1404650711_Strength of Materials 2-2 Set-2 (K)

S.11Strength of Materials (April/May-2013, Set-2) JNTU-Kakinada

( JNTU-Kakinada ) B.Tech. II-Year II-Sem.

Code No.: R22012/R10

II B.Tech. II Semester Regular Examinations

April/May - 2013

STRENGTH OF MATERIALS( Civil Engineering )

Time: 3 Hours Max. Marks: 75

Answer any FIVE Questions

All Questions carry Equal Marks

- - -1. A simply supported beam of span 6 m is subjected to a concentrated load of 18 kN at 4 m from left support. E = 200

GPa and I = 15 × 106 mm4. Calculate,(a) The position and value of the maximum deflection(b) Slope at mid-span(c) Deflection under load. [7+4+4] (Unit-I, Topic No. 1.4)

2. (a) Calculate the minimum thickness of the shell of a thin cylinder of 1 m diameter, if it is to withstand an internalpressure of 2 MPa and the longitudinal stress is not to exceed 30 MPa and the hoop stress is not to exceed 40MPa. (Unit-II, Topic No. 2.2)

(b) Derive the Lame’s equations for a thick cylinder using normal conventions. [7+8] (Unit-II, Topic No. 2.2.1)

3. A square element of a thin plate subjected to a compressive stress of 6 MPa in x-direction, a tensile stress of 12 MPain y-direction and a shear stress of 10 MPa (clockwise). Determine the principal stresses and their directions byconstructing Mohr’s circle. Also find the normal stress and shear stress on the diagonal plane of the square element.[15] (Unit-III, Topic No. 3.2)

4. A solid shaft transmits 250 kW at 100 r.p.m. If the shear stress is not to exceed 75 MPa, what should be the diameterof the shaft? If this shaft is to be replaced by a hollow one whose internal diameter is 0.6 times the outer diameter,determine the size and the percentage saving in weight, the maximum shear stress being the same. [15](Unit-IV, Topic No. 4.1.3)

5. A solid steel column and a hollow steel column, both having same length and cross-sectional area, are fixed at theends. If the internal diameter of hollow column is 2/3 of its external diameter, find the ratio of buckling strengths ofsolid column to hollow column. [15] (Unit-V, Topic No. 5.1)

6. A concrete wall of rectangular cross-section is 1 m thick and 2.5 m high. It has to retain water up to a height of 2.25m. Determine the stress intensities at the base, if the concrete weighs 25 kN/m3. [15] (Unit-VI, Topic No. 6.1)

7. A curved beam, rectangular in cross-section is subjected to pure bending with couple of 400 N-m. The beam has awidth of 20 mm and depth of 40 mm and is curved in a plane parallel to the depth. The mean radius of curvature is 50mm. Find the position of the neutral axis and the ratio of the maximum stress to the minimum stress. Also plot thevariation of the bending stress across the section. [15] (Unit-VII, Topic No. 7.2)

8. Determine the forces in all the members of the truss as shown in the figure. P1 = 40 kN and P2 = 20 kN. [15](Unit-VIII, Topic No. 8.1)

1.5 m 1.5 m 1.5 m 1.5 m

2 m

A EFG

B C D

P1 P2

1.5 m 1.5 m 1.5 m 1.5 m

2 m

A EFG

B C D

P1 P2Figure

Set-2Solutions

S.12 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014

B.Tech. II-Year II-Sem. ( JNTU-Kakinada )

SOLUTIONS TO APRIL/MAY-2013, SET-2, QPQ1. A simply supported beam of span 6 m is

subjected to a concentrated load of 18 kN at 4m from left support. E = 200 GPa and I = 15 ×106 mm4. Calculate,

(a) The position and value of the maximumdeflection

(b) Slope at mid-span

(c) Deflection under load.

