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**1. What is the moment of inertia of a circular section?**

a) πD^{4}/64

b) πD^{3}/32

c) πD^{3}/64

d) πD^{4}/32

Answer: a

Explanation: The moment of inertia of a circular section is πD^{4}/64.

**2. What is the moment of inertia of a rectangular section about an horizontal axis through C.G?**

a) bd^{3}/6

b) bd^{2}/12

c) b^{2}d^{2}/12

d) bd^{3}/12

Answer: d

Explanation: The moment of inertia of a rectangular section about an horizontal axis through C.G is bd^{3}/12.

**3. What is the moment of inertia of a rectangular section about an horizontal axis passing through base?**

a) bd^{3}/12

b) bd^{3}/6

c) bd^{3}/3

d) bd^{2}/3

Answer: c

Explanation: The moment of inertia of a rectangular section about an horizontal axis passing through base is bd^{3}/3.

**4. What is the moment of inertia of a triangular section about the base?**

a) bh^{2}/12

b) bh^{3}/12

c) bh^{3}/6

d) bh^{2}/6

Answer: b

Explanation: The moment of inertia of a triangular section about the base is bh^{3}/12.

**5. What is the moment of inertia of a triangular section about an axis passing through C.G. and parallel to the base?**

a) bh^{3}/12

b) bh^{3}/24

c) bh^{3}/36

d) bh^{3}/6

Answer: c

Explanation: The moment of inertia of a triangular section about an axis passing through C.G. and parallel to the base is bh^{3}/36.

**6. What is the product of the mass and the square of the distance of the center of gravity of the mass from an axis?**

a) Moment of inertia

b) Mass moment of inertia

c) Center of gravity

d) Product of inertia

Answer: b

Explanation: The product of the mass and the square of the distance of the center of gravity of the mass from an axis is known as the mass moment of inertia about that axis.

**7. What is the unit of mass moment of inertia?**

a) m^{4}

b) m^{6}

c) N

d) m^{2}

Answer: b

Explanation: The mass moment of inertia is the product of moment of inertia and area. So L^{4} x L^{2} = L^{6}. so its unit will be m^{6}.

**8. What is mass moment of inertia of circular plate?**

a) Md^{2}/3

b) Md^{2}/12

c) Mr^{2}/4

d) Mr^{2}/3

Answer: c

Explanation: The mass moment of inertia of circular plate is Mr^{2}/4.

**9. What is the mass MOI of a rectangular plate about x-axis passing through the C.G of the plate if the y-axis is parallel to d and perpendicular to b?**

a) Mb^{2}/12

b) Md^{2}/12

c) Md^{2}/6

d) Mb^{2}/6

Answer: b

Explanation: As the mass MOI is to be find along the x-axis, it would be Md^{2}/12.

**10. What is the mass MOI of right circular cone of radius R and height H about its axis?**

a) 4MR^{2}/10

b) MR^{2}/10

c) 3MR^{2}/10

d) MR^{2}/12

Answer: c

Explanation: The mass MOI of right circular cone of radius R and height H about its axis is 3MR^{2}/10.

**11. In the given figure a stepped column carries loads. What will be the maximum normal stress in the column at B in the larger diameter column if the ratio of P/A here is unity?**

a) 1/1.5

b) 1

c) 2/1.5

d) 2

Answer: c

Explanation: Normal stress at B = Total load acting at B / Area of a cross-section at B

= (P + P) / 1.5 A = 2P/ 1.5A = 2/1.5.

**12. The stress which acts in a direction perpendicular to the area is called ____________**

a) Shear stress

b) Normal stress

c) Thermal stress

d) None of the mentioned

Answer: b

Explanation: Normal stress acts in a direction perpendicular to the area. Normal stress is of two types tensile and compressive stress.

**13. Which of these are types of normal stresses?**

a) Tensile and compressive stresses

b) Tensile and thermal stresses

c) Shear and bending

d) Compressive and plane stresses

Answer: a

Explanation: The normal stress is divided into tensile stress and compressive stress.

**14. In a body loaded under plane stress conditions, what is the number of independent stress components?**

a) 1

b) 2

c) 3

d) 6

Answer: c

Explanation: In a body loaded under plane stress conditions, the number of independent stress components is 3 I.e. two normal components and one shear component.

**15. If a bar of large length when held vertically and subjected to a load at its lower end, its won-weight produces additional stress. The maximum stress will be ____________**

a) At the lower cross-section

b) At the built-in upper cross-section

c) At the central cross-section

d) At every point of the bar

Answer: b

Explanation: The stress is the load per unit area. After the addition of weight in the bar due to its loading on the lower end the force will increase in the upper cross-section resulting in the maximum stress at the built-in upper cross-section.

**16. The dimension of strain is?**

a) LT^{-2}

b) N/m^{2}

c) N

d) Dimensionless

Answer: d

Explanation: Strain is the ratio of change in dimension to original dimension. So it is dimensionless.

**17. What is tensile strain?**

a) The ratio of change in length to the original length

b) The ratio of original length to the change in length

c) The ratio of tensile force to the change in length

d) The ratio of change in length to the tensile force applied

Answer: a

Explanation: The tensile stress is the ratio of tensile force to the change i length. It is the stress induced in a body when subjected to two equal and opposite pulls. The ratio of change in length to the original length is the tensile strain.

**18. Find the strain of a brass rod of length 250mm which is subjected to a tensile load of 50kN when the extension of rod is equal to 0.3mm?**

a) 0.025

b) 0.0012

c) 0.0046

d) 0.0014

Answer: b

Explanation: Strain = dL/L = 0.3/250 = 0.0012.

**19. Find the elongation of an steel rod of 100mm length when it is subjected to a tensile strain of 0.005?**

a) 0.2mm

b) 0.3mm

c) 0.5mm

d) 0.1mm

Answer: c

Explanation: dL = strain x L = 0.005 x 100 = 0.5mm.

**20. A tensile test was conducted on a mild steel bar. The diameter and the gauge length of bat was 3cm and 20cm respectively. The extension was 0.21mm. What is the value to strain?**

a) 0.0010

b) 0.00105

c) 0.0105

d) 0.005

Answer: b

Explanation: Strain = dL/L = 0.21/200 = 0.0005.

**21. What will be the elastic modulus of a material if the Poisson’s ratio for that material is 0.5?**

a) Equal to its shear modulus

b) Three times its shear modulus

c) Four times its shear modulus

d) Not determinable

Answer: b

Explanation:

Explanation: Elastic modulus = E

Shear modulus = G

E = 2G ( 1 + μ )

Given, μ= 0.5, E = 2×1.5xG

E = 3G.

**22. A rigid beam ABCD is hinged at D and supported by two springs at A and B as shown in the given figure. The beam carries a vertical load P and C. the stiffness of spring at A is 2K and that of B is K.**

What will be the ratio of forces of spring at A and that of spring at B?

a) 4

b) 3

c) 2

d) 1

View Answer

Answer: b

Explanation: The rigid beam will rotate about point D, due to the load at C.

From similar triangle,

δ_{a}/2a = δ_{b}/3b

Force in spring A/Force in spring B = Pa/Pb

= 2k/k x 3/2 = 3.

Explanation: The rigid beam will rotate about point D, due to the load at C.

From similar triangle,

δ

Force in spring A/Force in spring B = Pa/Pb

= 2k/k x 3/2 = 3.

**23. A solid metal bat of uniform diameter D and length L is hung vertically from a ceiling. If the density of the material of the bar is 1 and the modulus of elasticity is E, then the total elongation of the bar due to its own weight will be ____________**

a) L/2E

b) L^{2}/2E

c) E/2L

d) E/2L^{2}

Answer: b

Explanation: The elongation of bar due to its own weight is δ= WL/2AE

Now W = ρAL

There fore δ= L^{2} / 2E.

**24. A bar of diameter 30mm is subjected to a tensile load such that the measured extension on a gauge length of 200mm is 0.09mm and the change in diameter is 0.0045mm. Calculate the Poissons ratio?**

a) 1/3

b) 1/4

c) 1/5

d) 1/6

Answer: a

Explanation: Longitudinal strain = 0.09/200

Lateral strain = – 0.0045/30

Poissons ratio = – lateral strain/ longitudinal strain

= 0.0045/30 x 200/0.09

= 1/3.

**25. What will be the ratio of Youngs modulus to the modulus of rigidity of a material having Poissons ratio 0.25?**

a) 3.75

b) 3.00

c) 1.5

d) 2.5

Answer: d

Explanation: Modulus of rigidity, G = E/2(1 + μ)

Therefore, E/G = 2x(1+0.25) = 2.5.

**1. _______ is a horizontal structural member subjected to transverse loads perpendicular to its axis.**

a) Strut

b) Column

c) Beam

d) Truss

Answer: c

Explanation: A beam is a horizontal structural member subjected to a transverse load perpendicular to its own axis. Beams are used to support weights of roof slabs, walls and staircases. The type of beam usually depends upon the span, type of load elasticity and type of structure.

**2. Example for cantilever beam is ______**

a) Portico slabs

b) Roof slab

c) Bridges

d) Railway sleepers

Answer: a

Explanation: A beam which is fixed at one end and is free at other end, it is called cantilever beam. The examples for it are portico slabs and sunshades.

**3. The diagram depicts _______ kind of beam.**

a) Cantilever

b) Continuous

c) Over hanging

d) Propped cantilever

Answer: d

Explanation: A beam which is fixed at one end and free at other end is called cantilever beam. In this case, some support other than the existing ones may be provided in order to avoid excessive deflection or to reduce the amount of bending moment, the additional support is known as a prop. The beam is known as a propped cantilever beam.

Explanation: A beam which is fixed at one end and free at other end is called cantilever beam. In this case, some support other than the existing ones may be provided in order to avoid excessive deflection or to reduce the amount of bending moment, the additional support is known as a prop. The beam is known as a propped cantilever beam.

**4. Fixed beam is also known as __________**

a) Encastered beam

b) Built on beam

c) Rigid beam

d) Tye beam

Answer: a

Explanation: A beam which is fixed at both supports is called fixed beam or encastered beam. All framed structures are examples of fixed beams.

**5. U.D.L stands for?**

a) Uniformly diluted length

b) Uniformly developed loads

c) Uniaxial distributed load

d) Uniformly distributed loads

Answer: d

Explanation: These loads are uniformly spread over a portion or whole area. They are generally represented as rate of load that is Kilo Newton per meter length (KN/m).

