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1113/SY/Pre_Pap/Mech/SOM_Soln 1

Vidyalankar S.Y. Diploma : Sem. III

[ME/MH/MI/PG/PT/AE/FE/PS] Strength of Materials

Prelim Question Paper Solution

(i) Elasticity : It is the property of a material by virtue of which it regains its original size and shape after deformation, when the loads causing deformation are removed.

Plasticity: Lack of elasticity is called plasticity. The plasticity of a material is the ability to change its shape without destruction under the action of external loads and to regain the shape given to it when the forces are removed.

(ii) Creep : Many structural members and machine parts sustain steady loads for long periods of time. For example, beams in a R.C.C. building, plastic mountings for the parts of electronic devices, blades of turbine rotor, etc. Under such conditions, the material may continue to deform and will ultimately break. Creep continues as long as the load is applied. Therefore, it is a time dependent phenomenon. The greater the time, the more will be the creep. The continuous deformation with time which the material undergoes due to application of external steady loads is called creep or time yield or plastic flow.

(iii) Resultant Stress : It is the resultant of normal stress and tangential stress

and is denoted by r. since n and t act at right angles to each other, 2

r = 2 2n t

r = 2 2n t

Angle of Obliquity () : It is the angle made

by the resultant stress with the normal stress and is denoted by . Refer to figure.

tan = t

n

= tan1 t

n

(iv) Types of Beams (1) Simply supported beam : A beam

which is freely supported on the walls or columns at its both the ends is called as a simply supported beam.

(2) Cantilever beam : A beam fixed at one end and free at the other is called as a cantilever beam.

1. (a)

1. (a)

1. (a)

n

x x

r

t

1. (a)

L

B A

Fig. 2: Cantilever beam

L

B A

Fig. 1: Simply supported beam

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(3) Overhanging beam : If the end portion of the beam extends beyond the support, it is called as an overhanging beam. A beam may be overhanging on one side or on both sides as shown in figure 3. A chajja of a room is an example of a overhanging beam.

(4) Fixed beam : A beam whose both the ends are rigidly fixed in walls is called a fixed beam, constrained beam, built-in beam or an encastre beam.

(5) Continuous beam : A beam which is supported on more than two

supports (i.e. at least three supports) is called a continuous beam. The end supports of a continuous beam may be simply supported or fixed.

In this topic, we shall study shear force and bending moment diagrams for simply supported, overhanging and cantilever beams only. The shear force and bending moment diagrams for the fixed beams and continuous beams will be studied in the subject ‘Theory of Structures”.

(v) M.I. of hollow circular section: Consider a hollow circular section of outer diameter ‘D’ and inner diameter ‘d’ as shown in figure. In this case, centroid of both the circles is the same i.e. the point G. The M.I. of the shaded area is to be calculated about the centroidal axes. Since the section is symmetrical about X-X and Y-Y axes, IXX and IYY are equal.

M.I. of a hollow

circular sec tion

= M.I. of a M.I. of a

outer circle inner circle

I = IXX = IYY = 4 4D d64 64

= 4 4D d64

Fig. 4: Fixed beam

(i) Overhanging on right side

(ii) Overhanging on both side

(iii) Overhanging on left side

Fig. 3

L1 L2

(i) Two span continuous beam

L1 L2

(ii) Three span continuous beam

L3

Fig. 5

1. (a)

X X

Y

D d

Y

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Prelim Question Paper


(vi) Assumptions in the theory of Simple Bending The following assumptions are made in the theory of simple bending while

deriving the flexural formula. 1) The material of the beam is homogeneous and isotropic. i.e. the beam is

made up of the same material throughout and it has the same elastic properties in all the directions.

2) The beam is straight before loading and is of uniform cross-section throughout.

3) The beam material is stressed within its elastic limit and thus obeys Hooke’s law.

4) The transverse sections which were plane before bending remain plane after bending.

5) The beam is subjected to pure bending i.e. the effect of shear stresses is totally neglected.

6) Each layer of the beam is free to expand or contract independently of the layer above or below it.

7) Young’s modulus E for the beam material has the same value in tension and compression.

