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Module5/Lesson3
1 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Module 5: Two Dimensional Problems in Cartesian Coordinate System 5.3.1 SOLUTIONS OF TWO-DIMENSIONAL PROBLEMS BY THE USE
OF POLYNOMIALS
The equation given by
÷÷ø
öççè
涶
+¶¶
2
2
2
2
yx ÷÷ø
öççè
涶
+¶¶
2
2
2
2
yx
ff =
4
4
x¶¶ f
+ 2 22
4
yx ¶¶¶ f
+ 4
4
y¶¶ f
= 0 (5.13)
will be satisfied by expressing Airy’s function ( )yx,f in the form of homogeneous polynomials.
(a) Polynomial of the First Degree
Let ybxa 111 +=f Now, the corresponding stresses are
xs = 21
2
y¶¶ f
= 0
ys = 21
2
x¶¶ f
= 0
xyt = yx¶¶
¶- 1
2f = 0
Therefore, this stress function gives a stress free body.
(b) Polynomial of the Second Degree
Let 2f = 222
22
22y
cxybx
a++
The corresponding stresses are
xs = 22
2
y¶¶ f
= 2c
ys = 22
2
x¶¶ f
= 2a
xyt =yx¶¶
¶-
f2
= 2b-
This shows that the above stress components do not depend upon the co-ordinates x and y, i.e., they are constant throughout the body representing a constant stress field. Thus, the
Module5/Lesson3
2 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
stress function 2f represents a state of uniform tensions (or compressions) in two perpendicular directions accompanied with uniform shear, as shown in the Figure 5.3 below.
Figure 5.3 State of stresses
Figure 5.3 Constant Stress field (c) Polynomial of the Third Degree
Let 3f = 33232333
6226y
dxy
cyx
bx
a+++
The corresponding stresses are
xs = 23
2
y¶¶ f
= ydxc 33 +
ys = 23
2
x¶¶ f
= ybxa 33 +
xyt = yx¶¶
¶- 3
2f = ycxb 33 --
This stress function gives a linearly varying stress field. It should be noted that the magnitudes of the coefficients 333 ,, cba and 3d are chosen freely since the expression for
3f is satisfied irrespective of values of these coefficients.
Now, if 0333 === cba except 3d , we get from the stress components
ydx 3=s
0=ys and 0=xyt
This corresponds to pure bending on the face perpendicular to the x-axis. \ At y = -h, hdx 3-=s
Module5/Lesson3
3 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
and At y = +h, hdx 3+=s
The variation of xs with y is linear as shown in the Figure 5.4.
Figure 5.4 Variation of Stresses
Similarly, if all the coefficients except 3b are zero, then we get
0=xs
yby 3=s
xbxy 3-=t
The stresses represented by the above stress field will vary as shown in the Figure 5.5.
Module5/Lesson3
4 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Figure 5.5 Variation of Stresses
In the Figure 5.5, the stress ys is constant with x (i.e. constant along the span L of
the beam), but varies with y at a particular section. At y = +h, hby 3=s (i.e., tensile),
while at y = -h, hby 3-=s (i.e. compressive). xs is zero throughout. Shear stress xyt is
zero at 0=x and is equal to Lb3- at x = L. At any other section, the shear stress is
proportional to x.
(d) Polynomial of the Fourth Degree
Let 4f = 44342243444
1262612y
exy
dyx
cyx
bx
a++++
The corresponding stresses are given by
244
24 yexydxcx ++=s
244
24 ycxybxay ++=s
244
24
22
2y
dxycx
bxy ÷
øö
çèæ--÷
øö
çèæ-=t
Now, taking all coefficients except 4d equal to zero, we find
,4 xydx =s ,0=ys 24
2y
dxy -=t
Module5/Lesson3
5 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Assuming 4d positive, the forces acting on the beam are shown in the Figure 5.6.
Figure 5.6 Stresses acting on the beam
On the longitudinal sides, y = ±h are uniformly distributed shearing forces. At the ends, the shearing forces are distributed according to a parabolic distribution. The shearing forces acting on the boundary of the beam reduce to the couple.
Therefore, M = hLhd
hLhd
223
12
2
24
24 -
Or M = Lhd 343
2
This couple balances the couple produced by the normal forces along the side x = L of the beam.
