Design of Beams (Unit  8)


 Eugene Marsh
 1 years ago
 Views:
Transcription
1 Design of Beams (Unit  8) Contents Introduction Beam types Lateral stability of beams Factors affecting lateral stability Behaviour of simple and built  up beams in bending (Without vertical stiffeners) Design strength of laterally supported beams Design strength of laterally unsupported beams Shear strength of beams Maximum deflection Design of Purlins
2 Introduction Beams are structural elements subjected to transverse loads in the plane of bending causing BMs and SFs. Symmetrical sections about zz axis are economical and geometrical properties of such sections are available in SP (6) The compression flange of the beams can be laterally supported (restrained) or laterally unsupported (unrestrained) depending upon whether restraints are provided are not. The beams are designed for maximum BM and checked for maximum SF, local effects such as vertical buckling and crippling of webs and deflection. Beams can be fabricated to form different types of c/s for the specific requirements of spans and loadings. Section 8 shall be followed in the design of such bending members. Types of beam cross sections Beams can be of different cross sections depending on the span and loadings and are shown below 
3 C / S of Plate Girders
4 Simple I sections are used for normal spans and loadings with all the geometrical properties available in IS 800 : All the other sections indicated in the figure are built up sections. These sections are used when the normal I sections become inadequate due to large spans and loadings. These sections are also used due to other functional requirements. I section with cover plates are used when the loads are heavy and the spans are large. If the depth of the beam is restricted due to functional reasons, smaller depth I sections with cover plates can be used Additional cover plates increases the lateral load resistance with increase in I YY. The properties of ISMB and ISWB sections with cover plates are available in SP (6). (Z PZ has to be obtained from calculations)
5 Two I sections with cover plates can be used when very heavy loads and spans act on the beam. The properties of these sections are not available in SP (6) and have to be calculated. Two I sections placed one above the other are used when the loads are light with large spans, where deflection is the main criteria. The properties of these sections are not available in SP (6) and have to be calculated Gantry girders are used in industrial buildings to lift loads and typical sections used are indicated in the figure. The properties of these sections are available in SP (6), (Z PZ has to be obtained from calculations) Plate girders are used where the spans exceed 20m and the loads are heavy. The properties of these sections are available in SP (6), (Z PZ has to be obtained from calculations) Box sections have large torsional rigidity and can be used as single cell, twin cell or multi  cell sections. The openings are advantageously used for service lines. Castellated beams are special sections fabricated from I sections and are used for light loads and large spans. The openings are advantageously used for service lines. In all built up beams, the fabrication cost is higher due to the provision of connections between the elements. Section Classification There are four classes of section namely Plastic, Compact, Semi  Compact and Slender sections as given in IS 800 : [cl pp  17] For design of beams, only Plastic and Compact sections are used. Lateral Stability of Beams A beam transversely loaded in its own plane can attain its full capacity (Plastic moment) only if local and lateral instabilities are prevented. Local buckling of beams can be due to web crippling and web buckling. They are avoided by proper dimensioning of the bearing plate and through secondary design checks. Flanges shall always satisfy the outstand to thickness ratio as per IS 800 : 2007 so that local failures of flanges are avoided. Plastic and Compact sections are used. Lateral buckling of beams is the out of plane bending and is due to compressive force in the flange and is controlled by providing sufficient lateral restraint to the compressive flange.
6 Lateral stability of beams is affected by span of the beam, moment of inertia and the support conditions. Local failures of flanges (Secondary design checks) The local failure of flanges (plates) reduces the plastic moment capacity of the section due to buckling and is avoided by limiting the outstand to thickness ratios as given in IS 800: Local failures of web (Secondary design checks) The web of a beam is thin and can fail locally at supports or where concentrated loads are acting. There are two types of web failure  Web Crippling (or Crimpling) 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.
7 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 = F wc Thickness of web = t w Yield stress in web = f yw Width of bearing plate = b 1 Width of dispersion = n 2 = 2.5 h 2 Depth of fillet = h 2 (from SP [6]) F wc = [(b 1 + n 2 ) t w f yw ] / γ mo Reaction, R U For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as F wc = [(b n 2 ) t w f yw ] / γ mo Concentrated load, W U
8 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.
9 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 = F wb Thickness of web = t w Design compressive stress in web = f cd Width of bearing plate = b 1 Width of dispersion = n 1 F wb = (b 1 + n 1 ) t w f cd Reaction, R U For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as F wc = [(b n 1 ) t w f cd ] Concentrated load, W U The design compressive stress f cd is calculated based on a effective slenderness ratio of 0.7 d / r y, where d = clear depth of web between the flanges. r y = radius of gyration about yy axis and is expressed as = (I yy / area) = [(t w ) 3 / 12] / t = (t w ) 2 / 12] kl / r y = (0.7 d) / (t w ) 2 / 12] = * d / t w Design compressive stress in web, f cd for the above slenderness ratio is obtained from curve, C (Buckling class C) (Table 9c, pp 42) Shear lag effects In simple theory of bending, plane sections remain plane before and after bending require that no SF is present in the beam. But in practice, SF influences the bending stress in the flanges and causes the section to warp. This results in non uniform distribution of flexural stresses in the flanges with stress being greater at the junction of flange and web. This is known as shear lag effect. In built up beams with wide flanges, this can be considerable, while in normal simple I sections, it is negligible. Shear lag depends on the width to span ratio and is specified in cl (pp  53 and 54)
10 Distribution of compressive stress in the flange Laterally Supported (Restrained) beams Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (I YY ) and its presence reduces the plastic moment capacity of the section. Beams where lateral buckling of the compression flange are prevented are called laterally restrained beams. Such continuous lateral supports are provided in two ways  i) The compression flange is connected to an RC slab throughout by shear connectors. ii) External lateral supports are provided at closer intervals to the compression flange so that it is as good continuous lateral support. Cl (pp  52 and 53) gives the specifications in this regard. Typical lateral supports are shown in the figure.
11 Typical lateral supports Design of such laterally supported beams are carried out using Clauses , , , 8.4, 8.4.1, , and (Deflection) In addition, the beams shall be checked for vertical buckling of web and web crippling. The design is simple, but lengthy and does not involve trial and error procedure. Design steps for laterally supported beams The design of laterally supported beams consists of selecting a section based on the plastic section modulus and checking for its shear capacity, deflection, web buckling and web crippling. Most of the equations are available in IS 800 : The steps are  i) The maximum BM and SF at collapse is calculated based on the service loads (characteristic loads) and the span of the beam. Factored load at collapse = 1.5 * characteristic loads.
