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# Multiple choice question for engineering

## Set 1

1. The design bending strength of laterally unsupported beams is governed by
a) torsion
b) bending
c) lateral torsional buckling
d) yield stress

Answer: c [Reason:] Beams with major axis bending and compression flange not restrained against lateral bending (or inadequate lateral support) fail by lateral torsional buckling before attaining their bending strength.

2. The effect of lateral-torsional buckling need not be considered when
a) λLT ≤ 0.4
b) λLT ≥0.4
c) λLT > 0.8
d) λLT = 0.8

Answer: a [Reason:] The effect of lateral-torsional buckling need not be considered when λLT ≤ 0.4, where λLT is the non dimensional slenderness ratio for lateral torsional buckling.

3. The bending strength of laterally unsupported beams is given by
a) Md = βbZp /fbd
b) Md = βb /Zpfbd
c) Md = βbZp
d) Md = βbZpfbd

Answer: d [Reason:] The bending strength of laterally unsupported beams is given by Md = βbZpfbd, where βb is a constant, Zp is plastic section modulus, fbd is design bending compressive stress.

4. The value of βb in the equation of design bending strength of laterally unsupported beams for plastic sections is
a) 0.5
b) 2.5
c) 1.0
d) 1.5

Answer: c [Reason:] The value of βb in the equation of design bending strength of laterally unsupported beams for plastic and compact sections is 1.0. This constant depends on elastic and plastic section modulus for semi-compact sections.

5. The value of βb in the equation of design bending strength of laterally unsupported beams for semi-compact sections is
a) Ze/Zp
b) ZeZp
c) Zp/Ze
d) Zp

Answer: a [Reason:] The value of βb in the equation of design bending strength of laterally unsupported beams for semi-compact sections is Ze/Zp, where Ze is elastic section modulus, Zp is plastic section modulus.

6. The value of design bending compressive stress fbd is
a) XLT fy
b) XLT fy /fy
c) XLT fy fy
d) XLT /fy

Answer: b [Reason:] The value of design bending compressive stress fbd is XLT fy /fy, where XLT is bending stress reduction factor to account for lateral torsional buckling, fy is yield stress, fy is partial safety factor for material (=1.10).

7. The bending stress reduction factor to account for lateral buckling is given by
a) XLT = 1/{φLT + (φ2LT – λ2LT)}
b) XLT = 1/{φLT – (φ2LT + λ2LT)}
c) XLT = 1/{φLT – (φ2LT + λ2LT)0.5}
d) XLT = 1/{φLT + (φ2LT – λ2LT)0.5}

Answer: d [Reason:] The bending stress reduction factor to account for lateral buckling is given by XLT = 1/{φLT + (φ2LT – λ2LT)0.5}, where φLT depends upon imperfection factor and non dimensional slenderness ratio, λLT is non dimensional slenderness ratio.

8. The value of φLT in bending stress reduction factor is given by
a) φLT = [ 1 – αLTLT + 0.2) + λ2LT].
b) φLT = [ 1 + αLTLT – 0.2) + λ2LT].
c) φLT = 0.5 [ 1 – αLTLT + 0.2) + λ2LT].
d) φLT = 0.5 [ 1 + αLTLT – 0.2) + λ2LT].

Answer: d [Reason:] The value of φLT in bending stress reduction factor is given by φLT = 0.5 [ 1 + αLTLT – 0.2) + λ2LT], where αLT is imperfection factor, λLT is non dimensional slenderness ratio.

## Set 2

1. For the calculation of net area of flat with staggered bolts, the area to be deducted from gross area is :
a) nd
b) n’p2t/8g
c) ndt – n’p2t/4g
d) nd + n’p2t/4g

Answer: c [Reason:] The net area of flat with staggered hole is given by : A = (b – ndh + n’p2/4g)t, where b = width of plate, n = number of holes in zig-zag line, n’ = number of staggered pitches, p = pitch distance, g= gauge distance, t = thickness of flat.

2. What is the net section area of steel plate 40cm wide and 10mm thick with one bolt if diameter of bolt hole is 18mm?
a) 38.2 cm2
b) 20 cm2
c) 240 mm2
d) 480 mm2

Answer: a [Reason:] b = 40cm = 400mm, t = 10mm, dh = 18mm Net section area = 400×10 – 16×10 = 3820mm2 = 38.2 cm2.

3. Which section to be considered in the design for the net area of flat? a) 1-5-6-3
b) 2-7-4
c) 1-5-7-4
d) 1-5-7-6-3

Answer: d [Reason:] The section giving minimum area of plate is considered for design. So, section 1-5-7-6-3 is used for net area of flat.