(vii) DIRECT LOAD Definition : A load whose line of action coincides with the axis of a member

is called an axial load or direct load. Let us consider a column of rectangular section of width b and thickness d

carrying an axial compressive load P as shown in figure (a). The load P acts exactly over the centroid G of the section. The load P causes a direct stress of compressive nature whose intensity is uniform throughout the cross section i.e. the stress at any point in the cross section is P/A. Hence the stress distribution diagram at the base of a column is a rectangle as shown in figure (c). Direct stress is denoted by O.

Cross-sectional area of column, A = b × d

Direct stress O = P PA b d

(Compressive)

(viii) In deriving the torsional formula, we make the following assumptions:

1) The shaft is straight having uniform circular cross-section. 2) The shaft is homogenous and isotropic. 3) Circular sections remain circular even after twisting. 4) Plain sections before twisting remain plain after twisting and do not twist

or warp. 5) A diameter in the section before deformation remains a diameter or

straight line after deformation.

1. (a)

1. (a)

1. (a)

P

(a) Elevation b

dG

X X

(b) Plan (c) Stress distribution

Y

Y

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6) Stresses do not exceed the proportional limit. 7) Shaft is loaded by twisting couples in the planes that are perpendicular to

the axis of the shaft. 8) Twist along the shaft is uniform.

(i) Data : Load W = 6 kN = 6 103 N, Stress = 110 N/mm2

To find : Diameter of steel cable (d).

Concept : = WA

, A = 2d4

Stress, = WA

110 = 36 10

A

A = 36 10

110

= 54.54 mm2

Now, A = 2d4

54.54 = 2d4

d = 8.33 mm

(ii) Reactions : MA = 0 gives,

(2 4 42

) + (5 5) (RB 7) = 0

RB = 5.86 kN

Fy = 0 gives, RA + RB (2 4) 5 = 0 RA = 13 RB = 13 5.86 = 7.14 kN S.F. calculations : S.F. at any section in AC at a distance x from A is given by Fx = 7.14 2x at x = 0, FA = 7.14 kN at x = 4 m, FC = 7.14 (2 4) = 0.86 kN

S.F. at any section in CD at a distance x from A, Fx = 7.14 (2 4) = 0.86 kN and it remains constant from C to D.

S.F. at any section in DB Fx = 7.14 (2 4) 5 = 5.86 kN and it remains constants from D to B.

FB = 5.86 kN = RB

1. (b)

1. (b)

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Check : The S.F.D. is as shown in Figure (b).

B.M. calculations : At simply supported ends, MA = MB = 0

MC = (7.14 4) (2 4 42

) = 12.56 kN-m

MD = 5.86 2 = 11.72 kN-m From the S.F.D. it can be seen that, S.F. changes its sign at E in portion AC.

Therefore, the maximum B.M. will occur at E. Let ‘x’ be the distance of E from A.

(d)

From the similar triangles in S.F.D. as shown in Figure (d)

x

7.14 =

4 x0.86

0.86 x = (4 x) 7.14 0.86 x = 28.56 7.14 x 8x = 28.56 x = 3.57 m

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Therefore B.M. at 3.57 m from A is given by,

ME = Mmax = (7.14 3.57) (2 3.57 3.57

2)

= 25.49 12.74 = 12.75 kN-m

From A to C, there is a u.d.l. and hence B.M.D. is a parabolic curve and the maximum B.M. means the highest point of the parabolic curve. Between the points C and D; D and B there is no load. Hence B.M.D. is a straight line. The B.M.D. is as shown in Figure (c).

(iii) Data: A square of side 200 mm.

To find: IP

Concept: (i) IP = IXX + IYY (ii) For a square section

IXX = IYY = 4a

12

(i) For a square of side a,

IXX = 3a a

12

= 4a

12=

4200

12= 1.33 × 108 mm4

(ii) For a square section IYY = IXX = 1.33 × 108 mm4 IP = IXX + IYY = 1.33 × 108 + 1.33 × 108 = 2.66 × 108 mm4 IP = 2.66 × 108 mm4 Relation between three elastic constants E, G and K We know that, E = 2 G (1 + ) … (1) and E = 3K (1 2 ) … (2)

From equation (1), 1 + = E

2G

= E

12G

(3)

Substituting the value of in equation (2).

