properties of materials chapter syllabus : 1

EDUZPHERE PUBLICATIONS | ©All Rights Reserved | www.eduzpherepublications.com

1. The stretch in a steel rod of circular section,

having a length L subjected to a tensile load

P and tapering uniformly from a diameter d1

at one end to a diameter d2 at the other end,

is given by

(A) PL

4E d1d2 (B)

PLπ

E d1d2

(C) PLπ

4E d1d2 (D)

4PL

πE d1d2

2. The total extension of the bar loaded as

shown in the figure is

A= area of cross-section

E= modulus of elasticity

10T 3T 2T

10mm 10mm 10mm

9T

(A) 10× 30/AE (B) 26× 10/AE

(C) 9× 30/AE (D) 30× 22/AE

3. For a composite bar consisting of a bar

enclosed inside a tube of another material

and when compressed under a load W as a

whole through rigid plate s at the end of

the bar. The equation of compatibility is

given by (suffixes 1 and 2 refer to bar and

tube respectively)

(A) W1 + W2

(B) W1 + W2 = Constant

(C) W 1

A1E1 =

W 2

A2E2

(D) W 1

A1E2 =

W 2

A2E1

4. A tapering bar ( diameter of end sections

being, d1 and d2) and a bar of uniform cross

section ‘d’ have the same length and are

subjected the same axial pull. Both the bars

will have the same extension if ‘d’ is equal to

(A) d1+d2

2

(B) d1d2

(C) d1+d2

2 (D)

d1+d2

2

5. The deformation of a bar under its own

weight as compared to that when subjected

to a direct axial load equal to its own weight

will be

(A) the same (B) one fourth

(C) half (D) double

Practice Problems Level - 1

Chapter

1

PROPERTIES OF

MATERIALS Syllabus : Classification of material, Mechanical properties of materials, Tests

for the mechanical properties of materials, Classification of loads, Effect of a

load on a member, Stress, Types of stresses, Strain, types of strains, Elasticity

and elastic limit, Hooke’s law, Relations between three elastic constants i.e. E, G

and K, Behavior of ductile metals in tensile test, Maximum or ultimate tensile

stress, Working stress, Factor of safety, Breaking stress, Proof stress.


2

SOM

6. A slender bar of 100 mm2 cross-section is

subjected to loading as shown in the figure

below. If the modulus of elasticity is taken as

200× 109 Pa, then the elongation produced in

the bar will be

200kN

200kN100kN 100kN

0.5m 1.0m 0.5m

(A) 10mm (B) 5 mm

(C) 1 mm (D) nil

7. A rigid beam of negligible weight is

supported in a horizontal position by two

rods of steel and copper, 2m and 1 m long

having values of cross-sectional area 1 cm2

and 2 cm2 and E of 200 GPa an 100 GPa

respectively. A load P is applied as shown in

the figure below.

2mSteel

1mCopper

Rigid Beam

P

It the rigid beam is to remain horizontal then

(A) the forces on both sides should be equal

(B) the force on copper rod should be twice

the force on steel

(C) the force on the steel rod should be

twice the force on copper

(D) the force P must be applied at the centre

of the beam

8. A straight bar is fixed at edges A and B. its

elastic modulus is E and cross- section is A.

There is a load P = 120 N acting at C. The

reactions at the ends are

A

2

CP = 120 N

B

(A) 60 N at A, 60 N at B

(B) 30 N at A, 90 N at B

(C) 40 N at A, 80 N at B

(D) 80 N at A, 40 N at B

9. A bar of length L tapers uniformly from

diameter 1. 1 D at one end of 0.9 D at the

other end. The elongation due to axial pull is

computed using mean diameter D. What is

the approximate error in computed

elongation?

(A) 10% (B) 5%

(C) 1% (D) 0.5%

10. A solid uniform metal bar of diameter D and

length L is hanging vertically from its upper

end. The elongation of the bar due to self

weight is

(A) Proportional to L and inversely

proportional to D2

(B) Proportional to L2 and inversely

proportional to D2

(C) Proportional to L but independent of D

(D) Proportional to L2 but independent of D

11. Two tapering bars of the same material are

subjected to a tensile load P. The lengths of

both- the bars are the same. The larger

diameter of each of the bars is D. The

diameter of the bar A at its smaller end is

D/2 and that of the bar B is D/3. What is the


3

SOM

ratio of elongation of the bar A to that of the

bar B?

(A) 3 : 2 (B) 2 : 3

(C) 4 : 9 (C) 1 : 3

12. Which one of the following expresses the

total elongation of a bar of length L with a

constant cross-section of A and modulus of

Elasticity E hanging vertically and subject to

its own weight W?

(A) WL

AE (B)

WL

2AE

(C) 2WL

AE (D)

WL

4AE

13. A prismatic bar, as shown in figure is

supported between rigid supports. The

support reactions will be

A

C

10 kN

B

1m 2m

(A) A B

10 10R kN and R kN

3 3

(B) A B

20 10R kN and R kN

3 3

(C) RA= 10 kN and RB = 10 kN

(D) RA =5 kN and RB = 5kN

14. In the arrangement as shown in the figure,

the stepped steel bar ABC is loaded by a load

P. The material has Young’s modulus

E= 200GPa and the two portions AB and BC

have area of cross section 1 cm2 and 2 cm

2

respectively. The magnitude of load P

required to fill up the gap of 0.75 mm is:

A B P C

1m 1m Gap0.75mm

(A) 10kN (B) 15 kN

(C) 20 kN (C) 25kN

15. If Poisson’s ratio of a material is 0.5, then

the elastic modulus for the material is

(A) three times its shear modulus

(B) four times its shear modulus

(C) equal to its shear modulus

(D) indeterminate

16. The number of independent elastic constants

required to express the stress-strain

relationship for a linearly elastic isotropic

material is

(A) one (B) two

(C) three (D) four

17. The number of elastic constants for a

completely anisotropic elastic material is

(A) 3 (B) 4

(C) 21 (D) 25

18. The poisson’s ratio of a material which has

young’s modulus of 120GPa and shear

modulus of 50 GPa, is

(A) 0.1 (B) 0.2

(C) 0.3 (D) 0.4

19. For a given material, the modulus of rigidity

is 100GPa and Poisson’s ratio is 0.25. The

value of modulus of elasticity in GPa is

(A) 125 (B) 150

(C) 200 (D) 250

20. The modulus of elasticity for a material is

200 GN/m2 and poisson's ratio is 0.25 What

is the modulus of rigidity?

(A) 80GN/m2 (B) 125 GN/m

2


4

SOM

(C) 250GN/m2 (C) 320 GN/m

2

21. E,G,K and μ represent the elastic modulus,

shear modulus, bulk modulus and Poisson’s

ratio respectively of a linearly elastic,

isotropic and homogeneous material. To

express the stress-strain relations completely

for this material, at least

(A) E,G and μ must be known

(B) E, K and μ must be known

(C) any two of the four must be known

(D) All the four must be known

22. If the ratio G/E (G =Rigidity modulus, E =

Young’s modulus of elasticity) is 0.4, then

what is the value of the Poisson ration?

(A) 0.20 (B) 0.25

(C) 0.30 (C) 0.33

23. What is the relationship between the linear

elastic properties; Young’s modulus (E),

rigidity modulus (G) and bulk modulus (K)?

(A)1

E=

9

K +

3

G (B)

3

E=

9

K +

1

G

(C) 9

E=

3

K +

1

G (D)

9

E=

1

K +

3

G

24. What is the relationship between elastic

constants E. G and K?

(A)E = KG

9K+G (B) E =

9KG

K+G

(C) E = 9KG

K+3G (D) E =

9KG

3K+G

25. A bar produces a lateral strain of magnitude

– 60× 10−5, when subjected to tensile stress

of magnitude 300 MPa along the axial

direction. What is the elastic modulus of the

material, if the Poisson’s ratio is 0.3?

(A) 100 GPa (B) 150GPa

(C) 200GPa (D) 400GPa

26. A copper rod 400 mm long is pulled in

tension to a length of 401.2 mm by applying

a tensile stress of 330 MPa. If the

deformation is entirely elastic, the Young’s

modulus of copper is

(A) 110 GPa (B) 110MPa

(C) 11GPa (D) 11MPa

27. Consider the following statements:

Modulus of rigidity and bulk modulus of a

material are found to be 60 GP a and 140

GPa respectively. Then

1. Elasticity modulus is nearly 200 GPa

2. Poisson’s ratio is nearly 0.3

3. Elasticity modulus is nearly 158 GPa

4. Poisson’s ratio is nearly 0.25

Which of these statements are correct?

(A) 1and 3 (B) 2 and 4

(C) 1 and 4 (D) 2 and 3

28. A 16 mm diameter bar elongates by 0.04%

under a tensile force of 16 kN. The average

decrease in diameter is found to be 0.01%

Then:

1. E =210 GPa and G =77 GPA

2. E =199 GPa and v =0.25

3. E =199 GPa and v =0.30

4. E = 199 GPa and G = 80 GPa

Which of these values are correct?

(A) 3and 4 (B) 2 and 4

(C) 1 and 3 (D) 1 and 4

29. The ratio of lateral strain to longitudinal

strain is called

(A) Poisson’s ratio

(B) Bulk modulus


5

SOM

(C) Modulus of rigidity

(D) Modulus of elasticity

30. Hook’s law holds good up to

(A) Proportional limit (B) Yield point

(C) Elastic limit (D) Plastic limit

31. A tensile fore (P) is acting on a body of

length (L) and area of cross-section (A). The

change in length would be

(A) P

LAE (B)

PE

AL

(C) PL

AE (D)

AL

PE

32. The modulus of elasticity (E) and bulk

modulus (K) are related by:

(A) k = mE

3(m−2) (B) k =

mE

2(m+1)

(C) k = 3(m−2)

mE (D) k =

2(m+1)

mE

Where 1

m = Poisson’s ratio

33. The strain energy stored per unit volume of

the material is known as

(A) Resilience (B) Ductility

(C) Elasticity (D) Plasticity

34. The elongation of a conical bar due to its

own weight is equal to

(w = Weight Density)

(A) Wl

2E (B)

Wl 2

6E

(C) Wl 3

6E (D)

Wl 4

6E

35. The limiting values of Poisson’s ratio are

(A) -1 and 0.5 (B) -1 and -0.5

(C) 1 and -0.5 (D) 0 and 0.5

36. If the Young’s modulus of elasticity (E) of a

material is 2 times that of modulus of

rigidity, then Poisson’s ratio of the material

(A) -1 (B) zero

(C) +1 (D) 2

37. Which of following has the highest value of

passion’s ratio

(A) Rubber (B) Steel

(C) Aluminum (D) Copper

38. Modulus of rigidity is defined as the ratio of

(A) Longitudinal stress to longitudinal strain

(B) shear stress to shear strain

(C) stress to strain

(D) Stress to volumetric strain.

39. Limit of proportionality depends upon

(A) Area of cross-section

(B) Type of loading

(C) Type of material

(D) All of the above

40. The relationship between Young’s modulus

of elasticity E, bulk modulus K and

Poisson’s ratio μis given by

(A) E = 2K(1-2μ) (B) E = 3K(1+μ)

(C) E = 3K(1-2μ) (D) E = 2K(1+μ)

41. The elongation of a conical bar under its own

weight is equal to

(A) That of a prismatic bar of same length

(B) One half that of a prismatic bar of same

length

(C) One third that of a prismatic bar of same

length


6

SOM

(D) One fourth that of a prismatic bar of

same length

42. If a material has identified properties in all

the directions, it is said to be

(A) Homogeneous (B) Isotropic

(C) Elastic (D) Orthotropic

43. If a composite bar of steel and copper is

heated, then the copper bar is under

(A) Tension (B) Compression

(C) Shear (D) Torsion

44. Proof resilience is the maximum energy

stored at

(A) Limit of proportionality

(B) Elastic limit

(C) Plastic Limit

(D) None of the above

45. Strain energy stored in a member is given by

(A) 0.5 ×stress × volume

(B) 0.5 ×strain × volume

(C) 0.5 ×stress × strain × volume

(D) 0.5 ×stress × strain

46. If the depth of a beam of rectangular section

is reduced to half, strain energy stored in the

beam becomes

(A) (1/4) time (B) (1/8) time

(C) 4 times (D) 8 times

47. The stress below which a material has a high

probability of not failing under the reversal

of stress is known as

(A) Tolerance limit (B) Elastic limit

(C) Proportional limit (D) Endurance limit

48. In terms of bulk modulus (k) and modulus of

rigidity (G), the Poisson’s ratio can be

expressed as

(A) 3k−4G

6k+4G (B)

3k+4G

6k−4G

(C) 3k−2G

6k+2G (D)

3k+2G

6k−2G

49. Match list 1 and list 2 and select the correct

answer using the codes given below the lists:

List- I List- II

(i) Ratio of lateral

strain to linear

strain

1. Strain

(ii) Ratio of stress to

strain

2. Poisson’s ratio

(iii) Ratio of extension

to original length

3. Tensile stress

(iv) Ratio of axial pull

to area of section

4. Young’s

modulus

Codes:

(i) (ii) (iii) (iv)

(A) 4 2 3 1

(B) 4 2 1 3

(C) 2 4 3 1

(D) 2 4 1 3

50. A round steel bar of overall length 40 cm

consists of two equal portions of 20 cm each

having diameters of 10 cm and 8 cm

respectively. If the rod is subjected to a

tensile load of 10 tonnes, the elongation will

be given by (E = 2 × 106 kg/cm

2)

(A) 1

10π {

1

25+

1

16} cm (B)

2

10π {

1

25+

1

16} cm

(C) 3

10π {

1

25+

1

16} cm (D)

4

10π {

1

25+

1

16} cm


7

SOM

51. A copper bar of 25 cm length is fixed by

means of supports at its ends. Supports can

yield (total) by 0.01 cm. If the temperature of

the bar is raised by 1000 C, then the stress

induced in the bar for α = 20 × 10-6

/0C and

Ec = 1 × 106 kg/cm

2 will be

(A) 2 × 102 kg/cm

2 (B) 4 × 10

2 kg/cm

2

(C) 8 × 102 kg/cm

2 (D) 16 × 10

2 kg/cm

2

52. A steel cube of volume 8000 cc is subjected

to an all round stress of 1330 kg/sq.cm. The

bulk modulus of the material is 1.33 × 106

kg/cm2. The volumetric change is

(A) 8 cc (B) 6 cc

(C) 0.8cc (D) 10-3

cc

53. A bar of circular cross section varies

uniformly from a cross section 2D to D. If

extension of the bar is calculated treating it

as a bar of average diameter, then the

percentage error will be

(A) 10 (B) 25

(C) 33.33 (D) 50

54. The length coefficient of thermal expansion

and Young’s modulus of bar ‘A’ are twice

that of bar ‘B’. If the temperature of both

bars is increased by the same amount while

preventing any expansion, then the ratio of

stress developed in bar A to that in bar B will

be

(A) 2 (B) 4

(C) 8 (D) 16

55. If all the dimensions of a prismatic bar of

square cross section suspended freely from

the ceiling of a roof are doubled, then the

total elongation produced by its own weight

will increase

(A) eight times (B) four times

(C) three times (D) two times

56. A 10 cm long and 5 cm diameter steel rod

fits snugly between two rigid walls 10 cm

apart at room temperature. Young’s modulus

of elasticity and coefficient of linear

expansion of steel are 2×106kgf /cm

2and

12×10-6/0

C respectively. The stress

developed in the rod due to a 1000C rise in

temperature wall be

(A) 6×10-10

kgf/cm2

(B) 6×10-9

kgf/cm2

(C) 2.4×103kgf/cm

2 (D) 2.4×10

4kgf/cm

2

57. A rod of material with E = 200× 103MP a

and =10–3

mm/ mm0C is fixed at both the

ends. It is uniformly heated such that the

increase in temperature is 300C. The stress

developed in the rod is

(A) 6000 N/mm2( tensile)

(B) 6000 N/mm2( compressive )

(C) 2000 N/mm2( tensile)

(D) 2000 N/mm2( compressive)

58. Strain energy stored in a body of volume V

subjected to uniform stress ς is

(A) ςE/V (B) ςE2/V

(C) ςV2/E (D) ς2

V/ 2E

59. A bar having length L and uniform cross-

section with area A is subjected to both

tensile force P and torque T. If G is the shear

modulus and E is the Young’s modulus, the

internal strain energy stored in the bar is

(A) T 2L

2GJ +

P2L

AE (B)

T 2L

GJ +

P2L

2AE


8

SOM

(C) T 2L

2GJ +

P2L

2AE (D)

T 2L

GJ +

P2L

AE

60. A bar of copper and steel form a composite

system. They are heated to a temperature of

400C. What type of stress is induced in the

copper bar?

