DARBHANGA COLLEGE OF ENGINEERING

DEPARTMENT OF CIVIL ENGINEERING

Subject Code: 011620 Subject Name: Design of Steel Structures

Subject teacher: Ahsan Rabbani

1. Unit mass of Steel is

a) 785 kg/m3
b) 450 kg/m3
c) 450 kg/cm3
d) 7850 kg/m3

Answer: d

2. Which of the following is a correct criterion to be considered while designing?

a) Structure should be aesthetically pleasing but structurally unsafe
b) Structure should be cheap in cost even though it may be structurally unsafe
c) Structure should be structurally safe but less durable
d) Structure should be adequately safe, should have adequate serviceability

Answer: d

3. The structure is statically indeterminate when

a) static equilibrium equations are insufficient for determining internal forces and
reactions on that structure
b) static equilibrium equations are sufficient for determining internal forces and
reactions on that structure
c) structure is economically viable
d) structure is environment friendly

Answer: a

4. Which of the following relation is correct?

a) Permissible Stress = Yield Stress x Factor of Safety
b) Permissible Stress = Yield Stress / Factor of Safety
c) Yield Stress = Permissible Stress / Factor of Safety
d) Permissible Stress = Yield Stress – Factor of Safety

Answer: b

5. Describe stress strain curve for the Mild Steel with neat sketch.

Answer:

When steel is curved, it is important to keep the stress-strain curve ratio for mild steel in
mind. Below is a stress-strain graph that reviews the properties of steel in detail.
If tensile force is applied to a steel bar, it will have some elongation. If the force is small
enough, the ratio of the stress and strain will remain proportional. This can be seen in the
graph as a straight line between zero and point A – also called the limit of proportionality.
If the force is greater, the material will experience elastic deformation, but the ratio of stress
and strain will not be proportional. This is between points A and B, known as the elastic limit.

Beyond the elastic limit, the mild steel will experience plastic deformation. This starts the
yield point – or the rolling point – which is point B, or the upper yield point. As seen in the
graph, from this point on the correlation between the stress and strain is no longer on a
straight trajectory. It curves from point C (lower yield point), to D (maximum ultimate stress),
ending at E (fracture stress).
Now, we’ll look at each individual measure on the graph above and explain how each is
derived.

 Stress: If an applied force causes a change in the dimension of the material, then the
material is in the state of stress. If we divide the applied force (F) by the cross-
sectional area (A), we get the stress.

The symbol of stress is σ (Greek letter sigma). For tensile (+) and compressive (-) forces.
The standard international unit of stress is the pascal (Pa), where 1 Pa = 1 N/m2. The
formula to derive the stress number is σ = F/A.
For tensile and compressive forces, the area taken is perpendicular to the applied force. For
sheer force, the area is taken parallel to the applied force. The symbol for shear stress is tau
(τ).

 Strain: Strain is the change in the dimension (L-L0) with respect to the original. It is
denoted by the symbol epsilon (ε). The formula is ε = (L-L0) / L0. For a shear force,
strain is expressed by γ (gamma)

 Elasticity: Elasticity is the property of the material which enables the material to
return to its original form after the external force is removed.

 Plasticity: This is a property that allows the material to remain deformed without
fracture even after the force is removed.
The definitions below are important for understanding the Stress-Strain interactions as seen
in the graph.

 Hooke’s Law: Within the proportional limit (straight line between zero and A), strain
is proportionate to stress.

 Young’s modulus of elasticity: Within the proportional limit, stress = E × strain. E

is a proportionality constant known as the modulus of elasticity or Young’s modulus
of elasticity. Young’s modulusis a measure of the ability of a material to withstand
changes in length when under lengthwise tension or compression. E has the same
unit as the unit of stress because the strain is dimensionless. The formula is E = σ / ε
Pa.

 Modulus of Resilience: The area under the curve which is marked by the yellow
area. It is the energy absorbed per volume unit up to the elastic limit. The formula for
the modulus of resilience is 1/2 x σ x ε = 0.5 x (FL/AE).

 Modulus of toughness: This is the area of the whole curve (point zero to E). Energy
absorbed at unit volume up to breaking point.

Chicago Metal Rolled Products sets the industry standard for adherence to the stress strain
curve for mild steel and other materials.

6. What are the Advantages and disadvantages of steel as structural materials?

Answer:
Steel is one of the most generally utilized materials of construction time. Without the use of
steel, the structure doesn't make a solid while seismic tremors like earthquakes etc. happen.
Steel structures are susceptible to various ecological conditions. There are a few properties
wherein solid structures are preferred over steel and the utilization of steel is consistently
expanding everywhere throughout the world in development projects and also in civil
engineering-related fields. According to 'Lorraine Farrelly', before the utilization of steel in
development building, became a common practice, the weight of the structure material and
the forces of gravity and pressure defined the endurance, chance of stability in structure, and
its architectural possibilities. Each steel structure has some advantages as well as
disadvantages. And now we are going to elaborate on the complete description regarding
steel here.

ADVANTAGES OF STEEL BUILDINGS

a. Steel is moderately cheap when compared with other structure materials
b. Steel structures are highly fire-resistant when contrasted to a wooden structure as wood
is a combustible material and less fire-resistant when contrasted with RCC structure.
c. One of the advantages of using a steel structure in development is the ability of steel to
span greater distances with steel ceiling joists. This enables architects to grow their
choices, enabling them to make new/huge space utilizing steel items that simply weren't
accessible with different materials.
d. Steel can be easily & effectively manufactured and delivered greatly. Steel structures
can be delivered off-site at shop floors and after that gathered nearby. This spares time
and increases the efficiency of the general development process.
e. Steel structures can withstand outside weights, for example, earthquakes, thunder
storms, and cyclones. A well-fabricated steel structure can last more than30 years
whenever looked after well.
f. Flexibility is one of the great advantages of steel structure, which means that it tends to
be planned according to the design requirements. This plans a steel structure so that it
can withstand heavy winds or earthquakes, especially in the case of the bridges or tall
towers.
g. Because of simple-to-make portions of a steel structure, it is hassle-free to install and
assemble them on-site, and furthermore, there is no need of estimating and cutting of
parts nearby.
h. Some of the common advantages of using steel buildings are Design, Strength and
Durability, Light in Weight, Easy Installation and Speed in Construction, Versatile,
Flexibility, Ductility, Easy Fabrication in Different Sizes, Fire Resistance, Pest and Insect
Resistant, Moisture and Weather Resistance, Adaptability, Cost-effective, Environment
Friendly, Energy Efficiency, Improved Construction Quality, Temporary Structures, Safe
and Resistant and Risk Index.

