Som Solution

7/29/2019 Som Solution

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1

CHAPTER 2

Simple Stresses and Strains

Problem 1. Tension test was conducted on a specimen and the following readings recorded:

Diameter = 22 mm

Gauge length of extensometer = 200 mm

Least count of extensometer = 0.001 mm

At a load of 22 kN, extensometer reading = 60

At a load of 36 kN, extensometer reading = 94

Yield load = 95 kN

Maximum load = 157 kN

Diameter at neck = 15 mm

Final length over 110 mm original length = 132 mm

Find Youngs Modulus, yield stress, ultimate stress, percentage elongation and percentage reduction inarea.

Solution: Diameter = 22 mm, Cross-sectional area =2

(22)4

= 380.132 mm2

Yield load = 95 kN = 95000 N

Yield stress = py =95000

380.132= 249.91 N/mm2. Ans.

Maximum load = 157 kN = 157000 N.

Ultimate stress =157000

380.132= 413.01 N/mm2. Ans.

For a load of (36 22) kN = 14 kN extension is (94 60) m = 34 0.001 mm.

From the relation

=PL

AE, we get

34 0.001 =314 10 200

380.132 E

E = 2.16643 105 N/mm2. Ans.

% Elongation =Increase in length

100Original length

=132 110

100110

= 20. Ans.

% Reduction in area =

2 2

2

(22) (15)4 4 100

(22)4

= 53.51. Ans.


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Problem 2. The composite bar shown in Fig. 1 is subjected to a tensile force of 30 kN. The extension

observed is 0.372 mm. Find the Youngs Modulus of brass, if Youngs Modulus of steel is 2 105 N/mm2.

Fig. 1

Solution: At any section in the bar load

P = 30 kN = 30,000 N

Total extension = s s b b

s s b b

P L P L

A E A E+

0.372 =3 3

2 5 2

30 10 400 30 10 300

30 2 10 204 4

bE

+

3

2

30 10 300

204

bE

=3

2 5

30 10 4000.372

30 2 104

= 0.2871

Eb =3

2

30 10 300

20 0.28714

= 0.99787 105 N/mm2. Ans.

Problem 3. The steel flat shown in Fig. 2 has uniform thickness of 20 mm. Under an axial load of 80 kN,

its extension is found to be 0.17 mm. Determine the Youngs Modulus of the material.

Fig. 2

Solution: t = 20 mm P = 80 kN = 80 103 N

= 0.17 mm L = 500 mmFrom the relation

= 1

1 2 2

log( )

PL b

tE b b b, we get

0.17 =380 10 80

log20 (80 40) 40E

E = 2.03867 105 N/mm2. Ans.Problem 4. Find the extension of the bar shown in Fig. 3 under an axial load of 20 kN.

TakeE= 200 GN/m2

.

Fig. 3


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Solution: P = 20 103 N

E =9

2

200 10

(1000)

N/mm2 = 2 105 N/mm2

= 1 + 2 + 3

=1 2 3

2 21 1 2 2

4 4 4

PL PL PL

d E d d E d E + + where d1 = 25 mm and d2 = 15 mm.

i.e. =2 2

4 400 400 400

25 1525 15

P

E

+ +

=3

5 2 2

4 20 10 400 400 400

25 152 10 25 15

+ +

= 0.444 mm. Ans.

Problem 5. The composite bar shown in Fig. 4 is made of steel in portionACand copper in portion CD.

The two materials are rigidly joined at C. Find the extension of the bar under the loading shown in Fig. 4.

TakeEs = 2 105 N/mm2 andEc = 1.1 10

5 N/mm2

Fig. 4

Solution: Taking sections throughAB,BCand CD, we find

P1 = 55 kN P2 = 25 kN and P3 = 25 kN.

Extension = 1 + 2 + 3

= 1 1 2 2 3 3

1 2 3s s c

P L P L P L

A E A E A E+ +

=3 3 3

2 5 2 5 2 5

55 10 300 25 10 300 25 10 400

30 2 10 30 2 10 20 1.1 104 4 4

+ +

= 0.459 mm. Ans.

Problem 6. Find the force P acting at C in the bar shown in Fig. 5. Find the extension of the bar if

E= 2 105MPa.

