1404650711_Strength of Materials 2-2 Set-1 (K)

S.1Strength of Materials (April/May-2013, Set-1) 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) Using the moment-area method, determine the deflection at the free end of a cantilever beam of span l subjectedto a load P at a distance a from the fixed end. El is constant. (Unit-I, Topic No. 1.5)

(b) Derive the differential equation of the elastic curve for a beam subjected to bending in a plane of symmetry.[5+10] (Unit-I, Topic No. 1.1)

2. (a) Derive expression for volumetric change in thin cylinder subjected to an internal pressure p.(Unit-II, Topic No. 2.1.1)

(b) A thick cylindrical pipe of outside diameter 300 mm and internal diameter 200 mm is subjected to an internal fluidpressure of 14 MPa. Determine the maximum hoop stress in the section. Draw the variation of the hoop stressacross the thickness of the pipe. [7+8] (Unit-II, Topic No. 2.2)

3. A square element of a thin plate subjected to a compressive stress of 5 MPa in x-direction, a tensile stress of 15 MPain y-direction and a shear stress of 10 MPa (clockwise). Determine the principal stresses and their directions by usinganalytical method. 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 closed coil helical spring is made of 12 mm diameter wire and is having mean diameter of 150 mm and 10 completeturns. The modulus of rigidity of the material of spring is 80 MPa. When a load of 500 N is applied, determine themaximum shear stress, strain energy stored, deflection produced and stiffness of the spring. [15](Unit-IV, Topic No. 4.2.2)

5. A built-up I section has the following dimensions,

Overall depth : 400 mm

Flanges : 300 mm × 50 mm

Web thickness : 30 mm

When it is used as a simply supported beam, it deflects by 10 mm when subjected to a load of 40 kN/m over its entirelength? Find the safe load, if the same section is used as a column with both ends hinged. Use Euler’s formula.Assume a factor of safety 1.75 and take E = 2 × 105 MPa. [15] (Unit-V, Topic No. 5.1.2)

6. The cross-section of a masonry pier is a hollow rectangle, the dimensions of external and internal rectangles being1200 mm × 800 mm and 900 mm × 500 mm respectively. A load of 300 kN in the vertical plane bisecting the 1200 mmwidth of pier is transmitted at an eccentricity of ‘e’. Calculate the maximum value of ‘e’ so that no tension is inducedin the section. [15] (Unit-VI, Topic No. 6.2)

Set-1Solutions

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014

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

7. Determine the maximum positive bending moment and maximum twisting moment in a semi-circular beam simplysupported on three equally spaced supports. Take radius of the centre line of the beam as ‘R’ and load per unit lengthof the beam as ‘w’. [15] (Unit-VII, Topic No. 7.2.2)

8. Determine the forces in all the members of the truss as shown in the figure. [15] (Unit-VIII, Topic No. 8.2)

30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

Figure



SOLUTIONS TO APRIL/MAY-2013, SET-1, QPQ1. (a) Using the moment-area method,

determine the deflection at the free endof a cantilever beam of span l subjectedto a load P at a distance a from thefixed end. El is constant.

Answer : April/May-13, Set-1, Q1(a) M[5]

Note : In the given reference replace W with P.

For answer refer Unit-I, Q24.

(b) Derive the differential equation of theelastic curve for a beam subjected tobending in a plane of symmetry.

Answer : April/May-13, Set-1, Q1(b) M[10]

For answer refer Unit-I, Q1.

Q2. (a) Derive expression for volumetricchange in thin cylinder subjected to aninternal pressure p.

Answer : April/May-13, Set-1, Q2(a) M[7]

For answer refer Unit-II, Q3.

(b) A thick cylindrical pipe of outsidediameter 300 mm and internal diameter200 mm is subjected to an internal fluidpressure of 14 MPa. Determine themaximum hoop stress in the section.Draw the variation of the hoop stressacross the thickness of the pipe.

Answer : April/May-13, Set-1, Q2(b) M[8]

Given that,

Outer diameter = 300 mm

Inner diameter = 200 mm

Internal pressure = 14 MPa

Required

(i) Maximum hoop stress in a section

(ii) Representation of variation of hoop stressacross thickness of the pipe.

150 mm

100 mm

150 mm

100 mm

Thickness

Let,

Px= Radial pressure

fx

= Hoop stress.