DISADVANTAGES OF STEEL BUILDINGS

a. Buckling is an issue with steel structures. As the length of the steel segment builds the
chances of buckling also increases.
b. Steel is available only at the steel plants where it is produced and should be transported
for long distances to the site of construction, not at all like concrete or different materials
that might be accessible right at the site of development.
c. Due to the activity of rust in steel, costly paints are required to re-establish from time to
time. So that resistance against serious conditions increments.
d. Despite the fact that steel is a flexible material, it is difficult to make field corrections if
one or more components do not fit appropriately. Large portions of the metal structure
makes perform adhere to strict quality assurance procedure guarantee all pieces of a
structure fit accurately. But in actual it is not possible. One can't form it or cut it in the
ideal shape on-site once it is fabricated.
e. Steel can't mold in any path you required. It must be utilized in structures in which areas
initially exist.
f. Steel is agood conductor of heat, touches off materials in contact and often causes fires,
which quickly spread to different segments of a structure. Hence, steel structures may
require extra fireproofing treatment.
g. If steel loses its great property of ductility and then there are more chances to increase
the fractures.
h. Some of the common disadvantages of using steel buildings are High Maintenance &
Capital Cost, Susceptibility to Buckling, Fatigue and Fracture, Fireproof Treatment, Fire
Damage and Fabrication Error.

7. Describe the various types of loads and Load Combinations as per IS code.

Answer:

Clause 3.2 of IS 800:2007 specifies the various loads and forces that has to be considered
while performing the design of steel structures. As per Cl. 3.2.1 of IS 800:2007, for the
purpose of designing any element, member or a structure, the following loads (actions) and
their effects shall be taken into account, where applicable, with partial safety factors and
combinations (Cl. 5.3.3 of IS 800:2007). (a) Dead loads; (b) Imposed loads (live load, crane
load, snow load, dust load, wave load, earth pressures, etc); (c) Wind loads; (d) Earthquake
loads; (e) Erection loads; (f) Accidental loads such as those due to blast, impact of vehicles,
etc; and (g) Secondary ffects due to contraction or expansion resulting from temperature
changes, differential settlements of the structure as a whole or of its components,eccentric
connections,rigidity of joints differing from design assumptions.
1. Dead loads (Cl. 3.2.1.1 of IS 800:2007)
Dead loads should be assumed in design as specified in IS 875 (Part 1).
2. Imposed Loads (Cl. 3.2.1.2 of IS 800:2007)
IS 800:2007 specifies in Cl.3.2.1.2 that imposed loads for different types of occupancy and
function of structures shall be taken as recommended in IS 875 (Part 2). Imposed loads
arising from equipment, such as cranes and machines should be assumed in design as per
manufacturers/suppliers data (Cl. 3.5.4 of IS 800:2007). Snow load shall be taken as per IS
875 (Part 4).
3. Wind loads (Cl. 3.2.1.3 of IS 800:2007)
Wind loads on structures shall be taken as per the recommendations of IS 875 (Part 3).
4. Earthquake loads (Cl. 3.2.1.4 of IS 800:2007)
Earthquake loads shall be assumed as per the recommendations of IS 1893 (Part 1).
5. Erection Loads (Cl. 3.3 of IS 800:2007)
All loads required to be carried by the structure or any part of it due to storage or positioning
of construction material and erection equipment, including all loads due to operation of such
equipment shall be considered as erection loads. The structure as a whole and all parts of
the structure in conjunction with the temporary bracings shall be capable of sustaining these
loads during erection.
6. Temperature Effects (Cl. 3.4 of IS 800:2007)
Expansion and contraction due to changes in temperature of the members and elements of
a structure shall be considered and adequate provision made for such effect. The co-efficient
of thermal expansion for steel is as given in Cl. 2.2.4.l of IS 800:2007.
7. Load Combinations
All structures must be designed to support their own weight along with any superimposed
forces, such as the dead loads from other materials, live loads, wind pressures, seismic
forces, snow and ice loads, and earth pressures (if buried underground). Because various
loads may act on a structure simultaneously, load combinations should be evaluated to
determine the most severe conditions for design (worst case scenario). These load
combinations vary from one document to another, depending upon the jurisdiction. There are
a set of combinations for the allowable stress design and another set that incorporates load
factors for strength design.

Load combinations for design purposes shall be those that produce maximum forces and
effects and consequently maximum stresses and deformations. The following combination of
loads with appropriate partial safety factors as given in Table 4 of IS 800:2007 may be
considered. The table is reproduced here as Table 2 for ready reference. a) Dead load +
imposed load, b) Dead load + imposed load + wind or earthquake load, c) Dead load + wind
or earthquake load, and d) Dead load+ erection load. The effect of wind load and earthquake
loads shall not be considered to act simultaneously. The load combinations are outlined in
detail in Cl. 3.5 of IS 800:2007.

8. Explain limit state of serviceability and limit state of collapse briefly.

Answer:

The most important limit states which are considered in design as are follows:

(i) Limit state of collapse.

(ii) Limit state of serviceability

Limit State of Collapse

This limit state is also called as strength limit state as it corresponds to the maximum
load carrying capacity i.e., the safety requirements of the structure. The limit state of
collapse is assessed from collapse of the whole or part of the structure. As per this
limit state, the resistance to bending, shear, torsion and axial loads at ev ery section
shall not be less than that produced by the most unfavorable combination of loads on
that structure. The following limit states of collapse are considered in design:

(i) Limit state of collapse in flexure (bending)

(ii) Limit state of collapse in compression

(iii) Limit state of collapse in shear

(iv) Limit state of collapse in torsion.

Limit State of Serviceability

A structure is of no use if it is not serviceable. Thus, this limit state is introduced to
prevent excessive deflection and cracking. It ensure the satisfactory performance of
the structure at working loads. It is estimated on the basis of elastic theory or working
stress method because deformation is of significance under working load and not at
collapse. Limit state of serviceability of following limit states:

(i) Limit state of deflection

(ii) Limit state of cracking

(iii) Limit state of vibration

The structure should be designed which considering all the appropriate limit state of
safety and serviceability and on the basis of most critical limit state and then checked
for all other limit states.

9. What are the factors to be considered in mechanical properties of structural steel?

Answer:

Steel derives its mechanical properties from a combination of chemical composition, heat
treatment and manufacturing processes. While the major constituent of steel is iron, the
addition of very small quantities of other elements can have a marked effect upon the
properties of the steel. The strength of steel can be increased by the addition of alloys such
as manganese, niobium and vanadium. However, these alloy additions can also adversely
affect other properties, such as ductility, toughness and weldability .

