Category: Design of Steel Structures Objective Questions

Design of Steel Structures Questions and Answers for Entrance exams on “Behaviour of Compression Members & Types of Sections”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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?
a) seismic load
b) initial lack of straightness
c) residual stress
d) variation of material properties
Answer: a
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Answer: d
Clarification: Equal angle are desirable and economical over unequal angles because least radius of gyration of equal angle is greater than that of unequal angle for same area of steel. Single angle sections are not suitable for long lengths. Least radius of gyration of single angle section is small compared to channel and I-sections.

Design of Steel Structures for Entrance exams,

Basic Design of Steel Structures Questions and Answers on “Design Strength of Laterally Supported Beams – III”.

1. What is shear lag effect?
a) the phenomenon of non uniform bending stress not due to influence of shear strain induced on bending stresses in flanges
b) the phenomenon of uniform bending stress not due to influence of shear strain induced on bending stresses in flanges
c) the phenomenon of uniform bending stress due to influence of shear strain induced on bending stresses in flanges
d) the phenomenon of non uniform bending stress due to influence of shear strain induced on bending stresses in flanges
Answer: d
Clarification: The shear strain induced influences bending stresses in flanges and causes sections to warp. This consequently modifies the bending stresses determined by simple bending theory and results in higher stresses near junction of web to flange elements with stress dropping as distance from beam web increases. The resultant stress distribution across flange is therefore non uniform and this phenomenon is known as shear lag.

2. As per IS 800:2007, shear lag effects in flanges may be disregarded for outstand elements if
a) b_o ≥ L₀ / 20
b) b_o ≤ L₀ / 20
c) b_o > L₀ / 20
d) b_o = L₀ / 10
Answer: b
Clarification: As per IS 800:2007, shear lag effects in flanges may be disregarded for outstand elements if b_o ≤ L₀ / 20, where b_o = width of flange outstand, L₀ = length between points of zero moment in the span.

3. As per IS 800:2007, shear lag effects in flanges may be disregarded for internal elements if
a) b_i ≤ L₀ / 10
b) b_i ≤ L₀ / 20
c) b_i > L₀ / 10
d) b_i = L₀ / 20
Answer: a
Clarification: As per IS 800:2007, shear lag effects in flanges may be disregarded for internal elements if b_i ≤ L₀ / 10, where b_i = width of internal element, L₀ = length between points of zero moment in the span.

4. Shear lag effect depends on
a) material of beam
b) width of beam only
c) width-to-span ratio
d) cost
Answer: c
Clarification: Shear lag effect depends upon width-to-span ratio, beam end restraints, and type of load.

5. Which of the following is true?
a) point load causes less shear lag than uniform load
b) point load causes more shear lag than uniform load
c) point load causes half times the shear lag than uniform load
d) point load causes equal shear lag as uniform load
Answer: b
Clarification: The resultant stress distribution across flanges is non-uniform and is called shear lag. Point load causes more shear lag than uniform load.

6. The moment capacity of plastic section for V > 0.6V_d is given by
a) M_dv = M_d – β(M_d – M_fd)
b) M_dv = M_d + β(M_d – M_fd)
c) M_dv = M_d – β(M_d + M_fd)
d) M_dv = M_d + β(M_d + M_fd)
Answer: a
Clarification: The moment capacity of plastic or compact section for V > 0.6V_d is given by M_dv = M_d – β(M_d – M_fd), where M_d = plastic design moment of whole section disregarding high shear force effect but considering web buckling effect, M_fd = plastic design strength of area of cross section excluding shear area, considering partial safety factor γ_m0, β is constant.

7. The value of β in equation of moment capacity of plastic section for V > 0.6V_d is given by
a) ([V_d/V] -1)²
b) (2[V_d/V] +1)²
c) (2[V_d/V] -1)²
d) (2[V_d/V] -1)
Answer: c
Clarification: The value of β in equation of moment capacity of plastic section for V > 0.6V_d is given by β = (2[V_d/V] -1)², where V_d = design shear strength as governed by web yielding or web buckling, V = factored applied shear force.

8. The check for moment capacity of plastic section for V > 0.6V_d is given by
a) M_dv ≥ 1.2Zef_y/γ_m0
b) M_dv ≤ 1.2Zef_y/γ_m0
c) M_dv > 1.2Zef_y/γ_m0
d) M_dv = 2.2Zef_y/γ_m0
Answer: b
Clarification: The check for moment capacity of plastic section for V > 0.6V_d is given by M_dv ≤ 1.2Zef_y/γ_m0, where Z_e = elastic section modulus of whole section, f_y = yield stress of material, γ_m0 = partial safety factor.

