Category: Design of Steel Structures Objective Questions

Design of Steel Structures Multiple Choice Questions on “Beam and Column Splices”.

1. Which of the following is the reason for beams, plate girders and columns being spliced?
a) full length is available from the mill
b) for easy transportation
c) for aesthetic appearance
d) for frictional resistance
Answer: b
Clarification: Rolled beams, plate girders and columns are spliced due to following reasons : (i)full length of the member may not be available from the mill, (ii)size of section which can be transported depends on size of truck, so for easy transportation, (iii)splice points may be used to camber the beam, (iv)when a change in section is required to fit variation in strength required along span of beam.

2. Which of the following is correct regarding splice plates used for beam splices?
a) plates on the flange should be designed to do the work of the web and plates on the web should be designed to do the work of the flange
b) plates on the flange should be designed to do the work of the web and plates on the web should be designed to do the work of the web
c) plates on the flange should be designed to do the work of the flange and plates on the web should be designed to do the work of the flange
d) plates on the flange should be designed to do the work of the flange and plates on the web should be designed to do the work of the web
Answer: d
Clarification: For beam splices, each element of the splice is designed to do the work the sections underlying the splice plates could do, if uncut. Plates on the flange should be designed to do the work of the flange and plates on the web should be designed to do the work of the web.

3. According to IS code, strength of spliced portion ________ of the effective strength of material spliced.
a) should not be less than 50%
b) should be less than 50%
c) should not be less than 80%
d) should be less than 80%
Answer: a
Clarification: As per IS code, strength of spliced portion should not be less than 50% of the effective strength of material spliced.

4. Choose the correct option from the following regarding basic forms of beam splices.
a) Flush end plates are used when bending moments to be resisted are high
b) Extended end plates are used when bending moments to be resisted are not high
c) Flush end plates are used when bending moments to be resisted are modest
d) Extended end plates are used when torsional moments to be resisted are not high
Answer: c
Clarification: Flush end plates are used when bending moments to be resisted are modest. Singly or doubly extended plates are used when resisting high moments of one sign or full reversal respectively.

5. When are longitudinal stiffeners introduced to beam splices?
a) when change in size between two sections of beam occurs
b) when change in size between two sections of beam does not occur
c) when change in moment between two sections of beam occurs
d) when change in moment between two sections of beam does not occur
Answer: a
Clarification: When change in size between two sections of beam occurs at an end plate splice, it can be easily accommodated by introducing longitudinal stiffeners to the larger beam.

6. In direct end bearing arrangement for column splices,
a) load is transferred through splices
b) splices are designed only to resist accidental tension
c) bending moment is transferred through splices
d) splices are designed only to resist bending moment
Answer: b
Clarification: End bearing arrangement may be used when the columns carry predominantly axial forces. In this load is transferred through contact area and splices are designed only to resist accidental tension due to some uplift loading or internal explosion in the building.

7. Which of the following is true regarding arrangement of leaving a gap between the ends for column splices?
a) load is transferred through splices
b) splices are designed only to resist accidental tension
c) load is transferred through contact area
d) splices are designed only to resist bending moment
Answer: a
Clarification: When the columns carry predominantly axial forces, leaving a gap between the ends may be used. In this case, the whole load is transmitted through means of splice plates. HSFG bolts may be used in the connections and changes in size of column may be accommodated using packing plates.

8. Which of the following is true when end plate splices is used for columns?
a) Short end plates are used for heavy moments
b) Extended end plates are used for moderate moments
c) Short end plates are used for moderate moments
d) Short end plates and extended end plates are used for moderate moments
Answer: c
Clarification: End plate splices can be used in column to provide load reversals in columns. Short end plates are used for moderate moments and extended end plates are used for heavy moments.

9. Position of splices should be _____ in normal practice.
a) at mid height of columns
b) at three fourth height of column from bottom of column
c) at three fourth height of column from top of column
d) just above the floor level
Answer: d
Clarification: In normal practice, splices are positioned just above the floor level to neglect the effects of flexing of the column. In regions of seismic activity, splices should be placed near mid-height of columns, where bending moments will be minimum.

Design of Steel Structures Multiple Choice Questions & Answers on “Theorem of Plastic Collapse & Methods of Plastic Analysis”.