(A) Tensile

(B) Compressive

(C) Both tensile end compressive

(D) Shear

61. A cube with a side length of 1 cm is heated

uniformly 10C above the room temperature

and all the sides are free to expand. What

will be the increase in volume of the cube?

(Given coefficient of thermal expansion is ∝

per0(C)

(A) 3∝ cm3

(B) 2∝ cm3

(C) ∝cm3

(D) zero

62. A metal rod is rigidly fixed at its both ends.

The temperature of the rod is increased by

1000C. If the coefficient of linear expansion

and elastic modulus of the metal rod are

12×10-6

C and 200 GPa respectively, then

what is the stress produced in the rod?

(A) 100 MPa (tensile)

(B) 240 MPa (tensile)

(C) 240 MPa (compressive)

(D) 100 MPa (compressive)

63. A 100 mm × 5 mm × 5mm steel bar free to

expand is heated from 150

C to 400

C. what

shall be developed?

(A) Tensile stress (B) Compressive stress

(C) Shear stress (D) No stress

64. A steel specimen 150 mm2 in cross- section

stretches by 0.05 mm over a 50 mm gauge

length under an axial load of 30 kN. What is

the strain energy stored in the specimen?

( Take E = 200 GP(A)

(A) 0.75 N-m (B) 1.00 N-m

(C) 1.50 N-m (D) 3.00 N-m

65. What is the expression for the strain energy

due to bending of a cantilever beam (length

L, modulus of elasticity E and moment of

inertiaI)?

(A) 2 3P L

3EI (B)

2 3P L

6EI

(C) 2 3P L

4EI (D)

2 3P L

48EI

66. A circular bar L m long and d m in diameter

is subjected to tensile force of F kN. Then

the strain energy, U will be (where, E is the

modulus of elasticity in kN/m2)

(A) 2

2

4F L

d E (B)

2

2

F L

d E

(C) 2

2

2F L

d E (D)

2

2

3F L

d E

67. A round bar made of same material consists

of 3 parts each of 100 mm length having

diameters of 40 mm, 50 mm and 60 mm

respectively. If the bar is subjected to an

axial load of 10 kN, the total elongation of

the bar would be (E is modulus of elasticity

in kN/mm2)

(A) 0.4

πE {

1

16 +

1

25 +

1

36} mm

(B) 4

πE {

1

16 +

1

25 +

1

36} mm

(C) 4 2

πE {

1

16 +

1

25 +

1

36} mm

(D) 40

πE {

1

16 +

1

25 +

1

36} mm

68. A cylindrical bar of 20 mm diameter and 1 m

length is subjected to a tensile test. Its

longitudinal strain is 4 times that of its lateral


9

SOM

strain. If the modulus of elasticity is 2 × 105

N/mm2, then its modulus of rigidity will be

(A) 8 × 106 N/mm

2 (B) 2 × 10

5 N/mm

2

(C) 0.8× 104 N/mm

2 (D) 0.8 × 10

5 N/mm

2

69.

A B C

D 2F D/2 3F D

D

2F

2L L 2L

F

For the compound bar shown in above fig.,

the ratio of stresses in the portions

AB:BC:CD will be

(A) 4: 1: 2 (B) 1: 2: 4

(C) 1: 4: 2 (D) 4: 2: 1

70. A straight wire 15 m long is subjected to a

tensile stress of 2000 kgf/cm2. Elastic

modulus is 1.5 × 106kgf/cm

2. Coefficient of

linear expansion for the material is 16.66 ×

10−6 /0F. The temperature change in

0F) to

produce the same elongation as due to the

2000 kgf/cm2 tensile stress in the material is

(A) 40 (B) 80

(C) 120 (D) 160

71. A composite section made of two materials

has moduli of elasticity in the ratio 1:2 and

lengths in the ratio 2:1. The ratio of

corresponding stresses under a direct load is

(A) 2:1 (B) 1:2

(C) 4:1 (D) 1:4

72. Two similar round bars A and B are each

30 cm long as shown in Fig.

2cm

(A) (B)

4cm

20 cm

10 cm

4cm

20 cm

10 cm

2cm

The ratio of the energies stored by the bars A

and B, U B

U A is

(A) 3/2 (B) 1.0

(C) 5/8 (D) 2/3

73. A mild steel specimen is under uniaxial

tensile stress. Young’s modulus and yield

stress for mild steel are 2 × 105 MPa and 250

MPa respectively. The maximum amount of

strain energy per unit volume that can be

stored in this specimen without permanent

set is

(A) 156 Nmm/mm3 (B) 15.6 Nmm/mm

3

(C) 1.56 Nmm/mm3 (D) 0.156 Nmm/mm

3

Directions: The following items consists of

two statements; one labeled as ' Statement

(I)' and the other as ' Statement (II)' You are

to examine these two statements carefully

and select the answers to these items using

the codes given below :

Codes:

(A) both A and R are true and R is the

correct explanation of A

(B) both A and R are true but R is not a

correct explanation of A

(C) A is true but R is false

(D) A is false but R is true


10

SOM

74. Assertion A: Strain is a fundamental

behavior of the material, while the stress is a

derived concept

Reason R: Strain does not have a unit while

the stress has a unit

(A) A (B) B

(C) C (D) D

75. Assertion A: A bar tapers from a diameter of

d1 to a diameter of d2 over its length ‘l’ and is

subjected to a tensile force ‘P’. If extension

is calculated based on treating it as a bar of

average diameter, the calculated extension

will be more than the actual extension.

Reason R: The actual extension in such bars

is given by △ = 4P

πd1d2

L

E

(A) A (B) B

(C) C (D) D

1. (D)

2. (B)

3. (C)

4. (B)

5. (C)

6. (D)

7. (B)

8. (D)

9. (C)

10. (D)

11. (B)

12. (B)

13. (B)

14. (B)

15. (A)

16. (B)

17. (C)

18. (B)

19. (D)

20. (A)

21. (C)

22. (B)

23. (D)

24. (D)

25. (B)

26. (A)

27. (D)

28. (B)

29. (A)

30. (A)

31. (C)

32. (A)

33. (A)

34. (B)

35. (A)

36. (B)

37. (A)

38. (B)

39. (C)

40. (C)

41. (C)

42. (B)

43. (B)

44. (B)

45. (C)

46. (D)

47. (D)

48. (c )

49. (D)

50. (A)

51. (D)

52. (A)

53. (A)

54. (B)

55. (B)

56. (C)

57. (B)

58. (D)

59. (C)

60. (B)

61. (A)

62. (C)

63. (D)

64. (A)

65. (B)

66. (C)

67. (D)

68. (D)

69. (C)

70. (B)

71. (D)

72. (A)

73. (D)

74. (B)

75. (D)

Answer key


11

SOM

[Sol] 2. (B)

l = δl1 + δl2 + δl3

=P1 l1

A E+

P2 l2

A E+

P3 l3

A E

1

10 10 7 10 9 10AE

=26×10

A×E

[Sol] 3. (C)

ε1 = ϵ2

W 1

A1E 1

=W 2

A2E 2

[Sol] 4. (B)

δL Taper =4PL

πEd1d 2

δL uniform =4PL

πEd2

As δL Taper = δL Uniform

d2 = d1d2

d = d1d2

[Sol] 5. (C)

δL own weight =WL

2AE

δL Axial load =WL

AE

[Sol] 6. (D)

δL Total =PL1

AE−

PL2

AE+

PL3

AE

=1

AE[(100 × 0.5) − (100 × 1) + (100

× 0.5)]

δ = 0

[Sol] 8. (D)

I

A BRA RA

RBRB

B C

2I

Both the ends are fixed, then

δl A + δl B = 0

RA l

AE+

RB 2l

AE= 0

RA + 2RB = 0 − − − − − −(1)

RB + 120 = RA

RA − RB = 120 − − − − − −(2)

Form equation (1) and (2)

−3RB = 120

RB = −40

RA = 80

[Sol] 9. (C)

δl uniform =4PL

πED .D=

4PL

πE.D2

δl Taper =4PL

πED1d2

δl =4PL

πE×1.1D×0.9D=

4PL

πE 0.99 D2

Percentage error =1

0.99−1

1

0.99

× 100 = 1%

Explanations


12

SOM

[Sol] 10. (D)

Deformation due to self-weight:

Let L = Length of bar

A= Area of cross section

γ = Weight density

E = Young’s Modulus of elasticity.

Consider a small section of length δx at a

distance x from free end.

The deformation δΔ is given by

δΔ =W xδx

AE

Wx = weight of the portion below the

section.

= A × x × γ

∵ δ\Detla = Axγ

Ae δx =

xγδ x

E

Total deformation of the rod Δ = xγ

Edx

L

0

Δ =γL2

2E=

ρgl2

2E

[Sol] 11. (B)

Elongation of tapered bar (δ) = 4PL

πEd1d2

δ ∝1

d1d2

δA

δB=

(d1d2)B

(d1d2)A =

d×D

3

D×D

2

=2

3

[Sol] 13. (B)

RA + RB = 10kN …… . (1)

δl AC + δl BC = 0

(As both ends are fixed)

RA 1

AE+

−RB (2)

AE= 0

RA − 2RB = 0 ……… 2

By solving

RA =20

3N, RB =

10

3N

[Sol] 14. (B)

Until the gap of 0.75mm is filled the load P

is taken by AB only.

P required to cause elongation of 0.75mm in

AB is

δl =Pl

AE

0.75 =P×1000

1×102×200×103

∴ P = 15 kN

[Sol] 19. (D)

E = 2G (1 + μ)

E = 2 × 100 1 + 0.25

E = 200 (1.25) = 250 GPa

[Sol] 25. (B)

μ = −Lateral strain

Longitudinal strain

0.3 =60×10−5

300

E

E=1.5 ×105 MPa = 150 GPa

[Sol] 26. (A)

ς = 330 MPa

δl = 401.2 − 400 = 1.2 mm

E =ς

ϵ=

ςl

δl=

3300×400

1.2= 110GPa


13

SOM

[Sol] 27. (D)

Given G = 60 GPa

K=140 GPa

E =9KG

3K+G=

9×140×60

3×140×60= 157.5 GPa.

E = 2G (1 + μ)

157.5 =2×60(1+μ)

μ = 0.3

[Sol] 28. (B)

δl

l= 0.04% =

0.04

100= 0.0004

Given P = 16 × 103N, diameter,

d = 16mm

ϵh =δD

D= 0.01% =

0.01

100= 0.0001

μ =ϵh

ϵl =

0.0001

0.0004= 0.25

ς =P

A=

16×103

π

4×162

= 80MPa

E =ς

ϵl=

80

0.0004= 2 × 105MPa

= 200 GPa

E = 2G 1 + μ

200 = 2G (1+0.25)

G=80GPa

[Sol] 56. (C)

ς = E αΔT

6 62 10 12 10 100

= 2400 kgf/cm2 = 2.4 × 103kgf/cm2

[Sol] 57. (B)

Thermal stress (ς)= -EαT

= −200 × 103 × 10−3 × 30

= 6000 N/mm2(compression)

Here at the supports compressive stresses are

developed.

[Sol] 61. (A)

Volumetric strain, ϵv = 3ϵ

= 3α ΔT L = 3α 1 1 = 3α

Change in volume = ϵV × V

= 3 α × 1 = 3 α cm3


1. Constant bending moment over span ‘l’

will occur in

(A)

W

(B)

W

(C)

W

(D)

W W

1 1

2. The given figure shows a beam BC

simply supported at C and hinged at B

(free end ) a cantilever AB. The beam and

the cantilever carry forces of 100 kg) and

200 kg respectively. The bending moment

at B is

A

200 kg

B

100 kg

C

1m 1m 1m 1m

(A) Zero (B) 100 kg-m

(C) 150 kg-m (D) 200 kg-m


If at a section distant from one of the ends

of the beam, M represents the bending

moment V the shear force and w the

intensity of loading, then

1) dM/ dx= V

2) dV/ dx = w

3) dw/dx = y

(the deflection of the beam at the section )


(A) 1 and 3 (B) 1 and 2

(C) 2 and 3 (D) 1, 2 and 3

4. A cantilever beam having 5 m length is so

loaded that it develops a shearing force of

20 T and a bending moment of 20 T-m at

a section 2 m from the free and. Max

shearing force and max, bending moment

developed in the beam under this load,

are respectively 50 T and 125 T-m. The

load on the beam is

(A) 25 T concentrated load at free end

(B) 20 T concentrated load at free end

(C) 5 T concentrated load at free end

2T/m load over entire length

(D) 10 T/ m udl over entire length

Practice Problems Level – 1

Chapter

2 SHEAR FORCE AND

BENDING MOMENT Syllabus : Beam, Types of beams, Types of end supports of beams,

Types of loads, Shear force and bending moment diagrams, Shear force

and bending moment diagrams for overhanging beams, Point of contra

flexure.


15

SOM

5. If the shear force acting at every section

of a beam is of the same magnitude and

of the same direction then it represents a

(A) simply supported beam with a

concentrated load at the centre

(B) overhung beam having equal

overhung at both supports and

carrying equal concentrated loads

acting in the same direction at the

free ends

(C) Cantilever subjected to concentrated

load at the free end

(D) simply supported beam having

concentrated loads of equal

magnitude and in the same direction

acting at equal distances from the

supports

6. For a cantilever beam of length 'L'

flexural rigidity El and loaded at its free

end by a concentrated load W, match

List-I with List- II and select the correct

answer using the codes below the lists:

List- I List- II

(i) Maximum

bending

1. WL moment

(ii) Strain energy 2. WL²/ 2EI

(iii) Maximum slope 3. WL3/ 3EI

(iv) Maximum

deflection

4. W2 L

3/6EI

Codes:

(i) (ii) (iii) (iv)

(A) 1 4 3 2

(B) 1 4 2 3

(C) 4 2 1 3

(D) 4 3 1 2

7. The given figure shows the shear force

diagram for the beam ABCD bending

moment in the portion BC of the beam

A B

C D

(A) is a non zero constant

(B) is zero

(C) varies linearly form B to C

(D) varies parabolically from B to C

8. The maximum bending moment in a

simply supported beam of length L

loaded by a concentrated load W at the

midpoint is given by

(A) WL (B) WL

2

(C) WL

4 (D)

WL

8

9. A beam, built in both ends, carries a

uniformly distributed load over its entire

span as shown in figure. Which one of the

diagrams given below represents bending

moment distribution along the length of

the beam?