To practice basic questions and answers on all areas of Design of Steel Structures,

Design of Steel Structures Interview Questions and Answers on “Determination of Wind Loads & Load Combinations”.

1. Which IS Code is used for design loads for buildings and structures for wind load?
a) IS 456
b) IS 875 Part 3
c) IS 500
d) IS 1280
Answer: b
Clarification: For design loads for buildings and structures for wind load, IS 875-Part 3 given by Bureau of Indian Standards is used.

2. IS Code gives basic wind speed averaged over a short interval of ______
a) 10 seconds
b) 20 seconds
c) 5 seconds
d) 3 seconds
Answer: d
Clarification: Basic wind speed is based on peak wind gust velocity averaged over a short interval of 3s and having return period of 50 years and corresponds to mean above ground level in open terrain.

3. Positive sign of pressure coefficient indicates ______________
a) pressure acting towards the surface
b) pressure acting away the surface
c) pressure acting above the surface
d) pressure acting below the surface
Answer: a
Clarification: Positive sign of pressure coefficient indicates pressure acting towards the surface and negative sign of pressure coefficient indicates pressure acting away the surface.

4. Which of the following relation is correct for pressure coefficient?
V_p = Actual wind speed at any point on structure at height corresponding to V_z (design wind speed)
a) [1+(V_p/V_z)²].
b) [1+(V_z/V_p)²].
c) [1-(V_z/V_p)²].
d) [1-(V_p/V_z)²].
Answer: d
Clarification: Pressure Coefficient is the ratio of difference between pressure acting at point on surface and static pressure of incident wind to the design wind pressure.

5. What is return period?
a) number of years, the reciprocal of which gives the probability of extreme wind exceeding given wind speed in any one year
b) number of years, the reciprocal of which gives the probability of extreme wind less than given wind speed in any one year
c) number of years, the reciprocal of which gives the probability of mild wind exceeding given wind speed in any one year
d) number of years, the reciprocal of which gives the probability of mild wind less than given wind speed in any one year
Answer: a
Clarification: Wind load acting on structure varies from year to year based on wind speed and maximum that can be expected to occur at a given location only once in so many years. This period is called return period.

6. Wind Pressure at any height of structure does not depend on _______
a) velocity and density of air
b) angle of wind attack
c) topography of ground surface
d) material of structure
Answer: d
Clarification: Wind Pressure at any height of structure depend on (i)velocity and density of air, (ii)height above ground level, (iii)shape and aspect ratio of building, (iv) topography of surrounding ground surface, (v)angle of wind attack, (vi)solidity ratio or openings in the structure.

7. Which of the following relation is correct for design wind speed (V_z) and basic wind speed (V_b) ?
a) V_z ∝ V_b²
b) V_z ∝ 1/V_b²
c) V_z ∝ V_b
d) V_z ∝ 1/V_b
Answer: c
Clarification: V_z = k₁k₂k₃V_b , where k₁=probability factor(risk coefficient), k₂=terrain, height and structure size factor, k₃=topography factor.

8. Calculate design wind speed for a site in a city with basic wind speed of 50 m/s, risk coefficient =1, topography factor = 1, terrain is with closely spaced buildings and height of building (class A) = 15m.
a) 40 m/s
b) 48.5 m/s
c) 50 m/s
d) 52.5 m/s
Answer: b
Clarification: V_b = 50m/s, k₁ = 1, k₃ = 1,
for terrain with closely spaced buildings, height of building=15m, class A : k₂=0.97 (from IS 875 Part 3)
V_z = k₁k₂k₃V_b = 1×0.97x1x50 = 48.5 m/s.

9. Which of the following relation between design pressure, pz and design wind speed, Vz is correct?
a) p_z ∝ V_z²
b) p_z ∝ 1/V_z²
c) p_z ∝ V_z
d) p_z ∝ 1/V_z
Answer: a
Clarification: p_z = 0.6V_z², where p_z is in N/m² and V_z is in m/s. 0.6 factor depends on number of factors and mainly on atmospheric pressure and air temperature.

