172
87.4k

100+ Structural Design and Drawing Solved MCQs

These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Civil Engineering .

51.

When a sliding joint is made what is interposed at the junction of wall and base?

A. rubber
B. timber
C. plastic
D. soil
Answer» A. rubber
Explanation: a sliding joint is made by interposing rubber or neoprene pads at the junction of the wall and the base and the preload engineering company has developed this type of sliding base in which a vertical water stop is inserted between two rubber strips and in the present state of art, single neoprene pads have also used and the main function of these pads is to allow for free horizontal movement of the wall relative to the base by shear deformation of the rubber joint, which does not exceed a critical value of 30 degrees.
52.

In the fixed base joint the junction is between the tank wall and                      

A. slab
B. footing
C. beams
D. columns
Answer» B. footing
Explanation: in fixed base joint the junction is between the tank wall and footing is the most vulnerable location as far as leakage is
53.

Calculate circumferential prestress of a cylindrical prestressed concrete water tank given that the thickness is 12mm, loss ratio is 0.75, the maximum stress under working pressure is 1n/mm2(Nd value is 720)?

A. 9.4n/mm2
B. 5.6n/mm2
C. 11.2n/mm2
D. 15.2n/mm2
Answer» A. 9.4n/mm2
Explanation: nd = 720, fmin.w = 1, ɳ = 0.75, t
54.

wires of 8mm diameter stressed to 1200n/mm2 are to be used for vertical prestressing?

A. 15
B. 12
C. 8
D. 4
Answer» B. 12
Explanation: 5mm diameter wires stress is 1000n/mm2, 12 wires of 8mm diameter are stressed to 1200n/mm2, fc = (12x1000x200)/(1000) = 2400kn.
55.

Calculate the spacing of 5mm wires having a loss ratio of 0.075, compressive stress is 10.75n/mm2, 5mm diameter wires stress is 1000n/mm2, 12 wires of 8mm diameter are stressed to 1200n/mm2(Nd = 840n/mm2)?

A. 15.4mm
B. 11.6mm
C. 12.4mm
D. 18.5mm
Answer» B. 11.6mm
Explanation: ɳ = 0.075, t = 120mm, internal diameter is 30×103, nd = 840
56.

Calculate the maximum vertical moment due to prestress if given self weight moment is 16.5kn/m, thickness is 0.115m and loss ratio is 0.0075?

A. 15.4
B. 21.5
C. 25.4
D. 2.6
Answer» C. 25.4
Explanation: mw = 16.5kn/m, t = 0.115, ɳ = 0.075
57.

Find vertical prestressing force if characteristic strength is 8.2, wires are stressed at 1000n/mm2, diameter is 150mm?

A. 1500kn
B. 1230kn
C. 4567kn
D. 8967kn
Answer» B. 1230kn
Explanation: fc = 8.2, stress = 1000, diameter
58.

Prestressed concrete tanks have been widely used for the storage of                      

A. gas
B. air
C. fluids
D. water
Answer» C. fluids
Explanation: prestressed concrete tanks have been widely used for storage of fluids, such as water, oil, gas, sewage, granular materials like cement, process liquids and chemicals, slurries and more recently cryogens water storage tanks of large capacity are invariably made of prestressed concrete recent applications include special forms of prestressed concrete tanks, which are triaxially prestressed and serve as containment vessels and biological shields for nuclear reactors.
59.

The metal linear concept in prestressed tanks has proved to be success in case of

A. air tanks
B. water tanks
C. fluid tanks
D. vapour tanks
Answer» B. water tanks
Explanation: the metal linear concept has proved so successful that it is being increasingly used in america, even for large water tanks and in the case of sanitary structures like sludge digestion tanks, spherical shapes are preferred and for practical reasons, the tank is made up of a top
60.

Calculate minimum wall thickness given a cylindrical prestressed water tank of internal diameter 30m over a depth of 7.5m and the permissible compressive stress at transfer is 13n/mm2 and the maximum compressive stress under working pressure is 1n/mm2 and the loss ratio is 0.75?

A. 43.8
B. 82.3
C. 64.5
D. 90.4
Answer» B. 82.3
Explanation: d = 30m, h = 7.5m, nd = 720n/mm, ɳ = 0.75, fct = 13n/mm2, pressure is 1n/mm2
61.

Calculate vertical prestressing force if wires of 5mm diameter with an initial stress of 1000n/mm2 are available for circumferential winding and Freyssinet cables made up of 12 wires of 8mm diameter stressed to 1200n/mm2 are to be used for vertical prestressing?

