1404650711_Strength of Materials 2-2 Set-3 (K).pdf

S.23Strength of Materials (April/May-2013, Set-3) 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) State and prove moment-area theorems. (Unit-I, Topic No. 1.5)

(b) Using the moment-area method, determine the deflection at the free end of a cantilever beam of span ‘l’subjected to a load ‘P’ at the fixed end. ‘EI’ is constant. [11+4] (Unit-I, Topic No. 1.5)

2. (a) Derive expression for volumetric change in a thin spherical shell subjected to an internal pressure P.(Unit-II, Topic No. 2.1)

(b) A thick cylindrical pipe of outside diameter 300 mm and internal diameter 200 mm is subjected to an internal fluidpressure of 15 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 tensile 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 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. Compare the weight of solid with that of a hollow one having the same material and of same lengths are to transmitsame power at a given speed, if the first shaft is of a solid circular section and the second shaft is of hollow circularsection, whose internal diameter is 0.6 times the outside diameter. [15] (Unit-IV, Topic No. 4.1.2)

5. A round vertical bar is clamped at the lower end and is free at the other. The effective length is 2 m. A horizontal forceof 30 N at the top produces a horizontal deflection of 15 mm. What is the buckling load for the bar in the givenconditions? [15] (Unit-V, Topic No. 5.2.1)

6. An unsymmetrical I-section with the following dimensions is subjected to a bending moment of 15 kN-m, the topflange being in compression. Draw the bending stress distribution across the depth marking the salient points andcompute the total moment resisted by the top flange.

Top flange : 240 mm × 10 mm

Bottom flange : 180 mm × 10 mm

Web : 300 mm × 10 mm. [15] (Unit-VI, Topic No. 6.2)

7. A curved beam, rectangular in cross-section is subjected to pure bending with couple of 300 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)

S e t - 3S o l u t i o n s

S.24 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014

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

8. Determine the forces in all the members of the truss as shown in figure. [15] (Unit-VIII, Topic No. 8.1)

A

B C D

EFH G

6 k 6 k

1.5 m 1.5 m 1.5 m 1.5 m

2 m

2 m

I J

Figure



Q1. (a) State and prove moment-area theorems.

Answer : April/May-13, Set-3, Q1(a) M[11]

For answer refer Unit-I, Q21 and Q23.

(b) Using the moment-area method,determine the deflection at the free endof a cantilever beam of span ‘l’subjected to a load ‘P’ at the fixed end.‘EI’ is constant.

Answer : April/May-13, Set-3, Q1(b) M[4]

Note: In a given reference, replace W with P.

For answer refer Unit-I, Q24.

Q2. (a) Derive expression for volumetricchange in a thin spherical shellsubjected to an internal pressure P.

Answer : April/May-13, Set-3, Q2(a) M[7]For answer refer Unit-II, Q3.(b) A thick cylindrical pipe of outside

diameter 300 mm and internal diameter200 mm is subjected to an internal fluidpressure of 15 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-3, Q2(b) M[8]Given that,

Outer diameter = 300 mmInner diameter = 200 mmInternal pressure = 15 MPa

100 mm

150 mm

100 mm

150 mm

Required:(i) Maximum hoop stress in a section(ii) Representation of hoop stress across thickness of

the pipe.Let,

Px = Radial pressurefx = Hoop stress

Px = Ax

B −2

fx = Ax

B +2

SOLUTIONS TO APRIL/MAY-2013, SET-3, QP

At,

x = 100 mm; P100 = 15 MPa

x = 0 mm; P100 = 0

x = 150 mm; P150 = 0

Radial stress Px at x = 100 is,

P100= AB −

2100

15 = AB −

2100... (1)

Px at x = 150 is,

P150= AB −

2150

0 = AB

−2150

... (2)

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

A= 12

B = 2,70,000

Hoop stress = Ax

B +2

f100 = 2

2N/mm3912

)100(

000,70,2 =+

f150= 212

)150(

270000 + = 24 N/mm2

The maximum hoop stress is developed at inner face.

f110 = 12)110(

2700002

+

= 34.31 N/mm2

Similarly,

f120= 30.75 N/mm2

f130= 27.98 N/mm2

f140= 25.78 N/mm2



150

25.75

24

27.98

30.75

34.31

39

39

fx

R = 100

mm

R = 150 mm

140 130 120 110 100

Q3. A square element of a thin plate subjected to a tensile stress of 5 MPa in x-direction, a tensilestress of 15 MPa in y-direction and a shear stress of 10 MPa (clockwise). Determine the principalstresses and their directions by using analytical method. Also find the normal stress and shearstress on the diagonal plane of the square element.

