[Marks : 100 Vidyalankardiploma.vidyalankar.org/.../SemIII/Civil/MOS_Soln.pdf4 S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: 3 Hrs.] Prelim Question Paper Solution

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S.Y. Diploma : Sem. III [CE/CS/CR/CV]

Mechanics of Structure Time: 3 Hrs.] Prelim Question Paper Solution [Marks : 100

Q.1 (a) Attempt any SIX of the following. [12]Q.1 (a) (i) State parallel axis theorem of moment of inertia. [2] (A) The moment of inertia of any object about an axis through its center of

mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about any axis parallel to that axis through the center of mass is given by

I = Icm + md2

Q.1(a) (ii) Define ductility and malleability. [2](A) Ductility It is the property of material to undergo a considerable deformation. Under

tension without rupture. OR

It is the property of material due to which it can be drawn into thin wires. Malleability It is the property of a material by virtue of which it gets. Permanently

deformed by compression without rupture. OR

It is the property of a material due to which it can be drawn into thin sheets. Q.1(a) (iii) State Hooke’s law. [2](A) Hooke’s law that the extent to which an elastic material will change size and

shape under stress is directly proportional to the amount of stress applied to it.

Q.1(a) (iv) Write the expression for strain energy due to any type of load. [2](A) For gradually load work done for load ‘P’ is given by area of the load

deformation diagram. Work done by Ext. load

= 12

P L = 1 PLP2 AE

= 2

21 P AL2 A E

= 21 AL

2 E

L

L

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Prelim Question Paper Solution

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= 21

2 E (Volume)

Volume A LPA

= 2

V2E

Strain Energy u = 2

V2E

Q.1(a) (v) Write Euler’s formula with the meaning of the terms involved. [2]

(A) F = 2

2n El

L (1)

where F = allowable load (lb, N) n = factor accounting for the end conditions

E = modulus of elasticity 22

lb , Pa N / min

L = Effective length of column (in, m) I = Moment of inertia (in4, m4) Q.1(a) (vi) Define moment of inertia. State its value for a semicircle about its

centroid. [2]

(A) (a) Moment of inertia of section is defined as the product of area of a section and the square of the distance between the centroid of section and reference axis. It is also called as second moment of area.

(b) Moment of inertia value for semi-circle IXX = 0.11 R4

IYY = 4R

8

Q.1(a) (vii) Differentiate between gradual and sudden applied load with

respect to stress produced. [2]

(A) Gradual load Sodden load

(i) Stress produced is = PA

Stress produced is = 2PA

(ii) Stress due to gradual load is lesser then sudden load

Stress due to sodden load is higher than the gradual load.

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Q.1(a) (viii) Define elastic body, giving two examples. [2](A) A body is said to be elastic if it regains its original size and shape when an

externally applied force causing deformation is entirely removed. Examples : (i) Rubber band (ii) Golf ball (iii) Soccer ball Q.1(b) Attempt any TWO of the following : [8] Q.1(b) (i) A beam 120 mm wide and 200 mm deep is subjected to a shear

force of 48 kN at a particular section. Calculate maximum shear stress and sketch shear stress distribution diagram across thesection.

[4]

(A) S = 48 103 N A = 200 120 max = 1.5 avg

= 1.5 SA

= 1.5 348 10

200 120

max = 3 N/mm2 Q.1 (b) (ii) Draw a bending stress distribution diagram for following cases:

(1) A beam of rectangular cross section used as a simplysupported beam.

(2) A beam of T section used as a cantilever.

[4]

(A) (1)

200

120D

C

3Shear stress

distribution diagram

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(2)

Q.1 (b) (iii) Define short columns and long columns. [4](A) Short column When the ratio of effective length to the least lateral dimension’s of the

column is less than 12, then it is called a short column. OR

When the ratio of effective length to the least radius of duration is less than 45, then it is called short column.

Long column When the ratio of effective length to the least radius of duration is

duration is greater than 45 then it is called a long column. Q.2 Attempt any TWO of the following : [16] Q.2(a) Determine M.I. about X-X and Y-Y axis as shown in Fig.

