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Multiple choice question for engineering

Set 1

1. The chemical pre stressing is also called as:
a) Self stressing
b) Whole stressing
c) Half stressing
d) Model stressing

View Answer

Answer: a [Reason:] Self stressing or chemical pre stressing of concrete was made possible by the development of expanding cements by Lossier of France in 1944 and generally expanding cements consist of 75% Portland cement, 15% alumina cement and 10 % gypsum which result in formation of calcium Sulpho aluminate.

2. The linear expansion of the cement in chemical pre stressing is about:
a) 5-10%
b) 2-3%
c) 3-4%
d) 6-7%

View Answer

Answer: c [Reason:] The linear expansion of the cement in chemical prestressing is about 3 to 4% Mikhailov has reported that expansive cements have been used for pre stressing purposes in erstwhile in USSR since 1949 and splices are adopted for circular prestressing of tanks and the advantage is that there is no reduction in the strength of wire.

3. The degree of expansion can be controlled by varying the conditions of:
a) Bending
b) Bonding
c) Heating
d) Curing

View Answer

Answer: d [Reason:] The degree of expansion can be controlled by varying the curing conditions in expanding cements and when the concrete is subjected to compression the amount and rate of expansion is controlled by controlling the curing conditions of the member.

4. The expansion of the concrete is restrained by:
a) High-tensile steel
b) High-strength concrete
c) Precast members
d) Concrete members

View Answer

Answer: a [Reason:] concrete is composite material consisting of an aggregate and a binder phase and the binder phase is hardened cement and the expansion of concrete is restrained by high tensile steel wires, the compressive stresses that develop in concrete and steel wires are subjected to tensile stresses.

5. The tensile stresses developed in steel by the expansion of concrete are about:
a) 850n/mm2
b) 250n/mm2
c) 600n/mm2
d) 500n/mm2

View Answer

Answer: a [Reason:] Tensile stresses of up to 850n/mm2 were developed in steel by the expansion concrete results of laboratory investigations of several types of chemically pre stressed elements, such as beams, slabs, frames, columns, pipes and hyperbolic parabolic sheets have demonstrated the feasibility of chemical pre stressing.

6. The chemical pre stressing is generally suited for elements like:
a) Pre cast beams and columns
b) Bonding elements
c) Breakage elements
d) Tensioning elements

View Answer

Answer: a [Reason:] It has been found that structural elements ideally suited for chemical pre stressing include pipes, thin walls and slabs, shells; folded plates and composite columns as well as pre cast beams and columns.

7. The method of chemical pre stressing is not suited for:
a) High degrees of prestress
b) High degrees of compression
c) High degrees of anchoring
d) High degrees of jacking

View Answer

Answer: a [Reason:] In the present state of art, chemical pre stressing can be applied to structural elements and systems in which the optimum amount of pre stress is relatively low and this method is most not suited for high degrees of prestress and high percentages of steel where mechanical pre stressing can be conveniently used.

8. The suitability of post tensioning is good for:
a) Long spans
b) Break spans
c) Edge spans
d) End spans

View Answer

Answer: a [Reason:] The suitability of post tensioning is good for medium to long span in situ work, where the cost of tensioning is very less and the major advantage of is the stopped off and curved cables are allowed by which designer can easily differ the prestresss distribution.

9. The long span decks are fabricated by:
a) Post tensioning
b) Pre tensioning
c) Thermo electric prestress
d) Chemical prestress

View Answer

Answer: a [Reason:] Post tensioning is ideally suited in concrete construction work involving stage prestressing most of the long span bridge structures are constructed using post tensioning system and long span bridge decks are also fabricated by the use of post tensioning.

10. The concrete dams are constructed using which:
a) Post tensioning
b) Thermo electric pre tensioning
c) Biological pre tensioning
d) Elongation pre tensioning

View Answer

Answer: a [Reason:] Concrete dams, biological shields of nuclear reactors and circular pre stressing of large concrete tanks are strengthened by using post tensioning and high tensile wire is wrapped under high tension using a wire serving machine developed by the inventors.

Set 2

1. The chords and struts of trusses are designed for convenience in:
a) Plastering
b) Fabricating
c) Rubbing
d) Forrowing

View Answer

Answer: b [Reason:] The chords and struts of trusses are designed to have the same width for convenience in fabricating the truss in a horizontal position such that it is protected by external moments and also smoothen the members.