Minimizing the sulphur level can enhance ductility , and toughness can be improved by the
addition of nickel. The chemical composition for each steel specification is therefore carefully
balanced and tested during its production to ensure that the appropriate properties are
achieved.
The alloying elements also produce a different response when the material is subjected to
heat treatments involving cooling at a prescribed rate from a particular peak temperature.
The manufacturing process may involve combinations of heat treatment and mechanical
working that are of critical importance to the performance of the steel.
Mechanical working takes place as the steel is being rolled or formed. The more steel is
rolled, the stronger it becomes. This effect is apparent in the material standards, which tend
to specify reducing levels of yield strength with increasing material thickness.
The effect of heat treatment is best explained by reference to the various production process
routes that can be used in steel manufacturing, the principal ones being:
 As-rolled steel
 Normalized steel
 Normalized-rolled steel
 Thermomechanically rolled (TMR) steel
 Quenched and tempered (Q&T) steel.

Steel cools as it is rolled, with a typical rolling finish temperature of around 750°C. Steel that
is then allowed to cool naturally is termed 'as-rolled' material. Normalizing takes place when
as-rolled material is heated back up to approximately 900°C, and held at that temperature
for a specific time, before being allowed to cool naturally. This process refines the grain size
and improves the mechanical properties, specifically toughness. Normalized-rolled is a
process where the temperature is above 900°C after rolling is completed. This has a similar
effect on the properties as normalizing, but it eliminates the extra process of reheating the
material. Normalized and normalized-rolled steels have an 'N' designation.
The use of high tensile steel can reduce the volume of steel needed but the steel needs to
be tough at operating temperatures, and it should also exhibit sufficient ductility to withstand
any ductile crack propagation. Therefore, higher strength steels require improved toughness
and ductility, which can be achieved only with low carbon clean steels and by maximizing
grain refinement. The implementation of the thermomechanical rolling process (TMR) is an
efficient way to achieve this.
Thermomechanically rolled steel utilises a particular chemistry of the steel to permit a lower
rolling finish temperature of around 700°C. Greater force is required to roll the steel at these
lower temperatures, and the properties are retained unless reheated above 650°C.
Thermomechanically rolled steel has an 'M' designation.
The process for Quenched and Tempered steel starts with a normalized material at 900°C. It
is rapidly cooled or 'quenched' to produce steel with high strength and hardness, but low
toughness. The toughness is restored by reheating it to 600°C, maintaining the temperature
for a specific time, and then allowing it to cool naturally (Tempering). Quenched and
tempered steels have a 'Q' designation.
Quenching involves cooling a product rapidly by immersion directly into water or oil. It is
frequently used in conjunction with tempering which is a second stage heat treatment to
temperatures below the austenitizing range. The effect of tempering is to soften previously
hardened structures and make them tougher and more ductile.

10. What are special features of limit state design method compare to other methods
of design of steel structures?
Answer:
Limit state design has advancement over the traditional design philosophies. It considers the
safety at the ultimate load and serviceability at the working load, sort of extension of the
WSM and ULM.
“Limit state is the state of impending failure, beyond which a structure ceases to perform its
intended function satisfactorily, in terms of either safety or serviceability.”
Unlike WSM which based calculations on service load conditions alone, and unlike ULM,
which based calculations on ultimate load conditions alone, LSM aims for a comprehensive
and rational solution to the design problem, by considering safety at ultimate loads and
serviceability at working loads.
The LSM philosophy uses a multiple safety factor format which attempts to provide adequate
safety at ultimate loads as well as adequate serviceability at service loads, by considering all
possible ‘Limit State’.

A limit state is a state of impending failure, beyond which a structure ceases to perform its

intended function satisfactorily, in terms of either safety of serviceability i.e. it either

collapses or becomes unserviceable.There are two types of limit states:

Ultimate limit states (limit states of collapse):- which deal with strength, overturning, sliding,

buckling, fatigue fracture etc.

Serviceability limit states: – which deals with discomfort to occupancy and/ or malfunction,

caused by excessive deflection, crack width, vibration leakage etc., and also loss of

durability etc.

11. Which of the following is advantage of HSFG bolts over bearing type bolts?
a) joints are not rigid
b) bolts are subjected to shearing and bearing stresses
c) high strength fatigue
d) low static strength

Answer: c

12. Tacking fasteners are used when

a) minimum distance between centre of two adjacent fasteners is exceeded
b) maximum distance between centre of two adjacent fasteners is exceeded
c) maximum distance between centre of two adjacent fasteners is not exceeded
d) for aesthetic appearance

Answer: b

13. 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

14. The types of welded joints does not depend on

a) size of members connected at joint
b) type of loading
c) area available for welding
d) size of weld

Answer: d

15. The design nominal strength of fillet weld is given by

a) fu
b) √3 fu
c) fu/√3
d) fu/(1.25 x √3)

Answer: c
16. Which of the following is not true regarding effective throat thickness of weld?
a) Effective throat thickness should not be less than 3mm
b) It should not exceed 0.7t or 1t, where t is thickness of thinner plate of elements
being welded
c) Effective throat thickness = K x size of weld, where K is a constant
d) Effective throat thickness = K x (size of weld)2 , where K is a constant

Answer: d

17. A lap joint consists of two plates 200 x 12 mm connected by means of 20 mm

diameter bolts of grade 4.6. All bolts are in one line. Calculate strength of single
bolt and no. of bolts to be provided in the joint.

Solution:
Given
18. Design the Lap joint for the plates of sizes 100 × 16 mm and 100 × 10 mm thick
connected so as to transmit a factored load of 100 kN using single row of 16 mm
diameter bolts of grade 4.6 and plate of 410 grade.

Solution: Given
19. State types of bolted joints and types of failure in case of bolted joints.
Answer:
i) Types of bolted joint

20. State various advantages of welded joints and disadvantages of bolted joints.
21. List the values of partial safety factor for material strength in case of resistance by
yielding, buckling and ultimate stress in bolted connection.