12 ii) The plastic section modulus, Z PZ is calculated using Z PZ = M U / (f Y / 1.1) M U = Maximum BM f Y = Yield stress of the given grade of steel 1.1 = Partial safety factor in yielding iii) A trial section having the appropriate plastic section modulus is adopted using IS 800 or SP (6) depending upon the type of section required. The section shall be plastic or compact section. iv) The beam shall be checked for shear lag and design bending strength as given in cl and (pp 53) v) The beam is checked for deflection using appropriate formula depending on the type of loadings. vi) The section is checked for shear as given in cl. 8.4, and If V U 0.6 V d it is a case of high shear or otherwise low shear. For high shear, the design bending strength is calculated from cl. 9.2 vii) The section is checked for web buckling and crippling using appropriate formula. Laterally unsupported beam Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. Lateral buckling of beams involves three kinds of deformations namely lateral bending, twisting and warping. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (I YY ) and can reduce the plastic moment capacity of the section.
13 Factors affecting lateral stability Type of C/S  The lateral buckling strength can be improved by choosing an appropriate c/s where I YY is large. Box sections satisfies this and also has large torsional rigidity as it is a closed section. Open sections like I sections have low torsional rigidity and are more susceptible to lateral instability. Cl pp  54 mentions that hollow sections need not be checked for lateral buckling strength. Support conditions  The lateral restraint provided depends on the restraint provided by the supports. The effect of various support conditions is taken into account using the concept of effective length as given in Table 15  pp 58 for simply supported beams and Table 16  pp 61 for cantilever beams. Effective length  This concept incorporates the various types of restraints to the flanges and for simply supported beams Table 15  pp 58 can be used. The same information for cantilever beams is given in Table 16  pp 61. Beams without proper restraint of the compression flange undergoes lateral buckling resulting in lesser load carrying capacity. Behavior of beams The actual behavior of beams depends on whether the beam is allowed to reach its plastic moment capacity. If the beam is prevented from local and lateral buckling, the beam reaches its full plastic capacity with plastic hinges formed at critical points of maximum BM which has been described in plastic analysis. If the beam is not restrained laterally, the beam can undergo elastic lateral torsional buckling and can fail due to instability with large lateral deflections, rotations and warping. If the web is too thin, the beam can fail in shear due to diagonal compression. The beam can also fail due to local effects such as web buckling, web crippling or distortion of flanges if these effects are not considered in the design. Lateral Torsional Buckling of Beams The presence of compression in the flanges causes lateral deflection (side sway) along with rotation known as Lateral Torsional Buckling of Beams. The assumptions made in the analysis are  i) The beam is initially undistorted without residual stresses. ii) The beam behaves elastically upto failure. iii) The beam is subjected to pure bending in the plane of web
14 If lateral restraint is not provided to the compression flange, buckling of a beam takes place about its minor axis, accompanied by twisting moment and warping. The load at which the beam buckles will be much less than the load causing full plastic moment. The design bending compressive stress is dependent on a factor called non  dimensional slenderness ratio, λ LT, which in turn is dependent on the elastic lateral torsional buckling moment, M cr. The value of M cr can be obtained by solving a fourth order differential equation. The value of M cr can b calculated using the equations given in cl pp 54 for doubly symmetric c/s and annex E (pp ) for c/s symmetrical about the minor axis. The design bending compressive strength can be calculated using a set of equations as specified in cl (Table 13a and 13b) pp 54 to 57. Design steps for laterally unsupported beams The design of laterally unsupported beams consists of selecting a section based on the plastic section modulus and checking for its shear capacity, deflection, web buckling and web crippling. Most of the equations are available in IS 800 : The steps are  All the steps given in design of laterally supported beams shall be used here. Also, The plastic section modulus required is increased by 2550%. The design bending strength is calculated using the appropriate provisions in the code for lateral supported beams. Other checks like deflection, shear and local criteria will be same.
15 Design strength of laterally supported and unsupported beams These are analysis problems where the strength of the beam is required. The design strength will be based on flexural or bending strength and shear strength. Bending strength of laterally supported beams are calculated using the provisions given (pp 53) by knowing the plastic section modulus Z pz Bending strength of laterally unsupported beams are calculated using the provisions given and (pp 54) by knowing the plastic section modulus Z pz The shear strength of the c/s is obtained from cl. 8.4(pp 59). Shear strength of Beams Shear forces always exists with BMs and the maximum shear stress has to be checked with the shear yield stress. Shear stresses can become important if the depth of the beams are restricted and when beams are subjected to large concentrated loads near the supports. The distribution of shear stress at limit state (plastic) is shown below  The nominal shear yielding strength is based on the Von Mises yield criteria which assumes wide and thin webs without any local failures. The shear strength is expressed as τ y = f yw / 3 f yw = yield strength of the web The design strength V d = [A v f yw / 3] / 1.1 A v = Shear area as specified in cl pp  59
