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design of steel structures
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UNIT 1
1)Shape factor for rectangular section?
Z= (b*h*h)/6
Zp=A/2(y1’+y2’)
Zp=b.h/2((h/4) + (h/4))
S=Zp/Z=1.5
2)Assumptions made for fully plastic moment of a section?
The material obeys Hooke’s law until the stress reaches the upper yield value.
THE upper and lower yield stresses and the modules of elasticity have the same value in compression as well in tension.
The material is homogenous and isotropic in both the elastic and plastic states.
There is no resultant axial force on the beam.
3)Define the term mechanism in plastic analysis of structure.
When any elastic body is subjected to a system of external loads and deformation takes place and the resistance is set up against the deformation, then the elastic body is known as structure. In contradiction to this, if no resistance is set up in the body against the deformation, then it is known s a mechanical mechanism.
4)Write down the formula for finding the moment capacity of plastic section ?
MP = σYZP where ZP = A/2(Y1 + Y2)
Where ,
σY = yield stress .
ZP = plastic modulus of a section .
Y1 = distance of CG of area A1 from EAA .
Y2 = distance of CG of area A2 from EAA.
5)What is the Formula for check for moment capacity ?
b/3 of flange, d /t of web, b/t of flange of channel.
- In IS800, slender section classification belongs to class 4.
6)Write down the formula for finding moment capacity of plastic or compact section
Mfr = 0.25 bf tf2 fyf [1-{ Nf /(bf fyf/ ym)}2]
where bf, t~ = width and thickness of the relevant flange respectively fyf = yield stress of the flange
7)Advantages of plastic analysis of steel structures:
This method is rapid and provides a rational approach for theanalysis of the structure.
It also provides striking economy as regards the weight of steel since the sections required by this method are smaller in size than those required by the method of elastic analysis.
8)Draw idealised stress-strain curve for mild steel: Refer book for the figure Where,
A-limit of proportionality B-elastic limit C-upper yield point D-lower yield point E-plastic region F-ultimate load point G-breaking or fracture point9)Different stages of bending of beams reacting full plastic analysis: Refer book for figure.10)Static theorem or lower bound theorem
The static theorem or lower bound theorem states that for a given frame and loading, if there exists any distribution of bending moments, throughout the frame, which is safe and statically admissible with a set of loads, w, the value must be less than or equal to the collapse loads,(W<Wc).
11)Kinematic theorem or upper bound theorem
The kinematic theorem or upper bound theorem states that for a given frame subjected to a set of loads, w, the value of w, which is found to correspond to any assumed mechanism, must be either greater than or equal to the collapse load (W>Wc)
12)Uniqueness theorem or combined theorem
It states that if for a given frame and loading at least one safe and statically admissible bending moment distribution can be found and in this distribution, the bending moment is equal to the fully plastic moment at sufficient cross-sections to cause failure of the frame as a mechanism due to rotations of the plastic hinges at these sections, the corresponding load would be the collapse load(W=Wc)
Unit 2
1)What is the codal provision for maximum flat with ratio for different elements?
As per code IS 800 1975
Maximum allowable flat width ratio
For stiffened elements sttifend by
Simple lip 60
Any other stiffner 90
For unstiffened element 30
2)What is the maximum allowable web depth as per code IS 801?
For members with stiifened webs – 150
For members which are provided with adequate means of transmitting conc loads or reaction or both into the web - 200
3)Define the term form factor?
The form factor Q is also known as column factor or shape factor is introduced for columns and compression members of light guage steel sections to account for effect of local buckling.
4)How will you determine the form factor for member consisting of both stiffened and unstiffened element?
The member consisting of both stiffened & unstiffened elements will attain its failure load when the weaker of the unstiffened element buckles at the critical stress.At this stress effective area of stiffened elements computed for critical stress.
Q = Ae / A * Fc / F
Where,
Ae - Effective area
A – Gross area
Fc – Allwable compressive stress
F – Bending stress
5)What is the maximum slenderness ratio for compressive element in cold formed section?
The slenderness ratio K L/r of compression member shall not exceed 200,except that during construction only,K L/r shall not exceed 300.
6)MEMBERS COMPOSED OF STIFFENED ELEMENTS
Q = effective design area
Total area
7)MEMBERS COMPOSED OF UNSTIFFENED ELEMENTS
Q= allowable stress
Basic design stress
8)MEMBERS COMPOSED OF BOTH STIFFENED AND UNSTIFFENED ELEMENTS:
Q= eff area of stiff elements
Total area of unstiff elements
9)DRAW A NEAT SKETCH OF THE BASE PLATE CONNECTIONS IN CHIMNEY.