Answer:

22. Explain what do you mean by shear lag?

Answer:
23. A discontinuous compression member consists of 2 ISA 90 × 90 × 10 mm
connected back to back on opposite sides of 12 mm thick gusset plate and
connected by welding. The length of strut is 3 m. It is welded on either side.
Calculate design compressive strength of strut.
For ISA 90 × 90 × 10, Cxx = Cyy = 25.9 mm Ixx = Iyy = 126.7 × 104 mm4, rzz = 27.3
mm values of fcd are

KL/r 90 100 110 120

fcd (N/mm2) 121 107 94.6 83.7

Solution:
24. Check whether ISMB250@37.4 kg/m is suitable or not as a simply supported beam
over an effective span of 6 m. The compression flange of beam is laterally
supported throughout the span. It carries udl of 15 kN/m (including self wt.).
Properties of ISMB 250 are bf = 125 mm, tf = 12.5 mm, tw = 6.9 mm, Ixx =
5131.6×104 mm4, Zxx = 410×103 mm3, r1 = 13.0 mm, Zpx = 465.71 × 103 mm3, ym0 =
1.1, βb = 1 and fy = 250 MPa.

Solution:
25. State types of bolted joints and types of failure in case of bolted joints.

Answer:
26. Draw sketches of Howe type and Pratt type truss showing pitch, rise, panel point,
panel, principal rafters and all members in one of the above types.

Answer:

27. Sketch different sections used as built-up strut and built-up column.

Answer:

Built-up strut
28. State with a sketch the effective length for a compression member as per IS 800 –
2007 having end conditions as
(i) Translation restrained at both ends and rotation free at both ends
(ii) Translation and rotation restrained at both ends

Answer:
29. State the function of lacing and battening.

Answer:

30. Limiting width to thickness ratio for single beam section of plastic class is 9.4 and
d/tw = 84. State whether ISMB 500 @ 852 N/m is of plastic class or not. For ISMB
500; h = 500 mm, bf = 180 mm, tf = 17.2 mm, tw = 10.2 mm, r1 = 17.0 mm, fy = 250
MPa.

Solution:
31. 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

Answer: a

32. 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

Answer: c

33. What is the maximum effective slenderness ratio for members always in
tension?
a) 400
b) 200
c) 350
d) 150

Answer: a

34. 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
Answer: a

35. Which section to be considered in the design for the net area of flat?

a) 1-5-6-3
b) 2-7-4
c) 1-5-7-4
d) 1-5-7-6-3
Answer: d
36. What is the net section area of steel plate 40cm wide and 10mm thick with one
bolt if diameter of bolt hole is 18mm?

a) 38.2 cm2
b) 20 cm2
c) 240 mm2
d) 480 mm2

Answer: a

37. Which of the following is property of compression member?

a) member must be sufficiently rigid to prevent general buckling
b) member must not be sufficiently rigid to prevent local buckling
c) elements of member should be thin to prevent local buckling
d) elements of member need not prevent local buckling

Answer: a
38. Which of the following is true about tubular section?
a) tubes have low buckling strength
b) tubes have same radius of gyration in all direction
c) tubes do not have torsional resistance
d) weight of tubular section is more than the weight required for open profile sections

Answer: b
39. Effective length of compression member is
a) distance between ends of members
b) distance between end point and midpoint of member
c) distance between points of contraflexure
d) distance between end point and centroid of member

Answer: c
40. What is slenderness ratio of compression member?
a) ratio of effective length to radius of gyration
b) ratio of radius of gyration to effective length
c) difference of radius of gyration and effective length
d) product of radius of gyration and effective length

Answer: a
41. Which of the following is true?
a) built up column lacings or battens are uneconomical if load carrying members
permit greater reduction in weight than what is added by lacing or batten
 b) built up column lacings or battens are economical if load carrying members permit
greater reduction in weight than what is added by lacing or batten
c) no related shear stress force in plane of cross section
d) built up column designed as axially loaded column can never be eccentrically
loaded

Answer: b

42. Which of the following is true?

a) in case of rolled section, less thickness of plate is adopted to prevent local
buckling
b) for built-up section and cold formed section, longitudinal stiffeners are not
provided to reduce width to smaller sizes
c) local buckling cannot be prevented by limiting width-thickness ratio
d) in case of rolled section, high thickness of plate is adopted to prevent local
buckling

Answer: d

43. What are laterally restrained beams?

a) adequate restraints are provided to beam
b) adequate restraints are not provided to beam
c) economically not viable
d) unstable beams

Answer: a
44. Critical bending moment capacity of a beam undergoing lateral torsional
buckling is a function of
a) does not depend on anything
b) pure torsional resistance only
c) warping torsional resistance only
d) pure torsional resistance and warping torsional resistance

Answer: d
45. Which of the following statement is not correct?
a) Hollow circular tube has more efficiency as flexural member
b) Hollow circular tube has lesser efficiency as flexural member
c) It is the most efficient shape for torsional resistance
d) It is rarely used as a beam element

Answer: a
46. Which of following statement is correct?
a) elastic buckling stress may be decreased by using longitudinal stiffeners
b) elastic buckling stress may be decreased by using intermediate stiffeners
c) elastic buckling stress may be increased by using intermediate transverse
 stiffeners
d) elastic buckling stress is not affected by intermediate or longitudinal stiffeners

47. Structural members subjected to bending and large axial compressive loads
are known as
a) Strut
b) Purlin
c) beam-column
d) lintel

Answer: c
48. Which of the following assumptions were not made while deriving expression
for elastic critical moment?
a) beam is initially undisturbed and without imperfections
b) behaviour of beam is elastic
c) load acts in plane of web only
d) ends of beam are fixed support

Answer: d
49. The web is susceptible to shear buckling when d/tw
a) <67ε
b) < 2×67ε
c) >67ε
d) < 70ε

Answer: c

50. Find the value of permissible stress in axial tension (σat) for fy = 250 MPa. State
why unequal angles with long legs connected are more efficient?
51. Design a tension member consisting of single unequal angle section to carry a
tensile load of 340 kN. Assume single row 20 mm bolted connection. The length of
member is 2.4 m. Take fu = 410 MPa, α = 0.80
2
Section available (mm) Area (mm )
ISA 100 × 75 × 8 1336
ISA 125 × 75 × 8 1538
ISA 150 × 75 × 8 1748

Solution:
52. A hall of size 12m x 18m is provided with Fink type trusses at 3 m c/c. Calculate
panel point load in case of Dead load and live load from following data.
a. Unit weight of roofing = 150 N/m2
b. Self-weight of purlin = 220 N/m2
c. Weight of bracing = 80 N/m2
d. Rise to span ratio = 1/5
e. No. of panels = 6

Solution:
53. An industrial building has trusses for 14 m span. Trusses are spaced at 4m c/c
and rise of truss in 3.6m. Calculate panel point load in case of live load and wind
load using following data :
a. Coefficient of external wind pressure = - 0.7
b. Coefficient of internal wind pressure = ± 0.2
c. Design wind pressure = 1.5 kPa
d. Number of panels = 08
54. Design a slab base for column ISHB 400 @ 82.2 kg/m to carry factored axial
compressive load of 2000 kN. The base rests on concrete pedestal of grade M20.
For ISHB 400, bf = 250 mm, fy = 250 MPa, fu = 410 MPa, ymo = 1.1, tf = 12.7 mm.