Answer : April/May-13, Set-2, Q1 M[7+4+4]

Given that,

Concentrated load, P = 18 kN

Clear span, l = 6 m

Moment of inertia, I = 15 × 106 mm4

Modulus of elasticity, E = 200 GPa

(a) Position and Value of Maximum Deflection

Position of load on the span,

h = 3

2l =

3

l

6 m

M N4 m 2 m

z

x y

P = 18 kN

6 m

M N4 m 2 m

z

x y

P = 18 kN

h = 3

6

46.3=h from left support

Maximum deflection,

δmax

= EIl

ylPy

29

)–( 2/322

= 6101520039

)1000()2–6(2186

32/322

×××××××

mm224.23max =δ

(b) Slope at Mid-span

Slope at mid span,

iy

= EIl

ylPy

6

)( 22 −

= 2 2

6

18 2 (6 – 2 )

6 200 15 10 6

× ×× × × ×

× (1000)3

10.667yi∴ =

(c) Deflection Under Load

yz=

EIl

yPx

3

22

= 610152003

24186

22

××××××

× (1000)3

21.33 mmzy∴ =

Result

(a) Position and deflection at maximum,

h = 3.16 m from left support

δmax

= 23.224 mm

(b) Slope at mid span,

iy

= 10.667

(c) Deflection under load,

yz= 21.33 mm.

Q2. (a) Calculate the minimum thickness of theshell of a thin cylinder of 1 m diameter,if it is to withstand an internal pressureof 2 MPa and the longitudinal stress isnot to exceed 30 MPa and the hoopstress is not to exceed 40 MPa.

Answer : April/May-13, Set-2, Q2(a) M[7]

Given that,

Diameter of thin cylinder shell, d = 1 m

Internal pressure, P = 2 MPa

Longitudinal stress, f2 = 30 MPa

Hoop stress, f1 = 40 MPa



We known that,

Internal pressure, Px =

2x

b – a

Hoop stress, fx

= 2x

b + a

At x= r1 = 0.5 m, P

x = 2 MPa

⇒ 2 = 25.0

b – a

⇒ a = 25.0

b – 2 .. (1)

At x = r2, P

x = 0

⇒ 22r

b – a = 0

a = 22r

b... (2)

Hoop stress at x = r1 = 0.5, f

1 = 40 MPa

f0.5

= 2x

b + a

⇒ 40 = 25.0

b + a ... (3)

Substituting equation (1) in equation (3), we get,

⇒ 40 = 25.0

b +

25.0

b – 2

⇒ 2

2

0.5

b= 42

⇒ 2b = 42 × 0.52

b = 5.25 m

Substituting b = 5.25 m in equation (3), we get,

40 = 25.0

b + a

40 = 2

5.25

0.5 + a

a = 40 – 21

a = 19 m

Substituting a = 19 m, b = 5.25 m in equation (2), weget,

a = 22r

b

19 = 22

25.5

r

22r = 0.275

r2= 0.525

Thickness (t) = r2 – r

1

= 0.525 – 0.5= 0.025 m= 25 mm thickness

25 mmt∴ =

(b) Derive the Lame’s equations for a thickcylinder using normal conventions.

Answer : April/May-13, Set-2, Q2(b) M[8]For answer refer Unit-II, Q25.

Q3. A square element of a thin plate subjectedto a compressive stress of 6 MPa in x-direction, a tensile stress of 12 MPa in y-direction and a shear stress of 10 MPa(clockwise). Determine the principal stressesand their directions by constructing Mohr’scircle. Also find the normal stress and shearstress on the diagonal plane of the squareelement.

Answer : April/May-13, Set-2, Q3 M[15]Given that,

Compressive stress in x-direction,σ

x = 6 MPa

Tensile stress in y-direction,σ

y = 12 MPa

Shear stress acting in clockwise direction,τxy = 10 MPa

Principal StressesMajor Stresses

σ1=

2yx σ+σ

+ 2

2

2

–xy

yx τ+

σσ

= 2

126– + +

22

102

12–6– +

σ1= 16.453 MPa (Tension).