(e) Polynomial of the Fifth Degree
5 4 3 2 2 3 4 55 5 5 5 5 55 20 12 6 6 12 20
a b c d e fLet x x y x y x y xy yj = + + + + +
The corresponding stress components are given by
355
255
25
3525
2
)2(31
)32(3
ydbxyacyxdxc
yx +-+-+=¶¶
=f
s
3525
25
352
52
3y
dxycyxbxa
xy +++=¶¶
=f
s
355
25
25
35
52
)32(31
31
yacxydyxcxbyxxy ++---=
¶¶¶
-=f
t
Module5/Lesson3
6 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Here the coefficients 5555 ,,, dcba are arbitrary, and in adjusting them we obtain solutions
for various loading conditions of the beam.
Now, if all coefficients, except 5d , equal to zero, we find
÷øö
çèæ -= 32
5 32
yyxdxs
25
353
1
xyd
yd
xy
y
-=
=
t
s
Case (i) The normal forces are uniformly distributed along the longitudinal sides of the beam.
Case (ii) Along the side x = L, the normal forces consist of two parts, one following a linear law and
the other following the law of a cubic parabola. The shearing forces are proportional to x on
the longitudinal sides of the beam and follow a parabolic law along the side x = L. The distribution of the stresses for the Case (i) and Case (ii) are shown in the Figure 5.7.
Case (i)
Module5/Lesson3
7 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Case (ii)
Figure 5.7 Distribution of forces on the beam
5.3.2 BENDING OF A NARROW CANTILEVER BEAM SUBJECTED TO
END LOAD
Consider a cantilever beam of narrow rectangular cross-section carrying a load P at the end as shown in Figure 5.8.
Figure 5.8 Cantilever subjected to an end load
The above problems may be considered as a case of plane stress provided that the thickness of the beam t is small relative to the depth 2h.
Module5/Lesson3
8 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Boundary Conditions
( ) 0=±= hyAtxyt
( ) 0=±= hyAtys (5.14)
These conditions express the fact that the top and bottom edges of the beam are not loaded. Further, the applied load P must be equal to the resultant of the shearing forces distributed across the free end.
Therefore, P = - ò+
-
h
h xy dyb2t (5.14a)
By Inverse Method
As the bending moment varies linearly with x, and xs at any section depends upon y, it is
reasonable to assume a general expression of the form
xs = xycy 12
2
=¶¶ f
(5.14b)
where c1 = constant. Integrating the above twice with respect to y, we get
f = )()(61
213
1 xfxyfxyc ++ (5.14c)
where f1(x) and f2(x) are functions of x to be determined. Introducing the f thus obtained into Equation (5.12), we have
y 042
4
41
4
=+dx
fddx
fd (5.14d)
Since the second term is independent of y, there exists a solution for all x and y provided that
041
4
=dx
fd and 04
24
=dx
fd
Integrating the above, we get f1(x) = c2x3+c3x2+c4x+c5
f2(x) = c6x3+c7x2+c8x+c9
where c2, c3……., c9 are constants of integration.
Therefore, (5.14c) becomes
f = 982
73
6542
33
23
1 )(61
cxcxcxcycxcxcxcxyc ++++++++
Now, by definition,
( ) ( )73622
2
26 cycxcycxy +++=
¶¶
=fs
Module5/Lesson3
9 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
txy = 432
22
1
2
2321
cxcxcycyx
----=÷÷ø
öççè
涶
¶-
f (5.14e)
Now, applying boundary conditions to (5.14e), we get
c2 = c3 = c6 = c7 = 0 and c4 = 21
- c1h2
Also, ò ò+
- -=-=-
h
h
h
hxy Pdyhybcdyb )(221
2 221t
Solving, c1 = - ÷øö
çèæ-=÷÷
ø
öççè
æIP
hbP
343
where I = 34 bh3 is the moment of inertia of the cross-section about the neutral axis.