10. Calculate the design wind pressure if the basic wind speed is 44 m/s, risk coefficient is 1, topography factor is 1, terrain is with closely spaced buildings and height of building(class A) = 20m .
a) 1285 N/m²
b) 1580 N/m²
c) 1085 N/m²
d) 1185 N/m²
Answer: d
Clarification: V_b = 44m/s, k₁ = 1, k₃ = 1,
for terrain with closely spaced buildings, height of building=20m, class A: k₂=1.01 (from IS 875 Part 3)
V_z = k₁k₂k₃V_b = 1×1.01x1x44 = 44.44 m/s
p_z = 0.6Vz2 = 0.6x(44.44)2 = 1184.95 N/m².

11. What is the partial safety factor for combination of DL+LL for limit state of strength, where DL=Dead load, LL=imposed load?
a) 1.2
b) 1.0
c) 0.8
d) 1.5
Answer: d
Clarification: For limit state of strength, the load combination is 1.5(DL+LL), for limit state of serviceability, the load combination is 1.0(DL+LL), where DL=Dead load, LL=imposed load.

12. Which of the following load combination is not possible?
a) Dead load + imposed load + wind load
b) Dead load + imposed load + earthquake load
c) Dead load + wind load + earthquake load
d) Dead load + imposed load
Answer: c
Clarification: According to IS code, it is assumed that maximum wind load and earthquake load will not occur simultaneously on a structure. The following combination of loads with appropriate partial safety factors may be considered : (i)Dead load + imposed load, (ii) Dead load + imposed load + earthquake load or wind load, (iii) Dead load + wind load or earthquake load, (iv) Dead load + erection load.

13. What is the partial safety factor for dead load in combination of DL+LL+WL/EL for limit state of serviceability, where DL=Dead load, LL=imposed load , WL=wind load, EL=earthquake load ?
a) 1.0
b) 0.8
c) 1.5
d) 1.2
Answer: b
Clarification: For limit state of strength, the load combination is 1.2(DL+LL+WL/EL), for limit state of serviceability, the load combination is 1.0DL+0.8LL+0.8WL/EL, where DL=Dead load, LL=imposed load, WL=wind load, EL=earthquake load.

14. What is the partial safety factor for dead load in combination of DL+ WL/EL for limit state of serviceability, where DL=Dead load, WL=wind load, EL=earthquake load ?
a) 1.0
b) 1.5
c) 1.2
d) 0.8
Answer: a
Clarification: For limit state of serviceability, the load combination is 1.0(DL +WL/EL), for limit state of strength, the load combination is 1.5(DL +WL/EL), where DL=Dead load, WL=wind load, EL=earthquake load.

15. What is the partial safety factor for imposed load in combination of DL+LL+AL , where DL=Dead load, WL=wind load, AL=Accidental load ?
a) 1.0
b) 0.5
c) 0.4
d) 0.35
Answer: d
Clarification: The load combination is 1.0DL+ 0.35LL+ 1.0AL, where DL=Dead load, WL=wind load, AL=Accidental load for limit state of strength.

Design of Steel Structures Interview Questions and Answers for Experienced people on “Modes of Failure, Slenderness Ratio and displacement”.

1. What is slenderness ratio of a tension member?
a) ratio of its least radius of gyration to its unsupported length
b) ratio of its unsupported length to its least radius of gyration
c) ratio of its maximum radius of gyration to its unsupported length
d) ratio of its unsupported length to its maximum radius of gyration
Answer: b
Clarification: Slenderness ratio of tension member is ratio of its unsupported length to its least radius of gyration. This limiting slenderness ratio is required in order to prevent undesirable lateral movement or excessive vibration.

2. What is the maximum effective slenderness ratio for a tension member in which stress reversal occurs?
a) 180
b) 200
c) 280
d) 300
Answer: a
Clarification: The maximum effective slenderness ratio for a tension member in which stress reversal occurs due to loads other than wind or seismic forces is 180.

3. What is the maximum effective slenderness ratio for a member subjected to compressive forces resulting only from combination of wind/earthquake actions?
a) 180
b) 200
c) 340
d) 250
Answer: d
Clarification: The maximum effective slenderness ratio for a member subjected to compressive forces resulting only from combination of wind or earthquake actions, such that the deformation of such member does not adversely affect stresses in any part of structure is 250.