Solution:
55. Write steps to calculate the thickness of base plate used in slab base. Why anchor
bolts are used in slab base?

Answer:
56. Differentiate between Laterally supported and unsupported beams with a neat
sketch.

57. Define Gusseted base. Also draw its labelled sketch showing all details.

Answer:
58. How beam sections are classified for bending as per IS : 800- 2007. Describe any two of
them.

Answer:

59. A simply supported beam of 6 m span supports on R. C. C. slab where in compression

flange is embedded. The beam is subjected to a dead load of 25 kN/m and super imposed
load of 20 kN/m, over entire span. Calculate plastic and elastic modulus required.
Assume rf = 1.5, ym = 1.1 fy = 250 N/mm .
2

Solution:
60. Which of the following are true about roof trusses?

a) principal rafter are compression members used in buildings

b) principal rafter is bottom chord member of roof truss
c) struts are compression members used in roof trusses
d) struts are tension members used in roof trusses
Answer: c

61. Which of the following is not a load on columns in buildings?

a) load from floors
b) load from foundation
c) load from roofs
d) load from walls

Answer: b

62. What are loads on columns in industrial buildings?

a) wind load only
b) crane load only
c) wind and crane load
d) load from foundation

Answer: c

63. 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

64. 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

65. 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: c
66. Which of the following is true?
a) ultimate load is reached when a mechanism is formed
b) ultimate load is not reached when a mechanism is formed
c) plastic hinges are not required for beam to form a mechanism
d) frictionless hinges are not required for beam to form a mechanism

Answer: a

67. Which of the following condition is true for kinematic theorem?

a) load must be greater than collapse load
b) load must be less than collapse load
c) load must be not equal to collapse load
d) load cannot be related to collapse load

Answer: a

68. Design a suitable ‘I’ beam for a simply supported span of 3 m and carrying a dead
or permanent load of 17.78 kN/m and an imposed load of 40 kN/m. Assume full
lateral restraint and stiff support bearing of 100 mm.

Solution:
69. Obtain factored axial load on the column section ISHB400. The height of the
column is 3.0 m and it is pin-ended. Use fy = 250 N/mm2, E = 2 x 105 N/mm2, ym=
1.15.

Solution:
70. Design a simple base plate for a ISHB400 @ 0.822 kN/m column to carry a factored
load of 1800 kN. Use fcu = 40 N/mm2 ; fy = 250 N/mm2 ; γm = 1.15

Solution:

71. Check the adequacy of ISMB 450 to carry a uniformly distributed load of 24 kN / m
over a span of 6 m. Both ends of the beam are attached to the flanges of columns
by double web cleat.
72. In a roof truss, a tie member ISA 110 mm X 110mm X 8 mm carries a factored
tension of value 210 kN. The tie is connected to a gusset plate 8 mm thick. Design
the welded joint. Factored yield strength of steel is 217.4 N/mm2 and shear
strength of weld is 125 N/mm2.
Solution:
73. Design a bolted connection between a bracket 8 mm thick and the flange of an
ISHB 400 column using HSFG bolts, so as to carry a vertical load of 100 kN at a
distance of 200 mm from the face of the column as shown in Fig. E1.

Solution:

74. Design a bolted web cleat beam-to-column connection between an ISMB 400 beam
and an ISHB 200 @ 40 kg/m column. The connection has to transfer a factored
shear of 150 kN. Use bolts of diameter 20 mm and grade 4.6.

Solution:
75. Design a double web cleat connection for an ISMB 400 coped beam to an ISMB
600 main beam so as to transfer a factored load of 300 kN using HSFG bolts of
20mm diameter and grade 8.8.
76. The shear lag width for ISA 75X75X10 is (Assume gauge distance = 40 mm).

Solution:

The length of outstanding leg will be w = 75 mm and w1 = 40 mm.

So the shear lag width, bs = w + w1 – t = 75 + 40 – 10 = 105 mm.

77. Explain various types of standard rolled steel sections.

Answer:

Various types of standard rolled steel sections

i) Rolled steel I-sections (Beam sections)
ii) Rolled steel channel sections
iii) Rolled steel Tee sections
iv) Rolled steel angle sections

v) Rolled steel bars

vi) Rolled steel flats
vii) Rolled steel plates
viii) Rolled steel sheets
ix) Rolled steel strips
x) Rolled steel tubular sections

(a) Rolled steel I – sections (Beam sections)

• Indian Standard Junior Beam (ISJB)
• Indian Standard Light Beam (ISLB)
• Indian Standard Medium weight Beam (ISMB)
• Indian Standard Wide flange Beam (ISWB)
• Indian Standard Heavy Beam (ISHB)
• An I – Section is designated by its depth and weight

Eg: An ISLB 500 @ 735.8 N/m means, An I – section is 500 mm deep and self weight is
735.8 N per meter length.
• Special beam section available from Indian rolling mill is Indian Column Section (ISC)

(b) Rolled Steel Channel Sections

• Indian Standard Junior Channel (ISJC)
• Indian Standard Light Channel (ISLC)
• Indian Standard Medium Weight Channel with Sloping Flange
(ISMC)
• Indian Standard Medium Weight Channel with parallel flange
(ISMCP)
• Indian Standard Gate Channel (ISGC)
• Designated by its depth and weight

Ex: ISLC 350 @ 380.63 N/m

(c) Rolled Steel T – Sections

• Indian Standard rolled Normal T – section (ISNT)
• Indian Standard rolled Deep legged T – (ISDT)
• Indian Standard rolled silt Light weight T – bars (ISLT)
• Indian Standard rolled silt Medium weight T – bars (ISMT)
• Indian Standard rolled silt T – bars from H – section (ISHT)
• Designated by its depth and weight

Ex : ISNT 125 @ 274 N/m

(d) Rolled Steel Angle sections

• Indian standard equal angles, Indian standard unequal angles and Indian standard bulb
angles
• Designated by abbreviation ISA along with widths of both legs and thickness.
• Indian equal angles are designated as ISA or ISEA (Ex. ISEA 100 x 100 x 10 mm), Indian
standard unequal angles are designated as ISA (Ex. ISA 125 x 75 x 10 mm) and Indian
standard bulb angles are designated as ISBA.