From Equations (5.14b) and (5.14e), together with the values of constants, the stresses are found to be
xs = - )(2
,0, 22 yhIP
IPxy
xyy --
==÷øö
çèæ ts
The distribution of these stresses at sections away from the ends is shown in Figure 5.8 b
By Semi-Inverse Method
Beginning with bending moment Mz = Px, we may assume a stress field similar to the case of pure bending:
xs = yI
Px÷øö
çèæ-
xyt = ( )yxxy ,t (5.14f)
0==== yzxzzy ttss
The equations of compatibility are satisfied by these equations. On the basis of equation (5.14f), the equations of equilibrium lead to
0=¶
¶+
¶¶
yxxyx
ts, 0=
¶
¶
xxyt
(5.14g)
From the second expression above, txy depends only upon y. The first equation of (5.14g) together with equation (5.14f) gives
IPy
dy
d xy =t
Module5/Lesson3
10 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
or txy = c
IPy
+2
2
Here c is determined on the basis of (txy)y=±h = 0
Therefore, c = - I
Ph2
2
Hence, txy = I
PhI
Py22
22
-
Or txy = - )(2
22 yhI
P-
The above expression satisfies equation (5.14a) and is identical with the result previously obtained. 5.3.3 PURE BENDING OF A BEAM
Consider a rectangular beam, length L, width 2b, depth 2h, subjected to a pure couple M along its length as shown in the Figure 5.9
Figure 5.9 Beam under pure bending
Consider a second order polynomial such that its any term gives only a constant state of stress. Therefore
f = 2a22
22
2
2 ycxyb
x++
By definition,
Module5/Lesson3
11 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
xs = 2
2
y¶¶ f
, ys = 2
2
x¶¶ f
, txy = ÷÷ø
öççè
涶
¶-
yxf2
\ Differentiating the function, we get
sx = 2
2
y¶¶ f
= c2, sy = 2
2
x¶¶ f
= a2 and txy = ÷÷ø
öççè
涶
¶-
yxf2
= -b2
Considering the plane stress case, sz = txz = tyz = 0
Boundary Conditions
(a) At y = ± h, =ys 0
(b) At y = ± h, txy = 0 (c) At x = any value,
2b ò+
-
a
a x ydys = bending moment = constant
\2bx ò+
-
+
-
=úû
ùêë
é=h
h
h
h
yxbcydyc 0
22
2
22
Therefore, this clearly does not fit the problem of pure bending.
Now, consider a third-order equation
f = 6226
33
2323
33 ydxyc
yxbxa
+++
Now, xs = ydxcy 332
2
+=¶¶ f
(a)
ys = a3x + b3y (b)
txy = -b3x - c3y (c)
From (b) and boundary condition (a) above,
0 = a3x ± b3a for any value of x \ a3 = b3 = 0
From (c) and the above boundary condition (b), 0 = -b3x ± c3a for any value of x
therefore c3 = 0 hence, xs = d3y
ys = 0
txy = 0
Obviously, Biharmonic equation is also satisfied.
Module5/Lesson3
12 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
i.e., 024
4
22
4
4
4
=¶¶
+¶¶
¶+
¶¶
yyxxfff
Now, bending moment = M = 2b ò+
-
h
h x ydys
i.e. M = 2b ò+
-
h
hdyyd 2
3
= 2bd3 ò+
-
h
hdyy 2
= 2bd3
h
h
y+
-úû
ùêë
é3
3
M = 4bd3 3
3h
Or d3 = 34
3bhM
I
Md =3 where
34 3bh
I =
Therefore, xs = yI
M
Module5/Lesson3
13 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
5.3.4 BENDING OF A SIMPLY SUPPORTED BEAM BY A DISTRIBUTED
LOADING (UDL)
Figure 5.10 Beam subjected to Uniform load
Consider a beam of rectangular cross-section having unit width, supported at the ends and subjected to a uniformly distributed load of intensity q as shown in the Figure 5.10.
It is to be noted that the bending moment is maximum at position x = 0 and decreases with change in x in either positive or negative directions. This is possible only if the stress function contains even functions of x. Also, it should be noted that ys various from zero at
y = -c to a maximum value of -q at y = +c. Hence the stress function must contain odd functions of y.
