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Multiple choice question for engineering

Set 1

1. What are steel tension members?
a) Structural elements that are subjected to direct compressive loads
b) Structural elements that are subjected to direct tensile loads
c) Structural elements that are subjected to indirect compressive loads
d) Structural elements that are subjected to indirect tensile loads

View Answer

Answer: b [Reason:] Steel tension members are those structural elements that are subjected to direct axial tensile loads, which tend to elongate the members. A member in pure tension can be stressed up to and beyond the yield limit and does not buckle locally or overall.

2. The strength of tensile members is not influenced by :
a) length of connection
b) net area of cross section
c) type of fabrication
d) length of plate

View Answer

Answer: d [Reason:] The strength of tensile members is influenced by factors such as length of connection, size and spacing of fasteners, size and spacing of fasteners, net area of cross section, type of fabrication, connection eccentricity, and shear lag at the end connection.

3. Which of the following statement is correct?
a) single angle section with bolted connection produce eccentricity about both planes
b) single angle section with bolted connection produce eccentricity about one plane only
c) single angle section with welded connection produce eccentricity about both planes
d) single angle section with welded connection does not produce eccentricity about one plane

View Answer

Answer: a [Reason:] Single angle section with bolted connection produce eccentricity about both planes, whereas single angle section with welded connection may produce eccentricity about one plane only.

4. Which of the following statement is correct?
a) Single angle members are used where members are subjected to reversal of stresses
b) Double angle members are used in towers
c) Single angle members are used as web members in trusses
d) Double angle members are used as web members in trusses

View Answer

Answer: c [Reason:] Single angle members are used in towers and as web members in trusses. Double angle sections are used as chord members in light roof trusses or in situations where some rigidity is required and where members are subjected to reversal of stresses.

5. What is the difference between strand and wire rope?
a) Strand consists of individual wires wound helically around a central core, wire rope is made of several strand laid helically around a core
b) Wire rope consists of individual wires wound helically around a central core, strand is made of several wire ropes laid helically around a core
c) Strand consists of individual wires wound straight around a central core, wire rope is made of several strand laid helically around a core
d) Wire rope consists of individual wires wound straight around a central core, strand is made of several wire ropes laid helically around a core

View Answer

Answer: a [Reason:] Strand consists of individual wires wound helically around a central core, wire rope is made of several strand laid helically around a core. Wire ropes are exclusively used for hoisting purposes and as guy wires in steel stacks and towers.

6. Which of the following statement is not correct?
a) Cables in form of wires ropes and strands are used in application where high strength is required
b) Cables are generally long and their flexural rigidity is negligible
c) They are flexible
d) They are recommended in bracing systems

View Answer

Answer: d [Reason:] Cables used as floor suspenders in suspension bridges are made from individual strands wound together in rope like fashion. Cables in form of wires ropes and strands are used in application where high strength is required and flexural rigidity is unimportant. Cables are generally long and their flexural rigidity is negligible. They are flexible. They are not recommended in bracing systems as they cannot resist compression.

7. Bars and rods are not used as :
a) tension members in bracing systems
b) friction resistant members
c) sag rods to support purlin
d) to support girts in industrial buildings

View Answer

Answer: b [Reason:] Bars and rods are used as tension members in bracing systems, sag rods to support purlin between trusses, to support girts in industrial buildings, where light structure is desirable. Rods are also used in arches to resist thrust of arch.

8. Sagging of members by built up bars and rods may be minimised by
a) increasing length diameter
b) increasing thickness ratio
c) fabricating rod/bar short of its required theoretical length
d) fabricating rod/bar more than its required theoretical length

View Answer

Answer: c [Reason:] Sagging of members by built up bars and rods may be minimised by limiting length diameter or thickness ratio or by fabricating the rod/bar short of its required theoretical length by some arbitrary amount and drawing into place to provide an initial tension. The same effect may be produced by providing turnbuckle in the rod.

