__ Board Exam GUIDE inUnit 1: STEEL BEAMS Chapler qa grandest, and the * Look for the highest, and the highest you shall see! Go for the shall bel ~ Ralph Waldo Emerson grandest you ALLOWABLE FLEXURAL STRESS (NSCP Section 506) The NSCP requirements in solving the allowable flexural stress. are simplified in the following table and chart. by the author for easier understanding and memorization of the code LATERALLY | LATERALLY | LATERALLY | SHORT INTERMEDIATE SUPPORTED-| SUPPORTED- | SUPPORTED-| BEAM! BEAM! COMPACT SEMI NON- ADEQUATELY} SECTION COMPACT COMPACT SUPPORTED SECTION ‘SECTION 8 8 e@ @ 8 CONDITION: i Lose and | LysLeand | LysLeand | LesLesbs | LusLoand 2. <8 and Ww 170 «bu < 250 ar ALLOWABLE FLEXURAL STRESS: Fo= O66F, | Fo= [079 - 0.00076% (bat) Fy IFr Check for Flange Compaction Compare bu. ci with 170 and 250 cs Sws Fo=0.6Fy ‘Check for Web Compaction Compare = wath 8 ni s( 2) <3 F=0.6F Fret)? HO.S5xt0°Cs]Fy ; And Fase = BITADAVCy td Check for | Check for Lateral Slenderness Suppoit Ratio Compare Ls | Compare with Le and Ly 2 ue (2) with n ‘SR: and SRz LONG BEAM/ LATERALLY UNSUPPORTED Fe = 82740A;Cp Uyd J ‘STEEL DESIGN CamScanner3516330C), aa SR2 y — ty z d me ptt a ‘ 3 “J ri Fu=o Fy Fog = f ; [0.79 ree A or O.66Fy | 9.90076 Foo=0.6Fy | FoiO.6Fy | 3i9 = © (lt) Fy IFy ~Fyltsry)® | s1721000, 140.55x10°Cs]Fy, (Lin? ee , = 200b gR2 - 35163300, fh 5 137900A, 703270C, t, RRMA ISR, = 270C 1-3.74 ze | Fyd Fy Note: 1. Inregion 5, Allowable bending stress Fy is the bigger of Fas and Foes = SOAs Cv 2. In region 6, Allowable bending stress Fy is the bigger of Fxs and Fees = STAC Cv 3. In computing Sw, falFymax= 0.16 r= area of one flange = bit "r= radius of gyration of the compression flange + 1/3 of compression web bend about the minor axis . ‘CamScannerin finding allowable flexural stress of steel shapes The flowchart below shows the proce: ALLOWABLE FLEXURAL STRESS (NSCP Section 506) ‘a test for unstiffened elements only Note: This section i {for doubs (compact) I tetas this es {coneiton, .66F, | >\F> = 075 Fy (along! |minor _axis)unstifened ‘compression flange sen - corr te STEEL DESIGN CamScannerstructural Design and Construction Review Manual by Perfocto B, Pada Jt {for compression flange loded in the plane of {het wel» nul having ap ass of syrmnnety nthe plane ofthe wel Cre vert ple) jms» lager moment |H6 Mend moments [ (=) forsingle ieurvatere bending | But when bending within ptt & pt 2 is greater thin Mor Ms | Gt | Ir = radius of gyration = “t of compression web area of compression flange 4 82740C, and ————® but smatier Lydia, f than O.6F, y No 1172 100C, (2J bg 82740Cg Fy =the bigger of > gg ( q)~ but smaller than 0.6Fy (S39 CamScannerOTHER AISC REQUIREMENTS: —_| T. Tension members F, = 0.6 Fy except at pinholes F, = 0.45 Fy (for pinholes) where: Fy = minimum yield point of tension member IT. Safe load that tension members could carry: P= Aotocive Fe Actoctve = Smaller of Aner & O.B5Agass } x (hole diameter +3 mm) ickness of plate) ross width x plate thickness £85A, is only applied for tension | with holes. When under compression, Asis taken to be fully © effective. III. Slenderness Ratio Slendemess ratio is investigated fo assure members will be effective in resisting tension as it will not vibrate or deflect excessively. IL =<240 for main members r £ £300 for secoridary members where L = unsupported length least radius of gyration IV. Shear Stress (Vertical & Horizontal) F<0.4 F, MPa | *Please see problem #6 for complete data where \ctual vertical shearing stress {n= actual horizontal shearing stress \V= Vertical shear at the vertical section d= depth of member tw = thickness of web R = Resultant horizontal shear force acting on tributary horizontal length of beam. s. = taken as the rivet spacing Q = Ay = statical moment of area