Px=

2x

B – A and f

x = 2x

B + A

At, x = 100 mm , P100

= 14 MPa

x = 0 mm , P0

= 0

x = 150 mm , P150

= 0

Radial stress Px at x = 100 is,

P100

= 2)100(

B – A

14 = 2)100(

B – A ... (1)

Px at x = 150 is,

P150

= 2)150(

B – A

0 = 2)150(

B – A ... (2)

On solving equations (1) and (2) we get,

A = 11.2

B = 252000

Hoop stress,

fx

= 2x

B + A

f100

= 2)100(

252000 + 11.2

= 36.4 N/mm2

f150

= 2)150(

252000 + 11.2

= 22.4 N/mm2



The maximum hoop stress is developed at the innersurface.

f110

= 2)110(

252000 + 11.2 = 32.026 N/mm2

f120

= 2)120(

252000 + 11.2 = 28.7 N/mm2

f130

= 2)130(

252000 + 11.2 = 26.11 N/mm2

f140

= 2)140(

252000 + 11.2 = 24.054 N/mm2

36.4

36.422.424.126.11

28.732

22.4

140

130

120110150 100

Thickness 100 mm

150 mm

fx

36.4

36.422.424.126.11

28.732

22.4

140

130

120110150 100

Thickness 100 mm

150 mm

fx

Figure: Representation of Hoop Stress Across the Thickness of

Pipe

Q3. A square element of a thin plate subjectedto a compressive stress of 5 MPa in x-direction, a tensile stress of 15 MPa in y-direction and a shear stress of 10 MPa(clockwise). Determine the principal stressesand their directions by using analyticalmethod. Also find the normal stress andshear stress on the diagonal plane of thesquare element.

Answer : April/May-13, Set-1, Q3 M[15]

Given that,

Stress acting on x-axis, σx = – 5 N/mm2

Stress acting on y-axis, σy = 15 N/mm2

Shear stress τxy

= 10 N/mm2 (Clockwise)

Required,

(i) Principal stress and their directions

(ii) Normal stress and shear stress on the diagonalplane of the square elements.

(i) Principal Stress and their Direction

Principal stress,

(σ1, σ

2) =

2yx σ+σ

± 2

2

2

–xy

yx τ+

σσ

σmax

= σ1

= 2

yx σ+σ +

22

2

–xy

yx τ+

σσ

= 2

15)5(– + +

22

)10(2

15–)5(– +

= 5 + 100100 += 19.14 N/mm2

σmin

= σ2

=

σ+σ2

yx –

22

2

–xy

yx τ+

σσ

= 2

155– + –

22

)10(2

15–5– +

= 5 – 100100×

σ2

= – 9.14 N/mm2

The angle made by principal plane and shear stresswith x-axis is,

tan2θ = yx

xy

σστ–

2–

= 15–5–

102– ×

tan2θ = 1

θ = 2

)1(tan 1–

θ = 22°30'00'

θ1= 90 + θ

θ1= 112°30'00'



(ii) Normal Stress and Shear Stress on the DiagonalPlane of the Square Element

Normal stress on the diagonal of the square element(σ

n) (i.e., θ = 45°) is,

σn

=

σ+σ2

yx +

σσ2

– yxcos2θ + τ

xy sin2θ

=

+

2

15)5(–+

2

15–)5(–cos(2 × 45) + 10 sin(2 × 45)

= 5 + (– 10) cos90 + 10 sin90

= 15 N/mm2

Shear stress on the diagonal of square element is,

τxy

=

σσ2

– yxsin2θ + τ

xy cos2θ

=

2

15–)5(–sin(2 × 45) + 10 cos(2 × 45)

= – 10 sin90 + 10 cos90

= – 10 N/mm2

Q4. A closed coil helical spring is made of 12mm diameter wire and is having meandiameter of 150 mm and 10 complete turns.The modulus of rigidity of the material ofspring is 80 MPa. When a load of 500 N isapplied, determine the maximum shearstress, strain energy stored, deflectionproduced and stiffness of the spring.


Given that,

Diameter of steel wire, d = 12 mm

Number of turns in spring, n = 10

Mean diameter, D = 150 mm

Mean radius, R = 75 mm

Modulus of rigidity, G = 80 N/mm2

Load, W = 500 N

Required,

(i) Maximum shear stress

(ii) Deflection produced

(iii) Strain energy stored

(iv) Stiffness of the spring.