2. The precast roof slabs are used for:
a) Roof coverings
b) Slab coverings
c) Column coverings
d) Beam coverings

View Answer

Answer: a [Reason:] If precast roof slabs are used for roof covering the upper chord panels are made equal to the width of the precast slabs which is usually about 3m and the lower tension chord is prestressed using bunched high strength wires or cables running in perforated holes.

3. For spans in the range of 18-24m, the trusses are made in how many pieces:
a) 4 pieces
b) 2 pieces
c) 1 piece
d) 5 pieces

View Answer

Answer: c [Reason:] For spans in the range of 18-24m, the trusses are made in one piece but when spans run from 24-30m, they are made in two pieces with a joint at mid span and beyond them they are divided into equal spans and joints are made.

4. The polygonal trusses with inclined top chords are generally made of:
a) 8m Blocks
b) 6m blocks
c) 12m block
d) 65m block

View Answer

Answer: b [Reason:] Polygonal trusses with inclined top chords are generally made of 6m blocks or half trusses with 3m panels and due to higher tensions developed in the diagonal members of large span trusses, prestressing them becomes inevitable.

5. The polygonal trusses are less economical than:
a) Circular
b) Bow type
c) Curved
d) Oval

View Answer

Answer: a [Reason:] In general polygonal trusses are less economical than the bow type with regard to material and labour costs and various types of trusses like circular, cylindrical, hollow etc used for various structures are not much economical because of their shapes and freeness to work.

6. The steel bearing plates which are anchored serves as:
a) Loading
b) Bearing
c) Stressing
d) Deforming

View Answer

Answer: b [Reason:] At the ends of trusses near supports, 10-12mm steel bearing plates are anchored and embedded while casting, which serve as bearing and for fixing the trusses on neoprene pad bearings located on the columns.

7. Calculate the area of concrete section such that loss ratio is 0.18 and compressive strength is 15n/mm2 (Nd = 377×103)?
a) 9.43n/mm2
b) 6.54n/mm2
c) 8.5n/mm2
d) 9.34n/mm2

View Answer

Answer: a [Reason:] Nd = 377×103, ɳ = 0.18, fct = 15n/mm2 Area of concrete section = (Nd/ɳfct) = (377×103)/0.8×15) = 3.316mm2.

8. Calculate the number of wires which are subjected to prestressing force of 471.5kn, the section adopted is 50000 using 7mm diameter high tensile wires initially stressed at 1100n/mm2?
a) 13.2
b) 11.13
c) 24.5
d) 34.2

View Answer

Answer: b [Reason:] Given section adopted = 50000, prestressing force 471.5kn, using 7mm diameter high tensile wires initially stressed so 1100n/mm2 N = (471.5×103/38.5×1100) = 11.13.

9. Calculate the cracking load such that section adopted is 50000, the loss ratio is 0.8, compressive strength of concrete is 9.43 and minimum reinforcement is 4.0mm2?
a) 455kn
b) 324kn
c) 577.2kn
d) 456.6kn

View Answer

Answer: c [Reason:] The cracking load = (50000 (0.8 x 9.43) + 4.0 / 1000), Section adapted = 50000, ɳ = 0.8, fct = 9.43, As = 4.0 (As is the minimum reinforcement of 0.8 percent in the section).

10. Which type of analysis should be done which will lead to an optimal design of planning a structure?
a) Comparative analysis
b) Strength analysis
c) Transferred analysis
d) Global analysis

View Answer

Answer: a [Reason:] It is important to note that there is no single form of design which would be most economical in a given situation and to arrive at an economical design, several alternatives using different materials and structural configurations should be examined and a comparative analysis made which will lead to an optimal design.

Set 3

1. Cantilever construction eliminates the use of:
a) Steel
b) Cement
c) Formwork
d) Workman

View Answer

Answer: c [Reason:] Most of the long span bridges are built using prestressed concrete and those built by the cantilever method developed by Finister walder, demonstrate the latest refinements of the construction techniques and this method eliminates the use of expensive formwork and scaffolding especially for bridges in deep valleys and rivers with large depth of water.

2. The major methods of cantilever construction are classified into how many types:
a) 2 tyres
b) 3 types
c) 4 types
d) 6 types

View Answer

Answer: a [Reason:] There are two major methods of cantilever construction techniques are classified as: Cast in situ construction: In this method the bridge in cat in situ with sections 3-6m long, cantilevering symmetrically on both sides of pier Construction using precast segments: In this method type of construction, the bridge segments comprising structural elements (mainly segmental single or multi celled box girders) are cast in a casing yard using special forms and they are transported to the work site.