E = 3KE

1 2 12G

= 3KE

1 2G

E = 3KE

3G

= 3K

3G EG

EG = 3K (3G E) = 9KG 3KE EG + 3KE = 9KG E (G + 3K) = 9KG

E = 9KG

G 3K

1. (b)

2. (a)

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Data : A = 200 mm2 Axial pull = P max = 30 MPa = 30 N/mm2. Elongation L = 1.2 mm Gauge length L = 3 m = 3 × 103mm.

To find: P, E.

Concept: Use of standard formulae.

Stress, = PA

30 = P

200

P = 30 × 200 = 6000 N = 6 kN

Strain, e = LL

= 3

1.2

3 10 = 0.4 × 103

Young’s modulus, E = 3

30e 0.4 10

= 7.5 × 104 N/mm2

(i) P = 6 kN, (ii) E = 7.5 104 N/mm2 Data : Steel rod, L = 4 m = 4000 mm d = 20 mm P = 45 kN = 45 103 N E = 2 105 N/mm2

= 14

= 0.25

To find : (i) L (ii) d

Concept : (i) L = PLAE

(ii) Lateral strain = e = dd

Solution : A = 2d4

= 4 (20)2 = 314.16 mm2

L = PLAE

= 3

5

(45 10 ) 4000314.16 (2 10 )

= 2.86 mm (increase)

Note : Since the rod is subjected to an axial tensile load, there will be increase in the length of the bar and decrease in the diameter.

Linear strain, e = LL

= 2.864000

= 0.000715

Lateral strain = e = 0.25 0.000715 = 0.00017875

dd

= 0.00017875 d

lateral strain = d

2. (c)

2. (b)

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= 30 mm = 25 mm = 20 mm

P kN 5 kN

0.5 m 0.7 m 0.9 m

20 kN

d

20

= 0.00017875

d = 20 0.00017875 = 0.003575 mm (decrease) (i) L = 2.86 mm (increase), (ii) d = 0.003575 mm (decrease) Data : A bar as shown in Figure, E = 210 GPa = 210 103 N/mm2

Fig. (a) To Find : (i) P (ii) L

Concept : Applying principle of superposition i.e. L = L1 + L2 + L3

Solution : For the equilibrium of the entire bar, Fx = 0 ( +, ) 10 + 5 P + 20 = 0 P = 15 kN ()

F.B.D.s of each section are shown in Figure (b)

In this case, each section is under tension. Therefore, there will be increase in length of each section.

L1 = 1 1

1

PLA E

= 3 3

2 3

(10 10 ) (0.5 10 )

(20) (210 10 )4

= 0.076 mm

L2 = 2 2

2

P LA E

= 3 3

2 3

(5 10 ) (0.7 10 )

(25) (210 10 )4

= 0.034 mm

L3 = 3 3

3

P LA E

= 3 3

2 3

(20 10 ) (0.9 10 )

(30) (210 10 )4

= 0.121 mm

2. (d)

FIg. (b)

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L = L1 + L2 + L3 = 0.076 + 0.034 + 0.121 = 0.231 mm Data : AC = 20 mm2 AS = 30 mm2 L = 1 m = 1000 mm P = 8 kN = 8 103 N ES = 20 105 MPa = 20 105 N/mm2 EC = 1 105 MPa = 1 105 N/mm2 To Find : C, S Concept : Two equations for composite bar.

Solution : Using the relation, 1 = 12

2

EE

S = SC

C

EE

= 5

5

20 101 10

= 20 C (1)

Using the relation, P = 1 A1 + 2 A2 we have, P = S AS + C AC 8 103 = (20 C 30) + (C 20) = 600 C + 20 C = 620 C

C = 38 10

620

= 12.9 N/mm2

Substituting this value in eq. (1) S = 20 12.9 = 258.1 N/mm2 S.F.D. and B.M.D for Cantilever Beams A cantilever beam carrying a point load at its free end : Let us consider a cantilever beam AB of span L carrying a point load W at its free end B as shown in figure (a). Take a section XX at a distance x from the free end B. Fx = S.F. at section XX = +W (positive since right downward)

2. (e)