(i) Maximum Shear StressThe maximum shear stress (τ) is given by equation,

WR = 16

π × τ × d3

500 × 75 = 16

π × τ × (12)3

τ = 110.52 N/mm2

(ii) Deflection ProducedDeflection in the spring,

δ = 4

364

Gd

nWR

δ = 4

3

1280

107550064

××××

= 81380.20 mm

∴ δ = 81.3 m

(iii) Strain Energy StoredStrain energy stored,

u = 2

1Wδ

= 2

1 × 500 × 81.3

∴ u = 20325 N-m

(iv) Stiffness of the Spring Stiffness of the spring,

S = δ

W

= m3.81

500

= 6.15 N/m

Q5. A built-up I section has the followingdimensions.

Overall depth : 400 mmFlanges : 300 mm × 50 mmWeb thickness : 30 mm

When it is used as a simply supported beam,it deflects by 10 mm when subjected to aload of 40 kN/m over its entire length. Findthe safe load, if the same section is used asa column with both end hinged. Use Euler’sformula. Assume a factor of safety as 1.75and take E = 2 × 105 MPa.




300 mm

300 mm

400 mm 30mm

300 mm

50 mm

50 mm

135 mm

300 mm

300 mm

400 mm 30mm

300 mm

50 mm

50 mm

135 mm

Given that,

Load, W = 40 kN/m

Deflection, δ = 10 mm

Factor of safety = 1.75

E = 2 × 105 N/mm2

Required

Safe load for the beam

Assume, length of the beam = L

Moment of inertia I-section about x-axis is,

Ix=

×12

400300 3

– 2 ×

×12

300135 3

= 16 × 108 – 60.75 × 107

= 99.25 × 107 mm4

Moment of inertia of the section about y-axis is,

Iy= 2

Flange

3

12

db +

Web

3

12

db

Iy= 2

×12

30050 3

+

×12

30300 3

= 22.5 × 107 + 67 500

= 22.5067500

= 22.5 × 107

Deflection, δ = EI

WL

384

5 4

I : Greatest between Ix and I

y of the section.

(Here I = Ix)

10 = 75

46

1025.99102384

10405

××××××× L

l = 248.47 mm

∴ Length of the beam = 248.47 mm

Finding the safe load when beam is used as a columnwith both the ends hinged. When both the ends are hinged,using Eulers formula,

PEulers

= 2

2

l

EI yπ

= 2

752

)47.248(

105.22102 ××××π × 10– 6 kN

∴ Load, P = 7193.8 kN

–~ 7194 kN

Safe load = safety ofFactor

EulerP

= 75.1

7194

= 4110.80 kN

Q6. The cross-section of a masonry pier is ahollow rectangle, the dimensions of externaland internal rectangles being 1200 mm × 800mm and 900 mm × 500 mm respectively. Aload of 300 kN in the vertical plane bisectingthe 1200 mm width of pier is transmitted atan eccentricity of ‘e’. Calculate the maximumvalue of ‘e’ so that no tension is induced inthe section.


Given that,

In a masonry pier of hollow rectangular section

Dimensions of external rectangle,

B × D = 1200 × 800 mm

Dimensions of internal rectangle,

b × d = 900 mm × 500 mm



In case of hollow rectangular section,

Area = BD – db

= (1200 × 800) – (900 × 500)

A = 51 × 104 mm2

1200 mm

800 mm500 mm

e

900 mm

1200 mm

800 mm500 mm

e

900 mm

Moment of inertia at XX,

IXX

= 12

3BD –

12

3bd

= 12

8001200 3× –

12

500900 3×

= 4.1825 × 1010 mm4

IXX

= 41825 × 106 mm4

IYY

= 12

3DB –

12

3db

= 12

1200800 3× –

12

900500 3×

IYY

= 8.4825 × 1010 mm4

= 84825 × 106 mm4

ZXX

= maxy

I XX

= D

bdBD

6

– 33

= 8006

500900–8001200 33

×××

ZXX

= 10456.25 × 104 mm4

ZYY

= maxy

IYY

= B

dbDB

6

– 33

= 12006

900500–1200800 33

×××

ZYY

= 14137.5 × 104 mm4

Condition for no tension,

e ≤ b

KYY22

Radius of gyration,

2YYK =

A

IYY

= 4

6

1051

1084825

××

2YYK = 1663.25 × 102 mm

Eccentricity, e ≤ b

KYY22×

≤ 900

1025.16632 2××

≤ 369.61 –~ 370

e ≤ 370 mm

Result

Maximum value of eccentricity,

e = 370 mm.