9. Which of the following type of tension member is not mainly used in modern practice?
a) open section such as angles
b) flat bars
c) double angles
d) circular section

View Answer

Answer: b [Reason:] Tension members were generally made of flat bars earlier. But modern practice is to use mainly the following sections for tension members: (i)open sections such as angles, channels and I-sections, (ii)compound and built up sections such as double angle and double channels with or without additional plates, (iii)closed sections such as circular, square, rectangular or hollow sections.

10. Which among the following comparison between angle and flat bars is not true?
a) for light loads, angles are preferred over flat bars
b) flat bar tension members tend to vibrate during passage of load in light bridges
c) flat bars are used instead of angles in case of stress reversal
d) angles are used instead of flat bars in case of stress reversal

View Answer

Answer: c [Reason:] For light loads, angles are preferred over flat bars. In many light bridges, flat bar tension members tend to vibrate during passage of load. In case of stress reversal angles are more suitable whereas flat bars are unfit to carry compressive load on reversal due to their small radius of gyration in one direction.

11. Which of the following statement is correct?
a) angles placed on same side of gusset plate produce eccentricity about one plane only
b) angles placed on same side of gusset plate produce eccentricity about two planes
c) angles placed on opposite side of gusset plate produce eccentricity about one plane only
d) angles placed on opposite side of gusset plate produce eccentricity about two planes

View Answer

Answer: a [Reason:] Two angle sections can either be placed back-to-back on the same side of gusset plate, or back-to-back on the opposite side of gusset plate. When angles are connected on the same side of gusset plate, the eccentricity is about one plane only, which can be almost eliminated when the same angles are connected on opposite side of gusset plate.

12. Which of the following is true about built up section?
a) Built up members are less rigid than single rolled section
b) Single rolled section are formed to meet required area which cannot be provided by built up members
c) Built up members can be made sufficiently stiff
d) Built up sections are not desirable when stress reversal occurs

View Answer

Answer: c [Reason:] Built-up members, made up of two or more plates or shapes and connected to act as single member, are formed primarily to meet required area which cannot be provided by single rolled section. Built up members are more rigid because for same area much greater moment of inertia can be obtained than single rolled section. Built up members can be made sufficiently stiff to carry compression and tension thus desirable when stress reversal occurs.

Set 2

1. Which of the following statement is correct?
a) stress and strain calculated using initial cross section area and initial gauge length are referred to as true stress and true strain
b) stress and strain calculated using current cross section area and initial gauge length are referred to as true stress and engineering strain
c) stress and strain calculated using initial cross section area and initial gauge length are referred to as engineering stress and engineering strain
d) stress and strain calculated using current cross section area and gauge length are referred to as engineering stress and engineering strain

View Answer

Answer: c [Reason:] Stress and strain calculated using initial cross section area and initial gauge length are referred to as engineering stress and engineering strain. Stress and strain calculated using current cross section area and gauge length are referred to as true stress and true strain.

2. Arrange the regions of engineering stress-strain curve in order from right to left as in graph
a) strain softening region, strain hardening region, yield plateau, linear elastic region
b) strain hardening region, strain softening region, linear elastic region, yield plateau
c) strain softening region, yield plateau, linear elastic region, strain hardening region
d) strain hardening region, linear elastic region, yield plateau, strain softening region

View Answer

Answer: a [Reason:] The engineering stress-strain curve is typically represented by four regions : linear elastic region, yield plateau, strain hardening region, strain softening (unloading)region.

3. Which of the following is true regarding engineering stress-strain curve?
a) it gives true indication of deformation characteristics of metal because it is entirely based on true dimensions of specimen
b) it does not gives true indication of deformation characteristics of metal because it is entirely based on true dimensions of specimen
c) it gives true indication of deformation characteristics of metal because it is not entirely based on true dimensions of specimen
d) it does not gives true indication of deformation characteristics of metal because it is not entirely based on true dimensions of specimen

View Answer

Answer: b [Reason:] The engineering stress-strain curve does not provide true indication of deformation characteristics of metal. It is entirely based on true dimensions of specimen and these dimensions change continuously as the load increases.