3. (a)

x

B A

RA

L

X

W

X

(a) Cantilever beam

W

(b) SFD

WL

(c) BMD

Wx

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Thus, S.F. is constant for all sections between A and B and it is W. Hence S.F. diagram is horizontal as shown in figure (b). Now, MX = B.M. at section XX = W.x (negative since clockwise moment to the right i.e. hogging) Now, since x is measured from B, B is taken as origin. At B, x = 0 MB = W × 0 = 0 At A, x = L MA = W.L

Since in the expression of Mx, the power of x is one, therefore the variation of B.M. from 0 to WL is linear i.e. B.M.D. is a straight line as shown in figure (c). Note : (i) In this case, the maximum bending moment occurs at the fixed end

A and the bending moment at the free end B is zero. Mmax = MA = WL (Hogging)

(ii) From Fy = 0, RA W = 0 i.e. RA = W and FA = RA = W Thus, S.F. at A = Reaction at A

(iii) S.F. at B = Load acting at B

(iv) If the S.F. diagram is a horizontal line, the B.M. diagram is an inclined straight line.

Reactions : Taking moments about A,

3 2

2 3 3 3 1 2 3 5 52 2

= RB 7

RB = 517

= 7.28 kN

RA = (2 3 + 3 + 1 2 + 5) 7.28 = 8.72 kN

S.F. calculations : At the points C and D, u.d.l. ends and the point load acts. In such cases, it is convenient to take a section just to the left and just to the right of the point under consideration.

FA = + RA = 8.72 kN

LCF = 8.72 2 3 = 2.72 kN

RCF = 8.72 2 3 3 = 0.28 kN

LDF = 8.72 2 3 3 1 2 = 2.28 kN

RDF = 8.72 2 3 3 1 2 5 = 7.28 kN

FB = RB = 7.28 kN

3. (b)

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Note :

LCF means S.F. at C assuming that section XX is just to the left of C but

very near to C. RCF means S.F. at C assuming that the section XX is just to the

right of C but very near to C. The S.F.D. is as shown in Fig. (b)

B.M. calculations :

MA = MB = 0

MC = 3

8.72 × 3 2 × 3 ×2

= 17.16 kN.m

MD = 7.28 2 = 14.56 kN.m At C, S.F. changes its sign from positive to negative. Hence the maximum B.M. occurs at C. The B.M.D. is as shown in Fig. (c). Reactions : Taking moments about A,

6

4 2 5 4 (1.5 6)2

= RB 6

RB = 9.17 kN RA = 4 + 5 + (1.5 6) 9.17 = 8.83 kN

3. (c) Vidyala

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Shear force calculations : FA = + RA = 8.83 kN

LCF = 8.83 1.5 2 = 5.83 kN

RCF = 8.83 1.5 2 4 = 1.83 kN

LDF = 8.83 1.5 2 4 1.5 2 = 1.17 kN

RDF = 8.83 1.5 2 4 1.5 2 5 = 6.17 kN

FB = 8.83 1.5 2 4 1.5 2 5 1.5 2 = 6.17 3 = 9.17 kN = RB

The S.F.D. is as shown in Fig. (b).

B.M. calculations : MA = MB = 0

MC = 2

8.83 2 (1.5 2)2

= 14.66 kN.m

MD = 2

9.17 2 (1.5 2)2

= 15.34 kN.m

To find Mmax : At point E in S.F.D., S.F. changes its sign from positive to negative. Hence the maximum B.M. will occur at E. Let x be the distance of E from A. From similar triangles in SFD.,

x1.83

= 2 x1.17

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1.17x = 1.83 (2 x) = 3.66 1.83 x 3x = 3.66 x = 1.22 m

Distance of E from A = 2 + x = 2 + 1.22 = 3.22 m B.M. at 3.22 m from A is given by,

ME = Mmax = 8.83 3.22 4 1.22 (1.5 3.22)3.22

2

= 15.78 kN.m The B.M.D. is an shown in Fig. (c).

Support reactions : Taking moments @ A,

RB 4 = 5

10 52

= 125

RB = 31.25 kN RA = Total load RB = 10 5 31.25 = 50 31.25 = 18.75 kN S.F. calculations :

FA = 18.75 kN

LBF = 18.75 10 4 = 21.25 kN

RBF = 18.75 10 4 + 31.25 = 10 kN

FC = 18.75 10 4 + 31.25 10 1 = 0 (Check : FC = load at C = 0 )

The S.F.D. is as shown in Fig. (b).

