# 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

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2. The effect of lateral-torsional buckling need not be considered when

a) λ_{LT} ≤ 0.4

b) λ_{LT} ≥0.4

c) λ_{LT} > 0.8

d) λ_{LT} = 0.8

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_{LT}≤ 0.4, where λ

_{LT}is the non dimensional slenderness ratio for lateral torsional buckling.

3. The bending strength of laterally unsupported beams is given by

a) M_{d} = β_{b}Z_{p} /f_{bd}

b) M_{d} = β_{b} /Z_{p}f_{bd}

c) M_{d} = β_{b}Z_{p}

d) M_{d} = β_{b}Z_{p}f_{bd}

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_{d}= β

_{b}Z

_{p}f

_{bd}, where β

_{b}is a constant, Z

_{p}is plastic section modulus, f

_{bd}is design bending compressive stress.

4. The value of β_{b} 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

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_{b}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 β_{b} in the equation of design bending strength of laterally unsupported beams for semi-compact sections is

a) Z_{e}/Z_{p}

b) Z_{e}Z_{p}

c) Z_{p}/Z_{e}

d) Z_{p}

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_{b}in the equation of design bending strength of laterally unsupported beams for semi-compact sections is Z

_{e}/Z

_{p}, where Z

_{e}is elastic section modulus, Z

_{p}is plastic section modulus.

6. The value of design bending compressive stress f_{bd} is

a) X_{LT} f_{y}

b) X_{LT} f_{y} /f_{y}

c) X_{LT} f_{y} f_{y}

d) X_{LT} /f_{y}

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_{bd}is X

_{LT}f

_{y}/f

_{y}, where X

_{LT}is bending stress reduction factor to account for lateral torsional buckling, f

_{y}is yield stress, f

_{y}is partial safety factor for material (=1.10).

7. The bending stress reduction factor to account for lateral buckling is given by

a) X_{LT} = 1/{φ_{LT} + (φ^{2}_{LT} – λ^{2}_{LT})}

b) X_{LT} = 1/{φ_{LT} – (φ^{2}_{LT} + λ^{2}_{LT})}

c) X_{LT} = 1/{φ_{LT} – (φ^{2}_{LT} + λ^{2}_{LT})0.5}

d) X_{LT} = 1/{φ_{LT} + (φ^{2}_{LT} – λ^{2}_{LT})0.5}

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_{LT}= 1/{φ

_{LT}+ (φ

^{2}

_{LT}– λ

^{2}

_{LT})0.5}, where φ

_{LT}depends upon imperfection factor and non dimensional slenderness ratio, λ

_{LT}is non dimensional slenderness ratio.

8. The value of φ_{LT} in bending stress reduction factor is given by

a) φ_{LT} = [ 1 – α_{LT} (λ_{LT} + 0.2) + λ^{2}_{LT}].

b) φ_{LT} = [ 1 + α_{LT} (λ_{LT} – 0.2) + λ^{2}_{LT}].

c) φ_{LT} = 0.5 [ 1 – α_{LT} (λ_{LT} + 0.2) + λ^{2}_{LT}].

d) φ_{LT} = 0.5 [ 1 + α_{LT} (λ_{LT} – 0.2) + λ^{2}_{LT}].

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_{LT}in bending stress reduction factor is given by φ

_{LT}= 0.5 [ 1 + α

_{LT}(λ

_{LT}– 0.2) + λ

^{2}

_{LT}], where α

_{LT}is imperfection factor, λ

_{LT}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’p^{2}t/8g

c) ndt – n’p^{2}t/4g

d) nd + n’p^{2}t/4g

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^{2}/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 cm^{2}

b) 20 cm^{2}

c) 240 mm^{2}

d) 480 mm^{2}

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_{h}= 18mm Net section area = 400×10 – 16×10 = 3820mm

^{2}= 38.2 cm

^{2}.

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

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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) 2796mm^{2}

b) 2681mm^{2}

c) 2861mm^{2}

d) 3055mm^{2}

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_{h}= 18+2 =20mm, n = 3, n’ = 1, p = 75mm, g = 50mm Effective net area = (B-nd

_{h}+n’p

^{2}/4g)t = (300 – 3×20 + 1×75

^{2}/4×50)x10 = 2681.25 mm

^{2}.

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 mm^{2}

b) 756 mm^{2}

c) 800 mm^{2}

d) 640 mm^{2}

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_{h}= 20mm Net Area A

_{n}= A

_{g}– d

_{h}t = 100 x 8 – 20 x 8 = 640mm

^{2}.

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

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_{nc}= (100 – 12/2) x 12 = 1128 mm

^{2}Net area of outstanding leg, A

_{go}= (75 – 12/2) x 12 = 828 mm

^{2}Total net area = 1128 + 828 = 1956 mm

^{2}.

