Multiple choice question for engineering
Set 1
1. What is plastic-collapse load?
a) load at which sufficient number of elastic hinges are formed
b) load at which sufficient number of plastic hinges are not formed
c) load at which sufficient number of plastic hinges are formed
d) load at which structure fails
Answer
Answer: c [Reason:] The load at which sufficient number of plastic hinges are formed in a structure such that a collapse mechanism is created is called plastic-collapse load or plastic-limit load.
2. What is difference between plastic design and elastic design?
a) In plastic design, redistribution of bending moment is considered
b) In plastic design, redistribution of bending moment is not considered
c) In elastic design, redistribution of bending moment is considered
d) Both in plastic and elastic design, redistribution of bending moment is considered
Answer
Answer: a [Reason:] The difference between plastic design and elastic design is that plastic design takes into account the favourable redistribution of bending moment which takes place in indeterminate structure after first hinge forms at the point of maximum bending moment.
3. Which of the following is true in a fixed beam having concentrated load at one-third point?
a) first hinge is formed at centre of beam
b) after first hinge, moment at that point increases
c) after first hinge, moment at that point decreases
d) after first hinge, moment at that hinge remains constant
Answer
Answer: d [Reason:] In a fixed beam having concentrated load at one-third point, first hinge forms at one of the end supports. As load is increased further, moment at this hinge remains constant at Mp, while the moments at the other support and load point increase until second hinge is formed. When load is further increased, the moment at these two hinges remain constant at Mp, until third and final hinge is formed to make the beam a mechanism. The final ultimate load will be 33% higher than first hinge load.
4. In a fixed beam having concentrated load at one-third point, final ultimate load will be ____ than first hinge load.
a) 33% lower
b) 33% higher
c) 50% higher
d) 50% lower
Answer
Answer: b [Reason:] In a fixed beam having concentrated load at one-third point, final ultimate load will be 33% higher than first hinge load.
5. Which of the following statement is correct?
a) plastic limit load is obtained by multiplying working load with load factor
b) plastic limit load is obtained by dividing working load with load factor
c) working load is obtained by multiplying plastic limit load with load factor
d) working load is obtained by multiplying working load with load factor
Answer
Answer: a [Reason:] Plastic limit load is obtained by multiplying working load with load factor. Depending on combination of loads and their probability of acting at same time, different load factors are used.
6. Cantilevers and over hanging beams collapse as _____
a) single-bar mechanism
b) double-bar mechanism
c) three-bar mechanism
d) does not collapse
Answer
Answer: a [Reason:] Cantilevers and over hanging beams generally collapse as single-bar mechanisms with a single plastic hinge at one of the supports.
7. Single-span beams collapse as ________
a) single-bar mechanism
b) two-bar mechanism
c) three-bar mechanism
d) does not collapse
Answer
Answer: b [Reason:] Single-span beams generally collapse as two-bar mechanisms with a hinge at each support and plastic hinge within the span.
8. Multi-span beams collapse in one span as ___________
a) does not collapse
b) single-bar mechanism
c) two-bar mechanism
d) three-bar mechanism
Answer
Answer: c [Reason:] Multi-span beams generally collapse in one span as a local two-bar mechanism, within the span and a hinge at each support. It is possible to have a three-bar mechanism, where two adjacent spans combine to form a mechanism.
9. Among which of the following is the location of plastic hinge?
a) at supports
b) at centre of beam
c) at points away from concentrated load
d) at centre for uniformly distributed load
Answer
Answer: a [Reason:] Plastic hinges normally occur at supports, points of concentrated load, and points where cross section change. The location of plastic hinge in a beam with uniformly distributed load is not well defined.
Set 2
1. Structures designed using elastic analysis may be ______ than those designed using plastic analysis
a) lighter
b) heavier
c) of same weight
d) almost half times the weight
Answer
Answer: b [Reason:] In elastic design structures are designed for allowable stress level well below the elastic limit, whereas in plastic design structures are designed using ultimate load rather than yield stress, Hence, structures designed using elastic analysis may be heavier than those designed using plastic analysis.
