Category: Design of Steel Structures Objective Questions

Design of Steel Structures Multiple Choice Questions on “Design of Compression Members – I”.

1. A column that can support same load in compression as it can in tension is called
a) intermediate column
b) long column
c) short column
d) cannot be determined
Answer: c
Clarification: A column that can support same load in compression as it can in tension is called short column. Short column usually fail by crushing.

2. The strength of compression members subjected to axial compression is defined by curves corresponding to _______ classes
a) a, b, c and d
b) a, d
c) b, e, f
d) e, f, g
Answer: a
Clarification: The strength of compression members subjected to axial compression is defined by curves corresponding to a, b, c and d classes. The value of imperfection factor depends on type of buckling curve.

3. Which of the following is not a compression member?
a) strut
b) boom
c) tie
d) rafter
Answer: c
Clarification: Strut, boom and rafter are compression members, whereas tie is a tension member.

4. The best compression member section generally used is
a) single angle section
b) I-section
c) double angle section
d) channel section
Answer: b
Clarification: Generally, ISHB sections are used as compression members.

5. The best double-angle compression member section is
a) unequal angles with short leg connected
b) unequal angles with long leg connected
c) unequal angles on opposite side of gusset plate
d) unequal angles on same side of gusset plate
Answer: a
Clarification: Unequal angles with short leg connected are preferred as compression member section.

6. The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 10.5ε
c) 15.7ε
d) 9.4ε
Answer: c
Clarification: The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than 15.7ε and for a welded section, less than 13.6ε.

7. The flange is classified as plastic if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 9.4ε
c) 10.5ε
d) 15.7ε
Answer: b
Clarification: The flange is classified as plastic if outstand element of compression flange of rolled section is less than 9.4ε and for a welded section, less than 8.4ε.

8. The outstand element of compression flange of a rolled section is 10.2 (ε=1). The flange will be classified as
a) compact
b) plastic
c) semi-compact
d) slender
Answer: a
Clarification: The flange is classified as compact if outstand element of compression flange of rolled section is less than 10.5ε and for a welded section, less than 9.4ε.

9. The design compressive stress of compression member in IS 800 is given by
a) Rankine Formula
b) Euler Formula
c) Perry-Robertson formula
d) Secant-Rankine formula
Answer: c
Clarification: The design compressive stress of axially loaded compression member in IS 800 is given by Perry-Robertson formula. IS 800:2007 proposes multiple columns curves in nin-dimensional form based on Perry-Robertson approach.

Design of Steel Structures Multiple Choice Questions on “Castellated Beams & Lintels”.

1. What is castellated beam?
a) beam with no openings in web
b) beam with number of regular openings in web and flange
c) beam with number of regular openings in web
d) beam with number of regular openings in flange
Answer: c
Clarification: A beam with number of regular openings in its web is called castellated beam. A castellated beam is formed by flame cutting a single rolled wide flange beam in a definite predetermined pattern and then rejoining the segments by welding to form a regular pattern of holes in the web.

2. The new rolled section of castellated beam will have depth
a) 50% more than original section
b) 50% less than original section
c) 25% less than original section
d) depth does not change
Answer: a
Clarification: The new rolled section of castellated beam will have depth at least 50% more and its section modulus increases by 2.25 times the original section. This allows the beam to span further than parent rolled section.

3. Castellated beams have ______ shear capacity than original beams
a) shear capacity does not change
b) twice
c) increased
d) reduced
Answer: d
Clarification: Castellated beams have reduced shear capacity. It has reduced shear capacity due to stress concentrations near the openings.

4. Which of the following measures can be taken to improve shear capacity of castellated beams?
a) openings can be made away from neutral axis
b) openings can be made close to neutral axis
c) making cuts in straight manner
d) by not using stiffenings
Answer: b
Clarification: Shear capacity of castellated beams can be improved by making openings close to neutral axis and making cuts in a wavy manner. Stiffening can be provided at load concentrations and reaction points to improve its shear carrying capacity.

5. Which of the following is not an advantage of castellated beam?
a) light in weight
b) can be assembled fast
c) cheaper
d) high fire resistance than original rolled section
Answer: d
Clarification: Castellated beams are light in weight, cheaper, they have relatively high resistance and can be assembled fast at the construction site. They are less fire resistant than normal rolled sections. Castellated beams can very easily be cambered and cranked.

6. In which of the following cases are castellated beam desirable?
a) when more span to be covered than rolled section
b) when beam subjected to substantial concentrated loads
c) when beam to be used as continuous beam
d) when higher fire resistance than rolled section required
Answer: a
Clarification: The section of castellated beam will have more depth and section modulus than original rolled section. This allows the beam to span further than parent rolled section. Castellated beams may not be desirable when beam is subjected to substantial concentrated loads, or when castellated beam is used as a continuous beam across several supports. Castellated beams are less attractive when very high requirements for fire resistance are required because the fire resistant coating has to be around 20% thicker than for rolled sections in order to obtain the same fire resistance as rolled section.

7. Match the pairs

   (A) Vierendeel mechanism		  (i) caused by heavy loading and short span
   (B) Lateral torsional buckling of web  (ii) caused due to excessive horizontal shear
   (C) Rupture of welded joint in web	  (iii) due to excessive deformation across openings in web
   (D) Web Buckling			  (iv)caused by large shear

a) A-i, B-ii, C-iii, D-iv
b) A-iv, B-iii, C-ii, D-i
c) A-iii, B-iv, C-ii, D-i
d) A-i, B-iv, C-iii, D-ii
Answer: c
Clarification: There are number of possible modes of failure for castellated beams. Some of them are as follows: (i) Vierendeel mechanism – occurs due to excessive deformation across one of the openings in web and formation of hinges in corners of castellation, (ii) Lateral torsional buckling of web – caused by large shear at welded joint, (iii) Rupture of welded joint in web – caused due to excessive horizontal shear at welded joint in the web, (iv) Web Buckling – caused by heavy loading and short span of beam, this may be avoided at support by filling firt castellation by welding plate in the hole.

8. What are lintels?
a) beams provided in foundation
b) beams on roof of building
c) columns above openings in wall
d) beams above openings in wall
Answer: d
Clarification: Beams provided above the openings in walls to support masonry that comes in between the opening and slab above are called as lintels. It is desirable that lintel is built flush from both the sides of the walls.

9. _____ section is suitable for small openings and _____ section is suitable for large openings
a) flat, I-section
b) I-section, flat
c) angles, flat
d) angles, angles
Answer: a
Clarification: Flats and plate sections are used for small openings. For openings of moderate dimension, back-to-back angles and inverted T-sections are best options. For large openings, channels, I-sections or built-up sections are preferred. If there is any doubt about lateral support from the wall, I-section with plates can be used.

10. Design of lintel is carried out for
a) weight of slab
b) no load is considered from masonry load above the opening
c) small portion of masonry load above the opening
d) large portion of masonry load above the opening
Answer: c
Clarification: Design loads for lintels are not well defined because it is not certain as how much load from masonry will come over lintel. It is assumed that after setting of mortar, load from masonry is distributed by arch action. Design of lintel is carried out for small portion of masonry load above the opening.

11. When the slab over lintel is above apex of equilateral triangle formed on lintel, load of masonry is considered as
a) rectangular load
b) triangular load
c) trapezoidal load
d) no load is considered
Answer: b
Clarification: When the slab over lintel is above apex of equilateral triangle formed on lintel, the load of masonry in the triangle thus formed is assumed to act over it. When the design load is from triangular portion of masonry , the maximum moment will be Wl/6, where W = triangular load from masonry and l = effective span of lintel.

12. When the slab over lintel is below apex of equilateral triangle formed on lintel, load of masonry is considered as
a) rectangular load
b) triangular load
c) trapezoidal load
d) no load is considered
Answer: a
Clarification: When the slab over lintel is below apex of equilateral triangle formed on lintel, the load of masonry in the rectangle is considered. The load of masonry in the rectangle is assumed to act over by taking length equal to span of lintel and height equal to clear height of slab above the lintel.

Design of Steel Structures Multiple Choice Questions on “Pin Connections”.

1. What are pin connections?
a) structural members connected by bolts
b) structural members connected by cylindrical pins
c) structural members connected by bolts and pins
d) structural members connected by welding
Answer: b
Clarification: When two structural members are connected by means of cylindrical-shaped pins, the connection is known as pinned connection. It resists horizontal and vertical movement, but not moment.