250mm

20 mm

220 mm

20 mm

180mm

20 mm

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(A) (1) Center of Gravity of Section. Due to symmetry at xy axis.

x = 2502

= 125 mm.

y = 1 1 2 2 3 3

1 2 3

a y a y a ya a a

y =80 20 10 20 220 130 250 20 250

180 20 20 220 250 20

y = 142.92mm (2) Moment of Inertia. (i) About x x axis IXX = IXX1 + IXX2 + IXX3

= 2 2 21 1 1 2 2 2 3 3 3IG A h IG A h IG A h

h1 = 142.92 10 = 132.92 mm h2 = 142.92 130 = 12.92 mm h3 = 250 142.92 = 107.08 mm IXX = [(120 103) + 63.60 106)] + [(17.75 106) + 734.47 103) [(166.67 103 + 57.33 106)] IXX = 139.40 106 mm4

Due to symmentry IYY = IG + IG2 + IG3 = [9.72 106 + 146.67 103 + 26.04 106] IYY = 35.908 106 mm4 Q.2(b) From a plate 4 cm 8 cm, a

triangular portion as shown in figure is cut. Determine the moment of inertia of the remainder about the horizontal axis passing through top of the lamina :

4 cm

8 cm

AB

E

D C

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(A)

Q1 = 4 8 = 32 cm

y1 = 82

= 4 cm

x1 = 2 cm

Q2 = 12

4 4 = 8 cm

y2 = 43

= 1.33 cm

x2 = 2 cm Now,

y = 1 1 2 2

1 2

a y a ya a

y = 32 4 8 1.3332 8

y = 4.89 cm x = Due to symmetry = 2 cm IXX = [M.I.]1 [M.I.]2

= 3 3

2 21 1 2 2

bd bha h a h12 36

= 3 32 24 8 4 432 4.89 4 8 4.89 1.33

12 36

= [196.01] [108.49] IXY = 87.54 cm4

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M.I. of section of top of lamina IAB = IXX + Ah2 A = A1 A2 = 32 8 = 24 cm2 h = 8 4.89 = 3.11 IAB = 87.51 + 24 (3.11)2 = 319.64 cm4 Q.2(c) Find the M.I. of section shown in Fig. about horizontal axis passing

through C.G.

(A)

(i) Area calculation A1 = 30 x 50 = 1500 mm2 A2 = 130 x 20 = 2600 mm2 A3 = 70 x 30 = 2100 mm2

Y

30mm

70mm

XX

Y

20mm

50mm

0

50 mm 50 mm

Y

30mm

70mm

XX

Y

20mm

0

50 mm 50 mm

y

x xB

B E2

1

2

3

50mm

A A

y 66.77mm

y 73.23mm

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(ii) Distances of centroid from the base AA y1 = 70 + 20 + 50/2 = 115 mm y2 = 70 + 20/2 = 70 + 10 = 80 mm y3 = 70/2 = 35mm

y = 1500 x 115 2600 x 80 2100 x 351500 2600 2100

y = 73.23 mm (iii) To find IXX

1GI =

33 430 x 50 312.5 x 10 mm

12

h1 = 66.77 50 41.772

h1 = 41.77 mm

IXX1 = 1

2 3 2G 1 1I D h 312.5 x 10 1500 x (41.77)

= 2.929 x 106mm4

2GI =

33 4130 x 20 86.67 x 10 mm

12

h2 = 66.77 20502

h2 = 6.77 mm IXX2 =

2

2G 2 2I A h

= 86.67 x 103 + 2600 x (6.77)2 IXX2 = 205.84 x 103 mm4

3GI =

33 430 x 70 875.5 x 10 mm

12

h3 = 73.23 702

= 38.23 mm

IXX3 = IG + A323h = 875.5 x 103 + 2100 x (38.23)2

IXX3 = 3.9447 x 106 mm4 Now, IXX = IXX1 + IXX2 + IXX3 = 2.929 x 106 + 86.67 x 103 + 3.9447 x 106 IXX = 6.96 x 106 mm4 Q.3 Attempt any TWO of the following : [16] Q.3(a) A bar of 30 mm diameter is subjected to a pull of 60 KN. The

measured extension on gauge length of 200 mm is 0.09 mm and changein diameter is 0.0039 mm. Calculate the Poisson’s ratio and the valuesof the three moduli.