Q7. Determine the maximum positive bendingmoment and maximum twisting moment ina semi-circular beam simply supported onthree equally spaced supports. Take radiusof the centre line of the beam as ‘R’ and loadper unit length of the beam as ‘w’.


For answer refer Unit-VII, Q25.



Q8. Determine the forces in all the members ofthe truss as shown in the figure.

30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

Figure


30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

30 kN20 kN 20 kN

40 kN

J I H G F

E

4 m

AB C D

16 mm 4 @ 4 m

Calculation of forces in the members of the truss are,

Reactions

RA + R

E= 30 + 20 + 20 + 40

= 110 kN ... (1)

Taking moments about ‘A’,

ΣME

= 0

RA × 16= 40 × 12 + 20 × 8 + 20 × 4

RA × 16= 480 + 160 + 80

RA × 16= 720

RA

= 16

720

RA= 45 kN

Substitute RA in equation (1), we get,

RA + R

E= 110 kN

45 kN + RE

= 110

RE

= 110 – 45

RE

= 65 kN

By using method, of joints.

At Joint ‘A’

Free body diagram of joint ‘A’,

45°

30 kN

20 kN

A B

J I

45°

30 kN

20 kN

A B

J I

Taking forces vertically,

ΣFy

= 0

FAJ

= 30 kN ; FAI

= 20 kN

Taking forces horizontally,

ΣFx

= 0

FAB

× cos45°= RA × 4

FAB

= °×45cos

445

FAB

= 254.55 kN

At Joint ‘B’

Free body diagram of joint ‘B’,

C

HI

BA C

HI

BA

Taking forces horizontally.

ΣFx

= 0

FAB

+ FBC

× cos45° = 0

254.55 + FBC

cos45° = 0

FBC

= °45cos

55.254–

FBC

= – 360 kN (Compression).




ΣFy

= 0

FBI

– FBH

sin45° = 0

FBH

= °45sin

20

= 28.284 kN

FBI

= 20 kN

FBH

= 28.284 kN

At Joint ‘C’

Free body diagram of joint ‘C’,

D

H

CB

20 kN

D

H

CB

20 kN

FCH

= – 20 kN (Compression)


FBC

+ FCD

× cos45° = 0

– 360 + FCD × cos45° = 0

FCD

= °45cos

360

FCD

= 509.11 kN

At Joint ‘D’

Free body diagram of joint ‘D’,

E

HG

DC E

HG

DC


ΣFx= 0

FCD

+ FDE

– RE × 4 = 0

509.11 + FDE

= 65 × 4

FDE

= – 249.11 kN


ΣFy

= 0

FGD

= 40 × 4

= 160 kN

– FHD

× sin45° – FGD

= 0

– FHD

= °45sin

160

= 226.27 kN

FHD

= – 226.27 kN

At Joint ‘E’

Free body diagram of joint ‘E’,

GF

ED

45°

65 kN

GF

ED

45°

65 kN

FDE

× 4 – 20 × 4 + RE × 4 + F

FE = 0

– 249.11 × 4 – 80 + 65 × 4 + FFE

= 0

– 816.44 + FFE

= 0

FFE

= 816.44 kN (Tension)

At Joint ‘F’

Free body diagram of joint ‘F’,

G F

E

G F

E



Taking moments,

– FFE

+ FGF

× 4 = 0

– 816.44 + 4FGF

= 0

FGF

= 4

44.816

FGF

= 204.11 kN (Tension)

At Joint ‘G’

Free body diagram of joint ‘G’,

GH F

D

FGD

GH F

D

FGD

Moment at ‘G’,

FGD

= 40 × 4

= 160 kN


FGF

+ FHG

× 4 = 0

204.11 + 4FHG

= 0

FHG

= 4

11.204–

FHG

= – 51.027 kN

At Joint H

Free body diagram of joint H,

HI G

C DB

HI G

C DB


FBI

= 20 kN

FIH

+ FHG

× cos45° = 0

FIH

= – 51.027 × cos45°

FIH

= – 36.081 kN (Compression)

At Joint I

Free body diagram at joint I,

IJ H

BA

IJ H

BA


ΣFx

= 0

FJI + F

IH cos45° = 0

FJI

= – 36.081 × cos45°

FJI

= – 25.513 kN (Compression)

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1404650711_Strength of Materials 2-2 Set-1 (K)