(Autonomous) Model Answer: Summer 2017 - msbte.engg-info ...msbte.engg-info.website/sites/default/files/s17mo s3/17311 2017 Summer... · 1) The answers should be examined by key words

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

(Autonomous) (ISO/IEC - 27001 - 2005 Certified)

Model Answer: Summer 2017 Subject: Mechanics of Structures ---------------------------------------------------------------------------------------------------------------

Page No 1 / 21

Sub. Code: - 17311

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model answer

scheme. 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the

understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and Communication Skills.)

4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by the candidate and those in the model answer may vary. The examiner may give credit for any equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate’s answers and the model answer.

6) In case of some questions credit may be given by judgment on part of examiner of relevant answer based on candidate’s understanding.

7) For programming language papers, credit may be given to any other program based on

equivalent concept. ----------------------------------------------------------------------------------------------------------------------------- ----

Que.

No.

Sub.

Que. Model Answers Marks

Total

Marks

1

i)

Ans.

ii)

Ans.

Attempt any six of the following :

Define moment of inertia. State MI of triangular section about its

base.

Moment of Inertia: -

It is the second moment of area which is equal to product of area of

the body and square of the distance of its centroid from that axis, is

called as moment of Inertia.

OR

Moment of inertia of a body about any axis is defined as the sum of

second moment of all elementary areas about that axis.

MI of triangular section about base 3

baseI12

bh

Where, b = Base of triangle and h = Height of triangle

Calculate polar MI of solid circular shaft section having Dia. ‘D’

4

xx yy

p xx yy

4 4

p

4 4

p p

I = I for soild circular section. 64

Polar moment of inertia ,

I = I + I

I = 64 64

I = D Or I = 0.098170D32

D

D D

01

01

01

01

12

02

02




Page No 2 / 21

Sub. Code: - 17311

Que. No.

Sub. Que.

Model Answers Marks Total

Marks

1 iii)

Ans.

iv)

Ans.

v)

Ans.

vi)

Ans.

vii)

Ans.

Define ‘Modulus of rigidity’. State its SI unit.

Modulus of rigidity : - It is ratio of shear stress to shear strain, is called as Modulus of

Rigidity. Unit: - N/m2 Or Pascal Or N/mm2

Draw stress-strain curve for a ductile material showing important

points.

(Note: One mark for curve and one mark for important point shown

in curve)

State the relationship between linear strain and lateral strain.

Lateral strain is directly proportional to linear strain. When a

homogeneous material is loaded within its elastic limit, the ratio of

lateral strain to linear strain is called as Poisson’s ratio.

Lateral strain produced in any material is equal to product of

Poission’s ratio and linear strain of same material.

Define slenderness ratio.

It is the ratio of effective length of column to minimum radius of gyration, is called as slenderness ratio.

Give an example of suddenly applied load. Also write equation for

stress developed due to suddenly applied load.

Example: -

01

01

02

02

02

02

02

02

02




Page No 3 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

1.

viii)

Ans.

B)

i)

Ans.

1. Weight on balance.

2. Load in truck.

3. Person standing on weighing balance.

4. Load lifted by a small height and dropped on platform.

Equation for stress developed due to suddenly applied load, 2P

A

Where, P = Load, A= Cross section area

Define resilience and modulus of resilience.

Resilience: -

It is the energy stored in the body or material, when loaded within

elastic limit is called as strain energy or resilience.

Modulus of Resilience: -

It is the proof resilience per unit volume, called as modulus of

resilience is called as modulus of resilience.

OR

It is the maximum strain energy stored in body per unit volume is

called modulus of resilience.

Attempt any two of the following :

State any four assumptions made in theory of pure bending.

1. The material of the beam is homogeneous and isotropic i.e. the

beam made of the same material throughout and it has the

same elastic properties in all the directions.

2. The beam is subjected to pure bending that is shear stress is

totally neglected.

3. The beam material is stressed within its elastic limit and thus

obeys Hooke’s law.

4. The transverse sections which where plane before bending

remains plane after bending.