Minor Stresses

σ2=

2yx σ+σ

– 2

2

2

–xy

yx τ+

σσ

= 2

126– + –

22

102

12–6– +

σ2= – 10.453 MPa (compression).

Angle

Tan 2θ = yx

xy

σστ–

2

= 12–6–

102×

tan 2θ = – 1.11 [Q tan(–θ) = – tanθ]

2θ = – 47°59' or 132°1' θ = – 23°59' or 66°0' θ

1= 23°59', θ

2 = 66° (Change in directions)

Maximum shear stress,

τmax

= 2

– 21 σσ

= 2

)453.10(––453.16

τmax

= 13.453 kNThe direction of maximum stress is,

τmax

= 13.453 kN

θ' = 45° + 23°91' θ' = 68°59'

Graphical Method of Mohr’s Circle

2θ1= 132°1'

2θ2

Z

Y

XF

τmax = 13.453 kN

+ 47°59'

D

G

E C BO

2θ1

σ2 = 10.45 MPa(Compressive)

σ1 = 16.45 MPa(Tensile)

A

2θ1= 132°1'

2θ2

Z

Y

XF

τmax = 13.453 kN

+ 47°59'

D

G

E C BO

2θ1

σ2 = 10.45 MPa(Compressive)

σ1 = 16.45 MPa(Tensile)

A

Construction Steps

1. From the figure drawn, we have plotted as,

OF = σ1 = 16.45 MPa which is under tension and

OE = σ2 = 10.45 MPa (Compression).

2. Draw perpendiculars AG and CD from A and C tobisect AC at B.

3. With radius BG and B as centre draw a circle.

4. The required Mohr circle is formed.

Q4. A solid shaft transmits 250 kW at 100 r.p.m. Ifthe shear stress is not to exceed 75 MPa, whatshould be the diameter of the shaft? If thisshaft is to be replaced by a hollow one whoseinternal diameter is 0.6 times the outerdiameter, determine the size and thepercentage saving in weight the maximumshear stress being the same.

Answer : April/May-13, Set-2, Q4 M[15]

Given that,

Power transmitted, P = 250 kN

Number of rotations, N = 100 r.p.m

Allowable shear stress,

τ = 75 MPa

or

= 75 N/mm2.

= 75 × 106 N/m2

(i) Diameter of Solid Shaft

We known that,

Power transmitted, P = 100060

2

×πNT

250= 31060

1002

×××π× T

T = 1002

2501060 3

×π×××

T = 23873.24 N-m

Torque, T = 23873.24 N-m

But,

Torque, T = τ.16

π × d

S3

23873.24= 75 × 106 × 16

π × d

S3



dS3= 610726.14

24.23873

×

dS3= 1.6211 × 10– 3

dS= 0.1174 m

dS= 117.4 mm

Diameter of solid shaft, dS = 117.4 mm

Hollow Shaft

Assuming that length, material and maximum shearstress at both the shafts are same.

Torque at hollow= Torque at the solid shaft,

TH

= TS

H

HH

D

dD )–(16

44πτ= τ .

16

πD

S3

DS3 =

H

HH

D

dD 44 –

But,

Given that,

dH

= 0.6 DH

... (1)

DS3 =

H

HH

D

DD 444 )6.0(–

1.6211 × 10– 3 = (1 – 0.64) 3HD

3HD = 1.86 × 10–3 [Q D

s = d

s]

DH

= 0.123 m

DH

= 123 mm

Substituting DH

= 123 mm in equation (1), we get,

dH

= 0.6 × 123

= 73.8 mm

(ii) Percentage Saving in Weight

% saving =

S

H

W

W–1 × 100

So,

S

H

W

W=

SSS

HHH

WLA

WLA

..

..