3. (d)

3. (e)

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B.M. calculations : At simple support A, MA = 0

MB = 1

(10 1)2

= 5 kN.m

At free end C, MC = 0.

Point of zero S.F. (Location of maximum B.M.) Let AD = x. S.F. at section at a distance x from A, Fx = 18.75 10 x = 0 10x = 18.75 x = 1.875 m (i.e. AD = 1.875 m) (x can also be calculated by considering similar triangles in S.F.D. as follows):

18.75

x =

21.254 x

18.75 (4 x) = 21.25x 75 18.75x = 21.25x 75 = 18.75 x + 21.25 x = 40 x

x = 7540

= 1.875 m

Hence maximum B.M. occurs at 1.875 m from A.

Magnitude of maximum B.M. :

Mmax = M1.875 = RA x 10 x x2

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= 1.875

18.75 1.875 10 1.8752

= 17.578 kN.m The B.M.D is as shown in Fig. (c). Point of contraflexure : Let the point of contraflexure ‘E’ is at a distance x from A.

Mx = x

18.75 x 10 x2

= 0

x = 3.75 m from A (i.e. AE = 3.75 m) It states, “The moment of inertia of a plane section about any axis parallel to the centroidal axis is equal to the moment of inertia of the section about the centroidal axis plus the product of the area of the section and the square of the distance between the two axis.” Consider an irregular area A as shown in Fig.(a) Let G be its centroid.

Fig.(a)

Let IG = M.I. of the area about its centroidal axis

IPQ = M.I. of the area about any axis PQ which is parallel to the centroidal axis XX

A = Area of the section and h = Distance between the two axes.

Then, by parallel axis theorem, IPQ = IG + A h2

In this case, IG = IXX

IPQ = A h2

Now, consider a rectangular section of width b and depth d as shown in Fig.(b). The positions of XX, YY, PQ and AB axes are as shown in Fig. (b). Let h = Distance between X-X axis and PQ axis and h = Distance between Y-Y axis and AB axis Now M.I. of a rectangle about the axis PQ is given by,

4. (a)

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Fig. (b)

IPQ = IG + Ah2

= IXX + Ah2

= 3

2bd(bd)h

12

M.I. of a rectangle about the axis AB is given by,

IAB = IG + A(h)2 = IYY + A(h)2

= 3

2db(db) (h )

12

Use : This theorem is used to find the M.I. of any plane figure about any axis located at some distance away from the centroidal axis. Data : A square of side ‘a’ as shown in fig.

To find : ICD

Concept : Theorem of parallel axis. (i) For a square of side ‘a’,

IXX = 3a a

12

= 4a

12

(ii) Area of section, A = a a = a2

(iii) The outer edge CD is parallel to XX axis. Distance between XX axis and outer edge CD,

h = a2

(iv) Using parallel axis theorem, M.I. about parallel axis = M.I. about centroidal axis + Ah2

ICD = IXX + Ah2

4. (b)

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= 24

2a aa

12 2

a

h2

= 4 4a a

12 4 =

4 4a 3a12

= 44a

12 =

4a3

ICD = 4a

3 =

4side of square

3

Data : A hollow rectangle as shown in Fig.

b = 160 mm, d = 260 mm, B = 200 mm, D = 300 mm

To find : IBase.

Concept : (i) M.I. of hollow rectangle. (ii) Use of parallel axis theorem.

We have to calculate the moment of inertia of a hollow rectangle about the base PQ. Base PQ is parallel to XX axis.

(i) IXX = 3 3BD bd

12 12 = 3 3

1BD bd12

= 3 31

200 300 160 26012

= 215653333.3 mm4 (ii) A = BD bd = 200 300 160 260

= 18400 mm2 (iii) Distance between base PQ and XX axis,

h = D2

= 300

2 = 150 mm

(iv) Applying the theorem of parallel axis, M.I. about parallel axis = M.I. about C.G. axis + A h2

IPQ = IXX + Ah2 = 215653333.3 + 18400 1502 = 435563333.3 mm4

IPQ = 435563333.3 mm4 Refer to Figure, In ABD,

tan 60 = ADBD

=

h2 / 2

h = 1.732 m M.I. of a triangle about its base,

IBase = IBC = 3bh

12 =

32 (1.732)12

= 0.865949 m4

4. (c)