UDL

(A)

(B)

(C)

(D)


16

SOM

10. If a beam is subject to a constant bending

moment along its length then the shear

force will

(A) also have a constant value

everywhere along its length

(B) be zero at all sections along the

beam

(C) be maximum at the centre and zero

at the ends

(D) zero at the centre and maximum at

the ends

11. A simply supported beam with width 'd'

carriesa central load W and undergoes

deflection δ at the centre. If the width and

depth and interchanged, the deflection at

the centre of the beam would attain the

value

(A) d

b δ (B)

d

b 2δ

(C) d

b 3δ (D)

d

b 3/2δ

12. For the beam shown in the figure below,

the elastic curve between the supports B

and C will be

P P

a 2b a

B C

(A) Circular (B) parabolic

(C) elliptic (D) a straight line

13. A simply supported beam is loaded as

shown in the figure below.

W 2W W

C C C C

The maximum shear force in the beam

will be

(A) Zero (B) W

(C) 2W (D) 4 W

14. A lever is supported on two hinges at A

and C. It carries a force of 3 kN as shown

in the figure below. The bending moment

at B will be

A B C

1m

3 kN

1m 1m 1m

(A) 3 k N-m (B) 2 kN-m

(C) 1 kN-m (D) Zero

15. The bending moment diagram shown in

figure-1 corresponds to the shear force

diagram in

(A)

(B)

(C)

(D)

16. Which one of the following portions of

the loaded beam shown in the given

figure is subjected to pure bending?

AB C D

W W

E

L L L L


17

SOM

(A) AB (B) DE

(C) AE (D) BD

17. Match List –I with List –II and select the

correct answer using the codes below the

lists:-

List- I List- II

(i) Bending moment

is constant

1. Point of

contraflexure

(ii) Bending moment

is maximum or

minimum

2. Shear force

changes sign

(iii) Bending moment

is zero

3. Slope of

shear force

diagram is

zero over the

portion of

the beam

(iv) Loading is

constant

4. Shear force

is zero over

the portion

of the beam

Codes:

(i) (ii) (iii) (iv)

(A) 4 1 2 3

(B) 3 2 1 4

(C) 4 2 1 3

(D) 3 1 2 4

18. A loaded beam is shown in the figure

below.

W W W

L L L L

The bending moment diagram of the

beam is best represented as

(A)

(B)

(C)

(D)

19. Bending moment distribution in a built

beam is shown in the figure below.

A B

C

D

The shear force distribution in the beam is

represented by

(A) A C E

(B) A E

(C) A C

E

(D)

A E

C

20. A horizontal beam carrying uniformly

distributed load is supported with equal

overhang as shown in the given figure.

a b a

The resultant bending moment at the

midspan shall be zero if a/b is

(A) 3/4 (B) 2/3

(C) 1/2 (D) 1/3


18

SOM

21. A simply supported beam of span I is

subjected to a uniformly varying load

having zero intensity at the left support

and w N/m at the right support. The

reaction at the right support is

(A) w l/2 (B) wl/5

(C) w l/4 (D) wl/3

22. A simply supported beam has equal

overhanging lengths and carries equal

concentrated loads P at ends. Bending

moment over the length between the

supports

(A) is zero

(B) is a non-zero constant

(C) varies uniformly form one support to

the other

(D) is maximum at mid-span

23. Consider the following statements : In a

cantilever subjected to a concentrated

load at free end.

1. The bending stress is maximum at

the free end

2. The maximum shear stress is

constant along the length of the

beam

3. The slope of the elastic curve is zero

at the fixed end


(A) 1,2 and 3 (B) 2 and 3

(C) 1 and 3 (D) 1 and 2

24. The shear stress distribution over a beam

cross-section is shown in the figure

below.

The beam is of

(A) equal flange 1- section

(B) unequal flange 1-section

(C) circular cross-section

(D) T- section

25. A beam of length 4L is simply supported

on two supports with equal overhangs of

L on either sides and carries three equal

loads, one each at free ends and the third

at the mid-span. Which one of the

following diagrams represents correct

distribution of shearing force on the

beam?

(A)

(B)

(C)

(D)

26. A simply supported beam is subjected to

a distributed loading as shown in the

diagram given below:

w N/m

L

The maximum shear force in the beam is

(A) wL/4 (B) wL/2

(C) wL/3 (D) wL/6


19

SOM

27. The point of contraflexure is a point

where:

(A) Shear force changes sign

(B) Bending moment changes sign

(C) Shear force is maximum

(D) Bending moment is maximum

28. The figure shown below represents the

BM diagram for a simply supported

beam.

+

–A

M

M

B

1/2 1/2

The beam is subjected to which one of the

following?

(A) A concentrated load at is mid-length

(B) A uniformly distributed load over its

length

(C) A couple at its mid-length

(D) Couple at one-fourth of the span

from each end

29. A beam is said to be of uniform strength

if

(A) The bending moment is the same

throughout the beam

(B) The shear stress is the same

throughout the beam

(C) The deflection is the same

throughout the beam

(D) The bending stress is the same at

every section along its longitudinal

axis]

30. The shearing force diagram for a beam is

shown in the figure below.

A SFD B

C

The bending moment diagram is

represented by which one of the

following?

(A)

A

C

B

(B)

A B

C

(C)

A B

C

(D)

A B

C

31. A uniformly distributed load w (in kN/m)

is acting over the entire length of a 3 m

long cantilever beam. If the shear force at

the midpoint of cantilever is 6 k N, what

is the value of W?

(A) 2 (B) 3

(C) 4 (D) 5

32. An overhanging beam ABC is supported

at points A and B, as shown in the figure

below.

A

2 kN

B

6 kN

C

1m 1m 1m


20

SOM

Find the maximum bending moment and

the point where it occurs.

(A) 6 kN-m at the right support

(B) 6 kN-m at the left support

(C) 4.5 kN-m at the right support

(D) 4.5 kN-m at the midpoint between

the supports

33. A freely supported beam at its ends

carries a central concentrated load, and

maximum bending moment is M. If the

same load be uniformly distributed over

the beam length, then what is the

maximum bending moment?

(A) M (B) M/2

(C) M/3 (D) 2M

34. Match List-1 (Cantilever loading) with

List –II (Shear Force Diagram ) and select

the correct answer using the codes given

below the lists:

List – I

(i) A B C

P1

PP2

(ii) A B C

P

M

(iii) A B C

P

P

(iv) A B

C

P

M

List – II

1.

A B C

2.

A B C

3.

A B C

4.

A B

5.

A B C

Codes:

(i) (ii) (iii) (iv)

(A) 1 5 2 4

(B) 4 5 2 3

(C) 1 3 4 5

(D) 4 2 5 3

35. Match List-I with List-II and select the

correct answer using the code given

below the Lists:

List- I List- II

(i) Subjected to

bending moment at

the end of a

cantilever

1. Triangle

(ii) Cantilever carrying

uniformly

distributed load over

the whole length

2. Cubic

parabola

(iii) Cantilever carrying

linearly varying load

from zero at the free

end to maximum at

the supports

3. Parabola

(iv) A beam having a

load at centre &

supported at the

ends

4. Rectangle

Codes:

(i) (ii) (iii) (iv)

(A) 1 2 3 4

(B) 4 3 2 1

(C) 1 3 2 4

(D) 4 2 3 1


21

SOM

36. A beam simply supported at equal

distance from the ends carries equal loads

at each end. Which of the following

statements is true?

(A) The bending moment is minimum at

the mid-span

(B) The bending moment is minimum at

the support

(C) The bending moment varies

gradually between the supports

(D) the bending moment is uniform

between the supports

37. Match List I with List II and select the

correct answer using the codes given

below the lists

List- I (Type of beam with type

of loading)

(i)

(ii)

(iii)

(iv)

List- II (S.F. Diagram)

1.

2.

3.

4.

Codes:

(i) (ii) (iii) (iv)

(A) 4 1 3 2

(B) 4 3 2 1

(C) 3 4 1 2

(D) 3 4 2 1

38. The bending moment diagram for an

overhanging beam is shown in Fig.

A B

C G D

E F

The point of contraflexure would include

(A) A and F (B) B and E

(C) C and D (D) A and D

39. In Fig shows

W

0.1

a beam of supported length 'I' and

overhang 0.11, carrying a concentrated

load W at the end of the overhang. Which

one of the following figures would

represent the correct shear force diagram

for the beam?

(A)

0.1

w

(B)

0.1

w

0.1

(C)

0.1

w


22

SOM

(D)

0.1w

0.1

w

40. In the following figure

ML

2 kN

N

2m 1m 2m

O


beam is

(A)

2 kNm

1.6 kNm

LM N

(B)

2.4 kNm

LM N

(C)

2.4 kNm

LM N

(D)

2.4 kNm

LM N

41. In the following fig,.

1m

1t 1t(+)

1mS.F.D

4t. m (–)2 t.m

1 t.m

B.M.D.

The SFC and BMD for a beam are shown

in 10.30 (A) and 10.30 (B). The

corresponding loading diagram would be

(A)

1t2 t/m

1m C 1mBA

(B)

2 t/m

1mA

1m

C

1t

B

(C)

1t

B1m1m

AC

2 t/m

(D)

1t

B1m1m

AC

2 t/m

42. For the shear force diagram shown in

figure the loaded beam will be

4m 8m 4mA

D B

C

4t

16t

S.F. Diagram

14t 9t

3t

(A)

18 t 1.5 t/m 3t

4m 8m 4m

(B)

1.5 t/m

4m 8m 4m

14 t 3 t

(C)

10t 1.5 t/m 3t

4m 8m 4m

(D)

1.5 t/m 14 t

4m 8m 4m

3 t


23

SOM



below the lists:

List- I

(Type of beam with

type Of loading)

List- II

(Max.B.M.

formula)

(i) w

L

1. 2wL

12

(ii) w/m

L

2. 2wL

6

(iii

) w

L

3. 2wL

2

(iv) W/m

L

4. 2wL

8

Codes:

(i) (ii) (iii) (iv)

(A) 2 3 1 4

(B) 1 2 3 4

(C) 4 3 1 2

(D) 2 1 4 3

44. A simply supported beam is shown in fig.

2m 2m

20 kN 10 kN/m

The corresponding SFD and BMD would

be

(A)

+

–

30 kN

30 kNSFD

BMD

40 kNm

+

(B)

+–

10 kN

10 kN 30 kN

30 kNSFD

BMD +

40 kNm

(C)

+–

10 kN

10 kN 30 kN

30 kNSFD

BMD +

40 kNm

(D)

+–

30 kNSFD 30 kN

BMD +

40 kNm

45. A beam' s S.F. D. and B.M.D. are shown

in fig. (A) and (B) respectively.

4m

(a)

10 kNm (–)

(b)

10 kNm

The corresponding load diagram will be

(A) 4m

10 kNm

(B) 4m

10 kNm10 kNm

(C)

10 kN 10 kN

1m 2m 1m

(D)

10 kN/m 10 kN/m

1m 2m 1m

46.

w/ unit length

DA

2

2

B C


beam shown in fig. is

(A)

A B C D

–w2

2

(B)

A B C D

w2

8

w2

8–


24

SOM

(C)

B C DA

w2

4–

w2

8–

(D)

A B C D

w2

8–

w2

8–

47. Figure shows a simply- supported beam

overhanging to the left. The beam carries

a uniformly distributed load of w/m

throughout.

w/ m

A

2

The correct bending moment diagram for

the beam is

(A)

w2

8

–w2

16

–

/4

/2

(B) w2

8

–

(C) w2

8

–

(D)

w2

8

– w2

8

/2



below the lists:

List- I (Beam with loading)

(i) M

Hinge Roller

L

A B

(ii) w/m

A B

(iii) w/m

A B

(iv) A B

W

List- II (B.M. diagram)

1.

A B

2.

A B

3. A B

4.

A B

Code:

(i) (ii) (iii) (iv)

(A) 3 4 2 1

(B) 1 2 3 4

(C) 1 3 4 2

(D) 2 1 4 3

49. Figure shows a beam cantilevering out at

one end. It carries a uniformly distributed,

load W over the cantilever.

w

0.2

Which one of the following figures

correctly represents the shear force

diagram for the beam?

(A)

w0.1w

0.2

(B) w

(C) w

(D)

w

0.1w


25

SOM

1. (D)

2. (A)

3. (B)

4. (D)

5. (C)

6. (B)

7. (A)

8. (C)

9. (D)

10. (B)

11. (B)

12. (A)

13. (C)

14. (C)

15. (B)

16. (D)

17. (C)

18. (A)

19. (A)

20. (C)

21. (D)

22. (B)

23. (B)

24. (B)

25. (D)

26. (A)

27. (B)

28. (C)

29. (D)

30. (B)

31. (C)

32. (A)

33. (B)

34. (B)

35. (B)

36. (D)

37. (B)

38. (C)

39. (B)

40. (A)

41. (C)

42. (A)

43. (A)

44. (B)

45. (B)

46. (D)

47. (A)

48. (B)

49. (A)

Answer key


26

SOM

\

[Sol] 1. (D)

Bending moment diagram for given loading.

Sol] 2. (A)

At internal Hinge bending moment is always Zero.

[Sol] 4. (D)

From the above similar Δle

At a distance of 2m, Shear force = 20,

At distance of 5m, SF…….?