Hence lengths are same,

S

H

A

A=

2

22

4

)–(4

S

HH

D

dD

π

π

][ SH LL =Q

= 2

2

4.117

8.73–123 2

S

H

A

A= 0.7025

S

H

W

W=

S

H

A

A= 0.7025

% Saving = (1 – 0.7025) × 100

= 29.75%

Result

Diameter of solid shaft, DS = 117.4 mm

% saving in weight = S

H

W

W

= 29.75%

Q5. A solid steel column and a hollow steelcolumn, both having same length and cross-sectional area, are fixed at the ends. If theinternal diameter of hollow column is 2/3 ofits external diameter, find the ratio ofbuckling strengths of solid column to hollowcolumn.


Given that,

Internal diameter of hollow column is 2/3 of itsexternal diameter.

Let,

d1 = External diameter of hollow column

d2 = Internal diameter of hollow column

d = Diameter of solid steel column

Area of solid steel column = 4

πd2

Area of hollow column = 4

π)–( 2

221 dd



But given,

Internal diameter, d2 = 3

2d1

Let us assume,

Area of steel column and hollow column are same.

d2 = )–( 22

21 dd

= 21d –

9

421d

d2 = 9

521d

In solid column,

Buckling strength = 2

22

2

2

l

EAk

l

EI SS π=π

In hollow column,

Buckling strength = 2

22

2

2

l

EAk

l

EI hh π=π

Where, ks = Radius of gyration at solid column

kh= Radius of gyration at hollow column.

Ratio of buckling strength,

s

h

P

P= 2

2

s

h

k

k

Radius of gyration at hollow column,

2hk =

column hollow of Area

inertia ofMoment

= )–(

4

)–(64

22

21

42

41

dd

dd

π

π

2hk =

16

)( 22

21 dd +

... (1)

We know that,

d2=

3

2d

1... (2)

Substituting equation (2) in equation (1), we get,

2hk =

1694 2

121 dd +

= 144

13 21d

2hk =

144

13 .

5

9d2

∴ 2hk =

2

80

13d

At solid column,Radius of gyration,

2sk =

2

4

4

64

d

d

π

π

= 16

2d

2sk =

16

2d

Ratio of buckling strength,

s

h

P

P= 2

2

s

h

k

k

=

2

2

1380

16

d

d

= 5

13

∴ s

h

P

P= 2.6

ResultThe ratio of bucking strength = 2.6.

Q6. A concrete wall of rectangular cross-sectionis 1 m thick and 2.5 m high. It has to retainwater up to a height of 2.25 m. Determinethe stress intensities at the base, if theconcrete weighs 25 kN/m3.

Answer : April/May-13, Set-2, Q6 M[15]Given that,

Height of concrete wall, h = 2.5 m



Height of water retained, hR = 2.25 m

Weight of concrete, p = 25 kN/m3

Thickness, t = 1 m

Let, weight of water be 10 kN/m3

1 m

2.5 m

Concretewall

2.25 m

Water

1 m

2.5 m

Concretewall

2.25 m

Water

Total Pressure per ‘m’ Length of Wall

P = 2

2RWh

= 2

25.210 2×

P = 25.312 kN

The resultant cuts the base at a point per ‘m’ lengthweight of concrete,

W = p × t × h

= 25 × 1 × 2.5

W = 62.5 kN

Let ‘a’ be the distance between the point where theresultant cuts the base and centre of gravity of the damsection is,

a = W

P ×

3Rh

= 5.62

312.25 ×

3

25.2

a = 0.303 m

We know that,

Eccentricity, ‘e’ = x = 0.303 m

Maximum Intensity of Stress at the Base

σmax = b

W

+

b

e61

= 1

5.62

×+

1

303.061

(Q t = b = 1)

= 176.125 kN/m2

σmax

= 176.125 kPa (Compression).

Minimum Intensity of Stress at the Base

σmin

= b

W

b

e6–1

= 1

5.62

×

1

303.06–1

= –51.125 kN/m2

σmin

= 51.125 kPa (Tension).