78. Which of the following is correct in case of angle members?

a) connection of lug angle to angle member should be capable of developing a strength
of 10% of excess of force of outstanding leg of angle
b) connection of lug angle to angle member should be capable of developing a strength
of 20% of excess of force of outstanding leg of angle
c) lug angles and their connection to gusset should be capable of developing a strength
of less than 20% of excess of force of outstanding leg of angle
d) lug angles and their connection to gusset should be capable of developing a strength
of not less than 20% of excess of force of outstanding leg of angle

Answer: d
79. The effective length of compression flange of simply supported beam not
restrained against torsion at ends is
a) 1.2 L
b) 1.0 L
c) 0.8 L
d) 0.5 L

Answer: a

80. Which of the following assumptions were not made while deriving expression for
elastic critical moment?
a) beam is initially undisturbed and without imperfections
b) behaviour of beam is elastic
c) load acts in plane of web only
d) ends of beam are fixed support

Answer: d

81. A single angle section 90X60X10 is connected with gusset plate with 7 bolts of 20
mm diameter in one line at pitch of 50 mm and edge distance of 30 mm. What is
the design tensile strength of the section for rupture of net section? (Assume the
section is connected with longer leg and gauge distance = 50 mm)

Solution:

Anc = (90 - 10/2 - 22) × 10 = 630 mm2

Ago = (60 - 10/2) × 10 = 550 mm2
An = 630 + 550 = 1180 mm2

The length of outstanding leg will be w = 60 mm and w1 = 50 mm. So the shear lag width,
bs = w + w1 – t = 60 + 50 – 10 = 100 mm.

Distance between end bolts , Lc = 6 × 50 = 300 mm.

82. A single ISA 75 × 50 × 8 is connected (longer leg) with gusset plate using use 4
bolts of 20 mm diameter in one line at pitch of 50 mm and edge distance of 30 mm.
What is the Design tensile strength due to block shear failure? (Assume gauge
distance = 35 mm)

Solution:
Avg = 8 × (3 × 50 + 30) = 1440 mm2
Avn = 8 × (3 × 50 + 30 – 3.5 × 22) = 824 mm2
Atg = 8 × 40 = 320 mm2 [assuming gauge g = 35 for 75 mm leg]
Atn = 8 × (40 – 0.5 × 22) = 232 mm2
83. An ISA 90 x 90 x 8 used as tension member is connected to a 10 mm gusset plate
by fillet weld of size 5 mm. The design strength of the member is 300 kN. Calculate
the length of the weld.

84. What are the various assumptions used in plastic analysis theory?
Answer:

The following are the assumptions are made in plastic design to simplify computations:

1) The material obeys Hooke, Law till the stress reaches fy.

2) The yield stress and modulus of elasticity have the same value in compression and
tension.

3) The material is homogeneous and isotropic in both the elastic and plastic states.

4) The material is assumed to be sufficiently ductile to permit large rotation of the section to
take place.

5) Plastic hinge rotation is large compare with the elastic deformations so that all the
rotations are concentrated at the plastic hinges. The segments between the plastic hinges
are rigid.

6) The magnitude of bending moment caused by the external loads will at the most be equal
to the plastic moment reached the capacity of the section.

7) The influence of normal and shear forces on plastic moments is not considered.
8) Plane sections remain plane even after bending and the effect of shear is neglected.

9) The equilibrium of forces at the time of collapse is considered for the undeformed state of
the structure.

10) No instability occurs in any member of the structure upto collapse.

85. As per IS: 800 what are the various conditions satisfied in order to use plastic method of
analysis?

Answer:

IS: 800 stipulates that the following conditions should be satisfied in order to use the plastic
method of analysis:

1) The yield stress of steel used should not be greater than 450MPa.

2) The stress-strain characteristic of the steel used should obey the following conditions, in
order to ensure plastic moment redistribution. a) The yield plateau (horizontal portion of the
stress-strain curve) should be greater than 6 times the yield strain. b) The ratio of the
ultimate tensile stress to the yield stress should be more than 1.2. c) The elongation on the
standard gauge length should be more than 15%. d) The steel should exhibit strain-
hardening capacity.

3) The members shall be hot-rolled or fabricated using hot-rolled plates.

4) The cross section of the members not containing plastic hinges should be ‘compact’ and
those of member containing plastic hinges should be ‘plastic’. 5) The cross-section should
be symmetrical about its axis perpendicular to the axis of the plastic hinge rotation.

These limitations are intended to ensure that there is a sufficiently long plastic plateau to
enabling a hinge to form and that the steel will not experience premature strain hardening.

86. Write down various advantages and disadvantages of plastic design.

Answer:

Advantages of Plastic Design

Plastic design methods offer the following advantages:

1) Realization of uniform and realistic F.O.S for all parts of the structures (in contrast to
elastic methods, where the safely factor varies)
2) Simplified analytical procedure and readily of obtaining design moments, since there is no
need to satisfy elastic strain compatibility conditions.
3) Saving of material over elastic methods resulting in lighter structures.
4) No effect due to temperature changes, settlement of supports, imperfection, erection
method, etc. (because their only effect is to change the amount of rotation required). This is
in contrast to the elastic method, where extra calculation are required. However, calculation
for instability and elastic deflection required careful considerations in plastic method. The
plastic design method is very popular for design of some structure, e.g, beams and portal
frames.
5) Gives some idea of collapse mode and strength of structure.
6) In the elastic method of design, the design process is repeated several times to obtain an
optimum solution, where the plastic method of design produces a balanced section in a
single attempt.

Dis-advantages of plastic design:

The disadvantages of plastic design method are the following:

1) Obtaining collapse load is difficult if the structure is reasonably complicated.
2) There is little saving in column design.
3) Difficult to design for fatigue.
4) Lateral bracing requirements are more than stringent than elastic design.
5) Calculations for elastic deformations require careful considerations.
6) When more than one loading condition occurs, it is necessary to perform separate
calculations, one for each loading condition; the section requiring the largest plastic moment
is selected. Unlike the elastic method of design, wherein the moment produced by different
loading condition can be added together, the plastic moment obtained by different loading
conditions cannot be combined(i.e, the plastic moment calculated for a given set of loads is
valid only for that loading condition). This is because the ‘principle of super position’
becomes invalid when certain parts of the structure have yielded.

87. Write a short note on plastic hinge and hinge length.

Answer:

Plastic Hinge:
A plastic hinge is a zone of yielding due to flexure in a structural member. Although hinges
do not actually form, it cal be seen that large changes of slope occurs over small length of
the member at position of maximum moments. A strain hardening action usually occurs at
these hinges so that large deflections are accompanied by a slight increase in load.