2 → 20

5 →? ⇒ SF =100

2= 50T

∴ 50T = 10T/m udl over enter length for 5m

[Sol] 8. (A)

Explanations


27

SOM

Bending moment between B and C is constant/uniform

[Sol] 14. (C)

[Sol] 15. (A)

RA + RC = 0, MD = 3 × 1 = 3 kN − m

Take moments about ‘A’


28

SOM

ΣMA = 0

⇒ RC × 3 − 3 = 0

⇒ RC = 1 kN, RA = −1 kN

[Sol] 19. (A)

RB + RD = 3W

ΣMB = 0

W 3l + Wl − RD2L = 0

RD =2W

2= 1.5W

RB = 1.5 W

RB = 1.5 W

BM A = 0 = BM E

BM B = −WL = BM D

BM C = −W × 2L + 1.5W × L

= −0.5 WL

[Sol] 21. (C)

By symmetry R1 = R2 =W

2(2a + b)

Bending moment at mid span

Mx = R2 ×b

2−

W

2 a +

b

2 a +

b

2

∴ 0 =w

2 2a + b ×

b

2−

w

2 a +

b

2

2

w

2 2a + b

b

2=

w

2 a +

b

2

2


29

SOM

(2a + b) b

2=

2a+b 2

4

⇒b

2=

2a+b

4⇒ 4b = 4a + 2b ⇒

a

b= 0.5

[Sol] 22. (A)

RA = RB =Total Load

2=

1

2wL

2=

wL

4

[Sol] 29. (C)


1. A rectangular section beam subjected to a

bending moment M varying along its length

is required to develop same maximum

bending stress at any cross-section. If the

depth of the section is constant, then its

width will vary as

(A) M (B) M

(C) M2

(D) M1

2. In a beam of circular cross-section, the shear

stress variation across a cross-section is

(A) (B)

(C) (D)

3. A wooden beam of rectangular cross-section

10cm deep by 5 cm wide carries maximum

shear force of 2000 kgf. stress at neutral axis

of the beam section is

(A) Zero (B) 40 kgf/cm2

(C) 60 kgf/ cm2

(D) 80 kgf/ cm2

4. A beam cross-section is used in two different

orientations as shown in the figure given

below:

(A)

(B)

b

b/2

bb/2

Bending moments applied to the beam in

both cases are same. The maximum bending

stresses induced in cases (A) and (B) are

related as

(A) ςA = ςB (B) ςA = 2ςB

(B) ςA = ςB

2 (D) ςA =

ςB

4

5. Two beams of equal cross-section area are

subjected to equal bending moment. If one

beam has square cross-section and the other

has circular section, and the other has

circular section, then

(A) both beams will be equally strong

(B) circular section beam will be stronger

(C) square section beam will be stronger

(D) the strength of the beam will depend on

the nature of loading

Practice Problem Level - 1

Chapter

3 BENDING STRESS &

SHEAR STRESS Syllabus : Bending stresses, Section modulus (modulus of section),

Neutral axis (neutral layer), Strength of sections, Moment of

resistance, Relation between bending stress and cetroid, Shear stress

distribution for beam sections of various shapes.


31

SOM

6. The distribution of shear stress of a beam is

shown in the given figure.

The cross-section of the beam is

(A) 1 (B) T

(C) (D)

7. The given figure (all dimensions are in mm)

shows an 1- section of the beam.

N N

20

Q

P 20

40

40

100

20

The shear stress at point P (very close to the

bottom of the flange) is 12 MPa. The stress

at point Q in the wed( very close to the

flange) is

(A) indeterminable due to incomplete data

(B) 60 MPa

(C) 18 MPa

(D) 12 MPa

8. The stiffness of the beam shown in the figure

below is ( I = 375× 10–6

m4., L = 0.5 m and

E = 200 GPa

2

P

(A) 12× 108 N/m (B) 10× 10

8 N/m

(C) 4× 108 N/m (D) 8× 10

8 N/m

9. Select the correct shear stress distribution

diagram for a square beam with a diagonal in

a vertical position:

(A) (B)

(C) (D)

10. What is the nature of distribution of shear

stress in a rectangular beam?

(A) Linear (B) Parabolic

(C) Hyperbolic (D) Elliptic

11. At a section of a beam, shear force is F with

zero BM. The cross-section is square with

side 'a'. Point A lies on neutral axis and point

B is mid way between neutral axis and top

edge, i.e. at distance a /4 above the neutral

axis. If τA and τB denote shear stresses at

points A and B, then what is the value of

τA / τB?

(A) 0 (B) 3/4

(C) 4/3 (C) None of above

12. If the area of cross-section of a circular

section beam is made four times, keeping the

loads. Length, support conditions and

material of the beam unchanged, then the

quantities (List-1) will change through

different factors (List-II). Match the list-1

with the list-II and select the correct answer

using the code given below the lists:

List- I List- II

(i) Maximum BM 1. 8

(ii) Deflection 2. 1

(iii) Bending Stress 3. 1/8


32

SOM

(iv) Section Modulus 4. 1/16

Codes:

(i) (ii) (iii) (iv)

(A) 3 1 2 4

(B) 2 4 3 1

(C) 3 4 2 1

(D) 2 1 3 4

13. Beams of uniform strength vary in section

such that

(A) bending moment remains constant

(B) deflection remains constant

(C) maximum bending stress remains

constant

(D) shear force remains constant

14. In the case of beams with circular cross-

section, what is the ratio of the maximum

shear stress to average shear stress?

(A) 3: 1 (B) 2:1

(C) 3:2 (D) 4 :3

15. When a rectangular section beam is loaded

transversely along the length shear stress

develops on

(A) Top fibre of rectangular beam

(B) Middle fibre of rectangular beam

(C) Bottom fibre of rectangle beam

(D) None of these

16. In I-section of a beam subjected to transverse

shear force, the maximum shear stress is

developed

(A) at the centre of the web

(B) at the top edge of the top flange

(C) at the bottom edge of the top flange


17. What is the shape of the shearing stress

distribution across a rectangular cross-

section beam?

(A) Triangular

(B) Parabolic only

(C) Rectangular only

(D) A combination of rectangular and

parabolic shape

18. A beam having rectangular cross-section is

subjected to an external loading. The average

shear stress developed due to the external

loading at a particular cross-section is

τ avg.What is the maximum shear stress

developed at the same cross-section due to

the same loading?

(A) 1

2τavg. (B) τavg.

(C) 3

2τavg. (D) 2 τavg.

19. Match List-I with List –II and select the

correct answer using the codes given below

the lists:

List- I List- II

(i) Point of

inflection

1. Strain energy

(ii) Shearing

strain

2. Equation of bending

(iii) Section

modulus

3. Equation of torsion

(iv) Modulus of

resilience

4. Bending moment

diagram

Codes:

(i) (ii) (iii) (iv)

(A) 1 3 2 4

(B) 4 3 2 1


33

SOM

(C) 1 2 3 4

(D) 4 2 3 1

20. A T- section beam is simply supported and

subjected to a uniformly distributed load

over its whole span. Maximum longitudinal

stress in the beam occurs at

(A) Top fibre of the flange

(B) The junction of web and flange

(C) The midsection of the web

(D) The bottom fibre of the web

21. The ratio of the section modulii of a square

Beam (Z) when square section is placed (i)

with two sides horizontal (Z1) and (ii) with a

diagonal horizontal (Z2 ) as shown is

b

A

X

D

b

X–C

B

C

D

X

b

A

X

b

B

(i) (ii)

(A) Z1

Z2= 1.0 (B)

Z1

Z2= 2.0

(C) Z1

Z2= 2 (D)

Z1

Z2= 1.5

22. Abeam with rectangular section of 120 mm ×

60 mm, designed to be placed vertically is

placed horizontally by mistake. If the

maximum stress is to be limited, The

reduction in load carrying capacity would be

(A) 1

4 (B)

1

3

(C) 1

2 (D)

1

6

23. If E = elasticity modulus, I = moment of

inertia about the neutral axis and M =

bending moment in pure bending under the

symmetric loading of a beam, the radius of

curvature of the beam :

1. Increases with E

2. Increases with M

3. Decreases with 1

4. Decreases with M

Which of these are correct?

(A) 1 and 3 (B) 2 and 3

(C) 3 and 4 (D) 1 and 4

24. For given shear force across a symmetrical 'I'

section, the intensity of shear stress is

maximum at the

(A) extreme fibres

(B) centroid of the section

(C) at the junction of the flange and the

web, but on the web

(D) at the junction of the flange and the

web, but on the flange

25. The maximum bending stress induced in a

steel wire of modulus of elasticity 200

kN/mm2 and diameter 1 mm when wound on

a drum of diameter 1 m is approximately

equal to

(A) 50 N /mm2

(B) 100 N /mm2

(C) 200 N /mm2

(D) 400 N /mm2

26. A homogeneous, simply supported prismatic

beam of width B, depth D and span L is

subjected to a concentrated load of

magnitude P. The load can be placed

anywhere along the span of the beam. The

maximum flexural stress developed in beam

is

(A) 2

3

PL

BD2 (B) 3

4

PL

BD2


34

SOM

(C) 4

3

PL

BD2 (D) 3

2

PL

BD2

27. The shear stress at the neutral axis in a beam

of triangular section with a base of 40 mm

and height 20 mm, subjected to a shear force

of 3 kN is

(A) 3 MPa (B) 6 M Pa

(C) 10 MPa (D) 20 MPa

28. A symmetric I-section (with width of each

flange = 50 mm, thickness of web = 10 mm )

of steel is subjected to a shear force of 100

kN. Find the magnitude of the shear stress

( in N/ mm2) in the web at its junction with

the top flange.

10

mm

10

0m

m

50mm

10mm

50mm

10mm

29. Beam fixed at both its ends, is called a

(A) fixed beam

(B) Built – in beam

(C) Any one of the above


30. For a beam, the term M / E I is:

(A) Stress (B) Rigidity

(C) Curvature (D) Shear force

31. Of the several prismatic beams of equal

lengths, the strongest in flexure is the one

having maximum

(A) Moment of inertia

(B) Section modulus

(C) Tensile strength

(D) Area of cross-section

32. A prismatic beam when subjected to pure

bending assumes the shape of

(A) Centenary (B) Cubic parabola

(C) Quadratic parabola (D) Arc of a circle

33. A beam of square section (with sides of the

square horizontal and vertical) is subjected to

a bending moment M and the maximum

stress developed is 100 MPa. If the diagonals

of the section take vertical and horizontal

directions, bending moment remaining the

same, the maximum stress developed will

become

(A) 100 2MPa (B) 100

2MPa

(C) 50 MPa (D) None of these

34. What diameter should the driving pulley

have on which a rubber belt runs so that

bending stress in belt is limited to 5MPa?

(The belt cross-section is a rectangle 15 mm

thick ×110 mm wide, E for belt material is

100 MPa

(A) 30 mm (B) 150 mm

(C) 300 mm (D) 15 mm

35. A tapered cantilever beam of constant

thickness is loaded as shown in the figure.

The bending stress will be

b

P

L

(A) maximum near the fixed end


35

SOM

(B) maximum at × =L/2

(C) maximum at × = 2L/3

(D) uniform throughout length

36. A rectangular beam of width 100 mm is

subjected to maximum shear force of 60 kN.

The corresponding maximum shear stress in

the cross-section is 4 MPa. The depth of the

beam should be

(A) 150 mm (B) 225 mm

(C) 200 mm (D) 100 mm

37. The shear stress distribution over a beam

cross- section is shown in the figure. The

beam is of

(A) equal flange I- section

(B) Unequal flange I-section

(C) circular cross-section

(D) 'T' cross- section

38. For a given stress, the ratio of the moment of

resistance of a beam of square section when

placed with one diagonal horizontal to the

moment of resistance of the same beam

when placed with two sides horizontal will

be

(A) 1/2 (B) 2

(C) 1.414 (D) 1

1.414

39. A cantilever beam of span L carries a

concentrated load 'W' at the free end. If the

width 'b' of the beam is constant throughout

the pan, then for the beam to have uniform

strength, the depth 'd ' at the fix end should

be

(A) 6WL

bf (B)

3WL

bf

(C) 3WL

bf (D)

6WL

bf

40. A beam has a solid circular cross-section

having diameter d. If a section of the beam is

subject to a shear force F, the maximum

shear stress in the cross-section is given by

(A) 4

3

F

πd2 (B) 16F

3πd2

(C) 8

3

F

πd2 (D) 3

16

F

πd2

41. A beam has a triangular cross-section having

base 40 mm and altitude 60 mm. If this

section is subjected to a shear force of

36000N, the maximum shear stress in the

cross-section would be

(A) 60 N/ mm2

(B) 36 N/ mm2

(C) 45 N/ mm2

(D) 30 N/ mm2


36

SOM

1. (A)

2. (A)

3. (C)

4. (B)

5. (C)

6.(B)

7. (B)

8. (C)

9. (D)

10. (B)

11. (C)

12. (B)

13. (C)

14. (D)

15. (B)

16. (A)

17. (B)

18. (C)

19. (B)

20. (D)

21. (C)

22. (C)

23. (D)

24. (B)

25. (C)

26. (D)

27. (C)

28. (A)

29. (C)

30. (C)

31. (B)

32. (D)

33. (A)

34. (C)

35. (D)

36. (B)

37. (B)

38. (D)

39. (D)

40. (B)

41. (C)

[Sol] 1. (A)

ςb =M

Z

M ∝ Z (Z =bd2

6…… . . for rec tan gle)

M ∝bd2

6⇒ b ∝ M

[Sol] 3. (C)

τmax =3

2τavg for Rectangular cross − section.

= 3

2

F

A =

3

2×

2000

10×5= 60 kgf/cm2

[Sol] 4. (B)

M

I=

ςb

y⇒ ςb ∝

1

Z

ςA

ςB=

b2 b/2

6×

6

b b

2

2 = 2

ςA = 2ςB

[Sol] 5. (C)

Area of circle =π

4d2

Area of square = a2

For equal cross − sectional area

a2 =π

4d2 ⇒ a =

π

4× d

ςb

y=

M

I⇒ ςb α

1

Z

Zc =π

64d4

d

2

=π

32d3 ……… for circle

Zs = a3/6 … . . for solid square

Zs

Zc=

a3

6×

32

πd3

=

π

4d2×

π

4d×32

6×πd3 =2 π

3

∴ Zs > SC

Hence square beam will be stronger the

circular beam in bending.

Explanations

Answer key


37

SOM

[Sol] 7. (B)

Total depth of flange dA = 40mm

Total depth of web dB = 80mm.

τ =FA Y

Ib τ ∝

1

b

τQ

τP=

bP

bQ=

100

20

τQ =100

20× 12 = 5 × 12 = 60 MPa

[Sol] 11. (C)

τA = τmax =3

2

F

a×a =

3F

2a2

τB =FA Y

Ib F

a

4×a

a

8+

a

4

a 4

12 a

=9F

8a2

τA

τB=

3

2

9

8

=4

3

[Sol] 13. (D)

Given M = constant

⇒dM

dx= 0 ⇒ SF = 0

If bending moment is constant then shear

force is zero.

[Sol] 16. (A)

[Sol] 18. (C)

τmax =FA Y

IB

=F× b×

d

2 ×

d

4

b d 3

12×b

=3

2

F

bd

Average shear stress τavgF

bd

∴τmax

τavg=

3

2= 1.5

[Sol] 21. (C)

z1 =I

ymax=

b b 3

1 2b

2

=b3

6

z2 =b b 3

1 2b

2

=b3

6 2

Z1

z2=

b3

6×

6 2

b3 = 2


1. Two identical cantilevers are loaded as

shown in the respective figures. If slope at

the free end of the cantilever in figure E is θ,

the slope at free end of the cantilever in

figure F will be

M = PL/2L

P

L

(A) 1

3θ (B)

1

2θ

(C) 2

3θ (D) θ

2. A simply supported beam of constant

flexural rigidity and length 2 L carries a

concentrated load P at its mid- span and the

deflection under the load is δ. If a cantilever

beam of the same flexural rigidity and length

L is subjected to a load P at its free end, then

the deflection at the free end will be

(A) δ/2 (B) δ

(C) 2δ (D) 4δ

3. A cantilever beam of rectangular cross

section is subjected to a load W at its free

end. If the depth of the beam is doubled and

the load is halved, the deflection of the free

end as compared to original deflection will

be

(A) half (B) one eighth

(C) one sixteenth (D) double

4. The two cantilevers A and B shown in the

given figure have the same uniform cross-

section and the same material. Free end

deflection of cantilever A is δ.The value of

mid-span deflection of the cantilever B is

P

L L

L L

P

A

B

(A) δ

2 (B)

2

3 δ

(C) δ (D) 2δ

5. A cantilever of length L, moment of inertia I,

and Young’s modulus E carries a

concentrated load W at the middle of its

length, the slope of cantilever at the free end

is

(A) WL²/2 EI (B) WL²/4 EI

(C) WL²/8 EI (D) WL²/16 EI

6. Maximum deflection of a cantilever beam of

length L carrying uniformly distributed load

'w' per unit length will be

(A) WL4/ EI (B) WL4/ 4E


Chapter

4 SLOPE AND

DEFLECTION OF BEAMS Syllabus : Introduction, Moment area method, Strain energy method,

Castigliano’s theorem.