of tha vertical section above (or below the fiber considered about the neutral axis. 1 = moment of inertia of the section about the neutral axis Ay= sheared area V. Torsion Torsion in flexural members exist when a flexural load is applied eccentrically with respect to the shear center. ‘STEEL DESIGN CamScanner\ ‘Structural Dest Construction Review Manual by Perfecto 8, Padilla Jt d= depth of member te = thickness of web by = width of flange CODE PROVISIONS: Me Vaby fy aE + Ts Fcouane) | iy not For value of Fyaiowase) , See flowchart on page 7 /, = shear at x distance from the shear center of torsional stress _ Vet. J thickriess. of element fy tm where torsional shearing stress is desired 3 = x : for webs & some flanges Structural Steel Designation 1, Wide flange (WW) Example: Wooo «27060 {920 = approximate depth ( mm ) 270.80 = weight of beam ( kgim ) 2. Channel (Cy Example: Cisox 1930 150 = approximate depth (mm ) 19.30 = weight (kglm) 3. Angle Bars (2) Example: 2229x102 x19 229 = longer leg (mm) 4102 = shorter leg ( mm) 19 = leg thickness (mm) for flange of channel SITUATION a CE Brd Exam-Nov 2006 (S39 CamScannerSITUATION 1: CE Brd Exam-Nov 2006 A simply supported beam with a span of 7.8 meters is subjected to a uniform vertical load of 52 kN/m, acting on the plane of the minor axis of the beam section. The beam is restrained against lateral buckling at the top and bottom flanges for the entire span, The material is A36 steel with Fy =248 MPa and modulus of elasticity of 200 GPa, The allowable flexural stress for laterally braced compact sections is 0.66Fy. Allowable shear stress is 0.40Fy and allowable deflection is 1/360 of the span. Assume the section is compact. Properties of steel section: Moment of inertia, lx = 0.00055 m*, total depth, d = 533 mm, and thickness of web, {, = 10 mm, bieam's wt = 90 kgm 1. Compute the maximum bending stress. a. 194.87 b, 154,32 c. 20040 » 4. 240.60 2. Compute the maximum deflection: a. 23.17 b, 15.60 ©. 1987 d. 25.70 3. Check the adequacy of the section: a. the section, adequate b. the section, inadequate in flexure and deflection c. the section, inadequate in flexure, deflection and shear d. the section, inadequate in shear and flexure, SOLUTION 1 - 3: Maximum bending stress: Mc fee w= 52 + 90(9.81/1000) = 52.883 kN/m wl? _ 52.883(7. 8) 8 8 M = 402.175 kN.m R = 52,883(7.8/2) = 206.24 KN | 0.00055 m* = 550 x 10° mm C = di2 = 533/2 = 266.5 mm fp = £02.175(10) (266.5) = 194,87 MPa 550x10° Moxinnum dofloction S5wi4 5(62.883\7800)' 384(200000}550x10° chock the adequacy Allow bending stress = 0.66Fy = 163,68MPa ‘Allow shear stress = 0.40Fy = 99.2 MPa Allow deflection = L/360 = 21.67 mm Actual shear stress = Vidty, Actual shear stress = 206240/533(10) Actual shear stress = 38.69 MPa Thus, the beam section, inadequate in flexure and deflection SITUATION I: CE Brd Exam-May 2000 A steel beam 8 m long is simply supported and carries a uniformly distributed load of 40 kNim (including the weight of the beam) acting on the plane of minor axis: Assume the beam is compact and braced against lateral buckling throughout the whole span. Allowable bending stress / for compact section is 0.66Fy. Allowable deflection is 1/360 of span. Modulus of elasticity E = 200 GPa and Fy = 248 MPa section 1m’) depth(m) W24x55 | 0.0005619 | 0.599 w2ix57 | 0.000480 | 0.535 W21x62 _ | 0.0005536 | 0.533 wiex71__[ 0.0004870 | 0.469 4, Which of the follwing most nearly gives the required section modulus so that allowable bending stress will not be exceeded? a. 0.000124m? —b._0.121432 m? ¢. 0.006321m* —d._0,001955 m? 5. Which of the following gives the required moment of inertia so that allowable deflection will not be exceeded? a.0.12143m‘ —b. 0.00048 m* .