Category: Design of Steel Structures Objective Questions

Design of Steel Structures online quiz on “Behaviour and Ultimate Strength of Plates and Possible Failure Modes”.

1. What is the fundamental path in graph?
a) line along load axis up to P > P_cr
b) line along load axis up to P < P_cr
c) line along load axis up to P = P_cr
d) line along load axis up to P ≥ P_cr
Answer: c
Clarification: If axial load verses lateral displacement is plotted, we get a line along the load axis up to P=P_cr, this is called fundamental path.

2. Which of the following is true about secondary path?
a) lateral displacement increases indefinitely at constant load
b) lateral displacement decreases indefinitely at constant load
c) lateral displacement remains same at constant load
d) lateral displacement increases indefinitely and decreases at constant load
Answer: a
Clarification: When the axial load reaches the Euler bucking load, the lateral displacement increases indefinitely at constant load. This is called secondary path, which bifurcates from fundamental path at the buckling load.

3. Which of the following statement is true?
a) plate cannot carry loads higher than elastic critical load
b) plate cannot carry loads lesser than elastic critical load
c) secondary path for a plate is unstable
d) secondary path for a plate is stable
Answer: d
Clarification: The secondary path shows that plate can carry loads higher than elastic critical load. The secondary path for a plate is stable.

4. What is apparent modulus of elasticity?
a) ratio of average strain carried by plate to average stress
b) ratio of average stress carried by plate to average strain
c) product of average strain carried by plate to average stress
d) product of average stress carried by plate to average strain
Answer: b
Clarification: Elastic post-buckling stiffness is measured in terms of apparent modulus of elasticity, E*. It is the ratio of average stress carried by plate to average strain. In most of the cases, the value of E* is in the range 0.408 – 0.5 E and may be approximately taken as 0.5E.

5. The effective width of hot-rolled and welded plates is given by
a) b_e/b = α √(f_y/f_cr)
b) b_e/b = α √(f_cr x f_y)
c) b_e/b = α √(f_cr +f_y)
d) b_e/b = α √(f_cr/f_y)
Answer: d
Clarification: The effective width of hot-rolled and welded plates is given by b_e/b = α √(f_cr/f_y), where α is the parameter that indicates the inclusions of influence of initial curvatures and residual stress. Generally α=0.65.

6. The effective width of cold-formed steel sections is given by
a) b_e/b = (f_cr/f_y)[1-0.22√(f_cr/f_y)].
b) b_e/b = (f_cr/f_y)[1+0.22√(f_cr/f_y)].
c) b_e/b = (f_y/f_cr)[1-0.22√(f_cr/f_y)].
d) b_e/b = (f_cr/f_y)[1+0.22√(f_y/f_cr)].
Answer: a
Clarification: The effective width based on tests on cold-formed steel sections is given by b_e/b = (f_cr/f_y)[1-0.22√(f_cr/f_y)]. This formula was first adopted in AISC specification for light gauge cold-formed sections.

7. Which of the following is true about local buckling?
a) failure occurs by twisting of one or more individual elements of member
b) failure occurs by buckling of one or more individual elements of member
c) failure occurs by both buckling and twisting of one or more individual elements of member
d) cannot be prevented by selecting suitable width-to-thickness ratio of elements
Answer: b
Clarification: Local buckling is failure which occurs by buckling of one or more individual elements of member. It can be prevented by selecting suitable width-to-thickness ratio of elements.

8. What is squash load?
a) yield stress + area of cross section
b) yield stress – area of cross section
c) yield stress / area of cross section
d) yield stress x area of cross section
Answer: d
Clarification: When length of column is relatively small and its component elements are prevented from local buckling, then column will be able to attain its full strength or squash load (squash load = yield stress x area of cross section).

9. What is overall flexural buckling?
a) failure occurs by excessive deflection in plane of weaker principal axis
b) failure occurs by excessive deflection in plane of stronger principal axis
c) failure occurs by twisting of member
d) failure caused by seismic load
Answer: a
Clarification: Overall flexural buckling is failure which occurs by excessive deflection caused by bending or flexure, about axis corresponding to weaker principal axis(minor) – one with smallest radius of gyration, largest slenderness ratio.