4. Calculate the minimum effective net area for the given section (300mm width, 10mm thick) connected to a 10 mm thick gusset plate by 18mm diameter bolts. a) 2796mm2
b) 2681mm2
c) 2861mm2
d) 3055mm2

Answer: b [Reason:] B = 300mm, t = 10mm, dh = 18+2 =20mm, n = 3, n’ = 1, p = 75mm, g = 50mm Effective net area = (B-ndh+n’p2/4g)t = (300 – 3×20 + 1×752/4×50)x10 = 2681.25 mm2.

5. What is the net area for the plate 100 x 8 mm bolted with a single bolt of 20mm diameter in case of drilled hole ?
a) 624 mm2
b) 756 mm2
c) 800 mm2
d) 640 mm2

Answer: d [Reason:] In case of drilled hole, dh = 20mm Net Area An = Ag – dht = 100 x 8 – 20 x 8 = 640mm2.

6. Determine the effective net area for angle section ISA 100 x 75 x 12 mm, when 100mm leg is connected to a gusset plate using weld of length 140mm.
a) 1795 mm2
b) 1812 mm2
c) 1956 mm2
d) 2100 mm2

Answer: c [Reason:] Net area of connected leg, Anc = (100 – 12/2) x 12 = 1128 mm2 Net area of outstanding leg, Ago = (75 – 12/2) x 12 = 828 mm2 Total net area = 1128 + 828 = 1956 mm2.

7. Calculate the value of β for the given angle section ISA 150x115x8mm of Fe410 grade of steel connected with gusset plate : Length of weld = 150mm
a) 0.89
b) 0.75
c) 0.5
d) 1

Answer: a [Reason:] w=115mm, t=8mm, b=115mm, Lc=150mm, fy=250MPa, fu=410MPa β = 1.4 – [0.076 (w/t)(fy/fu)(bs/Lc)] = 1.4 – [ 0.076 x (115/8) x (250/410) x (115/150)] = 0.89 (>0.7) .

8. Calculate the tensile strength due to gross section yielding of an angle section 125 x 75 x 10mm of Fe410 grade of steel connected with a gusset plate.
a) 780 kN
b) 586.95 kN
c) 432.27 kN
d) 225.36 kN

Answer: c [Reason:] fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25, For ISA 125 x 75 x 10mm : gross area Ag = 1902 mm2 Tensile strength due to gross section yielding, Tdg = Agfym0 = 1902 x 250 x 10-3 / 1.1 = 432.27 kN.

9. A single unequal angle 100 x 75 x 10 of Fe410 grade of steel is connected to a 10mm thick gusset plate at the ends with six 16mm diameter bolts with pitch of 40mm to transfer tension. Find the tensile strength due to net section rupture if gusset is connected to 100mm leg.
a) 526.83 kN
b) 385.74 kN
c) 450.98 kN
d) 416.62 kN

Answer: d [Reason:] dh=18 mm, fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25 Anc = (100 – 10/2 – 18) x 10 = 770 mm2, Ag0 = (75 – 10/2) x 10 = 700 mm2 β = 1.4 – 0.076(w/t)(fy/fu)(bs/Lc) = 1.4 – 0.076 [(75-5)/10] [250/410] [{(75-5)+(100-40)}/{40×5}] = 1.19 > 0.7 and < 1.44[(410/250)(1.1/1.25)] (=2.07) Tdn = 0.9fuAncm1 + βAg0fy m0 = [0.9x410x770/1.25 + 1.19x700x250/1.1] x 10-3 = 416.62 kN.

10. Determine block shear strength of tension member shown in figure if grade of steel is Fe410. a) 309.06 kN
b) 216.49 kN
c) 258.49 kN
d) 326.54 kN

Answer: b [Reason:] fu = 410 MPa, fy = 250 MPa, γm0 = 1.1, γm1 = 1.25 Avg = ( 1×100 + 50 ) x 8 = 1200 mm2 Avn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm2 Atg = 35 x 8 = 280 mm2 Atn = (35 – 1/2 x 20) x 8 = 200 mm2 Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfum1) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1) = [ (0.9x1440x410 / 1.25x√3) + (280×250/1.1) ] x 10-3 = 309.06 kN Block shear strength of tension member is 216.49 kN.

## Set 3

1. The design nominal strength of fillet weld is given by ____________
a) fu
b) √3 fu
c) fu/√3
d) fu/(1.25 x √3)

Answer: c [Reason:] Design nominal strength of fillet weld = fu/√3, where fu is smaller of ultimate stress of weld or parent metal.