16 The web can buckle elastically or inelastically depending on the ratio of d / t w. if this ratio exceeds 67 ε, where ε = (250 /f y ) and d = clear depth of the web between flanges, resistance to shear buckling has to be verified. Shear failure can occur due to excessive yielding of the web area if the shear capacity is exceeded. The beam will be a high shear condition if V U > 0.6 V d and the moment capacity of the section decreases and has to calculated using the provisions given in cl pp  70 Maximum Deflection A beam may have adequate strength in flexure and shear and can be unsuitable if it deflects excessively under the service loads. Excessive deflection causes problems in the functioning of the structure. It can harm floor finishes, cause cracks in partitions and excessive vibrations in industrial buildings and ponding of water in roofs. Cl.5.6.1, and Table 6 gives relevant specifications with respect to deflection. The beam size may have to be taken based on deflection, if the spans and loadings are large. Typical maximum deflection formulae for simple loadings are given below 
17 Example 1 Design a simply supported beam of span 8 m. The spacing of the beams are 4m with thickness of RC slab =150 mm, floor finishes = 1.4 kn / m 2 and light partitions = 1kN / m 2. The beam also carries a central concentrated load of 250 kn with all the loads being characteristic loads. The beam is laterally restrained. with grade of steel being Fe 490. Check the beam for deflection, shear, web buckling and crippling. Load on the RC slab = 0.15 * = = kn/m 2 To calculate the self weight of the beam  Total load on the beam = * 4 * = kn Self weight of the beam = (Total load on the beam) / 350 = / 350 = 1.64 kn/m Total UDL on the beam, w = * = kn/m The beam loaded is shown in the figure. Design for flexure  Maximum BM at centre, M U = (42.24 * 8 2 / * 8 / 4) * 1.5 = knm Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl pp 53) = ( x 10 6 * 1.1) / (1.0 * 350) = x 10 3 mm 3 Adopt ISWB kg/m (1.45 kn / m < 1.64 kn / m)
18 Thickness of flange, t f = 23.6 mm > 20mm, f Y = 330 N / mm 2, (Table 1, pp 14) Z P ) REQD = x10 3 mm 3 Z P ) PRO = x10 3 mm 3 (OK) Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * x10 3 * 330) / 1.1 = x 10 6 N mm = knm > M U (OK) To check the type of section  Outstand = b / 2 = 125 mm (fig 2, pp 19) Outstand / t f = 125 / 23.6 = 5.3 < 9.4 ε < 9.4 * (250 / 330) < 8.18 Hence the section is plastic (OK) Check for deflection  Maximum deflection = (5wL 4 ) / (384EI) + (WL 3 ) / (48EI) = (5 * * ) / (384 * 2 x 10 5 * x 10 4 ) + (250 x 10 3 * ) / (48 * 2 x 10 5 * x 10 4 ) = mm < L / 360 < 8000 / 360 < mm (OK) kn / m = N / mm (with actual self weight of the beam) E = 2 x 10 5 N / mm 2 and I = I ZZ = I XX (from SP  6, Table I) Check for shear lag effect  width of flange = 250 mm L / / 20 = 400 mm (cl , pp  53) (OK)
19 Check for shear  V U = (42.05 * 8 / / 2) * 1.5 = kn V d = (A V * f yw ) / ( 3 * 1.1) (cl. 8.4, and , pp  59) = (600 * 11.8 * 330) / ( 3 * 1.1) = x 10 3 N = kn V U < 0.6 * V d < kn (Low shear) (OK) d / t w = ( * 2) / 11.8 = < 67ε < (cl , pp 59) (OK) Check for web Buckling  kl/r y = * d / t w = * ( * 2) / 11.8 = From Table 9(c), pp  42, f cd = N / mm 2 (f cd can also be calculated using the equations given in cl as in compression members) F wb = (b 1 + n 1 ) t w f cd = (b / 200) * 11.8 * To get b 1, F wb = V U = x 10 3 b 1 = mm, say 90 mm 90 mm wide bearing plate is provided. (minimum of 75 mm shall be provided) Check for web crippling  F wc = [(b 1 + n 2 ) t w f yw ] / γ mo = [( * 46.05) * 11.8 * 330] / 1.1 = x 10 3 N = kn > V U (OK) Hence ISWB kg / m satisfies all the specifications and can be used for the given problem. Example 2 A floor plan has a series of secondary beams spaced at 2 m c/c supported on main beams spaced at 12 m c/c. The main beams are supported on columns spaced at 12 m c/c. The floor is used for commercial purpose. Design the main beam by assuming suitable loads. The beam is laterally restrained. with grade of steel being Fe 490. Check the beam for deflection, shear, web buckling and crippling. The layout for the given problem is down in the figure. SB = Secondary Beams MB = Main Beams C = Columns Design of Secondary Beam  Load on the RC slab = 0.1 * = 8.9 kn/m 2
20 To calculate the self weight of the beam  Total load on the beam = 8.9 * 2 * 12 = kn Self weight of the beam = (Total load on the beam) / 350 = / 350 = 0.61 kn/m Total UDL on the beam, w = 8.9 * = kn / m Design for flexure  Maximum BM at centre, M U = (18.41 * 12 2 / 8) * 1.5 = knm Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl pp 53) = ( x 10 6 * 1.1) / (1.0 * 350) = x 10 3 mm 3 Adopt ISLB 75 kg/m (0.75 kn / m >0.61 kn / m) Thickness of flange, t f = 15 mm < 20mm, f Y = 350 N / mm 2, (Table 1, pp 14) Z P ) REQD = x10 3 mm 3 ( with new self weight)
21 Z P ) PRO = x10 3 mm 3 (OK) Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * x10 3 * 350) / 1.1 = x 10 6 N mm = knm > M U (OK) Check for deflection  Maximum deflection = (5wL 4 ) / (384EI) = (5 * * ) / (384 * 2 x 10 5 * x 10 4 ) = mm > L / 360 < / 360 < mm (UNSAFE) E = 2 x 10 5 N / mm 2 and I = I ZZ = I XX (from SP  6, Table I) I ZZ = I XX ) REQD = (5wL 4 ) / (384E * 33.33) = (5 * * ) / (384 * 2 x 10 5 * 33.33) = x 10 4 mm 4 Adopt ISMB kg / m (1.23 kn/m) Maximum deflection = mm < mm (OK) NOTE  The chosen section is heavier than ISLB 75 kg/m and hence the design bending strength will be satisfactory. Other checks like shear lag effect, shear, web buckling and web crippling can be satisfied on similar lines. Design of Main Beam  Reaction from each secondary beam = (8.9 * ) * 12 = kn say, 230 kn Self weight of main beam = (Total load on the beam) / 350 = (230 * 5) / 300 = 3.8 kn/m The beam is shown in the figure below  Maximum BM at centre, M U = (597.8 * * * 2  [3.8 * 6 * 6] / 2) * 1.5 = kn m Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl pp 53) = ( x 10 6 * 1.1) / (1.0 * 320) = x 10 3 mm 3 Single I section is not possible and it is proposed to provide I section with cover plates on either side. Properties are available in Table XIV for ISMB and ISWB sections in SP  6
22 Z EZ = Z XX ) REQD = Z P ) REQD / 1.14 = 9672 x 10 3 mm 3 The section shall also be worked using deflection condition. δ MAX = (WL 3 / 24EI) * β * (34 * β 2 ) for each pair of concentrated loads + (WL 3 ) / (48EI) + (5wL 4 ) / (384EI) ; β = a / L Equating this to the maximum deflection, L / 360 = mm, we get = (230 * 1000 * ) / (24 * 2 x 10 5 * I ZZ ) * [ (2 / 12) * {34 * (2 / 12) 2 } + (4 / 12) * {34 * (4 / 12) 2 }] + (230 x 10 3 * ) / (48 * 2 x 10 5 * I ZZ ) + (5 * 3.8 * ) / (384 * 2 x 10 5 * I ZZ ) Solving for I ZZ, we get = (1.57 x10 11 ) / I ZZ I ZZ ) REQD = 4.71 x 10 9 mm 4 = cm 4 No section in Table XIV gives this value of I ZZ. Adopt ISMB kg/m with 40mm thick plates. To calculate the width of the plate x 10 9 = x [(b * 40 3 ) / 12 + b * 40 * ] * 2 b = mm say, 475 mm I ZZ ) PRO = 4.81 x 10 9 mm 4 > I ZZ ) REQD Defection is within limits.