A structure can support the computed ultimate load due to the formation of plastic hinges at
certain critical sections. The member remain elastic until the moment reaches a value Mp,
the maximum moment of resistance of a fully yielded cross section or fully plastic moment of
a section(Mp = fy Zp). Any additional moment will cause the beam to rotate with little
increase in stress. The rotation occurs at a constant moment (Mp). The zone acts as if it was
hinges except with a constant restraining moment (Mp). The plastic hinge, therefore, can be
defined as a yielded zone due to flexure in a structure in which infinite rotation can take
place at a constant restraining moment (Mp) of the section. It is represented normally by a
black dot. The value of the moment at the adjacent sections of the yield zone for a certain
length is more than the yield moment. This length is known as hinge length, depends upon
the loading and geometry of the section. To simplify the analysis, this small length is
neglected and the plastic hinge is assumed to be formed at discrete points of zero length.
But, it cannot be neglected for the calculation of deflections and the design of bracings as
the length over which yielding extends is quite important.

The plastic hinges are formed first at the sections subjected to the greatest deformation (
curvature). The possible places for plastic hinges in a structure with prismatic members are
points of concentrated loads, at the ends of member meeting at a connection involving a
change in geometry and at the point of zero shear in a span under distributed load.
Hinge Length:

Consider a simply supported rectangular beam subjected to a gradually increasing

concentrated load P, at the centre. A plastic hinge will be formed at the centre.
Mp = PL/4; My = fy x Ze = fy x bd^2/6 = fy (1/6) x {4x(1/4)} bd^2 = (2/3) x fy xbd^2/4 = (2/3) x
fy x Zp = (2/3)Mp, i.e, Mp is 1.5 times more than My.

From the BM diagram, Mp / (L/2) = My / (L/2-x/2) => x=L/3. Therefore, the hinge length of
the plasticity zone is equal to 1/3rd of the span.

Similarly, the hinge length of the plasticity zone for a simple beam subjected to uniformly
distributed load is L/sqrt(3).

88. Explain in detail shape factor and load factor.

Answer:

Shape Factor (v):

The ratio Mp / My is a property of a cross sectional shape and is independent of the material
properties. This ration is known as the shape factor v and is given by v = Mp / My = = fy Zp /
fy Ze.= Zp / Ze For wide-flange I-section in flexure about the strong axis, the shape factor
ranges from 1.09 to about 1.18 with the average value being 1.14. One may conservatively
take the plastic moment strength Mp of I-section bent about their strong axis to be at least
15% greater than the strength My. On the other hand, the shape factor for I-section bent
about their minor axis is about the same as for a rectangular section, i.e, about 1.5.

Load Factor:

Load factor is defined as the ratio of the collapse load to the working load(service load) and
is represented by F, i.e, F = Pc / Pw
If a collection of beams having different end conditions (free or fixed) and the working load W
were first design elastically and then plastically, the ratio Pc / Pw will not be identical. Only
the beams that are simply supported will produce a constant ratio of Pc / Pw and for these
cases the values of Pc / Pw will be the lowest.

From a practical point of view, a minimum acceptable and constant load factor is required,
and that found for a simply supported beam may be regarded satisfactory. For a simple
beam the variation of the bending moment with the load is linear.

In actual practice a load factor varying from 1.7 to 2.0 is assumed depending upon the
designer’s judgment. When the structures are subjected to wind the corresponding load
factor for plastic design is reduced by 25%. The prime function of the load factor is to ensure
that the structure will be safe under the collapse load. Therefore, it may be regarded as a
factor of safety based upon the collapse load. It depends upon the nature of loading, the
support conditions, and the geometrical shape of the structural members. Uncertainty of the
loads, imperfection in workmanship and error in fabrication are some of the other factors
which influence the choice of the load factor.
89. Explain various mechanism of plastic analysis.

Answer:
When a structure is subjected to a system of loads, it is stable and hence functional until a
sufficient number of plastic hinges have been formed to render the structure unstable. As
soon as the structure reaches an unstable condition, it is considered to have been failed.
The segments of the beams between the plastic hinges are able to move without an
increase of load. This condition in a member is called mechanism. The concept of
mechanism formation in a structure due to loading beyond the elastic limit and of virtual work
is used in the plastic analysis and design of steel structures. If an indeterminate structure
has the redundancy r, the insertion of r plastic hinges makes it statically determinate. Any
further hinge converts this statically determinate structure into mechanism. Hence, for
collapse, the numbers of plastic hinges required are (r+1).

Types of Mechanism: Various possible mechanism are listed below:

a) Beam mechanism
b) Panel / sway mechanism
c) Joint mechanism
d) Gable mechanism
e) Composite (combined) mechanism

Number of Independent Mechanism: Let,

N = number of possible plastic hinges

r = number of redundancies
n = possible independent mechanism.

Then, n = N – r

After finding out the number of independent mechanism all the possible combinations are
made in such a way so as to make the external works maximum or the internal work a
minimum. This is done to obtain the lowest load.

90. What are the different theorems of plastic analysis? Explain

Answer:
The plastic analysis of a structure is govern by three theorem, which are as follows:
1) The static or lower bound theorem: states that a load (P<Pc)computed on the basis of an
assumed equilibrium moment diagram, in which the moments(M) are nowhere greater than
the plastic moment(M<Mp), is less than, at the best equal to, the correct collapse load.
Hence the static method represents the lower limit to the true ultimate load and has a
maximum factor of safety. The static theorem satisfies the equilibrium and yield conditions.

2) Kinematic or Upper bound theorem: states that a load computed on the basis of an
assumed mechanism will always be greater than or at the best equal to , the true collapse
load (P>Pc). Hence the kinematic method represents an upper limit to the true ultimate load
and has a smaller factor of safety. The kinematic theorem satisfies the equilibrium and
continuity conditions.

3) Uniqueness theorem: The lower and upper bound theorems can be combined to produce
the uniqueness theorem, which states that the load that satisfies both the theorems at the
same time is the correct collapse load. When both the theorems are satisfied in a given
problem then the solution is said to be the correct (unique) one.
Using the principle of virtual work and the upper and lower bound theorems, a structure can
be analysed for its ultimate load by any of the following methods:

1) Static method: This consists of selecting the redundant forces, The free and redundant
bending moment diagram is drawn for the structure. A combined bending moment diagram
is drawn in such a way that a mechanism is formed. The collapse load is found by working
out the equilibrium equation. It is checked that the bending moment is not more than the fully
plastic moment at any section.