5. Each layer of the beam is free to expand or contact

independently of the layer above or below it.

6. Young’s modulus (E) for the material has the same value in

tension and compression.

½

mark

each

(any

two)

01

01

01

01

mark

each

(any

four)

02

02

08

04




Page No 4 / 21

Sub. Code: - 17311

Que. No.

Sub. Que.


Marks

1

ii)

Ans.

iii)

Ans.

7. The radius of curvature is large as compared to the dimensions

of the cross section

A circular section diameter 150 mm is subjected to shear force

10kN when used as a beam. Calculate average and maximum

shear stress and draw shear stress distribution diagram.

2 2 2

32

2

max

A= ( ) (150) 17671.458684 4

10 100.566 /

17671.458

4 40.566 0.754 /

3 3

avg

avg

D mm

Sq N mm

A

q q N mm

A column having diameter 300mm is 5 m long. Determine Euler’s

crippling load, if both end of column are fixed. Take E =

2x105N/mm2.

4 4

min

4

min

2

min

2

2 5

2

6

50002500

2 2

( ) (300)64 64

397607820.2

( )

2 10 397607820.2

(2500)

125575420.6

125.575 10

LLe mm

I D

I mm

EIP

Le

P

P N

P N

01

01

01

01

01

01

01

01

04

04




Page No 5 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

2.

a)

Ans.

b)

Ans.

Solve any two of the following :

A column having diameter 300 mm is 5 m long. Determine Euler’s

crippling load, if both ends of column are fixed. Take E=2x105

N/mm2.

PQ

3 2 4

1 1

3 2 4

8 8

10 4

I = MI of reactangle about PQ - MI of semicircle about PQ

1

12 128

1 = 600 400 (600 400) 200 (300)

12 128

128 10 1.99 10

1.26 10

PQ

PQ

PQ

I bd Ah D

I

I mm

Calculate Ixx for the T section having flange 200 x 20 mm and web

20 x 200 mm overall depth is 220 mm.

02

03

02

01

16

08




Page No 6 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

2

c)

Ans.

1 1 2 2

1 2

(200 20)10 (200 20) 120

(200 20) (200 20)

65 from top

a x a xY

a a

Y

Y mm

3 2 2 2

1 1 1 1 2 2 2 2

3 2 3 2

6 6

6 4

1 1

12 12

1 1 = 200 20 (200 20) 55 20 200 (20 200) 55

12 12

12.233 10 25.433 10

37.67 10

xx

xx

xx

xx

I b d Ah b d A h

I

I

I mm

i) Calculate the radius of gyration of steel pipe having external

diameter 22 mm and internal diameter 16 mm.

ii) Find the diameter of circular rod 2.4 m long when subjected to

an axial pull 15 kN, shows an elongation of 1 mm. Take E= 205

kN/mm2.

4 4

4 4

4

2 2

2 2

2

)

( )64

(22 16 )64

8282.024

,

( )4

(22 16 )4

= 179.07 mm

xxxx

yy

yy

xx yy

xx yy

i

IK

A

IK

A

I I D d

I I mm

Area

A D d

A

02

02

02

02

01

01

01

08

08




Page No 7 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

2 8282.024

179.07

= 6.80mm

xx yy

xx yy

K K

K K

3

2 3 2

2

2

3

3

)

15 15 10

2.4 2400

1

205 / 205 10 /

?

( )4

4

4

4 15 10 2400

205 10 1

223.5932859

14.953

ii

P kN N

t mm

l mm

E kN mm N mm

d

PLL

AE

PLA

E L

PLd

E L

PLd

E L

PLd

E L

d

d

d mm

01

01

01

02




Page No 8 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

3

a)

Ans.

Solve any two of the following:

A bar of uniform cross section area 100 mm2 is subjected to axial

forces as shown in fig 2. Calculate the net change in length of the

bar. Take E = 25 x 105 N/mm2.

2 5 2

3 3 3

3 3 3

100 2 10 ?

