sixthTH SEMESTER DIPLOMA EXAMINATION IN ENGINEERING 

/TECHNOLOGY October 2019

 

Subject:  Structural Design‐II 

Subject code: 6013 

Branch: ARCHITECTURE 

   

 

Prepared By  

        Name : UDAYAKUMAR V.D 

        Designation :Lecturer 

        Department :Civil Engineering 

        Mobile No.9605881664 

 

 

 

 

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PART A

I)

1. Bolted connection OR riveted connection OR Welded connection

2. Tension member or Tie member is a structural member which resists axial tension

3. It is the ratio of effective height to least lateral dimension, λ= L/ rmin

4. Effective Height - The height of a wall or column to be considered for calculating slenderness ratio.

5. A beam which does not laterally move nor rotate is known as Laterally-Supported Beam. It can be achieved by immersing compression flange of the beam in the RCC slab

PART B

II)

1. ( SAME AS GIVEN IN SECTION 2.2.4 OF IS 800)

Density- Structural steel has density of 7.75 to 8.1 g/cm3.

Elastic Modulus-Typical values for structural steel range from 190-210 gigapascals.

Poisson's Ratio-Acceptable values for structural steel are 0.27 to 0.3.

Tensile Strength

Yield Strength

Structural steel made of carbon has yield strengths of 187 to 758 MPa. Structural steel made of alloys has values from 366 to 1793 MPa.

Melting Point- There is no defined value for melting point due to the wide variations in types of structural steel. Melting point is the temperature at which object starts to melt when heated.

Specific Heat - Structural steel made of carbon has values from 450 to 2081 and that made from alloys has values ranging from 452 to 1499.

Hardness -Structural steel made by using alloys has hardness value between 149-627 Kg. Structural steels made of carbon has value of 86 to 388 Kg.

Maximum percentage of elongation

Notch toughness

2.  

Nominal diameter of bolt d= 16 mm

For grade Fe 410 bolt, fu =410 MPa

Assuming threads in the shear plane, nn=1, ns=0

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Shear Area of one bolt Anb = 0.8 Asb = 0.8 x π/4 x 162 = 156.83 mm2

Design shear strength per bolt Vdsb = Vnsb / γmb

But Vnsb = fu/ √3 x (nnAnb+nsAsb) = 37.12 KN

Vdsb = 37.12/1.25 = 29.7 KN

3.

Gross area (Ag) – It is the total cross sectional area of any member without deducting any hole area

Net area (An)- The net area at any section is equal to the gross area minus the deduction for holes at that section.

Net effective area – the equivalent area of imaginary axially loaded member of equal load carrying capacity is called net effective area

4.

Step 1. From the actual length of the compression member and the support conditions of the member, which are known, the effective length of the member is computed.

Step 2. From the radius of gyration about various axes of the section given in section tables, the minimum radius of gyration (rmin) is taken. rmin for a built up section is calculated.

Step 3. The maximum slenderness ratio (l/ rmin) is determined for the compression member.

Step 4. The allowable working stress (σac) in the direction of compression is found corresponding to the maximum slenderness ratio of the column from IS:800-1984.

Step 5. The effective sectional area (A) of the member is noted from structural steel section tables. For the built up members it can be calculated.

Step 6. The safe load carrying capacity of the member is determined as P=(σac.A), where P=safe load

5. ( Cl 3.7.2 of IS 800)

Plastic cross-sections: Plastic cross-sections are those which can develop their fullplastic moment Mp and allow sufficient rotation at or above this moment so that redistribution of bending moments can take place in the structure until complete failure mechanism is formed

• Compact cross-sections: Compact cross-sections are those which can develop their full-plastic moment Mp but where the local buckling prevents the required rotation at this moment to take place.

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• Semi-compact cross-sections: Semi-compact cross-sections are those in which the stress in the extreme fibers should be limited to yield stress because local buckling would prevent the development of the full-plastic moment Mp. Such sections can develop only yield moment My

• Slender cross-sections: Slender cross-sections are those in which yield in the extreme fibers cannot be attained because of premature local buckling in the elastic range .

6.