Refer Fig 13.3 of STEEL CHIMNEYS
10) GIVE THE FORMULA TO FIND THE THICKNESS OF BASE PLATE IN CHIMNEYS.
Thickness of base plate,
Where,
11)GIVE THE FORMULA TO DETERMINE THE NUMBER OF BOLTS REQUIRED TO RESIST THE UPLIFT OF WINDWARD SIDE OF CHIMNEY.
The ANCHOR BOLTS are provided on the base plate from both the inside and outside of the stack
Spacing of bolts ,S=
Where ,
12)What are the different methods of analysis of roof truss?
Method of sections .
Method of joints.
Analysis of by pin jointed truss.
13)Write down the design procedure for roof truss analysis?
Selection of configuration.
Load calculation.
Member forces are to be found using method of sections or method of joints.
Design of members (one compression & one tension).
Moment capacity, Shear force,Deflection checks are to be done.
14)How will you find design wind speed, wind pressure as per is 875.?
1)To find the design wind speed pressure in any place in India IS 875 divides the country into six zones.(pg no.9).for important cities wind pressure is already given in page no.53.
2)To find the wind speed we should go to page number 8 in IS 875 .The formula given in the page is Vz=k1k2k3Vb.
15)What is the classification of steel stacks ?
On basis of construction, the steel stacks can be classified into,
a) Self supporting stacksb) Guyed stacks
16)Draw a neat sketch of self supporting chimney. Refer text book material
If the volume of gas is in m3/sec and velocity of gas in m2/sec, then the minimum internal diameter of the chimney is, Sqrt [(4Q)/ (pi)(V)].
17)How will you determine the form factor for members consisting of both stiffened and unstiffened elements ?
The form factor Q ,is the product of stress factor Qs and area factor Qa. The effective area is computed by considering the area of both stiffened and unstiffened elements.
Q = Qs.Qa = (fc/fb).(Aeff/Agross)
18)Mention the different types of foundation in chimney.
Cylindrical block foundation of plain concrete Truncated cone foundation of plain concrete Raft foundation of R.C.C
o Full raft (circular slab)
o Annular raft
Piled foundation
19)What is the shape factor of cylindrical chimney?
k = Shape factor. It accounts for the shape of the structure; the shape factor for cylindrical portion is 0.7.
20)How will you find stress due to wind loads in a chimney design?
The stress due to wind is given by
Mw = P * (x/2)
Stress = (Mw* D) / ( l* 2) = Mw / Z
= (4 Mw)/ ( π D2 t)
Where,
Mw = moment due to wind.
P = wind force.
x = height of stack above the section under consideration.
Z = section modulus (π D2 t/4)
21)Explain the design procedure for calculating the moment and shear due to seismic loads for chimney.
The moments and shears at any section are computed from the following expressions.
Moment, Mx = αh W h { 0.6 ( x/H)1/2 + 0.4 (x/H)4 }
Shear, Vx = Cv αh W {[(5/3)*( x/H)1/2]– [(2/3)*(x/H)4] }
Where, h= height above the the base.
W = weight of the section.
Cv = coefficient form table 19.1.
22)What is the difference between stiffened and unstiffened compression elements?
STIFFENED ELEMENTS UNSTIFFENED ELEMENTSWhen any flat compression elementis stiffened on both edges parallel to the
direction of stress by connected stiffeners or by formed lips, or flanges offering specified
resistance to lateral deflection, then the element is considered as stiffened.
When an element is having one or both edges parallel to the direction of stress free from
lateral restraint then the element is considered as unstiffened.
In stiffened elements sudden buckling of the elements does not occur.
In an unstiffened element, the sudden local buckling occurs at a definite critical load.
23)What is the difference between multiple stiffened elements and sub element?
Multiple stiffened elements Sub elementWhen an element is stiffened web, or
between a web and an edge, by providing intermediate stiffeners parallel to the direction of stress, then the element is
considered multiple stiffened elements.
A portion of the element between adjacent stiffeners or between edge and stiffeners or between web and stiffener is known as sub
element.
24)Write down the formula for lateral torsional buckling as per IS 800
Mcr=pi^2 EIy hf/2LLT^2
[1+1/20(LLT/ry/hf/tf)^2]^0.5
25)What is the maximum slenderness ratio for compression elements in cold-formed section ?
200 and during construction it should not exceed 300.
26)formula check for bulking resistance? Md=betab zp fbd for plastic section betab=1
27)formula check for deflection? At working load, deflection is to be limited to L/750. 1/2total M/EI diagram.