©.0.00715 m* d. 0.00012 m* ‘STEEL DESIGN 7 | ‘CamScannerructurl Design and Construction Review Manual by Perfecto B. Padilla Jr Al 6. Which of the following gives the most economical section? a W245 cc W21x62, b. Watxs7 a Wiex71 I 8m I SOLUTION #4: Value of required section modulus: ovation 0.06F, > ct) M M = 320 kN.m 320x10° Sreqa Sreaiires 2 1955.03 x 10° mm? ‘Srequres 2 0.001955 m? 0.66(248) > 360 ~ 384EI 8000 , 5(40{8000)* 360 ~ 384(200000) | requves 2 480 x 10° mm* Trequred 2 0.00048 m* SOLUTION #6: Most economical section: Compare actual bending stress. foratowasie) = 0.66(248) foyotowabie) = 163.68 MPa Using W24 x 55: (= MC _ (g20xt0® |599/2) eel. Be1s0xi0® {y= 170.56 MPa > 163.68 MPa (failed) - and. allowable | ng W21x 57 0x10 }535,2) (aaixt0°) {= 177.06 MPa> 163.68 MPa (tailed) Using W2tx62: __ 820x108 }533/2) (623.60x10°) 20x10°}633/2) (6s, 6Oxt 0°) 154.05 MPa < 163.68 MPa (safe!) Using W18 x71 4, ~ 820x108 [aso/2) fy \487x10° fo = 154.09 MPa <163.68 MPa (safe!) Thus: ‘The most safe and economical section is W21x 62 beam ANOTHER SOLUTION: Yactual $ Yatow Bolte Lb 384EI 360 5(40000)(8)* _ 8 384(200x10°)| 360 1 0.00048 m* Considering |, all sections qualify Considering S 4 Cc diz _ 0.005619 1 0.599/2 0004810 0535/2 _0.0005536 S3°9 533/2 s, = 2.0004870 4 0.489/2 .001876 Sreqa =0,002077 >Sreaa W18x71 & W 21x62 qually but W21x62 is lighter thus = It is more economical section. CamScannerSITUATION III: CE Brd Exam-May 2010 Beams 3m long, spaced every 2m, supports a load P at midspan, DL= 5 kW/m (including beam weight, LL= 7.2 kN/m throughout its length, Said beams are supported in turn by an &m simply supported steel girder (at the latter's quarter points) and a hinge on the other end Properties of Steel Girder: Fy = 248 MP. w = 167.4 kalm A = 21300 mm d = 289 mm by = 265 mm t= 31.8 mm ty = 19.2 mm Axis x - x I 00 10° mm* S = 2.08 x 10° mm' r= 119mm Axis y - y 1 = 9.88 x 107 mm* S= 7.46 x 105 mmé r= 68.1 mm 7. Determine the maximum force P without exceeding the allowable shear stress a. 567 KN b. 421 kN c. 689 KN d. 464 kN 8. Determine the maximum force P without exceeding the allowable flexural stress in the girder of 0.66F, a. 86.71 KN b. 109.28 KN c. 119.63 KN d. 126.86 kN 9. If the girder is pre-cambered to its allowable deflection of 360, find the additional uniform live load that may be applied onto it to offset said camber before the application of load from superimposed beams. a. 20.31 kNim ©. 22 kNim b. 23.36 kNim d. 29.16 KN/m SOLUTION 7 - 9: | Consider reaction on 3m beam PULP 3 Fee iwe=Fs1223 22a 2 au F=ip+183+0 Consider the Girder- w= 167.4 kgim (9.81) R=34+0.75P By symmetry R= 1.64(4) + 4 GF) = 6.56 +1.5F Subst © R=656+15($P +18.3)=34 +075? 3 @ To be sale in shear Fyjactual) S Fy (atow) Vy ae < 0.41 ce (34+ 0.75P)x 10° 289(19.2) P< 689 kN To be safe in flexure: foractua $ Fo (stow) M 5 0.86Fy <0.4(248) M Zoe x 108 < 0:98(248) Ms 340 kNmi S| CamScannerStructural Des Cm + M= RQ) — WS -F(2) $340 kN (3440.75P)4 — “ 64) ~caPa18 3)2 5340 Ps 126.86 kN —— Pre-cambered am ‘Beam —— SESS. To be safe in deflection: Yeectual) S Y{atiow) 5 wil 384 El 360 5 __w(8000)* __ 8000 384 200000(3x 10") 360 wes 25 kim Wor + Wi < 25 kim 1.64 + wi < 25 kNim Wut $23.36 kim PROBLEM 10: For the beani with the following properties: W250 x 167 A 36 steel beam = 167.4kg./m A__=21 300 mm’ d br te tw Axis x—x 1 = 300x 10°mm* S=2080x10'mm ——_—i| r Axis y—y 7 1 =98.8x10°mm* = 746 x 10 rt =68.1 Find the allowable bending stress if the unsupported length is 3m. @. 163.68 MPa b. 153.68 MPa ©. 17368MPa — d. 143.68 MPa and Construction Review Man Perfecto B. Padilla Jr SOLUTION : step I: check for lateral support = 200 (265) 248 L, = 3365.5mm Ly = 137900Af 1° Eg 137900(265)31.8) bos "7481289 Lu= 16 214mm Ly = 3m Ly -» laterally unsupported. = 6214.mm STEEL DESIGN WAM CamScannerP) Structural Design and Construction Review Manual by Perfecto B, Padila Jt | . | 2 M, M, Cy 21.75 + 1.05 | Mt ]+0.3/ we 175+ 1.05 (M)0a(M) 2s Since no end moments given, assume simply supported member, thus, Cp = 1 for simply supported members) V2 ne ee byt, + dly "6 98.8x10°/2 265 (31. 