Now, consider a polynomial of second degree with 022 == cb
222 2
xa
=\f
a polynomial of third degree with 033 == ca
33233 62
yd
yxb
+=\f
and a polynomial of fifth degree with 05555 ==== ecba
úûù
êëé -=-=\ 55
553255 3
2306
dfyd
yxd Qf
532 ffff ++=\
or 55325332322
306622y
dyx
dy
dyx
bx
a-+++=f (1)
Now, by definition,
Module5/Lesson3
14 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
÷øö
çèæ -+=
¶¶
= 32532
2
32
yyxdydyx
fs (2)
35322
2
3y
dyba
xy ++=¶¶
=fs (3)
253 xydxbxy --=t (4)
The following boundary conditions must be satisfied. (i) ( ) 0=
±= cyxyt
(ii) ( ) 0=+= cyys
(iii) ( ) qcyy -=
-=s
(iv) ( )ò+
-±= =
c
cLxx dy 0s
(v) ( )ò+
-±=
±=c
cLxxy qLdyt
(vi) ( )ò+
-±= =
c
cLxx ydy 0s
The first three conditions when substituted in equations (3) and (4) give 02
53 =-- cdb
03
3532 =++ c
dcba
qcd
cba -=-- 3532 3
which gives on solving
3532 43
,43
,2 c
qd
cq
bq
a -==-=
Now, from condition (vi), we have
ò+
-
=úû
ùêë
é÷øö
çèæ -+
c
c
ydyyyxdyd 032 32
53
Simplifying,
÷øö
çèæ --= 22
53 52
hLdd
Module5/Lesson3
15 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
÷ø
öçè
æ -=52
43
2
2
hL
hq
÷øö
çèæ --÷
ø
öçè
æ -=\ 32
32
2
32
43
52
43
yyxhq
yhL
hq
xs
3
3443
2y
hq
yhqq
y -+÷øö
çèæ-=s
2
343
43
xyhq
xhq
xy +÷øö
çèæ-=t
Now, ( ) 333
32
128
1221
hhh
I ==´
=
where I = Moment of inertia of the unit width beam.
( ) ÷ø
öçè
æ -+-=\532
2322 yhy
Iq
yxLI
qxs
÷ø
öçè
æ +-÷øö
çèæ-= 32
3
32
32hyh
yI
qys
( )22
2yhx
Iq
xy -÷øö
çèæ-=t
5.3.5 NUMERICAL EXAMPLES
Example 5.1 Show that for a simply supported beam, length 2L, depth 2a and unit width, loaded by a concentrated load W at the centre, the stress function satisfying the loading condition
is cxyxyb
+= 2
6f the positive direction of y being upwards, and x = 0 at midspan.
Module5/Lesson3
16 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Figure 5.11 Simply supported beam
Treat the concentrated load as a shear stress suitably distributed to suit this function, and so
that ò+
-
÷øö
çèæ-=
a
a
x
Wdy
2s on each half-length of the beam. Show that the stresses are
÷øö
çèæ-= xy
aW
x 343s
0=ys
úû
ùêë
é÷÷ø
öççè
æ--= 2
2
183
ay
aW
xyt
Solution: The stress components obtained from the stress function are
bxyyx =
¶¶
=2
2fs
02
2
=¶¶
=xy
fs
cby
yxxy +÷÷ø
öççè
æ-=
¶¶¶
-=2
22ft
Module5/Lesson3
17 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Boundary conditions are
(i) ayfory ±== 0s
(ii) ayforxy ±== 0t
(iii) ò+
-
±==-a
axy Lxfor
Wdy
2t
(iv) ò+
-
±==a
ax Lxfordy 0s
(v) ò+
-
±==a
ax Lxforydy 0s
Now,
Condition (i) This condition is satisfied since 0=ys
Condition (ii)
cba
+÷÷ø
öççè
æ-=
20
2
2
2bac =\
Condition (iii)
( )ò+
-
--=a
a
dyyabW 22
22
÷÷ø
öççè
æ--=
32
22
33 a
ab
÷÷ø
öççè
æ-=\
32
2
3baW
or ÷øö
çèæ-= 34
3aW
b
and ÷øö
çèæ-=
aW
c83
Condition (iv)
Module5/Lesson3
18 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
ò+
-
=÷øö
çèæ-
a
a
xydyaW
043
3
Condition (v)
ò+
-
=a
ax ydyM s
ò+
-
÷øö
çèæ-=
a
a
dyxyaW 2
343
2
WxM =\
Hence stress components are
xyaW
x ÷øö
çèæ-= 34
3s
0=ys
÷øö
çèæ-÷÷
ø
öççè
æ=
aWy
aW
xy 83
243 2
3t
úû
ùêë
é÷÷ø
öççè
æ--=\ 2
2
183
ay
aW
xyt
Example 5.2
Given the stress function ÷øö
çèæ
÷øö
çèæ= -
zx
zH 1tanp
f . Determine whether stress function f is
admissible. If so determine the stresses.