23 Mean Thickness of flange, t f = (475 * * 20.8) / 475 =49.2 mm > 40 mm, f Y = 320 N / mm 2, (Table 1, pp 14) Z PZ = x * 40 * 320 * 2 = x 10 3 mm 3 Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * x10 3 * 320) / 1.1 = 4558 x 10 6 N mm = 4558 knm > M U (OK) Check for shear  Self weight = [(475 * 40 * 2) / ] * = 4.21 kn/m V U = (4.21 * 12 / * 5) / 2) * 1.5 = kn V d = (A V * f yw ) / ( 3 * 1.1) (cl. 8.4, and , pp  59) = (680 * 12 * 320) / ( 3 * 1.1) = x 10 3 N = kn > V U (OK) 0.6 * V d = kn V U > kn (High shear) Design bending strength has to be modified. d / t w = ( * 2) / 12 = < 67ε < (cl , pp 59) (OK) M dv = M d  β * (M d  M fd ) (cl , pp  70) β = [(2V U ) / V d  1] 2 = M fd = Z P ) fd * f y / 1.1 Z P ) fd = [(47512) * 40 * (21012) * 20.8 * 289.6] * 2 = x 10 6 mm 3 M fd = knm M dv = * ( ) = 4517 knm > M U (OK) The other checks for web buckling, crippling and shear lag can be calculated as earlier. Example 3 A simply supported beam has an effective span of 8 m and the beam has a c/s ISWB kg/m. Calculate the design BM and the safe UDL, the beam can support. The beam is laterally restrained against torsion but partially restrained against warping. The grade of the structural steel is Fe 490. From Table 15, L LT = 0.85 * 8000 = 6800 mm The value of the Elastic Lateral Torsional Buckling moment, M cr is calculated using all the equations given in IS 800 : 2007
24 i) M cr = { (π 2 EI Y ) / (L LT ) 2 }* [GI t + (π 2 EI w ) / (L LT ) 2 ]} (cl , pp 54) E = 2 x 10 5 N/mm 2, I Y = x 10 4 mm 4,G = x 10 5 N/mm 2 ( cl pp  12), The equations given below are from Annex E pp I t = Torsional constant = Σ b i (t i ) 3 / 3 = 2 * 250 * / 3 + ( * 2) * / 3 = x 10 6 mm 4 I w = warping constant = (1 β f ) β f I Y (h y ) 2 β f = I fc / (I fc + I ft ) = 0.5 ( for symmetrical sections about both the axis) h Y = c/c distance between the flanges = = mm I w = 4.4 x mm 6 Substituting, M cr = x10 6 N mm = kn m ii) M cr = (π 2 EI Y h f ) / [2 * (L LT ) 2 ] }* [1+ 1 / 20 { (L LT / r y ) / (h f / t f )} 2 ] 0.5 h f = c/c distance between the flanges = = mm r y = 53.5 mm and t f = 23.6 mm Substituting, M cr = x 10 6 N mm = kn m iii) f cr,b = (1.1π 2 E) / (L LT / r y ) 2 * [1+ 1 / 20 { (L LT / r y ) / (h f / t f )} 2 ] 0.5 = N/mm 2 Alternately, f cr,b can also be obtained from Table 14  pp 57 M cr = β b * Z P * f cr,b = 1.0 * x10 3 * = x10 6 N mm = kn m Each equation gives different values of M cr and the procedure given in Annex E is more accurate but lengthy and cumbersome. Consider the third approach as it gives the least values required and simple. If more accuracy is required, averaging can be done. M cr = x10 6 N mm and f cr,b = N/mm 2 will be used in further calculations.
25 λ LT = (f y / f cr,b ) = (330 / ) = Also, λ LT [(1.2 * Z E * f Y ) / M cr ] [(1.2 * x 10 3 * 330) / x Adopt λ LT = ( > 0.4 Hence lateral buckling analysis required) φ LT = 0.5 * [1+ α LT * (λ LT  0.2) + (λ LT ) 2 ] = 0.5 * [1 + 0,21 * ( ) ] = χ LT = 1/ { φ LT + [ (φ LT ) 2  (λ LT ) 2 ] 0.5 } Substituting, χ LT = 0.49 < 1.0 f bd = χ LT * f Y / 1.1 = 147 N/mm 2 M d = β b * Z P * f bd = 1.0 * x10 3 * 147 = x 10 6 Nmm = knm M SAFE = / 1.5 = knm w SAFE * L 2 / 8 = M SAFE w SAFE = kn/m (including self weight) Other calculations can be carried out using an effective span of 8m as in laterally supported beams. Design of laterally unsupported beam The section can be chosen based on two conditions  i) Z P ) REQD = (M U * 1.1) / (β b * f Y ) * 1.25 to 1.5 ii) δ) PER = L / 360 for Simply supported beams = L / 180 for cantilever beams Relevant expressions for deflection are used based on the loadings. The moment of Inertia required is calculated..