4. (d) Vidyala

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Data : A symmetrical I section as shown in Fig. To find : Ip Concept : Calculate IXX and IYY. Ip = IXX + IYY. (i) Since the section is symmetrical about XX axis, its M.I. can be determined by

considering a hollow rectangular section. B = 100 mm, D = 10 + 100 + 10 = 120 mm, b = (100 10) = 90 mm, d = 100 mm

(ii) IXX = 3 3BD bd

12 12 = 3 3

1BD bd12

= 3 31

100 120 90 10012

= 6.9 106 mm4

Note : M.I. about YY axis should not be

calculated as 3 31

DB db12 since the

c.g. of rectangles EFLM and GHJK are not lying on YY axis.

(iii) IYY = 2 M.I. of flanges + M.I. of web

= 3 310 100 10

2 1012 12

= 1.675 106 mm4 (iv) Polar M.I. of section,

Ip = Izz = IXX + IYY = 6.9 106 + 1.675 106

= 8.575 106 mm4 Ip = 8.575 106 mm4

Data : A rectangular section as shown in Figure 1, b = 50 mm, d = 150 mm M = 600 N.m = 600 × 103 N.mm To find :

Concept : Use of bending stress equation MI

= y

.

4. (e)

5. (a)

X N

X A

d = 150 mm

b = 50 mm

(i) Section (ii) Bending stress distribution

+

3.2 N/mm2

3.2 N/mm2

yt

yo

Fig. 1

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Step (i) : IXX = 3bd

12 =

350 15012´

= 14062500 mm4

Step (ii) : y = d2

= 150

2 = 75 mm

Step (iii) : Using the relation,

MI

= y

3600 10

14062500´

= 75

= 3600 10 75

14062500´ ´

= 3.2 N/mm2

= 3.2 N/mm2

Hint : (i) Find bottomy

bottomy = 71.32 mm

(ii) Find IXX IXX = 1.7996 × 106 mm4

(iii) M.R

I = max

maxy

6

M.R

1.7996 10´ =

15071.32

M.R. = 3.785 × 106 N.mm = 3.785 × 103 N.m Let ‘w’ be in N/m.

(iv) M.R. = Mmax = 2wL

8

3.785 × 103 = 3 210 10 L

8´

L = 1.74 m

5. (b)

(i) Section (ii) Bending stress distribution (N/mm2)

Fig. 1

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Here b = 100 mm, d = 100 mm, S = 30 kN = 30 × 103 N

M.I. of whole section about N.A. = I = 3bd

12 =

3100 10012´

= 8333333.33 mm4

A = Shaded area above layer PQ = 100 × 30 = 3000 mm2

y = Distance of C.G. of this area from the N.A. = 30

202

= 35 mm

b = Width at layer PQ = 100 mm Shear stress at layer PQ is given by

qPQ = SAyIb

=

330 10 3000 358333333.33 100´

´ = 3.78 N/mm2

Note : Maximum shear stress at N.A. is given by

qmax = 1.5 × qav = 1.5 Sbd

= 1.5 × 330 10

100 100

´´

= 4.5 N/mm2

The shear stress distribution diagram is as shown in figure 1(ii). qPQ = 3.78 N/mm2

Data : b = 200 mm, d = 100 mm, P = 60 kN, e = 40 mm

To find : max and min

Concept : The load is eccentric w.r.t. YY axis. So we have to consider ZYY.

A = b × d = 200 × 100 = 2 × 104 mm2

0 = PA

=

3

4

60 10

2 10

´´

= 3 N/mm2

b = YY

MZ

= 2

P e

db / 6

´ =

2

6Pe

db

=

3

2

6 60 10 40

100 200

´ ´ ´´

= 3.6 N/mm2

5. (c)

5. (d)

(i) Section (ii) Shear stress distribution

Fig. 1

Y

Y

X X

200 mm

P

e 100 mm

+

6.6 N/mm2

0.6 N/mm2

Fig. 1

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max = 0 + b = 3 + 3.6 = 6.6 N/mm2 (Compressive) min = 0 b = 3 3.6 = 0.6 N/mm2 (Tensile) Let b and t be in ‘mm’. A = t × b = t × 3t = 3t2