A. 15
B. 12
C. 8
D. 4
Answer» B. 12
Explanation: 5mm diameter wires stress is 1000n/mm2, 12 wires of 8mm diameter are stressed to 1200n/mm2, fc = (12x1000x200)/(1000) = 2400kn.
62.

What is beam?

A. structural member subjected to transverse loads
B. structural member subjected to axial loads only
C. structural member subjected to seismic loads only
D. structural member subjected to transverse loads only
Answer» A. structural member subjected to transverse loads
Explanation: beam is a structural member subjected to transverse loads that is loads perpendicular to its longitudinal axis. the mode of deflection of beam is primarily by bending.
63.

Structural members subjected to bending and large axial compressive loads are known as

A. strut
B. purlin
C. beam-column
D. lintel
Answer» C. beam-column
Explanation: structural members subjected to bending accompanied by large axial compressive loads at the same time are known as beam-column. a beam-column differs from column only by presence of eccentricity of load application, end moment, transverse load.
64.

Members used to carry wall loads over wall openings are called

A. purlin
B. rafter
C. girder
D. lintels
Answer» D. lintels
Explanation: lintels are beam members used to carry wall loads over wall openings for doors, windows, etc.
65.

Load transfer by a beam is primarily by

A. bending only
B. shear only
C. bending and shear
D. neither bending nor shear
Answer» C. bending and shear
Explanation: the load transfer by beam is primarily by bending and shear. the mode of deflection of beam is primarily by bending.
66.

What are spandrels?

A. exterior beams at floor level of buildings
B. interior beams at floor level of buildings
C. exterior columns
D. interior columns
Answer» A. exterior beams at floor level of buildings
Explanation: spandrels are exterior beams at floor level of buildings, which carry part of floor load and exterior wall.
67.

Members used in bridges parallel to traffic are called

A. spandrel
B. stringers
C. purlin
D. joist
Answer» B. stringers
Explanation: stringers are members used in bridges parallel to traffic to carry the deck slab. they will be connected by transverse floor beams.
68.

Which of the following statement is correct?

A. beams are termed as fixed beams when end condition do not carry end moments
B. beams are termed as simply supported beams when ends are rigidly connected to other members
C. beams are termed as fixed beams when ends are rigidly connected to other members
D. beams are termed as continuous beams when they do not extend across more than two support
Answer» C. beams are termed as fixed beams when ends are rigidly connected to other members
Explanation: beams may be termed as simply supported beams when end condition do not carry any end moments from any continuity developed by connection. a beam is called continuous beam when it extends continuously across more than two supports. a fixed beam has its ends rigidly connected to other members, so that moments can be carried across the connection.
69.

Complex stresses may occur when

A. loads are inclined to principal axes
B. loads are along principal axes
C. symmetrical section are used
D. small values of shear and bending moment occur at section
Answer» A. loads are inclined to principal axes
Explanation: complex stresses may arise when loads are inclined to principal axes, when unsymmetrical sections are used or when large values of shear and bending moment occur at section.
70.

Simple bending takes place if

A. loading passes above shear centre for single symmetric open section
B. loading passes below shear centre for single symmetric open section
C. loading plane coincides with one of the principal planes of doubly symmetric section
D. loading plane do not coincide with one of the principal planes of doubly symmetric section
Answer» C. loading plane coincides with one of the principal planes of doubly symmetric section
Explanation: simple bending takes place if loading plane coincides with one of the principal planes of doubly symmetric section such as i-section or in case of singly symmetric open section such as channel section, the loading passes through shear centre and is parallel to the principal plane. unsymmetrical bending occurs if loading does not pass through shear centre.
71.

Which of the following buckling does not occur in beam?

A. lateral buckling of whole beam
B. local buckling of web
C. local buckling of flanges
D. longitudinal buckling of web
Answer» B. local buckling of web
Explanation: buckling may take place in many ways : (i) lateral buckling of whole beam between supports, (ii) local buckling of flanges, (iii) longitudinal buckling of web and buckling in depth direction under concentrated loads.
72.

The diameter of longitudinal bars of a column should never be less than?

A. 12 mm
B. 6 mm
C. 10 mm
D. 8 mm
Answer» A. 12 mm
Explanation: minimum diameter of longitudinal bar in rcc column shall not be less than 12mm (is456:2000, cl 26.5.3.1 d). indian standards specify 12mm as the least diameter of a vertical bar and 5mm as the least diameter of lateral bar or stirrup.
73.

The number of treads in a flight is equal to

A. risers in the flight
B. risers plus one
C. risers minus one
D. risers plus three
Answer» C. risers minus one
Explanation: it is often not simply the sum of the individual tread lengths due to the nosing overlapping between treads. if there are n steps, the total run equals n-1 times the going: the tread of the last step is part of a landing and is not counted.
74.