2. When welds are subjected to compressive or tensile or shear force alone, the stress in weld is given by :
a) P/ttlw
b) Ptt/lw
c) Pttlw
d) Plw/tt

Answer: a [Reason:] When welds are subjected to compressive or tensile or shear force alone, the stress in weld is given by q = P/ttlw , where q=shear stress in N/mm2, P = force transmitted, tt = effective throat thickness of weld in mm, lw = effective length of weld in mm.

3. When fillet welds are subjected to combination of normal and shear stress, the equivalent stress is given by :
a) √(fa2+q2)
b) √(fa2+2q2)
c) √(3fa2+q2)
d) √(fa2+3q2)

Answer: d [Reason:] When fillet welds are subjected to combination of normal and shear stress, the equivalent stress is given by fe = √(fa2+3q2), where fa = normal stresses, compression or tension, due to axial force or bending moment, q = shear stress due to shear force or tension.

4. Two plates of 12mm and 16mm thickness are to be joined by groove weld. The joint is subjected to factored tensile load of 400kN. Due to some reasons the effective length of weld that could be provided was 150mm only. What is the safety of joint if single-V groove weld is provided?
a) Safe
b) Unsafe
c) Unsafe, but adequate
d) Safe, but adequate

Answer: b [Reason:] Lw = 150mm, Throat thickness, te = 5×12/8 = 7.5 mm Strength of weld = Lwtefy/1.25 = 150×7.5x250x10-3/1.25 = 225kN < 400kN. Therefore joint is unsafe and inadequate.

5. What is the effective throat thickness dimension of 10mm fillet weld made by shielded metal arc welding and submerged arc welding?
a) 4.6mm, 5mm
b) 5mm, 4.6mm
c) 8.6mm, 7mm
d) 7mm, 8.6mm

Answer: d [Reason:] a) Using shielded metal arc welding process, effective throat thickness = 0.7a = 0.7×10 = 7mm b) Using submerged arc welding (we get better penetration than shielded metal arc welding): a = 10+2.4 = 12.4mm effective throat thickness = 0.7a = 0.7×12.4 = 8.6mm.

6. What is the strength of weld per mm length used to connect two plates of 10mm thickness using a lap joint?
a) 795.36 N/mm
b) 295.5N/mm
c) 552.6 N/mm
d) 487.93 N/mm

Answer: a [Reason:] Minimum size of weld = 3mm Maximum size of weld = 10-1.5 = 8.5mm Assume weld size = 6mm Effective throat thickness, te = 0.7 x 6 = 4.2mm Strength of weld = te [fu/(√3 x 1.25)] = 410 x 4.2 /(√3 x 1.25) = 795.36 N/mm.

7. What is the overall length of fillet weld to be provided for lap joint to transmit a factored load of 100kN? Assume site welds and width and thickness of plate as 75mm and 8mm respectively, Fe410 steel.
a) 500mm
b) 382mm
c) 201mm
d) 468mm

Answer: c [Reason:] Minimum size of weld = 3mm Maximum size of weld = 8-1.5 = 6.5mm Assume size of weld = 5mm Effective throat thickness = 0.7 x 5 = 3.5mm Strength of weld = 3.5×410/(√3 x1.5) = 552.33 N/mm Required length of weld = 100 x 103/552.33 = 181.05 mm length to be provided on each side = 181/2 = 90.5mm End return = 2×5 = 10mm Overall length = 2 x (90.5 + 2 x 5) = 201mm.

8. The clear spacing between effective lengths of intermittent welds should not be ______
a) less than 16t in case of tension joint, where t is thickness of thinner plate
b) less than 12t in case of compression joint, where t is thickness of thinner plate
c) less than 20t in case of tension joint, where t is thickness of thinner plate
d) less than 20t in case of compression joint, where t is thickness of thinner plate

Answer: a [Reason:] The clear spacing between effective lengths of intermittent welds should not be less than 16 times and 12 times the thickness of thinner plate jointed in case of tension joint and compression joint respectively, and should never be more than 200mm.

## Set 4

1. Which IS Code is used for designing a structure considering earthquake loads?
a) IS 800
b) IS 875
c) IS 1893
d) IS 456

Answer: c [Reason:] IS 1893(Part 1) is used for designing a structure considering earthquake loads. For all other loads like dead load, wind load, IS 875 is used for design.