8+ 4 289}1922) = 72.68 ty _ 18,000 t 72.68 703270 C, | LaTAy f (506-6) (606-8) t In foregoing: | L = distance between cross section braced against twist or lateral displacement of the compression flange in mm. For cantilevers braced against twist only at the support, L may be conservatively be taken as the .ctual length. ; i s reds of gyration of a section Comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plane of the web in mm ‘Ay = area of the compression flange in mm? 1.75 + 1.05(Ms/Ma) + 0.30(My/Mz2)? but not more than 2.3, where My is the smaller and Mp is the larger bending moment at the ends of the unbraced length taken about the strong axis of the member where (MMi/Mz) is the ratio of the end moments, is Positive when M; and Mz have the same sign’ (reverse curvature bending) and negative when they are opposite sign(single curvature bending).When the bending moment at any point’ within an unbraced length is larger than at both ends of this length , the value Cp shall taken as unity. A simply supported steel beam 8 m long is subjected to a counterclockwise couple M at left end and 25% of M applied at right end in clockwise direction Propertios of W21 x 62 F, = 250 MPa 53m 533m 0.210m 016 m. ‘S,= 0.00207 m* 49. Which of the following most nearly gives the min, slenderness ratio for which the beam will be considered long. a. 140.62 b. 150.94 c. 169.06 d. 130.21 20. Which of the following most nearly gives the allowable flexural stress using the code a. 115.12MPa b, 150MPa c. 169.06MPa d. 132.77MPa 21. Which of the following most nearly gives the value of the couple M. a. 310.50 kN.m b. 281.12 KN.m c. 274,.83KN.m d. 110.12 kN.m Note: The problem is updated for the SCP 2001 code ‘STEEL DESIGN CamScannerStiuctural Design and Construction Review Manual by Perfecto B. Pacilla Jt SOLUTION 19: Lo 8 f 0.053 SOLUTION 20: p= 1.75 + 1.05(My/M,) + 0.30(M/Ma? Cy =1.75 + 1.05(0.25MM) + 0,30(0.25\m4? Cp = 2.03 « less than 23. Thus, take Cy= 203 703270C, _ [703270(2.03) JE = | EE) 75.57 yo 250 35163300, _ (5516390(2.03) | ee =168.98 Note: /708270C, -L . /35163906, 5, 5 Fy 0.60F, = 0.6 (250) = 150 g,= 52740C, __82,740(2.03) LavA, ~ 8000(633)/(210)16) Fo = 132.35 MPa fs [s-itt 3 10.55x 10°C, 2 _ 250(150.94 Fo= | -—A has . E 70.55. 10°(2.03) Fi = 159.65 MPa when [103270C, L . [35163300, YB use the bigger of the last two values but smaller than 0.6F, Thus: Fy = 150 MPa SOLUTION 21: To be safe in bending fo acualS Fo atovale Frama Wi0.00207 m? = 150 x 10° Nim? M= 310500 Nm M= 310.5 kNm ATION NSP 2001 Section 506.5 says: For hily < 998 / fF, , on the overall dopth times the web thickness, the allowable shear stress is 0.4F For hit, > 998 / Fe , the allowable shear stress on the clear distance between the flanges times the web thickness is Fu= Cy F/2.89 Where: y= 210264Ky hen Cy is less than 0.8 Fy(hit,)? aft when Gy is more than 0.8 ky= 4.004 aig when alh is less than 1 a 4 = when afh is more than (alhy ky = 5.34 + ickness of web, mm lear distance between traverse stiffeners; mm = clear distance between flanges at the section under investigation 22. Find the allowable shearing stress on effective web of an | - section with the following properties: d=970 mm ty = 15.2mm t= 40mm b= 204mm Fy = 250 MPa Distance between traverse web-stiffeners = Am a, 139.41Mpa 129.41 Mpa c. 149.41 Mpa d, 159.41 MPa 23. Find the maximum allowable shear force that the said section can handle. a. 1868.7 KN b. 1968.7 kN c. 1768.7KN — d. 1468.7 kN SOLUTION 22: billy = 970/1 3.81 198 / 250 (S39 CamScannertw > 998/ JF, alh = 1.000/970 = 1.03 Thus: 4 (1.03) 310,264(9.11) _ 