29)what are the additional impact load for E.O.T & H.O.T? E.O.T - 25% of max. static wheel loads.
H.O.T - 10% of max. wheel loads. @1
30)What is load factor?
The load factor Q, used in plastic design is defined as the ratio of the collapse load Wc to the working load Ww. Load factor depends upon the nature of loading, support conditions and the geometrical shape of the structural member.
31)Define partial, complete, over complete collapse.
Partial collapse – In the partial collapse for a structure, a part of the structure may fail, which makes the structure as a whole useless. The number of plastic hinges is less than ( r + 1) where r is the redundancy of the structure.
Complete collapse – When the number of plastic hinge formed is ( r + 1 ) then the complete collapse of the structure takes place. The complete collapse of the structure enables to determine all the redundancies of the structure together with the collapse load.
Over complete collapse - When the number of plastic hinges formed at a collapse in a structure is more than ( r + 1 ) then the over complete collapse takes place. The value of collapse corresponds to two or more number of mechanisms.
32)What is the purpose of gusset plate ? what is the minimum thickness of gusset plate?
(A) Gusset plates are used to connect beams and columns together or to connect truss members.The minimum thickness of gusset plate depends on the welded or other connections made. The minimum thickness of gusset plate are usually ¼ inch for inside protected structures and 3/8 inch for outside exposed building
33)Mention the advantages of cold formed structural members compared to hot rolled
sections .
• Cold rolling can be employed to produce almost any desired shape to any desired length.
• Pre-galvanised or pre-coated metals can be formed, so that high resistance to corrosion, besides
an attractive surface finish, can be achieved.
• All conventional jointing methods, (i.e. riveting, bolting, welding and adhesives) can be
employed.
• High strength to weight ratio is achieved in cold-rolled products.
• They are usually light making it easy to transport and erect.
34)What are the two main methods of manufacturing light gauge sections ?
The methods of manufacturing light gauge sections are
• rolling
• pressing
35)Draw the common shapes of cold formed light gauge structural members.
36)Different types of buckling:
i) Flexural buckling
ii) Torsional buckling
iii) Lateral buckling
37)Flexural Buckling:
It is a buckling mode in which a compression member deflects
laterally without twist or change in cross-sectional shape.
38)Slenderness ratio:
It is the ratio of the effective length to the appropriate
radius of gyration
? =Kl/r
39)What do u mean by torsional flexural buckling ?
In cases where bending and twisting occurs simultaneously the
elastic torsional flexural buckling stress is given by the
equation.
For doubly symmetric and point symmetric shapes as shown in
the fig,the overall column buckling may be either flexural type or
torsional type.
40)What is the difference between stiffened and unstiffened compression element ?
Stiffened Element :
In the direction of stress, the plate element of any
section is sufficiently supported by the two adjacent elements in the
longitudinal directions is called as stiffened element.
Un-stiffened Element :
In this case, in the direction of stress, unstiffened
element is supported only at one edge in the longitudinal direction.
41)What is the difference between multiple stiffened elements and sub elements ?
When an element is stiffened between webs and a
stiffened edge,by means of intermediate stiffners which are parallel
to the direction of stress,then it is known as multiple stiffened
element.
A sub-element is the portion between adjascent
stiffners and intermediate stiffener.
UNIT 4
1)What are the difference to be consider in design of gantry girder?
vertical loads from the cranes.
impact loads from Crane
longitudinal horizontal force along the Crane rail.
lateral thrust along the Crane rail.
2)checking fatigue effect for gantry girder ? The fatigue strength of the standard details for the normal orshear fatigue range not corrected for discussed in 13.2.1.Is _800:2007.clauses 13.2.2.1and 13.3.
3) DRAW A NEAT SKETCH OF ELASTROMERIC BEARING?
(or)
4) EXPLAIN DEIGN STEPS FOR ELASTROMERIC BEARING IN BRIDGE?
The steps/ guide lines to designing Elastomeric Bearings as per ISRC:83(part II)-1987
Standard plan dimensions compiled in IRC:83(part II)-1987 are to be preferred but these values should follow the IRC code conditions
the number of elastromers to be find by using the formula
o αd=βnαbi max
o where αd=angle of rotation which may be taken as (400MmaxL)/(EI)10-3,n =no of
elastromer layers,β=(αm)/ (αm.max)
o αm.max=10N/mm2, αbi.max=(0.5 αm hi)/(b.s2)
o αm =avg compressive stress
under critical loading conditions following limits shall be satisfied to ensure adequate friction
o stress strain <0.2+0.1 αm
o 10N/mm2>αm> 2N/mm2
The total shear stress due to normal and horizontal loads and rotation should be less than 5 N/mm2
Standard plan dimensions and design data specified in IRC:83 is compiled
5)DRAW THE DEFFERENT TYPE OF TRUSSED BRIDGES?