Live load

Dead load

Snow load

Wind load

Earth quake load

Snow load

7. a)

Stress reduction factor - To account for uncertainties in construction, material properties, calculated versus actual strengths and anticipated failure modes, the nominal strength of a masonry element is multiplied by an appropriate stress reduction factor, Φ. Strength reduction factors are used in conjunction with the load factors applied to the design loads. The values of the strength reduction factors for various types of loading conditions are:

for reinforced masonry elements subjected to flexure or axial loads; Φ = 0.90;

for unreinforced masonry elements subjected to flexure or axial loads; Φ = 0.60;

for masonry elements subjected to shear loads; Φ = 0.80;

for bearing on masonry elements; Φ = 0.60.

b)

Area reduction factor – This factor affects the permissible compressive stress that can be allowed in masonry construction. This factor takes into consideration smallness of the sectional area of the element and is applicable when sectional area of the element is less than O-2 m2.

The factor, k, = 0.7 + 1.5 A, A being the area of section in m2.

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PART –C UNIT –I

III a)

Structural Steel is superior in strength and it does not age or decay as quickly as other construction materials.

With a much higher strength-to-weight ratio (the strength of a material over its density) and tensile strength (ability to withstand stress from tension), structural steel will make for a lighter building that requires a less substantial and expensive foundation.

The strength, stiffness, toughness, and ductile properties of structural steel allow it to be fabricated into an endless variety of shapes.

Steel structures are assembled by bolting or welding the pieces together on site as soon as they are delivered as opposed to concrete, which takes weeks to cure before construction can continue.

Distribution of a building’s compression and tension stress among steel beams allows architects more freedom with design space and the ability to make last minute alterations.

Short construction periods also facilitate less of a disruption to the neighborhood of the site being developed.

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OR IV a)

A design philosophy is a set of assumptions and procedures which are used to meet the conditions of serviceability, safety, economy and functionality of the structure. Several design philosophies have been introduced from different parts of the world. Some of the design philosophies that has been used by engineers are

1. Working Stress Method(WSM)/ Allowable Stress Design (ASD) 2. Ultimate Load Method (ULM) 3. Limit State Method(LSM

All the members in the structure should have adequate strength, stiffness and toughness to ensure proper functioning during service life. Members should have adequate strength, stiffness and toughness to ensure proper functioning during service life.

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UNIT -II V a)

Lug angle is small piece of angle used to connect outstand legs of the members to the gusset plate. The purpose of lug angle is to reduce the length of connection to the gusset plate and to reduce shear lag effect . If lug angle is used then the unconnected length of main angle behaves like a connected leg and entire cross section  area of the angle become effective in resisting tension. So if lug angle is used, then efficiency of the tension member increases because it reduces shear lag effect.

If lug angle is used the resultant reaction at bolt location 1 and 2 pass through CG of cross section. Since action and reaction pass through CG of angle, stress and strain distribution are uniform hence no shear lag

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OR

VI a)

The main difference is in the purpose for which they are provided-Lacing flats carry no force but their sole purpose is to prevent buckling of column while Battens along with to check buckling plays a role in force transfer.

Moreover difference also exists in the way they are provided. Lacing flats are inclined while

battens are the horizontal plates. Lacings are provided in diagonal while battens are provided in

horizontal direction. And battens fails first because of less effective area (way they are provided) than the lacing

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UNIT –III

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VII b)

A plate girder is a built up I-beam section, used to carry heavy loads which cannot be carried economically by rolled I-sections. It is made by riveting or welding the steel plates in I-beam shape. Components of a typical Plate girder and their functions are given below

1. Web

The deep central vertical plate is called as a web in plate girder. It separates the two flange plates by a required distance. Web is responsible to resist shear developed in the plate girder.

2. Flanges

Flanges or flange plates are horizontal elements of plate girder which are provided at the top and bottom and they are separated by the web. The main purpose of flange plates is to resist the bending moment acting on the girder.

The top flange resists the bending moment by developing compression and the bottom flange resists the tensile force. They should be provided with a required width and thickness to offer good resistance against bending moment.

3. Stiffeners Stiffeners are classified into two types :

1. Vertical Stiffeners

2. Horizontal Stiffeners

Vertical Stiffeners Vertical stiffeners are provided at right angles to the flanges and they are also called as transverse stiffeners. These are again classified into two types namely end stiffeners and intermediate stiffeners based on their position in the plate girder.

End stiffeners

They are provided at both the ends of the girder. They receive the load from the beam and transfer it to the support. In plate girder, some part of the end portion of the web is subjected to compressive loads. Due to these loads, the web of the section may get crushed. Here, the end stiffeners play an important role by keeping the web safe from crushing. End stiffeners are also called as bearing stiffeners.