1. What is static theorem ?
a) load must be greater than collapse load
b) load must be less than collapse load
c) load must be not equal to collapse load
d) load cannot be related to collapse load
Answer: b
Clarification: For a given frame and loading if there exists any distribution of bending moments throughout the frame which is both safe and statically admissible with set of load P, then value of load P must be less than or equal to collapse load.

2. Which of the following is true about static theorem?
a) it represents upper limit to true ultimate load
b) it represents plastic load
c) it has minimum factor of safety
d) it satisfies equilibrium and yield conditions
Answer: d
Clarification: The static method represents the lower limit to the true ultimate load and has maximum factor of safety. The theorem satisfies equilibrium and yield conditions.

3. Which of the following condition is true for kinematic theorem?
a) load must be greater than collapse load
b) load must be less than collapse load
c) load must be not equal to collapse load
d) load cannot be related to collapse load
Answer: a
Clarification: For a given frame subjected to a set of loads P, the value of P which is found to correspond to any assumed mechanism, must be greater than or equal to the collapse load Pu.

4. Which of the following is true about kinematic theorem?
a) it represents lower limit to true ultimate load
b) it represents plastic load
c) it has small factor of safety
d) it satisfies equilibrium and yield conditions
Answer: c
Clarification: Load computed on basis of this mechanism will always be greater than or at least equal to true ultimate load. Hence, kinematic method represents an upper limit to the true ultimate load and has a smaller factor of safety. This theorem satisfies equilibrium and continuity conditions.

5. Which of the following condition is true for uniqueness theorem?
a) load must be greater than collapse load
b) load must be less than collapse load
c) load must be equal to collapse load
d) load cannot be related to collapse load
Answer: c
Clarification: For given frame and loading, if at least one safe and statically admissible bending moment distribution can be found, and if in this distribution bending moment is equal to fully plastic moment at sufficient cross sections to cause failure of frame as a mechanism due to rotation of plastic hinges, then corresponding load will be equal to collapse load.

6. Load is called as correct collapse load when
a) static theorem is not satisfied
b) kinematic theorem is not satisfied
c) only static theorem is satisfied
d) both static and kinematic theorem are satisfied
Answer: d
Clarification: Load that satisfies both static and kinematic theorem at the same time is called correct collapse load.

7. Which of the following is true about kinematic analysis?
a) virtual work equations are not used to determine collapse load
b) virtual work equations are used to determine collapse load
c) equilibrium condition is assumed
d) plasticity condition is assumed
Answer: b
Clarification: In kinematic method, a mechanism is assumed and virtual work equations are used to determine the collapse load.

8. The number of independent mechanism is related to number of possible plastic hinge locations by ________
a) n = h * r
b) n = h / r
c) n = h + r
d) n = h – r
Answer: d
Clarification: The number of independent mechanism (n) is related to number of possible plastic hinge locations (h) and number of degree of redundancy (r) of the frame by equation n = h-r.

9. In static method of analysis, moment at any section is _______ plastic moment capacity.
a) greater than
b) two times
c) less than
d) three times
Answer: c
Clarification: In static method of analysis, an equilibrium moment diagram is obtained such that the moment at any section is less than or equal to the plastic moment capacity.

10. Which of the following relation between load factor, collapse load(Wc) and working load (W)
a) F = W_c / W
b) F = W / W_c
c) F = W_c W
d) F = W_c + W
Answer: a
Clarification: Load factor is ratio of collapse load to working load. It is represented by, F = W_c / W.

11. Which of the following is load factor for simply supported beam with central point load?
a) (f_yf_bc)v
b) (f_bc/ f_y)v
c) (f_y/f_bc)v
d) (f_y + f_bc)v
Answer: c
Clarification: For a simply supported beam with central point load,
M = WL/4 = f_bcZ_e, M_p = W_cL/4 = f_yvZ_e, where v is shape factor
F = M_p/M = W_c/W = (f_y/f_bc)v.

12. What is the value of load factor for I-section when f_bc= 0.66f_y and mean value of v is 0.14?
a) 1.727
b) 2.7
c) 1.56
d) 3.98
Answer: a
Clarification: Load factor, F = (f_y/f_bc)v. F = (1/0.66) x 0.14 = 1.27.

To practice all areas of Design of Steel Structures,

Design of Steel Structures Multiple Choice Questions on “Lateral Stability of Beams”.