4. What is the maximum effective slenderness ratio for a member normally acting as a tie in roof truss or a bracing member?
a) 180
b) 200
c) 350
d) 400
Answer: c
Clarification: The maximum effective slenderness ratio for a member normally acting as a tie in roof truss or a bracing member, which is not considered when subject to stress reversal resulting from action of wind or earthquake forces is 350.

5. What is the maximum effective slenderness ratio for members always in tension?
a) 400
b) 200
c) 350
d) 150
Answer: a
Clarification: The maximum effective slenderness ratio for members always in tension other than pre-tensioned members is 400.

6. The limits specified for slenderness ratio are not
a) applicable to cables
b) applicable to angle sections
c) applicable to built-up sections
d) applicable to circular sections
Answer: a
Clarification: The limits specified for slenderness ratio in the IS code are not applicable to cables. They are applicable to angle sections, built-up sections, circular sections.

7. The displacement of tension member under service load is given by
a) PLEA_g
b) PLE/A_g
c) PL/EA_g
d) P/LEA_g
Answer: c
Clarification: The displacement, that is increase in length of tension member, under service load is given by Δ = PL/EA_g, where Δ = Elongation of member in mm, P= unfactored axial load in N, L = length of member in mm, E = elastic modulus = 2×10⁵MPa, A_g = gross cross sectional area of member in mm².

8. What is gross section yielding?
a) considerable deformation of the member in longitudinal direction may take place before it fractures, making the structure unserviceable
b) considerable deformation of the member in longitudinal direction may take place before it fractures, making the structure serviceable
c) considerable deformation of the member in lateral direction may take place before it fractures, making the structure unserviceable
d) considerable deformation of the member in lateral direction may take place before it fractures, making the structure serviceable
Answer: a
Clarification: Tension member without bolt holes can resist loads up to ultimate load without failure. But such a member will deform in longitudinal direction considerably(10-15% of its original length)before fracture and the structure becomes unserviceable.

9. What is net section rupture failure?
a) rupture of member when the cross section reaches yield stress
b) rupture of member when the cross section reaches ultimate stress
c) rupture of member when the cross section reaches less value than yield stress
d) rupture of member when the cross section is reaches very less value than ultimate stress
Answer: b
Clarification: The point adjacent to hole reaches yield stress first when tension member with hole is loaded statically. The stress at that point remains constant and each fibre away from hole progressively reaches yield stress on further loading. With increasing load, deformations continue until finally rupture of member occurs when entire net cross section of member reaches ultimate stress.

10. The tensile stress adjacent to hole will be ____________
a) about five times the average stress on the net area
b) about half the average stress on the net area
c) equal to average stress on the net area
d) about two to three times the average stress on the net area
Answer: d
Clarification: From the theory of elasticity, the tensile stress adjacent to hole will be about two to three times the average stress on the net area, depending upon the ratio of diameter of hole to the width of plate normal to direction of stress.

11. What is stress concentration factor?
a) ratio of average stress to maximum elastic stress
b) product of average stress and maximum elastic stress
c) ratio of maximum elastic stress to average stress
d) twice the average stress
Answer: c
Clarification: The ratio of maximum elastic stress to average stress (fmax/favg)is called as stress concentration factor. It becomes very significant when repeated applications of load may lead to fatigue failure or where there is possibility of brittle fracture of tension member under dynamic load. It may minimised by providing suitable joint and member details.

12. What is block shear failure?
a) failure of fasteners occurs along path involving tension on one plane and shear on perpendicular plane along fasteners
b) failure of member occurs along path involving tension on one plane and shear on perpendicular plane along fasteners
c) failure of member occurs along path involving tension on one plane and shear on parallel plane along fasteners
d) failure of fasteners occurs along path involving tension on one plane and shear on parallel plane along fasteners
Answer: b
Clarification: Failure of member occurs along path that involves (i) tension on one plane and (ii) shear on perpendicular plane along fasteners in block shear failure mode.

13. The possibility of block shear failure increases by
a) larger connection length
b) increasing the number of bolts per connection
c) with use of low strength bolts
d) with use of high bearing strength material
Answer: d
Clarification: The block shear failure becomes a possible mode of failure when material bearing strength and bolt shear strength are higher. When high bearing strength of material and high strength bolts are used, only few bolts are required in connection. Decreasing number of bolts per connection results in smaller connection length, but the possibility of block shear failure increases.

Design of Steel Structures Questions and Answers for Campus interviews on “Effective Length and Slenderness Ratio of Compression Members”.