2. Both elastic and plastic methods neglect ________
a) live load acting on structure
b) dead load acting on structure
c) deformations due to load
d) influence of stability
Answer
Answer: d [Reason:] Both elastic and plastic methods neglect the influence of stability, which may significantly affect the load carrying capacity of structures or elements which are slender and subjected to compressive stresses.
3. What is buckling?
a) Structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it
b) Structural behaviour in which a deformation does not develop in direction of plane perpendicular to that of load which produced it
c) Structural behaviour in which a deformation develop in direction of plane parallel to that of load which produced it
d) Structural behaviour in which a deformation develops in direction of plane along that of load which produced it
Answer
Answer: a [Reason:] Buckling may be defined as structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it. This deformation changes rapidly with variations in the applied load.
4. Which of the following relation about plastic moment is correct?
a) Mp = Zp /fy
b) Mp = Zp + fy
c) Mp = Zpfy
d) Mp = Zp – fy
Answer
Answer: c [Reason:] When every fibre of the section has a strain equal to or greater than εy=fy/Es , nominal moment strength is referred as plastic moment and is given by Mp = Zpfy , where Zp = ∫ydA is plastic section modulus and fy = yield stress.
5. What is plastic moment of resistance?
a) maximum moment in stress strain curve, the point where the curvature can increase indefinitely
b) maximum moment in stress strain curve, the point where the curvature can decrease indefinitely
c) minimum moment in stress strain curve, the point where the curvature can increase indefinitely
d) minimum moment in stress strain curve, the point where the curvature can decrease indefinitely
Answer
Answer: a [Reason:] In stress strain curve, the maximum moment is reached at a point where the curvature can increase indefinitely, neglecting the strain hardening benefits. This maximum moment is called plastic moment of resistance and the portion where this moment occurs is called as plastic hinge.
6. In elastic stage, equilibrium condition is achieved when neutral axis ___________ and in fully plastic stage, it is achieved when neutral axis ___________
a) is above centroid of the section, divides the section into two parts of one-third area and two-third area
b) is below centroid of the section, divides the section into two parts of one-third area and two-third area
c) is above centroid of the section, divides the section into two equal areas
d) passes through centroid of the section, divides the section into two equal areas
Answer
Answer: d [Reason:] In elastic stage, when bending stress varies from zero at neutral axis to maximum at extreme fibres, equilibrium condition is achieved when neutral axis passes through centroid of the section. In fully plastic stage, because the stress is uniformly equal to yield stress, equilibrium condition is achieved when neutral axis divides the section into two equal areas.
7. Which of the following relation is correct for plastic section modulus, Zo ?
a) Zp = 2A(y1+y2)
b) Zp = A(y1+y2)/2
c) Zp = A(y1+y2)/4
d) Zp = 4A(y1+y2)
Answer
Answer: b [Reason:] Zp = A(y1+y2)/2, where A= area of cross section, y1 and y2 are centroids of portion above and below neutral axis respectively. Plastic modulus is defined as combined statical moment of cross sectional area above and below the equal-area axis.
8. Which of the following relation is correct about shape factor, v?
a) v = Zp+Ze
b) v = ZpZe
c) v = Zp/Ze
d) v = Ze/Zp
Answer
Answer: c [Reason:] Shape factor, v = Zp/Ze = Mp/My , where Zp and Ze are plastic and elastic section modulus respectively, Mp and My are plastic and elastic moments respectively.
9. The shape factor does not depend on ___
a) material properties
b) cross sectional shape
c) moment of resistance
d) section modulus
Answer
Answer: a [Reason:] Shape factor, v = Zp/Ze = Mp/My, where Zp and Ze are plastic and elastic section modulus respectively, Mp and My are plastic and elastic moments respectively. This ratio is the property of cross-sectional shape and is independent of material properties.