Stress intensities at the base,

σmax

= 176.125 kPa

σmin

= 51.125 kPa

Q7. A curved beam, rectangular in cross-sectionis subjected to pure bending with couple of400 N-m. The beam has a width of 20 mmand depth of 40 mm and is curved in a planeparallel to the depth. The mean radius ofcurvature is 50 mm. Find the position of theneutral axis and the ratio of the maximumstress to the minimum stress. Also plot thevariation of the bending stress across thesection.


Given that,

Width of the beam = 20 mm

Depth of the beam, D = 40 mm

Mean radius, R = 50 mm

Magnitude of pure bending with couple = 400 N-m

Required,

(i) Position of N.A (neutral axis)

(ii) Ratio of maximum stress to minimum stress

(iii) Plot of variation of bending stress across the section.

Area of the cross section = 20 mm × 40 mm

= 800 mm2



(i) Position of Neutral Axis

Let y0 be the distance of N.A from centroidal axis,

y0

= – 22

2

hR

Rh

+

For a rectangular section h2 is given by,

h2 = D

R3

ln

+

DR

DR

–2

2– R2

= 40

)50( 3

ln

×+×

40–502

40502 – (50)2

= 147.8 mm

The distance of N.A from centroidal axis is,

y0

= 2

–50 147.8

50 147.8

×+

y0

= – 2.79 mm

The negative sign indicates that the N.A is below the centroidal axis.

(ii) Determination of Maximum Stress to Minimum Stress Ratio

The bending stress (σ) at a distance ‘y’ from centroidal axis at any layer is given by,

σ = RA

M

+

+yR

y

h

R2

2

1

The maximum stress occurs when y is negative for the given values of M, R, A.

∴ The maximum stress occurs at the extreme bottom layer where y = – 20 mm.

y = 2

D =

2

40 = 20 mm

σmax

= 80050

1000400

××

250 –201

147.8 50 – 20

+

= – 102.8 N/mm2

= 102.8 MPa (Compressive)

σmin

= 80050

1000400

××

250 201

147.8 50 20

+ +

= 58.3 N/mm2 (tensile)

The minimum bending stress occurs at the inner side where y = 20 mm.

∴ The ratio of σmax

and σmin

is,

min

max

σσ

= 3.58

8.102 = 1.763



(iii) Plotting of Stress Variation Across the Section

Bending stress, σ = RA

M

+

+yR

y

h

R2

2

1

At y = 0,

σ = 80050

1000400

××

++

050

0

8.147

501

2

σ0= 10[1 + 0] = 10 N/mm2

At y = 5 mm, σ = 10

++

550

591.161 = 25.37 N/mm2

At y = 10 mm, σ = 38.01 N/mm2

At y = 15 mm σ = 49.02 N/mm2

At y = 20 mm σ = 58.3 N/mm2

At y = – 5 mm σ = – 8.79 N/mm2

At y = – 10 mm σ = – 32.275 N/mm2

At y = – 15 mm σ = – 62.47 N/mm2

At y = – 20 mm σ = – 102.8 N/mm2

At N.A y = y0 = – 2.79 mm, σ = 0 N/mm2

20 mm

20 mm

20 mm

R = 50 mm

G

Centre of curvature

Axis of curvature

O

y0

58.3

– 102.8

y

49.02

38.01

25.37

– 8.79

– 62.47

– 32.37

– 20

– 15

– 10

– 5

0Stress

Centroidal axis

Neutral axis

20

15

10

5

20 mm

20 mm

20 mm

R = 50 mm

G

Centre of curvature

Axis of curvature

O

y0

58.3

– 102.8

y

49.02

38.01

25.37

– 8.79

– 62.47

– 32.37

– 20

– 15

– 10

– 5

0Stress

Centroidal axis

Neutral axis

20

15

10

5

Figure: Plotting of Stress Across the Section



Q8. Determine the forces in all the members ofthe truss as shown in the figure. P1 = 40 kNand P2 = 20 kN.