1. Effective length of compression member is ________
a) distance between ends of members
b) distance between end point and midpoint of member
c) distance between points of contraflexure
d) distance between end point and centroid of member
Answer: c
Clarification: Effective length of compression member is distance between points of contraflexure. It should be derived from actual length and end conditions.

2. Magnitude of effective length depends upon
a) material of member
b) rotational restraint supplied at end of compression member
c) load applied on member
d) location where member is used
Answer: b
Clarification: Magnitude of effective length depends upon rotational restraint supplied at end of compression member and upon resistance to lateral movement provided.

3. Which of the following is true?
a) greater the effective length, greater the load carrying capacity
b) smaller the effective length, smaller the load carrying capacity
c) smaller the effective length, more the danger of lateral buckling
d) smaller the effective length, smaller the danger of lateral buckling
Answer: d
Clarification: Smaller the effective length of particular compression member, smaller is the danger of lateral buckling and greater is the load carrying capacity.

4. What is the effective length when both ends of compression member are fixed?
a) 0.65L
b) 0.8L
c) L
d) 2L
Answer: a
Clarification: The effective length of compression member when both ends of compression member are fixed is 0.65L (i.e. L/√2), where L is the length of the member.

5. What is the effective length when both ends of compression member are hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L
Answer: c
Clarification: The effective length of compression member when both ends of compression member are hinged is L, where L is the length of the member.

6. What is the effective length when one end of compression member is fixed and other end is free?
a) 0.65L
b) 0.8L
c) L
d) 2L
Answer: d
Clarification: The effective length of compression member when one end is fixed and other end is free is 2L, where L is the length of the member.

7. What is the effective length when one end of compression member is fixed and other end is hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L
Answer: b
Clarification: The effective length of compression member when one end is fixed and other end is hinged is 0.8L, where L is the length of the member.

8. What is slenderness ratio of compression member?
a) ratio of effective length to radius of gyration
b) ratio of radius of gyration to effective length
c) difference of radius of gyration and effective length
d) product of radius of gyration and effective length
Answer: a
Clarification: The tendency of member to buckle is usually measured by its slenderness ratio. Slenderness ratio of member is ratio of effective length to appropriate radius of gyration (λ = kL/r). This is valid only when column has equal unbraced heights for both axes and end condition is same for particular section.

9. Maximum radius of gyration (minimum slenderness ratio) can be obtained by
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
Clarification: Maximum radius of gyration is obtained when material of section is farthest from centroid i.e. away from its axis.

Design of Steel Structures for Campus Interviews,

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

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
Clarification: 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) λ_LT ≤ 0.4
b) λ_LT ≥0.4
c) λ_LT > 0.8
d) λ_LT = 0.8
Answer: a
Clarification: The effect of lateral-torsional buckling need not be considered when λ_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_bZ_p /f_bd
b) M_d = β_b /Z_pf_bd
c) M_d = β_bZ_p
d) M_d = β_bZ_pf_bd
Answer: d
Clarification: The bending strength of laterally unsupported beams is given by M_d = β_bZ_pf_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
Answer: c
Clarification: The value of β_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_eZ_p
c) Z_p/Z_e
d) Z_p
Answer: a
Clarification: The value of β_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
Answer: b
Clarification: The value of design bending compressive stress f_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 + (φ²_LT – λ²_LT)}
b) X_LT = 1/{φ_LT – (φ²_LT + λ²_LT)}
c) X_LT = 1/{φ_LT – (φ²_LT + λ²_LT)0.5}
d) X_LT = 1/{φ_LT + (φ²_LT – λ²_LT)0.5}
Answer: d
Clarification: The bending stress reduction factor to account for lateral buckling is given by X_LT = 1/{φ_LT + (φ²_LT – λ²_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) + λ²_LT].
b) φ_LT = [ 1 + α_LT (λ_LT – 0.2) + λ²_LT].
c) φ_LT = 0.5 [ 1 – α_LT (λ_LT + 0.2) + λ²_LT].
d) φ_LT = 0.5 [ 1 + α_LT (λ_LT – 0.2) + λ²_LT].
Answer: d
Clarification: The value of φ_LT in bending stress reduction factor is given by φ_LT = 0.5 [ 1 + α_LT (λ_LT – 0.2) + λ²_LT], where α_LT is imperfection factor, λ_LT is non dimensional slenderness ratio.