6)Write the design procedure for Lug Angle? In the case of angle members , the lug angle and their connections to gusset or
other supporting members shall be capable of developing strength not less than 20 % in excess of force in outstanding leg and the attachment of lug angle to angle member shall be capable of developing 40% in excess of that force.
The effective connections of lug angle shall be as far as possible terminate at the end of member connected and the fastening of lug angle to the member preferably start in advance of direct connection of thw member to gusset or other supporting member.
where lug angles are used to connect an angle member the whole area of the member shall be taken as effective rather than net effective section for angles as per Art 5.3.
In no case , less than two rivets or bolts shall be used for attaching the lug to the gusset or t\other supporting member .
In the case of channel members and the like , the lug angles and their connections to the gusset shall be capable of developing a strength of not less than 10 % in excess of the force not accounted for by direct connection .
7)What is the purpose of lug angle?
When the load is heavy and the number of rivets /length of weld required for making the connection is large, the size of gusset plate required may be uneconomical. So, an additional angle called lug angle is provided for the length required to accommodate the connection
What is the maximum slenderness ratio for a tension member in which the reversal of direct stress due to other than wind or seismic force occurs?
Slenderness ratio is the ratio of unsupported length (l) to its least radius of gyration (r).
Slenderness ratio due to reversal of direct stress other than wind or seismic force occurs is 180
3.Design procedure of tension member subjected to axial load.
The following procedure may be followed in the design of an axially loaded tension member.
1. The net area required (Anet) to carry the design load P is obtained in the equation,P = (sigma)at x Anet
2. The net area calculated thus, is increased suitably (25% - 40%) to compute the gross sectional area. From I.S.handbook No.1 suitable section providing a cross-sectional area matching with the computed gross sectional area is selected.
3. The number of rivets required to make the connection is calculated. These are arranged in a suitable pattern and the net area of section provided is calculated. This should be more than the net area in step 1.
4. The slenderness ratio of the member is checked as per I.S.specifications.
8)Design procedure For Gantry Girder
MAXIMUM WHEEL LOAD MAXIMUM BENDING MOMENT MAXIMUM SHEAR FORCE SECTION MODULUS (Z)
MOMENT OF INERTIA MAXIMUM BENDING STRESS Checked for BENDING STRESS MAXIMUM BENDING MOMENT & SHEAR MAXIMUM COMPRESSIVE STRESS DUE TO LATERAL LOAD LONGITUDINAL FORCES
8)How will you find maximum moment by moving wheel on gantry girder?
Position of crane hook for maximum vertical load on gantry girder Position of crane wheels for maximum effects on gantry girder
9)What is the purpose of lug angle?
When the load is heavy and the number of rivets /length of weld required for making the connection is large, the size of gusset plate required may be uneconomical. So, an additional angle called lug angle is provided for the length required to accommodate the connection.
10)What is the purpose of purlins? Mention the different sections use as purlins.
Purlin is a member which rest between roof trusses and support roof sheeting. I-section channels, angles, cold formed C or Z-sections are commonly used as purlins. The purlins are spaced on main rafter of trusses such that the sheets overlap on them.
11)What is purpose of bearings in bridges?
Bearings are provided at the junction of the girders or slabs and the top of pier and abutments.
They transmit the load from the super structure to the sub structure in such a way that the bearing stresses developed are within safe permissible limits.
They also provide for small movements of the super structure. They are arranged to allow the deck to expand and contract, but retain the deck in its
correct position on the substructure.
12)Design the ANGLE SECTION to carry compressive load of 165 kN as per is 800.
Solution:
Assuming : fcd = 90 N/mm2
A= 165 x 103/90 = 2000mm2
Try ISA 9090, 12mm, which has A=2019mm2.
rmin = rcv = 17.4mm.
Assuming the struct will be connected to the gusset plate with at least 2 bolts ( Note: Strength of 20mm bolt in single shear is about 45 kN )
KL = 0.85L = 0.85 x 3000 = 2550mm
KL/r = 2550/17.4 = 146.55
From IS 800 : 2007, table 10 (clause 7.1.2.2)
For fy = 250 N/mm2
When KL/r = 140, fcd = 60.2
When KL/r = 150, fcd = 59.2
When KL/r = 164.55
fcd = 58.4 – 6.55/4 (58.4 – 52.6 ) = 54.6 N/mm2
Pd = Afcd = 2019 x 54.6 = 110239 < 180000 N
Hence revise the section
Try ISA 130130, 8mm.
Area provided = 2022 mm2 , r = rvv = 25.5
KL/r = 2550/25.5 = 100
fcd = 107 N/mm2
Pcd = 2022 x 107 = 216354 >180000 N
Provide ISA 130130, 8mm.
13)Channel, Angle, Tee and Solid section of buckling of any axis belong to which type of buckling curve?
CHANNEL, ANGLE, T AND SOLID SECTION - α = 0.0055(Curve C)
14)Design procedure for compression members composed of two components back to back as per IS 800 – 2007 ?
Compression members composed of 2 angles, channels, or tees back to back in contact or separated by a small distance, shall be connected together by riveting, bolting or welding so that the ratio of most unfavourable slenderness of each member between the intermediate connections is not greater than 40 or 0.6 times the most unfavourable ratio of slenderness of the strut as a whole, whichever is less.
In no case shall the ends of strut be connected with less than 2 rivets & not less than 2 additional connections in between, where the members are back to back the rivets through these connections shall pass through solid washers or packing in between. Where the legs of the connected angles or tees are 125 mm wide or more, or where
webs of channels are 150 mm wide or more, not less than 2 rivets or bolts shall be used in connection.
When spacing is more solid packing shall be used to effect the jointing else direct welding can be permitted for joining the members.
The rivets, bolts or welds in these connections shall be sufficient to carry the shear force and moments, and in no case shall the rivets or bolts be less than 16 mm diameter for members up to and including 10 mm thick, 20 mm diameter up to 16 mm thick and 22 mm diameter over 16 mm thick.
Compression members connected by such shall not be subjected to transverse loading in a plane perpendicular to the riveted, bolted or welded surfaces and where these components are back to back, the spacing of the rivet, bolts shall not exceed the maximum spacing for compression members.
15)How will you determine the effective length of member for different boundry condition?
The effective length of the member is determine by using their end condition.
Held in position @ both ends and not restrained against rotation-1.00L
Held in position @ both ends and restrained against rotation @ 1end-.80L
Held in position @ both ends and restrained against rotation @ both end-.65L
Held in position and restrained against rotation @ 1end and @ the other end restrained against rotation but not held in position-1.20L
Held in position and restrained against rotation @ 1end and @ the other end partially restrained against rotation but not held in position-1.50L
Held in position @ 1 end but not restrained against rotation and @ the other end restrained against rotation but not held in position-2.00L
16)What do you mean y slenderness ratio?
It is the ratio of the effective length to the appropriate radius of gyration.
λ =Kl/r
17)Define condition equilibrium .
The algebraic sum of all vertical forces acting over a structure should be zero . the algebraic sum of all horizontal forces acting over a structure should be zero. The moment of all the forces acting over a structure about any point is zero.
18)Define mechanism equilibrium .
The ultimate or collapse load is reached when a mechanism is formed .The number of plastic hinges formed should be just sufficient to form a mechanism
19)Define yield mechanism
The bending moment at any section of the structure should not be more than the fully plastic moment of the section ..
20)Design procedure of channel/I-section purlin?
The span of purlin is c/c distance of adjacent trusses
the gravity loads P,due to sheeting and live load etc, and the load H due to wind are computed
the max bending moments Muu and Mvv along the axis are calculated
Muu=PL/10
Mvv=HL/10
Max Bending stress in a purlin is given by Muu/Zuu + Mvv/Ivv
Assume a suitable value of Zuu/Zvv for channel and I section
Compute the value of max bending stress and it should be less than permissible value(1.33*.66*fy)
21)Design procedure of angle section purlin?
Ascertain the vertical loads and wind loads.These are assumed to be normal to roof trusses
compute the max bending moment by Wl2/2 and Wl/10
compute the modulus of section Z=M/1.33*permissible bending stress
the trial sections and angle purlins are arrived by assuming the depth as 1/45 of the span and width as 1/60 of the span.the depth and width should not be less than the permissible value.
The suitable angle section is selected from IS code for calculated leg lengths of angle section.The modulus of section should be more than the modulus of section calculated by step 3
22)Different types of end bearing in Bridges?
Depending upon the magnitude of end reaction, and the span of bridge, the different types of bearings used for the bridges are as follows:
1-Plate bearings (Sliding& hinged bearings).
2-Rocker bearing
3-Roller bearings
4-Bearing adopted by Railway Board.
23)For ISHB 350 @ 710.2 N/m , (K l /r) = 43.58 & fy = 250 N/mm2. Find design compressive strength ?
From IS 800 – 2007 pg no : 138, we get
H = 350 mm
tf = 11.6 mm then, H/ tf = 30.17
From IS 800 – 2007 pg no : 57, clause 8.2.2.1, we get
Critical stress fcr,b = 1269.6 N/mm2
From IS 800 – 2007 pg no : 55, clause 8.2.2, we get
Design bending compressive strength fbd = 213.85 N/mm2
23)What do you mean by shear lag effect?
The in plane shear deformation effect by which concentrated forces tangential to the surface of a plate gets distributed over the entire section perpendicular to the load over a finite length of the plate along the direction of the load.24) Draw different compound and built-up section for the design of tension members.
25)What are the different types of failure in tension members? Gross section yielding Net section rupture Block shear failure
26)What do you mean by tension members? Mention different types of it. Tension members are structural elements that are subjected to axial tensile forces. A tension member is also called as a tie member or simply a tie.
The various types of tension members used may be grouped into following four groups:
Wires and Cables Rods and Bars Single structural shapes and plates Built-up members
It includes Angle section, Channel section, I section, Tee section, Box section, Tubular section.
27)Vertical Deflection
Manually operated cranes span/500 Electrically operated up to 500 KN span/750
Electrically operated over 500 KN span/1000 28)What do you mean by torsional flexural buckling?
This type of buckling only occurs in compression members that have unsymmetrical cross-section with one axis of symmetry. Flexural-torsional buckling is the simultaneous bending and twisting of a member. This mostly occurs in channels, structural tees, double-angle shapes, and equal-leg single angles.
29) Explain torsional buckling?
. When a slender member is subjected to an axial force, failure takes place due to bending or torsion rather than direct compression of the material. This is called torsional buckling. This type of buckling only occurs in compression members that are doubly-symmetric and have very slender cross-sectional elements. Torsional buckling occurs mostly in built-up section.
30) Where flexural torsional buckling occurs?
This type of buckling only occurs in compression members that have unsymmetrical cross-section with one axis of symmetry. Flexural-torsional buckling is the simultaneous bending and twisting of a member. This mostly occurs in channels, structural tees, double-angle shapes, and equal-leg single angles.
31) What is the basis of column curves in IS 800 2007?
On the basis of stress reduction factor and design compressive stress. For different buckling class the column curves are drawn.
32) formula for euler’s buckling stress
σcr =[π2 E/ (l/r)2]
Other Important Qurestions
33) design procedure or axially loaded compression members:
The slenderness ratio for the compression member and the value of yield stress for the steel are assumed.
For rolled steel beam section compression members, the slenderness ratio varies from 70 to 90. For struts, the slenderness ratio varies from 110 to 130. For compression members carrying large loads, the ratio is 40.
The effective sectional area (A) required or compression is determined
A= (P/σac)
Where P = load to be carried by the member.
From the steel section tables, section for the compression member of the required area is selected. It is selected such that it has the largest possible radius of gyration for the required sectional area.It should also be the most economical section.
Knowing the geometrical properties of the section, slenderness ratio is computed and the allowable axial stress in compression is found from IS-800-1984 or the quantity of steel assumed.
The safe loading capacity of the compression member is determined.
34)What are the advantages of cold formed structural members? Compare with hot rolled sections.
Instead of hot rolled sections, cold formed members are used essentially in 2 situations
Where moderate loads and spans render the thicker hot rolled shapes uneconomical.
Where the load carrying member is designed to provide useful surface such as floor, wall panels and roof coverings.
35)reasons for using cold formed sections:
Economy of material Ease of mass production Versatility Provision for effective 2 simple connection
36)What are the main processes for manufacturing light gauge sections?
There are 2 main process of manufacturing light gauge sections:
COLD ROLLING: Economical for large scale production. PRESSING :Small quantities of special shape.
37)Draw common shapes of cold formed light gauge sections.
40)Draw the bending stress diagrams for the following elements
A.
41). Write down the formula for checking combined bending stress in web as per IS 801
A. Refer IS 801 page 16 clause 6.4.3
42) What do you mean by web crippling of beams?
The phenomenon of buckling of a web of a beam under excessive compressive stress especially under point loads.
43)Write down the formula for finding design strength in shear ???
[(σ’bw / σbw )2 + (τ’v / τv )2 ]1/2 > 1.00
Where
σ’bw = actual compression stress at junction of flange and web
σbw = allowable compression stress
44)What do you mean by web buckling ?
The light gauge steel members are cold frmed from the steel sheets or strips . In the plate element the ratio of width to thickness used to be large. The failure of plate elements occur by buckling . the buckling occurs at lower stresses resulting due to compression or bending or shear or bearing.
45)direct stress due to other than wind or seismic force occurs?
Slenderness ratio is the ratio of unsupported length (l) to its least radius of gyration (r).
Slenderness ratio due to reversal of direct stress other than wind or seismic force occurs is 180
46)Design procedure of tension member subjected to axial load.
The following procedure may be followed in the design of an axially loaded tension member.
The net area required (Anet) to carry the design load P is obtained in the equation,P = (sigma)at x Anet
The net area calculated thus, is increased suitably (25% - 40%) to compute the gross sectional area. From I.S.handbook No.1 suitable section providing a cross-sectional area matching with the computed gross sectional area is selected.
The number of rivets required to make the connection is calculated. These are arranged in a suitable pattern and the net area of section provided is calculated. This should be more than the net area in step 1.
The slenderness ratio of the member is checked as per I.S.specifications.
47)Different classification of cross section as per IS 800
Rolled steel I section Rolled steel channel section Rolled steel T section Rolled steel angle section Rolled steel tubes Rolled steel bars
48)Mention different methods to provide lateral restrain of beams
beam in which compression flange restrained laterally is called as laterally restrained beams. Different methods to provide effective lateral restrain on beam are
By embedding the compression flange on the beam into the concrete slab Shear connectors are welded to the compression flange and are embedded into the
concrete slab
49)Define shear centre and for channel section where it will act.
The shear centre is the point in the cross-section of a member through which forces must be applied to avoid any twisting. For channel section will be acting along the central line of the section
50)What are the different sections classification as per IS 800?
The different classifications of sections as per IS 800 are:
Rolled I-section Welded I-section Hollow section Welded box section Channel, Angle, T and Solid section Built-up section.
51)Define the terms- low shear and high shear.
When a material is stationary for example when it is cooling following extrusion or being stored on a shelf it is subjected to a very low shear stress essentially the only forces acting on it are gravitational. Steel has low shear strength.
A material's ability to resist forces that can cause the internal structure of the material to slide against itself is known as high shear.. Carbon nano-tubes, aluminum honeycomb and adhesives have high shear strength.
52)What do you mean by compression member?Mention classification
A compression member is a structural member which is straight and subjected to two equal and opposite compressive forces applied at its ends.It can be classified based on the position in structures i)strut-roof truss ii)column-supporting floors iii)principal rafter-roof truss.
53)What is beam. What are the main considerations made during the design of beams.
A beam is defined as Horizontal or inclined structural member running between one or more supports, carrying different loads transverse to its longitudinal axis, as a girder, joist, purlin, or rafter. The main function of these members is to transfer these loads to columns and to support the roof or a floor. The main considerations to be included are
Deflection caused due to Dead load Live load Other loads such as Wind load, Seismic or earthquake.
54)What is the maximum allowable deflection in beam in light gauge section?
The maximum deflection should not be more than 1/325 of the span. This limit may be extended in a case, where the greater deflection would not impair the strength or efficiency of the structure or lead to damage the finishings.
55)Write down design steps for determining the allowable load/m on the beam in light gauge section.
determination of effective width
determination of form factor
calculation of allowable stress
detrmination of allowable load/m
b
56)Find Ixx for the given section
d
c t
b
Ixx 1= bd3/12 =tc3/12 Ixx 2= (b-2t)t3/12 Ixx 3= t(d-t)3/12 Ixx 4= bt3/12 Ixx= tc3/12+ (b-2t)t3/12+ t(d-t)3/12+bt3/12
57)What is the formula for determining load and deflection in light gauge section?
The formula for determining the load and deflection in light gauge section are
Load determination: b/t = 658/ (f^0.5) [1-{145/ (w/t) f^0.5}]
Deflection determination: b/t = 2710/ (f^0.5) [1-{600/ (w/t) f^0.5}]
58)What are the provisions in IS Code regarding adequacy of edge stiffeners for the compression elements in the light gauge sections?
The IS: 801-1975 has provided certain requirements for edge stiffeners:
In order that a flat compression element may be considered stiffened compression member, it shall be stiffened along each longitudinal edge parallel to the direction of stress by a web, lip or other stiffening means, having the following minimum moment of inertia:
Imin = 1.83t4 [(w/t)2 – 281200/fy] (but not less than 9.2 t4)
dmin = 2.8t [(w/t)2 – 281200/fy]1/6 (but not less than 1.4 t)
59)What are the different grades of steel and their basic allowable design stress used in light gauge section?
The values of basic allowable design stress f for some of the grades covered in IS: 1079-1973 are gives as below:
MKS Units( as per IS: 801-1975)
SI Units( Converted Values)
fy (kg/mm2) fa (kg/cm2) fy (N/mm2) fa (N/mm2)21 1250 206 12324 1450 235 14130 1800 294 17636 2160 353 212
60)Write down the formula for load and deflection determination in light gauge sections?
For load determination: Flanges are fully effective (b=w) upto (w/t)lim = 1435/sqr.root(f)
For flanges with (w/t) larger than (w/t)lim b/t = 2120/sqr.root(f) [1-465/(w/t) * sqr.root(f)]
Exception:
Flanges of closed square and rectangular tubes are fully effective (b=w) upto (w/t) = 1540/sqr.root(f)
For flanges with (w/t) larger than (w/t)lim
b/t = 2120/sqr.root(f) [1-420/(w/t)*sqr.root(f ]
For deflection determination:
Flanges are fully effective upto (w/t)lim = 1850/sqr.root(f)
For flanges with (w/t) larger than (w/t)lim
b/t = 2710/sqr.root(f) [1-600/(w/t)*sqr.root(f ]
Exception:
Flanges of closed and square and rectangular tubes are fully effective upto (w/t)lim = 1990/sqr.root(f)
For flanges with (w/t) larger than (w/t)lim
b/t = 2710/sqr.root(f) [1-545/(w/t)*sqr.root(f)]61)What provisions are made in IS-code regarding adequacy of edge stiffeners for compression element in light gauge sections?
In order that a flat compression element may be considered a stiffened compression element,it shall be stiffened along each longitudinal edge parallel to the direction of stress by a web,lip,or other stiffening means,having the following minimum moment of inertia:
Imin = 1.83 t^4 [sqr.root(w/t)2-281200/Fy] but not less than 9.2 t^4 Where,
Imin = minimum allowable moment of inertia of stiffener about its own centroidal axis parallel to the stiffened element in cm^4 &,
(w/t) = flat-width ratio of stiffened element
Where the stiffener consists of a simple lip bent at right angles to the stiffened element,the required overall depth dmin of such lip may be determined as follows:
dmin = 2.8 t ^6 [sqr.root(w/t)^2-281200/Fy] but not less than 4.8 t
A simple lip shall not be used as an edge stiffener for any element having a flat-width ratio greater than 60.
62)What are the different grades of steel and their allowable design stress used in light gauge sections?
Basic allowable design stress: Stress on net section of tension members and compression on extreme fibres of flexural members shall not exceed the value F
F = 0.6Fy where Fy is the value of basic allowable design stress
When the increase in steel strength resulting fromcold work forming is utilized,the basic design stress shall be determined as follows,
F = 0.6FyB where FyB is the average yield point of full section
Grade yield stress(min)
St 21 1250kgf/cm^2 St 24 1450kgf/cm^2 St 30 1800kgf/cm^2 St 34 2100kgf/cm^2 St 36 2160kgf/cm^2 St 42 2400kgf/cm^2 St 50 3000kgf/cm^2 St 52 3600kgf/cm^2
63)What do you mean by shear lag effect?
The in plane shear deformation effect by which concentrated forces tangential to the surface of a plate gets distributed over the entire section perpendicular to the load over a finite length of the plate along the direction of the load..64)Define the terms- low shear and high shear.
When a material is stationary for example when it is cooling following extrusion or being stored on a shelf it is subjected to a very low shear stress essentially the only forces acting on it are gravitational. Steel has low shear strength.
A material's ability to resist forces that can cause the internal structure of the material to slide against itself is known as high shear.. Carbon nano-tubes, aluminum honeycomb and adhesives have high shear strength.
In IS800, slender section classification belongs to class 4.
65)Write down the formula for finding moment capacity of plastic or compact section
Mfr = 0.25 bf tf2 fyf [1-{ Nf /(bf fyf/ ym)}2]
where bf, t~ = width and thickness of the relevant flange respectively fyf = yield stress of the flange
66)What is the purpose of lining? What are the different types of materials?a.) To protect the chimney shell from heat.b.) To act as a protective covering thus reducing corrosion, andc.) To maintain the temperature of the flu gases.
Materials:Firebricks, Insulation refractory bricks, Solid grade distomaceous bricks, Acid
reisting bricks, Solid grade distomaceous concrete, refractoryconcrete, Sand and concrete mixtures and other materials.
67)What is breech opening?Breech opening also known as flue opening is provided for the entrance of flue gases.
68)What are the forces acting on steel stack?Self weight, Weight of lining, Wind pressure and Seismic forces