250(63.81)" 4 (ashy? _ 310,264ky 914 ky = 5.34 5.344 7708 v 500 [9.11 s = 1496 63.81 ¥ 250 Therefore: Fy = Cy Fy/2.89 = 1.496 (250) / 2.89 = 129.41 MPa. ee SOLUTION 23: Av= effective sheared area Ay= ty x (d— 2t) = 15.2 x (970 -2(10)) = 14 440 mm? To be safe in web shear fy actuat S Fy atowable V/ Ay $129.41 V114 440 < 129.41 V= 1868.7 KN SITUATION VII: CE Brd Exam-Nov 2001 National Structural Code of the Philippines states that on extreme fibers of flexural members not covered in section 4.5.1,5.1.4.2,4.5.1.4.3 oF 4.5.1.4.4: 1. Tension: Fp = 0.60 Fy 2. Compression: a) For members meeting this requirement of section, 4.9.1.2 having an axis of symmetry in and loaded in, in the plane of their web, and compression on extreme fibers of channel bent about their major axis: The larger value computed by the formula (4.5-6a) or (4-.5-6b) and (4.5-7) as applicable (unless a higher value of can be justified on the basis of a more precise analysis) but not more than 0.60Fy. Whon: A y270Cb [351690 Fyn fy 2 FyUny 3 40,55x10 °c ry (4.5 6a) Or, when the compression flange is solid and approximately rectangular in cross section and its area is not less than that of the tension flange: Fy= S2740Cy id In the foregoing: L= distance between cross section braced against twist or lateral displacement of the compression flange, mm. For cantilevers braced against twist only at the support, L may conservatively be taken as the actual length. 1. = radius of gyration of a section comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plane of the web,mm = ‘Ay = area of the compression flange, mm’ Cy = 1.75 + 1.05(My/Mz) + 0.3(Mr/Mz)* but not more than 2.3, where Mi is the smaller and Mp the larger bending moment at the end of the unbraced length , taken about the strong axis of the member and where Mi/Mp , the ratio of end moments, is positive when M; and Mz have the same sign(reverse curvature bending) and negative when they are of the same sign (single curvature bending). When the bending moment at any point within an unbraced length is larger than that at both ends of this length, the value of Cp shall be ‘STEEL DESIGN a Beomscanner\ ‘structural Design and Construction Review Manual by Perfecto B. Padilla Jt taken as unity. When computing Fas and Fy to be used in Formula(4,6 18), Cy may be ‘computed by the formula given above f or frames braced against joint translation, Cy, may conservatively taken as unity for cantilever beams A simply supported beam made up of Ass = Pe long with no steel with Fy = 248 MPa is 8 m n fateral supports against buckling. The properties of steel were as follows: 24. Which of the following gives the slendemess ratio? a. 150.94 b. 110.92 c. 119.54 d. 96.78 25,Which of the following gives the allowable bending stress? 2. 654MPa b. 85.4MPa c, 554MPa d. 75.4MPa 26.Which of the following gives the maximum uniform load that can be carried by the beam. a. 10.67 kN b. 12.56 kNIm c. 19.78 kNm —d. 16.98 kNim SOLUTION #24: Slendemess ratio: 1 - 2000 -150.04 5 83 coe ReneS SOLUTION #25: Allowable Bending Stress: Cs = 1.0 ( For simply supported) 703270C, _ {703270(1.0) _ BY aaa 78825 3516330C, _ [3516330(1) _ F, pag 11007 Since Ly Beresz0ce " Fy = 721006, 117210011) = 54 44 pa Fr —b = yn? (ts0.94F or Foe AMONG, 82740(2107 16K). 659 die ud 000(533) a © bigger of $1.44 and 65.2 but nt bigger than O.60Fy = 0.6(248) = 148.8 Mpa Thus, Use Fy = 65.2MPa < 0.60Fy of! SOLUTION #2 Maximum uniform load it can carry: Fojation) > Facts M 65.2x 10° > = . Ss M 0.002077 Mss 135.420.4 Nm M= 135. 4 kNm For simply supported beam: wi? Me we 8 65.2x 10° > 135.4 = w= 16.93 kim SITUATION VIII: CE Brd Exam-May 2003INov 2003 National Structural Code of the Philippines states that on extreme fibers of flexural members not covered in section 4.5.1,5.1.4.2,4.5.1.4.3 oF 4.5.1.4.4: 1. Tension: we)? 