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

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_{c}=150mm, f

_{y}=250MPa, f

_{u}=410MPa β = 1.4 – [0.076 (w/t)(f

_{y}/f

_{u})(b

_{s}/L

_{c})] = 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

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_{u}= 410MPa, f

_{y}= 250MPa, γ

_{m0}= 1.1, γ

_{m1 }= 1.25, For ISA 125 x 75 x 10mm : gross area A

_{g}= 1902 mm

^{2}Tensile strength due to gross section yielding, T

_{dg }= A

_{g}f

_{y}/γ

_{m0}= 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

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_{h}=18 mm, f

_{u}= 410MPa, f

_{y}= 250MPa, γ

_{m0}= 1.1, γ

_{m1}= 1.25 A

_{nc }= (100 – 10/2 – 18) x 10 = 770 mm

^{2}, A

_{g0}= (75 – 10/2) x 10 = 700 mm

^{2}β = 1.4 – 0.076(w/t)(f

_{y}/f

_{u})(b

_{s}/L

_{c}) = 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) T

_{dn}= 0.9f

_{u}A

_{nc}/γ

_{m1}+ βA

_{g0}f

_{y }/γ

_{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

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_{u}= 410 MPa, f

_{y}= 250 MPa, γ

_{m0 }= 1.1, γ

_{m1 }= 1.25 A

_{vg}= ( 1×100 + 50 ) x 8 = 1200 mm

^{2}A

_{vn}= (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm

^{2}A

_{tg}= 35 x 8 = 280 mm

^{2}A

_{tn}= (35 – 1/2 x 20) x 8 = 200 mm

^{2}T

_{db1}= (A

_{vg}f

_{y}/√3 γ

_{m0})+(0.9A

_{tn}f

_{u}/γ

_{m1}) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN T

_{db2}= (A

_{tg}f

_{y}/ γ

_{m0})+(0.9A

_{vn}f

_{u}/√3 γ

_{m1}) = [ (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) f_{u}

b) √3 f_{u}

c) f_{u}/√3

d) f_{u}/(1.25 x √3)

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_{u}/√3, where f

_{u}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/t_{t}l_{w}

b) Pt_{t}/l_{w}

c) Pt_{t}l_{w}

d) Pl_{w}/t_{t}

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_{t}l

_{w}, where q=shear stress in N/mm

^{2}, P = force transmitted, t

_{t}= effective throat thickness of weld in mm, l

_{w}= 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) √(f_{a}^{2}+q^{2})

b) √(f_{a}^{2}+2q^{2})

c) √(3f_{a}^{2}+q^{2})

d) √(f_{a}^{2}+3q^{2})

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_{e}= √(f

_{a}2+3q

^{2}), where f

_{a}= 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

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_{w}= 150mm, Throat thickness, t

_{e}= 5×12/8 = 7.5 mm Strength of weld = L

_{w}t

_{e}f

_{y}/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

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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

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_{e}= 0.7 x 6 = 4.2mm Strength of weld = t

_{e}[f

_{u}/(√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

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^{3}/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

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## 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

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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

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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

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4. For earthquake loads, axially loaded members have to resist ________________

a) tension only

b) compression only

c) both tension and compression

d) bending moment

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5. For earthquake loads, beams are designed to resist ________________

a) tension only

b) positive and negative bending moments

c) compression only

d) torsion

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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

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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

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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

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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

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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

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11. Which of the following relation is correct for design horizontal seismic coefficient?

a) A = ZIS_{a}*2Rg

b) A = ZIS_{a}/2Rg

c) A = ZIS_{a}-2Rg

d) A = ZIS_{a}+2Rg

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_{a}/2Rg, where Z is zone factor, R is response reduction factor, I is importance factor, S

_{a}/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

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_{a}/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 = [ (C_{pe} – C_{pi})A/p_{d}].

b) F = [ (C_{pe} + C_{pi})A/p_{d}].

c) F = [ (C_{pe} – C_{pi})Ap_{d}].

d) F = [ (C_{pe} – C_{pi})/Ap_{d}].

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_{pe}– C

_{pi})A pd , where F=net wind force on element, C

_{pi}= internal pressure coefficient, C

_{pe}=external pressure coefficient, A=surface area of element, p

_{d}=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

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## 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

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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

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3. Long compression members will ______

a) not buckle

b) buckle inelastically

c) buckle plastically

d) buckle elastically

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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

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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

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6. Which of the following is not a parameter for decrease in strength of slender member?

a) seismic load

b) initial lack of straightness

c) residual stress

d) variation of material properties

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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

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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

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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

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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

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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

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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