**6. Given below diagram is ______ load.**

a) Uniformly distributed load

b) Uniformly varying load

c) Uniformly decess load

d) Point load

Answer: b

Explanation: A load which varies uniformly on each unit length is known as uniformly varying load. Sometimes the load is zero at one end and increases uniformly to the other forms of uniformly varying loads.

**7. Moving train is an example of ____ load.**

a) Point load

b) Cantered load

c) Rolling load

d) Uniformly varying load

Answer: c

Explanation: As train’s wheels (rolling stock) move in rolling way. The upcoming load will be considered as rolling load.

**8. Continuous beams are _________**

a) Statically determinate beams

b) Statically indeterminate beams

c) Statically gravity beams

d) Framed beams

Answer: b

Explanation: Fixed beams and continuous beams are statically indeterminate beams which cannot be analyzed only by using static equations.

**9. A beam which extends beyond it supports can be termed as __________**

a) Over hang beam

b) Over span beam

c) Isolated beams

d) Tee beams

Answer: a

Explanation: A Beam extended beyond its support. And the position of extension is called as over hung portion.

**10. Units of U.D.L?**

a) KN/m

b) KN-m

c) KN-m×m

d) KN

Answer: a

Explanation: As these loads distribute over span the units for this kind of loads will be load per meter length i.e KN/m. It is denoted by “w”.

**11. Shear force is unbalanced _____ to the left or right of the section.**

a) Horizontal force

b) Vertical force

c) Inclined force

d) Conditional force

Answer: b

Explanation: The shear force at the cross section of a beam may also be defined as the unbalanced vertical force to the left or right of the section. It is also the algebraic sum of all the forces I get to the left to the right of the section.

**12. SI units of shear force is _______________**

a) kN/m

b) kN-m

c) kN

d) m/N

Answer: c

Explanation: As shear force at any section is equal to the algebraic sum of the forces, the units of the shear force are also in kilo newtons and it is denoted by kN.

**13. Shear force is diagram is _______ representation of shear force plotted as ordinate.**

a) Scalar

b) Aerial

c) Graphical

d) Statically

Answer: c

Explanation: Shear Force diagram is a graphical representation of the shear force plotted as ordinate on baseline representing the axis of the Beam.

**14. Hogging is________**

a) Negative bending moment

b) Positive shear force

c) Positive bending moment

d) Negative shear force

Answer: a

Explanation: The bending moment at a section is considered to be negative when it causes convexity upwards or concavity at bottom, such bending moment is called hogging bending moment or negative bending moment.

**15. At the point of contraflexure, the value of bending moment is ____________**

a) Zero

b) Maximum

c) Can’t be determined

d) Minimum

Answer: a

Explanation: A point at which bending moment changes its sign from positive to negative and vice versa. Such point is termed as point of contraflexure. At this point, the value of bending moment is zero (0).

**16. _________ positive/negative bending moments occur where shear force changes its sign.**

a) Minimum

b) Zero

c) Maximum

d) Remains same

Answer: c

Explanation: If shear force and bending moment values obtained are thus plotted as a diagram, the SF & BM relationship always behaves vice versa.

**17. Which of these is the correct way of sign convention for shear force?**

a) R U P

b) L U P

c) R U N

d) L D P

Answer: c

Explanation: According to the theoretical approach, there are many sign conventions to follow but the standard one is “right upwards negative” the sign convention is thoroughly followed unanimously.

**18. At hinge, the moments will be _________**

a) Maximum

b) Minimum

c) Uniform

d) Zero

Answer: d

Explanation: At the support of a member, there is no distance prevailing to take the upcoming load. As we know the moment is a product of force and perpendicular distance, but at hinge (end support) the distance is zero. Hence the moment developed is zero.

**19. What is variation in SFD, if the type of loading in the simply supported beam is U.D.L is ____**

a) Rectangle

b) Linear

c) Trapezoidal

d) Parabolic

Answer: b

Explanation: The shear force is defined as the algebraic sum of all the forces taken from any one of the section. If you figure out the SFD for a simply supported beam carrying U.D.L throughout its entire length, in the SFD we can observe that shear force is same at supports. In the centre, the shear force is zero. Hence the diagram varies linearly.

**20. The rate of change of shear force is equal to _____**

a) Direction of load

b) Change in BMD

c) Intensity of loading

d) Maximum bending

Answer: c

Explanation: Consider a simply supported beam subjected to udl for the entire span considered a free body diagram of small portion of elemental length dx.

Let the shear force at left of the section is = F

Let the increase in shear force in length of the dx = dF

Let the Indian city of load on this part of the beam = w

Total downward load in this elemental length = wdx

€V = 0

dF = -wdx

dF/dx = -w

This rate of change of shear force at any section is equal to the intensity of loading at that section.

**21. The shear force in a beam subjected to pure positive bending is _____**

a) Positive

b) Negative

c) Zero

d) Cannot determine

Answer: c

Explanation: In the determination of shear force and bending moment diagrams it is clear that shear force changes its sign when the bending moment in a beam is maximum and the shear force in a beam subjected to pure positive bending will be zero as the neutralizing effect comes under.

**22. A cantilever beam subjected to point load at its free end, the maximum bending moment develops at the ________ of the beam.**

a) Free end

b) Fixed end

c) Centre

d) Point of inflection

Answer: b

Explanation: As the moment is the product of perpendicular distance and force. In cantilever beam, at its free end the moment will be zero as there is no distance, but at the fixed end the moment is maximum that is W×l.

**23. Bending moment in a beam is maximum when the _________**

a) Shear force is minimum

b) Shear force is maximum

c) Shear force is zero

d) Shear force is constant

Answer: c

Explanation: The maximum bending moment occurs in a beam, when the shear force at that section is zero or changes the sign because at point of contra flexure the bending moment is zero.

**24. Positive bending moment is known as _______**

a) Hogging

b) Sagging

c) Ragging

d) Inflection

Answer: a

Explanation: The positive bending moment in a section is considered because it causes convexity downwards. Such bending moment is called a sagging bending moment or positive bending moment.

**25. A simply supported beam of span “x” meters carries a udl of “w” per unit length over the entire span, the maximum bending moment occurs at _____**

a) At point of contra flexure

b) Centre

c) End supports

d) Anywhere on the beam

Answer: b

Explanation: As we know that BM occurs at center. Because at supports the moment is obviously zero.

At the centre, maximum bending moment is wl^{2}/8.

**26. The maximum BM is ______**

a) 40 kNm

b) 50 kNm

c) 90 kNm

d) 75 kNm

Answer: c

Explanation: Above diagram depicts cantilever beam subjected to point load at the free end. The maximum bending moment at A is W × I

= 30 × 3

= 90 kNm.

**27. Bending moment can be denoted by ____**

a) K

b) M

c) N

d) F

Answer: b

Explanation: Bending moment is the product of force and perpendicular distance. Units are kNm

It is denoted by “M”. Whereas SF is denoted by “F”.

**28. Number of points of contra flexure for a double over hanging beam.**

a) 3

b) 2

c) 4

d) Infinite

Answer: b

Explanation: Point of contraflexure in a beam is a point at which bending moment changes its sign from positive to negative and vice versa. In the case of overhanging beam, there will be two points of contraflexure.

**29. Maximum bending moment in a cantilever beam subjected to udl (w)over the entire span (l).**

a) wl

b) wl^{3}

c) wl^{2}

d) w

Answer: c

Explanation: In a cantilever beam the maximum bending moment occurs at the fixed end. Moment at the free end is 0 and maximum at the fixed end. Maximum shear force is w×l.

**30. Determine the maximum bending moment for the below figure.**

a) wl/2

b) wl/3

c) wl/4

d) wl

Answer: c

Explanation: First of all, let’s assume the length between end supports be ”l” the maximum bending moment in a simply supported beam with point load at its centre is wl/4. We know that in simply supported beam the maximum bending moment occurs at the centre only.

**31. What is the variation in the BM, if the simply supported beam carries a point load at the centre.**

a) Triangular

b) Rectangular

c) Trapezoidal

d) Other quadrilateral

Answer: a

Explanation: For simply supported beam with point load at the centre, the maximum bending moment will be at the centre i.e. wl/4. The variation in bending moment is triangular.

**32. What is the bending moment at end supports of a simply supported beam?**

a) Maximum

b) Minimum

c) Zero

d) Uniform

Answer: c

Explanation: At the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. As the distance is zero, the moment is obviously zero.

**33. What is the maximum shear force, when a cantilever beam is loaded with udl throughout?**

a) w×l

b) w

c) w/l

d) w+l

Answer: a

Explanation: In cantilever beams, the maximum shear force occurs at the fixed end. In the free end, there is zero shear force. As we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. For maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l).

**34. Sagging, the bending moment occurs at the _____ of the beam.**

a) At supports

b) Mid span

c) Point of contraflexure

d) Point of emergence

Answer: b

Explanation: The positive bending moment is considered when it causes convexity downward or concavity at top. This is sagging. In simply supported beams, it occurs at mid span because the bending moment at the supports obviously will be zero hence the positive bending moment occurs in the mid span.

**35. What is the maximum bending moment for simply supported beam carrying a point load “W” kN at its centre?**

a) W kNm

b) W/m kNm

c) W×l kNm

d) W×l/4 kNm

Answer: d

Explanation: We know that in simply supported beams the maximum BM occurs at the central span.

Moment at A = Moment at B = 0

Moment at C = W/2 × l/2 = Wl/ 4 kNm (Sagging).

**1. In simple bending, ______ is constant.**

a) Shear force

b) Loading

c) Deformation

d) Bending moment

Answer: d

Explanation: If a beam is undergone with simple bending, the beam deforms under the action of bending moment. If this bending moment is constant and does not affect by any shear force, then the beam is in state of simple bending.

**2. If a beam is subjected to pure bending, then the deformation of the beam is_____**

a) Arc of circle

b) Triangular

c) Trapezoidal

d) Rectangular

Answer: a

Explanation: The beam being subjected to pure bending, there will be only bending moment and no shear force it results in the formation of an arc of circle with some radius known as radius of curvature.

**3. When a beam is subjected to simple bending, ____________ is the same in both tension and compression for the material.**

a) Modulus of rigidity

b) Modulus of elasticity

c) Poisson’s ratio

d) Modulus of section

Answer: b

Explanation: It is one of the most important assumptions made in the theory of simple bending that is the modulus of elasticity that is Young’s modulus [E] is same in both tension and compression for the material and the stress in a beam do not exceed the elastic limit.