26 A suitable section based on the above requirement is chosen and the design bending strength, M d is calculated as in Example 3. This shall be greater than M U. If required the section has to be modified for economy. Once the section is chosen, other checks shall be using the effective span as in laterally supported beams. Example1 can be treated as laterally unrestrained beam and worked. Design of Purlins Purlins are flexural members used in trusses to support the roof covering and spans between the trusses. Purlins are provided on the top rafter (top chord) at all the joints. The spacing of the purlins depends on the type of the roofing material and for normal materials, it ranges from 1.4 to 1.8 m. The sections used for purlins are usually angles (equal or unequal) as they are economical and variety of sections is available. A typical view of purlin is shown in the figure. The new code do not provide the design specifications. Therefore the specifications as per the old code IS: is followed. Cl. 8.9 pp  69 shall also be followed. Based on IS : 875 Part 2, LL on inclined roofs shall be taken as  LL = / of the slope for slopes > 10 subjected to a minimum of 0.4 kn / m 2 For slopes 10, LL = 0.75 kn / m 2 DL of AC sheets = 0.17 kn / m 2 and GI sheets = 0.13 kn / m 2
27 Example 4 Design a suitable single angle purlin having AC sheets as covering with spacing of trusses = 4.5m. Pitch of the truss is 1 in 5 with spacing of the purlins = 1.6m. Wind pressure normal to the roof is 1,3 kn / m 2. Span of the truss = 18m. Rise of the truss = 1/5 * 18 = 3.6m Slope of the truss, tan θ = 3.6 / 9.0 = 0.4 θ = 21.8 For the given problem, LL = * ( ) = kn / m 2 (DL + LL) on plan area = = kn / m 2 Vertical load on each purlin = * 1.6 = kn/m Self weight of the purlin = 0.1 kn/m Total vertical load = kn/m Load acting normal to the purlin = * cos θ = 1.12 kn/m (DL + LL) DL acting normal to the purlin = (0.17 * ) = 0.37 kn/m (Downwards) WL acting normal to the purlin = 1.3 * 1.6 = 2.08 kn/m (Upwards) (DL + WL) = 1.71 kn/m (Upwards) (DL + WL) is governing for the design. Maximum BM, M = wl 2 / 10 = (1.71 * ) / 10 = 3.46 knm (assumed as continuous spanned purlins) Z E ) REQD = M / 0.66 f Y = 3,46 x 10 6 / (0.66 * 250) assuming Fe410 grade steel = 21 x 10 3 mm 3 (21 cm 3 ) Minimum depth of angle = L / 45 =100 mm Minimum width of angle = L / 60 =75 mm From Table IV, SP  6, choosing an angle ISA 125 x 75 x 6 (9.2 Kg/m) Z E ) PRO = 22.2 x 10 3 mm 3 > 21 x 10 3 mm 3 (OK)
28 Maximum deflection, δ = (w * L 4 ) / (384EI) (Continuous span) = (1.71 * ) / (384 * 2 x 10 5 * x 10 4 ) = 4.86 mm < L / 360 = 12.5 mm (OK) The chosen section ISA 125 x 75 x 6 is OK
Structural Steelwork Eurocodes Development of A Transnational Approach
Structural Steelwork Eurocodes Development of A Transnational Approach Course: Eurocode Module 7 : Worked Examples Lecture 0 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads
More informationDesign of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar
6. BEAMS 6.1 Introduction One of the frequently used structural members is a beam whose main function is to transfer load principally by means of flexural or bending action. In a structural framework,
More informationStructural Steelwork Eurocodes Development of A Transnational Approach
Structural Steelwork Eurocodes Development of A Transnational Approach Course: Eurocode 3 Module 7 : Worked Examples Lecture 20 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic
More informationSteel Structures Design and Drawing Lecture Notes
Steel Structures Design and Drawing Lecture Notes INTRODUCTION When the need for a new structure arises, an individual or agency has to arrange the funds required for its construction. The individual or
More informationSECTION 7 DESIGN OF COMPRESSION MEMBERS
SECTION 7 DESIGN OF COMPRESSION MEMBERS 1 INTRODUCTION TO COLUMN BUCKLING Introduction Elastic buckling of an ideal column Strength curve for an ideal column Strength of practical column Concepts of effective
More informationPLATE GIRDERS II. Load. Web plate Welds A Longitudinal elevation. Fig. 1 A typical Plate Girder
16 PLATE GIRDERS II 1.0 INTRODUCTION This chapter describes the current practice for the design of plate girders adopting meaningful simplifications of the equations derived in the chapter on Plate Girders
More informationStructural Steelwork Eurocodes Development of A Transnational Approach
Structural Steelwork Eurocodes Development of A Transnational Approach Course: Eurocode Module 7 : Worked Examples Lecture 22 : Design of an unbraced sway frame with rigid joints Summary: NOTE This example
More informationKarbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi
Chapter 05 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Beams By Dr. Jawad Talib AlNasrawi University of Karbala Department of Civil Engineering 71 Introduction
More informationAccordingly, the nominal section strength [resistance] for initiation of yielding is calculated by using Equation CC3.1.
C3 Flexural Members C3.1 Bending The nominal flexural strength [moment resistance], Mn, shall be the smallest of the values calculated for the limit states of yielding, lateraltorsional buckling and distortional
More informationJob No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet. Member Design  Steel Composite Beam XX 22/09/2016
CONSULTING Engineering Calculation Sheet jxxx 1 Member Design  Steel Composite Beam XX Introduction Chd. 1 Grade 50 more common than Grade 43 because composite beam stiffness often 3 to 4 times non composite
More informationMODULE C: COMPRESSION MEMBERS
MODULE C: COMPRESSION MEMBERS This module of CIE 428 covers the following subjects Column theory Column design per AISC Effective length Torsional and flexuraltorsional buckling Builtup members READING:
More informationFLOW CHART FOR DESIGN OF BEAMS
FLOW CHART FOR DESIGN OF BEAMS Write Known Data Estimate selfweight of the member. a. The selfweight may be taken as 10 percent of the applied dead UDL or dead point load distributed over all the length.
More informationJob No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet
CONSULTING Engineering Calculation Sheet E N G I N E E R S Consulting Engineers jxxx 1 Structural Description The two pinned (at the bases) portal frame is stable in its plane due to the moment connection
More informationKarbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi
Chapter 04 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Compression Members By Dr. Jawad Talib AlNasrawi University of Karbala Department of Civil Engineering
More informationFailure in Flexure. Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas
Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VIII Dr. Jason E. Charalambides Failure in Flexure!