10. Which of the following is true about torsional buckling?
a) failure occurs by bending about shear centre in longitudinal axis
b) failure occurs when torsional rigidity of member is greater than bending rigidity
c) standard hot rolled shapes are not susceptible to torsional buckling
d) it cannot occur with doubly symmetric cross section
Answer: c
Clarification: In torsional buckling, failure occurs by twisting about shear centre in longitudinal axis. It occurs when torsional rigidity of member is appreciably smaller than bending rigidity. It can occur only with doubly symmetric cross section with very slender cross sectional elements. Standard hot rolled shapes are not susceptible to torsional buckling.

11. Which of the following is not a solution for torsional buckling?
a) increasing length of members subjected to torsion
b) by careful arrangement of members
c) by providing bracing to prevent lateral movement and twisting
d) box section fabricated by adding welding side plates to ISHB sections
Answer: a
Clarification: Torsional buckling can be prevented by careful arrangement of members, by providing bracing to prevent lateral movement and twisting. In situations where torsion is expected either a box section fabricated by adding welding side plates to ISHB sections or by shortening box section fabricated by adding welding side plates to ISHB sections becomes the solution.

12. Flexural torsional buckling cannot occur in ________
a) unsymmetrical members
b) cross section with one axis of symmetry
c) cross section with no axis of symmetry
d) doubly symmetric members
Answer: d
Clarification: Flexural torsional buckling is a combination of flexural and torsion buckling. The member bends and twists simultaneously. It can occur only with open sections that have unsymmetrical cross section – both with one axis of symmetry(eg: channels, double angled shapes) and those with no axis of symmetry (eg: unequal leg single angles). Since the shear centre and centroid coincide with each other, doubly symmetric or point symmetric open sections are not subjected to flexural torsional buckling. Close sections are also immune to flexural torsional buckling.

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Design of Steel Structures Multiple Choice Questions on “Design Strength of Laterally Supported Beams – I”.

1. Which of the following aspects need not be considered for beam design?
a) deflection
b) material of beam
c) buckling
d) lateral supports
Answer: b
Clarification: The important aspects which need to be considered for beam design are moments, shears, deflection, crippling, buckling, and lateral support.

2. The design bending strength of laterally supported beams is governed by
a) torsion
b) bending
c) lateral torsional buckling
d) yield stress
Answer: d
Clarification: The design bending strength of laterally supported beams is governed by yield stress and that of laterally unsupported beams is governed by lateral torsional buckling.

3. The web is susceptible to shear buckling when d/t_w
a) <67ε
b) < 2×67ε
c) >67ε
d) < 70ε
Answer: c
Clarification: For beams with plastic, compact, semi-compact flanges and slender web (d/t_w > 67ε), the web is susceptible to shear buckling before yielding.

4. When there is no shear buckling,
a) V_p = V_n
b) V_p > V_n
c) V_p < V_n
d) V_p = 2V_n
Answer: a
Clarification: When there is no shear buckling (d/t_w ≤ 67ε), the nominal shear resistance V_n equals plastic shear strength V_p.

5. Plastic shear resistance is given by
a) f_y/√3
b) shear area x f_y x √3
c) shear area x f_y/√3
d) shear area / (f_y/√3)
Answer: c
Clarification: Plastic shear resistance is given by V_p = shear area x f_y/√3.

6. The design shear strength is given by
a) V_n
b) V_n/γ_m0
c) V_n x γ_m0
d) γ_m0
Answer: b
Clarification: The design shear strength is given by V_d = V_n/γ_m0 , where V_n= plastic shear resistance, γ_m0 = partial factor of safety.

7. The web area will be fully effective when shear force V
a) ≥ 0.6V_d
b) < 0.6V_d
c) ≤ 0.6V_d
d) >2×0.6V_d
Answer: c
Clarification: When shear force V ≤ 0.6V_d, the web area will be fully effective and entire cross section of beam will be effective in resisting the moment.

8. Which of the following is true about sections with high shear case V>0.6V_d ?
a) web area is ineffective
b) web area is fully effective
c) flanges will not resist moment
d) moment is not reduced
Answer: a
Clarification: When shear exceed the limit V&gt0.6V_d, web area will be ineffective and only flanges will resist the moment. Because of this for high shear case, moment capacity of beam is reduced.

Design of Steel Structures Multiple Choice Questions on “Characteristic Loads”.

1. Which IS code is used for calculating different loads on different structures?
a) IS 800
b) IS 200
c) IS 300
d) IS 875
Answer: d
Clarification: IS 875 (all 5 parts) is recommended by Bureau of Indian Standards for calculating various types of loads on the structure. Part 1 is for dead loads, part 2 for imposed loads, part 3 for wind load, part 4 for snow loads and part 5 for special loads and combinations.