To calculate, P

1 4 - 2

5 - 2

3

- -

1 10 300 2 10 400 2 10 600- -

100 200 10 100 200 10 100 200 10

AB BC CD

AB BC CD

AB BC

A mm E N mm L

P

P

P kN

L L L L

PL PL PLL

AE AE AE

L

-0.015 0.04 - 0.06

-0.035

CD

L

L mm

01

01

02

03

01

16

08

.




Page No 9 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

3 b)

Ans.

A steel tube with 40 mm inside diameter and 4 mm thickness is

filled with concrete. Determine load shared by each material due

to axial thrust of 60 kN.

Take E steel = 210 N/mm2

E concrete = 14 x 103 N/mm2

3 2

3 2

2 2

2 2

2

2

2

2

3

3

:

40

4

60

210 10

14 10

2 40 2 4

48

4

48 404

552.92

4

404

1256.637

210 1015

14 10

..........( )

S

C

S

S

S

C

C

C

S

C

s c

s c

S C

Ss

C

Given

d mm

t mm

P KN

E N mm

E N mm

D d t

D mm

A D d

A

A mm

A d

A

A mm

Em

E

e e

iE E

E

E

15

c

s c

3

........( )

60 10 15 552.92 1256.637

S C

s S c C

c c

P P P ii

P A A

01

01

01

01

08




Page No 10 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

3

c)

Ans.

3

3

2

2

60 10 8293.8 1256.637

60 10

9550.437

6.2824

6.2824 1256.637

7894.74

7.89

15

15 6.2824

94.236

94.236 552.92

52104.96

52.104

c c

c

c

c c C

c

c

c

s c

s

s

s s S

s

s

s

N mm

P A

P

P N

P kN

N mm

P A

P

P N

P kN

i) A square rod 10 mm x 10 mm in cross section and 1 m long is at

200 C. Find free expansion of rod, if temperature to700 C. If this

expansion is prevented, find temperature stress developed in the

bar. Take E = 2 x 105N/mm2 and = 12 x 12-6 per 0C

ii) With a neat sketch show effective length of Column for various

end conditions. (min. four)

i)

3

0

1

0

2

5 2

6 0

0

2 1

1 1 10

20

70

2 10

12 10 /

70 20 50

L m mm

t C

t C

E N mm

C

t t t C

3 6

) ,

( )

1 10 12 10 50

( ) 0.6

i Free Expansion

L L T

L mm

01

01

01

01

01

01




Page No 11 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

3

5 6

2

) .

( )

2 10 12 10 50

( ) 120

ii Temp stress

E T

N mm

ii)

01

01

01

mark

each

08




Page No 12 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

4

a)

Ans.

b)

Ans.

Solve any two of the following

A metal rod 20 mm diameter and 2 m long when subjected to

tensile force of 60 kN shows an elongation of 2 mm and reduction

in diameter 0.006 mm. Calculate the modulus of elasticity and

modulus of rigidity.

3 5

2

5 2

5

PL

A L

60 10 2 10E=

20 24

1.91 10 N/mm

Lateral Strain

linear Strain

d 0.006

20 =

2

2000

0.3

E = 2G 1+

= 2G 1+ 0.3

1.91 10 2G 1+ 0.3

E

E

d

L

L

5

4 2

1.91 10

2 1.3

7.345 10 N/mm

G

G

A cube of 200 mm side is subjected to a compressive force of 3600

kN on all its faces. The change in the volume of cube is found to be

5000 mm3. Calculate the Bulk modules. If the = 0.28, Find the

Young’s modulus.

3

2

33

6 3

3600 10

200 200

90N/mm

200

8 10

P

A

V L

V mm

01

01

01

01

01

01

01

01

02

01

16

08




Page No 13 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

4

c)

Ans.