4. Choose the correct option
a) post ultimate strain softening in engineering stress strain curve is present in true stress strain curve
b) post ultimate strain softening in true stress strain curve is absent in engineering stress strain curve
c) post ultimate strain softening in true stress strain curve is present
d) post ultimate strain softening in engineering stress strain curve is absent in true stress strain curve

View Answer

Answer: d [Reason:] The post ultimate strain softening in engineering stress strain curve caused by necking of cross section is absent in true stress strain curve as engineering stress strain curve are based on true dimensions of specimen and true stress strain curve are based on actual cross sectional area of specimen.

5. What is the yield point for high strength steel?
a) 0.5% of offset load
b) 0.2% of offset load
c) 0.1% of offset load
d) 1.5% of offset load

View Answer

Answer: b [Reason:] High-strength steel tension members do not exhibit well defined yield point and yield plateau. Hence, 0.2% of offset load is usually taken as yield point for such high strength steel.

6. True stress strain curve is also known as
a) flow curve
b) un-flow curve
c) elastic curve
d) parabolic curve

View Answer

Answer: a [Reason:] True stress strain curve is also known as flow curve since it represents basic plastic flow characteristics if the material. Any point on the flow curve can be considered as local stress for metal strained in tension by magnitude shown on the curve.

7. The design strength of tension member corresponding to gross section yielding is given by :
a) γm0 fyAg
b) γm0fy/Ag
c) fy/Ag γm0
d) fyAg/ γm0

View Answer

Answer: d [Reason:] The design strength of tension member corresponding to gross section yielding is given by Tdg = fyAg/ γm0, where fy = yield strength of material in MPa, Ag = gross cross-sectional area in mm2, γm0 = partial safety factor for failure in tension by yielding = 1.10.

8. Which of the following relation is correct?
a) Net area = Gross area x deductions
b) Net area = Gross area + deductions
c) Net area = Gross area – deductions
d) Net area = Gross area / deductions

View Answer

Answer: c [Reason:] Net area = Gross area – deductions, that is net area of tensile members is calculated by deducting areal of holes from the gross area.

9. The design strength of tension member corresponding to net section rupture is given by :
a) Anfyγm1
b) 0.9Anfyγm1
c) 0.9An/fyγm1
d) 0.9Anfym1

View Answer

Answer: d [Reason:] The design strength of tension member corresponding to net section rupture is given by Tdn = 0.9Anfym1, where An = net effective area of cross section in mm2, fy = ultimate strength of material in MPa, γm1 = partial safety factor for failure due to rupture of cross section = 1.25.

10. The block shear strength at an end connection for shear yield and tension fracture is given by :
a) (Avgfy/√3 γm0)+(0.9Atnfum1)
b) (Atgfy/√3 γm0)+(0.9Avnfum1)
c) (0.9Atgfy/√3 γm0)+( Avnfum1)
d) (0.9Avgfy/√3 γm0)+(Atnfum1)

View Answer

Answer: a [Reason:] The block shear strength at an end connection for shear yield and tension fracture is given by Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfum1), where Avg = minimum gross area in shear along line of action of force, Atn = minimum net area of cross section in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.

11. The block shear strength at an end connection for shear fracture and tension yield is given by :
a) (Avgfy/√3 γm0)+(0.9Atnfum1)
b) (Atgfy/ γm0)+(0.9Avnfu/√3 γm1)
c) (0.9Avgfy/√3 γm0)+(Atnfum1)
d) (0.9Atgfy/√3 γm0)+( Avnfum1)

View Answer

Answer: b [Reason:] The block shear strength at an end connection for shear fracture and tension yield is given by Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1), where Avn = minimum net area in shear along line of action of force, Atg = minimum gross area in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.