[4]

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(A) d = 30 mm P = 6 104 N L = 200 mm L = 0.09 mm d = 0.0039 mm To find : , E, G

L = 4

2

6 10

304

= 84.88 N/mm2

rd = 0

L = 1E

[ L d]

LL

= L1E

0.09200

= 84.88E

E = 188.62 103 N/mm2

d = 1E

[ d L]

dd

= 1E

L

0.003930

= 2

84.88188.62 10

G = E2 1

G = 3188.62 10

2 1 0.288

G = 73.222 103 N/mm2 Q.3(b) A steel tube 40 mm inside diameter and 4 mm metal thickness is

filled with concrete. Determine the stress in each material due toan axial thrust of 100 kN. Take ES = 2.1 105 N/mm2 and EC = 0.14 105 N/mm2.

[4]

(A) d = 40 mm D = 40 + 4 + 4 = 48 mm ES = 2.1 105 N/mm2 EC = 0.14 105 N/mm2

(i) Asteel = 4

(D2 d2)

= 4

(482 402)

Asteel = 552.92 mm2 40 mm

4 mm 4 mm

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(ii) AC = Area of concrete = 2404

AC = 1256.63 mm2 (iii) To find stress in each material

S

SE = C

CE

S = S

C

EE C

S = 5

52.1 10

0.14 10C

S = 15 C

P = PS + PL = S AS + C AC 100 103 = S 552.92 + C 1256.03 100 103 = 15 C 552.92 + 1256.03 C 100 103 = 9.55 103 C

C = 3

3100 109.55 10

C = 10.47 N/mm2

S = 15 C S = 15 10.47 S = 157.07 N/mm2 Q.3(c) A composite bar comprising of aluminum and steel is as shown in Fig.

Find the value of ‘P’ if net elongation produced in the bar is 2 mm. Take Es = 20 104 N/mm2 and Eal = 7 104 N/mm2.

[4]

C/S Area 300 mm2

Steel2P Aluminium

C/S Area 75 mm2C/S Area 75 mm2

Steel 4P 4P 2P

1.5m 1m1m

Fig.

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(A) [1 Mark] Given data (i) Asteel = 75 mm2 ii) AAl = 300 mm2 L = 2mm L = L1 + L2 + L3 [1 mark]

L = steel Al steel

P.L P.L P.LA.E A.E A.E

[1 mark]

2 = 4 4 4

2p x 1000 2p x 1500 2p x 100075 x 20 x 10 3000 x 7 x 10 75 x 20 x 10

[1 mark]

2 = 6 6 62000P 3000P 2000P

15 x 10 21 x 10 15 x 10 [1 mark]

2 = 1.3333 x 10 4 + 1.4286 x 10 4 1.3333 x 10 4 [1 mark] 2 = 1.238 x 10 4 P

P = 42

1.238 x 10 [1 mark]

P = 16.15 x 103 N P = 16.15 kN ( ve sign indicate P is compressive in nature) [1 mark] Q.4 Attempt any TWO of the following : [16] Q.4(a) A steel rod, 1 m long is fixed at the ends and subjected to a pull of

9 kN. Determine the residual stress due to an increase of 20°C.Diameter of bar = 12 mm. E = 200 kN/mm2, = 16 × 10 6 / °C.

[8]

(A) L = 103 mm P = 9 103 kN T = 20 C d = 12 mm E = 2 105 N/mm2 = 16 10 6/ C

Total stress = Thermal stress + Compressive stress Thermal stress = E T = 2 105 16 10 6 20 = 64 N/mm2

4P 4P 2P

Steel

1.5m1m 1m

2P 2P Steel2P 2P

4P 2P

2P

Aluminum 4P 2P

2P

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Compressive stress = PA

= 3

2

9 10

124

= 79.577 N/m2 Total stress = 64 + 79.577 = 143.577 N/mm2 Q.4(b) Draw SFD and BMD of a beam as shown in figure. Also find the

point of contra flexure.