3. In cat insitu construction the formwork is supported by:
a) Aluminium frame work
b) Wood frame work
c) Steel frame work
d) Cloth formwork

View Answer

Answer: c [Reason:] The form work for cat in situ construction is supported by steel frame work attached to the completed part of the bridge and the formwork moves from one complete section to the next part and the sequential operations in this method are: fabrication of steel truss, placing of reinforcement, concreting using concrete mix, curing the concrete, threading the high tensile cables and anchoring and grouting of the cables, releasing formwork.

4. One of the example of cast insitu construction:
a) Boussens bridge, France
b) Larens bridge, Uganda
c) Lithcher bridge, Spain
d) Rouli bridge, Africa

View Answer

Answer: a [Reason:] Boussens bridge over the garnne river in France having spans of 49-96m and the supporting form work to facilitate the concreteing of the cantilever portion of the bridge abuts the previously constructed section and the typical cross section comprises box girders of constant or variable depth with cables running in the ribs and flanges and a notable example of cast in situ cantilever construction of the basin creek bridge by gammon India company at Bombay.

5. The main advantage of using precast segmental units is they can be cast on:
a) Beam
b) Slab
c) Edges
d) Ground

View Answer

Answer: d [Reason:] The main advantage of using precast segmental units is that they can be cast on ground near the work site well in advance and the quality of units will be better than those which are cast in situ and another advantage is the units can be curved to achieve their full strength before bringing them to assemble at worksite.

6. In the cost insitu method the least time required to move formwork is:
a) 7
b) 8
c) 3
d) 2

View Answer

Answer: a [Reason:] In the cast in situ method at least a week’s time is required to move the formwork to the next incremental length and the precast segment system, the units can be bought to site and lifted by cranes to join them to the previous units by using temporary stressing cables and the rate of construction will be faster in the precast method than in the cast in situ method and in both methods a typical cross section would be a box girder of constant or variable depth.

7. The cantilever method has been successfully used in the span range of:
a) 40-150
b) 30-150
c) 50-200
d) 40-200

View Answer

Answer: c [Reason:] The cantilever method has been successfully used in the span range of 50-200m and for small spans of less than 50m and for elevated roads or flyovers were scaffolding beneath the structure must be avoided precast segmental construction is preferred and at present for spans over 70m, prestressed concrete single or multi cell box girders complete successfully with steel construction.

8. Which one of the following method of construction is adopted when low clearance is required below the deck?
a) Erection method
b) Stagging method
c) Elongation method
d) Longitudinal method

View Answer

Answer: b [Reason:] The staging method is adopted when low clearances required below the deck and supporting formwork does not interfere with the traffic and this method facilitates rapid construction by maintaining correct geometry of the structure with relatively low cost.

9. Which type of construction is particularly advantageous in long via duct structures, a segmental?
a) Span to span
b) Edge to edge
c) Column to column
d) Beam to beam

View Answer

Answer: a [Reason:] In long span via duct structures, a segmental span by span (stage by stage construction, is particularly advantageous and the movable formwork may be supported from the ground and the traveler consists of a steel super structure which is moved from the completed portion of the structure to the next span to facilitate the casting or supporting of the precast units.

10. In push out technique each unit in cast directly against the:
a) Previous unit
b) Tensile unit
c) Last units
d) Progressive units

View Answer

Answer: a [Reason:] Segments of the bridge super structure are cast at site in lengths of 10-30m in stationary forms located behind the abutments and each unit is cast directly against the previous unit and after the concrete attains the desired strength the new unit is joined to the previous unit by post tensioning and the assembly of units is pushed forward in a step wise manner to permit casting of the succeeding segments.

11. The construction starts at one end of the structure and proceeds continuously to other end is:
a) Progressive placement method
b) Transverse placement method
c) Horizontal placement method
d) Linear placement method

View Answer

Answer: a [Reason:] In the method the construction starts at one end of the structure and proceeds continuously to the other end and in contrast to the balanced cantilevered construction in which the super structure proceeds or both sides of the pier, in the progressive placement technique, the precast segments are placed from one owned of the structure to the other in successive cantilevers on the same side of the various piers and at present this method has been found to be practicable and economical in the span range of 30-90m.