12. The block shear strength of connection is ________
a) block shear strength at an end connection for shear fracture and tension yield
b) block shear strength at an end connection for shear yield and tension fracture
c) larger of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)
d) smaller of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)

View Answer

Answer: d [Reason:] The block shear strength of connection is smaller of block shear strength at an end connection for shear yield, tension fracture Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfum1) and block shear strength at an end connection for shear fracture, tension yield Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γvm1).

13. The design tensile strength of tensile member is
a) minimum of strength due to gross yielding, net section rupture, block shear
b) maximum of strength due to gross yielding, net section rupture, block shear
c) strength due to gross yielding
d) strength due to block shear

View Answer

Answer: a [Reason:] The design tensile strength of tensile member is taken as the minimum of strength due to gross yielding (Tdg=fyAg/1.1), net section rupture(Tdn=0.9Anfy/γm1), block shear (Tdb1=(Avgfy/√3 γm0)+(0.9Atnfum1), Tdb2=(Atgfy/ γm0)+(0.9Avnfu/√3 γm1)).

Set 3

1. Arrange the following welds in ascending order as per their usage in structural engineering applications.
a) fillet weld, groove weld, slot and plug weld
b) slot and plug weld, groove weld, fillet weld
c) groove weld, fillet weld, slot and plug weld
d) fillet weld, slot and plug weld, groove weld

View Answer

Answer: b [Reason:] Fillet welds are used extensively (about 80%) followed by groove welds (15%). Slot and plug welds are rarely used (less than 5%) in structural engineering applications.

2. Which of the following type of weld is most suitable for lap and T-joints?
a) Fillet weld
b) Groove weld
c) Slot weld
d) Plug weld

View Answer

Answer: a [Reason:] Fillet welds are suitable for lap and T-joints and groove welds are suitable for butt, corner, and edge joints.

3. Which of the following is true about back-up strip provided at bottom of single-V grooves?
a) Back-up strips are commonly used when welding is done from both the sides
b) Back-up strips are commonly used when root opening is sufficient
c) It creates a problem of burn-through
d) It introduces a crevice into the weld geometry

View Answer

Answer: d [Reason:] Back-up strip is provided at the bottom of single-V/bevel/J or U grooves. It is commonly provided when welding is done from one side or when the root opening is excessive. It introduces a crevice into the weld geometry and prevents the problem of burn-through.

4. The size of root gap and root face for groove weld does not depend on :
a) type of welding process
b) welding position
c) type of metal plate
d) volume of deposited material

View Answer

Answer: c [Reason:] For groove weld, the root opening or gap is provided for the electrode to access the base of the joint. The size of root gap and root face depends on the following : (i) type of welding process, (ii) welding position, (iii) volume of deposited material, (iv)cost of preparing edges, (v)access for arc and electrode, (vi)shrinkage and distortion.

5. Which of the following groove weld is used for plates of thickness more than 40mm?
a) Double-bevel
b) Single-J
c) Single-U
d) Double-U

View Answer

Answer: d [Reason:] The groove is made of double-bevel or double-V for plates of thickness more than 12mm, and made of double-U or double-J for plates of thickness more than 40mm. For plates of thickness between 12-40mm, single-J and single-U grooves may be used.

6. Groove welds should have ________ strength as member they join.
a) same
b) less
c) greater
d) half

View Answer

Answer: a [Reason:] Groove welds will transmit full load of the members they join, so they should have the same strength as the members they join.

7. Which of the following is not true regarding fillet welds?
a) They require less precision in fitting up two sections
b) They are adopted in field as well as shop welding
c) They are assumed to fail in tension
d) They are cheaper than groove welds

View Answer

Answer: c [Reason:] Fillet welds require less precision in fitting up two sections. They are adopted in field as well as shop welding. They are assumed to fail in shear and are cheaper than groove welds.