2) Kinematic method: This consists of locating the possible places of plastic hinges. The
possible independent and combined mechanism are ascertained. The collapse load is found
by applying the principle of virtual work. A bending moment diagram corresponding to the
collapse mechanism is drawn and it is checked that the bending moment is not more than
the fully plastic moment at any section.

For complicated frames, the static method of analysis is more difficult, and finding the correct
equilibrium equation becomes illusive. In these cases, the kinematic method is more
practical.

91. A simply supported beam of span L supports a concentrated load W at its

midspan. If the cross-section of the beam is circular, then the length of elastic-
plastic zone of the plastic hinge will be

Solution:
92. A fixed beam made of steel is shown in the figure below. At collapse, the value of
load P will be equal to

Solution:

93. Explain various types of bolt briefly.

Answer:

Types of bolts
There are several types of bolts used to connect structural members. Some of them are
listed below

(a) Black bolts or unfinished bolts

• Black bolts are referred to as ordinary, rough or common bolts. They are least expensive
bolts and are made of low carbon steels (mild steel) with square or hexagonal head. The
diameter of the hole is about 1.0 to 2.0 mm larger than the bolt diameter for ease in fitting.
They are designated as Md× l, ‘d’ – shank diameter of bolt and , l – length of the bolt.

They are primarily used in light structures under static loads such as small trusses, purlins,
bracings. They are also used as temporary fasteners during erection where HSFG bolts or
welding are used as permanent fasteners.

• These bolts are not recommended for connections, which are subjected to impact load,
vibration and fatigue.
• For bolt of a grade or property class 4.6 represents the ultimate tensile strength is 400
N/mm2 and yield strength is 0.6 times 400 which is 240N/mm2.

• Ordinary bolted joints, the force transfer through interlocking and bearing of bolts and joint
is called bearing type joint.

(b) High Strength Friction Grip (HSFG) bolts

• High strength friction grip bolts are made from bars of medium carbon heat treated steel
(high tensile steel). The bolt property class 10.9S and 12.9S are commonly used in steel
connections.

• The HSFG bolts are available in sizes from 16mm to 36mm and are designated as M16,
M20, M24 and M30.

• These bolts tightened (by torque wrenches) until they have very high tensile stresses, so
that connected parts are clamped tightly together between the bolt head and nut, this
permits load to be transferred primarily by friction not by shear.

• These bolts are most suitable for bridges where the stress reversal may occur or slippage
is undesirable also for seismic loading and for fatigue load.

• High strength bolts have replaced rivets and black bolts are being used in structures, high
raised building, bridges etc.

94. Explain various steps involved in the design of laterally unsupported beam.

Answer:

Steps involved are:

95. Explain web crippling and web buckling with the help of sketch.

Answer:

Web crippling:

Web crippling causes local crushing failure of web due to large bearing stresses under
reactions at supports or concentrated loads. This occurs due to stress concentration
because of the bottle neck condition at the junction between flanges and web. It is due to the
large localized bearing stress caused by the transfer of compression from relatively wide
flange to narrow and thin web. Web crippling is the crushing failure of the metal at the
junction of flange and web. Web crippling causes local buckling of web at the junction of web
and flange.
For safety against web crippling, the resisting force shall be greater than the reaction or the
concentrated load. It will be assumed that the reaction or concentrated load is dispersed into
the web with a slope of 1 in 2.5 as shown in the figure

Let Resisting force = Fwc

Thickness of web = tw
Yield stress in web = fyw
Width of bearing plate = b1
Width of dispersion = n2 = 2.5 h2
Depth of fillet = h2 (from SP [6])
Fwc = [(b1 + n2) tw fyw ] / ymo ≥ Reaction, Ru

For concentrated loads, the dispersion is on both sides and the resisting force can be
expressed as

Fwc = [(b1 + 2 n2) tw fyw ] / ymo ≥Concentrated load, W u

Web Buckling:

The web of the beam is thin and can buckle under reactions and concentrated loads with the
web behaving like a short column fixed at the flanges. The unsupported length between the
fillet lines for I sections and the vertical distance between the flanges or flange angles in built
up sections can buckle due to reactions or concentrated loads. This is called web buckling.
For safety against web buckling, the resisting force shall be greater than the reaction or the
concentrated load. It will be assumed that the reaction or concentrated load is dispersed into
the web at 45° as shown in the figure.

Let Resisting force = Fwb

Thickness of web = tw
Design compressive stress in web = fcd
Width of bearing plate = b1
Width of dispersion = n1
Fwb = (b1 + n1) tw fcd ≥ Reaction, Ru

For concentrated loads, the dispersion is on both sides and the resisting force can be
expressed as

Fwc = [(b1 + 2 n1) tw fcd ] ≥Concentrated load, Wu

The design compressive stress fcd is calculated based on a effective slenderness ratio of
0.7 d / ry, where d = clear depth of web between the flanges.
ry = radius of gyration about y-y axis and is expressed as
= sqrt (Iyy / area) = sqrt [(tw)3 / 12/ t] = sqrt [(tw)2 / 12]

kl / ry = (0.7 d) / sqrt [(tw)2 / 12] = 2.425 * d / tw

96. Explain various steps involved in the design of laterally supported beam as per IS
800 (2007).

Answer:

Various steps involved are:

1. Calculate the factored load and the maximum bending moment and shear force
2. Obtain the plastic section modulus required
Select a suitable section for the beam-ISLB, ISMB, ISWB or suitable built up sections
(doubly symmetric only). (Doubly symmetric, singly symmetric and asymmetric- procedures
are different)
3. Check for section classification such as plastic, compact, semi-compact or slender. Most
of the sections are either plastic or compact. Flange and web criteria.

6. Check for buckling

7. Check for crippling or bearing
8. Check for deflection

97. At the location of plastic hinge

(a) Radius of curvature is infinite
(b) Curvature is infinite
(c) Moment is infinite
(d) Flexural stress is infinite

Answer: b

98. A ductile structure is defined as one for which the plastic deformation before
fracture
(a) is smaller than the elastic deformation
(b) vanishes
(c) is equal to the elastic deformation
(d) is much larger than the elastic deformation

Answer: d

99. Assertion (A): The shape factor of a circular section is less than that of a rectangular
section.
Reason (R): Compared to rectangular section, a circular section has more area near the
neutral axis than at the extreme fibre.
Of these statements:
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true but R is not a correct explantion of A
(c) A is true but R is false
(d) A is false but R is true

Answer: d

100. A fixed beam made of steel is shown in the figure below. At collapse, the value of
load P will be equal to

Solution: