 is equal to 0.8; cos θ 2 is equal to
432
00:56:14,970 --> 00:56:24,869
0.923, and sin θ 2
is equal to 0.385.
433
00:56:24,869 --> 00:56:32,210
Now if we take the summation of horizontal
forces as 0, that gives us the
434
00:56:32,210 --> 00:56:44,380
F 1 cos θ 1 is equal to F 2 cos θ 2 .
Now it has been indicated that the maximum
435
00:56:44,380 --> 00:56:48,119
tensile stress that the wires can withstand
as 100
436
00:56:48,119 --> 00:56:53,539
MPa and cross sectional area where the two
wires are given, thereby if we compute, F
437
00:56:53,539 --> 00:57:00,550
1 will gives
area times stress which is equals to 40 Kilo
438
00:57:00,550 --> 00:57:01,550
Newton.
439
00:57:01,550 --> 00:57:07,460
F 2 thereby is given 20 Kilo Newton.
440
00:57:07,460 --> 00:57:11,569
From
this expression if we substitute the values
441
00:57:11,569 --> 00:57:17,130
for cos θ 1 and cos θ 2 , we find that if
we substitute for
442
00:57:17,130 --> 00:57:33,369
F 2 for F 1 is equal to 40 Kilo Newton, and
F 2 is equal to 40 into 0.6 by 0.923 is equal
443
00:57:33,369 --> 00:57:41,920
to 26 Kilo
Newton which is greater than 20 Kilo Newton.
444
00:57:41,920 --> 00:57:48,190
So instead of using F 2 , let us use, F 1
so that if we
445
00:57:48,190 --> 00:58:01,440
represent F 1 in terms of F 2 this is 20 into
0.923 by 0.6 which is less than 40.
446
00:58:01,440 --> 00:58:11,609
Hence if we compute the values, if we take
the summation in the vertical direction, summation
447
00:58:11,609 --> 00:58:17,420
of
forces we get, that Mg is equal to F 1 sin
448
00:58:17,420 --> 00:58:24,180
θ 1 plus F 2 sin θ 2 and thereby the values
from M this you
449
00:58:24,180 --> 00:58:33,649
will get as 3293.4 Kg.
450
00:58:33,649 --> 00:58:34,649
..
451
00:58:34,649 --> 00:58:39,859
You can look into this particular problem,
you can compute the state of stress in transformation
452
00:58:39,859 --> 00:58:48,549
equations, also you can use the same problem
to solve through Mohr’s circle of stress.
453
00:58:48,549 --> 00:58:49,549
.
454
00:58:49,549 --> 00:58:55,050
These questions which we have posed in the
last time they quite straight forward, the
455
00:58:55,050 --> 00:58:57,700
first one is
the maximum normal stress and shear stress
456
00:58:57,700 --> 00:59:01,120
in an axially loaded bar, which we know as
P by A
457
00:59:01,120 --> 00:59:07,890
is the normal stress and P by 2A is the shear
stress and thereby tau the shear stress is
458
00:59:07,890 --> 00:59:12,359
equal to σ 2 ,
half the normal stress.
459
00:59:12,359 --> 00:59:15,000
We have discussed about stress variants and
we know that the value of
460
00:59:15,000 --> 00:59:17,799
shear stress on the principal plane is zero.
461
00:59:17,799 --> 00:59:18,799
..
462
00:59:18,799 --> 00:59:23,289
To summarize: what we have done today in this
particular lesson is developed the transformation
463
00:59:23,289 --> 00:59:30,319
equation for evaluation of stresses for two
dimensional problems; evaluation of principal
464
00:59:30,319 --> 00:59:34,750
stresses
and shear stresses, we have shown how to construct
465
00:59:34,750 --> 00:59:37,619
Mohr’s circle for the evaluation of stress
and
466
00:59:37,619 --> 00:59:45,180
some examples to demonstrate the evaluation
of stresses at a point.
467
00:59:45,180 --> 00:59:46,180
.
468
00:59:46,180 --> 00:59:47,180
These are the questions:
•
469
00:59:47,180 --> 00:59:49,719
What is meant X the state of pure shear?
470
00:59:49,719 --> 00:59:51,270
•
What is the co-ordinate of the centre of the
471
00:59:51,270 --> 00:59:53,519
Mohr’s circle and what is its radius?
472
00:59:53,519 --> 00:59:54,519
.•
473
00:59:54,519 --> 01:00:04,089
What happens when ( σ x plus σ y ) is equal
to 0 and σ 1 is equal to σ 2 ?
474
01:00:04,089 --> 01:00:04,089
.