2. Which of the following load is to be considered on liquid retaining structure?
a) hydrostatic load
b) wave and current load
c) earth pressure
d) dynamic load
Answer: a
Clarification: Hydrostatic load is considered on liquid retaining structures or hydraulic structures. Wave and current load is considered in marine and offshore structure. Earth pressure is considered in basements, retaining walls, column footings, etc. Dynamic load is due to earthquake and wind.

3. What is P-Δ effect?
a) earthquake load
b) second order moments arising from joint displaced
c) second order moments arising from member deflection
d) load due to shrinkage effect
Answer: b
Clarification: Second order moments arising from joint displaced is called P-Δ effect and second order moments arising from member deflection is called P-δ effect.

4. Match the pair :

	(A) Mass and gravitational effect		(i) wind load
	(B) Mass and acceleration effect 		(ii) load due to settlement
	(C) Environmental effects			(iii) imposed load

a) A-i, B-ii, C-iii
b) A-iii, B-ii, C-i
c) A-iii, B-i, C-ii
d) A-ii, B-iii, C-i
Answer: c
Clarification: Load on structure may be due to following :
1) Mass and gravitational effect : examples of such loads are dead loads, imposed loads, snow and ice loads, earth loads, etc.
2) Mass and acceleration effect : examples of such loads are those caused by earthquake, wind, impact, blasts, etc.
3) Environmental effects : examples of such loads are due to temperature difference, settlement, shrinkage, etc.

5. The probability that a specific load will be exceeded during life of structure depends on _______
a) wind
b) factor of safety
c) partial factor of safety
d) period of exposure
Answer: d
Clarification: The probability that a specific load will be exceeded during life of structure depends on period of exposure. It also depends on magnitude of design load.

6. What is characteristic load?
a) seismic load
b) load which will be exceeded by certain probability during life of structure
c) load which will not be exceeded by certain probability during life of structure
d) pressure load
Answer: c
Clarification: Characteristic load is the load which will not be exceeded by certain assumed or pre-assumed probability during life of structure. These loads are anticipated loads due to self weight, imposed load, snow, wind load, etc.

7. Which of the following is not included in imposed load classification?
a) Residential load
b) Earthquake load
c) Industrial load
d) Educational load
Answer: b
Clarification: Imposed loads are gravity loads other than dead load and cover factors such as occupancy by people, stored material etc. It is classified into following groups : (i)residential, (ii)educational, (iii)institutional, (iv)assembly halls, (v)office and business buildings, (vi)mercantile buildings, (vii)industrial, (viii)storage buildings.

8. What is the minimum imposed load on roof trusses as per IS code?
a) 0.5 kN/m²
b) 0.4 kN/m²
c) 0.9 kN/m²
d) 0.75 kN/m²
Answer: b
Clarification: As per IS 875, the minimum imposed load on roof truss should be 0.4 kN/m². For sloping roof upto 10˚, the imposed load is taken as 0.5 kN/m² if access is not provided and 0.75 kN/m² if access is provided.

9. For roofs of slope greater than 10˚, the imposed load is reduced by ____ for every degree rise in slope.
a) 0.02 kN/m²
b) 0.05 kN/m²
c) 0.75 kN/m²
d) 0.5 kN/m²
Answer: a
Clarification: As per IS 875, for roofs of slope greater than 10^o, the imposed load is taken as 0.75 kN/m² and reduced by 0.02 kN/m² for every degree rise in slope.

10. Calculate imposed load on roof truss of span 20m with slope of 20^o.
a) 0.75 kN/m²
b) 0.95 kN/m²
c) 0.45 kN/m²
d) 0.55 kN/m²
Answer: d
Clarification: As per IS 875, for roofs of slope greater than 10^o, the imposed load is reduced by 0.02 kN/m² for every degree rise in slope.
Therefore, Imposed load = 0.75 – 0.02*(20^o-10^o) = 0.55 kN/m².

Design of Steel Structures Multiple Choice Questions on “Design of Welds”.