Solution: For the stress function f to be admissible, it has to satisfy bihormonic equation. Bihormonic equation is given by
02 4
4
22
4
4
4
=¶¶
+¶¶
¶+
¶¶
zzxxfff
(i)
Now, úû
ùêë
é÷øö
çèæ+÷
øö
çèæ
+-=
¶¶ -
zx
zxxzH
z1
22 tanp
f
( )[ ]32322
2222
2
21
xxzxxzxzzx
Hz
----+
÷øö
çèæ=
¶¶
pf
( ) úúû
ù
êêë
é
+÷øö
çèæ-=
¶¶
\ 222
3
2
2 2
zx
xHz pf
Module5/Lesson3
19 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Also, ( ) úúû
ù
êêë
é
+=
¶¶
322
3
3
3 8
zx
zxHz pf
( ) úúû
ù
êêë
é
+
-=
¶¶
422
235
4
4 408
zx
zxxHz pf
( ) úúû
ù
êêë
é
+
-÷øö
çèæ-=
¶¶¶
322
423
2
3 32
zx
xzxHxz p
f
( ) úúû
ù
êêë
é
+
--=
¶¶¶
422
5423
22
4 82464
zx
xxzzxHxz p
f
Similarly,
( )úûù
êë
é+
=¶¶
22
2
zxzH
x pf
( ) úúû
ù
êêë
é
+÷øö
çèæ-=
¶¶
222
2
2
2 2
zx
xzHx pf
( )( ) ú
úû
ù
êêë
é
+
-=
¶¶
322
222
3
3 32
zx
zxz
Hx pf
( ) úúû
ù
êêë
é
-
-=
¶¶
422
234
4
4 2424
zx
zxxzHx pf
Substituting the above values in (i), we get
( )[ 5423234
422824642424
14xxzzxzxxz
zx--+-
+p] 0408 235 =-+ zxx
Hence, the given stress function is admissible. Therefore, the stresses are
( ) úúû
ù
êêë
é
+÷øö
çèæ-=
¶¶
= 222
3
2
2 24
zx
xzx pfs
( ) úúû
ù
êêë
é
+÷øö
çèæ-=
¶¶
= 222
2
2
2 24
zx
xxy pfs
and ( ) úúû
ù
êêë
é
+÷øö
çèæ-=
¶¶¶
= 222
22 24
zx
zxzxxy p
ft
Module5/Lesson3
20 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Example 5.3
Given the stress function: ( )zdxzdF
2323 -÷
øö
çèæ-=f .
Determine the stress components and sketch their variations in a region included in z = 0, z = d, x = 0, on the side x positive.
Solution: The given stress function may be written as
33
22
23xz
dF
xzdF
÷øö
çèæ+÷
øö
çèæ-=f
xzd
FdFx
z÷øö
çèæ+÷
øö
çèæ-=
¶¶
\322
2 126f
and 02
2
=¶¶
xf
also 232
2 66z
dF
dFz
zx÷øö
çèæ+÷
øö
çèæ-=
¶¶¶ f
Hence xzd
FdFx
x ÷øö
çèæ+÷
øö
çèæ-=
32
126s (i)
0=zs (ii)
232
2 66z
dF
dFz
zxxz ÷øö
çèæ+÷
øö
çèæ-=
¶¶¶
-=jt (iii)
VARIATION OF STRESSES AT CERTAIN BOUNDARY POINTS (a) Variation of xσ
From (i), it is clear that xs varies linearly with x, and at a given section it varies linearly with z. \ At x = 0 and z = ± d, xs = 0
At x = L and z = 0, ÷øö
çèæ-= 2
6dFL
xs
At x = L and z = +d, 232
6126dFL
Ldd
FdFL
x =÷øö
çèæ+÷
øö
çèæ-=s
At x = L and z = -d, ÷øö
çèæ-=÷
øö
çèæ-÷
øö
çèæ-=
232
18126dFL
Ldd
FdFL
xs
The variation of xs is shown in the figure below
Module5/Lesson3
21 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Figure 5.12 Variation of xσ
(b) Variation of zσ
zs is zero for all values of x.
(c) Variation of xzτ
We have xzt = 232 .
66z
dF
dFz
÷øö
çèæ-÷
øö
çèæ
From the above expression, it is clear that the variation of xzt is parabolic with z. However,
xzt is independent of x and is thus constant along the length, corresponding to a given value of z. \At z = 0, xzt = 0
At z = +d, 066 2
32 =÷øö
çèæ-÷
øö
çèæ= d
dF
dFd
xzt
At z = -d, ÷øö
çèæ-=-÷
øö
çèæ-÷
øö
çèæ-=
dF
ddF
ddF
xz
12)(
66 232t
The variation of xzt is shown in figure below.
Module5/Lesson3
22 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Figure 5.13 Variation of xzτ
Example 5.4 Investigate what problem of plane stress is satisfied by the stress function
32
2
34 3 2F xy p
xy yd d
jé ù
= - +ê úë û
applied to the region included in y = 0, y = d, x = 0 on the side x positive.