**4. E/R = M/I = f/y is a bending equation.**

a) True

b) False

Answer: a

Explanation: The above-mentioned equation is absolutely correct.

E/R = M/I = f/y is a bending equation. It is also known as flexure equation (or) equation for theory of simple bending.

Where,

E stands for Young’s modulus or modulus of elasticity.

R stands for radius of curvature.

M stands for bending moment

I stand for moment of inertia

f stands for bending stress

y stands for neutral axis.

**5. Maximum Shearing stress in a beam is at _____**

a) Neutral axis

b) Extreme fibres

c) Mid span

d) Action of loading

Answer: a

Explanation: Shearing stress is defined as the resistance offered by the internal stress to the shear force. Shearing stress in a beam is maximum at a neutral axis.

**6. At the neutral axis, bending stress is _____**

a) Minimum

b) Maximum

c) Zero

d) Constant

Answer: c

Explanation: Neutral axis is defined as a line of intersection of neutral plane or neutral layer on a cross section at the neutral axis of that section. At the NA, bending stress or bending strain is zero. The first moment of area of a beam section about neutral axis is also zero. The layer of neutral axis neither contracts nor extends.

**7. Curvature of the beam is __________ to bending moment.**

a) Equal

b) Directly proportion

c) Inversely proportion

d) Coincides

Answer: b

Explanation: From the flexural equation, we have 1/R is called as the “curvature of the beam”.

1 / R = M / EI

Hence the curvature of the beam is directly proportional to bending moment and inversely proportional to flexural rigidity (EI).

**8. What are the units of flexural rigidity?**

a) Nm^{2}

b) Nm

c) N/m

d) m/N^{3}

Answer: a

Explanation: The product of young’s modulus (E) of the material and moment of inertia (I) of the beam section about its neutral axis is called flexural rigidity.

Units for E are N/m^{2}

Units for I are m^{4}

Their product is Nm^{2}.

**9. What are the units for section modulus?**

a) m^{2}

b) m^{4}

c) m^{3}

d) m

Answer: c

Explanation: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section or section modulus. It is generally denoted by the letter Z. Section modulus is expressed in m^{3}

Z = I/y

= m^{4}/ m

= m^{3}.

**10. What are the units of axial stiffness?**

a) m^{3}

b) m^{2}

c) N/ m

d) -m

Answer: c

Explanation: Axial rigidity is a product of young’s modulus (E) and the cross-sectional area (A) of that section. Axial rigidity per unit length is known as axial stiffness the si units of axial stiffness are Newton per metre (N/m).

**11. Calculate the modulus of section of rectangle beam of size 240 mm × 400 mm.**

a) 5.4 × 10^{6} mm^{3}

b) 6.2 × 10^{6} mm^{3}

c) 5.5 × 10^{6} mm^{3}

d) 6.4 × 10^{6} mm^{3}

Answer: d

Explanation: b = 240 mm & d = 400 mm

Moment of inertia (I) = bd^{3}/12; y = d/2

Section modulus (Z) = I/y = bd^{2}/ 6

= 1/6 × 240 × 400 ×400

= 6.4 × 10^{6} mm^{3}.

**12. What is the product of force and radius?**

a) Twisting shear

b) Turning shear

c) Turning moment

d) Tilting moment

Answer: c

Explanation: Twisting moment will be equal to the product of the perpendicular force and existing radius. Denoted by letter T and SI units are Nm.

**13. Determine section modulus for beam of 100mm diameter.**

a) 785 × 10^{3} mm^{3}

b) 456 × 10^{3} mm^{3}

c) 87 × 10^{3} mm^{3}

d) 98 × 10^{3} mm^{3}

Answer: d

Explanation: d = 300mm

For circular sections; I = π / 64 × d^{4}

y= d/2

Z = π/32 × d^{3} (d = 100 mm)

Z = 98.17 × 10^{3}mm^{3}.

**14. What is the section modulus (Z) for a rectangular section?**

a) bd^{2}/6

b) a^{3}/6

c) BD^{3}-bd^{3}

d) D^{4}-d^{4}

Answer: a

Explanation: The modulus of section may be defined as the ratio of moment of inertia to the distance to the extreme fibre. It is denoted by Z.

Z= I/y ; For rectangular section, I = bd^{3}/12 & y = d/2.

Z= bd^{2}/6.

**15. Find the modulus of section of square beam of size 300×300 mm.**

a) 4.8 × 10^{6} mm^{3}

b) 4.5 × 10^{6} mm^{3}

c) 5.6 × 10^{6} mm^{3}

d) 4.2 × 10^{6} mm^{3}

Answer: b

Explanation: Here, a = side of square section = 300 mm.

I = a^{4}/12. y= a/2.

Z = I/y = a^{3}/6

= 300^{3}/6

= 4.5 × 10^{6} mm^{3}.

**16. _________ of a beam is a measure of its resistance against deflection.**

a) Strength

b) Stiffness

c) Deflection

d) Slope

Answer: b

Explanation: A beam is said to be a strength when the maximum induced bending and shear stresses are within the safe permissible stresses stiffness of a beam is a measure of its resistance against deflection.

**17. To what radius an Aluminium strip 300 mm wide and 40mm thick can be bent, if the maximum stress in a strip is not to exceed 40 N/mm ^{2}. Take young’s modulus for Aluminium is 7×105 N/mm^{2}.**

a) 45m

b) 52m

c) 35m

d) 65m

Answer: c

Explanation: Here, b = 300mm

d= 40mm. y= 20mm.

From the relation; E/R = f/y

R= E×y/f

=70×10

= 35m.

**18. The bending stress in a beam is ______ to bending moment.**

a) Less than

b) Directly proportionate

c) More than

d) Equal

Answer: b

Explanation: As we know, the bending stress is equal to bending moment per area. Hence, as the bending (flexure) moment increases/decreases the same is noticed in the bending stress too.

**19. The Poisson’s ratio for concrete is __________**

a) 0.4

b) 0.35

c) 0.12

d) 0.2

Answer: d

Explanation: The ratio of lateral strain to the corresponding longitudinal strain is called Poisson’s ratio. The value of poisons ratio for elastic materials usually lies between 0.25 and 0.33 and in no case exceeds 0.5. The Poisson’s ratio for concrete is 0.20.

**20. The term “Tenacity” means __________**

a) Working stress

b) Ultimate stress

c) Bulk modulus

d) Shear modulus

Answer: b

Explanation: The ultimate stress of a material is the greatest load required to fracture the material divided by the area of the original cross section in the point of fracture The ultimate stress is also known as tenacity.

**21. A steel rod of 25 mm diameter and 600 mm long is subjected to an axial pull of 40000. The intensity of stress is?**

a) 34.64 N/mm^{2}

b) 46.22 N/mm^{2}

c) 76.54 N/mm^{2}

d) 81.49 N/mm^{2}

Answer: d

Explanation: Cross sectional area of steel rod [Circular]be 490.87 mm^{2}.

The intensity of stress = P/A = 40000/490.87

= 81.49 N/mm^{2}.

**22. The bending strain is zero at _______**

a) Point of contraflexure

b) Neutral axis

c) Curvature

d) Line of action of loading

Answer: b

Explanation: The neutral axis is a line of intersection of neutral plane or neutral layer on a cross section. The neutral axis of a beam passes through the centroid of the section. At the neutral axis bending stress and bending strain is zero.

**23. Strength of the beam depends only on the cross section.**

a) True

b) False

Answer: b

Explanation: The strength of two beams of the same material can be compared by the section modulus values. The strength of beam depends on the material, size and shape of cross section. The beam is stronger when section modulus is more, strength of the beam depends on Z.

**24. A beam has a triangular cross-section, having altitude ”h” and base “b”. If the section is being subjected to a shear force “F”. Calculate the shear stress at the level of neutral axis in the cross section.**

a) 4F/5bh

b) 4F/3bh

c) 8F/3bh

d) 3F/4bh

Answer: c

Explanation: For a triangular section subjected to a shear force, the shear stress at neutral axis is

= 4/3 × average shear stress

= 4/3 × F/A/2 ; A = bh

= 8F/3bh.

**25. The maximum shear stress in the rectangular section is ______________ times the average shear stress.**

a) 3/4

b) 3/7

c) 5/3

d) 3/2

Answer: d

Explanation: The maximum shear stress occurs at the neutral axis. So, y = 0.

Maximum shear stress = 3/2 × F/bd (• Average shear stress = F/bd ).

= 3/2 × average shear stress.

**26.The modular ratio for M20 grade concrete is _____________**

a) 16

b) 13

c) 11

d) 07

Answer: b

Explanation: According to Indian Standards 456-2000 The modular ratio (m) = 280/3× cbc, For M20; compressive bearing capacity in concrete = 7 N/mm^{2} & tensile strength = 330 N/mm^{2}.

Modular ratio (m) = 280/21 = 13.33.

**27. In doubly reinforced beam, the maximum shear stress occurs ______________**

a) along the centroid

b) along the neutral axis

c) on the planes between neutral axis and tensile reinforcement

d) on the planes between neutral axis and compressive reinforcement

Answer: d

Explanation: In continuous beam the moments developed at the supports are greater than a moment’s developed at the mid span, show the maximum bending moment occurs at the supports.

For continuous beams, the maximum shear stress occurs at the planes intersecting the compressive reinforcement and the neutral axis.

**28. A cylindrical section having no joint is known as _______________**

a) Proof section

b) Seamless section

c) Target section

d) Mown section

Answer: b

Explanation: A cylindrical section having no joint is known as seamless section. Built up section is not that strong as a seamless section of the same thickness.

**29. The efficiency of cylindrical section is the ratio of the strength of joint to the strength of _______________**

a) Solid plate

b) Boilerplate

c) Circumferential plate

d) Longitudinal plate

Answer: a

Explanation: The strength of plate or strength of rivet whichever is less is called the strength of joint. The ratio of the strength of joint to the strength of steel plate is called the efficiency of the cylinder.

**30. Calculate the modulus of section for a hollow circular column of external diameter 60 mm and 10 mm thickness.**

a) 170 m

b) 190 m

c) 250 m

d) 300 m

Answer: a

Explanation: Given data :

D = 60 mm ; t = 10 mm & d = 60 – 2×10 = 40 mm

For hollow circular section, modulus of section( Z ) = 3.14 × D^{4} – d^{4} / 32 D.

= 17016.3 mm = 170 m.

**31. Determine the modulus of a section for an I section, given the distance from neutral axis is 50 mm and moment of inertia is 2.8×10 ^{6} mm^{4}.**

a) 59m

b) 51m

c) 58m

d) 63m

Answer: c

Explanation: The modulus of section is the ratio of the moment of inertia to the distance of the neutral axis.

Given y = 50 mm

I = 2.8×10

& Z = I/y = 2.8×10

= 57.76 ×10

= 57.7 ~ 58 m.

**32. A circular Beam of 0.25 m diameter is subjected to you shear force of 10 kN. Calculate the value of maximum shear stress. [Take area = 176 m ^{2}].**

a) 0.75 N/mm

b) 0.58 N/mm

c) 0.73 N/mm

d) 0.65 N/mm

Answer: a

Explanation: Given diameter = 0.25 m

Area (A) = 176 m

Shear Force (F) = 10 kN ~ 10000 N.

For circular cross section the maximum shear stress is equal to 4/3 times of average shear stress

Maximum shear force = 4/3 × F/A

= 4/3 × 10000/176

= 0.75 N/mm

**33. The maximum shear stress distribution is _____________ percentage more than average shear stress in circular section.**

a) 54 %

b) 60 %

c) 33 %

d) 50 %

Answer: c

Explanation: Maximum shear stress occurs at neutral axis; y =0

Maximum shear stress = 16/3 × average shear stress

But 4F / A is the average shear stress.

So, the maximum shear stress = 4/3 times the average shear stress.

Hence the maximum shear stress is 33% more than the average shear stress in a circular section.

**34. Shear stress at top most fibre of rectangular section is _____________**

a) Maximum

b) Minimum

c) Zero

d) Uniform through out

Answer: c

Explanation: In rectangular section,

The shear stress at a distance “y “ from NA = 6F/bd^{3} × u (u = d^{2}/4-y)

The maximum shear stress occurs at a neutral axis, in the above equation when y is equal to zero. q is max. Hence the shear stress topmost fibre of rectangular section is zero.

**35. 1 GPA = ____________ pa.**

a) 10^{5}

b) 10^{6}

c) 10^{8}

d) 10^{9}

Answer: d

Explanation: 1 Giga Pascal is equal to 10^{9}N/m^{2}(Pascal)

In the same way 1 kilo Pascal equal to 10^{3} pascals

1 mega Pascal is equal to 106 pascals.

**36. The maximum shear stress in an I section is __________**

a) F/8I ×[B/b (D^{2}-d^{2})+d^{2}]

b) F/6I ×[B/b (D^{2}-d^{2})+d^{2}]

c) F/8I ×[B/b (D^{3}-d^{3})+d^{2}]

d) F/4I ×[B/b (D^{2}-d^{2})+d^{2}]

Answer: a

Explanation: Shear stress at top flange of the I section is zero.

Shear stress at the junction of web and flange= B/b ×F/8I (D^{2}-d^{2}).

Shear stress at bottom of the flange = F/8I (D^{2}-d^{2}).

And shear force is maximum at neutral axis i.e F/8I ×[B/b (D^{2}-d^{2})+d^{2}].

**37. In steel sections, the junction between a flange and web is known as ________**

a) Edge

b) Fillet

c) Corner

d) Lug

Answer: b

Explanation: In a steel section, the junction between the flange and the web is known as fillet. The connections solve issues of complex geometry for joining the members of a central hub while they provide the standard connection through out. They are not readily available.

**38. The percentage of carbon in structural steel is __________**

a) 0.2 – 0.27 %

b) 0.6 – 0.85 %

c) 0.7 – 1.23 %

d) 1.23 – 1.45%

Answer: a

Explanation: The percentage of carbon in structural steel is 0. 2 to 0.27. Percentage of the carbon in steel increases the ductility of the Steel decreases.

**1. Torsional sectional modulus is also known as _________**

a) Polar modulus

b) Sectional modulus

c) Torsion modulus

d) Torsional rigidity

Answer: a

Explanation: The ratio of polar moment of inertia to radius of section is called Polar modulus or Torsional section modulus. Its units are mm^{3} or m^{3} (in SI).

**2. ________ is a measure of the strength of shaft in rotation.**

a) Torsional modulus

b) Sectional modulus

c) Polar modulus

d) Torsional rigidity

Answer: c

Explanation: The polar modulus is a measure of the strength of shaft in rotation. As the value of Polar modulus increases torsional strength increases.

**3. What are the units of torsional rigidity?**

a) Nmm^{2}

b) N/mm

c) N-mm

d) N

Answer: a

Explanation: The product of modulus of rigidity (C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity is a torque that produces a twist of one radian in a shaft of unit length.

**4. The angle of twist can be written as ________**

a) TL/J

b) CJ/TL

c) TL/CJ

d) T/J

Answer: c

Explanation: The angle of Twist = TL/CJ

Where T = Torque in Nm

L = Length of shaft

CJ = Torsional rigidity.

**5. The power transmitted by shaft SI system is given by __________**

a) 2πNT/60

b) 3πNT/60

c) 2πNT/45

d) NT/60 W

Answer: a

Explanation: In SI system, Power (P) is measured in watts (W) ; P = 2πNT/60

Where T = Average Torque in N.m

N = rpm

= 2πNT/ 45 1 watt = 1 Joule/sec = 1N.m/s.

**6. Area of catchment is measured in ___________**

a) mm^{3}

b) Km^{2}

c) Km

d) mm

Answer: b

Explanation: Catchment area can be defined as the area which contributes the surplus water present over it to the stream or river. It is an area which is responsible for maintaining flow in natural water bodies. It is expressed in square kilometres.

**7. ______ catchment area is a sum of free catchment area and intercepted catchment area.**

a) Total

b) Additional

c) Combined

d) Overall

Answer: c

Explanation: Combined catchment area is defined as the total catchment area which contributes the water in to stream or a tank. Combined Catchment area = Free catchment area + intercepted catchment area.

**8. ___________ has steep slopes and gives more run off.**

a) Intercepted Catchment Area

b) Good Catchment Area

c) Combined Catchment Area

d) Average Catchment Area

Answer: b

Explanation: Good catchment area consists of hills or rocky lands with steep slopes and little vegetation. It gives more run off.

**9. How many number of rain gauge stations should be installed an area between 250 to 500 km ^{2}.**

a) 2

b) 4

c) 3

d) 5

Answer: c

Explanation: 3 number of rain gauge stations should be installed an area between 250 to 500 km

Area of Basin(Km^{2}) |
Number of Rain gauge stations |

< 125 | 1 |

125 – 250 | 2 |

250 – 500 | 3 |

**10. Trend of rainfall can be studied from _______**

a) Rainfall graphs

b) Rainfall records

c) Rainfall curves

d) Rainfall cumulatives

Answer: b

Explanation: Rainfall records are useful for calculating run off over a basin. By using rainfall records estimate of design parameters of irrigation structures can be made. The maximum flow due to any storm can be calculated and predicted.

**11. Estimation of run off “R” is 0.85P-30.48.**

The above formula was coined by _____

a) Lacey

b) Darcy

c) Khosla

d) Ingli

Answer: d

Explanation: Run off can be estimated by

R= 0.85P-30.48

Where R = annual runoff in mm

P = annual rainfall in mm.

**12. Monsoon duration factor is denoted by ________**

a) P

b) S

c) F

d) T

Answer: c

Explanation: Monsoon duration factor is denoted by F.

Class of Monsoon | Monsoon Duration Factor (F) |

Very Short | 0.5 |

Standard length | 1.0 |

Very long | 1.5 |

**13. Runoff coefficient is denoted by _______**

a) P

b) N

c) K

d) H

Answer: c

Explanation: The runoff coefficient can be defined as the ratio of runoff to rainfall. Rainfall and runoff can be interrelated by runoff coefficient.

R = KP

K = R/P [K = is a runoff Coefficient depending on the surface of the catchment area].

**14. _________ is a graph showing variations of discharge with time.**

a) Rising limb graph

b) Crest graph

c) Hydraulic graph

d) Gauge graph

Answer: c

Explanation: Hydrograph is a graph showing variations of discharge with time at a particular point of the stream. The hydrograph shows the time distribution of total run off at a point of measurement. Maximum flood discharge can also be calculated by using hydrograph.

**15. Calculate the torque which a shaft of 300 mm diameter can safely transmit, if the shear stress is 48 N / mm ^{2}.**

a) 356 kNm

b) 254 kNm

c) 332 kNm

d) 564 kNm

Answer: b

Explanation: Given, the diameter of shaft D = 300 mm

Maximum shear stress fs = 48 N/mm

Torque = T = π/16 fs D

= 254469004.9 Nmm

= 254 kNm.

**16. The intensity of shear stress at a section is ______ to the distance of the section from the axis of the shaft.**

a) Inversely proportional

b) Directly proportional

c) Equal

d) Parallel

Answer: b

Explanation: The intensity of shear stress at a section is directly proportional to the distance of the section from axis of the shaft. The shear stress at a distance from the centre of the shaft is given by fs/R × r.

**17. The shear stress is ____________ at the axis of the shaft.**

a) Minimum

b) Maximum

c) Zero

d) Uniform

Answer: c

Explanation: The shear stress is zero at the axis of the shaft and the shear stress is linearly increasing to the maximum value at the surface of the shaft.

**18. The shear stress at the outer surface of hollow circular section is _________**

a) Zero

b) Maximum

c) Minimum

d) Can’t determined

Answer: b

Explanation: The shear stress in a hollow circular section varies from maximum at the outer surface to a minimum (but not zero) in the inner face. The minimum value should be greater than zero.

**19. The hollow shaft will transmit greater _______ then the solid shaft of the same weight.**

a) Bending moment

b) Shear stress

c) Torque

d) Sectional Modulus

Answer: c

Explanation: For the same maximum shear stress, the average shear stress in a hollow shaft is greater than that in a solid shaft of the same area. Hence the hollow shaft will transmit greater torque than the solid shaft of the same weight.

**20. The process of measurement of discharge and water level of a river is called _________**

a) Meandering

b) River coursing

c) River gauging

d) Scouring

Answer: c

Explanation: The process of measurement of discharge and water level of a river is known as river gauge. It helps in determining the characteristics of flow different times during the year.

**21. The quantity of losses in the river can be measured with an aid of ________**

a) Runoff coefficient

b) Hydrograph

c) River Coursing

d) River gauging

Answer: d

Explanation: By measuring river discharge for number of years, it is possible to know the available and dependable supply. The river gauging helps in measuring discharge in the river and the quantity of losses can also be known.

