Category: Design of Steel Structures Objective Questions

Design of Steel Structures Interview Questions and Answers for freshers on “Bolted Connections – III”.

1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1
b) (V_sb/V_db)2 + (T_sb/T_db)2 ≥ 1
c) (V_sb/V_db) + (T_sb/T_db) ≤ 1
d) (V_sb/V_db) + (T_sb/T_db) ≥ 1
Answer: a
Clarification: Bolt required to satisfy both shear and tension at the same time should satisfy (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1 , where V_sb= factored shear force, V_db = design shear capacity, T_sb = factored tensile force, T_db= design tensile capacity.

2. Shear Capacity of HSFG bolts is
a) μ_fn_ek_hF_o
b) μ_fn_ek_hF_oγ_mf
c) μ_fn_ek_h? /F_oγ_mf
d) μ_fn_ek_hF_o/γ_mf
Answer: d
Clarification: Shear Capacity of HSFG bolts is μ_fn_ek_hF_o/γ_mf, where μ_f = coefficient of friction(0.55), n_e = number of frictional interfaces offering frictional resistance to slip, k_h = 1 for fasteners in clearance holes, 0.85 for fasteners in over sized and short slotted holes, γ_mf = 1.1 (slip resistance designed at service load), 1.25 (slip resistance designed at ultimate load), F_o = minimum bolt tension = A_nbf₀ , where A_nb = net area of bolt, f₀ = 0.7fub , f_ub = ultimate tensile stress of bolt.

3. The maximum number of bolts of diameter 25mm that can be accomodated in one row in 200mm wide flat are:
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Minimum end distance = 2.5×25 = 62.5mm
Number of bolts that can be accommodated = (200-2×62.5)/25 = 3 bolts.

4. Calculate strength in shear of 16mm diameter of bolt of grade 4.6 for lap joint
a) 50 kN
b) 40 kN
c) 29 kN
d) 59 kN
Answer: c
Clarification: Bolts will be in single shear. Diameter of bolt = 16mm. Net area = 0.78x(π/4)x16²=156.83mm².
Strength of bolt in shear = Anbfub/(√3 x 1.25) = 156.83x400x10^-3/1.25x√3 = 28.97kN.

5. What is the value of kb in nominal bearing strength for a bolt of 20mm diameter of grade 4.6?
a) 0.5
b) 1
c) 0.97
d) 2
Answer: a
Clarification: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm
e=1.5×22=33mm, p=2.5×20=50mm
e/3d₀ = 33/(3×22) = 0.5, p/3d₀ -0.25 = 50/(3×22) -0.25=0.5, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀ , p/3d₀ -0.25, f_ub /f_b, 1) = 0.5.

6. Calculate bearing strength of 20mm diameter bolt of grade 4.6 for joining main plates of 10mm thick using cover plate of 8mm thick using single cover butt joint.
a) 70.26 kN
b) 109.82 kN
c) 50.18 kN
d) 29.56 kN
Answer: c
Clarification: diameter of bolt = 16mm, diameter of hole =16+2 =18mm
e=1.5×18=27mm, p=2.5×16=40mm
e/3d₀ = 27/(3×18) = 0.5, p/3d₀ -0.25 = 40/(3×18) -0.25=0.49, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀, p/3d₀ -0.25, f_ub /f_b,1) = 0.49
bearing strength = 2.5kbdtfu/1.25 = 2.5×0.49x16x8x400x10-3/1.25 = 50.18 kN.

7. Find the number of HSFG bolts of diameter 20mm, grade 88 for connection of member carrying factored tensile load of 200kN when no slip is permitted.
a) 5
b) 4
c) 3
d) 2
Answer: b
Clarification: F_o=0.7f_ubA_nb=0.7x800x0.78x(π/4)x20²x10^-3=137.22 kN
Assume μ_f=0.5, n_e=1, k_h=1
Slip resistance of bolt = μ_f n_e k_h F_o/1.25 = 0.5x1x1x137.22/1.25 =54.88 kN
Number of bolts required = 200/54.88 = 3.64 = 4(approximately).

8. What is the efficiency of joint when strength of bolt per pitch length is 60kN and strength of plate per pitch length is 150kN?
a) 25%
b) 30%
c) 35%
d) 40%
Answer: d
Clarification: Efficiency = (strength of bolt per pitch length/ strength of plate per pitch length)x100 = 60×100/150 = 40%.

9. Strength of bolt is
a) minimum of shear strength and bearing capacity of bolt
b) maximum of shear strength and bearing capacity of bolt
c) shear strength of bolt
d) bearing capacity of bolt
Answer: a
Clarification: Strength of bolt is minimum of shear strength and bearing capacity of bolt. Design shear strength = nominal shear capacity/1.25, Design bearing strength = nominal bearing capacity/1.25.

10. Prying forces are
a) friction forces
b) shear forced
c) tensile forces
d) bending forces
Answer: c
Clarification: In connections subjected to tensile stresses, the flexibility of connected parts can lead to deformations that increases tension applied to bolts. This additional tension is called prying force.

Design of Steel Structures Multiple Choice Questions on “Plastic Theory”.

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: b
Clarification: 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: d
Clarification: 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: a
Clarification: 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) M_p = Z_p /f_y
b) M_p = Z_p + f_y
c) M_p = Z_pf_y
d) M_p = Z_p – f_y
Answer: c
Clarification: When every fibre of the section has a strain equal to or greater than ε_y=f_y/E_s , nominal moment strength is referred as plastic moment and is given by M_p = Z_pf_y , where Z_p = ∫ydA is plastic section modulus and f_y = 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: a
Clarification: 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: d
Clarification: 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) Z_p = 2A(y₁+y₂)
b) Z_p = A(y₁+y₂)/2
c) Z_p = A(y₁+y₂)/4
d) Z_p = 4A(y₁+y₂)
Answer: b
Clarification: Z_p = 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 = Z_p+Z_e
b) v = Z_pZ_e
c) v = Z_p/Z_e
d) v = Z_e/Z_p
Answer: c
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y , where Z_p and Z_e are plastic and elastic section modulus respectively, M_p and M_y 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: a
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y, where Z_p and Z_e 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: c
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y. 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 .

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.

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