= 15 mm

60 kN

AB C

30 kN

= 20 mm

P

D

= 15 mm

80 kN

300 mm 400 mm 300 mm Fig. 5


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Solution: Since the body is in equilibrium, 60 + 30 P + 80 = 0

or P = 50 kN. Ans.

At any section inAB, considering the equilibrium of left hand side portion, we find

P1 = 60 kN

Similarly taking a section inBC, taking equilibrium of left side portion,

P2 = 30 kNTaking a section in CD and considering right hand portion, we find

P3 = 80 kN.

E = 2 105 N/mm2.

= 1 + 2 + 3

=3 3 3

2 5 2 5 2 5

60 10 300 30 10 400 80 10 300

15 2 10 20 2 10 15 2 104 4 4

+ +

= 1.379 mm. Ans.

Problem 7. A stepped bar of steel is held between two unyielding supports as shown in Fig. 6(a) and is

subjected to loads. P1 = 80 kN, P2 = 60 kN. Find the reactions developed at endsA andB.

300

300

P2

P1

150

C

A

D

B

(c)(b)(a)

RB

Fig. 6

Solution: Given P1 = 80 103 N P2 = 60 103 N

If support atB is removed (Fig. b)

Force in AC = 140 103 N

and Force in CD = 60 103 N, inDEit is zero.

Extension of the bar =3 3

2 2

140 10 60 100

40 404 4

E E

+ +

=2

200 150

404 E

... (1)

Shortening of the bar due to a forceRB [Fig. (c)],

=2 2

300 300

30 404 4

B BR R

E E

+


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=2 2

300 1 1

30 40

4

BR

E

+

... (2)

Equating equations (1) and (2), we get

3

2 22

200 10 150 300 1 1

30 40404 4

BR

E E

= +

RB = 36 103 N = 36 kN. Ans.

v = 0 RA + RB = 80 + 40, whereRA is reactor atA

RA = 120 36 = 104 kN. Ans.

Problem 8. A rigid bar ABCD is connected to steel bar at A andB and is having hinged support at C. At

free end a load of 40 kN is acting as shown in Fig. 7. Find the forces developed in the bars and deflections

of free end ifE= 2 105 N/mm2, diameter of rod atA is 30 mm and atB is 25 mm.

300 mm

B CA

400 mm

200 mm 200 mm 400 mm

40 kN

D

Fig. 7

Solution:

Let the rigid barABCD taking the positionA,B, CandD after the application of load.

Extension of the rod atB =BB = B

and extension of the rod atA =AA = A

A

B

=

400

200= 2

i.e. A = 2 B ... (1)

Let PA and PB be the forces in the rods atA andB respectively.

From equation (1), we get

2

400

304

AP

E

=2

3002

254

BP

E

i.e. PA = 2.16 PB ... (2)

Taking moment about C, we get


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PA 400 + PB 200 = 40 400

i.e. 2 PA + PB = 80 ... (3)

From equations (2) and (3), we get

2 2.16 PB + PB = 80

PB = 15.038 kN. Ans.

Hence PA = 2.16 PB = 32.482 kN. Ans.

A

=3

2 5

32.482 10 400

30 2 104

A A

A

P L

A E

=

= 0.0919 mm. Ans.

Problem 9. Three identical wires support a rigid bar ABCas shown

in Fig. 8. Determine the forces developed in each bar when a load of

15 kN is applied atD.

Solution: Let the loads shared by wires atA,B and Cbe PA, PB and

PCrespectively.

Since the bar is rigid,

Extension ofB =

1

2 (Extension ofA + Extension ofB)

i.e. B =2

A C +

BP L

AE=

1

2

A CP L P L

AE AE

+

i.e. PB =2

A CP P+ ... (1)

v = 0 PA + PB + PC= 15 ... (2)

i.e. 2 PB + PB = 15, since PA + PC= 2 PB

PB = 5 kN. Ans.

Taking moment aboutA, we get

PB a + PC 2a = 15 1.5a

i.e. PB + 2 PC = 22.5

PC =22.5

2

BP = 8.75 kN. Ans.

From equation (2), PA = 15 8.75 5 = 1.25 kN. Ans.