1.5 m 1.5 m 1.5 m 1.5 m

2 m

A EFG

B C D

P1 P2

1.5 m 1.5 m 1.5 m 1.5 m

2 m

A EFG

B C D

P1 P2

Figure


1.5 m 1.5 m 1.5 m 1.5 m

A EFG

B C D

P1 P2

1.5 m 1.5 m 1.5 m 1.5 m

A EFG

B C D

P1 P2

Given that,

Load, P1 = 40 kN

Load, P2 = 20 kN

Reactions

RA + R

E = 40 + 20

RA + R

E = 60 kN ... (1)

Taking moments about ‘E’ in clockwise direction,

40 × 1.5 + 20 × 4.5 – RA × 6 = 0

60 + 90 – RA = 0

RA = 25 kN

Substitute RA = 25 kN in equation (1), we get,

RA + R

E = 60

RE + 25 = 60

RE = 45 kN

By using method of joints.

At Joint A

θ = 53°8'A B

G

θ = 53°8'A B

G

Taking forces vertically ΣFy = 0

FAG

sin53°8' = RA

FAG

= '853sin

25

°

FAG

= 31.24 kN

Taking forcess horizontally

ΣFX = 0

FAB

= FAG

cos53°8'

= 31.24 × cos53°8'

FAB

= 18.74 kN

At Joint B

Free body diagram at joint ‘B’

A

G

CB

53°8'

A

G

CB

53°8'

Taking forces vertically,

FAB

sinθ + FBG

= 0

FBG

= – 18.74 × sin53°8'

FBG

= 14.99 kN (Compression)

Taking forces horizontally,

FAB

cosθ + FBC

= 0

FBC

= – 18.74 × cos53°8'




At Joint G

Free body diagram at joint G.

53°8'F

G

20 kN

A

CB

53°8'F

G

20 kN

A

CB

We known that,

FAG

= 31.24 kN


ΣFx= 0

FAG

+ FGF

= FGC

× cos53°8'

31.24 + FGF

= FGC

× cos53°8' ... (2)

Taking forces vertically,

ΣFY

= 0

P2

= FGC

× sin53°8'

[Q Load acting in FBG

direction P2 = 20 kN]

20 = FGC

× sin53°8'

FGC

= '853sin

20

°

FGC

= 24.99 kN (Tension)

Substituting FGC

in equation (2), we get,

31.24 – FGF

= 24.99 × cos53°8'

– FGF

= – 16.24 kN

FGF

= 16.24 kN (Tension)

At Joint F

Free body diagram of Joint ‘F’.

53°8'E

F

20

G

C D

53°8'E

F

20

G

C D

Taking forces vertically

ΣFY

= 0

⇒ P2 – F

FC sinθ = 0

[Q Load acting in FFD

direction P2 = 20 kN]

20 = FFC

sin θ

⇒ FFC

= '853sin

20

°

⇒ FFC

= 24.99 kN –~ 25

FFC

= 25 kN


ΣFX

= 0

FGF

+ FFE

+ FFC

cos 53° 8'

⇒ 16.24 + FFE

+ FFC

cos53°8' = 0

⇒ 16.24 + FFE

+ 25 cos53°8' = 0

FFE

= 31.24 kN (compression)

At Joint E

Free body diagram at joint ‘E’.

53°8'REF

D

53°8'REF

D


ΣFx

= 0

FFE

+ FED

× cos53°8' = 0

FED

= '853cos

–

°FEF

= '853cos

24.31–

°

FED


At Joint D

Free body diagram at joint ‘D’.

EF

C D

36°52'

EF

C D

36°52'




ΣFx= 0

– FCD

+ FED

× cos36°52' = 0

– FCD

= – 52.07 × cos36°52'

FCD

= + 41.657 kN (Tension)

At Joint C

Free body diagram at joint ‘C’.

θ

G F

B DC

θ

G F

B DC


ΣFx

= 0

FBC

+ FCD

× cos53°8'

FBC

= – FCD

× cos53°8'

FBC

= 24.99 kN (Compression).

Documents

1404650711_Strength of Materials 2-2 Set-2 (K)