A foundation rests on                      

A. base of the foundation
B. foundation soil
C. subgrade
D. foundation soil and subgrade
Answer» D. foundation soil and subgrade
Explanation: a foundation (or, more commonly, base) is the element of an architectural structure which connects it to the ground, and transfers loads from the structure to the ground. foundations are generally considered either shallow or deep. foundation engineering is the application of soil mechanics and rock mechanics (geotechnical engineering) in the design of foundation elements of structures.
75.

For initial estimate for a beam design, the width is assumed?

A. 1/10th of span
B. 1/30th of span
C. 1/15th of span
D. 1/5th of span
Answer» B. 1/30th of span
Explanation: design codes prescribe beam width limitations to minimise the shear lag effect on the formation of full-width plastic hinges and achieving the expected capacity. however, owing to insufficient experimental and analytical studies, empirical design formulas for the beam width limitation, with
76.

High strength concrete is used in prestressed member?

A. to ovecome bursting stresses at the ends
B. to provide high bond stresses
C. to overcome cracks due to shrinkage
D. to overcome bursting stresses, provide high bond stresses and overcome cracks
Answer» D. to overcome bursting stresses, provide high bond stresses and overcome cracks
Explanation: the primary difference between high-strength concrete and normal- strength concrete relates to the compressive strength that refers to the maximum resistance of a concrete sample to applied pressure.
77.

The advantage of reinforced concrete is due to                        

A. monolithic character
B. moulding in any desired shape
C. fire-resisting and durability
D. monolithic character, moulding any shape and fire-resisting
Answer» D. monolithic character, moulding any shape and fire-resisting
Explanation: reinforced concrete (rc) is a composite material in which concrete’s relatively low tensile strength and ductility are counteracted by the inclusion of reinforcement having higher tensile strength or ductility. the reinforcement is usually, though not necessarily, steel reinforcing bars (rebar) and is usually embedded passively in the concrete before the concrete sets.
78.

Cracking of the concrete section is nearly impossible to prevent.

A. true
B. false
Answer» A. true
Explanation: however, the size and location of cracks can be limited and controlled by appropriate reinforcement, control joints, curing methodology and concrete mix design. cracking can allow moisture to penetrate and corrode the reinforcement. this is a serviceability failure in limit state design.
79.

The architect is usually the lead designer on buildings, with a structural engineer employed as a sub-consultant.

A. false
B. true
Answer» B. true
Explanation: the degree to which each discipline actually leads the design depends heavily on the type of structure. many structures are structurally simple and led by architecture, such as multi-storey office buildings and housing, while other structures, such as tensile structures, shells and gridshells are heavily dependent on their form for their strength, and the engineer may have a more significant influence on the form, and hence much of the aesthetic, than the architect.
80.

Why plate girder is preferred over truss girder?

A. plate girder requires costly maintenance
B. higher vertical clearance required for plate girder than truss girder
C. cost of fabrication of plate girder is high
D. cost of fabrication of truss girder is high
Answer» D. cost of fabrication of truss girder is high
Explanation: when load is heavier and span is also large, either plate girder or truss girder can be used. but, plate girder is preferred because of the disadvantages of truss girder. the disadvantages of truss girder are higher cost of fabrication and erection, problem of vibration and impact, requirements of higher vertical clearance and costly maintenance.
81.

Bending resistance of plate girders can be increased by

A. decreasing distance between flanges
B. increasing distance between flanges
C. reducing distance between flanges to half
D. bending resistance cannot be increased
Answer» B. increasing distance between flanges
Explanation: plate girders are built-up flexural members. their bending resistance can be increased by increasing the distance between flanges. this also increases the shear resistance as web area increases.
82.

Which of the following is economical if depth is limited and loads are too large?

A. rolled section beam
B. truss girder
C. welded box plate girder
D. bolted box plate girder
Answer» C. welded box plate girder
Explanation: when the loads and span are large, plate girder sections either with riveted/bolted connections or welded connections may be provided. the number of flange plates can be increased depending upon the moment to be resisted. if depth is limited and loads are too large, welded box plate girder is provided. a box girder with riveted/bolted connections can b e provided but it is too costly as compared to welded one. box girders have great resistance to lateral buckling.
83.