39

SOM

(C) WL4/8EI (D)WL4/ 384EI

[Where E = modulus of elasticity of beam

material and I= moment of inertia of beam

cross-section]

7. Slope at the end of a simply supported beam

of span I with uniformly distributed load w/

unit length over the entire span is given by

(A) wl2

16EI (B) wl3

16EI

(C) wl3

24EI (D) wl2

24EI

8. If the deflection at the free end of a

uniformly loaded cantilever beam is 15 mm

and the slope of the deflection curve at the

free end is 0.02 radian, then the length of the

beam is

(A) 0.8 m (B) 1 m

(C) 1.2 m (D) 1.2 m

9. If the deflection at the free end of a

uniformly loaded cantilever beam of length 1

m is equal to 7.5 mm, then the slope at the

free end is

(A) 0.001 radian (B) 0.015 radian

(C) 0.01 radian (D) none of these

10. A cantilever beam carries a uniformly

distributed load from fixed end to the centre

of the beam in the first case and a uniformly

distributed load of same intensity form centre

of the beam to the free end in the second

case. The ratio of deflections in the two cases

is

(A) 1/2 (B) 3/11

(C) 5/24 (D) 7/41

11. If the length of a simply supported beam

carrying a concentrated load at the centre is

doubled, the defection at the centre is

doubled, the defection at the centre will

become

(A) two times (B) four times

(C) eight times (D) sixteen times

12. A simply supported beam with rectangular

cross-section is subjected to a central

concentrated load. If the width and depth of

the beam are doubled, then the deflection at

the centre of the beam will be reduced to

(A) 50% (B) 25%

(C) 12.5% (D) 6.25%

13. A simply supported beam of span 'L' and

uniform flexural rigidity EI, carries a central

load 'W' and total uniformly distributed load

'W' throughout the span. The maximum

deflection is given by

(A) 13 WL3/96 EI (B) 5 WL

3/96 EI

(C) 5 WL3/96 EI (D) 13 WL

3/384 EI

14. A beam simply supported at both the ends, of

length 'L' carries two equal unlike couples M

at two ends. If the flexural rigidity, EI=

constant, then the central deflection of beam

is given by

(A) ML 2

4EI (B)

ML 2

16EI

(C) ML 2

64EI (D)

ML 2

8EI

15. A simply supported rectangular beam of span

'L ' and depth 'd ' carries a central load 'W'.


40

SOM

The ratio of maximum deflection to

maximum bending stress is

(A) L2/6Ed (B) L2/8Ed

(C) L2/48Ed (D) L2/12Ed

16. The deflection at the free end of a cantilever

of rectangular cross- section due to certain

loading is 0.8 cm. If the depth of the section

is doubled keeping the width the same, then

the deflection at the free end due to the same

loading will be

(A) 0.1 cm (B) 0.4 cm

(C) 0.8 cm (D) 1.6 cm

17. System A is a simply supported beam with a

load P at mid span. System B is the same

beam but the load is replaced by audI of

intensity P/L wherein L is the span. The mid

span deflection of system B will

(A) be the same as that of system A at mid

span

(B) be less than that of system A at mid

span

(C) be more than that of system A at mid

span

(D) bear no relation to that of system A

18. Consider the following statements regarding

a simply supported beam subjected to a

uniformly distributed load over the entire

span :

1. The bending moment is maximum at the

central position.

2. The shear force is zero at the central

position.

3. The slope is maximum at the middle

position Of these statements

(A) 1, 2 and 3 are correct

(B) 1 and 2 are correct

(C) 2 and 3 are correct

(D) 1 and 3 are correct

19. Match List I with List Ii and select the

correct answer using the codes given below

the lists:

List- I

(Nature of beam)

List- II

(Max.defl

ection)

(i) Cantilever beam

subjected to a

concentrated load W

at free end

1. wL3

15EI

(ii) Simply supported

beam subjected to a

point load W at the

centre

2. wL3

3EI

(iii) Cantilever beam

subjected to a

Hydrostatic load with

zero intensity at the

free end and W at the

fixed end

3. wL3

60EI

(iv) Simply supported a

triangularly distributed

Load with its apex of

magnitude W at the

mid span

4. wL3

48EI

Codes :

(i) (ii) (iii) (iv)

(A) 2 1 4 3


41

SOM

(B) 3 1 2 4

(C) 2 4 1 3

(D) 1 4 2 3

20. The deflection at the free end of a cantilever

subjected to a couple M at its free end and

having a uniform flexural rigidity EI

throughout its length 'L ' is equal to

(A) ML2

2EI (B)

ML2

3EI

(C) ML2

6EI (D)

ML2

8EI

1. (D)

2. (C)

3. (C)

4. (C)

5. (C)

6. (C)

7. (C)

8. (B)

9. (C)

10. (D)

11. (C)

12. (D)

13. (D)

14. (D)

15. (A)

16. (A)

17. (B)

18. (B)

19. (C)

20. (A)

Sol. 2 (C)

W

L

W

2L

For simply supported beam

3 3

1

W 2L WL

48EI 6EI

For cantilever

3 3

2 1

WL 2WL2

3EI 6EI

Sol. 3 (C)

Deflection 3

A

WL

3EI

W

A

3

W

d

3bdI

12

313 1

2 2 1

33

1 1 2 1 1

Wd

W d 12

W d 16W 2d

12

16

Sol. 4 (C)

By Maxwell’s theorm

P Px

x

Explanations

Answer key


1. Match List -1 (End conditions of columns)

with List-II (Lowest critical load and select


below the lists:

List- I List- II

(i) Column with both

ends hinged

1. π2EI/L

2

(ii) Column with both

ends fixed

2. 2π2EI/L

2

(iii) Column with one ends

fixed and the other

end hinged

3. 4π2EI/L

2

(iv) Column with one ends

fixed and the Other

end free

4. π2EI/4L

2

(E is the Young’s modulus of elasticity of

column material, L is the length and I is the

second moment of area of cross-section of

the column.)

Codes:

(i) (ii) (iii) (iv)

(A) 1 2 3 4

(B) 3 2 1 4

(C) 1 3 2 4

(D) 2 4 3 1

2. The ratio of the compressive critical load for

a long column fixed at both the ends and a

column with one end fixed and the other and

free is

(A) 1 : 2 (B) 1 : 4

(C) 1 : 8 (D) 1 : 16

3. The Euler’s crippling load for a 2 m long

slender steel rod of uniform cross-section

hinged at both the ends is 1 kN. The Euler’s

crippling load for a 1 m along steel rod of the

same cross-section and hinged at both ends

will be

(A) 0.25 kN (B) 0.5 kN

(C) 2 kN (D) 4 kN

4. A short column of external diameter D and

internal diameter d carries an eccentric load

W. The greatest eccentricity which the load

can be applied without producing tension on

the cross-section of the column would be

(A) D+d

8

(B)

D²+d3

8d

(C)D²+d²

8d

(D)

D²+d²

8

Practice Problems level - 1

Chapter

5

COLUMNS AND STRUTS Syllabus : Columns and struts, Failure of a column, Slenderness

ratio, Buckling load, Safe load, Classification of columns, End

conditions of the column, Equivalent length of a column (effective

length), Formulae for finding buckling load in columns and struts,

Assumptions in the euler’s column theory (for long columns),

Euler’s formula.


43

SOM

5. If diameter of a long column is reduced by

20%, the percentage of reduction in Euler

buckling load is

(A) 4 (B) 36

(C) 49 (D) 59

6. A long slender bar having uniform acted

upon by an axial compressive force. The

sides B and H are parallel to x- and y- axes

respectively. The ends of the bar are fixed

such that they behave as pin- jointed when

the bar buckles in a plane normal to x-axis,

and they behave as built- in when the bar

buckles in plane normal to y-axis. If load

capacity in either mode of buckling is same,

the value of H/B will be

(A) 2 (B) 4

(C) 8 (D) 16

7. Match List-1 (End conditions of columns)

with List-II (Equivalent length in terms of

length of hinged- hinged column) and select


below the lists:

List- I List- II

(i) Both ends hinged 1. L

(ii) One end fixed and

other end free

2. L / 2

(iii) One end fixed and the

other hinged

3. L/2

(iv) Both ends fixed 4. 2L

Codes:

(i) (ii) (iii) (iv)

(A) 1 3 4 2

(B) 1 4 2 3

(C) 3 1 2 4

(D) 3 1 4 2

8. A short column of symmetric cross-section

made of a brittle material is subjected to an

eccentric vertical load P at an eccentricity 'e'.

To avoid tensile stress in the short column,

the eccentricity 'e'. To avoid tensile stress in

the short column, the eccentricity 'e'. should

be less than or equal to

e

A D

B Ch

b

L

Pe

(A) h/12 (B) h/6

(C) h/3 (D) h/2

9. With one fixed end and other free and, a

column of length L buckles at load

P1.Another column of same length and same

cross-section fixed at both ends buckles at

load P2./ P1. Is

(A) 1 (B) 2

(C) 4 (D) 16

10. Slenderness ration of a column is defined as

the ratio of its length to its

(A) Least radius of gyration

(B) Least lateral dimension

(C) Maximum lateral dimension

(D) Maximum radius of gyration

11. Four columns of same material and same

length are the rectangular cross-section of


44

SOM

same breadth 'b'. The depth of the cross-

section and the end conditions are, however

different are given as follows:

Column

Depth

End conditions

1. 0.6 b Fixed- Fixed

2. 0.8b Fixed – hinged

3. 10b Fixed – hinged

4. 2.6b Fixed–Free

Which of the above columns has maximum

value of Euler buckling load?

(A) Column 1 (B) Column 2

(C) Column 3 (D) Column 4

12. What is the expression for the crippling load

for a column of length / with one end fixed

and other end free?

(A) P = 2π²EI

l²

(B) P =

π²EI

4l²

(C) P = 4π²EI

l²

(D) P =

π2EI

l2

13. A structural member subjected to an axial

compressive force is called

(A) beam (B) strut

(C) frame (D) None of these

14. If one end of a hinged column is made fixed

and the other free, how much is the critical

load compared to the original value?

(A) One-fourth (B) Half

(C) Twice (D) Four times

15. The buckling load for a column hinged at

both ends is 10 kN. If the ends are fixed, the

buckling load changes to

(A) 40 kN (B) 2.5 kN

(C) 5 kN (D) 20 kN

16. If diameter of a long column is reduced by

20%, the percentage reduction in Euler’s

buckling load for the same end conditions is

(A) 4 (B) 36

(C) 49 (D) 60

17. The end conditions of a column for which

length of column is equal to the equivalent

length are :

(A) Both the ends are hinged

(B) Both are ends are fixed

(C) One end fixed and other end free

(D) One end fixed and other end hinged

18. Determine the ratio of the buckling strength

of a solid steel column to that of a hollow

column of the same material having the same

area of cross section. The internal diameter

of the hollow column is half of the external

diameter. Both column is half of the external

diameter. Both columns are of identical

length and are pinned or hinged at the ends:

(A) Ps

Ph=

2

5 (B)

Ps

Ph=

3

5

(C) Ps

Ph=

4

5 (D)

Ps

Ph= 1

19. Euler’s formula for a mild steel long column

hinged at both ends is not valid for

slenderness ratio

(A) greater than 80 (B) less than 80

(C) greater than 120 (D) greater than 120

20. A long column has maximum crippling load

when its

(A) both ends are hinged

(B) both ends are fixed

(C) one end is fixed and other end is hinged

(D) one end is fixed and other end is free


45

SOM

21. Slenderness ratio of a 5 m long column

hinged at both ends and having a circular

cross-section with diameter 160 mm is

(A) 31.25 (B) 62.5

(C) 100 (D) 125

22. Effective length of a column fixed at one end

and hinged at the other end is

(A) 1 /2 (B) 1 / 2

(C) 2 1 (D) 2 1

23. A short column of external diameter of 250

mm and internal diameter of 150 mm carries

an eccentric load of 1000 kN. The greatest

eccentricity which the load can have without

producing tension anywhere is

(A) 20 mm (B) 31.25

(C) 37.5 mm (D) 42.5

24. A masonry pier ABCD as shown in Fig.

supports a vertical load W at a point P. The

nature of bending stresses at A due to

eccentricity of load about X-X axis and y-y

respectively are

A B

CD

y

y

x x

P

(A) compressive and compressive

(B) tensile and tensile

(C) compressive and tensile

(D) tensile and compressive

1. (C)

2. (D)

3. (D)

4. (C)

5. (D)

6. (A)

7. (B)

8. (B)

9. (D)

10. (A)

11. (D)

12. (B)

13. (B)

14. (A)

15. (A)

16. (D)

17. (A)

18. (B)

19. (B)

20. (B)

21. (D)

22. (B)

23. (D)

24. (C)

Answer key


46

SOM

[Sol] 2. (D)

P =π2EI

le2 , P ∝

1

le

(∵ le = 2L for free fix ends)

∴ le = L/2 for fix fix ends

∴P fix −fix

P fixfix=

le free −fix

le free −fix=

4L 2

L

2

2 =16

1

[Sol] 3. (D)

Given : l1 = 2m, L2 = 1m

P =π2EI

le2

As P ∝1

le2 [le = l, if both ends are hinged]

P2

P1=

l12

l22 =

4

1

P2 = 4P1 = 4kN

[Sol] 5. (D)

Pcr1 =π2EI

Le2 =

π2E×π

64d4

Le2 ⇒ Pcr1 ∝ d4

% reduction in Euler load

=Pcr 1−Pcr 2

Pcr 1× 100 =

d4− 0.8d 4

d4

= d4−0.4096d4

d4 = 0.59 = 59%

[Sol] 8. (B)

Direct stress ς1 =P

b.h

Bending stress ς2 =M

Z=

6Pe

bh2

To avoid tensile stress,

Total stress = −ς1 + ς2 ≤ 0

⇒ −P

bh+

6Pe

bh2 ≤ 0

⇒ e ≤h

6

A

B C

D e

h

b

[Sol] 15. (A)

P =π2EI

l2 ⇒ p ∝1

l2

(∵ fix − fix ends le = L/2, hinge − hinge le = L)

Phinge −hinge

P fix −fix=

lfix −fix2

lhinge −hinge2 =

L/2 2

L 2 =1

4

Pfix−fix = 4 × 10 = 40 kN

Explanations


1. A 3 – meter long steel cylindrical shaft is

rigidly held at its two ends. A pulley is

mounted on the shaft at 1 meter from one

end; the shaft is twisted by applying torque

on the pulley. The maximum shearing

stresses developed in 1 m and 2 m lengths

are respectively τ2. The ratio τ2 :τ1 is

(A) 1/2 (B) 1

(C) 2 (D) 4

2. A round shaft of diameter 'd' and length 'l'

fixed at both ends 'A' and 'B', is subjected to

a twisting moment ' T' at ' C 'at a distance of

1/4 from A (see figure). The torsional

stresses in the parts AC and CB will be

AT

C

B

L/4

(A) equal;

(B) in the ratio of 1:3

(C) in the ratio of 3:1

(D) indeterminate

3. Maximum shear stress in a solid shaft of

diameter D and length L twisted through an

angle θis τ. A hollow shaft of same material

and length having outside and inside

diameters of D and D/2 respectively is also

twisted through the same angle of twist θThe

value of maximum shear in the hollow shaft

will be

(A) 16

15τ (B)

8

7τ

(C) 4

3τ (D) τ

4. Two hollow shafts of the same material have

the same length and outside diameter, Shaft 1

has internal diameter equal to one third of the

outer diameter and shaft 2 has internal

diameter equal to half of the outer diameter.