Problem 10. The compound bar shown in Fig. 9 consists of three materials and supports a rigid platform

weighing 80 kN. Find the stresses developed in each bar if the platform remains horizontal even after the

loading. Given,

Bar 1 Bar 2 Bar 3

Length 1000 mm 1500 mm 2000 mmC.S. Area 600 mm2 800 mm2 1000 mm2

Modulus of

Elasticity

2 105 N/mm2 1.2 105 N/mm2 1.0 105

N/mm2

3a

AB 15 kN

DC

a a/2 a/2 Fig. 8


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P = 80 kN

1 2 3

Fig. 9

Solution: Let the load shared by each bar be P1, P2 and P3.

P1 + P2 + P3 = 80 kN ... (1)

Since the platform remains horizontal, the compression of each bar will be the same. Let 1, 2 and 3

be the compressions of the bars respectively.

1 1

1 1

P L

A E= 2 2 3 3

2 2 3 3

P L P L

A E A E=

P1 =53 3 1 1 3

53 3 1

2000 600 2 10

10001000 1 10

P L A E P

A E L =

= 2.4 P3 ... (2)

Similarly P2 =5

3 3 2 2 35

3 3 2

2000 800 1.2 10

15001000 1 10

P L A E P

A E L

=

= 1.28 P3 ... (3)

From equations (1), (2) and (3), we get

2.4 P3 + 1.28 P3 + P3 = 80

P3 = 17.094 kN

Hence P1 = 2.4 17.094 = 41.025 kN

P2 = 1.28 17.094 = 21.88 kN

Stress in Bar 1 =

3

1

141.025 10

600P

A= = 68.376 N/mm2. Ans.

Stress in Bar 2 =3

2

2

21.88 10

800

P

A

= = 27.350 N/mm2. Ans.

and Stress in Bar 3 =3

3

3

17.094 10

1000

P

A

= = 17.094 N/mm2. Ans.

Problem 11. The barABCD is very rigid. It has pinpointed support atA and is supported by a steel wireBE

and copper wire CFas shown in Fig. 10. Find the stresses produced in steel and copper wires when a load

of 12 kN acts at free end. Find also the deflection of free end.

Take: AS

= 400 mm2, AC

= 600 mm2

ES

= 2 105 N/mm2, EC

= 1.2 105 N/mm2


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Fig. 10

Solution:

Let Ps and Pc be the forces in kN in the steel and copper wires respectively.

Taking moment about A,

Ps 200 + Pc 500 = 12 700

2 Ps + 5 Pc = 84 ... (1)From compatibility condition,

s

c

=

2

5or s =

2

5c

i.e. s s

s s

P L

A E=

2

5

c c

c c

P L

A E

5 5

250 2 400

5400 2 10 600 1.2 10

s cP P =

Ps = 0.71 Pc ... (2)

From equations (1) and (2),

2 0.71 Pc + 5 Pc = 84.0

Pc = 13.084 kN

Hence Ps = 0.71 13.084 = 9.2897 kN

Stress in copper wire =13.084 1000

600

= 21.8 N/mm2. Ans.

Stress in steel wire =9.2897 1000

400

= 23.25 N/mm2. Ans.

From compatibility condition,

d

s

=

700

200

or d = 72

s or d= 72

s s

s s

P LA E

i.e. s

=3

5

7 9.289 10 250

2 400 2 10

= 0.1018 mm. Ans.


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Problem 12. The composite bar shown in Fig. 11 is 0.2 mm short of

distance between the rigid supports at room temperature. What is the

maximum temperature rise which will not produce stresses in the

bar? Find the stresses induced when temperature rise is 40C.

Given: As:Ac = 4 : 3

s = 12 106/C c = 17.5 10

6/C

Es = 2 105 N/mm2 Ec = 1.2 10

5 N/mm2

Solution: Let maximum temperature rise be t

(i) Maximum free expansion permitted without stresses, being

induced = 0.2 mm.

0.2 = s t Ls + c t Lc

= t(12 106 300 + 17.5 106 200)

t = 28.169C. Ans.