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Intermediate stiffeners

They are required when there are concentrated loads acting on the plate girder. When the thickness of the web is very less (less than 1/85th of the depth of the web), then the web may buckle due to shear. In that scenario, intermediate stiffeners are provided in order to improve the buckling strength of the web.

Horizontal Stiffeners Horizontal stiffeners are provided in parallel to the flange plates. They are also called as longitudinal stiffeners. These stiffeners will improve the buckling strength of the web portion. Horizontal stiffeners are either continuous or discontinuous.

OR

VIII a)

Steel beam is a structural member having length much greater than its cross sectional dimensions and are subjected to lateral loads. Beams are primarily designed to resist both BM and SF due to lateral loads. Angles, channels, Tees and I sections are commonly used beam sections. For heavy spans and loads built up beam sections known as plate girders are used. There are mainly two types of beams.Laterally supported and unsupported beams. A laterally supported beam is one where the compression flange is supported and prevented from buckling in the horizontal plane due to the compressive forces in the top flange. Laterally supported beams are mainly subjected to SF and BM while laterally unsupported beam is having fewer loads bearing capacity due lateral buckling of compression flange

Design procedure

Choose a trial section assuming as it as plastic section

Section checked for class it belongs

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Check for bending strength

Check for shear strength

Check for deflection

Section is revised if it needed

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UNIT –IV

IX a)

King Post Truss

Pratt Truss

Queen Post Truss

Howe Truss

Fan Truss

North Light Roof Truss

Quadrangular Roof Trusses

Parallel Chord Roof Truss

Scissor Roof Truss

Raised Heel Roof Truss

b) Given P = 10 KN, e = 30 mm, t =290 mm, S.l ratio – 12, L =2.5 m

Eccentricity ratio, e/t = 30/290 =0.103

Eccentricity ratio lies between 1/24 and 1/6

Area of the wall = L X t = 2.5x .29 = 0.725 m2 > 0.2 m2 ( Cl 5.4.1.2)

Area reduction factor Ka = 1

Stress reduction factor: from table 9 of IS 1905

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For Sl .Ratio = 12,

e/t =1/12, Ks = 0.81

e/t =1/6, Ks = 0.78

Hence for e/t = 0.103,Ks =0.803

Shape modification factor: (Kp): from table 10

For height to width ratio 1 and crushing strength of bricks 5N/mm2 Kp = 1.2

For M1 mortar type from table 8 , fb = 0.5N/mm2

Permissible compressive stress = 1.25x fb x Kp x Ks x Ka

= 1.25 x 0.5 x 1 x 0.803 x 1.2 = 0.602N/mm2

Maximum compressive stresses = (p/1000xt ) + (6Pe/1000t2) = 0.059 N/mm2

Minimum compressive stresses = (p/1000xt ) - (6Pe/1000t2) = 0.021 N/mm2

Permissible stress is less than actual stress

OR

X a)

Masonry structures gain stability from the support offered by cross walls, floor, roofs and other elements such as piers and buttress. Load bearing walls are structurally more efficient when the load is uniformly distributed and the structure is planned such that the eccentricity of the loading the member is small as possible. Eccentric loading is to be avoided by providing adequate bearing of floor or roof slab on the wall. Adequate stiffness in slab is to be provided and fixity at supports is to be avoided. The strength of wall is measured in terms of its resistance to the c0mbination of self weight, super imposed load and lateral pressure. The stability of wall is indicated caused by excessive slenderness ratio. The strength of a masonry wall depends upon the strength of masonry unit and strength of mortar. The quality of workmanship and the method of bonding also have a bearing on the strength of masonry wall. The mix proportions and strength to commonly used mortars for masonry may be obtained from IS 1905-1987.e thickness of load bearing wall should be such that it is sufficient at all points to ensure the resulting stresses due to the loading for which the wall is designed as within the prescribed limits for that type of wall

b) From wind zone map (IS875-Part3),

The basic wind Speed in Luknow is Vb=47m/s

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Risk co efficient K1, from table 12.1 for general building with

Terrain height and structure size factor K2

Dimension being 40 m, It belong to class B structure. K2=0.88 for h=10m, K2=0.94 if h=15m

There for h=12m, K2=0.904

Topography factor K3, the ground near shad may be assumed

Pain K3= 1+Cs

Cs=Z/L = 0, there for K3=1.0

Design wind speed V2=K1.K2.K3.Vb = 1.0x0.904x1.0x47=42.488 m/s

Basic wind pressure Pz=0.60 Vz2 =0.60x (42.488)2 = 1083N/m2

 

 

      

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