Solution: The given stress function may be written as 3
23
3 14 4 2F Fxy p
xy yd d
jæ öæ ö æ ö= - +ç ÷ç ÷ ç ÷
è ø è øè ø
02
2
=¶¶
\xf
2
2 3 3
3 2 2. 1.5
4 2Fxy p F
p xyy d dj¶ ´æ ö æ ö= - + = -ç ÷ ç ÷¶ è ø è ø
and 3
22
43
43
dFy
dF
yx-=
¶¶¶ f
Hence the stress components are 2
2 31.5x
Fp xy
y djs ¶
= = -¶
02
2
=¶¶
=xy
fs
Module5/Lesson3
23 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
dF
dFy
yxxy 43
43
3
22
-=¶¶
¶-=
ft
(a) Variation of xσ
31.5x
Fp xy
ds æ ö= - ç ÷
è ø
When x = 0 and y = 0 or , xd ps± = (i.e., constant across the section)
When x = L and y = 0, x ps =
When x = L and y = +d, 21.5x
FLp
ds æ ö= - ç ÷
è ø
When x = L and y = -d, 2
5.1dFL
Px +=s
Thus, at x = L, the variation of xs is linear with y.
The variation of xs is shown in the figure below.
Figure 5.14 Variation of stress xσ
(b) Variation of zσ
02
2
=¶¶
=xy
fs
\ ys is zero for all value of x and y
Module5/Lesson3
24 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
(c) Variation of xyτ
÷øö
çèæ-÷÷
ø
öççè
æ=
dF
d
Fyxy 4
343
3
2
t
Thus, xyt varies parabolically with z. However, it is independent of x, i.e., it's value is the
same for all values of x.
\At ÷øö
çèæ-==
dF
y xy 43
,0 t
At 043
)(43
, 23
=úûù
êëé-úû
ùêëé=±=
dF
ddF
dy xyt
Figure 5.15 Variation of shear stress xyτ
The stress function therefore solves the problem of a cantilever beam subjected to point load F at its free end along with an axial stress of p. Example 5.5 Show that the following stress function satisfies the boundary condition in a beam of rectangular cross-section of width 2h and depth d under a total shear force W.
úûù
êëé --= )23(2
23 ydxy
hdWf
Solution: 2
2
yx ¶¶
=fs
Y
d
d
L
X o
t xy
- 3F 4d
- 3F 4d
Module5/Lesson3
25 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Now, [ ]23
662
xyxydhdW
y--=
¶¶f
[ ]xyxdhdW
y126
2 32
2
--=¶¶ f
[ ]xyxdhdW
x 633
--=\s
02
2
=¶¶
=xy
fs
and yxxy ¶¶
¶-=
ft2
= [ ]23
662
yydhdW
-
= [ ]23
33 yydhdW
-
Also, 02
22
4
4
4
4
44 =ú
û
ùêë
鶶¶
+¶¶
+¶¶
=Ñ ffyxyx
Boundary conditions are
(a) dandyfory 00 ==s
(b) dandyforxy 00 ==t
(c) LandxforWdyhd
xy 0.2.0
==òt
(d) WLMLxandxfordyhMd
x ===== ò ,00.2.0
s
(e) Lxandxfordyyhd
x ===ò 00..2.0
s
Now, Condition (a)
This condition is satisfied since 0=ys
Condition (b)
[ ] 033 223
=- ddhdW
Module5/Lesson3
26 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
Hence satisfied.
Condition (c)
[ ] hdyyydhd
Wd
2330
23ò -
[ ]dyyydd
Wd
ò -=0
23
332
d
ydy
dW
0
32
3 232
úû
ùêë
é-=
úû
ùêë
é-= 3
3
3 232
dd
d
W
2.
2 3
3
ddW
=
= W
Hence satisfied.
Condition (d)
[ ] hdyxyxdhd
Wd
2630
3ò --
[ ]dxyxyd
dW
02
3 332
--=
= 0
Hence satisfied.
Condition (e)
[ ] ydyhxyxdhdWd
.2630 3ò --
d
xyxdy
dW
0
32
32
232
úû
ùêë
é--=
úû
ùêë
é--= 3
3
32
232
xdxd
d
W
Module5/Lesson3
27 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju
úûù
êëé--= 3
3 212
xddW
Wx=
Hence satisfied