1. Which of the following assumptions is not an ideal beam behaviour?
a) local and lateral instabilities of beam are prevented
b) any form of local buckling is prevented
c) compression flange of beam is restrained from moving laterally
d) compression flange of beam is not restrained from moving laterally
Answer: d
Clarification: Two important assumptions are made to achieve ideal beam behaviour: (i) compression flange of beam is restrained from moving laterally, (ii) any form of local buckling is prevented. A beam loaded predominantly in flexure would attain its full moment capacity if local and lateral instabilities of beam are prevented.

2. In beam design, sections are proportioned as such that _____ to achieve economy.
a) moment of inertia about principal axis parallel to the web is equal to moment of inertia about principal axis normal to the web
b) moment of inertia about principal axis parallel to the web is considerable larger than moment of inertia about principal axis normal to the web
c) moment of inertia about principal axis normal to the web is considerable larger than moment of inertia about principal axis parallel to the web
d) moment of inertia about principal axis normal to the web is considerable lesser than moment of inertia about principal axis parallel to the web
Answer: c
Clarification: In beam design, sections are proportioned as such that moment of inertia about principal axis normal to the web is considerable larger than moment of inertia about principal axis parallel to the web to achieve economy. Such sections are relatively weak in bending resistance.

3. To ensure that compression flange of beam is restrained from moving laterally, the cross section must be
a) plastic
b) semi-compact
c) slender
d) thin
Answer: a
Clarification: To ensure that compression flange of beam is restrained from moving laterally, the cross section must be plastic or compact. if significant ductility is required, section must invariably be plastic.

4. What are laterally restrained beams?
a) adequate restraints are provided to beam
b) adequate restraints are not provided to beam
c) economically not viable
d) unstable beams
Answer: a
Clarification: In laterally restrained beams, adequate restraints are provided to beam in plane of compression flange.

5. Characteristic feature if lateral buckling is ___________
a) entire cross section do not rotate as rigid disc without any cross sectional distortion
b) entire cross section rotates as rigid disc without any cross sectional distortion
c) entire cross section rotates as rigid disc with cross sectional distortion
d) entire cross section do not rotate as rigid disc
Answer: b
Clarification: The characteristic feature if lateral buckling is entire cross section rotates as rigid disc without any cross sectional distortion. This behaviour is similar to axially compresses long column which after initial shortening in axial direction, deflects laterally when it buckles.

6. Lateral buckling in beam is _________
a) does not occur in beam
b) one dimensional
c) two dimensional
d) three dimensional
Answer: d
Clarification: Lateral buckling in beam is three dimensional in nature. It involves coupled lateral deflection and twists that is when beam deflects laterally, the applied moment exerts a torque about the deflected longitudinal axis, which causes the beam to twist.

7. What is elastic critical moment?
a) bending moment at which beam do not fail by lateral buckling
b) bending moment at which beam fails by lateral buckling
c) shear force at which beam do not fail by lateral buckling
d) shear force at which beam fails by lateral buckling
Answer: b
Clarification: Bending moment at which beam fails by lateral buckling when subjected to a uniform end moment is called elastic critical moment.

8. Which of the following condition causes lateral instabilities?
a) section possesses different stiffness in two principal planes
b) section possesses same stiffness in two principal planes
c) applied loading does not induce bending in stiffer plane
d) applied loading induce twisting in stiffer plane
Answer: a
Clarification: Lateral instabilities occurs only if following conditions are satisfied : (i) section possesses different stiffness in two principal planes, (ii) applied loading induces bending in stiffer plane (about major axis).

9. Which of the following is not a method for providing effective lateral restraints?
(i) by embedding compression flange inside slab concrete
(ii) by providing shear connectors in compression flange
(iii) by bracing compression flanges of adjacent beams
a) i only
b) i, iii
c) ii, iii
d) i, ii, iii
Answer: d
Clarification: Effective lateral restraints can be provided by embedding compression flange inside slab concrete, by providing shear connectors in compression flange and embedding in concrete slab, by providing torsional bracings in the compression flanges of adjacent beams preventing twists directly.

Design of Steel Structures Multiple Choice Questions on “Web Panel subjected to Shear”.