[8]

(A) (i) To find reactions RA and RB RA + RB = 20 10 RA + RB = 200 (i)

MA = 20 10 102

RB 8

MA = 1000 8 RB 8RB = 1000 RB = 125 kN RA = 200 Rb RA = 200 125 RA = 75 kN (ii) Step 2 : Shear force calculation FC = 0 FBR = 20 2 = 40 kN FBL = 40 125 FBL = 85 kN FA = 85 + 20 8 FA = 85 + 160 FA = 75 kN (iii) B.M. Calculation MC = 0 kN-m

MB = (20 2) 22

= 20 2 1 = 40 kN-m

MA = 0 kN-m

8 m 2 m

C AB

20 kN/m

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(iv) To locate the position of point of contrashear

858 x

= 75x

85x = 75 (8 x) 85x = 600 75x 160x = 600 x = 3.75 m (from left)

MXX = 75 x 20 x x2

MXX = 75 3.75 20 3.75 3.752

MXX = 281.25 140.625 MXX = 140.625 kN-m (sagging)

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To locate the point of contraflexure (PCF)

1XM = 75 x1 20 x1 1x

2

0 = x1 (75 10x1) 75 = 10x1 = 0 75 = 10x1 x1 = 7.5 m from A Point of contraflexure (Pcf) = 75 m from ‘A’ Q.4(c) A cube of 250 mm side is subjected to a compressive, force of

3.8 MN on each face. The change in volume is found to be5200 mm3. Find E and K if (μ) = 0.25.

[8]

(A) Px = Py = Pz = 3.8 106 N V = 5200 mm3 = 0.25 side = 250 mm

x = 6

23.8 10

250 = 60.8 = y = z

Volumetric strain V = x + y + z V = 3 x

VV

= 13E

[ x ( y + z)]

VV

= 3E

[ 1 2 ] x

3

5200250

= 3E

[ 1 2 0.25] . 60.8

E = 274.038 103 N/mm2

K = x

V =

3

60.85200250

K = 182.692 103 N/mm2 Q.5 Attempt any TWO of the following : [16] Q.5(a) A T section beam having flange 180 mm wide and 20 mm thick and

web 150 mm long and 20 mm thick carries u.d.l. of 80 kN/m overan effective span of 8 m. Calculate the maximum bending stress.

[8]

(A) W = 80 kN/m L = 8 m To find B max

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By Bending formula

y = 1 1 2 2

1 2

A y A YA A

= 180 20 160 150 20 753600 3000

y = 121.36 mm assuming Beam to be simply supported yt = 121.36 ye = 48.64

M = BMmax = 2 2WL 80 8

8 8 = 640 kNm

= 640 106 Nmm I = Ixx1 + Ixx2

= 3 3180 20 20 150

12 12 +3600 (38.64)2 + 3000 46.362

I = 17.567 106 mm4 For max bending stress y = 121.36

6

t6

640 10121.3617.567 10

t = 4.421 kN/mm2 Q.5(b) A Simply supported beam 5 m long carries a point load of 20 KN and

anticlockwise moment of 8 kN m at a distance of 3 m from the lefthand support. Draw SF and BM diagrams.

[8]

(A)

1) Support reactions a) Fy = 0; RA + RB = 20KN b) m@A = 0; (20 x 3) – 8 – RB x 5 = 0

52 = 5 RB RB = 10.4KN RB = 9.6KN

2) S.F. Calculation i) S.F. at just left of A = 0 ii) S.F. at just right of A = RA = 9.6KN iii) S.F. at just left of C = 9.6 KN

180 mm20 mm

20 mm

150 mm

A 3m 2m B

20KC 8KN.

9.6K 9.6K

(-) S.F.D

10.4K10.4K

0(+)

B.M.D

20.8KN.28.8KN.

0

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iv) S.F. at just right of C = 9.6 20 = 10.4KN v) S.F. at just left of B = 10.4KN vi) S.F. at just right of B = 10.4 + RB = 0KN 3) B.M calculation i) B.M at A = B.M at B = 0 KN.m s.s end ii) B.M just left of C MCL = RA x 3 = 9.6 x 3 = 28.8 KN.m iii) B.M at just right of C MCR = RA x 3 8 = 9.6 x 3 8 = 20.8 KN.m Q.5(c) A cantilever beam is loaded as shown in figure. Draw the S.F.D.

and B.M.D.