10. Match the pairs with correct shape factor
Cross section Shape factor (average or maximum) A) Circular (i) 1.8 B) Rectangular (ii) 1.14 C) wide flange I-section (about major axis) (iii) 1.7 D) Channels (about minor axis) (iv) 1.5
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-iii, B-ii, C-iv, D-i
Answer
Answer: c [Reason:] Shape factor, v = Zp/Ze = Mp/My. The shape factor for various cross section are (i) for circular = 1.7, (ii) for rectangular = 1.5, (iii) wide flange I-section (about major axis) = 1.09-1.18, average is 1.14, (iv) wide flange I-section (about minor axis) = 1.67, (v) channels (about major axis) = 1.16-1.22, average is 1.18, (vi) channels (about minor axis) = 1.8 .
Set 3
1. Steel is mainly an alloy of
a) Iron and Carbon
b) Sulphur and Zinc
c) Zinc and tin
d) Phosphorous and Tin
Answer
Answer: a [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: d [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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
Answer: d [Reason:] 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
Answer: b [Reason:] 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
Answer: d [Reason:] 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
Answer: c [Reason:] 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
Answer: a [Reason:] Steel is weak in fire and corrosion resistance. So, to improve corrosion resistance and high temperature resistance, chromium and nickel are added to steel.
Set 4
1. The depth-to-thickness ratio of web connected to flanges along one longitudinal edge only when transverse stiffeners are not provided is _____ to meet serviceability criteria.
a) >180ε
b) ≥90ε
c) ≤90ε
d) >90ε
Answer
Answer: c [Reason:] When transverse stiffeners are not provided, the depth-to-thickness ratio of web connected to flanges along one longitudinal edge only is ≤ 90ε and when web connected to flanges along both longitudinal edges is ≤ 200ε to meet the serviceability criteria.
2. The depth-to-thickness ratio of web when only transverse stiffeners are provided and 3d ≥ c ≥d, where c is clear distance between stiffeners and d is depth of web is_____ to meet serviceability criteria
a) ≤ 200εw
b) ≥ 200 εw
c) > 200 εw
d) ≤ 400 εw
Answer
Answer: a [Reason:] For serviceability criteria, when only transverse stiffeners are provided, the depth-to-thickness ratio of web should be ≤ 200 εw, when 3d ≥ c ≥d and for c > 3d, the web is considered as unstiffened, where c is clear distance between stiffeners and d is depth of web, εw = √(250/fyw), fyw is yield stress of web.
3. The depth-to-thickness ratio of web when only transverse stiffeners are provided and c < 0.74d, where c is clear distance between stiffeners and d is depth of web is_____ to meet serviceability criteria
a) ≤ 200 εw
b) ≥ 270 εw
c) > 200 εw
d) ≤ 270 εw
Answer
Answer: d [Reason:] For serviceability criteria, when only transverse stiffeners are provided, the depth-to-thickness ratio of web should be ≤ 270 εw when c < 0.74d and c/tw ≤ 200 εw, when 0.74d ≤ c ≤d, where c is clear distance between stiffeners and d is depth of web, εw = √(250/fyw), fyw is yield stress of web.
4. What is the range of c to meet serviceability criteria when transverse and longitudinal stiffeners are provided at one level only, at 0.2d from compression flange and c/tw ≤ 250 εw ?
a) c < 0.74d
b) 0.74d ≤ c ≤ d
c) c ≥ d
d) c > 2d
Answer
Answer: b [Reason:] For serviceability criteria, when transverse and longitudinal stiffeners are provided at one level only, at 0.2d from compression flange d/tw ≤ 250 εw for 2.4d ≥c ≥ d, c/tw ≤ 250 εw for 0.74d ≤ c ≤ d and d/tw ≤ 340 εw for c < 0.74d, where c is clear distance between stiffeners and d is depth of web, εw = √(250/fyw), fyw is yield stress of web.
5. When second longitudinal stiffener is provided, d/tw to meet serviceability criteria is
a) ≤ 400 εw
b) ≥ 400 εw
c) > 800 εw
d) ≤ 800 εw
Answer
Answer: a [Reason:] When second longitudinal stiffener is provided at neutral axis, d/tw ≤ 400 εw to meet serviceability criteria, where d is depth of web, εw = √(250/fyw), fyw is yield stress of web.
6. When a plate girder bends, vertical compression in web is due to
a) downward vertical component of compression flange bending stress only
b) downward vertical component of tension flange bending stress only
c) downward vertical component of compression flange and upward vertical component of tension flange bending stress
d) upward vertical component of compression flange and downward vertical component of tension flange bending stress
Answer
Answer: c [Reason:] When a plate girder bends, its curvature creates vertical compression in web is due to downward vertical component of compression flange and upward vertical component of tension flange bending stress. The web must have sufficient vertical buckling strength to withstand this squeezing effect.
7. The d/tw should be ___ to avoid buckling of compression flange into web when transverse stiffeners are not provided
a) ≥ 500 εf2
b) ≤ 345 εf2
c) ≥ 345 εf2
d) ≤ 500 εf2
Answer
Answer: b [Reason:] To avoid buckling of compression flange into web when transverse stiffeners are not provided, d/tw ≤ 345 εf2, where d is depth of web, εf = √(250/fyf), fyf is yield stress of compression flange.
8. When only transverse stiffeners are provided and d/tw < 345 εf to meet compression flange buckling criteria, the range of c should be
a) c ≥ 4.5d
b) c > 3d
c) c > 1.5d
d) c < 1.5d
Answer
Answer: d [Reason:] To avoid buckling of compression flange into web, d/tw ≤ 345 εf2 for c ≥ 1.5d, d/tw < 345 εf for c < 1.5d when only transverse stiffeners are provided, where c is clear distance between stiffeners, d is depth of web, εf = √(250/fyf), fyf is yield stress of compression flange.
9. The optimum depth of web of plate girder is given by
a) (k/fy)0.33
b) (Mzk/fy)
c) (Mzk/fy)0.33
d) (Mzk/fy)2
Answer
Answer: c [Reason:] The depth of plate girder for which area of steel used is minimum and will have minimum weight is called optimum depth. It is given by (Mzk/fy)0.33, Mz is moment resisted entirely by flanges, k = d/tw, d is depth of web, tw is thickness of web, fy = design strength of flanges.
10. The minimum area of flange angles with cover plate for riveted/ bolted plate girder should be
a) 1/6th of calculated flange area
b) 1/3rd of calculated flange area
c) 1/8th of calculated flange area
d) 1/4th of calculated flange area
Answer
Answer: b [Reason:] Flanges of riveted/ bolted plate girder consists of pair of angles with or without cover plates. The flange angles should form as large part of the area of the flange as practicable and preferably not less than 1/3rd of the calculated flange area to keep the centre of gravity of flange within back of angles and not in the flange cover plate else stability is affected.
11. Which of the following angle should be ideally used in bolted plate girder flange?
a) bulb angle
b) equal angle
c) unequal angle with short leg horizontal
d) unequal angle with long leg horizontal
Answer
Answer: d [Reason:] Unequal angles with long leg horizontal are preferred. It is because the moment of inertia of the section will be more and a large length will be available for making the connection with the flange plate.
12. For which of the following cases are equal angles preferred in bolted plate girder flange?
a) when large number of connectors are required to connect flange angle to web
b) when very few number of connectors are required to connect flange angle to web
c) for reducing cost
d) for aesthetic appearance
Answer
Answer: a [Reason:] When shear is heavy, large number of connectors are required to connect flange angle to web. In such case, equal angles may be preferred, if unequal angles are used, the longer leg msy be placed parallel to the web.
13. Flange cover plates are used in plate girder when
a) flange cover plates are not used
b) for aesthetic appearance
c) when moment resisting capacity has to be increased
d) when moment resisting capacity has to be decreased
Answer
Answer: c [Reason:] When moment resisting capacity of the plate girder has to be increased, flange cover plates are provided over flange angles. The moment of inertia and consequently the moment resisting capacity of the girder is increased considerably as the flange cover plates are at the greatest distance from neutral axis.
14. The thickness of flange cover plate should be ______ flange angle in bolted connections
a) less
b) more
c) twice
d) can be more or less
Answer
Answer: a [Reason:] The flange cover plates should not be thicker than the flange angles in riveted/bolted connections. Hence, more than one plate may be required. It is preferred that all cover plates should have same thickness. If they are of different thickness, then outer plates should not be thicker thin the inner plates.
Set 5
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
Answer: c [Reason:] 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
Answer: a [Reason:] 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
Answer: d [Reason:] 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
Answer: d [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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 Myy
c) these rods increase the moment Myy
d) these rods result in smaller purlin sections
Answer
Answer: c [Reason:] 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 Myy 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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 Mz = Pl/10 and My = 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) Zpz = Myγm0/fy + (b/d)(Mzγm0/fy)
b) Zpz = Mzγm0/fy + (b/d)(Myγm0/fy)
c) Zpz = Mzγm0/fy + 2.5(b/d)(Myγm0/fy)
d) Zpz = Myγm0/fy + 2.5(b/d)(Mzγm0/fy)
Answer
Answer: c [Reason:] The required section modulus of the purlin section can be determined by Zpz = Mzγm0/fy + 2.5(b/d)(Myγm0/fy ), where γm0 is partial safety factor for material = 1.1, d is depth of trial section, b is the breadth of the trial section, Mz and My are factored bending moments about Z and Y axes, respectively, and fy 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 = Zp/fy
b) M = Zpγm0fy
c) M = Zpγm0/fy
d) M = γm0/fy
Answer
Answer: b [Reason:] The design capacity of channel/I-section purlin is given by Mdz = Zpzγm0/fy and Mdy = Zpyγm0/fy , Mdz and Mdy are design moment capacity about Z and Y axes, respectively, Zpz and Zpy are plastic section modulus about Z and Y axes, respectively and fy 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) Mdz ≤ 1.2Zeyfy/γm0, Mdy ≤ 2.4Zezfy/γm0
b) Mdz ≤ Zezfy/γm0, Mdy ≤ 1.2Zeyfy/γm0
c) Mdz ≤ γfZeyfy/γm0, Mdy ≤ 1.2Zezfy/γm0
d) Mdz ≤ 1.2Zezfy/γm0, Mdy ≤ γfZeyfy/γm0
Answer
Answer: d [Reason:] The check for design capacity of channel/I-section purlin is given by Mdz ≤ 1.2Zezfy/γm0 , Mdy ≤ γyZeyfy/γm0 , where Mdz and Mdy are design moment capacity about Z and Y axes, respectively, Zez and Zey are elastic section modulus about Z and Y axes, respectively and fy is yield stress of steel. Since in y-direction, the shape factor Zp/Ze 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) (Mz/Mdz) + (My/Mdy) ≥ 1
b) (Mz/Mdz) + (My/Mdy) ≤ 1
c) (Mdz/Mz) + (My/Mdy) ≤ 1
d) (Mdz/Mz) + (Mdy/My) ≥ 1
Answer
Answer: b [Reason:] The local capacity of the section is checked by interaction equation. It is given by (Mz/Mdz) + (My/Mdy) ≤ 1 , where Mdz and Mdy are design moment capacity about Z and Y axes, respectively, and Mz and My 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
Answer: a [Reason:] 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/(0.66xfy)
b) Z = M/(1.33×0.66xfy)
c) Z = M/(1.33×0.66xfy)
d) Z = M/(1.33xfy)
Answer
Answer: c [Reason:] The modulus of section required for angle section purlin is given by Z = M/(1.33×0.66xfy), M = maximum bending moment = wl2/10, w = unfactored uniformly distributed load, l = span of purlin, fy 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.