1. The shear capacity of web comprises of strength
a) before onset of buckling strength only
b) post buckling strength only
c) before onset of buckling strength and post buckling strength
d) compression strength
Answer: c
Clarification: The shear capacity of web comprises of strength before onset of buckling strength and post buckling strength. strength before onset of buckling is contributed because of elastic behaviour wherein stresses are entirely elastic and the only requirement for the stiffeners is to keep the web flat.

2. What will happen when d/tw is sufficiently low?
a) web will yield under buckling before shear
b) web will yield under shear before buckling
c) web will not yield under shear
d) web will not yield under both shear and buckling
Answer: b
Clarification: When d/tw ratio is sufficiently low, the elastic critical stress increases above the value of yield shear stress and the web will yield under shear before buckling.

3. The nominal shear strength according to simple post-critical method is given by
a) A_v
b) A_vτ_b
c) τ_b
d) A_v /τ_b
Answer: b
Clarification: Simple post-critical method based on the shear buckling strength can be used for web of I-section girders, with or without intermediate transverse stiffeners, provided that web has transverse stiffeners at the supports. The nominal shear strength is given by V_n= Avτb, where A_v = area of web, τ_b = shear stress corresponding to web buckling.

4. The value of τb in the nominal shear strength equation according to simple post-critical method is given by
a) f_yw / √ λ_w
b) f_yw/λ_w
c) f_yw/λ_w²
d) f_yw/(√3 λ_w²)
Answer: d
Clarification: The value of τb in the nominal shear strength equation according to simple post-critical method is given by τ_b = f_yw/√3 for λ_w ≤0.8, [1-0.8(λ_w -0.8)](f_yw/√3) for 0.8 < λwyw/(√3 λ_w²) for λ_w ≥1.2, where f_yw is yield strength of web, λ_w is non-dimensional web slenderness ratio for shear buckling stress.

5. The value of non-dimensional web slenderness ratio in the nominal shear strength equation according to simple post-critical method is given by
a) √(f_yw/(√3τ_cr,e))
b) (f_yw/(√3τ_cr,e))
c) (f_yw/(τ_cr,e))
d) (f_yw/(√3τ_cr,e))²
Answer: a
Clarification: The value of non-dimensional web slenderness ratio in the nominal shear strength equation according to simple post-critical method is given by λ_w =√(f_yw/(√3τ_cr,e)), where f_yw is yield strength of web, τ_cr,e is elastic critical shear stress of the web.

6. The elastic critical shear stress of the web is given by
a) k_vπ²/[12(1+μ²)(d/t_w)²].
b) k_vπ²E/[12(1+μ²)(d/t_w)²].
c) k_vπ²E/[12(1-μ²)(d/t_w)²].
d) k_vE/[12(1-μ²)(d/t_w)].
Answer: c
Clarification: The elastic critical shear stress of the web is given by τ_cr,e = k_vπ²E/[12(1-μ²)(d/t_w)²], where E is elastic modulus, μ is Poisson’s ratio, k_v is constant which depends on spacing of transverse stiffeners and depth of web.

7. The value of k_v in the elastic critical shear stress equation for c/d < 1 is given by
a) 4.0 – [5.35/(c/d)].
b) 4.0 + [5.35/(c/d)²].
c) 5.35 + [4/(c/d)²].
d) 5.35 – [4/(c/d)].
Answer: b
Clarification: The value of k_v in the elastic critical shear stress equation is given by k_v = 5.35 when transverse stiffeners are provided only at supports, k_v = 4.0 + [5.35/(c/d)²] for c/d < 1.0, k_v = 5.35 + [4/(c/d)²] for c/d ≥ 1.0, where c and d are spacing of transverse stiffeners and depth of web respectively.

8. Which of the following conditions are true when tension field method is used?
a) it is based on pre-buckling strength
b) c/d < 1.0
c) it may not be used for webs with intermediate stiffeners
d) it may be used for webs with intermediate stiffeners
Answer: d
Clarification: The tension field method, based on the post-shear buckling strength, may be used for webs with intermediate transverse stiffeners at supports, provided the panels adjacent to the panel angle tension field action or the end posts provide anchorage for the tension field and is c/d>1.0.

9. What is the value of nominal shear strength according to tension field method?
a) A_vτ_b
b) 0.9w_tft_wf_vsinφ
c) A_vτ_b – 0.9w_tft_wf_vsinφ
d) A_vτ_b + 0.9w_tft_wf_vsinφ
Answer: d
Clarification: In tension field method, the nominal shear strength is given by V_n = A_vτ_b + 0.9w_tftf_vsinφ, where A_v is area of web, τ_b is buckling strength or shear stress corresponding to web buckling, f_v is yield strength of tension field which depends on inclination of tension field, w_tf is the width of tension field, t_w is width of web.