(ii) When the rise of temperature = 40C,

Free expansion of the bar = c t lc + s t ls

= 17.5 106 40 200 + 12 106 40 300

= 0.244 mm. Expansion prevented = 0.284 0.20 = .084 mm

Since the same force, sayp, exists in the two portions of the rod,ps As= pcAc

ps =3

4

cc c

s

Ap p

A=

c s c c

s c

p l p l

E E+ = 0.084

5 5

3/ 4 300 200

2 10 1.2 10

c cp p +

= 0.084

i.e. pc = 30.090 N/mm2. Ans.

ps = 34

pc = 22.567 N/mm2. Ans.

Problem 13. A steel tube of 50 mm outer diameter and 10 mm thick is fitted into a copper tube of inner

diameter 50 mm and 10 mm thick. They are connected by using 20 mm diameter pins at the ends. If the

length of compound bar is 600 mm find the stresses produced in the tubes and pins when temperature is

raised by 25C.

Take: s = 12 106/C, c = 17.5 10

6/C

Es = 2 105 N/mm2, Ec = 1.2 10

5 N/mm2

Solution: t= 25C As =2 2

(50 30 )4

= 1256.637 mm2

Ac =2 2(70 50 )

4

= 1884.55 mm2

Since the free expansion of copper (c t l) is more than free expansion of steel (s t l), compressive

force Pc develops in copper and tensile force Ps develops in steel to keep them in the same position

(Referring to the Fig. 12). Say c.c.

300 mmSteel

Copper

200 mm

0.2 mm

Fig. 11


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Forces along longitudinal direction = 0, gives

Pc = Ps = P (say)

From the figure it is clear that,

s + c = c t l s t l

where s and c are the changes in lengths due to Ps and Pc.

i.e.s s c c

PL PL

A E A E+ = (17.5 12) 106 25 600

5 5

1 1600

1256.637 2 10 1884.55 1.2 10P

+

= 5.5 106 25 600

P = 16369.04 N

Stress in steel =16369.04

1256.637

s

s s

P P

A A= = = 13.026 N/mm2

Stress in copper =16369.04

1884.55

c

c c

P P

A A= = = 8.684 N/mm2

Shear stress in pins =2

16369.04

22 20

4p

P

A=

= 26.05 N/mm2. Ans.

Problem 14. At room temperature the gap between bar A and barB shown in Fig. 13 is 0.25 mm. What

are the stresses induced in the bars, if temperature rise is 35C?

400 mm 400 mm

BA

0.25 mm

Fig. 13

Given: Aa = 1000 mm2, Ab = 800 mm

2

Ea = 2 105 mm2, Eb = 1 10

5 mm2

a = 12 106/C, b = 23 10

6/C

La = 400 mm, Lb = 300 mm.

Solution: Free expansion of barA = A = a t LA

= 12 106 35 400

Free expansion of barB = B = b tLB

= 23 106 35 300.

A + B = 0.4095 mm.


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But gap is only 0.25 mm.

Expansion prevented = 0.4095 0.25 = 0.1595 mm.

Let the force produced due to this prevention be P which is compressive in bothA andB.

a b

a a b b

PL PL

A E A E+ = 0.1595

5 5400 300

1000 2 10 800 1 10P P +

= 0.1595

P = 27739.13

Hencepa =27739.13

1000 pa = 27.739 N/mm

2. Ans.

and pb =27739.13

800= 34.674 N/mm2. Ans.

Problem 15.AB is a rigid bar and has hinged support at Cas shown in Fig. 14. A steel and an aluminium

bar support it at ends A and B respectively. The bars were stress free at room temperature. What are the

stresses induced when temperature rises by 40C.

Given: As = 1200 mm2, Aa = 800 mm

2

Es = 2 105 N/mm2, Ea = 1 105 mm2

s = 12 106/C, a = 23 10

6/C

1000 mm

A B

1000 mm 800 mm

Steel

Aluminium500 mm

C

Fig. 14

Solution:

A

B

a a atL

s s s tL

Let Ps and Pa be the forces developed in the steel and aluminium respectively. Taking moment about C,

we get

Ps 1000 = Pa 800

or Ps = 0.8 Pa ... (1)

Free expansion of steel = s t Ls

= 12 106 40 1000 = 0.48 mmFree expansion of aluminium = a t La

= 23 106 40 500 = 0.46 mm

Since force expansion of aluminium is more than that of steel, tensile force develops in steel bar and

compressive force develops in aluminium bar. Deformed shape will be as shown in figure.


