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Preface In the recent years PSUs like BHEL, BEL, GAIL, IOCL, HPCL, ONGC and other Government Sectors like DMRC, DRDO, RRB, Staff Selection Commission are preferring to hire Junior Engineers and Technicians which has resulted in a large job opportunities for diploma holders. As these PSUs also offer job security and decent perks, many candidates are attracted towards it, gradually increasing the competition level. Unlike other competitive examinations, preparing for vacancy based exams of these PSUs is not an easy task as each exam has its own pattern, syllabus and trend. Most of these exams like to throw surprises based on the vacancies, leaving students amazed during their exam preparation with very less time for preparation. However, you can certainly guarantee yourself a smooth journey if you start early and plan it well. As the technical section of these exams hold the maximum scoring scope; it is very important to pay enough attention to this section. This series focuses on the technical section of various exams conducted by PSUs such as DRDO, BHEL and other Government Sectors like DMRC and Railways for the Diploma Engineers. These books help to prepare for all the exams at a single go which saves time instead of preparing separately for each PSU’s exam. Every chapter contains a brief theory followed by large number of practice questions. There is another section that includes questions asked in previous PSU exams. As a suggestion, please make sure that you spend enough time in understanding the fundamentals and concepts before going to practice exercise and previous years’ solved questions. GKP has also launched an Android App to provide you an update on all upcoming vacancies in the technical segment and it also has a lot of added content to aid your preparation. We hope this little effort of ours will be helpful in achieving your dreams. If you have any suggestions on improvement of this book, you can write to us at [email protected].

A ll the Best! Team GKP

Contents 1.

Strength of Material

1.1 – 1.22

2.

Building Material

2.1 – 2.26

3.

Concrete Technology

3.1 – 3.20

4.

Concrete Structures

4.1 – 4.14

5.

Steel Structures

5.1 – 5.14

6.

Soil Mechanics and Foundation Engineering

6.1 – 6.18

7.

Fluid Mechanics and Hydraulics

7.1 – 7.16

8.

Hydrology

8.1 – 8.8

9.

Irrigation

9.1 – 9.8

10.

Transportation Engineering

10.1 – 10.10

11.

Surveying

11.1 – 11.20

1

Strength of Material

C HAPTER

SIMPLE STRESS When some external system of forces acts on a body, it undergoes deformation. The deformation increases, as the forces increases. Sometimes, when external forces causing the deformation, are removed, the body springs back to its original position. This property of certain materials, of returning back to their original position after removing the forces is called elasticity. The body is said to be perfectly elastic, if it returns back completely to its original position after removal of the external forces. But it said to be partially elastic if it does not return back completely to its original position, after removal of the load. As the body undergoes deformations, it sets up some resistance to deformation. This resistance per unit area to deformation is called stress. P stress, = A where P is applied load and A is the cross-sectional area. Unit : N/m2. One newton per square meter is called one pascal (Pa). In S.I. 1 MPa = 106 N/m2 TYPES OF STRESSES (1) Tensile Stress : When a section is subjected to two equal and opposite pulls, as a result of which the body tends to lengthen, as shown in the figure, the stress induced is called tensile stress. P

P

(2) Compressive Stress : When a section is subjected to two equal and opposite pushes, as a result of which the body tends to shorten its length as shown in figure, the stress induced is called compressive stress. P

P

(3) Shear Stress : When a section is subjected to two equal and opposite forces acting tangentially across the resisting section, as a result of which the body tends to undergo an angular deformation, the stress is called shear stress.

For equilibrium of the system, shear resistance R should be equal to tangential load P, i.e. R =P Intensity of shear resistance along the section X–X is called shear stress. Therefore, R Shear resistance Shear stress,  = = A Shear area STRAIN Deformation per unit length is called Strain.  e= L where e is strain,  is change in length and L is original length. Average strain in a length is so small that the strain must be constant over that length. However, under certain conditions, the strain may be assumed constant and its value is computed from equation (1). These conditions are as follows : (i) Specimen must be of constant cross-section. (ii) Material must be homogeneous. (iii) Load must be axial, so that it produce uniform stress. Unit : Strain is a dimensionless quantity. SHEARING STRAIN An element subjected to tension undergoes an increase in length, whereas an element subject to shear does not change the length of its sides, but it undergoes a change in shape, from rectangle to parallelogram as shown in the figure.