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)
Answer: c
Clarification: Design nominal strength of fillet weld = f_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_tl_w
b) Pt_t/l_w
c) Pt_tl_w
d) Pl_w/t_t
Answer: a
Clarification: When welds are subjected to compressive or tensile or shear force alone, the stress in weld is given by q = P/t_tl_w , where q=shear stress in N/mm², 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_a²+q²)
b) √(f_a²+2q²)
c) √(3f_a²+q²)
d) √(f_a²+3q²)
Answer: d
Clarification: When fillet welds are subjected to combination of normal and shear stress, the equivalent stress is given by f_e = √(f_a2+3q²), 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
Answer: b
Clarification: L_w = 150mm, Throat thickness, t_e = 5×12/8 = 7.5 mm
Strength of weld = L_wt_ef_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
Answer: d
Clarification: 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
Clarification: Minimum size of weld = 3mm
Maximum size of weld = 10-1.5 = 8.5mm
Assume weld size = 6mm
Effective throat thickness, t_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
Answer: c
Clarification: 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
Clarification: 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.

Design of Steel Structures Question Bank on “Compression Members and Loads on Compression Members”.

1. What is compression member?
a) structural member subjected to tensile force
b) structural member subjected to compressive force
c) structural member subjected to bending moment
d) structural member subjected to torsion
Answer: b
Clarification: Structural member which is subjected to compressive forces along its axis is called compression member. Compression members are subjected to loads that tend to decrease their lengths.

2. Which of the following is true about axially loaded column?
a) member subjected to bending moment
b) member subjected to axial force and bending moment
c) net end moments are not zero
d) net end moments are zero
Answer: d
Clarification: if the net end moments are zero, the compression member is required to resist load acting concentric to original longitudinal axis of member and is called axially loaded column or simply column.

3. Which of the following is true about beam column?
a) member subjected to bending moment
b) member subjected to axial force only
c) member subjected to axial force and bending moment
d) net end moments are zero
Answer: c
Clarification: If the net end moments are not zero, the member will be subjected to axial force and bending moments along its length. Such members are called beam-columns.

4. What are columns?
a) vertical compression members in a building supporting floors or girders
b) vertical tension members in a building supporting floors or girders
c) horizontal compression members in a building supporting floors or girders
d) horizontal tension members in a building supporting floors or girders
Answer: a
Clarification: The vertical compression members in a building supporting floors or girders are normally called as columns. They are sometimes called as stanchions. They are subjected to heavy loads. Vertical compression members are sometimes called posts.

5. Which of the following are true about roof trusses?
a) principal rafter are compression members used in buildings
b) principal rafter is bottom chord member of roof truss
c) struts are compression members used in roof trusses
d) struts are tension members used in roof trusses
Answer: c
Clarification: The compression members used in roof trusses and bracings are called as struts. They may be vertical or inclined and normally have small lengths. the top chord members of a roof truss are called principal rafter.

6. Knee braces are __________
a) long compression members
b) short compression members
c) long tension members
d) short tension members
Answer: b
Clarification: Short compression members at junction of columns and roof trusses or beams are called knee braces. They are provided to avoid moment.

7. Which of the following is not a load on columns in buildings?
a) load from floors
b) load from foundation
c) load from roofs
d) load from walls
Answer: b
Clarification: Axial loading on columns in buildings is due to loads from roofs, floors, and walls transmitted to the column through beams and also due to its own self weight.

8. Which of the following is correct?
a) moment due to wind loads is not considered in unbraced buildings
b) wind load cause large moments in braced buildings
c) wind loads in multi-storey buildings are not usually applied at respective floor levels
d) wind loads in multi-storey buildings are usually applied at respective floor levels
Answer: d
Clarification: Wind loads in multi-storey buildings are usually applied at respective floor levels and are assumed to be resisted by bracings. Hence in braced buildings wind loads do not cause large moments. But, in unbraced rigid framed buildings, the moment due to wind loads should also be taken into account in the design of columns.

9. What are loads on columns in industrial buildings?
a) wind load only
b) crane load only
c) wind and crane load
d) load from foundation
Answer: c
Clarification: In industrial buildings, loads from crane and wind cause moments in columns. In such cases, wind load is applied to the column through sheeting rails and may be taken as uniformly distributed throughout the length of column.

10. The strength of column does not depend on
a) width of building
b) material of column
c) cross sectional configuration
d) length of column
Answer: a
Clarification: The strength of column depends on material of column, cross sectional configuration, length of column, support conditions at the ends, residual stresses, imperfections.

11. Which of the following is not an imperfection in column?
a) material not being isotropic
b) geometric variations of columns
c) material being homogenous
d) eccentricity of load
Answer: c
Clarification: Imperfections in column include material not being isotropic and homogenous, geometric variations of columns and eccentricity of load.