2. Pin connections are provided when _______ required.
a) hinge joint
b) fixed joint
c) irrotational joint
d) rigid joint
Answer: a
Clarification: Pin connections are provided when hinged joints are required , where zero moments or free rotation is desired and horizontal and vertical movement are not desired.

3. Pins used for the connection _________
a) does not affect secondary stresses
b) increase secondary stresses
c) reduce secondary stresses
d) doubles secondary stresses
Answer: c
Clarification: Pins used for the connection reduce secondary stresses. It serves the same purpose as shank of bolt.

4. Forces acting on pin are ______ those on bolt
a) less than
d) equal to
c) half the force
d) greater than
Answer: d
Clarification: Since only one pin is present in the connection, forces acting on pin are greater than those on bolt.

5. In which of the following cases pin connections are not used?
a) truss bridge girders
b) hinged arches
c) tall buildings
d) diagonal bracing connection
Answer: c
Clarification: Pin connections are used in following cases : (i) truss bridge girders, (ii) hinged arches, (iii)tie rod connection in water tanks, (iv)as diagonal bracing connections in beams and columns, (v)chain-link cables suspension bridges.

6. Shear capacity of pin when rotation is allowed is given by
a) 0.5f_ypA
b) 0.6f_ypA
c) 0.7f_ypA
d) 0.8f_ypA
Answer: a
Clarification: Shear capacity of pin is given by (i) 0.5f_ypA, when rotation is required, (ii) 0.6f_ypA, when rotation is not required, where f_yp=design strength of pin, A = cross sectional area of pin.

7. Bearing capacity of pin when rotation is not allowed is given by
a) 0.8f_ypdt
b) 0.6f_ypdt
c) 0.7f_ypdt
d) 1.5f_ypdt
Answer: d
Clarification: Bearing capacity of pin given by (i) 1.5f_ypA, when rotation is not required, (ii) 0.8f_ypdt, when rotation is not required, where f_yp=design strength of pin.

8. Moment capcity of pin when rotation is not allowed is given by
a) 0.8f_ypZ
b) 0.6f_ypZ
c) 1.5f_ypZ
d) 2.0f_ypZ
Answer: c
Clarification: Moment capacity of pin given by (i) 1.5f_ypZ, when rotation is not required, (ii) 1.0fypdt, when rotation is not required, where f_yp=design strength of pin, Z=section modulus of the pin.

9. Members joined by pin connections are separated some distance _____
a) to allow friction
b) to allow for bolt heads
c) to allow bending
d) to allow to be removed
Answer: b
Clarification: Members joined by pin connections are separated some distance (i) to prevent friction, (ii) to allow for bolt heads, if the members are built up, (iii) to facilitate painting.

10. Design of pin connections is primarily governed by
a) shear
b) bending
c) flexure
d) friction
Answer: c
Clarification: Large bending moments are generated since members joined by pin connections are separated some distance. So the pin diameter is generally governed by flexure.

Design of Steel Structures MCQs on “General Requirements for Plastic Design and Plastic Hinge”.

1. Which of the following assumptions is correct for plastic design?
a) material obeys Hooke’s law before the stress reaches f_y
b) yield stress and modulus of elasticity does not have same value in compression and tension
c) material is homogenous and isotropic in both elastic and plastic states.
d) material is not sufficiently ductile to permit large rotations
Answer: c
Clarification: The material obeys Hooke’s law till the stress reaches f_y. The yield stress and modulus of elasticity have the same value in compression and tension. The material is homogeneous and isotropic in both elastic and plastic states. The material is assumed to be sufficiently ductile to permit large rotations of section to take place.

2. Which of the following assumptions is not correct for plastic design?
a) plastic hinge rotations are small compared with elastic deformations so all the rotations are concentrated at plastic hinges
b) segments between plastic hinges are rigid
c) influence of normal and shear forces on plastic moments is not considered
d) plane section remains plane after bending and the effect of shear is neglected
Answer: a
Clarification: The plastic hinge rotations are large compared with elastic deformations so all the rotations are concentrated at plastic hinges. The segments between plastic hinges are rigid. The influence of normal and shear forces on plastic moments is not considered. The plane section remains plane even after bending and the effect of shear is neglected.

3. Which of the following are the conditions that stress strain characteristics of steel should obey to ensure plastic moment redistribution?
a) yield plateau should be less than 6 times the yield strain
b) ratio of ultimate tensile stress to yield stress should be less than 1.2
c) steel should not exhibit strain-hardening capacity
d) elongation on standard gauge length should be more than 15%
Answer: d
Clarification: The conditions that stress strain characteristics of steel should obey to ensure plastic moment redistribution are: (i) yield plateau(horizontal portion of stress strain curve) should be greater than 6 times the yield strain, (ii) ratio of ultimate tensile stress to yield stress should be more than 1.2, (iii) elongation on standard gauge length should be more than 15%, (iv) steel should exhibit strain-hardening capacity.

4. Which of the following conditions are true for using plastic method of analysis as per IS 800?
a) members shall not be hot-rolled or fabricated using hot-plates
b) yield stress of steel should not be greater than 450MPa
c) cross section should be unsymmetrical about its axis perpendicular to axis of plastic hinge rotation
d) cross section of members not containing plastic hinges should be ‘plastic’ and those members containing plastic hinges should be ‘compact’
Answer: b
Clarification: The following are conditions for using plastic method of analysis as per IS 800: (i) yield stress of steel should not be greater than 450MPa, (ii) members shall be hot-rolled or fabricated using hot-plates, (iii) cross section should be symmetrical about its axis perpendicular to the axis of plastic hinge rotation, (iv) cross section of members not containing plastic hinges should be ‘compact’ and those members containing plastic hinges should be ‘plastic’.

5. Which of the following is true regarding plastic design methods?
a) design needs to satisfy elastic strain compatibility conditions
b) different factor of safety for all parts of the structure
c) saving of material over elastic methods resulting in lighter structures
d) design is effected by temperature changes, settlement of support, etc
Answer: c
Clarification: The following are the advantages of plastic design methods: (i) realisation of uniform and realistic factor of safety for all parts of the structure , (ii)simplified analytical procedure and rapidity of obtaining moments since no need to satisfy elastic strain compatibility conditions, (iii) no effect due to temperature changes, settlement of support, etc, (iv) saving of material over elastic methods resulting in lighter structures.

6. Which of the following is true regarding plastic design methods?
a) difficult to design for fatigue
b) more saving in column design
c) lateral bracing requirements are less stringent than for elastic design
d) moments produced by different loading conditions can be added together
Answer: a
Clarification: (i) There is little saving in column design, (ii) lateral bracing requirements are more stringent than for elastic design, (iii) difficult to design for fatigue,(iv) moments produced by different loading conditions needs to be calculated separately(cannot be added together) and the largest plastic moment is selected.

7. What is plastic hinge?
a) zone of bending due to flexure in a structural member
b) zone of yielding due to flexure in a structural member
c) zone of non-yielding due to flexure in a structural member
d) zone of yielding due to twisting in a structural member
Answer: b
Clarification: Plastic hinge is a zone of yielding due to flexure in a structural member. It is used to describe a deformation of a section when plastic bending occurs.

8. Plastic hinge behaves like a ______
a) friction mechanical hinge except that there is always a fixed moment constraint
b) frictionless mechanical hinge except that there is no fixed moment constraint
c) friction mechanical hinge except that there is no fixed moment constraint
d) frictionless mechanical hinge except that there is always a fixed moment constraint
Answer: d
Clarification: Plastic hinge behaves like a frictionless mechanical hinge except that there is always a fixed moment constraint which is equal to plastic moment capacity of the section.

9. Which of the following is true about hinged length?
a) value of moment adjacent to yield zone is more than yield moment up to hinged length of structural member
b) value of moment adjacent to yield zone is less than yield moment up to hinged length of structural member
c) value of moment adjacent to yield zone is half the yield moment up to hinged length of structural member
d) value of moment adjacent to yield zone is equal to yield moment up to hinged length of structural member
Answer: a
Clarification: The value of moment adjacent to yield zone is more than yield moment up to a certain length of structural member. This length is called hinged length.

10. Hinged length depends upon
a) weight of member
b) type of connection
c) type of loading
d) number of bolts used in connection
Answer: c
Clarification: The hinged length depends upon the type of loading and the geometry of cross section of structural member.

11. What is the hinged length for simply supported rectangular beam of span L with central concentrated load?
a) L/√2
b) 2L
c) L/2
d) L/3
Answer: d
Clarification: For simply supported rectangular beam with central concentrated load, the hinged length is equal to one-third of the span.

12. What is the hinged length for simply supported rectangular beam of span L with uniformly distributed load?
a) L/√3
b) L/√2
c) L/2
d) L/5
Answer: a
Clarification: For simply supported rectangular beam with uniformly distributed load, the hinged length is equal to span/√3.

To practice MCQs on all areas of Design of Steel Structures,

Design of Steel Structures Assessment Questions and Answers on “Design of Compression Members – II”.

1. The design compressive strength of member is given by
a) A_ef_cd
b) A_e /f_cd
c) f_cd
d) 0.5A_ef_cd
Answer: a
Clarification: The design compressive strength of member is given by P_d = A_ef_cd, where A_e is effective sectional area, f_cd is design compressive stress.

2. The design compressive stress, f_cd of column is given by
a) [f_y / γ_m0]/ [φ – (φ²-λ²)²].
b) [f_y / γ_m0] / [φ + (φ²-λ²)].
c) [f_y / γ_m0]/[φ – (φ²-λ²)^0.5].
d) [f_y / γ_m0] / [φ + (φ²-λ²)^0.5].
Answer: d
Clarification: The design compressive stress, f_cd of column is given by f_cd = [f_y / γ_m0] / [φ + (φ²-λ²)^0.5], where f_y is yield stress of material, φ is dependent on imperfection factor, λ is non dimensional effective slenderness ratio.

3. What is the value of imperfection factor for buckling class a?
a) 0.34
b) 0.75
c) 0.21
d) 0.5
Answer: c
Clarification: The value of imperfection factor, α for buckling class a is 0.21. The imperfection factor considers all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

4. If imperfection factor α = 0.49, then what is the buckling class?
a) a
b) c
c) b
d) g
Answer: b
Clarification: For buckling class c, the value of imperfection factor is 0.49. The imperfection factor takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

5. The value of φ in the equation of design compressive strength is given by
a) φ = 0.5[1-α(λ-0.2)+λ²].
b) φ = 0.5[1-α(λ-0.2)-+λ²].
c) φ = 0.5[1+α(λ+0.2)-λ²].
d) φ = 0.5[1+α(λ-0.2)+λ²].
Answer: d
Clarification: The value of φ in the equation of design compressive strength is given by φ = 0.5[1+α(λ-0.2)+λ²], where α is imperfection factor(depends on buckling class) and λ is non-dimensional effective slenderness ratio.

6. Euler buckling stress f_cc is given by
a) (π²E)/(KL/r)²
b) (π²E KL/r)²
c) (π²E)/(KL/r)
d) (π²E)/(KLr)²
Answer: a
Clarification: Euler buckling stress f_cc is given by f_cc = (π²E)/(KL/r)², where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length, KL to appropriate radius of gyration, r.

7. What is the value of non dimensional slenderness ratio λ in the equation of design compressive strength?
a) (f_y /f_cc)
b) √(f_y f_cc)
c) √(f_y /f_cc)
d) (f_y f_cc)
Answer: c
Clarification: The value of non dimensional slenderness ratio λ in the equation of design compressive strength is given by λ = √(f_y /f_cc) , where f_y is yield stress of material and f_cc = (π²E)/(KL/r)², where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length.

8. The design compressive strength in terms of stress reduction factor is given by
a) Xf_y
b) Xf_y / γ_m0
c) X /f_y γ_m0
d) Xf_y γ_m0
Answer: b
Clarification: The design compressive strength in terms of stress reduction factor is given by f_cd = Xf_y / γ_m0 , where X = stress reduction factor for different buckling class, slenderness ratio and yield stress = 1/ [φ + (φ²-λ²)^0.5], f_y is yield stress of material and γ_m0 is partial safety factor for material strength.

9. The value of design compressive strength is limited to
a) f_y + γ_m0
b) f_y
c) f_y γ_m0
d) f_y / γ_m0
Answer: d
Clarification: The value of design compressive strength is given by f_cd = [f_y / γ_m0] / [φ + (φ²-λ²)^0.5] ≤ f_y / γ_m0 i.e. f_cd should be less than or equal to f_y / γ_m0.