1.2

Strength of Material

= G where G = Modulus of elasticity in shear or Modulus of rigidity.

Transverse displacement Distance from the lower face dl  = l

Shear strain  =

STRESS-STRAIN DIAGRAM In tensile test, a specimen is gripped between jaws of a testing machine. The elongation in a specified length called gauge length is observed simultaneously. These data are then plotted on a graph with the ordinate representing the load and the abscissa representing the elongation. Figure represents such a graph for structural steel. In this stress is plotted against the unit elongation (i.e. strain) only by reducing observed values to a unit basis. The properties of one specimen is compared with those of other specimens. The diagram shown in figure is called a stress strain diagram.

HOOKE'S LAW Consider straight line portion of the stress-strain diagram. Slope of that line is the ratio of stress to strain. It is called modulus of elasticity and is denoted by E. Slope of stress-strain curve,  E= e or  = Ee This is called Hooke's law. Units : Modulus of elasticity is expressed in N/m2. It is also denoted by using prefix M or G, i.e. MPa, GPa (106 and 109 N/m2 respectively).  L P PL = E , or = = Now L E A AE Relation bewteen Shearing stress and Shearing strain, assuming Hooke's law to apply to shear is given by

POISSON’S RATIO Ratio of the lateral strain to longitudinal strain is constant within the elastic limit for a homogeneous material and is called Poisson’s ratio, denoted by  or 1/m ey e 1 So,  or = – – z ex ex m where ex = strain due to stress in the x direction and ey and ez = strains induced in the perpendicular directions. The minus sign indicates a decrease in the transverse direction, when ex is positive, as in case of tensile elongation. Relation Between Modulus of Elasticity, Modulus of Rigidity and Bulk Modulus 2 E = 2G 1  1 , E = 3 K 1  m m 9KG and E = G  3K

FG H

FG H

IJ K

IJ K

STRAIN ENERGY AND RESILIENCE Consider a bar carrying a normal stress  (compressive or tensile) producing a normal strain e. Then 1 Strain energy per unit volume, u = . E 2 (area of the shaded portion)  but strain, e = E 2  u= 2E Total strain energy, 2 U=  volume of the specimen 2E The strain energy absorbed by the specimen is also called Resilience. Units : Strain energy has units of work done, i.e. Nmm or kg-cm etc. Proof Resilience Maximum strain energy absorbed by the body up to its elastic limit is called Proof resilience. 2 Proof resilience = e  Volume 2E  e A L.  e  2e (A.L. ) = = 2 E 2E Pe  e . L where eA = load at elastic limit = , 2 E  e .L where is change in length upto elastic limit E 1 Pe (L e ) Proof resilience = 2

FG H

IJ FG KH

FG IJ b g H K

IJ K FG IJ FG H KH

IJ K

Strength of Material

1.3

Proof resilience per unit volume is called modulus of resilience. 2

e 2E Similarly, Shear strain energy per unit volume, 

Modulus of resilience =

us =

1 2 2   2 2G

Total shear strain energy, U =

2  Volume 2G

where,  = shear stress and G is modulus of rigidity or shear modulus. TEMPERATURE STRESSES Change in the temperature causes change in the dimension of the material section. When a member is free for expansion or contraction, no stresses will be induced due to rise or fall of the temperature.

While designing a structure, a designer has to determine critical region of the member, where the stress developed due to loads is of maximum magnitude. There always exists a set of three orthogonal planes. The normal stresses on these three planes are called principal stresses and the planes are called principal planes. Member subjected two Normal and Tangential stresses

But if natural expansion or contraction is prevented, then due to rise or fall of temperature, temperature stress will be developed. Consider a bar of length L, fixed at A and B. Let temperature rise by TC. Let the rod is premitted to expand only , then temperature strain, TL –  Expansion prevented = et = L original length and, temperature stress, t=