8. Which of the following is true about slot and plug welds?
a) They are extensively used in steel construction
b) They are assumed to fail in shear
c) The inspection of these welds is easy
d) They are normally used to connect members carrying tensile loads

View Answer

Answer: b [Reason:] Slot and plug welds are not extensively used in steel construction. They are used to fill up holes in connections. They are assumed to fail in shear. The inspection of these welds is difficult. They are useful in preventing overlapping parts from buckling.

9. Choose the correct option regarding weld metal.
a) Weld metal is same as parent metal
b) Weld metal is same as steel
c) It has higher yield to ultimate ratio
d) It has higher ductility compared to structural steel

View Answer

Answer: c [Reason:] Weld metal is a mixture of parent metal and steel melted from electrode. The solidified weld metal has properties characteristic of cast steel. It has higher yield to ultimate ratio but lower ductility compared to structural steel.

10. Which of the following is not true regarding pre-heating of heat affected zone ?
a) Pre-heating does not help to reduce heat affected zone cracks
b) Pre-heating increases the cost of welding
c) It is done to remove surface moisture in highly humid conditions
d) It is done to disperse hydrogen away from weld pool and heat affected zone

View Answer

Answer: a [Reason:] Pre-heating of joints help to reduce heat affected zone cracks but increases the cost of welding. It is done to remove surface moisture in highly humid conditions, to disperse hydrogen away from weld pool and heat affected zone, to bring steel to ambient temperature in cold climates.

Set 4

1. Which of the following is the property of high carbon steel?
a) high toughness
b) reduced ductility
c) high strength
d) reduced strength

View Answer

Answer: b [Reason:] High carbon steel contains high carbon content. Hence it has reduced ductility, toughness and weldability.

2. High carbon steel is used in ______________
a) transmission lines and microwave towers
b) structural buildings
c) fire resistant buildings
d) for waterproofing

View Answer

Answer: a [Reason:] High carbon steel is used in transmission lines and microwave towers where relatively light members are joint by bolting.

3. What is the permissible percentage of micro-alloys in medium and high strength micro-alloyed steel?
a) 0.1%
b) 0.5%
c) 0.25%
d) 1.0%

View Answer

Answer: c [Reason:] Medium and High strength micro-alloyed steel have low carbon content, but alloys such as niobium, vanadium, titanium or boron are added to achieve high strength.

4. Fire resistant steels are also called as ____________
a) Stainless steel
b) Weathering steel
c) High strength steel
d) Thermomechanically treated steel

View Answer

Answer: d [Reason:] Fire resistant steels are also called as thermomechanically treated steel. They perform better than ordinary steel under fire.

5. What is the minimum percentage of chromium and nickel added to stainless steel?
a) 0.5%, 10.5%
b) 2%, 20%
c) 10.5%, 0.5%
d) 30%, 50%

View Answer

Answer: c [Reason:] Stainless steel are low carbon steels to which a minimum of 10.5% chromium (maximum 20%) and 0.5% nickel is added.

6. Match the pair of Type of steel with its ultimate tensile strength :

	TYPE OF STEEL					ULTIMATE TENSILE CAPACITY
(A) Carbon Steel 					(i) 700-950 MPa
(B) High Strength Carbon Steel				(ii) 440-590 MPa
(C) Weathering Steel					(iii) 410-440 MPa
(D) High Strength quenched & tempered steel		(iv) 480 MPa
(E) Medium and High strength microalloyed steel	        (v) 480-550 MPa

a) A-i, B-ii, C-iii, D-iv, E-v
b) A-v, B-iv, C-iii, D-ii, E-i
c) A-iii, B-v, C-iv, D-i, E-ii
d) A-ii, B-iii, C-v, D-i, E-iv

View Answer

Answer: c [Reason:] Ultimate tensile strength is the capacity of material to withstand loads tending to elongate. It is the maximum stress that a material can withstand while being stretched or pulled. The ultimate tensile strength for Carbon Steel is 410-440 MPa, 480-550 MPa for High Strength Carbon Steel, 480 MPa for Weathering Steel, 700-950 MPa for High Strength quenched & tempered steel, 440-590 MPa for Medium and High strength microalloyed steel.