2. What is the load factor considered for steel structures when combination of dead load and earthquake load is considered?
a) 1.5
b) 1.3
c) 1.2
d) 1.7

Answer: d [Reason:] In plastic design of steel structures, load factor is 1.7 when combination of dead load and earthquake load is considered i.e. 1.7(DL + EL) or 1.7(DL – EL).

3. What is the load factor considered for steel structures when combination of dead load, imposed load and earthquake load is considered?
a) 1.5
b) 1.3
c) 1.2
d) 1.7

Answer: b [Reason:] In plastic design of steel structures, load factor is 1.3 when combination of dead load, imposed load and earthquake load is considered i.e. 1.3(DL +/- IL +/- EL ).

4. For earthquake loads, axially loaded members have to resist ________________
a) tension only
b) compression only
c) both tension and compression
d) bending moment

Answer: c [Reason:] Earthquake are cyclic and induce reversal of stresses. Hence as per seismic philosophy, axially loaded members have to resist both tension and compression.

5. For earthquake loads, beams are designed to resist ________________
a) tension only
b) positive and negative bending moments
c) compression only
d) torsion

Answer: b [Reason:] Beams are designed to resist positive and negative bending moments at the same section for earthquake loads.

6. Structures are designed for seismic forces which is ____ than expected seismic force.
a) lesser
b) greater
c) equal to
d) seismic forces are not considered

Answer: a [Reason:] Structures are designed for seismic forces which is less than expected seismic force under strong earthquakes, if the structures were to remain linearly elastic.

7. Which of the following factors does not influence earthquake resistance design?
a) geographical location of structure
b) wind of location
c) site soil
d) strength of structure

Answer: b [Reason:] The factors which influence earthquake resistance design are : (i) geographical location of structure, (ii) site soil and foundation condition, (iii) importance of structure, (iv) dynamic characteristics of structure such as strength, stiffness, ductility and energy dissipation factor.

8. Structures should be designed such that ___________
a) Minor and frequent earthquakes can collapse the structure
b) Moderate earthquakes can cause damage to the structure
c) Major earthquakes should not cause any damage to the structure and the structure should be functional
d) Minor earthquake should not cause any damage to the structure and the structure should be functional

Answer: d [Reason:] As per IS 1893, the following seismic philosophy is adopted (i) Minor and frequent earthquakes should not cause any damage to the structure (ii) Moderate earthquake should not cause significant structural damage but could have some non structural damage(the structure should become operational once the damage is repaired) (iii) Major and infrequent earthquake should not cause collapse( structure will be dysfunctional for further use, but will stand so that people can be evacuated and property can be recovered).

9. Which of the following assumption is correct for earthquake design resistant structure?
a) Earthquake will not occur simultaneously with wind
b) Earthquake will occur simultaneously with maximum flood
c) Earthquake will occur simultaneously with maximum sea waves
d) Earthquake will occur simultaneously with wind

Answer: a [Reason:] As per IS 1893, it is assumed that earthquake is not likely to occur simultaneously with wind or maximum flood or maximum sea waves.

10. Which analysis is used to obtain design seismic force?
a) Elastic Analysis
b) Plastic Analysis
c) Dynamic Analysis
d) Both elastic and plastic analysis

Answer: c [Reason:] Dynamic analysis is used to obtain design seismic force, and its distribution to different level along the height of the building and to various lateral load resisting members.

11. Which of the following relation is correct for design horizontal seismic coefficient?
a) A = ZISa*2Rg
b) A = ZISa/2Rg
c) A = ZISa-2Rg
d) A = ZISa+2Rg

Answer: b [Reason:] Design horizontal seismic coefficient is given by A = ZISa/2Rg, where Z is zone factor, R is response reduction factor, I is importance factor, Sa/g is average response acceleration coefficient.

12. What is structural response factor?
a) factor denoting the acceleration response spectrum of the structure subjected to earthquake ground vibrations
b) factor by which the actual base shear force is reduced
c) factor to obtain the design spectrum
d) factor used to obtain the design seismic force

Answer: a [Reason:] Structural response factor (Sa/g) is the factor denoting the acceleration response spectrum of the structure subjected to earthquake ground vibrations. It depends on natural period of vibration and damping of the structure. Response Reduction Factor(R) is the factor by which the actual base shear force is reduced. Zone Factor(Z) is factor to obtain the design spectrum. Importance Factor (I) is the factor used to obtain the design seismic force.

13. Wind pressure acting normal to individual is element or claddity unit is _________
a) F = [ (Cpe – Cpi)A/pd].
b) F = [ (Cpe + Cpi)A/pd].
c) F = [ (Cpe – Cpi)Apd].
d) F = [ (Cpe – Cpi)/Apd].