E (TL  ) L

Consider two normal stresses x and y acting at two mutually perpendicular planes, as shown in the figure. Then normal stress on the plane plane EF

x  y

n =

2



x  y 2

cos 2   sin 2

And tangential stresses on the plane EF

t =

x  y 2

sin 2   cos 2

Principal stresses are given by

1 =

and 2 =

PRINCIPAL STRESS AND STRAINS The inclined fracture indicates failure by shear. This shows that, compressive force along an axis not only induces compressive stresses on the cross sectional area perpendicular to the load but also induces shear stress in the inclined plane.

x  y 2

x  y 2

2

 x  y     2  2   2

 x  y     2 2  

Principal stresses calculated by these equations may be like or unlike. After determining principal stresses and principal planes, the planes of maximum shear stress are easily determined. Direction of principal plance and stress are shown in the figure.

1.4

Strength of Material

(3) Propped cantilever beam: These beams are statically determinate, as the unknowns are three and the conditions available are also three.

Maximum shear stress, max =

1   2 c 2

BENDING MOMENT AND SHEAR FORCES Beam is structural member, which is designed to support any type of load but in general, it is required to carry load at right angle to its longitudinal axis. After the application of loads, beam bends and the amount of bending depends upon type of the loads, length of the beam, elasticity of the beam, and type of support of the beam. CLASSIFICATION OF BEAMS In general the beams are classified as : (1) Cantilever beam : A cantilever beam is statically determinate, as reaction of these beam at their support can be determined by the use of static equilibrium, i.e. V = 0 and M = 0 at free end

Fixed end Slope = 0 Deflection = 0 Fixed end moment = Exists Support reaction = Exists

Free end Slope exists Deflection exists End moment = 0 End reaction = 0

(2) Simply supported beam: In this beam both ends are simply supported, and they may be hinged support or a combination of roller support or a combination of roller and hinged supports. These beams are statically determinate, as total unknowns are two and conditions available are also two V = 0 and M = 0

Fixed Slope = 0 Deflection = 0 End moment = Exists End reaction = Exists

(4) Overhanging beam: These beams are statically determinate, since in these beams total unknowns are four and the conditons available are also four.

Fixed Slop = 0 Deflection = 0 End moment = Exists End reaction = Exists

Continuous Deflection = 0 Continuing moment = Exists

Free Slop = Exist Deflection = Exists End moment = 0

Continuing reaction= Exists

End Reaction = 0

(5) Fixed beam: These beams are statically indeterminate, since reaction at support cannot be determined by the use of equation of static equilibrium.

Fixed Slope = 0 Deflection = 0 End moment = Exists End reaction = Exists

Hinged support Deflection = 0 End moment = 0 Slope = Exist End reaction = Exists

Roller support Deflection = 0 End moment = 0 Slope = Exists End reaction = Exists

Simply supported Deflection = 0 End moment = 0 Slope = Exists End reaction = Exists

Fixed Slope = 0 Deflection = 0 End moment = Exists End reaction = Exists

(6) Continuous beam: When a beam is supported on more than two spants, it is defined as continuous beam.

Strength of Material

1.5

TYPE OF LOADING A beam may be subjected to the following types of loads or combination of loads : These beams are statically indeterminate. SPAN OF BEAM The horizontal distance between the supports is called clear span of the beam. The horizontal distance between the lines of action of the supports is called effective span. Note : IS 456, 1978 has defined the effective span for various R.C.C beams and IS 800, 1982 has defined the effective span for various steel structures.

(1) Point loads or Concentrated loads or Isolated loads : A load acting at a point is described as point load or concentrated load.

Practically it is not possible to apply load at a point, as it must have some contact area. But this area being so small as compared to the length of beam, it can be neglected. Units : N, kN or MN

(2) Uniformly distributed loads : Loads which is distributed over a beam uniformly, so that each unit length is loaded to the same extent, is called uniformly distributed load.

Units: N/m, kN/m or MN/m. (3) Uniformly varying loads : Load which is distributed over a beam such that its magnitude varies uniformly on each unit length of the beam, is called uniformly varying loads. Depending upon the nature of variation, these loads may be classified as triangular loads, or parabolic loads etc.