**22. The site for the river gauging station should not be liable to ____________**

a) Silting

b) Coursing

c) Meandering

d) Runoff

Answer: a

Explanation: River gauging station site should be selected in such a way that the site should be stable and there should not be any choice of scouring and silting. At the gauge site, the river section should be at right angles to the flow of the river.

**23. Stage discharge relationship method is also known as ________ method.**

a) Velocity Volume

b) Velocity Area

c) Distance Area

d) Displacement Momentum

Answer: b

Explanation: Stage discharge relationship method is a direct method of computing a discharge in a stream by measuring velocity and area of flow. The place where such measurements are taken is known as velocity area station and the method is known as the velocity area method.

**24. Velocity in a river flow can be calculated by using _________**

a) By current meter

b) By emperical formulae

c) By infiltration method

d) By hydrograph

Answer: a

Explanation: The velocity flow at any point in an open channel or in a river can be most accurately and conveniently determined by a mechanical device called current metre in this device the velocity of flow can be read from rating table.

**25. Which of the following method is not used in measuring the velocity of a stream?**

a) By floats

b) By rod float

c) By hydrograph

d) By colour

Answer: c

Explanation: Hydrograph is a method of estimation of runoff. While the rest of the methods used in measuring the velocity of a stream/ river or canal. Hydrograph is a graph which shows the variations of discharge with respect to time.

**26. The maximum flood discharge is also known as ___________**

a) Peak flow

b) Maximum flow

c) Peak discharge

d) Peak flood

Answer: a

Explanation: The maximum rate of discharge during a period of runoff, which is caused by a storm, is called a peak flow maximum flood discharge. Estimation of maximum flood discharge is a first step in planning for flood regulation.

**27. Which of the following method is used to estimate maximum flood discharge?**

a) By travelling screen

b) By current meter

c) By physical indication of past floods

d) By salt velocity

Answer: c

Explanation: The results obtained by the physical indication of past floods methods are somewhat reliable. By oral enquiry in the villages situated on the banks of the river, the maximum water level attained in the past 35 years can be obtained. But this method is out-dated.

**28. ________formula is used only in southern India for calculating maximum flood discharge.**

a) Dickens

b) Ryve’s

c) Lacey’s

d) Francis

Answer: b

Explanation: Ryve’s (1884) formula is used only in Southern India.

Q = C(A)^{2/3}.

The coefficient “C” depends on the maximum intensity of rainfall and other factors such as shape slopes exedra of the catchment.

**29. A catchment area of 30.5 km ^{2} is situated in Central India calculate the maximum discharge coming from the catchment area.**

a) 253.08 cumecs

b) 341.06 cumecs

c) 457.88 cumecs

d) 485.66 cumecs

Answer: a

Explanation: As the catchment area is situated in central India. Dicken’s formula is suitable and a maximum value of Dickens Coefficient is taken as 19.5

Q = CA

Q = 19.5 × (30.5)

Q = 253.08 cumecs.

**30. If the catchment area is situated in north India, then what is the flood coefficient?**

a) 10.45

b) 11.37

c) 12.6

d) 19.4

Answer: b

Explanation: Dicken’s formula (1865)

Region | Value of C |

North India | 11.37 |

Central India | 11.77 – 19.28 |

Western India | 22.04 |

**31. A circular shaft of diameter 30 mm is tested under torsion the gauge length of test specimen is 300 mm. A torque of 2kNm produces an angle twist of 1°. Calculate CJ.**

a) 0.432 × 10^{6} N/mm^{2}

b) 0.324 × 10^{6} N/mm^{2}

c) 0.46 × 10^{6} N/mm^{2}

d) 0.532 × 10^{6} N/mm^{2}

Answer: a

Explanation: Angle of twist = 1° = π/180 radians.

Polar moment of inertia = π/32 × 30^{4}mm^{4}.

To find CJ: T/J =C× twist angle

C = Tl/J×twist angle = 2×10^{6} ×300 / π/32 × 30^{4}mm^{4}× π/180.

C = 0.4323 × 10^{6} N/mm^{2}.

**32. __________ has perfect control on river flow.**

a) Barrage

b) Weir

c) Marginal bunds

d) Guide banks

Answer: a

Explanation: Barrages are much more costly than weirs. Gates are raised off the high flood to pass floods. They have perfect control of the river flow.

**33. When the gross length is more than 6 metres between the face of abutment it is called as ________**

a) Cause way

b) Bridge

c) Culvert

d) Cassion

Answer: b

Explanation: When the gross length is more than 6 m between the faces of abutment measured at right angles is called a bridge. If a bridge supports a road way over a railway then it is called Road over a bridge.

**34. The minimum straight approach provided on either side of bridge is ___________**

a) 12 m

b) 15 m

c) 20 m

d) 22 m

Answer: b

Explanation: The bridge site should be far away from the confluence of tributaries as far as possible the straight approaches are to be provided on either side of the bridge for at least 15m.

**35. ________ should be taken below the deepest scour level.**

a) Foundation

b) Sub structure

c) Structure

d) Parapet

Answer: a

Explanation: Foundations to be provided for approaches abutments, piers etc., by considering the water in the river, sub soil conditions etc., a foundation should be taken below the deepest scour level.

**36. ___________ formula is used for calculating the depth of the foundation.**

a) Gordon’s

b) Rankine’s

c) WH Smith’s

d) Falcon

Answer: b

Explanation: Rankine’s formula is used for calculating the depth of the foundation.

h=P/w × (1-sin/1+sin). Minimum depth is restricted to 90 cms.

**37. What is the strain energy stored in a body due to gradually applied load?**

a) σE/V

b) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

Answer: d

Explanation: Strain energy when load is applied gradually = σ^{2}V/2E.

**38. Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is __________**

a) s^{2}V/2E

b) sV/E

c) sV^{2}/E

d) sV/2E

Answer: a

Explanation: Strain energy = s^{2}V/2E.

**39. In a material of pure shear stress τthe strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and v will be:**

a) τ^{2}/E x (1+ v)

b) τ^{2}/E x (1+ v)

c) τ^{2}/2E x (1+ v)

d) τ^{2}/E x (2+ v)

Answer: a

Explanation: σ_{1}=τ, σ_{2}= -τσ_{3}=0

U = (τ^{2}+- τ^{2}-2μτ(-τ))V = τ^{2}/E x (1+ v)V.

**40. PL ^{3}/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?**

a) P

b) P

c) P

d) P

Answer: b

Explanation: We may do it taking average

Strain energy = Average force x displacement = (P/2) x PL

**41. A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm ^{2}. If the modulus of rigidity of the material is 1×10^{6} kg/cm^{2}, the strain energy will be __________**

a) 125 kg-cm

b) 1000 kg-cm

c) 500 kg-cm

d) 100 kg-cm

Answer: a

Explanation: Strain energy stored = τ

**42. The ability of a material to absorb energy when elastically deformed and to return it when unloaded is called __________**

a) Elasticity

b) Resilience

c) Plasticity

d) Strain resistance

Answer: b

Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape.

**43. The strain energy stored in a specimen when stained within the elastic limit is known as __________**

a) Resilience

b) Plasticity

c) Malleability

d) Stain energy

Answer: a

Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

**44. The maximum strain energy stored at elastic limit is __________**

a) Resilience

b) Proof resilience

c) Elasticity

d) Malleability

Answer: b

Explanation: Proof resilience is the maximum stored energy at the elastic limit. Resilience is the ability of material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

**45. The ability of a material to absorb energy when elastically deformed and to return it when unloaded is called __________**

a) Elasticity

b) Resilience

c) Plasticity

d) Strain resistance

Answer: b

Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape.

**46. The strain energy stored in a specimen when stained within the elastic limit is known as __________**

a) Resilience

b) Plasticity

c) Malleability

d) Stain energy

Answer: a

Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

**47. The maximum strain energy stored at elastic limit is __________**

a) Resilience

b) Proof resilience

c) Elasticity

d) Malleability

Answer: b

Explanation: Proof resilience is the maximum stored energy at the elastic limit. Resilience is the ability of material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

**48. The mathematical expression for resilience ‘U’ is __________**

a) U = σ^{2}/E x volume

b) U = σ^{2}/3E x volume

c) U = σ^{2}/2E x volume

d) U = σ/2E x volume

Answer: c

Explanation: The resilience is the strain energy stored in a specimen so it will be

U = σ^{2}/2E x volume.

**49. What is the modulus of resilience?**

a) The ratio of resilience to volume

b) The ratio of proof resilience to the modulus of elasticity

c) The ratio of proof resilience to the strain energy

d) The ratio of proof resilience to volume

Answer: d

Explanation: The modulus of resilience is the proof resilience per unit volume. It is denoted by σ.

**50. What is the strain energy stored in a body when the load is applied gradually?**

a) σE/V

b) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

Answer: d

Explanation: Strain energy in gradual loading = σ^{2}V/2E.

**51. What is strain energy?**

a) The work done by the applied load In stretching the body

b) The strain per unit volume

c) The force applied in stretching the body

d) The stress per unit are

Answer: a

Explanation: The strain energy stored in a body is equal to the work done by the applied load in stretching the body.

**52. What is the relation between maximum stress induced due to gradual load to maximum stress the sudden load?**

a) Maximum stress in gradual load is equal to the maximum stress in sudden load

b) Maximum stress in gradual load is half to the maximum stress in sudden load

c) Maximum stress in gradual load is twice to the maximum stress in sudden load

d) Maximum stress in gradual load is four times to the maximum stress in sudden load

Answer: b

Explanation: Maximum stress in gradual loading = P/A

Maximum stress in sudden loading = 2P/A.

**53. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5cm long. What will be the stress in the rod if E=1×10 ^{5} N/mm^{2}?**

a) 47.746 N/mm

b) 34.15 N/mm

c) 48.456 N/mm

d) 71.02 N/mm

Answer: a

Explanation: Stress = Load/ area = 60,000 / (π/4 D

**54. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the stress in the rod if E=1×10 ^{5} N/mm^{2}?**

a) 1.19mm

b) 2.14mm

c) 3.45mm

d) 4.77mm

Answer: d

Explanation: Stress = Load/ area = 60,000 / (π/4 D

So stretch = stress x length / E = 4.77mm.

**55. A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm. What will be the stress in bar?**

a) 100 N/mm^{2}

b) 120 N/mm^{2}

c) 125 N/mm^{2}

d) 150 N/mm^{2}

Answer: c

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm^{2}.