More informationDesign of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar
5.4 Beams As stated previousl, the effect of local buckling should invariabl be taken into account in thin walled members, using methods described alread. Laterall stable beams are beams, which do not
More informationDEPARTMENT OF CIVIL ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SUBJECT: CE 2252 STRENGTH OF MATERIALS UNIT: I ENERGY METHODS 1. Define: Strain Energy When an elastic body is under the action of external
More informationEurocode 3 for Dummies The Opportunities and Traps
Eurocode 3 for Dummies The Opportunities and Traps a brief guide on element design to EC3 Tim McCarthy Email tim.mccarthy@umist.ac.uk Slides available on the web http://www2.umist.ac.uk/construction/staff/
More informationSTEEL MEMBER DESIGN (EN :2005)
GEODOMISI Ltd.  Dr. Costas Sachpazis Consulting Company for App'd by STEEL MEMBER DESIGN (EN199311:2005) In accordance with EN199311:2005 incorporating Corrigenda February 2006 and April details type;
More informationUNIT I Thin plate theory, Structural Instability:
UNIT I Thin plate theory, Structural Instability: Analysis of thin rectangular plates subject to bending, twisting, distributed transverse load, combined bending and inplane loading Thin plates having
More informationDesign of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar
5.10 Examples 5.10.1 Analysis of effective section under compression To illustrate the evaluation of reduced section properties of a section under axial compression. Section: 00 x 80 x 5 x 4.0 mm Using
More informationFundamentals of Structural Design Part of Steel Structures
Fundamentals of Structural Design Part of Steel Structures Civil Engineering for Bachelors 133FSTD Teacher: Zdeněk Sokol Office number: B619 1 Syllabus of lectures 1. Introduction, history of steel structures,
More informationQUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A
DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State
More informationDESIGN AND DETAILING OF COUNTERFORT RETAINING WALL
DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL When the height of the retaining wall exceeds about 6 m, the thickness of the stem and heel slab works out to be sufficiently large and the design becomes
More informationDesign of Steel Structures Dr. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati
Design of Steel Structures Dr. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati Module  6 Flexural Members Lecture 5 Hello today I am going to deliver the lecture
More informationDesign of a MultiStoried RC Building
Design of a MultiStoried RC Building 16 14 14 3 C 1 B 1 C 2 B 2 C 3 B 3 C 4 13 B 15 (S 1 ) B 16 (S 2 ) B 17 (S 3 ) B 18 7 B 4 B 5 B 6 B 7 C 5 C 6 C 7 C 8 C 9 7 B 20 B 22 14 B 19 (S 4 ) C 10 C 11 B 23
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
More informationUniversity of Sheffield. Department of Civil Structural Engineering. Member checks  Rafter 44.6
Member checks  Rafter 34 6.4Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic rafter sections, as it is the normal practice to use a UB cutting of
More informationUNITI STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2
UNITI STRESS, STRAIN 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2 Young s modulus E= 2 x10 5 N/mm 2 Area1=900mm 2 Area2=400mm 2 Area3=625mm
More informationPERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR  VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK
PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR  VALLAM  613 403  THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310
More informationUnit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir
Unit III Theory of columns 1 Unit III Theory of Columns References: Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength of Materials", Tata
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)
More information2012 MECHANICS OF SOLIDS
R10 SET  1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~
More informationMechanics of Structure
S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: Hrs.] Prelim Question Paper Solution [Marks : 70 Q.1(a) Attempt any SIX of the following. [1] Q.1(a) Define moment of Inertia. State MI
More informationUNIT III DEFLECTION OF BEAMS 1. What are the methods for finding out the slope and deflection at a section? The important methods used for finding out the slope and deflection at a section in a loaded
More informationCHAPTER 4. Design of R C Beams
CHAPTER 4 Design of R C Beams Learning Objectives Identify the data, formulae and procedures for design of R C beams Design simplysupported and continuous R C beams by integrating the following processes
More informationSabah Shawkat Cabinet of Structural Engineering Walls carrying vertical loads should be designed as columns. Basically walls are designed in
Sabah Shawkat Cabinet of Structural Engineering 17 3.6 Shear walls Walls carrying vertical loads should be designed as columns. Basically walls are designed in the same manner as columns, but there are
More informationChapter 8: Bending and Shear Stresses in Beams
Chapter 8: Bending and Shear Stresses in Beams Introduction One of the earliest studies concerned with the strength and deflection of beams was conducted by Galileo Galilei. Galileo was the first to discuss
More informationto introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling
to introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling In the case of elements subjected to compressive forces, secondary bending effects caused by,
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationBeam Design and Deflections
Beam Design and Deflections tation: a = name for width dimension A = name for area Areq dadj = area required at allowable stress when shear is adjusted to include self weight Aweb = area of the web of
More informationSERVICEABILITY LIMIT STATE DESIGN
CHAPTER 11 SERVICEABILITY LIMIT STATE DESIGN Article 49. Cracking Limit State 49.1 General considerations In the case of verifications relating to Cracking Limit State, the effects of actions comprise
More informationBPSC Main Exam 2019 ASSISTANT ENGINEER. Test 11. CIVIL ENGINEERING Subjective PaperI. Detailed Solutions. Detailed Solutions
etailed Solutions SC Main Exam 19 SSISTNT ENGINEER CIVI ENGINEERING Subjective aperi Test 11 Q.1 (a) Solution: (i) Calculation of maximum moment at 8 m. For maximum moment Case 1: Case : Case : Case 4:
More informationLecture08 Gravity Load Analysis of RC Structures
Lecture08 Gravity Load Analysis of RC Structures By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar www.drqaisarali.com 1 Contents Analysis Approaches Point of Inflection Method Equivalent
More informationUNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.
UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude
More information7.3 Design of members subjected to combined forces
7.3 Design of members subjected to combined forces 7.3.1 General In the previous chapters of Draft IS: 800 LSM version, we have stipulated the codal provisions for determining the stress distribution in
More informationLongitudinal strength standard
(1989) (Rev. 1 199) (Rev. Nov. 001) Longitudinal strength standard.1 Application This requirement applies only to steel ships of length 90 m and greater in unrestricted service. For ships having one or
More informationMechanics of Materials Primer
Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus
More informationMade by SMH Date Aug Checked by NRB Date Dec Revised by MEB Date April 2006
Job No. OSM 4 Sheet 1 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 DESIGN EXAMPLE 9  BEAM WITH UNRESTRAINED COMPRESSION
More informationLecture04 Design of RC Members for Shear and Torsion
Lecture04 Design of RC Members for Shear and Torsion By: Prof. Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com 1 Topics Addressed Design of
More informationmportant nstructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationAPPENDIX 1 MODEL CALCULATION OF VARIOUS CODES
163 APPENDIX 1 MODEL CALCULATION OF VARIOUS CODES A1.1 DESIGN AS PER NORTH AMERICAN SPECIFICATION OF COLD FORMED STEEL (AISI S100: 2007) 1. Based on Initiation of Yielding: Effective yield moment, M n
More information2. Determine the deflection at C of the beam given in fig below. Use principal of virtual work. W L/2 B A L C
CE1259, Strength of Materials UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS Part A 1. Define strain energy density. 2. State Maxwell s reciprocal theorem. 3. Define proof resilience. 4. State Castigliano
More informationε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram
CHAPTER NINE COLUMNS 4 b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case
More informationEquivalent Uniform Moment Factor for Lateral Torsional Buckling of Steel Beams
University of Alberta Department of Civil & Environmental Engineering Master of Engineering Report in Structural Engineering Equivalent Uniform Moment Factor for Lateral Torsional Buckling of Steel Beams
More informationDesign of AAC wall panel according to EN 12602
Design of wall panel according to EN 160 Example 3: Wall panel with wind load 1.1 Issue Design of a wall panel at an industrial building Materials with a compressive strength 3,5, density class 500, welded
More informationSRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA
SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA (Declared as Deemedtobe University under Section 3 of the UGC Act, 1956, Vide notification No.F.9.9/92U3 dated 26 th May 1993 of the Govt. of
More informationMAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAALLI  6113. QUESTION WITH ANSWERS DEARTMENT : CIVIL SEMESTER: V SUB.CODE/ NAME: CE 5 / Strength of Materials UNIT 3 COULMNS ART  A ( marks) 1. Define columns
More informationSUMMARY FOR COMPRESSION MEMBERS. Determine the factored design loads (AISC/LRFD Specification A4).