1.6

Strength of Material

SHEAR FORCE Shear force in a beam may be defined as algebraic sum of all the forces on its any side (either to the left, or to the right), or shear force is a vertical force at a section, which one part of the beam exerts on its other part at that section. BENDING MOMENT Bending moment in a beam may be defined as algebraic sum of the moments of the forces, to the right or to the left of the section. Note : Bending moment at any section of beam is also equal to the area of shearing force diagram either to the right or to the left of the section. SIGN CONVENTION (a) Shear force: The shear force is said to be positive at a section when the right hand portion tends to slide downwards with respect to left hand portion or. In other words, all the forces acting upwards to the left of the section causes positive shear force and these acting downwards will cause negative shear force.

(b) Bending moment: The bending moments which tend to bend the beam with concavity at top or upwards is taken as positive bending moment.

When the bending moment, which tend to bend the beam with concavity at bottom or downwards, is taken as negative bending moment.

CALCULATIONS OF S.F. AND B.M. Calculation of Shearing force at any section 1. Select either the right or the left hand portion of the beam from the section. 2. Calculate algebraic sum of all the forces including reactions, on the selected portion of the beam. 3. The algebraic sum of the forces on any one side of the section of the beam is the required shearing force at that section of the beam. Calculation of Bending moments at any section 1. Select either the right or the left hand portion of the beam from the section. 2. Calculate algebraic sum of the moments of all forces acting on the chosen portion of the beam, about the section. 3. Calculated algebaric sum of moments is the required bending moment at the considered section of the beam. RELATION BETWEEN SHEAR FORCE AND BENDING MOMENT 1. Maximum B.M. occurs at a section, where S.F. is either zero or changes sign. This is an important relation which helps in obtaining maximum values of B.M. 2. The rate of change of B.M. at a section is equal to the shear force at the section. 3. The rate of change of S.F. at a section is equal to the rate of loading at the section. With the help of (2) and (3), one can predict the S.F. and the loading, if rate of B.M. is given. SHEAR FORCE AND BENDING MOMENT DIAGRAMS Shear force and Bending moment can be calculated numerically at any particular section. But sometimes it is required to know the manner, in which these values vary along the length of the beam. This can be done by plotting S.F. or the B.M. as ordinate and position of the cross-section as abscissa. These diagrams are very useful, as they give a clear picture of the distribution of shear force and bending moment all along the beam. COLUMNS Structural member which is generally used to resist the compressive load is called column. If the column becomes slender compared to its length, it may fail by buckling, at load considerably less then those required to cause failure by crushing.

Strength of Material

Types of columns Column are usually subdivided into three groups based on the failure behavior as : (a) Short column : Generally fails by crushing of the material. (b) Intermediate column : Generally fail by a combination of crushing and buckling. (c) Long column : Generally fail by buckling or excessive lateral bending. AXIALLY LOADED COLUMNS Figure shows a short column subjected to axial compression P at both the ends.

1.7

Extreme stresses on the middle section are given by

P P.δ ± = o  b A Z where Z = section modulus of the section about axis of the bending O = stresses due to direct load b = stresses due to bending  = deflection of column at middle.  max =

σ o + σ b and, min = O + b

Failure of the column will occur either by max reaching the ultimate crushing strength of the material or by min reaching the ultimate tensile strength of the material. Sign convention for Bending moment. A moment which bends the column so as to cause convexity towards the initial centre line of the column will be positive bending moment, as shown in Fig. (a).

P A Now let the load P be gradually inceased till the column fails by crushing. Therefore, crushing load, P = c A. Safe load can be calculated by dividing the crushing load by a suitable factor of safety. But in case of intermediate columns and long columns, length is much more than the lateral dimensions, hence these members start bending which is called buckling of the columns. Once a column shows sign of buckling it will lead to the failure. The load at which the column just buckles is called buckling load (critical load or crippling load) and this is always less than the crushing load. Bucking load for a column depends on the length of the member and least lateral dimension. When an axially loaded column just buckles, it is said to develop an elastic instability. Figure shows a column axially loaded and just buckles due to load P. Stresses induced in the column section,  =

Maximum lateral deflection occurs at mid span.