If both the shafts are subjected to the same

toque, the ratio of their twists θ1/ θ2 will be

equal to

(A) 16 /18 (B) 8/27

(C) 19/27 (D) 243/256

5. A solid shaft of diameter 100 mm, length

1000mm is subjected to a twisting moment

'T’ the maximum shear stress developed in

the shaft to 60 N / mm2. A hole of 50 mm

diameter is now drilled throughout the length

of the shaft. To develop a maximum shear

stress to 60 N /mm2 in the hollow shaft, the

torque 'T' must be reduced by

(A) T /4 (B) T/8

(C) T /12 (D) T /16


Chapter

6 TORSION OF SHAFTS

Syllabus : Introduction, Theory of pure torsion, Shaft in series, Shafts

in parallel.


48

SOM

6. The diameter of shaft A is twice the diameter

of shaft B and both are made of the same

material. Assuming both the shafts to rotate

at the same speed, the maximum power

transmitted by B is it maximum shear stress

in both shafts remains the same

(A) The same as that of A

(B) half of A

(C) 1/8th of A

(D) 1/4th of A

7. The outside diameter of a hollow shaft is

twice that of its inside diameter. The torque-

carrying capacity of this shaft is Mt1. A solid

shaft of the same material has the diameter

equal to the outside diameter of the hollow

shaft. The solid shaft can carry a torque of

M12. The ratio M11/ M12 is

(A) 15/16 (B) 3/4

(C) 1/2 (D) 1/16

8. One-half length of 50mm diameter steel rod

is solid while the remaining half is hollow

having abore of 25 mm. The rod is subjected

to equal and opposite torque at its ends. If the

maximum shear stress in solid portion is τ,

the maximum shear stress in the hollow

portion is

(A) 15

16τ (B) τ

(C) 4

3τ (D)

16

15τ

9. A solid circular rod AB of diameter D and

length L is fixed at both ends. A torque T is

applied at a section X such that AX= 1/4 and

BX = 3L/4. What is the maximum shear

stress developed in the rod?

(A) 16T

πD3 (B) 12T

πD3

(C) 8T

πD3 (D) 4T

πD3

10. A hollow shaft of the same cross-section area

and material as that of a solid shaft,

transmits:

(A) Same torque

(B) Lesser torque

(C) More torque

(D) Cannot be predicated without more data

11. What is the total angle of twist of the stepped

shaft subject to torque T shown in figure

given below?

2d

2lT

l d

(A) 16Tl

πGd4 (B) 38Tl

πGd 4

(C) 64Tl

πGd 4 (D) 66Tl

πGD 4

12. For a power transmission shaft transmitting

power P at N rpm, its diameter is

proportional to

(A)

1/3P

N

(B)

1/2P

N

(C)

2/3P

N

(D)

P

N

13. While transmitting the same power by a

shaft, if its speed is doubled, what should be

its new diameter if the maximum shear stress

induced in the shaft remains same?

(A) 1

2 of the original diameter


49

SOM

(B) 1

2 of the original diameter

(C) 2 of the original diameter

(D) 1

(2)1/3 of the original diameter

14. What is the maximum torque transmitted by

a hollow shaft of external radius R and

internal radius r?

(A) π

16 (R3 − r3)fs

(B) π

2R (R4 − r4)fs

(C) π

8R (R4 − r4)fs

(D) π

32 (R4 − r43)fs

15. In power transmission shafts, if the polar

moment of inertia of a shaft is doubled, then

what is the torque required to produce the

same angle of twist?

(A) 1

4 of the original value

(B) 1

2 of the original value

(C) Same as the original value

(D) Double the original value

16. The diameter of a solid shaft is D. The inside

and outside diameters of a hollow shaft of

same material and length are D

3 and

2D

3

respectively. What is the ratio of the weight

of the hollow shaft to that of the solid shaft?

(A) 1 :1 (B) 1 : 3

(C) 1 :2 (D) 1 :3


Maximum shear stress induced in a power

transmitting shaft is

1. directly proportional to torque being

transmitted.

2. inversely proportional to the cube of its

diameter.

3. directly proportional to its polar

moment of inertia.


(A) 1, 2 and 3 (B) 1 and 3 only

(C) 2 and 3 only (D) 1 and 2 only

18. The ratio of torque carrying capacity of a

solid shaft to that of a hollow shaft is given

by

(A) (1 – K4) (B) (1 – K

4)

-1

(C) K4

(D) 1

K4

Where K = D1

D0

D1 = Inside diameter of hollow shaft D0 =

Outside diameter of hollow shaft material are

the same.

19. A solid shaft transmits a torque T. The

allowable shearing stress is τ . What is the

diameter of the shaft?

(A) 3 16T

πτ (B) 3

32T

πτ

(C) 3 16T

τ (D) 3

T

τ

20. A Solid steel shaft of diameter d and length I

is subjected to twisting moment T. Another

shaft B of brass having same diameter d, but

length 1/2 is also subjected to the same

moment. If shear modulus of steel is two

times that of brass, the ratio of the angular

twist of steel to that of brass shaft is

(A) 1 :2 (B) 1 : 1

(C) 2 :1 (D) 4 :1

21. For the two shafts connected in parallel


50

SOM

(A) Torque in each shaft is the same

(B) Shear stress in shaft is the same

(C) Angle of twist of each shaft is the same

(D) Torsional stiffness of each shaft is the

same

22. The magnitude of shear induced in a shaft

due to applied torque varies

(A) From maximum at the centre to zero at

the circumference

(B) From zero at the centre to maximum at

the circumference

(C) From maximum at the centre to

minimum but not zero at the

circumference

(D) From minimum but not zero at the

centre, to maximum at the

circumference.

23. If a solid circular shaft of steel 2 cm in

diameter is subjected to a permissible shear

stress 10 kN/cm2, then the value of the

twisting moment (Tr) will be

(A) 10π kN − cm (B) 20π kN − cm

(C) 15π kN − cm (D) 5 π kN − cm

24. The ratio of maximum shear stress developed

in a solid shaft of diameter D and a hollow

shaft of external diameter D and internal

diameter d for the same torque is give by

(A) D2+d2

D2 (B) D2−d2

D2

(C) D4−d4

D4 (D) D4−d4

d4

25. If a shaft of diameter d is subjected to a

torque, T the maximum shear stress is

(A) 32 T

πd3 (B) 16 T

πd2

(C) 16 T

πd3 (D) 64 T

πd4

26. A solid circular shaft of 6 m length is built in

at its ends and subjected to an externally

applied torque 60 kN-m at a distance of 2 m

from left end. The reactive torques at the left

end and the rights end are respectively

(A) 20 kN.m and 40 kN.m

(B) 40 kN.m and 20 kN.m

(C) 15 kN. M and 45 kN.m

(D) 30 kN. M and 30 kN.m

27. The ratio of strain energy stored by a hollow

shaft of external diameter D and internal

diameter d and strain energy stored by a solid

shaft of diameter D is

(A) D2+d2

D2 (B) D2−d2

D2

(C) D4−d4

D4 (D) D4+d4

d4

28. If the internal radius of a hollow shaft is n

times the external radius, then ratio of

torques carried by the hollow shaft and solid

shaft of same cross-sectional area and

subjected to the same maximum shearing

stress is

(A) 1–n4

(B) 1+n2

1–n2

(C) 1+n2

1−n2 (D) 1+n2

1−n2

29. If a circular shaft is subjected to a torque T

and bending moment M, the ratio of

maximum bending stress and maximum

(A) 2M

T (B)

M

2T

(C) M

T (D)

2T

M


51

SOM

30. Two circular bars A and B of Same material

and same length are of diameters DA and DB

respectively. The bars are subjected to the

same torque T. The ratio of strain energies

stored in the bars A and B is proportional to

(A) A

B

D

D (B) B

A

D

D

(C)

2

A

B

D

D

(D)

4

B

A

D

D

31. A circular shaft subjected to torsion

undergoes a twist of 10 in a length of 120 is

limited to 1000 kg/cm2 and if modulus of

rigidity G = 0.8 ×106 kg/ cm

2, then the radius

of the shaft should be

(A) π/18 (B) 18 / π

(C) π/27 (D) 27/π

32. If the diameter of a shaft subjected to torque

alone is doubled, then the horse power P can

be increased to

(A) 16 P (B) 8 P

(C) 4 P (D) 2P

33. A shaft turns at 150 rpm under a torque of

1500 Nm. Power transmitted is

(A) 15 π kW (B) 10 π Kw

(C) 7.5 π kW (D) 5 π kW

34. A hollow steel shaft of external diameter 100

mm and internal diameter 50 mm is to be

replaced by a solid alloy shaft. Assuming

the same value of polar modulus for both, the

diameter of the solid alloy shaft will be

(A) 10× 93753

mm

(B) 10× 93753

× 10mm

(C) 10× 9375

10

3 mm

(D) 93753

mm

35. In order to produce a maximum shearing

stress of 75 MN/m2 in the material of a

hollow circular shaft of 25 cm outer diameter

and 17.5 cm inside diameter, the torque that

should be applied to the shaft is

(A) 87.4 k N.m (B) 17.49 kN.m

(C) 174.9 kN.m (D) 349.7 kN.m

1. (A)

2. (C)

3. (D)

4. (D)

5. (D)

6. (C)

7. (A)

8. (D)

9. (B)

10. (C)

11. (D)

12. (A)

13. (D)

14. (B)

15. (D)

16. (A)

17. (D)

18. (B)

19. (A)

20. (B)

21. (C)

22. (B)

23. (D)

24. (C)

25. (C)

26. (B)

27. (A)

28. (D)

29. (A)

30. (D)

31. (D)

32. (B)

33. (C)

34. (C)

35. (C)

Answer key


52

SOM

[Sol] 1. (A)

τ

R=

Gθ

L

τ ∝1

L ∵ G, R, θ are constant

τ1

τ2=

L2

L1=

2

1

τ2

τ1=

1

2

[Sol] 2. (C)

τ

R=

Gθ

l

τ ∝1

l ∵ G, R, θare constant

τAC

τBC=

lBC

lAC=

3l/4

l/4=

3

1

[Sol] 3. (A)

τMAX =T

Zp

τH

τS=

ZP S

ZP H=

π

16D3

π

16D d4−

d

2

4

τH

τs=

16

15

[Sol] 4. (D)

T

J=

Gθ

l

For same Torque

θ ∝I

J

J1 =π

32 D4 −

D

3

4 =

80

81×

π

32D4

J2 =π

32 D4 −

D

2

4 =

15

16×

π

32D4

∴θ1

θ2=

243

256

[Sol] 7. (A)

Given D = 2d & τHollow = τsolid

Tensional shear stress in hollow shaft

τHollow =16Mt1

πD3 1−K4 Where K =

d

D

∴ τHollow =16Mt1

πD3 1−1

16

=16Mt1

πD3 ×16

15

τSolid =16Mt2

πD3

As τHollow = τSolid

∴16Mt1

πD3 ×16

15=

16Mt1

πD3 ⇒Mt1

Mt2

=15

16

[Sol] 11. (D)

T

J=

Gθ

l⇒

Tl

GJ

Total angle of twist θ = θ1 + θ2

=Tl

G×π

32× 2d 4

+ T×2l

G×π

32×d4

=66Tl

Gπd4

[Sol] 14. (B)

T

J=

fs

R

TMAX =fs

R

π

32 D4 − d4

=fs

R

π

3216 R4 − r4

=π

2R R4 − r4 fs

[Sol] 16. (A)

ds = D, dh =D

3, Dh =

2D

3

If materials & length of shaft are same then,

Weight of shaft ∝ Area of shaft

W h

W s=

Ah

As=

2d

3

2−

D

3

2

D2 = 1

Explanations


1. Match List-1 (stress) with List-II (Kind of

loading) and select the correct answer using

the codes given below the lists :

List- I List- II

(i)

1. Combined bending

and torsion of

circular shaft.

(ii)

2. Torsion of circular

shaft.

(iii)

3. Thin cylinder

subjected to

internal pressure.

(iv)

4. Tie bar subjected

to tensile force

Codes:

(i) (ii) (iii) (iv)

(A) 1 2 3 4

(B) 2 3 4 1

(C) 2 4 3 1

(D) 3 4 1 2

2. Consider in following statements.

State of stress at a point when completely

specified, enables one to determine the

1. Principal stresses at the point

2. Maximum shearing stress at the point

3. Stress components on any arbitrary place

containing the point


(A) 1,2 and 3 (B) 1 and 3

(C) 2 and 3 (D) 1 and 2

3. State of stress at a point in a strained body is

shown in Figure A. Which one of the figure

given below represents correctly the Mohr’s

circle for the state of stress?

xy

xy

(A)

y

x

(B)

y

x

(C)

x

y

(D)

y

x

4. Consider the following statements: State of

stress in two dimensions at a point in a

loaded component can be completely

specified by indicating the normal and shear

stresses on

1. a plane containing the point

Practice Problem

Chapter

7 PRINCIPAL STRESES,

STRAINS AND MOHR’S

CIRCLE Syllabus : Compound stress, Principal plane and principal stress.


54

SOM

2. any two planes passing through the

point

3. two mutually perpendicular planes

passing through the point


(A) 1 and 3 (B) 2 only

(C) 1 only (D) 3 only

5. Plane stress at a point in a body is defined by

principal stresses 3ςand ς. The ratio of the

normal stress to the maximum shear stress on

the plane of maximum shear stress is

(A) 1 (B) 2

(C) 3 (D) 4

6. Which one of the following Mohr’s circles

represents the state of pure shear?

(A)

O

(B)

O

(C)

O

(D)

7. In a two dimensional problem, the state of

pure shear at a point is characterized by

(A) εx= εy and γxy = 0

(B) εx= −εy and γxy ≠ 0

(C) εx= 2εy and γxy ≠ 0

(D) εx= 0.5εy and γxy ≠ 0

8. A cantilever is loaded by a concentrated load

P at the free end as shown. The shear stress

in the element LMNOPQRS is under

consideration.