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From the property of similar triangles,

1000

s s st L =800

a a at L

4 (s s t Ls) = 5 (a t La a)

4 s + 5 a = 5 a t La + 4 s t Ls

4 5s s a a

s s a a

P L P L

E A E A + = 40 (5 23 106 500 + 4 12 106 1000)

5 5

4 1000 5 5004

5 2 10 1200 1 10 800

a aP P +

= 40 (57.5 + 48) 103

Pa = 94654.5 N

Ps =4

94654.55

= 75723.6 N

Stress in steel =75723.6

1200 1200

sP = = 63.103 N/mm2. Ans.

Stress in aluminium =94654.5

800 800

aP = = 118.318 N/mm2. Ans.

Problem 16. A bar of rectangular cross-section 20 mm 50 mm is 400 mm long and is subjected to an

axial tensile load of 80 kN. If modulus of elasticity and modulus of rigidity of the material of bar are

1 105 N/mm2 and 0.4 105 N/mm2, find the bulk modulus and changes in dimensions and volume.

Solution: E= 1 105 N/mm2 G = 0.4 105 N/mm2

Tensile load P = 80 103 N

From the relation

E = 2G (1 + ), we get

1 105 = 2 0.4 105 (1 + )

1 + = 1.25 or = 0.25. Ans.

Let Kbe bulk modulus,

From the relation,

E = 3K(1 2), we get

K =51 10

3(1 2 ) 3(1 2 0.25)

E =

= 0.667 105 N/mm2. Ans.

Axial load = P = 80,000 N

x = 580,000 400

20 50 1 10

PL

AE

=

= 0.32 mm

ex =0.32 0.32

400L= = 0.0008

ey; ex = ez = cx = 0.25 0.0008 = 0.0002 (tensile)

y

= cy

Ly

= 0.0002 50 = 0.01 mm (comp). Ans.

z = czLz = 0.0002 20 = 0.004 mm (comp). Ans.

Change in volume dv = b tx l bz l t y

= 50 20 0.32 400 50 0.004 400 20 0.01

= 160 mm3.


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Fig. 15

Problem 17. A bar 600 mm long is having square cross-section of size 50 mm 50 mm. If the bar is

subjected to an axial tensile load of 120 kN and lateral compression of 600 kN on faces of sizes

50 mm 600 mm, find the changes in size and volume.

Given:E= 2 105 N/mm2 and = 0.3

Solution:

Fig. 16

L = 600 mm b = 50 mm t= 50 mm

px =3120 10

50 50

= 48 N/mm2 (tensile)

py =3600 10

50 600

= 20 N/mm2 (compressive)

pz =3600 10

50 600

= 20 N/mm2 (compressive)

Volumetric strainV

V

=

1

E(px + py + pz) (1 2 )

=1

E(48 20 20) (1 2 0.3)

=5

18 0.4

2 10

V =5

1

2 10 8 0.4 600 50 50 = 24 mm3. Ans.

ex =yx z

pp p

E E E

=48 ( 20) ( 20)

0.3 0.3E E E

=5 5

1 60(48 6 6)

2 10 2 10+ + =


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x = cx Lx = 560

6002 10

= 0.18 mm. Ans.

ey =y x z

p p p

E E E

=1

E[ 20 0.3 48 0.3 ( 20)] =

28.4

E

y = 528.4

502 10

= 0.0071 mm

= 0.0071 mmreduction. Ans.

z = 0.0071 mm reduction. Ans.

Problem 18. A bar of 25 mm diameter is tightly fitted into a tube. Find the stresses in the bar and changes

in its volume due to a compressive force of 60 kN in the bar if the tube restrains 50 per cent of expansion in

diameter. Take length of the bar = 400 mm,E= 2 105 N/mm2 and = 0.3.

