Numerical Examples 4

Module5/Lesson3

1 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju

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

6 the positive direction of y being upwards, and x = 0 at midspan.

Figure 5.11 Simply supported beam

Treat the concentrated load as a shear stress suitably distributed to suit this function, and so

that

a

a

x

Wdy

2 on each half-length of the beam. Show that the stresses are

xya

Wx 34

3

Module5/Lesson3


0y

2

2

18

3

a

y

a

Wxy

Solution: The stress components obtained from the stress function are

bxyy

x 2

2

02

2

xy

cby

yxxy

2

22

Boundary conditions are

(i) ayfory 0

(ii) ayforxy 0

(iii)

a

a

xy LxforW

dy2

(iv)

a

a

x Lxfordy 0

(v)

a

a

x Lxforydy 0

Now,

Condition (i)

This condition is satisfied since 0y

Condition (ii)

cba

20

2

2

2bac

Condition (iii)

Module5/Lesson3


a

a

dyyabW 22

22

3

22

2

33 aa

b

3

2

2

3baW

or 34

3

a

Wb

and a

Wc

8

3

Condition (iv)

a

a

xydya

W0

4

33

Condition (v)

a

a

x ydyM

a

a

dyxya

W 234

3

2

WxM

Hence stress components are

xya

Wx 34

3

0y

a

Wy

a

Wxy

8

3

24

3 2

3

2

2

18

3

a

y

a

Wxy

Example 5.2

Module5/Lesson3


Given the stress function z

xz

H 1tan . Determine whether stress function is

admissible. If so determine the stresses.

Solution: For the stress function to be admissible, it has to satisfy bihormonic equation.

Bihormonic equation is given by

024

4

22

4

4

4

zzxx (i)

Now, z

x

zx

xzH

z

1

22tan

32322

2222

2

21

xxzxxzxzzx

H

z

222

3

2

2 2

zx

xH

z

Also, 322

3

3

3 8

zx

zxH

z

422

235

4

4 408

zx

zxxH

z

322

423

2

3 32

zx

xzxH

xz

422

5423

22

4 82464

zx

xxzzxH

xz

Similarly,

22

2

zx

zH

x

222

2

2

2 2

zx

xzH

x

322

222

3

3 32

zx

zxz

H

x

422

234

4

4 2424

zx

zxxzH

x

Module5/Lesson3


Substituting the above values in (i), we get

5423234

422824642424

14xxzzxzxxz

zx0408 235 zxx

Hence, the given stress function is admissible.

Therefore, the stresses are

222

3

2

2 24

zx

x

zx

222

2

2

2 24

zx

x

xy

and 222

22 24

zx

zx

zxxy

Example 5.3

Given the stress function: zdxzd

F232

3.

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

3

3

2

2

23xz

d

Fxz

d

F

xzd

F

d

Fx

z 322

2 126

and 02

2

x

also 2

32

2 66z

d

F

d

Fz

zx

Hence xzd

F

d

Fxx 32

126 (i)

0z (ii)

2

32

2 66z

d

F

d

Fz

zxxz (iii)

VARIATION OF STRESSES AT CERTAIN BOUNDARY POINTS

(a) Variation of x

Module5/Lesson3


From (i), it is clear that x

varies linearly with x, and at a given section it varies linearly

with z.

At x = 0 and z = d, x

= 0

At x = L and z = 0, 2

6

d

FLx

At x = L and z = +d, 232

6126

d

FLLd

d

F

d

FLx

At x = L and z = -d, 232

18126

d

FLLd

d

F

d

FLx

The variation of x

is shown in the figure below

Figure 5.12 Variation of x

(b) Variation of z

z is zero for all values of x.

(c) Variation of xz

We have xz= 2

32.

66z

d

F

d

Fz

From the above expression, it is clear that the variation of xz is parabolic with z. However,

xz is independent of x and is thus constant along the length, corresponding to a given value

of z.

At z = 0, xz= 0

Module5/Lesson3


At z = +d, 066 2

32d

d

F

d

Fdxz

At z = -d, d

Fd

d

Fd

d

Fxz

12)(

66 232

The variation of xz

is shown in figure below.

Figure 5.13 Variation of

xz

Example 5.4

Investigate what problem of plane stress is satisfied by the stress function 3

2

2

3

4 3 2

F xy pxy y

d d

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

2

3

3 1

4 4 2

F Fxy pxy y

d d

02

2

x

Module5/Lesson3


2

2 3 3

3 2 2. 1.5

4 2

Fxy p Fp xy

y d d

and 3

22

4

3

4

3

d

Fy

d

F

yx

Hence the stress components are 2

2 31.5x

Fp xy

y d

02

2

xy

d

F

d

Fy

yxxy

4

3

4

33

22

(a) Variation of x

31.5x

Fp xy

d

When x = 0 and y = 0 or , xd p(i.e., constant across the section)

When x = L and y = 0, x p

When x = L and y = +d, 2

1.5xFL

pd

When x = L and y = -d, 2

5.1d

FLPx

Thus, at x = L, the variation of x

is linear with y.

The variation of x

is shown in the figure below.

Module5/Lesson3


Figure 5.14 Variation of stress x

(b) Variation ofz

02

2

xy

y is zero for all value of x and y

(c) Variation of xy

d

F

d

Fyxy

4

3

4

33

2

Thus, xy varies parabolically with z. However, it is independent of x, i.e., it's value is the

same for all values of x.

At d

Fy xy

4

3,0

At 04

3)(

4

3, 2

3 d

Fd

d

Fdy xy

Module5/Lesson3


Figure 5.15 Variation of shear stress xy

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.

)23(2

2

3ydxy

hd

W

Solution: 2

2

yx

Now, 2

366

2xyxyd

hd

W

y

xyxdhd

W

y126

2 32

2

xyxdhd

Wx 633

02

2

xy

and yx

xy

2

Y

d

d

L

X o

xy

3F

4d 3F

4d

Module5/Lesson3


= 2

366

2yyd

hd

W

= 2

333 yyd

hd

W

Also, 02

22

4

4

4

4

44

yxyx

Boundary conditions are

(a) dandyfory 00

(b) dandyforxy 00

(c) LandxforWdyh

d

xy 0.2.0

(d) WLMLxandxfordyhM

d

x ,00.2.0

(e) Lxandxfordyyh

d

x 00..2.0

Now, Condition (a)

This condition is satisfied since 0y

Condition (b)

033 223

ddhd

W

Hence satisfied.

Condition (c)

hdyyydhd

Wd

2330

2

3

dyyydd

Wd

0

2

333

2

d

ydy

d

W

0

32

3 2

32

Module5/Lesson3


33

3 2

32d

d

d

W

2.

2 3

3

d

d

W

= W

Hence satisfied.

Condition (d)

hdyxyxdhd

Wd

2630

3

dxyxyd

d

W0

2

333

2

= 0

Hence satisfied.

Condition (e)

ydyhxyxdhd

Wd.263

0 3

d

xyxdy

d

W

0

32

32

2

32

33

32

2

32xd

xd

d

W

3

3 2

12xd

d

W

Wx

Hence satisfied

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Numerical Examples 4