8 Fy=0.60 Fy 2. Compression: a) For members meeting this requirement of section, 4.9.1.2 having an axis of symmetry in and loaded in, in the plane of their web, and compression on extreme fibers of channel bent about their major axis: The larger value computed by the formula (4.5-6a) or (4-.5-6b) and (4.5-7) as applicable (unless a higher value of can be sid (S39 CamScannerjustified on the basis of a more precise analysis) but not more than 0.60Fy. When: 7os270C 351633000 Von i Fy _ (2. Fla? Fp = [2-H 3” 1055x10 cb When P (4.5 6a) Fy= 11721000 (4.5 gb) ati? Or, when the compression flange is solid and approximately rectangular in cross section and its area is not less than that of the tension flange: = 82740 Cy 2 ie In the foregoing: istance between cross section braced against twist or lateral displacement of the compression flange, mm. For cantilevers braced against twist only at the support, L may conservatively be taken as the actual length = radius of gyration of a section comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plane of the web, mm A; = area of the compression flange, mm? Cp = 1.75 + 1.05(Mi/t) + 0.3(My/M2)" but not.more than 2.3, where My is the smaller and Mz the larger bending moment at the end of the unbraced length , taken about the strong axis of the member and where My/Mz , the“ratio of end moments, is positive when M; and M, have'the same sign(reverse curvature bending) and negative when they are of the same sign (single curvature bending). When the bending moment at any point within an unbraced length is larger than that at both ends of this length, the value of Ce shall be taken as unity. When computing Fes and Fr, to be used in Formula(4.6 ~1a), Cpmay be computed by the formula given above f or frames braced against joint translation, Cp may conservatively taken as unity for " cantilever beams, A simply supported steel beam subjected toa clockwise moment aie ond and clockwise moment at the right end equal fo 25% of the moment at the left end. The beam is A36 steel with Fy = 250 MPa and is not restrained against lateral buckling. The beam is W21x 62 whose relevant properties are: = 0.053m d=0.533m b= 0.210m = 0.016 m Sx = 0.00207 m* 27. Which of the following gives the slendemness ratio? a. 174.86 b. 170.51 c. 153.67 d. 150.94 1. of the following gives the 28. W allowable "ending stress? a, 144.04MPa —b. 132.35 MPa c. 128.88MPa_— d. 145.05 MPa 29. Which of the’ following most nearly gives the value of the couple M_ in kiloNewton meter. a. 178.7 kNm c. 273.96 kNm SOLUTION #27: 75 + 1.05(Ms/Ma) + 0.30(Ms/Mz)* 75 +1.05(0.25MIM) +0.30(0.25MIM)? 03 < 2.3 ok )= —8_ = 150.94 h a) 0.053 [703270C, _ [703270(2.03) _ Fy 250 ae [3516330C, _ [3516330(2.03) ) 250 b. 239.3 kNm d. 289.4 KNm 169.02 STEEL DESIGN 138 — not adequate Solve for width (b) of cover plate. 33x10" whe (2p +6(12)6262| L J 1= 1.33 x 10° +.2550912b To be safe: factua $ faiowable: Me 138 — 092.35 x 10°(982) 1.33 x 10° +2550912 b b> 320.2 mm say b= 321mm 138 PROBLEM 2: In PROBLEM 1 above, find the theoretical length of the cover plate: a. 467 b. 5.67 c. 367 d. 3.17 SOLUTION 2 : Solve for theoretical length of cover plate Theorecal [cute pin a 4 33.3878 iN —————— A. ASCoverplates =f L=10m 256999 KN 256.939 4 FIGURE: Beam Loading with cover lates note: Cover plate is theoretically needed only at pts where flexural stress is greater than 138 MPa Let X= cutoff pts, ‘Without cover plate, solve for Mezpaciy To be safe: factuat $ fasowabie Me 138 — G20) a3 1.33 x 109 M$ 73.5625 x 10° Nmm note: Beyond distance x, moment is greater than 573.5625 x 108, thus cover plate is already needed. Take ‘moment a ‘x ostance fom face of support : 299788) or aes x 18 c Simply and solve by quadratic formula x < 2666.62 mm [2soaos iB CamScannerSay, x = 2666 mm Theoretical length of plate, L L= 10 - 2x = 10 ~2(2.666) = 4.668 m. SITUATION VII: CE Brd Exam-Nov 2005 A 300 mm x 12 mm cover plate is attached at the top and bottom of a wide flange 21x62 by two rows of 20 millimeter diameter rivets. The beam has a simple span of 6.0 m and carries @ uniform load of 270 kN/m. The depth of beam is 0.56 m and | = 0.00151 m 49. Which of the following most nearly gives the distance from neutral axis to the centroid of the cover plate? a, 286mm c. 256mm b. 296 mm d, 290mm 20. Which of the following most nearly gives the centroidal moment of inertia? a. 0.002099 m* b. 0.008122 m* cc. 0.01245m* d. 0.00084 m* 21. Which of the following most nearly gives the value of pitch of rivets(s) if capacity of each rivets is 30 KN? ‘a. 200 mm| c. 180mm = d b. 100mm 150 mm SOLUTION #19-21: 300 Distance from N.A to the centroid of cover plate. = 560/2 +6 = 286 mm Moment of inertia with respect to neutral axis: Ina = Lewe + Ix (cover sai) Ine ='0.00151 m* Iya = 0.00151 + (0.3)(0.012) +(0.3Y0.012Y0 246) | 7 Ina = 0.002099 m* Pitch of rivets/ spacing of rivets: 270kNim For the rivet to be safe in shear: Respociyl = VAyS — equation 1 Reapacity = 2X30 = 60 kN Vmex = R = 270(3) = 810 kN ‘Ay = 0.3(0.012)0.286 ‘Ay = 0.0010296 m?—> Substitute o eqn 1 60(0.002099) = 810(0.0010296)S $< 0.151m = 151mm SITUATION 1: CE Brd Exam-May 2001 ‘built up steel section is made up of a wide flange and two cover plates. 320 mm x 12 mm placed at the top and at the bottom flanges. The beam carries a total uniform load of 360 N/m and has a simple span of 6m. Allowable stresses are as follows: Allowable shearing stress = 0.4Fy Allowable bending stress = 0.6Fy ‘Allowable deflection =1/ 360 of span Properties of Wide Flange Section 1, = 0.001568 m* 56m 0.018 m 250 MPa 100 GPa STEEL DESIGN ‘CamScannerStructural Design and Construction Review Manual by Perfecto 8. Padilla Ir 560 mm vom | 30mm_ 3. Which of the following gives the section modulus? 3 '& 0n7s21Sm? —b 00022885? c. 0.0124563m° 4. 0,0042931 m? 4. Which of the following gives the maximum deflection of the beam? a. 10.67mm b. 19.34 mm c. 13.83mm d. 16.67 mm 5. The section is a. inadequate, fails in bending b. adequate c. inadequate , fails in shear d, Inadequate for shear and bending SOLUTION #3: Allow. shear stress = Allow. bending stres: MPa Allow deflection 4(250) = 100 MPa {60(250) = 150 = 600/360 = 16.667 mm Ta, = 0.001568 3 See)” 5 o3e x00rq0.2857 Ina = 0.00219628544 m*=2196.3%10° mm gz b= 9200219628644 c 0.292 $= 0.0075215 m* SOLUTION #4: Max. deflection Sul ( a (5)360(6000)* Yrax = 334(200 000)(2196.3% 10°) Ymax = 13.83 mm ¢ less than 16.667 mm safe in deflection SOLUTION #5: Tobe sae in flexure, fo salina) $ FO oow) ¢ < 0.6Fy > but Mzwl2#8 260(60007)/8 9 6,250) 7.6215x10" 218.38 < 150 < Not True! ‘Actual exural stress exceeds allowable, Thus, beam is not safe in flexure. To be safe in Shear. |, fv actual (max) $ Fv (atow) we < O.4Fy > but Vos =R = WL/2 =360(3) = 1080 AN 1080 000 << 0.4(250) 584(18) 102.74 < 100 < Not True! Actual Shoatng sess exceeds allowable. Thus, ‘beam is not safe in shear. There is no need to check ‘adequacy for horizontal shear. CONCLUSION: The beam inadequate bth shear and bending. (SCP Section 5043) “The longitudinal spacing of connectors between elements in contac consisting of two pales or pate and a shape shal not exceed 2) 24 times the thickness of the thinner plate nr 300 mm. {or painted members or unpainted members but nat subject to corrosion b) 14 times the thickness ofthe thinner plate nor 175 mm for unpainted members. (53 CamScannerBUILT-UP BEAMS PROBLEM 15 : A channel and a wide flange beam form a T— beam of 9m span. What concentrated load P can be supported at the center of the T-beam if the shear in the rivet is not to exceed 100 MPa: The fiber stress in the WF is not to exceed 138 MPa in tension: The fibers stress in the channel is not to exceed 118 MPa and the deflection for the concentrated load only is not to exceed 1/360 of the span. Use 16 mm diameter rivets spaced at 150 mm on centers longitudinally. E, = 2x 10° MPa [Properties of C380 x 30 | ‘A= 6426 mm d= 361mm 131 x 10° mm* Properties of W46 x 143 | A= 18190 mm_ d= 461mm —| Ix 699 x 10° mr mm’ [iy = 93.7 10%mm w= 143 kg. 7m a. 260 KN b. 307 kN ©. 