(a)

(b)

EQUIVALENT LENGTH Equivalent length of column with different end conditions can be expressed as the length, where both ends of the column are pin jointed (or hinged). Various equivalent length are shown in the figure below

(a)

(b)

(c)

(d)

1.8

Strength of Material

Formula for the Euler’s buckling load can be modified as π EI Le 2 2

Euler’s buckling load, Pe =

where I is minimum moment of inertia and Le depends upon end conditions. LIMITATIONS OF EULER’S THEORY Assumptions made in the Euler’s theory are : 1. Column has buckled under the load P, and the load P is determined. 2. Column fails by nuckling, i.e. only long columns. Above assumptions of this thoery cannot be applied for short or intermediate columns. π EI Le 2 where I Ak 2 , k being minimum radius of gyration. 2

Now from Euler’s theory, Pe =

or

Pe =

π 2 EAk2 π 2 AE = Le2 (L e / k) 2

or

π2 E < u (L e / k) 2

E Le > π σu k Now for mild steel column section, E = 2.1  105 MPa. Ultimate strength in compression, u = 330 MPa. or

Le 2.1  105 , i.e. 79.75  80 >  300 k This shows that for a mild steel column, Euler’s theory is applicable only when the slenderness ratio is greater than 80 (for both end hinged column). 

Le where is called slenderness ratio. k 

Above equation clearly indicates that the buckling load is inversely proportional to the square of slenderness ratio, i.e. buckling load goes on increasing as length of column decreases. Now consider u as the ultimate compressive strength of the material of the column. Pe  < u A [Condition for the column to fail by buckling]

2 E Pe = (L e / k) 2 A

EXERCISE – I 1. A body having similar properties throughout its volume is called (a) Homogeneous (b) Isotropic (c) Isentropic (d) Anisotropic 2. Strain at a point is (a) Scalar (b) vector (c) Tensor (d) none of these 3. Shear strain is defined as (a) change in angle between planes at right angles (b) distortion of fibre (c) change in angle between two angles (d) strain that normally occurs 4. Limit of proportionality depends upon (a) type of loading (b) type of material (c) area of cross–section (d) none of these 5. For an isotropic, homogeneous and elastic material obeying Hooke’s law, the number of independent elastic constants is (a) 1 (b) 2 (c) 4 (d) 6 6. Limiting value of Poisson’s ratio are (a) – 1 and 0.5 (b) 1 and 0.5 (c) 0 and 0.5 (d) none of these

7. The relationship between modulus of elasticity E and bulk modulus K is 2 2 (a) E  K 1 – (b) E  2K 1 + m m

FG H

FG H

(c) E  3K 1 –

IJ K

2 m

IJ K

(d)

FG IJ H K F 2I E  3K G 1 + J H mK

8. The relation between modulus of elasiticity E, modulus of rigidity G and Poisson’s ratio 1/m is 1 1 (b) E  G 1 – (a) E  G 1 + m m 1 1 (c) E  2G 1 + (d) E  2G 1 – m m 9. If Young’s modulus of elasticity of a material is twice its modulus of rigidity, then Poisson’s ratio of the material is (a) – 1 (b) – 0.5 (c) 0.5 (d) zero 10. The ratio of shear modulus to the modulus of elasticity when poisson’s ratio is 0.25, will be (a) 2 (b) 1.4 (c) 0.4 (d) zero 11. A bar of copper and steel form a composite system, which is heated to a temperature at 45C. The stress in the copper bar will be (a) tensile (b) compressive (c) shear stress (d) none of these