Answer : April/May-13, Set-3, Q3 M[15]Given that,

σx = 5 MPa (Tensile) σy = 15 MPa (Tensile) τ = 10 MPa (Clockwise)

Angle of inclination = 45° to verticalRequired

(i) Principal stress and their direction(ii) Normal stress and shear stress on the diagonal plane of the square element.Procedure(i) Principal Stress and Direction

σ1=

σ+σ2

yx +

22–

2x y

xyIσ σ

+

σ1=

+

2

155+

22

102

155 +

−

σ1= 10 + 5 5 ~ 21.18 N/mm2

σ2=

σ+σ2

yx – 2

2

2 xyyx τ+

σ−σ

σ2= 5 15

2

+ –

225 15

102

− +

= – 1.18 N/mm2



The angle of principal plane and shear stress made with x-axis.

tan 2θ = yx

xy

σ−στ−2

⇒ tan 2θ= 155

102–

−×

θ= 31° 43' 3''(ii) Normal Stree and Shear Stress

Normal stress on diagonal of the square element (σn)⇒ tan 2θ = 2

θ= tan–1

2

2

σn=

σ+σ2

yx +

σ−σ2

yx cos 2θ + τxy sin 2θ

= 20 N/mm2 (θ = 45° here)Shear stress on the diagonal of square element is,

τxy =

σ−σ2

yx sin 2θ + τxy cos 2θ

=

−

2

155 sin (2 × 45) + 10 cos (2 × 45)

= 5 N/mm2 (compressive)Q4. Compare the weight of solid with that of a hollow one having the same material and of same

lengths are to transmit same power at a given speed, if the first shaft is of a solid circular sectionand the second shaft is of hollow circular section, whose internal diameter is 0.6 times the outsidediameter.

Answer : April/May-13, Set-3, Q4 M[15]For answer refer Unit-IV, Q6.

Q5. A round vertical bar is clamped at the lower end and is free at the other. The effective length is 2m. A horizontal force of 30 N at the top produces a horizontal deflection of 15 mm. What is thebuckling load for the bar in the given conditions?

Answer : April/May-13, Set-3, Q5 M[15]

L = 2 me

30 N

15 mm

Figure



Given that,

Effective length, Le = 2 m

Deflection, S = 15 mm = 0.015 m

Horizontal force = 30 N

Deflection, δ= EI

WL

3

3

0.015 = EI

WL

3

3

Effective length, Le= 2 m

Length = 1 m

L = 2

2 = 1 m

0.015 = EI×

×3

130 3

EI = 666.66

Buckling load for column= 2

2

eL

EIπ

P = 2

2

2

66.666×π

= 1644.93 N

Q6. An unsymmetrical I-section with the following dimensions is subjected to a bending moment of 15kN-m, the top flange being in compression. Draw the bending stress distribution across the depthmarking the salient points and compute the total moment resisted by the top flange.

Top flange : 240 mm × 10 mm

Bottom flange : 180 mm × 10 mm

Web : 300 mm × 10 mm.


Given that,

Bending moment, B.M = 15 kN-m

Top flange is in compression

Required

Moment of resistance by the top flange diagram of bending stress distribution across the depth.

A = (240 × 10) + (300 × 10) + (180 × 10)

= 7200 mm2



y =

××+××+××

7200

3151018016010300510240

= 147.08 mm

I = 3240 10

12

×

+

×12

30010 3

+

×12

10180 3

= 22.5 × 106 mm4

yc= y = 147.08 mm

yt = 320 – 147.08 = 172.92 mm

σc = I

M × yt

= 6

6

15 10

22.5 10

××

× 147.08

σc = 98.05 N/mm2

σt = 6

6

15 10

22.5 10

××

× 172.92

= 115.28 N/mm2

Moment of resistance (Mr) = σc.z about top flange

z = 6

2bd (For rect.)

z = 6

10240 2× +

6

30010 2× +

6

10180 2×

= 157 × 103 mm3

Mr = 98.05 × 157 × 103

= 15393850

= 15.4 × 106 kN-mm = 15.6 kN-m

240 mm

10

10 300

10

N.A

180 mm

σt = 115.28

172.92

0 = σc

147.08

98.05h

I J

Figure



Q7. A curved beam, rectangular in cross-section is subjected to pure bending with couple of 300 N-m.The beam has a width of 20 mm and depth of 40 mm and is curved in a plane parallel to thedepth. The mean radius of curvature is 50 mm. Find the position of the neutral axis and the ratioof the maximum stress to the minimum stress. Also plot the variation of the bending stress acrossthe section.