12. The main feature of this method comprises a moveable temporary arrangement to limit the:
a) Feasible stress
b) Cantilever stress
c) Rigidity stress
d) Bond stress

View Answer

Answer: b [Reason:] The main feature of this method comprises moveable temporary arrangement to limit the cantilever stresses during construction to a reasonable level and the precast segmental units are transported over the completed portion of the deck to the tip of the cantilever span under construction, where they are positioned by swivel crane that moves over the deck.

13. The construction procedure must be planned using:
a) Sequential computations
b) Reverse computations
c) Complicated computations
d) Aligned computations

View Answer

Answer: a [Reason:] Construction techniques developed have shown major progress towards simplification and reduction of erection equipment and the construction procedure must however be planned using sequential computations for the alignment forces, exact lengths and angles considering temperature and creep influence which depends on seasonal, climatic and daily environmental conditions.

14. In incremental launching or pull out technique to allow the super structure to move forward, what are provided?
a) Sliding bearings
b) Rolling bearings
c) Strut bearings
d) Tensile bearings

View Answer

Answer: a [Reason:] Normally a work cycle of one week is required for casting and launching the segments and to allow the super structure to move forward, special low friction sliding bearings are provided at the top of various piers with proper lateral guides and the main problem is to ensure the safety of stresses in the super structure under its own self weight during all stages of launching at various critical sections.

15. If the spans are large, they can be sub divided by means of:
a) Struts
b) Piers
c) Ridges
d) Walls

View Answer

Answer: b [Reason:] If the spans are large, they can be sub divided by means of temporary piers to control the magnitude of bending moments with in safe limits and according to Raina this construction technique has been applied to spans up to 60m without the use of temporary false work bents and also spans up to 100m have been built using temporary supporting bents and the main girders must have a constant depth generally varying from 1/12 to 1/16 of the longest span.

Set 4

1. Imperfection factor for rolled section is
a) 0.1
b) 0.21
c) 2.1
d) 4.9

View Answer

Answer: b [Reason:] Imperfection factor for rolled section is 0.21. The imperfection factor takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses. It depends on the buckling curve.

2. Imperfection factor for welded section is
a) 4.9
b) 0.21
c) 2.1
d) 0.49

View Answer

Answer: d [Reason:] Imperfection factor for welded section is 0.49. The imperfection factor depends on the buckling curve and takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

3. Non-dimensional slenderness ratio is given by
a) λLT = √(βbZpfy/Mcr)
b) λLT = √(βbZpfyMcr)
c) λLT = √(βbZp/Mcr)
d) λLT = √(βbZpfy)

View Answer

Answer: a [Reason:] Non-dimensional slenderness ratio is given by λLT = √(βbZpfy/Mcr), where βb = 1 for plastic and compact sections, βb = Ze/Zp for semi-compact sections, Ze = elastic section modulus, Zp = plastic section modulus, Mcr is elastic critical moment.

4. The check for non- dimensional slenderness ratio is given by
a) λLT = 2.4 √(Zefy/Mcr)
b) λLT > 2 .4 √(Zefy/Mcr)
c) λLT ≤ 1.2 √(Zefy/Mcr)
d) λLT ≥ 1.2 √(Zefy/Mcr)

View Answer

Answer: c [Reason:] The non- dimensional slenderness ratio is given by λLT = √(βbZpfy/Mcr). The check for it is given by λLT ≤ 1.2 √(Zefy/Mcr), where Ze = elastic section modulus, Mcr is elastic critical moment.

5. Which of the following relation is correct?
a) λLT = √(fy/fcr,b)
b) λLT = fy/fcr,b
c) λLT = (fy/fcr,b)2
d) λLT = √(fy fcr,b)

View Answer

Answer: a [Reason:] λLT = √(βbZpfy/Mcr) = √(fy/fcr,b), where βb = 1 for plastic and compact sections, βb = Ze/Zp for semi-compact sections, Ze = elastic section modulus, Zp = plastic section modulus, Mcr is elastic critical moment, fcr,b is extreme compressive elastic buckling stress.

6. The elastic critical moment is given by
a) Mcr = βb fcr,b
b) Mcr = βbZp / fcr,b
c) Mcr = βbZp
d) Mcr = βbZp fcr,b

View Answer

Answer: d [Reason:] The elastic critical moment is given by Mcr = √{[π2EIy/ L2MLT][ GIt + (π2EIw/L2LT)]} = βbZp fcr,b , Iy = moment of inertia about minor axis, Iw = warping constant, It = St. Venant’s constant, G = Shear modulus.