An ideal bolted plate girder section consists of

A. flange angles and cover plates for both compression flange and tension flange
B. flange angles and cover plates for compression flange and only flange angle for tension flange
C. only flange angle for compression flange and flange angles and cover plates for tension flange
D. flange angles for both compression flange and tension flange
Answer» B. flange angles and cover plates for compression flange and only flange angle for tension flange
Explanation: an ideal bolted plate girder section consists of flange angles and cover plates for compression flange and only flange angle for tension flange. the various elements of riveted/bolted/welded plate girder are : web plate, flange angles with or without cover plates for riveted/bolted plate girder and only flange angles for welded plate girder, stiffeners – bearing, transverse and longitudinal, splices for web and flange.
84.

The modes of failure of plate girder are

A. by yielding of compression flange only
B. by buckling of tension flange only
C. by yielding of tension flange and buckling of compression flange
D. by yielding of compression flange and buckling of tension flange
Answer» C. by yielding of tension flange and buckling of compression flange
Explanation: the limit states considered for plate girder are yielding of tension flange and buckling of compression flange. the compression flange buckling can take place in various ways, such as vertical buckling into the web or flange local buckling. flange buckling can also be caused due to lateral- torsional buckling.
85.

At high shear locations in the girder web, principal plane will be              to longitudinal axis of member

A. inclined
B. parallel
C. perpendicular
D. coincides
Answer» A. inclined
Explanation: at high shear locations in the girder web, usually near supports and neutral axis, the principal planes will be inclined to longitudinal axis of the member. the principal stresses will be diagonal tension and diagonal compression along the principal planes.
86.

Which of the following causes web buckling in plate girder?

A. diagonal tension
B. diagonal compression
C. diagonal tension and diagonal compression
D. neither diagonal tension nor diagonal compression
Answer» B. diagonal compression
Explanation: the principal planes will be inclined to longitudinal axis of the member at high shear locations in the girder web. along the principal planes, the principal stresses will be diagonal tension and diagonal compression. diagonal tension does not cause
87.

Which of the following statement is correct for reducing web buckling due to diagonal compression?

A. not providing web stiffeners to increase shear strength
B. providing web stiffeners to reduce shear strength
C. increasing depth-to-thickness ratio
D. reducing depth-to-thickness ratio
Answer» D. reducing depth-to-thickness ratio
Explanation: diagonal compression causes web to buckle in the direction perpendicular to its action. this problem can be solved by any of the following ways : (i) reduce depth- to-thickness ratio of web such that problem is eliminated, (ii) provide web stiffeners to form panels that would enhance shear strength of web, (iii) provide web stiffeners to form panels that would develop tension field action to resist diagonal compression.
88.

Which of the following is correct during tension field action?

A. web can resist diagonal compression
B. horizontal component of diagonal compression is supported by flanges
C. vertical component of diagonal compression is supported by flanges
D. vertical component of diagonal compression is supported by stiffeners
Answer» B. horizontal component of diagonal compression is supported by flanges
Explanation: as web begins to buckle , the web loses its ability to resist diagonal compression. the diagonal compression is transferred to transverse stiffeners and flanges. the vertical component of diagonal compression is supported by stiffeners and horizontal component is resisted by flanges. the web resists only diagonal tension and this behaviour of web is called tension field action.
89.

Which of the statement is not true about intermediate stiffeners?

A. they reduce shear capacity of web
B. they improve shear capacity of web
C. they can be used to develop tension field action
D. their main purpose is to provide stiffness to the web
Answer» A. they reduce shear capacity of web
Explanation: intermediate stiffeners can be used to develop tension field action and improve shear capacity of web. the main purpose of these stiffeners is to provide stiffness to the web rather than to resist the applied loads. additional stiffeners called bearing stiffeners are provided at points of concentrated loads to protect the web from the direct compressive loads.
90.

Which of the following is correct regarding gantry girders?

A. it is laterally supported except at the columns
B. it is subjected to impact load
C. it should not be analysed for unsymmetrical bending
D. it is not subjected to longitudinal load
Answer» B. it is subjected to impact load
Explanation: gantry girder are different from beams in buildings. it is generally laterally unsupported except at the columns. it is subjected to impact load. it must be analysed for unsymmetrical bending because of lateral thrust from the starting and stopping of the crab. it is subjected to longitudinal load due to starting and stopping of crane bridge itself. they are always simply supported.
91.

Which of the following loads are not considered in the design of gantry girders?

A. longitudinal loads
B. gravity loads
C. lateral loads
D. wind loads
Answer» D. wind loads
Explanation: the loads considered in the design of gantry girders are vertical loads or gravity loads, longitudinal loads, lateral loads and impact loads. the vertical force is the reaction from crane girder, acting vertically downward. the longitudinal thrust is due to starting and stopping of crane acting in longitudinal direction. the lateral thrust is due to starting and stopping of the crab acting horizontally normal to the gantry girder.
92.