**56. What is the strain energy stored in a body when the load is applied with impact?**

a) σE/V

b) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

Answer: d

Explanation: Strain energy in impact loading = σ^{2}V/2E.

**57. What is the value of stress induced in the rod due to impact load?**

a) P/A (1 + (1 + 2AEh/PL)^{1/2})

b) P/A (2 + 2AEh/PL)

c) P/A (1 + (1 + AEh/PL)^{1/2})

d) P/A ((1 + 2AEh/PL)^{1/2})

Answer: a

Explanation: The value of stress is calculated by equating the strain energy equation and the work done equation.

**58. What will be the stress induced in the rod if the height through which load is dropped is zero?**

a) P/A

b) 2P/A

c) P/E

d) 2P/E

Answer: b

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)^{1/2})

Putting h=0, we get stress = 2P/A.

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)

Putting h=0, we get stress = 2P/A.

**59. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm ^{2} in section. What will be the instantaneous stress (E=210GPa)?**

a) 149.4 N/mm

b) 179.24 N/mm

c) 187.7 N/mm

d) 156.1 N/mm

Answer: c

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)

Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm

**60. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm ^{2} cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous stress induced in the vertical bar if E = 200GPa?**

a) 50.87 N/mm

b) 60.23 N/mm

c) 45.24 N/mm

d) 63.14 N/mm

Answer: b

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)

Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm

**1. Units of deflection are _________**

a) kNm

b) kN/m

c) kN

d) m

Answer: d

Explanation: The term “deflection” is defined as the transverse displacement of a point on any straight axis to the curved axis. It is expressed in metres (m).

**2. Which of the following method is used to determine the slope and deflection at a point?**

a) Arithmetic increase method

b) Mathematical curve setting

c) Macaulay’s method

d) Lacey’s method

Answer: c

Explanation: Macaulay’s method was devised by Mr WH Macaulay.

Advantages:

i. Gives one continuous expression for bending moment

ii. Constants of integration can be found by using end conditions

iii. By using this method, slope and deflection at any section can be determined throughout the length of the beam.

**3. Deflection is denoted by _______**

a) i

b) y

c) h

d) e

Answer: b

Explanation: The deflection of a point on the axis of the deflected beam is defined as the angle developed in radians with tangent at the section makes with the original axis of the beam.

**4. In cantilever beams, the deflection is zero at ___________**

a) Free and

b) Fixed end

c) At supports

d) Through out

Answer: b

Explanation: The deflection in cantilever beam is always zero at the fixed end and deflection in the cantilever beam at the free end is maximum.

**5. Mohr’s theorem -¡¡ states?**

a) Ax/EI

b) A/Ex

c) A/EI

d) Ae=Ix

Answer: a

Explanation: Mohr’s theorem -¡¡ states “the intercept taken on a vertical reference line of the tangent at any two points on an elastic line is equal to the moment of BMD between these points, about the reference line divided by flexural rigidity (EI).

**6. Calculate the deflection if the slope is 0.0225 radians. Take the distance of centre of gravity of bending moment to free end as 2 metres.**

a) 45mm

b) 35mm

c) 28mm

d) 49mm

Answer: a

Explanation: The deflection at any point on the elastic curve equal to Ax/EI

But, we know that A/EI is already slope equation.

So, slope × (the distance of centre of gravity of bending moment to free end = 2m).

0.0225 × 2

0.045m ~ 45 mm.

**7. In simply supported beams, deflection is zero at _________**

a) Mid span

b) Supports

c) Through out

d) Point of action of load

Answer: b

Explanation: The deflection is always zero at the supports and the deflection is maximum at the mid span of a symmetrically loaded simply supported beam.

**8. Which of the following is not a cross drainage work?**

a) Aqueduct

b) Level crossing

c) Head regulator

d) Super passage

Answer: c

Explanation: The head regulator is one of the canal regulation works. It can control the entry of silt into the canal. It can be used as a metre for measuring the discharge. It can shut out river floods.

**9. Tail escape is also called as ___________**

a) Outlet

b) Cross regulator

c) Weir type escape

d) Surplus escape

Answer: c

Explanation: The crest of the weir is fixed at canal FSL. When the water level rises above FSL, it is disposed of into the natural drain. Hence, the tale escape is also known as weir type escape.

**10. The land where all the water comes from ___________**

a) Ridge dam

b) Watershed

c) Meander

d) Groynes

Answer: b

Explanation: A watershed can be defined as an interconnected area of land which receives the water from surrounding ridge tops and transports it to a common point such as a lake or stream. All lands and waterways can be found within one watershed or another.

**11. ____________ reduces storm water discharge.**

a) Rain water harvesting

b) Water harvesting

c) Watershed

d) Watershed management

Answer: b

Explanation: The water harvesting is defined as the process of capturing rain where it falls. The objectives of water harvesting are 1) To provide drinking water 2) To provide irrigation water 3) To increase groundwater recharge to reduce storm water discharge.

**12. Which of the following is not a soil moisture conservation method?**

a) Spreading manure

b) Crop rotation

c) Recharge to ground water

d) By mulches

Answer: c

Explanation: The methods which are adopted for preserving the water in the soil from being lost are called as soil moisture conservation methods. The major part of the water is lost through evapotranspiration. The recharge to groundwater is one of the techniques in rainwater harvesting.

**13. Nutrients like ca, mg, si, al, S, K are lost due to ____________**

a) Soil erosion

b) Percolation

c) Water logging

d) Watershed

Answer: b

Explanation: The percolation is defined as a downward movement of water through the soil due to the force of gravity. The rapid percolation of water results in loss of plant nutrients and makes the soil acidic.

**14. Warabandi has been practiced in India for more than ____________ years.**

a) 130 years

b) 125 years

c) 140 years

d) 145 years

Answer: b

Explanation: Warabandi is a rotational method for allocation of the available water equally in an irrigation system. It provides continuous rotation of water generally lasts 7 days. It has been effectively practiced in India for more than 125 years.

**15. Gold, Copper and lead are the examples of ______**

a) Ductility

b) Creep

c) Plasticity

d) Malleability

Answer: c

Explanation: Plasticity in the property of Material by which the material can undergo permanent deformation and fails to regain its original shape on removal of load. Examples are gold, lead, etc.

**16. The ratio of maximum deflection of a beam to its ___________ is called stiffness of the beam.**

a) Load

b) Slope

c) Span

d) Reaction at the support

Answer: c

Explanation: The stiffness of a beam is a measure of it’s resistance against deflection. The ratio of the maximum deflection of a beam to its span can be termed as stiffness of the beam.

**17. Stiffness of the beam is inversely proportional to the _____ of the beam.**

a) Slope

b) Support reaction

c) Deflection

d) Load

Answer: c

Explanation: Stiffness of a beam is inversely proportional to the deflection. Smaller the deflection in a beam due to given external load, greater is its stiffness.

**18. The maximum ____ should not exceed the permissible limit to the span of the beam.**

a) Slope

b) Deflection

c) Load

dl Bending moment

Answer: b

Explanation: The maximum deflection of a loaded beam should not exceed the permissible limit in relation to the span of a beam. While designing the beam the designer should be keep in mind that both strength and stiffness criteria.

**19. In cantilever beam the deflection occurs at ______**

a) Free end

b) Point of loading

c) Through out

d) Fixed end

Answer: a

Explanation: Deflection can be defined as the perpendicular displacement of a point on straight access to the curved axis. In cantilever beams, the maximum deflection occurs at free end.

**20. The maximum deflection in cantilever beam of span “l”m and loading at free end is “W” kN.**

a) Wl^{3}/2EI

b) Wl^{3}/3EI

c) Wl^{3}/4EI

d) Wl^{2}/2EI

Answer: b

Explanation: Maximum deflection occurs at free end distance between centre of gravity of bending moment diagram and free end is x = 2l/3.

As deflection is equal to the slope × “x”. The slope = Wl2/2EI radians

Maximum deflection (y) = Ax/EI = Wl^{3}/3EI.

**21. In an ideal fluid, the ____________ stresses are pretend to be absent.**

a) Bending

b) Shearing

c) Tensile

d) Compressive

Answer: b

Explanation: An ideal fluid is a fluid where there is no resistance to the deformation. Ideal Fluids are those Fluids which have no viscosity surface tension. The shear stress is also absent. This fluid is also called as perfect fluid.

**22. Air and water are the examples of ___________**

a) Non Newtonian fluids

b) Vortex fluids

c) Real fluids

d) Ideal fluids

Answer: d

Explanation: The ideal Fluids are imaginary fluids in nature, they are incompressible. These fluids possess low viscosity. Air and water are considered as ideal fluids.

**23. _______ fluids are practical fluids**

a) Ideal

b) Real

c) Vortex

d) Newtonian

Answer: b

Explanation: These fluids possess properties such as viscosity, surface tension. They are compressible in nature. The certain amount of resistance is always offered by the fluids, they also possess shear stress. They are also known as practical fluids.

**24. Specific weight of water at 4°C is ____________ N/m ^{3}.**

a) 9810

b) 9760

c) 9950

d) 9865

Answer: a

Explanation: The specific weight (weight density) of a fluid is weight per unit volume. It is represented by symbol w & it is expressed in Newton per metre cube (N/m

**25. The inverse of specific weight of a fluid is __________**

a) Specific gravity

b) Specific Volume

c) Compressibility

d) Viscosity

Answer: b

Explanation: Specific volume is the volume of the fluid by Unit Weight it is the reciprocal of specific weight is denoted by “v”. SI units are m^{3}/N.

v= 1/specific weight.

**26. Calculate the specific gravity of mercury.**

a) 12.5

b) 14.7

c) 13.6

d) 11.8

Answer: c

Explanation: The specific gravity of any fluid is the ratio of the specific weight of fluid by specific weight of water. For mercury, the specific weight is 133416 N/m^{3}. For water, w = 9810 N/m^{3}.

S = 133416/9810

S= 13.6.

**27. Specific gravity of water is __________**

a) 0.8

b) 1

c) 1.2

d) 1.5

Answer: b

Explanation: The specific gravity is also called as relative density. It is dimensionless quantity and it has no units. The specific gravity of water is the ratio of specific weight of fluid to specific weight of water, as both the numerator and denominator are same. The value is 1.

**28. Compute the maximum deflection at free end of a cantilever beam subjected to udl for entire span of l metres.**

a) wl^{4}/8EI

b) wl^{4}/4EI

c) wl^{3}/8EI

d) wl^{2}/6EI

Answer: a

Explanation: The slope at free end = A/EI = wl^{3}/6EI

Maximum deflection at free end is Ax/EI; [x= ¾ l] y= wl^{3}/6EI × ¾ l = wl^{4}/8EI.