7. Warping constant in elastic critical moment is given by
a) (1+βff Iy h2f
b) (1-βff Iy h2f
c) βf Iy h2f
d) (1-βf)/βf Iy h2f

View Answer

Answer: b [Reason:] Warping constant in elastic critical moment is given by Iw = (1-βff Iy h2f , where βf is ratio of moment of inertia of compression flange to sum of moments of inertia of compression and tension flanges, Iy = moment of inertia about minor axis, hf = centre-to-centre distance between flanges.

8. St. Venant’s constant is given by
a) ∑biti2/3
b) ∑biti2
c) ∑biti3/3
d) ∑biti

View Answer

Answer: c [Reason:] St. Venant’s constant is given by It = ∑biti3/3. For open section (e.g. I -section) : It = 2bft3f/3 + bft3w/3.

9. The value of fcr,b is given by
a) fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}
b) fcr,b = [1.1π2E/(LLT/ry)]{1-1/20[(LLT/ry)/(hf/tf)]}
c) fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}0.5
d) fcr,b = [1.1π2E/(LLT/ry)2]{1-1/20[(LLT/ry)/(hf/tf)]2}0.5

View Answer

Answer: c [Reason:] The value of fcr,b is given by fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}0.5, where ry = radius of gyration about weaker axis, LLT = effective length for lateral-torsional buckling, tf = thickness of flange, hf = centre-to-centre distance between flanges.

Set 5

1. In analytical instrumentation, little circuitry is concerned with the amplification and processing of signals.
a) True
b) False

View Answer

Answer: b [Reason:] In analytical instrumentation, most circuitry is concerned with the amplification and processing of signals. The signals are usually in analog form.

2. Which of the following samples the analog signals to obtain their digital equivalent?
a) Analog to digital converter
b) Electronic counter
c) Comparator
d) Digital to analog converter

View Answer

Answer: a [Reason:] Analog to digital converter samples the analog signals to obtain their digital equivalent. There are different types of analog to digital converters.

3. Which of the following can be used to store information?
a) Gates
b) ADC
c) DAC
d) Flip-flops

View Answer

Answer: d [Reason:] Flip-flops can be used to store information. They are used in sequential circuits.

4. Which of the following is known as nibble?
a) 1 bit
b) 4 bits
c) 8 bits
d) 16 bits

View Answer

Answer: b [Reason:] 4 bit units are known as nibbles. A byte consists of two nibbles.

5. Which of the following is the binary equivalent of the decimal number 10?
a) 1100
b) 1010
c) 1011
d) 1110

View Answer

Answer: b [Reason:] The binary equivalent of the decimal number 10 is 1010. The base of binary numbers is 2.

6. 1001 is a data presented on a set of binary coded decimal output lines. What would be the decimal equivalent of this number in the case of negative logic?
a) 9
b) 6
c) 1
d) 8

View Answer

Answer: b [Reason:] The decimal equivalent of the number 1001 in the case of negative logic is 6. In the case of positive logic, the decimal equivalent will be 9.

7. Given below is the equivalent circuit of a logic gate. Identify the logic gate.
analytical-instrumentation-questions-answers-digital-circuits-q7
a) AND gate
b) OR gate
c) EX-OR gate
d) NOT gate

View Answer

Answer: a [Reason:] The given circuit is the AND gate. The output of the gate is, Y=X1.X2

8. Two switches are in parallel and are connected with a lamp and a supply in series. This represents which of the following logic gate?
a) AND gate
b) OR gate
c) EX-OR gate
d) NOT gate

View Answer

Answer: b [Reason:] Two switches are in parallel and are connected with a lamp and a supply in series. This forms the OR gate.

9. The output will be logic ‘1’ only if all the inputs are ‘0’. Which gate must be used to execute this condition?
a) AND gate
b) NAND gate
c) NOR gate
d) EX-OR gate

View Answer

Answer: b [Reason:] The output will be logic ‘1’ only if all the inputs are ‘0’. NAND gate must be used to execute this condition.

10. INHIBIT gate is ________ gate with an inhibiting input.
a) AND gate
b) OR gate
c) EX-OR gate
d) NOT gate

View Answer

Answer: b [Reason:] INHIBIT gate is OR gate with an inhibiting input. The gate is useful for controlling the inputs with an inhibiting signal.

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