7. What is weathering steel?
a) low-alloy atmospheric corrosion-resistant steel
b) low-carbon steel
c) high strength quenched and tempered steel
d) fire resistant steel

View Answer

Answer: a [Reason:] Weathering steel are low-alloy atmospheric corrosion-resistant steel. They are often left unpainted. They have an ultimate tensile strength of 480 MPa.

8. Match the pair of Type of steel with its yield strength:

	TYPE OF STEEL					YIELD STRENGTH
(A) Carbon Steel 					(i) 300-450 MPa
(B) High Strength Carbon Steel				(ii) 350 MPa
(C) Weathering Steel					(iii) 350-400 MPa
(D) High Strength quenched & tempered steel		(iv) 230-300MPa
(E) Medium and High strength microalloyed steel	        (v) 550-700 MPa

a) A-i, B-ii, C-iii, D-iv, E-v
b) A-i, B-iii, C-v, D-iv, E-ii
c) A-v, B-iv, C-iii, D-ii, E-i
d) A-iv, B-iii, C-ii, D-v, E-i

View Answer

Answer: d [Reason:] Yield Strength is the stress that a material can withstand without any permanent deformation i.e. the point of stress at which any material starts to deform plastically. The yield strength of carbon steel is 230-300MPa, 350-400 MPa for High Strength Carbon Steel, 350 MPa for Weathering Steel, 550-700 MPa for High Strength quenched & tempered steel, 300-450 MPa for Medium and High strength microalloyed steel.

Set 5

1. Which of the following is correct?
a) web in rolled section behaves like a column when not placed under concentrated loads
b) web in rolled section behaves like a column when placed under concentrated loads
c) web in rolled section does not behave like a column when placed under concentrated loads
d) web in rolled section cannot be compared with column

View Answer

Answer: b [Reason:] The web in rolled section behaves like a column when placed under concentrated loads. The web is quite thin and is therefore, subjected to buckling.

2. The effective depth when top flanges are restrained against lateral deflection and rotation is
a) d/3
b) d
c) 2d
d) d/2

View Answer

Answer: d [Reason:] Bottom flange is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered. The effective depth when top flanges are restrained against lateral deflection and rotation is d/2, where d is depth of web.

3. The effective depth when top flanges are restrained against lateral deflection but not against rotation is
a) 2d/3
b) d
c) 2d
d) d/2

View Answer

Answer: a [Reason:] The effective depth when top flanges are restrained against lateral deflection but not against rotation is 2d/3, where d is depth of web. Bottom flange is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered.

4. The effective depth when top flanges are restrained against rotation but not against lateral deflection is
a) 2d/3
b) 2d
c) d
d) d/2

View Answer

Answer: c [Reason:] Bottom flange is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered. The effective depth when top flanges are restrained against rotation but not against lateral deflection is d, where d is depth of web.

5. The effective depth when top flanges are not restrained against rotation and lateral deflection is
a) 2d/3
b) 2d
c) d
d) d/2

View Answer

Answer: b [Reason:] The effective depth when top flanges are not restrained against rotation and lateral deflection is 2d, where d is depth of web. Bottom flange is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered.

6. The maximum diagonal compression in plate girder simply supported occurs
a) does not occur
b) above neutral axis
c) below neutral axis
d) at neutral axis

View Answer

Answer: d [Reason:] The maximum diagonal compression in plate girder simply supported occurs at neutral axis . It will be inclined at 45˚ to the neutral axis.