[8]

(A)

3 m 4 m

AC B

2 kN1kN/m

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(i) Support reaction Fy = 0 RA = 2 + (1 4) = 6 kN (ii) Shear force calculation [ +ve ve] S.F. at just left of A = 0 kN S.F. at just right of A = RA = 6 kN S.F. at just left of C = 6 kN S.F. at just right of C = 6 2 = 4 kN S.F. at B = 4 (1 4) = 0 kN (iii) Bending moment calculation , Bm@B = 0 free end Bm@C = (1 4 2) = 8 kN.m Bm@A = (1 4 5) (2 3) = 26 kN.m Q.6 Attempt any TWO of the following : [16] Q.6(a) Find the crippling load by Rankine’s formula for a hollow circular

column of 200 mm external diameter and 150 mm internal diameter. Length of the column is 5 m. If (i) Both ends are fixed (ii) One end is fixed and other free (iii) One end is fixed and other is hinged (iv) Both ends are hinged. Take fc = 550 N/mm2, a=1/1600

[8]

(A) d1 = 200 mm d2 = 150 mm L = 5000 mm 2 = 550 N/mm2

a = 11600

k = IA

= 4 41 2

2 21 2

d d6

d d4

PR = c2

6 .A1 a

k = 62.5

A = 13.744 103 mm2

(i) for end fixed Le = L2

= 2500

= eLk

= 250062.5

= 40

ve +ve

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PR = 3

2550 13.744 10

4011600

= 3.779 kN

(ii) For one end fixed other free, Le = 2L = 104.

= 410

62.5 = 160

PR = 2

550 1374416011600

= 444.65 kN

(iii) for one end fixed and other end free, Le = L2

= 3535.5

= 3535.562.5

= 56.568

PR = 2

550 1374456.5611000

= 2.52 MN

(iv) PR = 2

550 137448011000

= 1.511 MN

Q.6(b) A steel rod of 25 mm diameter and 1500 mm long is subjected to a

load of 30 kN applied suddenly. Calculate the strain energy stored andmodulus of resilience along with change in length. Take E = 2.1 105 N/mm2.

[8]

(A) Given : D = 25mm, L = 1500mm, P = 30KN, E = 2.1 x 105 N/mm2

(1) Area of rod A = 2 2D (25)4 4

A = 490.87 mm2 [½ mark] (2) Volume of rod V = A x L = 490.87 x 1500 V = 736305 mm3 [½ mark] V = 736.305 x 10 6 m3

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(3) Stress for suddenly applied load

= 3

22P 2 x 30 x10 122.23N / mmA 490.87

[1 mark]

(4) Strain energy stored

U = 2 2

5122.23V x736305

2E 2 x2.1 x10 [1 mark]

U = 26191.72 N.mm = 26.19 N.m or joule [1 mark]

(5) Modulus of resilience = 6U 26.19v 736.305 x10

[1 mark]

= 35569.5 Joule/m3 [1 mark] (6) Change in length ( L)

L = 5L 122.23 x1500

E 2.1 x10 [1 mark]

L = 0.873 mm [1 mark] Q.6(c) A symmetrical I-section has the following dimensions,

Flanges : 150 20 mm Web : 300 10mm Find the maximum shearing stress developed in the section of thebeam for shearing force of 100kN.

[8]

(A)

Since section is symmetrical the N.A. will be at a distance of 170 mm from

the base. (i) M.I. of section

IXX = 3 3BD bd

12 12

IXX = 3 3150 340 140 300

12 12 = 176.3 106 mm4

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Let us consider portion of beam above N.A. (2) Shear stress at the junction of flange and web by considering width of

flange. (b = 150 mm)

q1 = SAybI

= 3

6100 10 150 20 160

150 176.3 10

q1 = 1.81 N/mm2 3) Shear stress at the junction of flange and web by considering width of

web (b = 10 mm)

q2 = SAybI

= 3

6100 10 150 20 160

10 176.3 10

q2 = 27.22 N/mm2 4) Additional shear stress due to web area above N.A.

qdditional = SAybI

= 3

6100 10 150 10 75

10 176.3 10

qdditional = 6.38 N/mm2 5) The maximum shear stress is at N.A. and is given by qmax = qN.A. = q2 + qadditional

qmax = 27.22 + 6.38 = 33.60 N/mm2

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Documents

[Marks : 100 Vidyalankardiploma.vidyalankar.org/.../SemIII/Civil/MOS_Soln.pdf4 S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: 3 Hrs.] Prelim Question Paper Solution