FG H

FG H

IJ K

IJ K

FG H

FG H

IJ K

IJ K

Strength of Material

12. A solid cube is subjected to equal forces of same type normally on all faces. The volumetric strain will be equal to (a) twice the linear strain (b) thrice the linear stain (c) half the linear strain (d) none 13. Shear modulus of most materials with respect to modulus of elasticity is (a) more than half (b) less than half (c) equal to half (d) none of these 14. The ratio of Young modulus to the bulk modulus is about 1 1 (a)  (b)  3 3 1 (c)  (d) none of these 2 15. Proof resilience is the greatest stored energy at (a) elastic limit (b) plastic limit (c) limit of proportionality (d) none of these 16. The modulus of resilience is (a) maximum energy stored at elastic limit (b) maximum strain energy stored at elastic limit per unit volume (c) strain energy per unit area (d) none of these 17. Shear stress on principal plane is (a) maximum (b) minimum (c) zero (d) none of these 18. Maximum shear stress is equal to (a) sum of the principal stresses (b) difference of the principal stresses (c) half of the difference of the principal stresses (d) half of the sum of the principal stresses 19. Sum of the normal stresses is (a) variable (b) invariant (c) depends on planes (d) none of these 20. Principal planes are the planes on which the resultant stress is (a) shear stress (b) normal stress (c) tangential stress (d) none of these 21. If principal stresses at a point in a strained body are P1 and P2 (P1 > P2), then resultant stress on a plane carrying maximum shear stress is equal to (a)

P12 + P22

(c)

P12 + P22 2

(b)

P12 − P22

(d)

P12 − P22 2

1.9

22. Extermities of any diameter on Mohr’s circle represent (a) principal stresses (b) normal stresses on planes at 45 (c) shear stresses on planes at 45 (d) none of these 23. If x and y are normal stresses and  is tan gential stress, then maximum shear stress will be 

(a)

 x  y   2  2 

(c)

 x  y   4 2  2 

2

(b)

 x   y   2  2 

(d)

 x  y   4 2  2 

2

2

24. If x and y are normal stresses and  is the tangential stress, then maximum principal stress will be (a)

(b)

(c)

(d)

x  y 2

x  y 2

x  y 2

x  y 2

2

 x  y     2 2   2

 x  y  2     2   2

 x  y     2  2   2

 x  y     2 2  

25. If x and y and normal stresses and  is the tangential stresses, then minimum principal stress will be (a)

(b)

(c)

x  y 2

x  y 2

x  y 2

(d)

x  y 2

2

 x  y     2  2   2

 x  y     2 2   2

 x  y     2  2   2

 x  y     2 2  

26. An element subjected to normal stress 1 and orthogonal normal stresses and all tangential stresses are zero. Then radius of Mohr’s stress circle will be

σ1 2 (c) 0

(a)

(b) 1 (d) none of these

1.10

Strength of Material

27. Reultant of forces parallel to the section of the beam carrying transverse loads on the left or on the right side of the section is called (a) Bending moment (b) Twisting moment (c) Shear force d) Torsional forces 28. Resultant moment of the forces on the left or on the right side of a section is called (a) Bending moment (b) Shearing moment (c) Twisting moment (d) Torsional moment 29. Maximum bending moment occurs at a point where (a) S.F is either zero or changes sign (b) S.F is maximum (c) transverse loading is zero (d) at the centre of the beam span 30. A point of contraflexure in a beam occurs at a point where (a) B.M changes sign (b) S.F changes sign (c) Loading becomes zero (d) B.M and S.F becomes zero 31. For a beam carrying transverse point loads and distributed loads, the rate of change of S.F at a particuler section is equal to (a) bending moment (b) rate of loading (c) twisting moment (d) none of these 32. Shear force at the free end of a cantilever beam is (a) maximum (b) zero

33.

34.

35.

36.

wL2 wL2 (c) (d) 8 4 If S.F.D is parabolic curve between two points, it indicates that between the points, there is (a) uniformly distributed load (b) uniformly varying load (c) point load (d) none of these For a beam carrying transverse point loads and distributed loads, rate of change of bending moment at a particular section is equal to (a) loading (b) shear force (c) slope (d) none of these If for a certain portion of a beam bending moment is constant the shear force in that section is (a) Minimum (b) Maximum (c) Zero (d) none of these Along the neutral axis of a simply supported beam, fibers (a) do not undergo strain (b) undergo minimum strain (c) undergo maximum strain (d) none of these

37. The slope of curve of shear force diagram at any section will be equal to (a) slope of loading (b) area of loading diagram (c) ordinate of loading diagram (d) bending moment 38. Which of the following sections is most efficient in carrying bending moments (a) Rectangular (b) Circular (c) I–section (d) T–section 39. Short compression members generally fail by (a) crushing (b) bending (c) buckling (d) both (a) and (c) 40. A compression member whose unsupported length is more than 10 times its least lateral dimension is called (a) beam member (b) Grillage member (c) columns (d) short strut 41. Critical loads on log columns by Euler’s formula is given by

42.