7. Web buckling strength at support of simply supported plate girder is given by
a) Fwb =Btwfcd
b) Fwb =twfcd
c) Fwb =Btw
d) Fwb =Bfcd

View Answer

Answer: a [Reason:] Web buckling strength at support of simply supported plate girder is given by Fwb =Btwfcd , where B is length of stiff portion of bearing plus additional length given by dispersion at 45˚to the level of neutral axis, tw is thickness of web, fcd is allowable compressive stress corresponding to assumed web strut according to buckling curve c.

8. Slenderness ratio when ends are assumed to be fixed for a plate girder is
a) 2.45 d
b) 8.5 d/t
c) 2.45 d/t
d) 8.5 t

View Answer

Answer: c [Reason:] When ends are assumed to be fixed, effective length = d√2/2 = d/√2, minimum radius of gyration = t/√12. Therefore slenderness ratio = (d/√2)/( t/√12) = 2.45d/t, where d = depth of web, t = thickness of web.

9. What is web crippling ?
a) web is of large thickness
b) flange near portion of stress concentration tends to fold over web
c) web near portion of stress concentration tends to fold over flange
d) flange is of large thickness

View Answer

Answer: c [Reason:] Webs of rolled section are subjected to large amount of stresses just below concentrated loads and above reactions from support. Stress concentration occurs at junction of web and flange. As a result, large bearing stresses are developed below concentrated load. Consequently, the web near portion of stress concentration tends to fold over flange. This type of local buckling phenomenon is called crippling or crimpling of web.

10. Which of the following is true?
a) web crippling is buckling of web caused by compressive force delivered through flange
b) web crippling is buckling of flange caused by compressive force delivered through web
c) web crippling is buckling of web caused by tensile force delivered through flange
d) web crippling is buckling of flange caused by tensile force delivered through web

View Answer

Answer: a [Reason:] Web crippling is buckling of web caused by compressive force delivered through flange. To keep bearing stresses within permissible limits, the concentrated load should be transferred from flanges to web on sufficiently large bearing areas.

11. The most critical location for failure due to web crippling is
a) flange cross section
b) middle of web
c) start of fillet
d) root of fillet

View Answer

Answer: d [Reason:] The most critical location for failure due to web crippling is root of fillet since resisting area has the smallest value here.

12. The angle of dispersion of load for web crippling is assumed to be
a) 2:1
b) 1:2.5
c) 4:5
d) 2:3

View Answer

Answer: b [Reason:] The angle of dispersion of load for web crippling is assumed to be 1:2.5 .With reference to this, bearing length is calculated.

13. The crippling strength of web at supports is given by
a) Fcrip = (b+n1)fyw
b) Fcrip = (b+n1) t
c) Fcrip = (b+n1)/fywt
d) Fcrip = (b+n1)fywt

View Answer

Answer: d [Reason:] The crippling strength of web at supports is given by Fcrip = (b+n1)fywt, where b+n1 is length obtained by dispersion through flange, t is thickness of web, fyw is design yield strength of web.

14. The crippling strength of web at interior point where concentrated load is acting is given by
a) Fcrip = (b+2n1)fyw
b) Fcrip = (b+2n1) t
c) Fcrip = (b+2n1)fywt
d) Fcrip = (b+n1)/fywt

View Answer

Answer: c [Reason:] The crippling strength of web at interior point where concentrated load is acting is given by Fcrip = (b+2n1)fywt, where b+2n1 is length obtained by dispersion through flange, t is thickness of web, fyw is design yield strength of web.

15. Which off the following will be a remedy to web crippling?
a) spreading load over small portion of flange
b) provide stiffeners which bear against flanges at load points
c) provide web of small thickness
d) web crippling cannot be prevented

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Answer: b [Reason:] Web crippling can be prevented by spreading load over large portion of flange. The other remedy is provide stiffeners which bear against flanges at load points and are connected to web to transfer force to it gradually. The other remedy is to make the web thicker.