**29. Calculate the maximum deflection of a cantilever beam with udl on entire span of 3m the intensity of you udl be 25 kN/m. Take EI as 4000 kN/m ^{2}.**

a) 0.052m

b) 0.063m

c) 0.076m

d) 0.09m

Answer: b

Explanation: For cantilever beams with udl on entire span, the maximum deflection = wl

y = wl

**30. Which of the following is not an example of Malleability?**

a) Wrought Iron

b) Ornamental silver

c) Torsteel

d) Ornamental gold

Answer: c

Explanation: Torsteel is an example of mechanical property ductility. The ductility is a property of a material by which material can be fractured into thin wires after undergoing a considerable deformation without any rupture.

**31. __________ of a beam is a measure of its resistance against deflection.**

a) Strength

b) Stiffness

c) Slope

d) Maximum bending

Answer: b

Explanation: The ratio of maximum deflection of a beam to its corresponding span is termed as the stiffness of the beam. It is the measure of resistance against the deflection.

**32. The maximum induced ___________ stresses should be within the safe permissible stresses to ensure strength of the beam.**

a) Tensile

b) Compressive

c) Bending

d) Lateral

Answer: c

Explanation: A beam is said to be strengthy when the maximum induced bending and shear stresses are within the safe permissible stresses of the beam material.

**33. Elastic line is also called as ___________**

a) Deflection curve

b) Plastic curve

c) Linear curve

d) Hooke’s curve

Answer: a

Explanation: The deflection curve is defined as the line to which the longitudinal axis of a beam deflects or bends under given load. This curve is also known as elastic line or elastic axis.

**34. In simply supported beams, the slope is _____________ at supports.**

a) Minimum

b) Zero

c) Maximum

d) Uniform

Answer: c

Explanation: The slope at any section in the deflected beam is defined as the angle developed in radians which the tangent at the section makes with the actual axis of the proposed beam. In simply supported beams, the slope is maximum at the supports.

**35. In simply supported beam deflection is maximum at ____________**

a) Midspan

b) Supports

c) Point of loading

d) Through out

Answer: a

Explanation: In simply supported beams, deflection is maximum at the mid span of a symmetrically loaded beam.

**36. Calculate the maximum deflection of a simply supported beam if the maximum slope at A is 0.0075 radians and the distance of centre of gravity of bending moment diagram to support A is 1.33 metres.**

a) 9.975 mm

b) 9.5 mm

c) 9.25 mm

d) 9.785 mm

Answer: a

Explanation: The deflection occurs at support A = A/EI = 0.0075 radians

Maximum deflection = Ax/EI = 0.0075 × 1.33

y = 9.975 mm.

**1. Column is a tension member.**

a) True

b) False

Answer: b

Explanation: Compression members are the structural elements that are pushed together or carrying a load; more technically they are subjected to axial compressive forces. Example: Column, strut etc.

**2. __________ is a vertical member subjected to direct compressive force.**

a) Strut

b) Beam

c) Column

d) Post

Answer: c

Explanation: A vertical member subjected to direct compressive forces is called a column or pillar. The column transfers the load from the beams or slab to the footings and foundations.

**3. The inclined member carrying compressive loads is __________**

a) Post

b) Stanchion

c) Strut

d) Column

Answer: c

Explanation: The inclined member carrying compressive load in case of frames and trusses is called as a strut. A strut is a member of a structure in any position carrying an axial load. Strut may be horizontal, inclined or even vertical.

**4. A built up rolled steel section carrying compressive force is called ___________**

a) Post

b) Pillar

c) Strut

d) Stanchion

Answer: d

Explanation: A built up rolled Steel section carrying compressive force is known as “stanchion”. A wood member carrying compressive force is called a “post”.

**5. The process of removing chlorine from water is known as ____________**

a) De chlorination

b) Re chlorination

c) Post chlorination

d) Pre chlorination

Answer: a

Explanation: De chlorination means removing the chlorine from the water this is generally required when super chlorination has been practiced.

**6. The organic impurities in the water from a layer on the top of a filtering media are called _______**

a) Filter layer

b) Permeable layer

c) Impermeable layer

d) Dirty skin

Answer: d

Explanation: The water from the sedimentation tank is allowed to enter over a bed of sand through the inlet chamber. The water percolates through sand bed during the dry skin is formed. The organic and bacterial impurities are removed by this layer.

**7. The rate of filtration in slow sand filter is ___________**

a) 100 to 150 **lit/hr/m**^{2}

b) 150 to 200 lit/hr/m^{2}

c) 250 to 350 lit/hr/m^{2}

d) 100 to 200 lit/hr/m^{2}

Answer: d

Explanation: The slow sand filter is effective in bacterial removal and it is preferable for uniform quality of treated water. It is simple to construct and supervise. The rate of filtration is small and ranges from 100 to 200 lit/hr/m^{2}.

**8. The sand used for filtration should not lose weight more than ________ when placed in Hcl for 24 hours.**

a) 5 %

b) 10 %

c) 15 %

d) 20 %

Answer: a

Explanation: According to the board of Indian Standards [BIS], the sand which is used for filtration process should not lose weight more than 5% when placed in hydrochloric acid for one day (24 hours).

**9. Cleaning period of slow sand filter is taken as __________**

a) 1 to 3 weeks

b) 1 to 3 days

c) 1 to 3 months

d) 1 to 2 hours

Answer: c

Explanation: For the purpose of cleaning the top layer of sand is removed to a depth of 15 mm to 25 mm. The water is admitted to the filter the cleaning interval varies from 1 to 3 months.

**10. The efficiency of slow sand filter is about ___________**

a) 99 %

b) 95 %

c) 85 %

d) 90 %

Answer: a

Explanation: The slow sand filters remove suspended and bacterial impurities to an extent of 98 to 99%. It requires a large area and unsuitable for treating high turbid water.

**11. What is the uniformity coefficient of sand used in the rapid sand filter?**

a) 1.5

b) 1.35

c) 1.75

d) 1.6

Answer: c

Explanation: The effective size of sand used in rapid sand filter is 1.5 mm and the uniformity Coefficient varies from 1.25 to 1.35.

**12. The dosage of ozone is about ________ ppm residual ozone.**

a) 2 to 3 ppm

b) 2 to 4 ppm

c) 1 to 5 ppm

d) Zero

Answer: a

Explanation: Ozone easily breaks down with oxygen and releases nascent oxygen which is powerful in killing bacteria. It also reduces organic matter present in the water the dosage of ozone is about 2 to 3 ppm.

**13. Which of the following process is also known as Ion exchange process?**

a) Lime soda process

b) Base exchange process

c) Demineralization process

d) Cation exchange process

Answer: a

Explanation: Softening of water can be done by the demineralization process which is also known as deionized water. In, each method minerals are removed my pass in the water through a bed of cation exchange resin.

**14. Aeration is effective in removing of _________ odours.**

a) 60 %

b) 75 %

c) 30 %

d) 40

Answer: b

Explanation: Aeration is effective in removing 75% of odours. This process also removes carbon dioxide to a great extent. Aeration is affected by filtering it through perforated trays through different methods.

**15. Which of the following is a control measure of removal of colour, taste and order?**

a) Ozone treatment

b) Silver treatment

c) Copper sulphate treatment

d) Use of chloramines

Answer: c

Explanation: Copper sulphate also helps in the removal of colour taste and odour it prevents the growth of algae and also acts as disinfectant. It is used for swimming pool water to give play then colour.

**16. _____________ of column mainly depends upon end conditions.**

a) Radius of gyration

b) Slenderness ratio

c) Factored load

d) Effective length

Answer: d

Explanation: The effective length of a column with given end conditions is a length of an equivalent column of the same material and cross section with hinged ends. The effective length of the column mainly depends upon end conditions.

**17. The hinged end is also known as ___________**

a) Fixed end

b) Pinned end

c) Rigid end

d) Free end

Answer: b

Explanation: In hinged end case, the end is fixed in position only (but the direction is free). The deflection in the case of this end is zero. (y = 0). It is also known as “Pinned end”.

**18. Long columns fail due to ____________**

a) Direct stress

b) Buckling stress

c) Lateral stress

d) Tensile stress

Answer: b

Explanation: In long columns, direct stress is very small compared to the bending stresses. The long column commonly fails because of bending stress.

**19. In short columns, the slenderness ratio is less than __________**

a) 32

b) 64

c) 56

d) 28

Answer: a

Explanation: The short column fails primarily due to direct stress. In short columns, the buckling stresses are very small compared to direct stresses. The short column is a column whose slenderness ratio is less than 32.

**20. For ___________ columns, the slenderness ratio is more than 32 and less than 120.**

a) Long

b) Short

c) Average

d) Medium

Answer: d

Explanation: Medium column is a column which fails either due to direct stress or buckling stress. For medium columns, the slenderness ratio is more than 32 and less than 120. For medium columns, the length is more than 8 times but less than 30 times their least lateral dimension.

**21. Radius of gyration is denoted by __________**

a) k

b) g

c) y

d) s

Answer: a

Explanation: The ratio of square root of the moment of inertia (I) to the cross sectional area(A) is called “radius of gyration”. It is denoted by “k” or “r”.

K = (I/A)^{1/2}.

**22. The __________ is the distance between Centres to centre of effective lateral ends.**

a) Mean length

b) Stripped length

c) True length

d) Actual length

Answer: d

Explanation: The actual length of a column is defined as the distance between the centre to centre of effective lateral restraints (L).

**23. Long axially loaded columns tends to deflect about ___________**

a) Moment of inertia

b) Effective length

c) Core

d) Safe loading

Answer: a

Explanation: A long axially loaded column tends to deflect about the axis of least moment of inertia the least radius of gyration and it should be used for determining the slenderness ratio.

**24. What is the effective length of a column at both ends fixed?**

a) l/3

b) l

c) l/2

d) 2×l

Answer: c

Explanation: The effective length of a column at both ends fixed is L =l/2.

End condition | Effective length |

Both ends hinged | L = l |

Both ends fixed | L =l/2 |

**25. Which of the following is the method of removing the temporary hardness of water?**

a) Lime soda method

b) Base exchange process

c) Boiling

d) Chlorination

Answer: c

Explanation: When the water is boiled up to a temperature of 80 degree, most of the bacteria will be killed and bicarbonates of calcium and magnesium are also eliminated.