10. The value of fv in the nominal shear strength according to tension field method is given by
a) [f_yw²+3 τ_b²+Ψ²]^0.5+Ψ
b) [f_yw²-3 τ_b²+Ψ²]^0.5-Ψ
c) [f_yw²-3 τ_b²-Ψ²] -Ψ
d) [f_yw²+3 τ_b²+Ψ²]+Ψ
Answer: b
Clarification: The value of f_v in the nominal shear strength according to tension field method is given by f_v = [f_yw²-3 τ_b²+Ψ²]^0.5-Ψ , where f_yw is yield stress of web, τ_b is buckling strength or shear stress corresponding to web buckling, Ψ is a parameter which depends on inclination of tension field and buckling strength.

11. What is the expression for Ψ in the f_v for nominal shear strength according to tension field method is given by
a) 1.5 τ_b sin2φ
b) sin2φ
c) 1.5 τ_b
d) 1.5 τ_b /sin2φ
Answer: a
Clarification: The value of Ψ in the f_v for nominal shear strength according to tension field method is given by Ψ =1.5 τ_b sin2φ, where τb is buckling strength or shear stress corresponding to web buckling, φ is inclination of tension field which depends on depth of web and spacing of stiffeners.

12. The inclination of tension field is
a) tan(c/d)
b) tan(d/c)
c) tan^-1(c/d)
d) tan^-1(d/c)
Answer: d
Clarification: The inclination of tension field is given by φ = tan^-1(d/c), where d is depth of web and c is spacing between stiffeners. The change in angle of inclination of tension field affects the width and yield strength of tension field.

13. Which of the following is an expression for width of tension field?
a) w_tf = d sinφ + (c-s_c-s_t)cosφ
b) w_tf = d cosφ + (c-s_c-s_t)sinφ
c) w_tf = d cosφ – (c+s_c-s_t)sinφ
d) w_tf = d sinφ – (c+s_c-s_t)cosφ
Answer: b
Clarification: The width of tension field in the tension field action method is given by w_tf = d cosφ + (c-s_c-s_t)sinφ, where d is depth of beam, φ = inclination of tension field = tan^-1(d/c), c is spacing between stiffeners, s_c and s_t are anchorage lengths of tension field along the compression and tension flanges respectively and depends on reduced plastic moment capacity, inclination of tension field, thickness of web and yield stress of web.

14. The anchorage length of tension field is
a) s = (2 sinφ)(M_fr/f_ywt_w)^0.5
b) s = (2/ sinφ)(M_fr/f_ywt_w)
c) s = (2/ sinφ)(M_fr/f_ywt_w)^0.5
d) s = (2/ sinφ)(M_frf_ywt_w)
Answer: c
Clarification: The anchorage length of tension field is s = (2/ sinφ)(M_fr/f_ywt_w)^0.5 where φ = inclination of tension field = tan^-1(d/c), c is spacing bet_ween stiffeners, d is depth of beam, M_fr is reduced plastic moment capacity of the respective flange plate, f_yw is yield stress of web and t_w is thickness of web. The anchorage length should be less than or equal to spacing bet_ween stiffeners.

15. Which of the following expression for reduced plastic moment capacity is correct?
a) M_fr = 0.25b_ft_f²f_yf {1-[N_f/( b_ft_ff_yf/γ_m0)²]}
b) M_fr = b_ft_ff_yf {1-[N_f/( b_ft_ff_yf/γ_m0)]}
c) M_fr = 0.25b_ft_f {1+[N_f/( b_ft_ff_yf/γ_m0)]}
d) M_fr = 0.25b_f {1+[N_f( b_ft_ff_yfγ_m0)²]}
Answer: a
Clarification: The reduced plastic moment capacity of respective flange plate is calculated after accounting for axial force in flange, due to overall bending and any external axial force in the cross section. It is given by M_fr = 0.25b_ft_f²f_yf {1-[N_f/( b_ft_ff_yf/γ_m0)²]}, where b_f and t_f are width and thickness of flange respectively, f_yf is yield stress of flange and N_f is the axial force in flange.