SUMMARY FOR COMPRESSION MEMBERS Columns with Pinned Supports Step 1: Step : Determine the factored design loads (AISC/LRFD Specification A4). From the column tables, determine the effective length KL using
More informationENG1001 Engineering Design 1
ENG1001 Engineering Design 1 Structure & Loads Determine forces that act on structures causing it to deform, bend, and stretch Forces push/pull on objects Structures are loaded by: > Dead loads permanent
More informationNUMERICAL EVALUATION OF THE ROTATIONAL CAPACITY OF STEEL BEAMS AT ELEVATED TEMPERATURES
8 th GRACM International Congress on Computational Mechanics Volos, 12 July 15 July 2015 NUMERICAL EVALUATION OF THE ROTATIONAL CAPACITY OF STEEL BEAMS AT ELEVATED TEMPERATURES Savvas Akritidis, Daphne
More information2. (a) Explain different types of wing structures. (b) Explain the advantages and disadvantages of different materials used for aircraft
Code No: 07A62102 R07 Set No. 2 III B.Tech II Semester Regular/Supplementary Examinations,May 2010 Aerospace Vehicle Structures II Aeronautical Engineering Time: 3 hours Max Marks: 80 Answer any FIVE
More informationSTRUCTURAL ANALYSIS CHAPTER 2. Introduction
CHAPTER 2 STRUCTURAL ANALYSIS Introduction The primary purpose of structural analysis is to establish the distribution of internal forces and moments over the whole part of a structure and to identify
More informationJob No. Sheet 1 of 7 Rev A. Made by ER/EM Date Feb Checked by HB Date March 2006
Job No. Sheet of 7 Rev A Design Example Design of a lipped channel in a Made by ER/EM Date Feb 006 Checked by HB Date March 006 DESIGN EXAMPLE DESIGN OF A LIPPED CHANNEL IN AN EXPOSED FLOOR Design a simply
More informationVTU EDUSAT PROGRAMME Lecture Notes on Design of Columns
VTU EDUSAT PROGRAMME 17 2012 Lecture Notes on Design of Columns DESIGN OF RCC STRUCTURAL ELEMENTS  10CV52 (PART B, UNIT 6) Dr. M. C. Nataraja Professor, Civil Engineering Department, Sri Jayachamarajendra
More informationDesign of Reinforced Concrete Structures (II)
Design of Reinforced Concrete Structures (II) Discussion Eng. Mohammed R. Kuheil Review The thickness of oneway ribbed slabs After finding the value of total load (Dead and live loads), the elements are
More informationENCE 455 Design of Steel Structures. III. Compression Members
ENCE 455 Design of Steel Structures III. Compression Members C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland Compression Members Following subjects are covered:
More informationFINITE ELEMENT ANALYSIS OF TAPERED COMPOSITE PLATE GIRDER WITH A NONLINEAR VARYING WEB DEPTH
Journal of Engineering Science and Technology Vol. 12, No. 11 (2017) 28392854 School of Engineering, Taylor s University FINITE ELEMENT ANALYSIS OF TAPERED COMPOSITE PLATE GIRDER WITH A NONLINEAR VARYING
More informationFinite Element Modelling with Plastic Hinges
01/02/2016 Marco Donà Finite Element Modelling with Plastic Hinges 1 Plastic hinge approach A plastic hinge represents a concentrated postyield behaviour in one or more degrees of freedom. Hinges only
More informationChapter Objectives. Design a beam to resist both bendingand shear loads
Chapter Objectives Design a beam to resist both bendingand shear loads A Bridge Deck under Bending Action Castellated Beams Posttensioned Concrete Beam Lateral Distortion of a Beam Due to Lateral Load
More informationCE5510 Advanced Structural Concrete Design  Design & Detailing of Openings in RC Flexural Members
CE5510 Advanced Structural Concrete Design  Design & Detailing Openings in RC Flexural Members Assoc Pr Tan Kiang Hwee Department Civil Engineering National In this lecture DEPARTMENT OF CIVIL ENGINEERING
More informationBUCKLING STRENGTH ANALYSIS OF BARS AND FRAMES, AND SPHERICAL SHELLS
CLASSIFICATION NOTES No. 30.1 BUCKLING STRENGTH ANALYSIS OF BARS AND FRAMES, AND SPHERICAL SHELLS APRIL 004 Veritasveien 1, NO13 Høvik, Norway Tel.: +47 67 57 99 00 Fax: +47 67 57 99 11 FOREWORD is an
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More information71 Laxmi Nagar (South), Niwaru Road, Jhotwara, Jaipur ,India. Phone: Mob. : /
www.aarekh.com 71 Laxmi Nagar (South), Niwaru Road, Jhotwara, Jaipur 302 012,India. Phone: 01412348647 Mob. : +919799435640 / 9166936207 1. Limiting values of poisson s ratio are (a) 1 and 0.5 (b)
More informationCE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR
CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR 20142015 UNIT  1 STRESS, STRAIN AND DEFORMATION OF SOLIDS PART A 1. Define tensile stress and tensile strain. The stress induced
More informationSlenderness Effects for Concrete Columns in Sway Frame  Moment Magnification Method (CSA A )
Slenderness Effects for Concrete Columns in Sway Frame  Moment Magnification Method (CSA A23.394) Slender Concrete Column Design in Sway Frame Buildings Evaluate slenderness effect for columns in a
More informationDesign of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar. Local buckling is an extremely important facet of cold formed steel
5.3 Local buckling Local buckling is an extremely important facet of cold formed steel sections on account of the fact that the very thin elements used will invariably buckle before yielding. Thinner the
More informationAn Increase in Elastic Buckling Strength of Plate Girder by the Influence of Transverse Stiffeners
GRD Journals Global Research and Development Journal for Engineering Volume 2 Issue 6 May 2017 ISSN: 24555703 An Increase in Elastic Buckling Strength of Plate Girder by the Influence of Transverse Stiffeners
More information8 Deflectionmax. = 5WL 3 384EI
8 max. = 5WL 3 384EI 1 salesinfo@mechanicalsupport.co.nz PO Box 204336 Highbrook Auckland www.mechanicalsupport.co.nz 2 Engineering Data  s and Columns Structural Data 1. Properties properties have been
More informationCIVL473 Fundamentals of Steel Design
CIVL473 Fundamentals of Steel Design CHAPTER 4 Design of Columns embers with Aial Loads and oments Prepared B Asst.Prof.Dr. urude Celikag 4.1 Braced ultistore Buildings  Combined tension and oments Interaction
More informationServiceability Deflection calculation
Chp6:Lecture Goals Serviceability Deflection calculation Deflection example Structural Design Profession is concerned with: Limit States Philosophy: Strength Limit State (safetyfracture, fatigue, overturning
More informationMade by PTY/AAT Date Jan 2006
Job No. VALCOSS Sheet of 9 Rev A P.O. Box 000, FI0044 VTT Tel. +358 0 7 Fax +358 0 7 700 Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb
More informationSTEEL BUILDINGS IN EUROPE. MultiStorey Steel Buildings Part 10: Technical Software Specification for Composite Beams
STEEL BUILDINGS IN EUROPE MultiStorey Steel Buildings Part 10: Technical Software Specification for Composite Beams MultiStorey Steel Buildings Part 10: Technical Software Specification for Composite
More informationDesign of Compression Members
Design of Compression Members 2.1 Classification of cross sections Classifying crosssections may mainly depend on four critical factors: 1 Width to thickness (c/t) ratio. 2 Support condition. 3 Yield
More information9.5 Compression Members
9.5 Compression Members This section covers the following topics. Introduction Analysis Development of Interaction Diagram Effect of Prestressing Force 9.5.1 Introduction Prestressing is meaningful when
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More informationDESIGN REINFORCED CEMENT CONCRETE STRUCTURAL MEMBERS
DESIGN OF REINFORCED CEMENT CONCRETE STRUCTURAL MEMBERS Contents CHAPTER 1.0 General: 1.1 Symbols 1. Materials 1..1 Cement 1.. Aggregate 1..3 Water 1..4 Admixtures 1..5 Reinforcement CHAPTER.0 Concrete:.1
More informationCHAPTER 6: ULTIMATE LIMIT STATE
CHAPTER 6: ULTIMATE LIMIT STATE 6.1 GENERAL It shall be in accordance with JSCE Standard Specification (Design), 6.1. The collapse mechanism in statically indeterminate structures shall not be considered.
More informationBridge deck modelling and design process for bridges
EURussia Regulatory Dialogue Construction Sector Subgroup 1 Bridge deck modelling and design process for bridges Application to a composite twingirder bridge according to Eurocode 4 Laurence Davaine
More informationC6 Advanced design of steel structures
C6 Advanced design of steel structures prepared b Josef achacek List of lessons 1) Lateraltorsional instabilit of beams. ) Buckling of plates. 3) Thinwalled steel members. 4) Torsion of members. 5) Fatigue
More informationD : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each.
GTE 2016 Q. 1 Q. 9 carry one mark each. D : SOLID MECHNICS Q.1 single degree of freedom vibrating system has mass of 5 kg, stiffness of 500 N/m and damping coefficient of 100 Ns/m. To make the system
More informationCritical Load columns buckling critical load
Buckling of Columns Buckling of Columns Critical Load Some member may be subjected to compressive loadings, and if these members are long enough to cause the member to deflect laterally or sideway. To
More informationINFLUENCE OF FLANGE STIFFNESS ON DUCTILITY BEHAVIOUR OF PLATE GIRDER
International Journal of Civil Structural 6 Environmental And Infrastructure Engineering Research Vol.1, Issue.1 (2011) 115 TJPRC Pvt. Ltd.,. INFLUENCE OF FLANGE STIFFNESS ON DUCTILITY BEHAVIOUR OF PLATE
More informationFLEXIBILITY METHOD FOR INDETERMINATE FRAMES
UNIT  I FLEXIBILITY METHOD FOR INDETERMINATE FRAMES 1. What is meant by indeterminate structures? Structures that do not satisfy the conditions of equilibrium are called indeterminate structure. These
More informationDesign of Steel Structures Prof. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati
Design of Steel Structures Prof. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati Module 7 Gantry Girders and Plate Girders Lecture  3 Introduction to Plate girders
More informationJob No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet. Member Design  Reinforced Concrete Beam BS8110 v Member Design  RC Beam XX
CONSULTING Engineering Calculation Sheet E N G I N E E R S Consulting Engineers jxxx 1 Effects From Structural Analysis Design axial force, F (tension ve and compression +ve) (ensure < 0.1f cu b w h 0
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CENEW)/SEM3/CE301/ SOLID MECHANICS
Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers
More informationA Parametric Study on Lateral Torsional Buckling of European IPN and IPE Cantilevers H. Ozbasaran
Vol:8, No:7, 214 A Parametric Study on Lateral Torsional Buckling of European IPN and IPE Cantilevers H. Ozbasaran Abstract IPN and IPE sections, which are commonly used European I shapes, are widely used
More informationDESIGN OF A STEEL FOOT OVER BRIDGE IN A RAILWAY STATION
International Journal of Civil Engineering and Technology (IJCIET) Volume 8, Issue 8, August 2017, pp. 1533 1548, Article ID: IJCIET_08_08_167 Available online at http://http://www.iaeme.com/ijciet/issues.asp?jtype=ijciet&vtype=8&itype=8
More information