Unit 4 Class Notes -2019 pat..pptx

Unit 4 Class Notes -2019 pat

1. DESIGN OF STEEL STRUCTURES TE Civil UNIT- 4 Design of Flexural Members P r o f . R. M. Raut ME (Structural Engineering) 1

2. UNIT- 4 Design of Flexural Members 2

3. I n tr o d u c t i o n Part a) Laterally Supported Beams (3 Lectures) Part b) Laterally Unsupported Beams (3 Lectures)

4. Introduction :- Beams are structural element subjected to transverse load in the plane of bending causing bending mo m e nt s and s h e a r f o r ce s . Symme t r i c al section ab out z - z ax i s (ma j o r a xi s ) a r e econ o mic a l and geometrical properties of such section are available in SP 6. The compression flange of the beam can be laterally supported (restrained) or laterally unsupported (unrestrained) depending upon weather restrained are provided or not. The beams are designed for maximum BM and checked for maximum SF, local effect such as vertical buckling and crippling of webs and deflection. Section 8/PN 52-69/IS 800:2007 shall be followed in the design of such bending members. 4

5. Type of sections :- Beams can be of different cross sections depending on the span and loadings are shown in below. 5

6. Functional classification of flexural members 6 Flexural members are labeled in different names, such as Purlins – Member carrying load from sheeting of roof truss Girt – Member carrying load from side sheeting Joist – Light member carrying load from floor in a building Girder – Mostly member carrying heavy loads Lintel – Member carrying load over window, door opening Stringer – Mostly member in bridges aligned in the direction of traffic. Spandrel – Member on periphery

7. L A T E R A L L Y S U PP O R T E D B E A M S 7 Conditions to qualify as laterally supported beam Section is symmetric about yy axis Loading is in plane containing yy axis For sections unsymmetrical about yy axis load acts through shear center None of its elements i.e. flange and web should buckle until a desired limit state is achieved

8. Full Lateral Restraint 8

9. Local Lateral Restraint 9

10. Load Through Centroid 10

11. Load Through Shear Center 11

12. Gradual Collapse of beam 12

13. 13

14. Table 2 (Page no 18) for Section Classification 14

15. 15 Limit states for beams Limit state of flexure Limit state of shear Limit state of bearing Limit state of serviceability

16. Modes of failure 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 beam can be due to web crippling and web buckling. They are avoided by proper dimensioning of the bearing plate and through secondary design checks. 1. Shear strength of Beams Ref. Cl. No. 8.4/PN 59/IS 800:2007 V d = Design shear strength of web γ m0 = = V n A v fyw √3 γ m0 16

17. 2. Deflection Limits:- Re f . C l . N o . 8 . 4 /PN 59 /IS 800 : 2007 17

18. 3 . Loca l F ai l u r e s o f Fla n g e s : - Re f . T a b l e . N o . 2 /PN 18 /IS 800 : 2007 The local failure of flanges reduces the plastic moment capacity of the section due to buckling and is avoided by limiting the outstand to thickness ratios as given in table 2 of IS 800 : 2007. 18

19. 3 . Loca l F ai l u r e s o f W e b : - Re f . T a b l e . N o . 2 /PN 18 /IS 800 : 2007 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 Buckling Web Crippling 19

20. Web Buckling : 20 Q . State and e xpl a in the term s with nea t s k etc h : W e b b u ckling D ec . 20 1 4, Dec. 2 1 5 . The web of the beam is thin and can buckle under reaction and concentrated loads with the web behaving like a short column fixed at the flanges. For safety against web buckling, the resisting force shall be greater than the reaction or the concentrated load is dispersed in to web at 45 as shown in fig. F wb = (b 1 + n 1 ) t w f cd ≥ Reaction, R F wb = Resisting force t w = Thickness of web f cd = Design compressive stress in web b 1 = Width o f bea r ing pl a te n 1 = Width of dis per s ion For concentrated loads, the dispersion is on both side and the resisting force can be expressed as, F wd = [(b 1 + 2 n 1 ) t w f cd ] ≥ Concentrated load, P R R must be less than f cd (b 1 +n 1 )t w

21. Web Crippling : Q. State and explain the term with neat sketch : Web crippling Dec. 2014, Dec. 2015 May 2017. Web crippling causes local crushing failure of web due to large bearing stress under reaction at supports or concentration loads. This occurs due to stress concentration loads. 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 f lan g e . F o r s a f et y a g ain s t w e b c r i p plin g , the r esi s ti n g F o r ce sha l l b e g r eate r t ha n the r e ac t ion o r the concentrated load. It will be assumed that the reaction or concentrated load is dispersed in to the web with a slope of 1 in 2.5 as shown in fig. F wc = Resisting force t w = Thickness of web f y w = Yie l d st r ess in w e b b 1 = Width o f bea r ing pl a te n 2 = Width o f di s pe r si on F wc = (b 1 + n 2 ) t w f yw γ m0 ≥ Reaction, R F wc = (b 1 + 2 n 2 ) t w f yw γ m0 ≥ Concent r ated lo a d, P R R R must be less than (b 1 + n 2 ) t w f yw / γ m0 21

22. Laterally supported beam using single rolled steel section with and without flange plate: Q. Explain laterally supported beam with suitable sketch. Dec. 2015, Dec. 2016, May 2017. Beam subjected to BM develop compressive and tensile force 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 beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of inertial 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 The compression flange is connected to an RC slab throughout by shear connectors External lateral supports are provided at closer interval to the compression flange so that it as good continuous lateral support Ref cl no 8.2.1 Design of such laterally supported beams are ca r r ied ou t u sing clauses 8. 2 .1. 2 , 8. 2 .1. 3 ,8. 2 .1. 5 , 8.4.8.4.1,8.4.1.1,8.4.2.1, and 5.6.1. 22

23. Design steps for laterally supported beams Determine maximum factored shear force ‘V’ and bending moment ‘M’ for a given loading and support condition Selection of cross section Find p Z required = M.γ m0 f y Take β b = 1 for assuming plastic or compact section Using Annex H IS 800 : 2007 plastic properties of section. Select required section. Or e Z required = 𝑍 𝑝 23 𝑠ℎ𝑎𝑝𝑒 factor (𝑠) Shape factor I and channel section – 1.11 to 1.15 Rectangular section – 1.5 Circular section – 1.69 Angle and T – section – 1.8 Square or diamond section – 2 ∴ Z p provided > Z p required

24. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Using Table no 2 / PN 18 / IS 800 : 2007 4. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n γ m0 A v f y √ 3 γ m V d = = > V ∴ ok & Safe 5. Design bending strength (M d ) i. If V 𝑘 . 6 V d or V < . 6 V d Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007 then M d = < β b .f y Z P 1.2 Ze .f y . γ m0 γ m0 ........for simply supported beams < 1.5 Z P . f y ………….for cantilever beams γ m0 β b = 1 for Plastic and compact section Whe r e 24

25. ii. I f V > . 6 V d then M d = M dv Where M d = Design bending strength under high shear. 𝜙 = M d - β (M d - M fd γ m ) ≤ 1.2 Ze .f y ……… For plastic and compact section Ze .f y ……… For semi compact section γ m0 If M d > M ∴ ok & Safe 25

26. 6 . Che c ks a) Web buckling Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007 F wb = (b 1 + n 1 ) t w f cd ≥ Reaction, R F w d = [( b 1 + 2 n 1 ) t w f c d ] ≥ Concent r ated lo a d, P ∴ ok & Safe Where f cd = is calculated using slenderness ratio 0.7 d r 𝑦 = 2.425 × d t 𝑤 and table 9 c b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007 F wc = (b 1 + n 2 ) t w f yw γ m0 ≥ Reaction, R F wc = (b 1 + 2 n 2 ) t w f yw γ m0 c) Deflection Limit 26 ≥ Con c en t r ated l oad , P ∴ ok & Safe Re f . T a ble no 6 / PN 3 1 / IS 800 : 2007 ∴ ok & Safe δ actual < δ Limiting

27. Design of laterally supported beams Example 1 A Simply supported beam of effective span 4 m carries a factored point load of 350 kN mid span. The section is laterally supported throughout the span. Design cross section using I-section. SPPU - Dec. 20 1 0, 10 m a r ks 1. Determine maximum factored shear force ‘V’ and bending moment ‘M’ Maximum Shear force V = Reaction = W = 175 kN-m 2 Maximum Bending Moment M = WL 4 = 350 kN-m 2. Selection of cross section p Z required = M.γ m0 f y = 350 ×10 6 ×1.10 250 = 1540 × 10 3 mm 3 Or 𝑍 Z e required = 𝑝 = 𝑠ℎ𝑎𝑝𝑒 factor (𝑠) Select an ISLB 500 @ 75 kg/m 1540 ×10 3 27 1 . 14 = 1350.8 × 10 3 mm 3 (Using annex ‘H’ IS 800:2007) d= 500 mm, b f = 180 mm, t f = 14.1 mm, t w = 9.2 mm, Z p = 1773 . 7 × 10 3 mm 3 > Z p r equ i r e d , Z e = 1545 . 2 × 10 3 mm 3 I zz = 38549 × 10 4 mm 4 ,

28. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 18 / 2 14.1 = 6.38 < 9.4 ∈ ∴ Flange is plastic W e b = b t w = h 1 t w = 430.2 9.2 = 46.76 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 28

29. 4. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n γ m0 A v f y √ 3 γ m V d = = > V = A v f y √ 3 γ m = 50 × 9 .2 × 250 √3 × 1.10 = 603 . 3 kN > V = 17 5 kN - m ∴ ok & Safe 29

30. 5. Design bending strength (M d ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007 i. If V 𝑘 . 6 V d or V < . 6 V d 0.6 V d = 361.8 kN ∴ V < 0.6 V d ∴ M d = . β b .fy Z P γ m M d = β b .fy Z P . γ m0 = 1×1773.7 ×10 3 × 250 1 . 10 =403.11 kN –m > M = 350 kN-m ∴ ok & Safe 30

31. 6 . Checks a) Web buckling f wb = (b 1 + n 1 ) t w f cd ≥ Reaction, R Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007 f w b = [( b 1 + 2 n 1 ) t w f c d ] ≥ Concent r ated lo a d, P Where f cd = is calculated using slenderness ratio 0.7 d r 𝑦 = 2.425 × d t 𝑤 9. 2 = 2.425 × 500 =131.78 B u ckl in g Cl a ss ‘c’ By interpolation cd f = 74.3 − 74.3−66.2 × (131.78 − 130 ) 140−130 = 72.85 N/mm 2 f cd At Concentrated load ∴ f wb = (180 + 2 × 250 ) × 9.2 × 72.85 = 455.74 kN > P = 350 kN At Re action ‘ R ’ ∴ f wb = (180 + 250 ) × 9.2 × 72.85 = 288.19 kN > R = 175 kN ∴ ok & Safe ∴ ok & Safe 31

32. b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007 F wc A t concent r ated lo a d ‘W’ = (b 1 + 2 n 2 ) tw fyw γ m0 = (180 +2 ×2.5 ×14.1 × 9.2 ×250 1.10 > Concent r ated lo a d, P = 3 5 kN = 523.77 kN c) Deflection Limit ∴ ok & Safe Ref. Table no 6 / PN 31 / IS 800:2007 working load = 350 1 . 5 = 233.33 kN δ actual WL 3 = 48 EI = 233.33 ×10 3 ×4000 3 48 ×2×10 5 × 38549 × 10 4 = 4.03 mm δ Limiting = span 300 = 4000 300 32 = 13.33 mm ∴ δ actual δ Limiting < ∴ ok & Safe Provide an ISLB 500 @ 75 kg/m beam. …………………………………………..Ans

33. Design of laterally supported beams Example 2 Design a laterally supported simply supported beam of 7 m effective span. It carries a load 0f 250 kN which is uniformly distributed load over the whole span. In addition the beam carries a point load of 100 kN at mid spa n . SPPU - D ec . 2010 , 1 ma r ks Solution : Given : UDL = 250 kN, UDL = 250 = 35.71 kN/m 7 Point load at centre = 100 kN 1. Determine maximum factored shear force ‘V’ and bending moment ‘M’ Maximum Shear force V = Reaction = 1.5 [ 35.71× 7 + 100 ] = 262.48 kN 2 2 Maximum Bending Moment 2 M = 1.5 [ 35.71 × 7 8 4 + 100 × 7 ] = 590.81 kN-m 2. Selection of cross section p Z required = M.γ m0 f y = 590.81 ×10 6 ×1.10 33 250 = 2598.55 × 10 3 mm 3 Select an ISLB 600 @ 99.5 kg/m (Using annex ‘H’ IS 800:2007) d= 600 mm, b f = 210 mm, t f = 15.5 mm, t w = 10.5 mm, Z p = 2798 . 5 6 × 10 3 mm 3 > Z p r equ i r e d , Z e = 2428 . 9 × 10 3 mm 3 I zz = 72867.6 × 10 4 mm 4 ,

34. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 21 / 2 15.5 = 6.77 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 520 10.5 = 49.52 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 34

35. 4. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n γ m0 A v f y √ 3 γ m V d = = > V = A v f y √ 3 γ m = 6 × 10 .5 × 250 √3 × 1.10 = 826.66 kN > V = 262.48 kN-m ∴ ok & Safe 35

36. 5. Design bending strength (M d ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007 i. If V 𝑘 . 6 V d or V < . 6 V d 0.6 V d = 495.99 kN ∴ V < 0.6 V d ∴ M d = . β b .fy Z P γ m M d = β b .fy Z P . γ m0 = 1×2798.56 ×10 3 × 250 1 . 10 = 636.04 kN –m > M = 590.81 kN-m ∴ ok & Safe 36

37. 6 . Checks a) Web buckling f wb = (b 1 + n 1 ) t w f cd ≥ Reaction, R Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007 f w d = [( b 1 + 2 n 1 ) t w f c d ] ≥ Concent r ated lo a d, P Where f cd = is calculated using slenderness ratio 0.7 d r 𝑦 = 2.425 × d t 𝑤 10 . 5 = 2.425 × 600 =138.57 B u ckl in g Cl a ss ‘c’ By interpolation cd f = 74.3 − 74.3−66.2 × (138.57 − 130 ) 140−130 = 67.35 N/mm 2 f cd At Rea c t ion ‘R’ ∴ f wb = (210 + 300) × 10.5 × 67.35 = 361.46 kN > R = 262.48 kN ∴ ok & Safe 37

38. b) Web Crippling A t concent r ated lo a d ‘W’ Ref. cl no 8.7.4 / PN 67 / IS 800:2007 F wc = (b 1 + 2 n 2 ) t w f yw γ m0 = (210 +2 ×2.5 ×15.5 × 10.5 ×250 1.10 > R = 262.48 kN = 593.60 kN c) Deflection Limit ∴ ok & Safe Re f . T a ble no 6 / PN 3 1 / IS 800 : 2007 δ actual = WL 3 + 5 WL 4 48 EI 348 EI = 12.56 mm = span = 7000 δ L imi t ing 300 300 38 = 23.33 mm ∴ δ actual δ Limiting < ∴ ok & Safe Provide an ISLB 600 @ 99.5 kg/m beam. …………………………………………..Ans

39. Design of laterally supported beams Example 3 An ISLB 600 @ 99.5 kg/m has been used a simply supported beam over 7.2 m span. Determine the safe uniformly distributed load ‘W’ so that the beam can carry in flexure. Assuming compressive flange is restrained throughout the span against lateral buckling and Fe 410 steel. SPP U - Dec. 2010, 10 m a r ks Solution : 1. Sectional properties 39 ISLB 600 @ 99.5 kg/m (Us ing ann e x ‘H’ IS 800 : 2007 ) d= 600 mm, b f = 210 mm, t f = 15.5 mm, Z p = 2798.56 × 10 3 mm 3 , t w = 10.5 mm, Z e = 2428.9 × 10 3 mm 3 , h 1 = 520.2 mm

40. 2. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 21 / 2 14.1 = 6.77 < 9.4 ∈ ∴ Flange is plastic W e b = b t w = h 1 t w = 520.2 10.5 = 49.54 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 40

41. 3. Maximum Bending Moment M (factored) M = WL 2 = 8 1.5 × W × 7200 2 8 = 9 . 7 2 W 4. Bending Strength of section M d = P y . γ m0 = β b . Z f 1× 2798.56 ×10 3 × 250 1 . 10 = 636.03 kN-m 5 . U DL ‘W’ equ ating M d and f acto r e d ‘M’ 9.72 W = 636.03 kN-m W = 65.43 kN/m Now, self weight of section = 99.5 × 9.81 41 1 = 0.976 kN/m Henc e , U DL W = 65 . 4 3 – . 97 6 = 64 . 4 5 kN /m ∴ W = 64.45 kN/m ………………………Ans

42. Design of laterally supported beams Example 4 Design a laterally supported beam of effective span 6 m for the following data : Maximum bending moment M = 150 kN-m Maximum Shear force V = 210 kN. Grade of steel Fe 410 Solution : Given : SPP U - De c . 2011 , 1 3 m a r ks , M a y 2014 , 1 5 m a r k s . 1. Determine maximum factored shear force ‘V’ and bending moment ‘M’ Maximum Shear force = 210 kN Maximum Bending Moment = 150 kN-m 2. Selection of cross section p Z required = M.γ m0 f y = 150 ×10 6 ×1.10 42 250 = 660 × 10 3 mm 3 Select an ISLB 350 @ 49.5 kg/m (Using annex ‘H’ IS 800:2007) d= 350 mm, b f = 165 mm, t f = 11.4 mm, t w = 7.4 mm, h 1 = 288.3 mm , Z p = 851.11 × 10 3 mm 3 > Z p required.

43. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 165 /2 11.4 = 7.23 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 288.3 7.4 = 38.95 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 43

44. 4. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n γ m0 A v f y √ 3 γ m V d = = > V = A v f y √ 3 γ m = 35 × 7 .4 × 250 √3 × 1.10 = 339.84 kN > V = 262.48 kN-m ∴ ok & Safe 44

45. 5. Design bending strength (M d ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007 i. If V 𝑘 . 6 V d or V < . 6 V d 0.6 V d = 203.9 kN ∴ V > 0.6 V d High Shear M d = M dv 2 β = ( - 1) = ( 2V 2 ×210 V d 339.84 2 - 1) = 0.055 β b .ZP f y . γ m0 M d = = 1× 851.11 ×10 3 × 250 1 . 10 = 193.43 kN-m Z fd = Z p - w A .d 4 = 85 1.11 × 10 3 - 7.4×288.3×350 4 = 664.43 × 10 3 β b .Zfd f y . γ m0 ∴ M fd = = 1 ×664.43 ×10 3 ×250 1 . 10 = 151 kN-m ∴ M dv = M d - β (M d - M fd ) = 193.43 - 0.055 (193.43 - 151) = 15 kN - m ≥ M = 15 kN - m ………………… … .. Ans 45

46. Laterally Unsupported Beam :- Q . Ex pl a in l a te r a l l y U nsuppo r ted bea m wi th suita b le s k etch . De c . 2 1 6 , M a y 2 1 7 . Beam subjected to BM develop compressive and tensile force 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 beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of inertial and its presence reduces the plastic moment capacity of the section. The value of M cr can be calculated using the equation given in cl. 8.2.2.1 / PN 54 / IS 800 : 2007 The design of bending compressive strength can be calculated using a set of equation as specified in cl 8.2.2 (Table 13 a and 13 b / PN 54 to 57 / IS 800:2007. The design of laterally unsupported consist of selecting a section based on the plastic sectional modulus and checking for its shear capacity & deflection. 46

47. Modes of Failure of Beam :- Q . Ex pl a in mo d e s o f f a i lu r e o f bea m with suita b le s k etches . De c . 2 1 6 . Bending Failure : Bending failure may be done due to fracture of tension flange and due to crashing of compression flange. Shear failure : It occurs when buckling of web of beam near location of high shear failure. T o rsi ona l b u c kli n g : D u e t o combin e d mo m e n t on the bea m b u ckle ab out b o th a x i s c a l led t o rsi ona l buckling. Lateral buckling : due to large span of beams deflection takes place beyond limits. Local failure of Web : due to heavy shear in some regions of beam web fails in crippling or buckling. 47

48. Design steps for laterally Unsupported beams Determine maximum factored shear force ‘V’ and bending moment ‘M’ for a given loading and support condition Selection of cross section Find p Z required = M d β b . f bd Take β b = 1 for assuming plastic or compact section Assume, f cd = 120 to 140 N/mm 2 for I section Using Annex H IS 800 : 2007 plastic properties of section. Select required section. Or e Z required = Z p 48 shape factor (s) Shape factor I and channel section – 1.11 to 1.15 Rectangular section – 1.5 Circular section – 1.69 Angle and T – section – 1.8 Square or diamond section – 2 ∴ Z p provided > Z p required

49. 3. Classification of Section Re f . T a ble no 2 / PN 1 8 / IS 80 : 2007 Using Table no 2 / PN 18 / IS 800 : 2007 4. Effective length of beam Re f . 8 . 3 . 1 / T a ble No 1 5 / PN 5 8 / IS 80 : 2007 Depending upon support condition, L LT calculated using Table – 15. 5. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n γ m0 A v f y 49 √ 3 γ m V d = = > V ∴ ok & Safe 6. Design bending strength (M d ) M d = Z p . β b . f bd Where, Z p = plastic sectional moduli f bd = design bending compressive stress M d > M ∴ ok & Safe Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007

50. 6. Design bending strength (M d ) Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007 7 . Check : De flection Limi t 50 Re f . T a ble no 6 / PN 3 1 / IS 800 : 2007 < δ actual δ Limiting ∴ ok & Safe

51. Examples on Laterally Unsupported Beams Example 1 Design a suitable I section for and simply supported beam of span 5 m carrying a dead load of 20 kN/m and imposed load of 40 kN/m. The beam is laterally unsupported throughout the span. Take f y = 250 MPa. SPPU- Dec. 2010, 15 Marks Solution : Factored Load = 1.5 × (20 + 40) = 90 kN/m Determine maximum factored shear force ‘V’ and bending moment ‘M’ Maximum shear force : V = WL = 90 × 5000 = 225 kN 2 2 Maximum bending force: 8 2 2 2 M = WL = 90 × 5000 = 281.25 kN-m 2. Selection of cross section Assuming f bd = 130 N/mm 2 p Z required = M d β b . f bd 6 = 281.25 ×10 130 51 = 2163.46 × 10 3 mm 3 Select ISWB 500 @ 95.2 kg/m d= 500 mm, b f = 250 mm, t f = 14.7 mm, Z p = 2351.35 × 10 3 mm 3 > Z p required, t w = 9.9 mm, r min = 49.6 mm, = 52290 . 9 × 10 4 m m 4 I xx

52. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 250 /2 14.7 = 8.50 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 431 9.9 = 43.53 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 52

53. 4. Effective length of beam Effective length KL = 5000 mm. Re f . 8 . 3 . 1 / T a ble No 1 5 / PN 5 8 / IS 80 : 2007 As end are restrained against torsion but compression flange is laterally unsupported. 5. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n V d = = γ m √ 3 γ A v f y m0 = 649 . 5 1 kN > V = 22 5 kN = 500×9.9×250 √3 ×1.1 ∴ ok & Safe 6. Design bending strength (M d ) M d = Z p . β b . f bd Where, Z p = plastic sectional moduli f bd = design bending compressive stress Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007 KL = 5000 r min 4 9 . 6 500 53 14 . 7 = 100.86, h = = 34.01 t f

54. From Table no14 / PN 57 / IS 800 :2007 F or kL r min = 100.86, h t 𝐹 = 34.01 By interpolation 35 − 30 A = 270.9 − 270.9 −257.7 × (34.01 − 30 ) = 260.31 N/mm 2 35 − 30 B = 231.1 − 231.1 −219.3 × (34.01 − 30 ) c r , b f = 260.31 − 260.31 −221.56 × (100.86 − 100 ) = 221.56 N/mm 2 = 256.97 N/mm 2 110−100 = 256.97 N/mm 2 f c r , b 54

55. Now, for rolled section α LT = 0.21 f bd = ? Cl N o . 8. 2. 2 / P N 5 4 / IS 8 0:2 07 Ref. Table no 13a / PN 55 / IS 800:2007 By interpolation bd f = 152 +( 300 − 250 163.6 −152.3 ) × (256.57 − 250 ) f bd = 153.48 N/mm 2 M d = Z p . β b . f bd = 1 × 2351.35 × 10 3 × 153.48 = 360.88 kN-m > M = 281.25 ∴ ok & Safe 55

56. 7 ) Check : De flection Limi t Re f . T a ble no 6 / PN 3 1 / IS 800 : 2007 δ actual 5 W L 4 = 384 EI = 5 ×60 ×5000 4 384× 2×10 5 × 52290.9 ×10 4 = 4.66 mm δ Limiting = sp a n 300 = 5000 300 56 = 16.67 mm < δ actual δ Limiting ∴ ok & Safe Hence, Provide ISWB 500 @ 95.2 kg/m. ……………………………………………Ans.

57. Examples on Laterally Unsupported Beams Example 2 A floor beam in a building has a span of 6 m. it is simply supported over supports and carries a uniformly distributed load of 40 kN/m, inclusive of self weight. Design the beam if the compression flange is unrestrained throughout the span against lateral buckling and Fe 410 steel. SPPU- Dec. 2011, 15 Mark Solution : Factored Load = 1.5 × 40 = 60 kN/m Determine maximum factored shear force ‘V’ and bending moment ‘M’ Maximum shear force : V = WL = 60 × 6000 = 180 kN 2 2 Maximum bending force: 8 2 2 2 M = WL = 60 ×6000 = 270 kN-m 2. Selection of cross section Assuming f bd = 130 N/mm 2 p Z required = M d β b . f bd 6 = 281.25 ×10 130 57 = 2163.46 × 10 3 mm 3 Select ISWB 500 @ 95.2 kg/m d= 500 mm, b f = 250 mm, t f = 14.7 mm, Z p = 2351.35 × 10 3 mm 3 > Z p required, t w = 9.9 mm, r min = 49.6 mm, = 52290 . 9 × 10 4 m m 4 I xx

58. 3. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 250 /2 14.7 = 8.50 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 431 9.9 = 43.53 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 58

59. 4. Effective length of beam Effective length KL = 6000 mm. Re f . 8 . 3 . 1 / T a ble No 1 5 / PN 5 8 / IS 80 : 2007 As end are restrained against torsion but compression flange is laterally unsupported. 5. Design Shear Strength ( v d ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007 V n V d = = γ m √ 3 γ A v f y m0 = 649 . 5 1 kN > V = 22 5 kN = 500×9.9×250 √3 ×1.1 ∴ ok & Safe 6. Design bending strength (M d ) M d = Z p . β b . f bd Where, Z p = plastic sectional moduli f bd = design bending compressive stress Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007 KL = 6000 r min 4 9 . 6 500 59 14 . 7 = 120.96, h = = 34.01 t f

60. From Table no14 / PN 57 / IS 800 :2007 F or kL r min = 120.96 , h t 𝐹 = 34.01 By interpolation A = 20 2.4 − 35 − 30 202.4 −190.1 × (34.01 − 30 ) = 192.53 N/mm 2 35 − 30 B = 179.0 − 179.0 −167.1 × (34.01 − 30 ) c r , b f = 192.53 − 192.53 −169.45 × (120.96 − 120 ) = 169.45 N/mm 2 = 190.31 N/mm 2 130−120 = 190.31 N/mm 2 f c r , b 60

61. Now, for rolled section α LT = 0.21 f bd = ? Cl N o . 8. 2. 2 / P N 5 4 / IS 8 0:2 07 Ref. Table no 13a / PN 55 / IS 800:2007 By interpolation bd f = 106.8 +( 134.1 −106.8 ) × (190.31 − 150 ) 200−150 128.80 N/mm 2 f bd = M d = Z p . β b . f bd = 1 × 2351.35 × 10 3 × 128.80 = 302.85 kN-m > M = 281.25 ∴ ok & Safe 61

62. 7 ) Ch ec k : De flection Limi t R e f . T a ble no 6 / PN 3 1 / IS 800 : 2007 δ actual = 5 W L 4 384 EI = 5 ×40 ×6000 4 384× 2×10 5 × 52290.9 ×10 4 = 6.45 mm δ Limiting = sp a n 300 = 6000 300 62 = 20 mm < δ actual δ Limiting ∴ ok & Safe Hence, Provide ISWB 500 @ 95.2 kg/m. ……………………………………………Ans.

63. Examples on Laterally Unsupported Beams Example 3 Determine the safe uniformly load that the beam ISLB 600 @ 99.5 kg/m has been used as a simply supported over 7.2 m span. The compression flange of beam is not restrained against lateral buckling. At the ends beam is fully restrained in torsion but both the flange are free to warp at the ends. SPPU- Dec. 2012, 15 Marks, May 2016, Dec. 2016, 16 Marks Solution : 1. Properties of ISLB 600 @ 99.5 kg/m A = 126.69 × 10 2 mm 2 , D = 600 mm, 63 b f = 210 mm, t w = 10.5 mm, h 1 = 520.2 mm, Z e = 2428.9 × 10 3 mm 3 t f = 15.5 mm, r min = 37.9 mm, Z p = 2798.56 × 10 3 mm 3 , I xx = 52290 . 9 × 10 4 mm 4

64. 2. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 210 /2 15.5 = 6.77 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 520.2 10.5 = 49.54 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 64

65. 3. Design bending strength (M d ) M d = Z p . β b . f bd From Table no14 / PN 57 / IS 800 :2007 Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007 = 189.97, 37.9 t f KL = 7200 h = 600 r min 15 . 5 = 38.71 By interpolation 40 − 35 A = 1 02.2 − 102.2 −95.2 × (38.71 − 35 ) = 97.01 N/mm 2 40 − 35 B = 94.6 − 94.6 −87.8 × (38.71 − 35 ) = 89.50 N/mm 2 = 89.52 N/mm 2 c r , b f = 99.62 − 99.62 −92.15 × (189.97 − 180 ) 190−180 = 89.52 N/mm 2 f c r , b 65

66. Now, for rolled section α LT = 0.21 for f cr, b = 92.17 N/mm 2 f bd = ? Cl N o . 8. 2. 2 / P N 5 4 / IS 8 0:2 07 Ref. Table no 13a / PN 55 / IS 800:2007 By interpolation bd f = 63.6 +( 70.5 −63.6 ) × (89.50 − 80 ) 90−80 70.15 N/mm 2 f bd = M d = Z p . β b . f bd = 1 × 2798.56 × 10 3 × 70.15 = 196.32 kN-m 66

67. 4) Safe uniformly distributed load (w) M = M d 8 wL 2 = 196.32 w = 30.30 1 . 5 67 Safe UDL w = 30.30 = 20.2 kN/m Including self weight. w = 20.2 kN ……………………………………………Ans.

68. Examples on Laterally Unsupported Beams Example 4 A simply supported beam of effective spam 8 m carries uniformly distributed load w kN/m throughout the span. The compression flange is laterally unsupported throughout the span. Determine intensity of uniformly distributed load ‘w’ so that ISMB 400 @ 61.6 kg/m provided for beam can carry safely. SPPU- May 2017, 10 Marks Solution : 1. Properties of ISMB 400 @ 61.6 kg/m A = 7846 mm 2 , d= 400 mm, 68 b f = 140 mm, t w = 8.9 mm, h 1 = 334.4 mm, t f = 16 mm, r min = 28.2 mm, Z p = 1176.18 × 10 3 mm 3 ,

69. 2. Classification of Section Re f . T a b le no 2 / P N 1 8 / I S 80 : 2007 Flange = b = b f /2 t f t f = 140 /2 15.5 = 4.375 < 9.4 ∈ ∴ Flange is plastic W e b = d t w = h 1 t w = 334.4 8.9 = 37.57 < 84 ∈ ∴ Web is plastic. ∴ Section is Plastic 69

70. 3. Design bending strength (M d ) M d = Z p . β b . f bd From Table no14 / PN 57 / IS 800 :2007 Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007 = 283.68, KL = 8000 h = 400 r min 28.2 t f 16 = 25. By interpolation c r , b 290−280 f = 74.7 − 74.7 −71.8 × (283.68 − 280 ) = 73.63 N/mm 2 f c r , b = 73..63 N/mm 2 70

71. Now, for rolled section α LT = 0.21 for f cr, b = 73.63 N/mm 2 f bd = ? Cl N o . 8. 2. 2 / P N 5 4 / IS 8 0:2 07 Ref. Table no 13a / PN 55 / IS 800:2007 By interpolation bd f = 56.8 +( 63..6 −56.8 ) × (73.63 − 70 ) 80−70 59.27 N/mm 2 f bd = M d = Z p . β b . f bd = 1 × 1176.18 × 10 3 × 59.27 = 69.71 kN-m 71

72. 4) Safe uniformly distributed load (w) M = M d 8 wL 2 = 69.71 w = 8.71 1 . 5 72 Safe UDL w = 8.71 = 5.80 kN/m Including self weight. w = 5.80 kN ……………………………………………Ans.

73. 73