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

1. Steel is mainly an alloy of
a) Iron and Carbon
b) Sulphur and Zinc
c) Zinc and tin
d) Phosphorous and Tin
Answer: a
Clarification: Steel is alloy of iron and carbon. Apart from carbon, a small percent of manganese, sulphur, phosphorous, chrome, nickel, and copper are added to give special properties to steel.

2. Which of the following is a disadvantage of Steel?
a) High strength per unit mass
b) High durability
c) Fire and corrosion resistance
d) Reusable
Answer: c
Clarification: Steel has high strength per unit mass, highly durable, and is reusable. But steel is poor in fire and corrosion resistance, it needs to be protected.

3. Elastic Modulus of Steel is __________
a) 1.5 x 10⁹ N/mm²
b) 2.0 x 10⁵ N/mm²
c) 2.0 x 10⁵ N/m²
d) 1.5 x 10⁹ N/m²
Answer: b
Clarification: Elastic modulus = Stress/Strain. As per IS 800-2007, elastic modulus of steel is 2.0 x 10⁵ N/mm².

4. Unit mass of Steel = ________
a) 785 kg/m³
b) 450 kg/m³
c) 450 kg/cm³
d) 7850 kg/m³
Answer: d
Clarification: As per IS 800-2007, unit mass of steel is 7850 kg/m³. A steel member with small section which has little self-weight is able to resist heavy loads because steel members have high strength per unit weight.

5. Poisson’s ratio of steel is ________
a) 0.1
b) 1.0
c) 0.3
d) 2.0
Answer: c
Clarification: Poisson’s ratio = transverse strain/axial strain. As per IS 800-2007, Poisson’s ratio of steel is 0.3 in elastic range and it is 0.5 in plastic range.

6. Structural Steel normally has carbon content less than _______
a) 1.0%
b) 0.6%
c) 3.0%
d) 5.0%
Answer: b
Clarification: Structural Steel normally have a carbon content less than 0.6%. Carbon content increases hardness, yield and tensile strength of steel but it decreases ductility and toughness.

7. What is the permissible percentage of Sulphur and Phosphorous content in steel?
a) 0.1%, 0.12%
b) 1.0%, 3.0%
c) 3.0%, 1.0%
d) 1.0%, 1.0%
Answer: a
Clarification: Sulphur content is generally between 0.02 – 0.1%. If more than 0.1%, it decreases strength and ductility of steel. If Phosphorous is more than 0.12%, it reduces shock resistance, ductility and strength of steel.

8. What happens when Manganese is added to steel?
a) decreases strength and hardness of steel
b) improves corrosion resistance
c) decreases ductility
d) improves strength and hardness of steel
Answer: d
Clarification: Manganese is added to improve strength and hardness of steel . Based on Manganese content, steel are classified as Carbon Manganese steel (Mn >1%) and Carbon Steel (Mn <1%). If its content exceeds 1.5%, it increases the formation of martensite and hence decreases ductility and toughness.

9. Which of the following is the effect of increased content of Sulphur and Phosphorous in Steel ?
a) yields high strength
b) affects weldability
c) increases resistance to corrosion
d) improves resistance to high temperature
Answer: b
Clarification: When sulphur and phosphorous is used beyond 0.06%, it imparts brittleness and affects weldability and fatigue strength.

10. Which of the following is added to steel to increase resistance to corrosion?
a) Carbon
b) Manganese
c) Sulphur
d) Copper
Answer: d
Clarification: Addition of small quantity of copper increases resistance to corrosion. Even Chrome and Nickel are added to impart corrosion resistance property to steel.

11. Which of the following properties are affected due to addition of carbon and manganese to steel?
(i) tensile strength and yield property (ii) Ductility (iii) Welding (iv) Corrosion resistance
a) i and ii only
b) i and iii only
c) i, ii, iii
d) i and iv only
Answer: c
Clarification: Increased quantity of carbon and manganese imparts higher tensile strength and yield properties but lowers ductility which is more difficult to weld.

12. Chrome and Nickel are added to Steel to improve _________
a) corrosion resistance and high temperature resistance
b) strength
c) ductility
d) weldablity
Answer: a
Clarification: Steel is weak in fire and corrosion resistance. So, to improve corrosion resistance and high temperature resistance, chromium and nickel are added to steel.