P

Which of the following figures represents the

shear stress direction in the cantilever?

(A)

P

L

SR

N

M

OQ

(B)

P

S R

N

ML

OQ

(C)

S

P

OQ

L M

N

R

(D)

P

S

OQ

L M

N

R

9. At a point in two –dimensional stress system

ςx= 100 N / mm2, ςy= τxy = 40 N/ mm

2 .

what is the radius of the Mohr circle for

stress drawn with a scale of 1 cm = 10 N /

mm2

(A) 3 cm (B) 4 cm

(C) 5 cm (D) 6 cm

10. Normal stresses of equal magnitude ς, but of

opposite signs, act at a point of a strained

material in perpendicular direction. What is

the magnitude of the stress on a plane

inclined at 450 to the applied stresses?

(A) 2 ς (B) ς/2

(C) ς/4 (D) Zero


55

SOM

11. A body is subjected to pure tensile stress of

100 units. What is the maximum shear

produced in the body at some oblique plane

due to the above ?

(A) 100 units (B) 75 units

(C) 50 units (D) 0 units

12. Principal strains at a point are + 100× 10-6

and -200 × 10-6

What is the maximum shear

strain at the point?

(A) 300 × 10-6

(B) 200 × 10-6

(C) 150 × 10-6

(D) 100 × 10-6

13. In a strained material one of the principal

stresses is twice the other. The maximum

shear stress in the same case is τmax. Then,

what is the value of the maximum principal

stress ?

(A) τmax. (B) 2τmax.

(C) 4τmax. (D) 8τmax.

14. For a general two dimensional stress system,

what are the coordinates of the centre of

Mohr’s circle?

(A) ςx−ςy

2, 0 (B) 0,

ςx +ςy

2

(C) ςx +ςy

2, 0 (D) 0,

ςx−ςy

2

15. Maximum shear stress in a Mohr’s Circle

(A) is equal to radius of Mohr’s Circle

(B) is greater than radius of Mohr’s circle

(C) is less than radius of Mohr’s circle

(D) could be any the above

16. A point in a two dimensional state of strain is

subjected to pure shearing strain of

magnitude γ xy radians. Which one of the

following is the maximum principal strain?

(A) γxy (B) γxy

2

(C) γxy

2 (D) 2γxy

17. Consider the Mohr’s circle shown below:

0n

What is the state of stress represented by this

circle?

(A) ςx = ςy ≠0,τxy = 0

(B) ςx = ςy =0,τxy ≠ 0

(C) ςx = 0, ςy = τxy≠ 0

(D) ςx ≠ 0, ςy = τxy =0


1. Two – dimensional stresses applied to a

thin plate in its own plane represent the

plane stress condition .

2. Under plane stress condition, the strain

in the direction perpendicular to the

plane is zero.

3. Normal and shear stresses may occur

simultaneously on a plane.

Which of these statements is / are correct?

(A) 1 only (B) 1 and 2

(C) 2 and 3 (D) 1 and 3

19. The principal strains at a point in a body,

under biaxial state of stress, are 1000× 10-6

and – 600 × 10-6

. What is the maximum

shear strain at that point?

(A) 200 × 10-6

(B) 800 × 10-6


56

SOM

(C) 100 × 10-6

(D) 1600 × 10-6

20.

A

B

A point in two –dimensional stress state, is

subjected to biaxial stress as shown in the

above figure. The shear stress acting on the

plane AB is

(A) Zero (B) ς

(C) ςcos2θ

(D) ς sin θ. Cosθ

21. If the principal stresses and maximum

shearing stresses are of equal numerical

value at a point in a stressed body, the state

of stress can be termed as

(A) Isotropic

(B) Uni-axial

(C) Pure shear

(D) Generalized plane state of stress

22. What are the normal and shear stresses on

the 450 planes shown?

45°

45°

= 400 MPa

(A) ς1 = −ς2=400 MPa and τ = 0

(B) ς1 = ς2=400 MPa and τ = 0

(C) ς1 = ς2= – 400 MPa and τ = 0

(D) ς1 = ς2= τ = ± 200 MPa

23. A piece of material is subjected to two

perpendicular tensile stresses of 70 MPa and

10 MPa. The magnitude of the resultant

stress on a plane in which the maximum

shear stress occurs is

(A) 70 MPa (B) 60 MPa

(C) 50 MPa (D) 10 MPa

24. The state of plane stress at a point in a loaded

member is given by

ςx = + 800 MPa

ςy = + 200 MPa

τxy =± 400 MPa

The maximum principal stress and maximum

shear stress are given by :

(A) ςmax = 800 MPa and τmax = 400 MPa

(B) ςmax = 800 MPa and τmax = 500 MPa

(C) ςmax = 1000 MPa and τmax = 500 MPa

(D) ςmax = 1000 MPa and τmax = 400 MPa

25. A failure theory postulated for metals is

shown in a two dimensional stress plane. The

theory is called

(A) Maximum distortion energy theory

(B) Maximum normal stress theory

(C) Maximum shear stress theory

(D) Maximum strain theory

26. If an element of a stressed body is in a state

of pure shear with a magnitude of 80 N/mm2

the magnitude of maximum principal stress

at that location is


57

SOM

(A) 80 N/mm2 (B) 113.14N/mm

2

(C) 120 N/mm2 (D) 56.57 N/mm

2

27. Pick the incorrect statement from the

following four statements

(A) On the plane which carries maximum

normal stress, the shear stress is zero

(B) Principal planes are mutually orthogonal

(C) On the plane which carries maximum

shear stress, the normal is zero

(D) The principal stress axes and principal

strain axes coincide for an isotropic

material

28. The state of two dimensional stresses acting

on a concrete lamina consists of a direct

tensile stress, ςx = 1.5N/mm 2, and shear

stress, τ = 1.20 N/mm 2, which cause

cracking of concrete. Then the tensile

strength of the concrete in N/mm2 is

(A) 1.50 (B) 2.08

(C) 2.17 (D) 2.29

29. In a two dimensional stress analysis, the state

of stress at a point is shown below, if ς=120

MPa and τ =70 MPa, ςx and ςy′ are

respectively,

y

1

CA

xB

y

AB = 4BC = 3AC = 5

(A) 26.7 MPa and 172.5 MPa

(B) 54 MPa and 128 MPa

(C) 67.5 MPa and 213.3 MPa

(D) 16 MPa and 138 MPa

30. If principal stresses in a two-dimensional

case are -10 MPa and 20 MPa respectively,

then maximum shear stress at the point is

(A) 10 MPa (B) 15 MPa

(C) 20 MPa (D) 30 MPa

31. Mohr’s circle for the state of stress defined

by 30 00 30

MPa is a circle with

(A) center at (0,0) and radius 30 MPa

(B) center at (0,0) and radius 60 MPa

(C) center at (30,0) and radius 30 MPa

(D) center at (30,0) and zero radius

32. An axially loaded bar is subjected to a

normal stress of 173 MPa. The shear stress in

the bar is

(A) 75 MPa (B) 86.5 MPa

(C) 100 MPa (D) 122.3 MPa


1. On a principal plane, only normal stress acts.

2. On a principal plane, both normal and

shear stresses act.

3. On a principal plane, only shear stress

acts.

4. Isotropic state of stress is independent

of frame of reference.

Which of these statements is / are correct ?

(A) 1 and 4 (B) 2 only

(C) 2 and 4 (D) 2 and 3

34. The major and minor principal stresses at a

point are 3 MPa and – 3MPa respectively.

The maximum shear stress at the point is

(A) Zero (B) 3 MPa

(C) 6 MPa (D) 9 MPa


58

SOM

35. 2D stress at a point is given by a matrix

ςxx τxy

τyx ςyy =

100 3030 20

MPa

The maximum shear stress in MPa is

(A) 50 (B) 75

(C) 100 (D) 110

36. When a member is subjected to axial tensile

load, the greatest normal stress is equal to

(A) half the maximum shear stress

(B) maximum shear stress

(C) twice the maximum shear stress

(D) none of the above

37. At a point in a strained body carrying two

unequal unlike principal stresses p1 and p2 (p1 >

p2) the maximum shear stress is given by

(A) p1/ 2 (B) p2/ 2

(C) (p1 –p2) /2 (D) (p1 +p2) / 2

38. If the principal stresses at a point in a

strained body are p1 and p2 (p1 >p2), then the

resultant stress on a plan e carrying the

maximum shear stress is equal to

(A)2 2p1 p2 (B)

P12+p22

2

(C) P12−P22

2 (D)

P12+P22

2

39. A point in a strained body is subjected to a

tensile stress of 100 MPa on one plane and a

tensile stress of 50 MPa on a plane at right

angle to it. If these planes are carrying shear

stresses of 50 MPa, then the principal

stresses are inclined to the larger normal

stress at an angle

(A) tan−1 2 (B) 1

2tan−1(2)

(C) 1

2tan−1

2

3 (D)

1

2tan−1

1

3

40. If a prismatic member with area of cross

section A is subjected to a tensile load P,

then the maximum shear stress and its

inclination with the direction of load

respectively are

(A) P/A and 450

(B) P/2A and 450

(C) P /2A and 600

(D) P /A and 300

41. The radius of Mohr’s circle for two equal

unlike principal stresses of magnitude p is

(A) P (B) p / 2

(C) zero (D) none of these

42. Shear stress on principal planes is

(A) zero (B) maximum

(C) minimum (D) none of these


In a uni-dimensional stress system, the

principal plane is defined as one of which the

1. shear stress is zero

2. normal stress is zero

3. shear stress is maximum

4. normal stress is maximum

Of these statements

(A) 1 and 2 are correct




44. If an element is subjected to pure shearing

stress τxy . the maximum principal stress is

equal to

(A) 2 (B) τxy

2

(C) τxy (D) 1 − (τxy )2


59

SOM

45. A cast iron block of 5 sq. cm. cross section

carries an axial tensile load of 10 t. Then

maximum shear stress in the block is given

by

(A) 2000 kg /cm2

(B) 1000 kg /cm2

(C) 500 kg /cm2

(D) 200 kg / cm2

46. A mothr’s circle reduces to a point when the

body is subjected to

(A) pure shear

(B) uniaxial stress only

(C) equal and opposite axial stresses on two

mutually perpendicular planes, the

planes being free of shear

(D) equal axial stresses on two mutually

perpendicular planes, the planes being

free of shear


If two planes at right angles carry only shear

stress of magnitude ‘q’, then the

1. diameter of Mohr’s circle would equal 2q.

2. centre of the Mohr’s circle would lie at

the origin

3. principal stresses are unlike and have

magnitude ‘q’,

4. angle between the principal plane and

the plane of maximum shear would be

equal to 450

Of these statements

(A) 1 and 2 are correct



(D) 1,2,3 and 4 are correct

48. A bar of square section is subjected to a pull

of 10,000 kg. If the maximum allowable

shear stress on any section is 500 kg /cm2 ,

then the side of the square section will be

(A) 5 cm (B) 10 cm

(C) 15 cm (D) 20 cm

49. The cross-section of a bar is subjected to a

uniaxial tensile stress p. The tangential stress

on a plane inclined at θ to the cross-section

of the bar would be

(A) p sin 2 θ

2 (B) p sin 2 θ

(C) p cos 2 θ

2 (D) p cos 2 θ


1. On planes having maximum and

minimum principal stresses, there will

be no tangential stress.

2. Shear stresses on mutually perpendicular

planes are numerically equal.

3. Maximum shear stress is equal to half

the usm of the maximum and minimum

principal stresses.

Of the statements

(A) 1,2 and 3 are correct





60

SOM

1. (C)

2. (A)

3. (C)

4. (D)

5. (B)

6. (C)

7. (B)

8. (A)

9. (C)

10. (D)

11. (C)

12. (A)

13. (C)

14. (C)

15. (A)

16. (C)

17. (B)

18. (D)

19. (D)

20. (A)

21. (C)

22. (A)

23. (C)

24. (C)

25. (C)

26. (A)

27. (C)

28. (C)

29. (C)

30. (B)

31. (D)

32. (B)

33. (A)

34. (B)

35. (A)

36. (C)

37. (D)

38. (B)

39. (B)

40. (B)

41. (A)

42. (A)

43. (C)

44. (C)

45. (B)

46. (B)

47. (B)

48. (c )

49. (A)

50. (B)

Sol. 1 (C)

1. Combined bending and torsion will

have bending stress (tensile or

compressive) along with torsional shear

stresses.

2. A shaft subjected torsion will have only

torsional shear stresses.

3. Thin cylinder will have tensile hoop and

circumferential stresses

4. A tie member will have only axial

tensile stress.

Sol. 5 (B)

n

max

3

2 23

2

Sol. 9 (C)

x

10010cm

10

y xy

404cm

10

2

x y 2

xyR2

2

2 2 210 44 3 4 5cm

2

Sol. 12 (A)

6

1 100 10 , 6

2 200 10

Maximum shear strain 1 2

6 6100 10 200 10

6300 10

Sol. 13 (C)

1 22

Explanations

Answer key


61

SOM

1 2 2 2 2max

2

2 2 2

2 max2

Maximum principal stress 1 22

max max2 2 4

Sol. 19 (B)

xy 1 2

6 61000 10 600 10

61600 10

Sol. 22 (A)

On 45° plane 1 2 and 0

Sol. 23 (C)

x 70MPa, y 10MPa,

xy 0

2

x y 2

max 30MPa2

x y1 2n

2 2

70 1040MPa

2

Resultant Stress 2 2

n max

2 240 30 50MPa

Sol. 24 (C)

Maximum (Major) principal stress

2

x y x y 2

1 xy2 2

2

2

1

800 200 800 200400

2 2

1 100MPa

Maximum shear stress:

2

x y 2

max xy2

2

2800 200400

2

500MPa


1. Hoops stress and longitudinal stress in a

boiler shell under internal pressure are 100

M N/ m2 and 50 M N/ m

2 respectively.

Young’s modulus of elasticity and Poisson’s

ratio of the shell material are 200 GN/m2

and 0.3 respectively. The hoop strain in

boiler shell is

(A) 0.425× 10-3

(B) 0.5× 10-3

(C) 0.585× 10-3

(D) 0.75 × 10-3

2. From design point of view, spherical

pressure vessels are preferred over

cylindrical pressure vessels because they

(A) are cost effective in fabrication

(B) have uniform higher circumferential

stress

(C) uniform lower circumferential stress

(D) have a larger volume for the same

quantity of material used

3. When a thin cylinder of diameter 'd' and

thickness 't' is pressurized with an internal

pressure of 'p' (1/m is the Poisson’s ratio and

E is the modulus of elasticity ), then out of

the following, which statement is correct

(A) The circumferential strain will be equal

to pd

etE

1

2

1

m

(B) The longitudinal stress will be equal to

pd

2tE 1 −

1

m

(C) The longitudinal stress will be equal to

pd

2t

(D) m−2

2m−1

4. Circumferential stress in cylindrical steel

boiler shell under internal pressure is 80

MPa. Young’s modulus of elasticity and

Poisson’s ratio are respectively 2× 105 MPa

and 0.28. The magnitude of circumferential

strain in the boiler shall be

(A) 3.44× 10-4

(B) 3.84× 10-4

(C) 4× 10-4

(D) 4.56 × 10-4

5. A thin cylinder with closed lids is subjected

to internal pressure and supported at the ends

as shown in figure-1

X X

The state of stress at point × is as represented

as

6. A thin cylinder with both ends closed is

subjected to internal pressure p. The

longitudinal stress at the surface has been


Chapter

8 THIN CYLINDRICAL

SHELLS Syllabus : Thin cylindrical shells, Expression for circumferential

stress (or hoop stress), Expression for longitudinal stress (or axial

stress), Relation between circumferential stress and longitudinal stress,

Efficiency of a joint, Thin spherical shells.