Answer: c [Reason:] F = (Cpe – Cpi)A pd , where F=net wind force on element, Cpi= internal pressure coefficient, Cpe=external pressure coefficient, A=surface area of element, pd=design wind pressure.

14. Internal pressure coefficient in a building is positive if acting from ________ and external pressure coefficient in a building is positive if acting from ___________
a) outside to inside, inside to outside
b) inside to outside, outside to inside
c) outside to inside, outside to inside
d) inside to outside, inside to outside

Answer: b [Reason:] Internal pressure coefficient in a building is positive if acting from inside to outside and external pressure coefficient in a building is positive if acting from outside to inside. The pressure depends on degree of permeability of cladding and direction of wind.

## Set 5

1. For very short compression member
a) failure stress will be greater than yield stress
b) failure stress will be less than yield stress
c) failure stress will equal yield stress
d) failure stress will be twice the yield stress

Answer: c [Reason:] For very short compression members, the failure stress will the equal yield stress and no buckling will occur.

2. The length of member should be _________ for a short column
a) L ≤ 88.5r
b) L ≥ 88.5r
c) L ≥ 125r
d) L > 150r

Answer: a [Reason:] For a member to be classified as short column, length of member should be L ≤ 88.5r , where r is radius of gyration. The slenderness ratio of column defines the column as short or long column.

3. Long compression members will ______
a) not buckle
b) buckle inelastically
c) buckle plastically
d) buckle elastically

Answer: d [Reason:] Long compression members will buckle elastically where axial buckling stress remains below proportional limit.

4. Which of the following is true about intermediate length compression members?
a) members will fail by yielding only
b) members will fail by both yielding and buckling
c) their behaviour is elastic
d) all fibres of the members will be elastic during failure

Answer: b [Reason:] For intermediate length compression members, some fibres would have yielded and some fibres will still be elastic. They will fail by both yielding and buckling and their behaviour is said to be inelastic.

5. What is squash load?
a) load at which member will not deform axially
b) load at which member deforms laterally
c) load at which member deforms axially
d) load at which member will not deform axially

Answer: c [Reason:] Large deformation is possible only when fc reached the yield stress. At this stage the member deforms axially. The value of axial force at which this deformation occurs is called squash load.

6. Which of the following is not a parameter for decrease in strength of slender member?
b) initial lack of straightness
c) residual stress
d) variation of material properties

Answer: a [Reason:] The decrease in strength of slender member is due to following parameter : imperfections- initial lack of straightness, accidental eccentricities of loading, residual stress, and variation of material properties over the cross section.

7. Which of the following is property of compression member?
a) member must be sufficiently rigid to prevent general buckling
b) member must not be sufficiently rigid to prevent local buckling
c) elements of member should be thin to prevent local buckling
d) elements of member need not prevent local buckling

Answer: a [Reason:] Member must be sufficiently rigid to prevent general buckling in any possible direction, and each element of member must be thick enough to prevent local buckling.

8. How can moment of inertia be increased?
a) by increasing load
b) by spreading material of section towards its axis
c) by spreading material of section away from its axis
d) by spreading material of section at its axis

Answer: c [Reason:] Most important property of section in compression member is radius of gyration and thus moment of inertia. it can be increased by spreading material of section away from its axis.

9. Which is an ideal section for compression member?
a) one having different moment of inertia about any axis through its centre of gravity
b) one having same moment of inertia about any axis through its centre of gravity
c) one having larger length
d) one made up of costly material

Answer: b [Reason:] Ideal section is the one which has same moment of inertia about any axis through its centre of gravity.

10. Rods and bars are recommended when length is ___________
a) greater than 4m
b) greater than 5m
c) greater than 3m
d) less than 3m

Answer: d [Reason:] Rods and bars withstand very little compression when length is more. Hence these are recommended for lengths less than 3m only.

11. Which of the following is true about tubular section?
a) tubes have low buckling strength
b) tubes have same radius of gyration in all direction
c) tubes do not have torsional resistance
d) weight of tubular section is more than the weight required for open profile sections

Answer: b [Reason:] Tubes have same radius of gyration in all direction. They have high buckling strength and have excellent torsional resistance. Weight of tubular section is less than one half the weight required for open profile sections.

12. Which of the following statement is true?
a) unequal angles are desirable over equal angles
b) least radius of gyration of equal angle is less than that of unequal angle for same area of steel
c) single angle sections are suitable for long lengths
d) least radius of gyration of single angle section is small compared to channel and I-sections