10. The compressive strength for ISMB 400 used as a column for length 5m with both ends hinged is
a) 275 kN
b) 375.4 kN
c) 453 kN
d) 382 kN
Answer: b
Clarification: K = 1 for both ends hinged, KL = 1×5000 = 5000, r = 28.2mm (from steel table), A_e = 7846 mm²(from steel table)
KL/r = 5000/28.2 = 177.3
h/bf = 400/140 = 2.82, t = 16mm Therefore, buckling class = b
From table in IS code, f_cd = 47.85MPa
P_d = A_e f_cd = 7846 x 47.85 = 375.43 kN.

Design of Steel Structures Assessment Questions,

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

1. What are purlins?
a) beams provided in foundation
b) beams provided above openings
c) beams provided over trusses to support roofing
d) beams provided on plinth level
Answer: c
Clarification: Purlins are beams provided over trusses to support sloping roof system between adjacent trusses. Channels, angle sections, and old formed Z-sections are widely used as purlins.

2. Theoretically, purlins are generally placed at
a) only at panel points
b) only at edges
c) only at mid span
d) only at corners of roof
Answer: a
Clarification: Theoretically, it is desirable to place purlins only at panel points. They are placed at panel points to avoid bending in the top chords of roof trusses. For large trusses, it is more economical to space purlins at closer intervals.

3. Purlin section is subjected to
a) not subjected to bending or twisting
b) twisting only
c) symmetrical bending
d) unsymmetrical bending
Answer: d
Clarification: The wind force is assumed to act normal to roof truss and gravity load pass through centre of gravity of purlin section. Hence, the purlin section is subjected to twisting in addition to bending. Such bending is called unsymmetrical bending.

4. If purlins are assumed to be simply supported, the moments will be
a) wl²/10
b) wl/8
c) wl/10
d) wl²/8
Answer: d
Clarification: Purlins can be designed simple, continuous or cantilever beams. If purlins are assumed to be simply supported, the moments will be wl²/8. If they are assumed to be continuous, the moments will be slightly less and taken as wl2/10. IS 800 recommends the purlins to be designed as continuous beams.

5. While erecting channel section purlins, it is desirable that they are erected over rafter with their flange
a) facing down slope
b) facing up slope
c) does not depend whether up slope or down slope
d) flanges are placed randomly
Answer: b
Clarification: While erecting angle, channel or I- section purlins, it is desirable that they are erected over rafter with their flange facing up slope. In this position, the twisting moment does not cause any instability. The twisting moment will cause instability if the purlins are kept in such a way that the flanges face the downward slope.

6. Sag rods are provided at
a) one-third points between roof trusses
b) end of span
c) two-third points between roof trusses
d) are never provided
Answer: a
Clarification: Purlin sections have tendency to sag in the direction of sloping roof . So, sag rods are provided midway or at one-third points between roof trusses to take up the sag.

7. Which of the following is not true about sag rods?
a) sag rods are provided at midway or at one-third points between roof trusses
b) these rods reduce the moment M_yy
c) these rods increase the moment M_yy
d) these rods result in smaller purlin sections
Answer: c
Clarification: Sag rods are provided midway or at one-third points between roof trusses to take up the sag in the direction of sloping roof by purlins. These rods provide lateral support with resprct to y-axis bending. Consequently, moment M_yy is reduced and thereby result in smaller purlin section. they are useful in keeping the purlins in proper alignment during erection until roofing is installed and connected to purlins.

8. When one sag rod is used, the moment about web axis
a) reduces by 50%
b) increases by 50%
c) increases by 75%
d) reduces by 75%
Answer: d
Clarification: If sag rods are not used, the maximum moment about web axis would be wl2/8. When one sag rod is used, the moments are reduced by 75% and when two sag rods are used at one-third points, the moments are reduced by 91%.

9. The maximum bending moment for design of channel/I-section purlin is calculated by
a) Wl/10, where W= concentrated load
b) Wl/8, where W= concentrated load
c) W/10, where W= concentrated load
d) W/8, where W= concentrated load
Answer: a
Clarification: The gravity load, P1 and load due to wind component, H1 are computed. The loads are multiplied by load factors. Thus, P = γfP1, H = γfH1 . The maximum bending moment are calculated as M_z = Pl/10 and M_y = Hl/10, where P= factored load along z-axis, H = factored load along y-axis, l= span of purlin (c/c distance between adjacent trusses).