**26. The application of chlorine beyond the stage of break point is _________**

a) Double chlorination

b) Post pollination

c) Super chlorination

d) Breakpoint chlorination

Answer: c

Explanation: Super chlorination is a term which indicates the addition of an excessive amount of chlorine that is 5 to 15 mg / l to the water that is the application of chlorine beyond the stage of break point.

**27. Which of the following methods of disinfection is usually adopted in swimming pools?**

a) Excess lime treatment

b) Iodine – Bromine method

c) Pottasium permanganate method

d) Ultraviolet rays method

Answer: d

Explanation: Ultraviolet rays are highly disinfectants and kill the bacteria. The rays penetrate in water and kill the bacteria. This process is very costly and requires technical skill and costly equipment. This method is adapted generally in swimming pools.

**28. Hardness due to calcium bi carbonate can be removed by ___________**

a) Boiling

b) Excessive lime

c) Zeolite

d) Soda treatment

Answer: b

Explanation: Lime is generally used as a water purification material the excess lime treatment of water about 99.9% to 100%. The lime excess line l is the pH value of water making extremely alkaline.

**29. __________ is used for spans ranging from 9 m to 15 m.**

a) King post truss

b) Queen post truss

c) Coral truss

d) Roof truss

Answer: b

Explanation: Queen post truss is used for spans 9 m to 15 m. It consists of principal rafters, common rafters, purlins. The queen posts are connected with the help of a straining beam.

**30. __________ is provided to prevent the movement of the post due to loads in Queen post truss.**

a) Purlin

b) Eaves board

c) Straining beam

d) Ridge beam

Answer: c

Explanation: The queen posts are connected with the aid of straining been of the upper ends and by a straining sill at the lower end to prevent the movement of post due to loads. In this truss, the straining beam acts as a compression member. Whereas the queen post act as a tension member.

**31. ___________ is a combination of king post truss and queen post truss.**

a) Steel slope truss

b) Pratt truss

c) Mansard roof truss

d) Fan truss

Answer: c

Explanation: Mansard truss is a combination of king post truss and queen post truss. This truss is used to obtain the maximum space for living purposes. The general height of the roof is comparatively kept low.

**32. In mansard truss, the upper slope is _________**

a) 45°

b) 30°

c) 60°

d) 90°

Answer: b

Explanation: The mansard has two different slopes, the lower slope should not be steeper than 75° and upper slope not greater than 30 degrees. The construction of the various joints is similar to the king post and queen post trusses.

**33. ___________ formula can be used only for long columns.**

a) Euler’s

b) Rankine’s

c) Swift’s

d) Johnson’s

Answer: a

Explanation: Euler’s formula is used only for long columns and l/k > 80 for mild steel columns.

Where l = effective length of column

k = radius of gyration.

**34. In Euler’s formula, the column fails due to __________ alone.**

a) Shear

b) Torsion

c) Tension

d) Bending

Answer: d

Explanation: In Euler’s formula, the column material is perfectly elastic, homogeneous, isotropic and obeys Hooke’s law. The self weight of column is ignorable and column fails due to buckling alone.

**35. The __________ joints are friction less.**

a) Free

b) Pin

c) Roller

d) Fixed

Answer: b

Explanation: A pinned joint offers resistance against horizontal and vertical movements but not against rotation. The deflection developed is zero (y = 0) and fixed ends are rigid.

**36. __________ formula is used for determining short as well as long columns.**

a) Gilbert’s

b) Rankine’s

c) Johnson’s

d) Euler’s

Answer: b

Explanation: The Rankine’s formula for crushing load = Pcr = fA/1+€(l/k)^{2}

Where; f = allowable crushing stress

A = area of cross section

K = least radius of gyration

€ = Rankine’s constant

• Rankine formula can be used for short columns as well as long columns.

**37. ________ attached to a Framework suspended from the main structure.**

a) Cantering

b) Shuttering

c) Bracing

d) Ceiling

Answer: d

Explanation: A suspended ceiling attached to a framework suspended from the main structure. It provides a void between the underside of the main structure and ceiling. General it is provided to conceal the unevenness of roof.

**38. _________ type of ceiling is adopted in modern hotels and auditorium.**

a) Plaster board

b) Fibre board

c) Decorative

d) Joint less

Answer: a

Explanation: Plaster board ceiling is adopted because of its ease of fixing and elimination of plaster mixer for the over head work. Plasterboard consists of gypsum plaster form in sheets 2.5 m × 0.75 m and compressed give strength.

**39. Upper floor is also known as ________**

a) Basement floor

b) Suspended floor

c) Supported floor

d) Rigid floor

Answer: b

Explanation: Any floor above the level of ground floor is termed as upper or suspended floor. Floors are named as ascending order. The name of the building in respect of the storeys is governed by the number of floors it possess.

**40. Paving is also known as ___________**

a) Floor covering

b) Sub floor

c) Sub grade

d) Wearing course

Answer: a

Explanation: The upper position of a floor structure consisting of base and topping is called floor covering or paving. The purpose of floor covering is to have a clean, smooth, non absorbent and durable surface.

**41. _________ floors are used in modern residential and religious buildings?**

a) Cement concrete

b) Terrazzo

c) Mosaic

d) Timber

Answer: b

Explanation: Terrazzo floors consist of the terrazzo finish at top. Generally it consists of wearing layer of the terrazzo mixture about 6 mm thick laid on and under layer. It furnishes attractive and durable floor.

**42. Scaffolding has to be done, if the height of structure is above _________**

a) 1.2

b) 1.4

c) 1.5

d) 1.8

Answer: c

Explanation: A temporary platform provided with necessary supports close to the work, provides limited space for the workers, building materials, tools etc. is termed as scaffolding. It is generally adopted for construction of masonry work above ground level 1.5 m.

**43. Calculate the Euler’s crippling load, if the effective length of column is 10 m take flexural rigidity as 6.14 × 10 ^{10} Nmm^{2}.**

a) 6 kN

b) 8 kN

c) 10 kN

d) 12 kN

Answer: a

Explanation: To find P:

P = π

P = π

P = 6.055 kN ~ 6 kN.

**44. A fine grained material is mostly ________**

a) Homogeneous

b) Isotropic

c) Isomeric

d) Elastic

Answer: b

Explanation: A material is said to be isotropic if at any point it has identical elastic properties in all directions. A fine grained material is mostly isotropic in nature.

**45. The tangential force per unit area is _________**

a) Shear strain

b) Shear stress

c) Modulus of rigidity

d) Torsion

Answer: b

Explanation: The tangential force acting along the section of the body is termed as shear force and the stress in the section due to shear force is called shear stress and it is denoted by fs.

**46. Which of the following is also known as pushing force?**

a) Tensile stress

b) Compressive stress

c) Shear stress

d) Temperature stress

Answer: b

Explanation: When an external force cause shortening of the body in the direction of the force it is termed as compressive force. The stress developed in the body due to the compressive force is called compressive stress.

**47. Which of the following is also known as pulling force?**

a) Tensile stress

b) Shear stress

c) Lateral stress

d) Axial stress

Answer: a

Explanation: When an external force produces elongation of the body in its direction, it is termed as a tensile force. The stress developed in a cross section of the body due to a tensile force is called tensile stress.

**48. Longitudinal strain is also known as ___________**

a) Direct strain

b) Axial strain

c) Indirect strain

d) Shear strain

Answer: a

Explanation: Direct strain is a measure of deformation produced by the application of the external forces. It is the ratio of change in dimension to the original dimension. It is also known as longitudinal strain.

**49. Which of the following is also known as transverse strain?**

a) Tensile strain

b) Compressive strain

c) Shear strain

d) Volumetric strain

Answer: c

Explanation: Shear Strain is a measure of the angle through which a body is this distorted with applied forces. Shear Strain is also known as the transverse strain.

Shear strain = ds/L.

**50. The hooks law is valid only for _________**

a) Uni axial forces

b) Bi axial forces

c) Tri axial forces

d) Lateral forces

Answer: a

Explanation: Hooke’s law: (Given by Sir Robert Hooke in 1678): stress is directly proportional to strain within limit of proportionality. It is valid for uniaxial force only.

**51. Which of the following is also known as endurance limit?**

a) Proportionality limit

b) Rupture limit

c) Elastic limit

d) Fatigue limit

Answer: d

Explanation: The greatest stress applied an infinite number of times that a material can take without causing Failure is known as endurance or fatigue limit.

**52. The ultimate strength in flexure is known as modulus of ________**

a) Toughness

b) Rupture

c) Resilience

d) Hardening

Answer: b

Explanation: The ultimate strength in flexure or torsion is known as modulus of rupture and the modulus of resilience is defined as the energy stored per unit volume at the elastic limit.

**53. A material which ruptures with little or no plastic deformation is said to be ____________**

a) Ductile material

b) Elastic material

c) Plastic material

d) Brittle material

Answer: d

Explanation: A material is said to be brittle, if it ruptures with little or no plastic deformation and a material said to be ductile if it undergoes deformation without rupture.

**54. The stress which is just sufficient to cause a permanent set is known as ___________**

a) tenacity

b) ultimate stress

c) proof stress

d) working stress

Answer: c

Explanation: Proof stress is a stress which is just sufficient to cause a permanent set equal to a specified percentage of the original gauge length. The stress corresponding to 0.2% of strain in the stress strain curve of mild steel is also known as proof stress.

**55. For engineering materials, the poison’s ratio lies in the range of ___________**

a) 0 and 1

b) -1 and 1

c) -2 and 2

d) 0 and 1/2

Answer: d

Explanation: The ratio of lateral strain to the corresponding longitudinal or linear strain is called poison’s ratio and it is denoted by 1 / m. The value of poison’s ratio for elastic materials usually lies between 0.25 and 0.33 in no case the value doesn’t exceed 0.5.

**56. For ductile materials, the factor of safety is the ratio of yield stress to ___________**

a) tenacity

b) ultimate stress

c) working stress

d) shear stress

Answer: c

Explanation: Factor of safety,; for ductile materials, F.O.S = yield stress/Working stress

For brittle materials; F.O.S = ultimate stress / working stress.

**57. A material having three mutually perpendicular planes of elastic symmetry is said to be _________**

a) Isotropic

b) Autotrophic

c) Orthotropic

d) Anisotropic

Answer: c

Explanation: If the material has three mutually perpendicular planes of elastic symmetry, then the material is said to be orthotropic material. The number of an independent constant is 9 in this case.

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