Given that,

Width of the beam = 20 mm

Depth of beam, D = 40 mm

Mean radius, R = 50 mm

Magnitude of pure bending

With couple action = 300 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 rectangular section h2 is given by,

h2 = ln3

D

R 2

2

2R

DR

DR −

−+

h2 = 40

)50( 3

ln2 50 40

2 50 40

× + × −

– (50)2

= 147.8 mm

y0 = – 2

50 147.8

50 147.8

×+

= – 2.79 mm

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

(ii) Ratio of Maximum Stress to Minimum Stress

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. Therefore, the maximum stress occursat the extreme bottom layer where y = – 20 mm

y = 2

D = 20 mm

σmax = 80050

1000300

××

−−+

2050

20

8.147

501

2

= – 77.07 N/mm2

= 77.07 MPa (compressive)

σmin = 3300 10

50 800

××

++

2050

20

8.147

501

2

= 43.76 N/mm2

= 43.76 MPa (Tensile)

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

The ratio of σmax to σmin is,

min

max

σσ

= 76.43

07.77 = 1.76

(iii) Plotting of Stress Variation Across the Section

Bending stress, σ = RA

M

+

+yR

y

h

R2

2

1 at any section

Where y : Distance of the section from centroidal axis

At y = 0, σ = 80050

10300 3

××

++

050

0

8.147

501

2

= 7.5 MPa

At y = 5 mm, σ = 19.03 MPa [Q 1 MPa = 1 N/mm2]

At y = 10 mm, σ = 28.64 MPa

At y = 15 mm, σ = 36.78 MPa

At y = 20 mm, σ = 43.76 MPa

At y = – 5 mm, σ = – 6.5 MPa

At y = – 10 mm, σ = – 24.22 MPa

At y = – 15 mm, σ = – 46.78 MPa

At y = – 20 mm, σ = – 77.07 MPa



− 46.87

− 24.2

6.5

2015

10

5

05

10

1520

− 77.07

43.76

σ = 0

36.78

28.64

19.03

y = 2.79 mm0

STRESS

Neutral axis

Centroidal axis

y

R = 50 mm

STRESS

40 mm

20 mm

20 mm

20 mm

Axis of Curvature

0

O : Centre of Curvature

Figure: Plotting of Stress Across the Section

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

A

B C D

EFH G

6 kN 6 kN

1.5 m 1.5 m 1.5 m 1.5 m

2 m

2 m

I J

FigureAnswer : April/May-13, Set-3, Q8 M[15]

A

B C D

EFGH

6 kN 6 kN

1.5 m 1.5 m 1.5 m 1.5 m

2 m

2 m

I J

Figure



ΣFY = 0

RA + RE= 6 + 6

ΣMA = 0

(6 × 3) + (6 × 4.5) – RE × 6 = 0

RE = 7.5 kN

RA = 4.5 kN

θ = tan–1

5.1

2

θ = 53° 8'

Joint A

AH

I

B

PAB = – 4.5 kN

ΣFY = 0

PAI sin 53° 8' = 4.5

PAI = 5.62 kN

ΣFX = 0

PAH + PAI cos 53°8' = 0

PAH = 3.37 kN

Join H

A G

I

H

ΣFX = 0

PHA = PHG = 3.37 kN

ΣFY = 0

PHI = 0

Joint G

PGC = – 6 kN

PGI = 6 sin 53°8'

PGI = 4.8 kN

ΣFX = 0

PGI cos θ + PGJ cos θ + PGH + PGF = 0

4.8 cos 53°8' + 4.8 cos 53°8' + 3.37 + PGF = 6

PGF = 9.14 kN (comp)

Joint F

G F

J

F

6 k

ΣFY = 0

PFJ = 6 kN

ΣFX = 0

PGF = PFE = 9.14 kN

Joint E

F E

JD

PED = – 7.5

ΣFY = 0

7.5 + PEJ sin θ = 0

PEJ = 9.37 kN

Joint D

C

E

D

PDC = 0 (Q FX = 0)



Joint C

B DC

IG

J

θ θ = 53° 8’

PBC = PDC = 0

ΣFY = 0

PCG + PCI (3 sin θ) + PC (3 sin θ) = 0

– 6 + PCI sin 53° 8' + PCE sin 53° 8' = 0

PCI + PCE = 7.499 kN

Joint I

I

A

HG

C

36° 52’

36° 52’

ΣFY = 0

PIA sin θ + PIH + PIG sin θ + PIC = 0

5.62 sin (36° 52') + 4.8 (sin 36°52') + PIC = 0

PIC = 13.21 kN

PCE = – 13.21 + 7.499

= 5.71 kN

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1404650711_Strength of Materials 2-2 Set-3 (K).pdf