10. The required section modulus of the channel/I-section purlin can be determined by
a) Z_pz = M_yγ_m0/f_y + (b/d)(M_zγ_m0/f_y)
b) Z_pz = M_zγ_m0/f_y + (b/d)(M_yγ_m0/f_y)
c) Z_pz = M_zγ_m0/f_y + 2.5(b/d)(M_yγ_m0/f_y)
d) Z_pz = M_yγ_m0/f_y + 2.5(b/d)(M_zγ_m0/f_y)
Answer: c
Clarification: The required section modulus of the purlin section can be determined by Z_pz = M_zγ_m0/f_y + 2.5(b/d)(M_yγ_m0/f_y ), where γ_m0 is partial safety factor for material = 1.1, d is depth of trial section, b is the breadth of the trial section, M_z and M_y are factored bending moments about Z and Y axes, respectively, and f_y is yield stress of steel. Since the above equation involves b and d of a section, trial section must be used and from the above equation , it is checked whether chosen section is adequate or not.

11. The design capacity of channel/I-section purlin is given by
a) M = Z_p/f_y
b) M = Z_pγ_m0f_y
c) M = Z_pγ_m0/f_y
d) M = γ_m0/f_y
Answer: b
Clarification: The design capacity of channel/I-section purlin is given by M_dz = Z_pzγ_m0/f_y and M_dy = Z_pyγ_m0/f_y , M_dz and M_dy are design moment capacity about Z and Y axes, respectively, Z_pz and Z_py are plastic section modulus about Z and Y axes, respectively and f_y is yield stress of steel. For safety, design moment capacity should be always greater than or equal to factored bending moments.

12. The check for design capacity of channel/I-section purlin is given by
a) M_dz ≤ 1.2Z_eyf_y/γ_m0, M_dy ≤ 2.4Z_ezf_y/γ_m0
b) M_dz ≤ Z_ezf_y/γ_m0, M_dy ≤ 1.2Z_eyf_y/γ_m0
c) M_dz ≤ γfZ_eyf_y/γ_m0, M_dy ≤ 1.2Z_ezf_y/γ_m0
d) M_dz ≤ 1.2Z_ezf_y/γ_m0, M_dy ≤ γfZ_eyf_y/γ_m0
Answer: d
Clarification: The check for design capacity of channel/I-section purlin is given by M_dz ≤ 1.2Z_ezf_y/γ_m0 , M_dy ≤ γyZ_eyf_y/γ_m0 , where M_dz and M_dy are design moment capacity about Z and Y axes, respectively, Z_ez and Z_ey are elastic section modulus about Z and Y axes, respectively and f_y is yield stress of steel. Since in y-direction, the shape factor Z_p/Z_e will be greater than 1.2, γf is used instead of 1.2. If 1.2 is used the onset of yielding under unfactored loads cannot be prevented.

13. Which of the following relation is correct for design of channel/I-section purlin?
a) (M_z/M_dz) + (M_y/M_dy) ≥ 1
b) (M_z/M_dz) + (M_y/M_dy) ≤ 1
c) (M_dz/M_z) + (M_y/M_dy) ≤ 1
d) (M_dz/M_z) + (M_dy/M_y) ≥ 1
Answer: b
Clarification: The local capacity of the section is checked by interaction equation. It is given by (M_z/M_dz) + (M_y/M_dy) ≤ 1 , where M_dz and M_dy are design moment capacity about Z and Y axes, respectively, and M_z and M_y are factored bending moments about Z and Y axes, respectively.

14. For which of the following slope of roof truss, angle section purlin can be used?
a) 25˚
b) 50˚
c) 75˚
d) 60˚
Answer: a
Clarification: Angle sections are unsymmetrical about both the axes. Angle sections can be used as purlin section. provided slope of the roof truss is less than 30˚.

15. The modulus of section required for angle section purlin is given by
a) Z = M/(0.66xf_y)
b) Z = M/(1.33×0.66xf_y)
c) Z = M/(1.33×0.66xf_y)
d) Z = M/(1.33xf_y)
Answer: c
Clarification: The modulus of section required for angle section purlin is given by Z = M/(1.33×0.66xf_y), M = maximum bending moment = wl²/10, w = unfactored uniformly distributed load, l = span of purlin, f_y is yield stress. The gravity and wind loads are determined to calculate bending moment and both loads are assumed to be normal to roof truss.