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Design of Steel Structures Problems on “Design Strength of Laterally Supported Beams – II”.

1. The design bending strength of beams when V ≤ 0.6Vd is given by
a) β_b /Z_pf_y γ_m0
b) β_bZ_pf_y / γ_m0
c) β_bZ_p /f_y γ_m0
d) β_bZ_pf_y γ_m0
Answer: b
Clarification: The design bending strength of beams when V ≤ 0.6Vd is given by M_d = β_bZ_pf_y / γ_m0 , where β_b is a constant, Z_p = plastic section modulus of cross section, f_y is yiled stress of material, γ_m0 = 1.1,partial safety factor.

2. The value of β_b in the equation of design bending strength for plastic section is given by
a) 1.5
b) 2.0
c) 0.5
d) 1.0
Answer: d
Clarification: The value of β_b in the equation of design bending strength is 1 for plastic and compact sections. The value of β_b in the equation of design bending strength for semi-compact section depends on section modulus.

3. The value of β_b in the equation of design bending strength for semi-compact section is given by
a) Z_e/Z_p
b) Z_eZ_p
c) Z_p/ Z_e
d) Z_e+Z_p
Answer: a
Clarification: The value of β_b in the equation of design bending strength for semi-compact section is given by β_b = Z_e/Z_p , where Z_e, Z_p are elastic and plastic moduli of the cross section.

4. The check for design bending strength for simply supported beams is given by
a) M_d = 2.4Z_pf_y/γ_m0
b) M_d < 1.2Z_pf_y/γ_m0
c) M_d ≤ 1.2Z_pf_y/γ_m0
d) M_d ≥ 1.2Z_pf_y/γ_m0
Answer: c
Clarification: The check for design bending strength for simply supported beams is given by M_d ≤ 1.2Z_pf_y/γ_m0 to ensure that onset of plasticity under unfactored loads is prevented.

5. The check for design bending strength for cantilever beams is given by
a) M_d = 2.4Z_pf_y/γ_m0
b) M_d ≤ 1.5Z_pf_y/γ_m0
c) M_d ≤ 1.2Z_pf_y/γ_m0
d) M_d ≥ 1.5Z_pf_y/γ_m0
Answer: b
Clarification: The check for design bending strength for cantilever beams is given by M_d ≤ 1.5Z_pf_y/γ_m0 to ensure that onset of plasticity under unfactored loads – dead loads, imposed loads and wind load- is prevented.

6. The design bending strength for slender sections is given by
a) M_d = Z_ef_y‘
b) M_d = f_y‘
c) M_d = Z_e /f_y‘
d) M_d = Z_e +f_y‘
Answer: a
Clarification: The design bending strength for slender sections is given by M_d = Z_ef_y‘ , where Z_e is elastic section modulus of cross section and f_y‘ is reduced design strength for slender sections.

7. IS 800 permits bolt holes in the flanges to be ignored when
a) 0.9f_uA_nf/γ_m1 ≤ 2f_yAgf/γ_m0
b) 0.9f_uA_nf/γ_m1 ≤ f_yAgf/γ_m0
c) 0.9f_uA_nf/γ_m1 ≥ f_yAgf/γ_m0
d) 0.9f_uA_nf/γ_m1 ≤ 0.5f_yAgf/γ_m0
Answer: c
Clarification: IS 800 permits bolt holes in the flanges to be ignored when the tensile fracture strength of flange is at least equal to tensile yield strength i.e. when 0.9f_uA_nf/γ_m1 ≥ f_yAgf/γ_m0 or (A_nf/Agf) ≥ (f_y/f_u)x(γ_m1 /γ_m0)x(1/0.9), where A_nf/Agf = ratio of net area to gross area of tension flange, f_y/f_u = ratio of yield stress to ultimate stress of material, γ_m1 /γ_m0 = ratio of partial safety factors against ultimate stress to yield stress.

8. The moment capacity of semi-compact section for V > 0.6V_d is given by
a) M_d = Zef_yγ_m0
b) M_d = Zef_y
c) M_d = f_y/γ_m0
d) M_d = Zef_y/γ_m0
Answer: d
Clarification: Few I-and channel sections are semi-compact because of width-thickness ratio. The moment capacity of semi-compact section for V > 0.6V_d is given by M_d = Zef_y/γ_m0, where Ze = elastic section modulus of whole section, f_y = yield stress of material, γ_m0 = partial safety factor.

Design of Steel Structures Problems,