The wheel load transferred from trolley to gantry girder is given by

A. w1 = [wt(lc+l1)]/(2lc)
B. w1 = [wt(lc-l1)](2lc)
C. w1 = [wt(lc-l1)]/(2lc)
D. w1 = [wt(lc+l1)]/( lc)
Answer» C. w1 = [wt(lc-l1)]/(2lc)
Explanation: since trolley moves on the crane girder along the span of truss, its weight is transferred to the crane wheels as the axle load and finally to gantry girder. the wheel load transferred from trolley to gantry girder is given by w1 = [wt(lc-l1)]/(2lc), where w1 is load of each wheel on gantry girder, wt is weight of trolley or crab car, lc is distance between gantry giders, l1 is distance between centre of gravity of trolley and gantry.
93.

The maximum wheel load is obtained when

A. crane crab is farthest to gantry girder
B. crane crab is closest to gantry girder
C. crane crab is not attached
D. crane crab is at mid span
Answer» B. crane crab is closest to gantry girder
Explanation: the maximum wheel load is obtained when crane crab is closest to gantry girder. the crab in such position on the crane girder gives maximum reaction on the gantry girder. the vertical reaction of crane girder is transferred through its two wheels on to the gantry girder. therefore, the maximum wheel load is half of this reaction. this maximum wheel load is then increased for impact and used for design of gantry girder.
94.

The bending moment due to dead load of girder is maximum at

A. one-third distance at span
B. two-third distance at span
C. end of span
D. centre of span
Answer» D. centre of span
Explanation: the bending moment considered in the design of gantry girder are the bending moment due to maximum wheel loads (with impact) and the bending moment
95.

The minimum recommended rise of trusses with Galvanised Iron sheets is

A. 1 in 12
B. 1 in 6
C. 1 in 10
D. 1 in 18
Answer» B. 1 in 6
Explanation: the pitch of truss depends upon the roofing material. the minimum recommended rise of trusses with galvanised iron sheets is 1 in 6 and with asbestos cement sheets is 1 in 12.
96.

The economic spacing of roof truss depends on

A. cost of purlins only
B. cost of purlins and cost of roof covering
C. dead loads
D. cost of roof covering and dead loads
Answer» B. cost of purlins and cost of roof covering
Explanation: the economic spacing of the truss is the spacing that makes the overall cost of trusses, purlins, roof coverings, columns, etc. the minimum. it depends upon the relative cost of trusses, purlins, roof coverings, spacing of columns, etc. if the spacing is large, the cost of these trusses per unit area decreases but the cost of purlin increases. but if the spacing of trusses is small, the cost of trusses per unit area increases. roof coverings cost more if the spacing of trusses is large.
97.

Which of the following is true for economic spacing?

A. cost of trusses should be equal to twice the cost of purlins
B. cost of trusses should be equal to twice the cost of purlins minus cost of roof coverings
C. cost of trusses should be equal to the cost of purlins plus cost of roof coverings
D. cost of trusses should be equal to twice the cost of purlins plus cost of roof coverings
Answer» D. cost of trusses should be equal to twice the cost of purlins plus cost of roof coverings
Explanation: for economic spacing of roof trusses, the cost of trusses should be equal to twice the cost of purlins plus cost of roof coverings. this equation is used for checking the spacing of trusses and not for design of trusses.
98.

Which of the following load combination is not considered for design of roof trusses?

A. dead load + crane load
B. dead load + wind load
C. dead load + earthquake load
D. dead load + live load + wind load
Answer» C. dead load + earthquake load
Explanation: earthquake loads are not significant for roof trusses because of the small self weights. the following load combinations can be considered : (i) dead load + snow load, (ii) dead load + partial/full live load, (iii) dead load + live load + internal positive air pressure, (iv) dead load + live load + internal suction air pressure, (v) dead load + live load + wind load.
99.

Live load for roof truss should not be less than

A. 0.4kn/m2
B. 0.2kn/m2
C. 0.75kn/m2
D. 0.8kn/m2
Answer» A. 0.4kn/m2
Explanation: the live load for roof truss should not be less than 0.4kn/m2. for roof slopes ≤ 10o and access provided, the live load to be taken is 1.5kn/m2 of plan area. for roof slopes > 10o and access is not provided , the live load to be taken is 0.75kn/m2 of plan area.
100.

Simple connections are used to transmit

A. forces
B. moments
C. stresses
D. both force and moment
Answer» A. forces
Explanation: simple connection is required to transmit force only and there may not be any moment acting on the group of connectors. this connection may be capable of transmitting some amount of moment.

Done Studing? Take A Test.

Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.