6

5 2

1 2

31 2

3 901 2 0.28

5000

8 10

E = 1.9 10 N/mm

x y zV

V E

EV

V

E

Draw SFD and BMD for the cantilever beam loaded as shown in

Fig. 3

L

R

L

SF Calculations:

SF at A = +50kN

C = +50kN

C = +50-30 =20 kN

B = +20kN

B = +20-20 = 0

BM Calculations:

BM at B = -15 kN-m

BM at C = -15-20 1.5 = - 45kN-m

BM at A =

-15-20 2.5-30 1 = - 95kN-m

01

01

02

01

02

03

01

02

08

08




Page No 14 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

5

a)

Ans.

Solve any two of the following.

A timber beam 15mm wide and 300mm deep is simply supported

over a span of 4m. It carries udl 10kN/m over entire span. Find

the maximum bending stress induced in the section. Draw bending

stress distribution diagram. Also find radius of curvature if E=1.4

kN/mm2.

2 3 2

max

3

3

6 4

E =1.4 kN/mm 1.4 10 /

150 , 300 , 4 , 10 /

: ;

1

12

1150 300

12

337.5 10

b

xx NA

xx NA

xx NA

N mm

b mm d mm L m w kN m

Find R

I I bd

I I

I I mm

2 2

max

6

max

300

2 2

150

10 420

8 8

20 10

dY

Y mm

wlM kN m

M N mm

b

Using relation,

M

I Y

01

½

01

½

1 ½

01

16

08




Page No 15 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

5

b)

Ans.

max

6

max 6

2

max

M

(20 10 )150

337.5 10

8.89 /

b

b

b

YI

N mm

b

b

3 36

6

M

E E OR

(1.4×10 ) (1.4×10 )= ×150 OR = ×337.5 10

8.89 (20 10 )

E

I Y R

R Y R IM

R R

23625

23.625

R mm

R m

A beam ABC supported at A and B such that BC as overhang. AB

= 3m, BC = 1m, span AB carried udl 10 kN/m and point load of 6

kN acts at point C. Draw shear force and bending moment

diagrams. Also locate point of contra flexure, if any.

y

A B

A B

B

L

R

Step1

Calculation of Reaction,

F =0

R + R = (10 3+6)

R + R 36

0

3R 3 10 3 6 4

2

23

36 23 13

Step 2

Shear force calculation ,

SF at A= +13kN

B = +13-10 3 17

B 17 23 6

A

B

A

M

R kN

R kN

kN

kN

L C 6

6 6 0

kN

C

01

½

01

01

01




Page No 16 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

5

A C

B

2

max

Step 3

Calculation of zero SF from A,

13 17

3

1.3 point

sec 0

13 10 0

1.3 support

Step 4

Bending Moment calculation,

M =M =0

M =-6 1= - 6kN-m

1.3M =13 1.3-10 8.45kN-m

2

M

X X

X m From A

SF at tion XX

x

x m from A

max

2

=8.45kN-m

Step 5

To locate point of contraflexure at section Y-Y,

0

13 10 02

13 5 0

Y= 2.6m from support A.

BM

YY

Y

01

01

01

01

01

01

08




Page No 17 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

5 C

Ans.

i) A simply supported beam of span ‘L’ carries central point load

‘W’. Draw SFD and BMD.

ii) Define shear force and bending moment. Write unit of each.

Also state relation between them.

i)

A B

x A

Step 1

Calculation of Reaction,

As, the load is at centre so,

support reaction are equal,

WR = R =

2

Step 2

Shear force calculation

a) S.F. at any section between A and C is,

WF =+R =

2

b) S.F. at any section

x B

between B and C is,

WF = -R = -

2

A B

max

Step-3 Bending Moment Calculation,

Beam is simply supported at the end A and B,

M =M =0

WM =Mc=+

2 2

WL Mc=

4

L

01

01

01

01

04




Page No 18 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

5

6

a)

Ans.

ii)

Shear force: - Shear force at any cross section of the beam is the

algebraic sum of vertical forces on the beam acting on right side or

left side of the section is called as shear force.

OR

A shear force is the resultant vertical force acting on the either side of

a section of a beam.

Unit :- kN or N

Bending Moment: - Bending moment at any section at any cross

section is the algebraic sum of the moment of all forces acting on the

right or left side of section is called as bending moment.