= PA

= 3

2

80 10

3t

´

N/mm2

But = 70 Mpa = 70 N/mm2

3

2

80 10

3t

´ = 70

t2 = 380 10

3 70´´

= 380.95 t = 19.52 mm say 20 mm b = 3t = 3 × 20 = 60 mm

Data : D = 30 mm, ex = 40 + 302

= 55 mm,

max = 80 MPa (tensile)

To find : P

Concept: MXX = P . ex and b = XX

XX

M

Z where ZXX = 3D

32

Direct stress, O = 2

P PA (30)

4

= 0.001415 P

Section modulus, ZXX =

4

3 3(D )

64 D (30 )D 32 322

= 2650.72 mm3

Bending moment, MXX = P . ex = P × 55 N.mm

Bending stress, b = XX

XX

M P 55Z 2650.72

= 0.020749 P max = O + b

80 = 0.001415 P + 0.020749 P = 0.022164 P

5. (e)

6. (a)

Vidyala

nkar



P = 80

0.022164

= 3609.46 N = 3.6 kN

Data : Solid shaft, D = 100 mm, = 2.75 = 2.75180

´ rad,

L = 6m = 6 × 103 mm, C = 80 kN/mm2 = 80 × 103 N/mm2 To find : T Concept : Use of torsional formula. Using the relation,

p

TI

= θC

L

4

T

D32

= θC

L

4T

10032

= 3

3

80 10 2.75180

6 10

´ ´

´

T = 6282734.283 N.mm = 6282.734 N.m = 6.282 kN.m Data: Solid shaft, D = 100 mm, P = 100 kW = 100 × 103 W, N = 150 r.p.m. To find: fs Concept: Use equation of power first, find T and then use equation of T to find fs

(i) Using the relation, P = 2 NT

60

100 × 103 = 2 150 T

60

T = 6366.1977 N.m = 6366.1977 × 103 N.mm

(ii) Now, using the relation, T = 3sf D

16

6366.1977 × 103 = 3sf (100)

16

fs = 32.42 ?N/mm2 Data : Let D = Diameter of solid shaft

D1 = Outer diameter of hollow shaft

d = Inner diameter of hollow shaft = 12

D3

6. (d)

6. (c)

6. (b)

Vidyala

nkar



To find : Compare weights of two shafts. Concept : (i) Zp for solid shaft = Zp for hollow shaft.

(ii) Weight of hollow shaftWeight of solid shaft

= Area of h.s.Area of s.s.

Polar modulus for hollow shaft :

Zp = 4 41D d

16 D

=

441 1

1

2D D

316 D

= 31

16D 1

16 81

= 31

65D

16 81 …(i)

Polar modulus for solid shaft :

Zp = 3D16 …(ii)

Equating (i) and (ii), we get

31

65D

16 81 = 3D

16

313

D

D =

8165

= 1.246

1D

D = 3 1.246 = 1.076

Weight of solid shaftWeight of hollow shaft

= Area of h.s.Area of s.s.

=

221 1

2

2D D

4 3

D4

=

21D5

9 D

= 251.076

9 = 0.643

Weight of hollow shaftWeight of solid shaft

= 0.643

Data: Solid shaft, D = 120 mm Power P = 100 kW = 100 × 103 W,

Speed N = 150 rpm, Tmax = 1.4 Tavg = 1.4 Tmean.

To find : fs

Concept: (i) Use equation of power first to find Tmean. (ii) Find Tmax by knowing Tmean. (iii) Use equation of Tmax to find fs.

6. (e) Vidyala

nkar



(i) Power P = mean2 N TW

60

100 × 103 = mean2 150 T

60

Tmean = 6366.197 N.m (ii) Tmax = 1.4 Tmean = 1.4 × 6366.197 = 8912.68 N.m = 8912.68 × 103 N.mm

(iii) Tmax = 3sf D

16

8912.68 × 103 = 3sf (120)

16

fs = 26.268 N/mm2 Note: fs can also be calculated by using the relation,

P

TI

= sf

R

3

4

8912.68 10

(120)32

= sf120

2

fs = 26.268 N/mm2

Vidyala

nkar

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