43.

44.

45.

46.

47.

2 Pe = n

EIπ 2 . L2

The minimum value of n for a minimum physical value of Pe is equal to (a) 0 (b) 1 (c) 2 (d) 3 If a steel column has a proportional limit of 200 MPa and E = 200 GPa, then for the hinged column, limiting slenderness ratio (L/k) will be (a) 10 (b) 50 (c) 100 (d) 1000 Euler’s formula for long column gives (a) critical load (b) working load (c) failure load (d) ultimate load The ratio of equivalent length of column with both its ends fixed to its own length is (a) 2.0 (b) 1.5 (c) 1.0 (d) 0.50 The ratio of equivalent length of column with one end fixed and one end free to its own length is (a) 2.0 (b) 1.44 (c) 1.0 (d) 0.50 The ratio of equivalent length of column with one fixed and one end is hinged to its own length is (a) 2.0 (b) 1.414 (c) 1.0 (d) 0.707 The ratio of equivalent length of column with both its ends hinged to its own length is (a) 2.0 (b) 1.414 (c) 1.0 (d) 0.707

Strength of Material 1.11

EXERCISE – II 1. Shear strain is defined as (a) change in angle between planes at right angles (b) distortion of fibre (c) change in angle between two angles (d) strain that normally occurs RRB 2012 2. If Young’s modulus of elasticity of a material is twice its modulus of rigidity, then Poisson’s ratio of the material is (a) – 1

(b) – 0.5

(c) 0.5

(d) zero

(RRB 2012)

3. Shear stress on principal plane is (RRB 2012) (a) maximum (c) zero

(b) minimum (d) none of these

4. Principal planes are the planes on which the resultant stress is RRB 2012 (a) shear stress (b) normal stress (c) tangential stress

(d) none of these

5. Resultant of forces parallel to the section of the beam carrying transverse loads on the left or on the right side of the section is called (a) bending moment (b) twisting moment (c) shear force (d) torsional forces RRB 2012 6. Maximum bending moment occurs at a point where (a) S.F is either zero or changes sign (b) S.F is maximum (c) transverse loading is zero (d) at the centre of the beam span RRB 2012 7. Euler’s formula for long column gives (a) critical load (b) working load (c) failure load (d) ultimate load RRB 2012 8. The ratio of equivalent length of column with both its ends hinged to its own length is (a) 2.0 (b) 1.414 (c) 1.0 (d) 0.707 RRB 2012 9. Modulus of rupture is a measure of (a) Direct tensile strength (b) Split tensile strength (c) Flexural tensile strength (d) Direct compressive strength

RRB 2012

10. The ratio of Young modulus to the bulk modulus is about RRB 2012 1 1 (a)  (b)  3 3 1 (c)  (d) none of these 2 11. Shear strength of a soil in the plastic limit state is (a) zero RRB 2012 (b) reasonable (c) small (d) close to saturated soil strength 12. Pure shear force on a beam causes (a) (b) (c) (d)

shear stress only RRB the nature of the stress depends on the plane shear as well as normal stress shear as well as normal stress on many planes

13. The relation between the deflection (v) and bending moment (M) is RRB dv 1 d2M (b) v = – (a) M = EI dx EI dx2 d2 v (c) v =  M dx (d) M = –EI dx 2 14. The maximum bending moment caused by three equal loads (each = W/3) symmetrically spaced apart at equal intervals of L/3 on a simply supported beam is RRB WL 5WL (a) (b) 6 36 WL 3WL (c) (d) 8 20 where, L = span and W = total of the loads. 15. The bending stress at a point on the cross section caused by a moment M is equal to RRB (a)

M A

(b)

My Z

My ME (d) I I where, M = bending moment, A = area, I = moment of inertia, Y= distance of the point from the neutral axis, Z = modulus of the section, and E = Young’s modulus 16. The maximum deflection of a beam occurs at (a) zero bending moment location (b) zero shear force location (c) zero slope location (d) none of these RRB (c)