63

SOM

calculated as ς0 . Maximum shear stress at

the surface will be equal to

(A) 2ς0 . (B) 1.5ς0 .

(C) 0 (D) None of these

7. A thin cylinder contains fluid at a pressure of

500 N/m2

, the internal diameter of the shell

is 0.6 m and the tensile stress in the material

is to be limited to 9000 N/ m2, The shell

must have a minimum wall the thickness of

nearly

(A) 9 mm (B) 11 mm

(C) 17mm (D) 21mm

8. A thin cylindrical shell is subjected to

internal pressure ' P '. The Poisson’s ratio of

the material of the shell is 0.3 Due to internal

pressure, the shell is subjected to

circumferential strain and axial strain. The

ratio of circumferential strain to axial strain

is

(A) 0.425 (B) 2.25

(C) 0.225 (D) 4.25

9. The commonly used technique of

strengthening thin pressure vessels is

(A) Wire winding

(B) Shrink fitting

(C) Auto –frettage

(D) Multi-layered construction

10. Match List-I with List-II and select the

correct answer Using the codes given below

the lists:

List- I List- II

(i) Wire

winding

1. Hydrostatic stress

(ii) Lame’s

theory

2. Strengthening of

thin cylindrical

shell

(iii) Solid sphere

subjected to

uniform

pressure on

the surface

3. Strengthening of

thick cylindrical

shell

(iv) Autofrettage 4. Thick cylinders

Codes :

(i) (ii) (iii) (iv)

(A) 4 2 1 3

(B) 4 2 3 1

(C) 2 4 3 1

(D) 2 4 1 3

11. A thin cylindrical shell of diameter 'd', length

ι and thickness 't' is subjected to an internal

pressure 'P' What is the ratio of longitudinal

strain to hoop strain in terms of Poisson’s

ratio (1/m) ?

(A) m−2

2m+1 (B)

m−2

2m−1

(C) 2m−1

m−2 (D)

2m+1

m−2

12. A water main of 1 m diameter contains water

at a pressure head of 100 meters. The

permissible tensile stress in the material of

the water main is 25 MPa. What is the

minimum thickness of the water main ?

( Take g = 10 m/s2)

(A) 10mm (B) 20mm

(C) 50mm (D) 60mm

13. A seamless pipe of diameter d m is to carry

fluid under a pressure of p kN/ cm2 The


64

SOM

necessary thickness t of metal in cm, if the

maximum stress is not to exceed ς kN/cm2 is

(A) t ≥pd

2ςcm (B) t ≥

100pd

2ςcm

(C) t ≥pd

2ςcm (D) t ≥

100 pd

2ςcm

14. What is the safe working pressure for a

spherical pressure vessel 1.5 m internal

diameter and 1.5 cm wall thickness. If the

maximum allowable tensile stress is 45

MPa?

(A) 0.9 MPa (B) 3.6 MPa

(C) 2.7 Mpa (D) 1.8 MPa

15. The design for thin cylindrical shells is made

on the basis of

(A) Mean of the hoop stress and

longitudinal stress

(B) Longitudinal strain

(C) Geometric mean of the hoop stress and

longitudinal stress

(D) Hoop stress.

16. A vessel is said to be thin walled, when

(A) Wall thickness is equal to or less than

1

20 of the internal diameter

(B) Vessel wall thickness is less than 4 mm

(C) Vessel wall thickness is 1

10 of outer

diameter

(D) None of the above.

17. Longitudinal stress developed in thin

cylinder or 'd' diameter 't' wall thickness and

'p' internal pressure as

(A) pd

2t (B)

pd

4t

(C) pd

8t (D)

pd

t

18. A thin cylinder of inner radius 500 mm and

thickness 10 mm subjected to an internal

pressure of 5 MPa. The average

circumferential (hoop) stress in MPa is

(A) 100 (B) 250

(C) 500 (D) 1000

19. A thin walled spherical shell is subjected to

an internal pressure. It the radius of the shell

is increases by 1% and the thickness is

reduced by 1% with the internal pressure

remaining the same, the percentage change

in the circumferential (hoop) stress is

(A) 0 (B) 1

(C) 1.08 (D) 2.02

20. A thin walled cylindrical pressure vessel

having a radius of 0.5 m and wall thickness

25 mm subjected to an internal pressure of

7.00 kPa. The hoop stress developed is

(A) 14 MPa (B) 1.4 MPa

(B) 0.14 MPa (C) 0.014 MPa

21. A water main of 1 m diameter contains

`water at a prevent head of 100 m. The

permissible tensile stress in the material of

the water main is 25 MPa. What is the

minimum thickness of the water main ?

( Take G = 10 m/s2)

(A) 10 mm (B) 20 mm

(C) 50 mm (D) 60 mm


65

SOM

1. (A)

2. (C)

3. (D)

4. (A)

5. (A)

6. (C)

7. (C)

8. (D)

9. (A)

10. (D)

11. (B)

12. (B)

13. (B)

14. (D)

15. (D)

16. (A)

17. (B)

18. (B)

19. (D)

20. (A)

21. (B)

[Sol] 1. (A)

∈h =ςh

E− μ.

ς l

E

=1

200×103 100 − 0.3 × 50

= 0.425 × 10−3

[Sol] 4. (A)

Circumferential Strain ∈h =ςh

E− μ.

ς l

E

=ςh

E− μ.

ςh

2E=

ςh

2E 1 −

μ

2

=80

2×105 1 −0.28

2 = 3.44 × 10−4

[Sol] 6. (C)

ςl = ς0; ςh = 2ς0

τmax = ςmax −ςmin

2

∴ τmax =ςh −0

2=

2ς0

2= ς0

[Sol] 8. (D)

∈h =Pd

2t− μ

Pd

4t

∈l=Pd

4t− μ

Pd

2t

∈h

∈l=

Pd

2t 1−

μ

2

Pd

2t

1

2−μ

=1−

0.3

21

2−0.3

= 4.25

[Sol] 12. (B)

Hoop stress =Pd

2t

25 × 106 =100×104×1

2×t ∵ P = pgh =

106Pa

t =1

50= 0.02m = 20mm

[Sol] 14. (D)

For a thin pressure vessel with internal

pressure.

ςh =PD

4t

45 =P×1500

4×15⇒ P = 1.8MPa

Explanations

Answer key


1. According to Rankine’s hypothesis, the

criterion of failure of a brittle material is

(A) maximum principal stress

(B) maximum shear stress

(C) maximum strain energy

(D) maximum shear strain energy

2. At a point in a steel member, the major

principal stress in 2000 kg/cm2 and the minor

principal stress is compressive, If the uni-

axial tensile yield stress is 2500 kg/ cm2,

then the magnitude of the minor principal

stress at which yielding will commence,

according to the maximum shearing stress

theory, is

(A) 1000 kg/ cm2

(B) 2000 kg/ cm2

(C) 2500 kg/ cm2

(D) 500 kg/ cm2

3. For the design of a cast iron member, the

most appropriate theory of failure is

(A) Mohr’s theory

(B) Rankine’s theory

(C) Maximum strain theory

(D) Maximum shear energy theory

4. At the point in a structure, there are two

mutually perpendicular tensile stresses of

800 kg/cm2 and 400 kg/cm

2. If the Poisson’s

ratio is μ = 0.25, what would be the

equivalent stress in simple tension according

to Maximum Principal Strain Theory?

(A) 1200 kg/cm2

(B) 900 kg/cm2

(C) 700 kg/cm2

(D) 400 kg/cm2

5. According to maximum shear stress failure

criterion, yielding in material occurs when

(A) maximum shear stress = 1/2 yield stress

(B) maximum shear stress = 2 × yield

stress

(C) maximum shear stress = 2

3× yield

stress

(D) maximum shear stress = 2 × yield stress

6. A certain steel has proportionality limit of

3000 kg/cm2 in simple tension. It is subjected

to principal stresses of 1200 kg/cm2

(tensile), 600 kg/cm2 (tensile) and 300

kg/cm2 (compressive ). What would be the

factor of safety according to maximum shear

stress theory?

(A) 1.50 (B) 1.75

(C) 1.80 (D) 2.00

7. In a strained body, three principal stresses at

a point are denoted by ς1, ς2 and ς3 such that


Chapter

9 THEORIES OF

FAILURE Syllabus : Introduction, Maximum principal stress theory, Maximum

principal strain theory, Maximum shear stress theory, Maximum strain

energy theory, Maximum shear strain energy theory, Limitations of the

theories of failure.


67

SOM

ς1, > ς2 > ς3. If ς0 denoted yield stress, then

according to the maximum shear stress

theory

(A) ς1−ς2 = ς0 (B) ς1−ς3 = ς0

(C) 2 – 3 = 0 (D) ς1+ς3

2 = ς0

8. At a point in a steel member, the major

principal stress is 200 MPa (tensile) and the

minor principal stress is compressive. If the

uniaxial tensile Yield stress is 250 MPa

,then according to the maximum shear stress

theory, the magnitude of the minor principal

stress (compressive) at which yielding will

commence is

(A) 200 MPa (B) 100 MPa

(C) 50 MPa (D) 25 Mpa

9. The limit of proportionality of a certain

sample is 300 MPa in simple tension. It is

subjected to principal stresses of 150 MPa

( tensile), 60 MPa (tensile ) and 30 MPa

(tensile). According to the maximum

principal stress theory, the factor of safety in

this case would be

(A) 10 (B) 5

(C) 4 (D) 2

10. A material of Young’s modulus 'E' and

Poisson’s ratio ' μ′ is subjected to two

principal stressς1and ς2 at a point in a two-

dimensional stress system. The strain energy

per unit volume of the material is

(A) 1

2E(ς 2

1+ ς 2

2− 2μς1ς2)

(B) 1

2E(ς 2

1+ ς 2

2− 2μς1ς2)

(C) 1

2E(ς 2

1− ς 2

2+ 2μς1ς2)

(D) 1

2E(ς 2

1− ς 2

2− 2μς1ς2)

11. If shaft made from ductile material is

subjected to combined bending and twisting

moments. Calculations based on which one

of the following failure theories would give

the most conservative value?

(A) Maximum principal stress theory

(B) Maximum shear stress theory.

(C) Maximum strain energy theory

(D) Maximum distortion energy theory

12. According to the maximum shear stress

theory of failure, permissible twisting

moment in a circular shaft is T. The

permissible twisting moment in the same

shaft as per the maximum principal stress

theory of failure will be

(A) T/2 (B) T

(C) 2T (D) 2T

13. A rod with cross-sectional area 100×10-

6m

2Tresca failure criterion, if the uniaxial

yield stress of the material is 200 MPa, the

failure load is

(A) 10kN (B) 20kN

(C) 100kN (D) 200 kN

14. A cold rolled steel shaft is designed on the

basis of maximum shear stress theory. The

principal stresses induced at its critical

section are 60 MPa and -60 MPa

respectively. If the yield stress for the shaft

material is 360 MPa, the factor of safety of

the design is

(A) 2 (B) 3

(C) 4 (D) 6


68

SOM

15. The maximum distortion energy theory of

failure is suitable to predict the failure of

which one of the following types of

materials?

(A) Brittle materials

(B) Ductile materials

(C) Plastics

(D) Composite materials

16. Match List-I (Theory of Failure) with List-II

(Predicted Ratio of Shear stress to Direct

Stress at yield Condition for Steel Specimen)

select the correct answer using the code

given below the lists:

List- I List- II

(i) Maximum shear Stress

theory

1. 1.0

(ii) Maximum energy of

distortion theory

2. 0.77

(iii) Maximum principal

stress theory

3. 0.62

(iv) Maximum principal

strain theory

4. 0.50

Codes :

(i) (ii) (iii) (iv)

(A) 1 2 4 3

(B) 4 3 1 2

(C) 1 3 4 2

(D) 4 2 1 3

17. Who postulated the maximum distortion

energy theory?

(A) Tresca (B) Rankine

(C) St. Venant (D) Mises-Henky

1. (A)

2. (D)

3. (B)

4. (C)

5. (A)

6. (D)

7. (B)

8. (C)

9. (D)

10. (A)

11. (B)

12. (B)

13. (B)

14. (B)

15. (B)

16. (B)

17. (D)

Answer key


1. What is the strain energy caused due to self

weight in a cylindrical bar?

(A) (W2 L) / (6 AE) (B) (W L) / (8 AE)

(C) (τ2/ 2G)V (D) (τ

2/ G)V

2. What is the maximum stress induced in a bar

2500 mm2, when a load of 2000 kN is

applied suddenly?

(A) 400 N/mm2

(B) 800 N/mm2

(C) 1600 N/mm2

(D) Insufficient data

3. Strain energy stored in a uniform bar is

given as ______

(A) (σ E/ 2A) (B) (σ L/ 2AE)

(C) (σ2 AL/ 4E) (D) (σ

2 AL/ 2E)

4. PL3/3EI is the deflection under the load P of

a cantilever beam. What will be the strain

energy?

(A) P2L

3/3EI (B) P

2L

3/6EI

(C) P2L

3/4EI (D) P

2L

3/24EI

5. Stress on an object due to sudden load is

_________ the stress induced when the load

is applied gradually.

(A) equal to (B) half

(C) twice (D) thrice

6. What is the strain energy stored in a simply

supported beam due to bending moment?

(A) ∫ (M2/EI) (B) ∫ (M

2/2EI)

(C) ∫ (M/2EI) (D) ∫ (2M/EI)

7. What is the proof resilience of a square bar

of 2500 mm2 and 200 mm long, when a load

of 150 kN is induced gradually? (Take E =

150 103 Mpa)

(A) 45 J (B) 8 J

(C) 5.3 J (D) 6 J

8. Modulus of resilience is the ratio of ______

(A) minimum strain energy and unit volume

(B) maximum stress energy and unit volume

(C) proof resilience and unit volume

(D) resilience and unit area

9. What is the strain energy stored in a cube of

50 mm, when it is subjected to shear stress

of 200 Mpa.

(G = 100 Gpa)

(A) 25 Nm (B) 75 Nm

(C) 125 Nm (D) 150 Nm

Practice Problem

Chapter

11 STRAIN ENERGY

Syllabus : Strain energy or resilience, Proof resilience, Modulus of

resilience, Strain energy in simple tension and compression, Power

transmitted by the shafts, Strength of a shaft, Strength of a hollow

circular shaft., Torsional rigidity (stiffness).


70

SOM

10. Energy stored in a body within an elastic

limit is called as _____

(A) resilience (B) strain energy

(C) both a. and b. (D) none of these

1. (A)

2. (C)

3. (D)

4. (B)

5. (C)

6. (B)

7. (D)

8. (C)

9. (A)

10. (C)

Answer key

properties of materials chapter syllabus : 1

Documents