Solution: Increase in diameter, if not restrained

= xp

dE

... (1)

If restrained by tube, let the stresses developed iny andz directions be p

Increase in diameter = x xp p p

dE E E

+

... (2)

The condition specified is eqn. (2) is half of eqn. (1).

i.e. xp p p

E E E + =

1

2

xp

E

Now px =2

60 1000

154

= 122.231 = 0.3 (given)

0.3 122.231 p + 0.3p =1

0.3 122.2312

0.7p = (0.3 0.15) 122.231

p = 26.192 N/mm2. Ans.

v

v

=

1

E(px +py +pz) (1 2 )

=5

1

2 10= (122.231 + 26.192 + 26.191) (1 2 0.3)

=69.846

E

v =5

69.846

2 10 /4 252 400

i.e. v = 68.571 mm3. Ans.

Problem 19. The diameter of a bar of 500 mm length varies uniformly from 60 mm at one end to 100 mm

at the other end. Find the strain energy stored in the bar when an axial force of 75 kN acts.

TakeE= 2 105 N/mm2

Solution: L = 500 mm, d1 = 100 mm d2 = 60 mm


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E= 2 105 N/mm2 P = 75 103 N.

Extension of the bar =1 2

4 PL

d d E

=3

5

4 75 10 500

100 60 2 10

= 0.03978 mm.

Strain energy stored in the bar

=1

2P

=1

2 75 103 0.03978

= 1492.08 N-mm. Ans.

Problem 20. Compare strain energy stored in barA with that of barB in Fig. 17 when

(a) same load P acts on them.

(b) the maximum stress produced is the same.

Assume both are made up of same material.d = 30 mm1 d = 20 mm

2 d = 10 mm3

400 mm 400 mm 400 mm

Bar A

d = 25 mm1

d = 15 mm2

d = 25 mm3

400 mm 400 mm 400 mm

Bar B

Fig. 17

Solution: (a) When bar applied load is same:

SE stored in bar A = SEstored in the three portions

= SE1 + SE2 + SE3

=2

2 21 12 2 3 3

1 1 1

2 2 2

p Vp V p V

E E E+ +

=

2

2 2

2 2

130 400 20 400

2 4 430 20

4 4

P P

E

+

+

22

2

10 400410

4

P


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=26.93

2

P

E

SE stored in barB:

=

2 22 2

2 2

12 25 400 15 400

2 4 425 154 4

P P

E

+

=2

3.8922

P

E

of bar

of bar

SE A

SE B=

6.93

3.892= 1.7805. Ans.

(b) When maximum stress produced is the same:

Let p be the maximum stress produced. It is in the 10 mm diameter portion of barA.

Stress in 20 mm diameter portion

=

2

2

104

420

4

pp

=

and stress in 30 mm portion

=

2

2

104

9304

pp

=

SE in barA

=2

2 4

p

E

102 400 +

221 20 400

4 2 4

p

E

+

2

21 30 4009 2 4p

E

=242760.566

2

p

E

For bar B:

Maximum stress produced is in 15 mm diameter portion. It is p.

Stress in 25 mm portion = .

2

2

1594

25254

p

p

=

SE in barB

=22

2 2915 400 (25) 400 22 4 25 4

pp

E

+

=2121579.63

2

p

E


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SE in bar A

SE in bar B=

2

2

42760.566

2

121579.63

2

p

E

p

E

= 0.3517. Ans.

Problem 21. A steel rod 20 mm diameter and 400 mm long has a collar at lower end and is fixed at top. A

load of 60 N falls freely along the rod and strikes the collar. If the instantaneous stress is not to exceed 250

N/mm2, find the maximum height from which the load can be allowed to strike the collar. How much is the

stress produced, if the load falls from a height of 200 mm? TakeE= 2 105 N/mm2.

Solution: Given diameter of the rod = 20 mm

Cross-sectional area = 2204

= 314.15 mm2 = 314.15 mm2

Length of the bar = 400 mm

Falling weight P = 60 N.

Induced stressf= 250 N/mm2

Let maximum height be equal to h

f =2

1 1P AE h

A PL

+ +

250 =560 2 314.15 2 10

1314.15 60 400

h +

or250 314.15

60

= 1 1 5235.83h+ +

h = 326.75 mm. Ans.

(b) If the height of fall is = 200 mm. Then

p =2

1 1P EAh

A PL

+ +

= 560 2 2 10 314.15 2001 1314.15 60 400

+ +

=60

[1 1023.312]314.15

+

= 195.63 N/mm2.

Som Solution

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