Unit: - kN-m or N-m

Relation between shear force and bending moment

dM = F

dx

The rate of change of bending moment at any section

is equal to the shear force at that section.

Solve any two of the following

A channel section as shown in Fig. carries shear force of 100kN at

a particular section. Calculate the ratio of average shear stress to

maximum shear stress.

max

2

3

2

3 3

3 3

7 4

:

100 ,

A = 2×80×20 + 160×20

A= 6400 mm

100 10

6400

15.625 / mm

1( )

12

1(80 200 60 160 )

12

3.285 10

avg

avg

avg

xx NA

xx NA

xx NA

Given

qSF kN Find

q

Sq

A

q N

I I BD bd

I I

I I mm

01

½

01

½

½

½

½

02

02

04

16

08




Page No 19 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

6

b)

Ans.

1 1 2 2

max

3

max 7

2

max

max

max

(80 20) 90 (20 80) 40

144000 64000

208000

100 10 208000

20 3.285 10

31.656 /

15.6250.493

31.656

0.493 0.50

avg

avg

AY a y a y

AY

AY

AY

SAYq

bI

q

q N mm

qRatio

q

q

q

A cast iron column 100 mm external diameter is an 80 mm

internal diameter 2m long. It is fixed at one end and hinged at

other end. Calculate the safe axial load by Rankine’s formula

taking factor of safety 3. Assume σc = 550 N/mm2 and Rankine’s

constant α= 1/1600.

2

min. xx. yy.

100 , 80 , 2 2000 ,

13, 550 / ,

1600

As, the column is fixed at one end and hinged at other end.

2000Efeective length,

2 2

1414.2

For hollow circular column,

I I I

c

Given

D mm d mm L m mm

FOS N mm

LLe

Le mm

4 4

4

min. xx. yy.

(100 80 )64

I I I 2898119.223mm

2 2

2

,

(100 80 )4

2827.433

Area

A

A mm

½

½

01

01

½

01

01

01




Page No 20 / 21

Sub. Code: - 17311

Que.

No.

Sub.


Total

Marks

C)

Ans.

2

2

2 2

2

2

2

cR 2

2

R

2898119.223

2827.433

32.0156

( ) 1 (1414.2)1 1

1600 32.0156

( ) 1 2.2195

By using Rankine's formula,

.AP =

( )1

550 2827.433P

2.2195

P 707644.2077

RankiSafe Load=

R

IK

A

K

K mm

LeaK

LeaK

LeaK

N

ne's crippling load

Factor of safety

707644.2077Safe Load=

3

Safe Load=233548.0692

Safe Load=233.548k

N

N

A weight of 2 kN falls on a collar attached at the lower end of a

vertical bar 3 m long and 25 mm in diameter. Calculate the height

of drop if the instantaneous stress developed is 120 N/mm2. Also

calculate corresponding elongation and strain energy stored in the

bar. Take E=2 X 105 N/mm2.

2

5 2

Given:

W = 2kN = 2000N

L=3m=3000mm

d=25mm

=120N/mm

2 10 N/mm

?, ?, ?

E

Find h l U

01

01

01

01

01

08




Page No 21 / 21

Sub. Code: - 17311

Que. No.

Sub. Que.


Marks

2

2

5

2 2 2

2

2

2

5

2

W W 2=

A A

2000 2000 2 2000 2 10120=

25 25 25 30004 4 4

120 4.074 (4.074) 543.2

120 4.074 (4.074) 543.2

(115.9256) 16.597 543.2

24.71

120 3000

2 10

1.8

2

WEh

AL

h

h

h

h

h mm

Ll

E

l mm

U

22

5

12025 3000

2 2 10 4

53014.376

53.014

ALE

U

U N mm

U N m

01

03

02

01

01

Documents

(Autonomous) Model Answer: Summer 2017 - msbte.engg-info ...msbte.engg-info.website/sites/default/files/s17mo s3/17311 2017 Summer... · 1) The answers should be examined by key words