317 KN d. 190 kN The moment of Inertia of the gross section about the neutral axis ig a. 43 672 x 10% mm, a. 38672 x 10 mm* a. 33 672 x 10° mm’ a. 31 672 x 10° mm 4 SOLUTION : kg 9.81 KN w= (50.4143) 1m 1000 kg 1.897 kMin Pp W469 Hilin et i Leon i Pre-+wid.)= 05P «A5(te97) | R = 0.5P «8.5356 HL (PR FIGURE’ Leet Diagram@nd Beam Section), (NSCP Section 504.3) The longitudinal spacing of connectors between ‘elements tn contact consisting of two plates ora plate and a shapo shall not exceed: 2) 24 times the thickness ofthe thinner plate nor 300 ‘mm for painted mombers or unpainted members but not subject to corrosion +) 14 times the thickness of the thinner plate nor 175 ‘mm for unpainted members. Locate the N. A.by taking the moment at the top: Ag = Ar + Ae Ac = 6426 + 18190 ‘Ag = 24616 mm? ' By Varignon 's Thoorem (about a - a) Ag Cr= Arcs + Aveo 24616 Cr = 6426 (20) +1100 (431 10. 2) = 183.1 mm. Cu 461 + 10,20 - Cr STEEL DESIGN WL CamScanner3 Structural Design and Construction Review Manual by Perfecto B. Padilla Jr y 26 | Cy= 461 + 10.20 - 183.1 = 288.1 mm y= Cr— 20 =183.1~ 20 =163.1 a= C, -“8t= 288.1 -481- 57.6 Solve for moment of Inertia: Ina = 3.38x10° + 6426(163.1) + 699x10° + 18190(57.6)? Ina = 9.33672 x 10° mm* Check fiber stress : = Me t= PQ), 1.897 OF 4 8 M=2.25P + 19.21 kNm M= (2.25P + 19.21)x10° Nmm For bottom flange to be safe in flexure: fo imaxybonom SF cacti) MCp I 5138 (2.25P +19.21)x10°(288.1) 9.3367210% P< 190 230N Fortop flange to be sae in flexure fo imax) top $F (atowable) MC. I 118 (2.25P +19.21)x10°(183.1) _ 9.33672«10° * P< 258 689.5 N Check for deflection Yoctual S Yatiow L Vocus $ 555 360 PL? 9000 ‘BEI * 360 P 000 <25 -48(200000)(8. 33672108) P< 307 381.7N Check Shear in Rivets: For the rivet to be safe in shear: Reapseiyl 2 VAyS —+ equation 1 V= Vmax = R = 0.5P + 8.5365 Repay = Fu capaci Aare x oe 10022167} = 40.212 KN Q-= Ayna = (6426)(163.1) = 1.048x10° ‘Substitute values to equation 1. 40.212 2 (9,33672x10")(0.5P + 8.5365) (1.048%10°)(150) P< 460.5kN — please see also similar ~ problem on welded connection Thus: Safe P < 190 230 N (ansiver) Ifthe channel is welded to the WF instead of being riveted, the same approach will be used except that = 41mm (or any value will do) (S39 CamScannerPLATE GIRDERS (NSCP Section 506.1) ‘Beams shall be distinguished from plate girders on the basis Cf the web slendemess ratio hit. When this value exceed ig stress is governed by 2547 | fy. the allowable b Section 507 and are already classified as Plate girders, s& Note: h absve is the over-all Height of the member cross-section smn [tecommende) + ef 2 T k z FIGURE: Plate Girders Considerations in the design : 1. Web plate solely resist vertical shear 2. Only 1/6 of web area is effective for flexure (tension and compression) 3. Rivets if used must be spaced by not more than 150 mm on center (AISC requirement) However for tension side : ‘If rivets holes exceeds 15% of flange gross area, deductions to web area must be made. Otherwise only 1/8 is effective. a . but not less than 0.26 inch PROBLEM 1 : Calculate the safe concentrated load at midspan which the given plate gifder below could carry aside from a uniform load of 20 kMim (including its weight). Allowable \ bending is 124 MPa. Beam has @ simple span of 12 m. Rivets are 20-mm. diameter Properties of 2 - 200 x 100 x 22 mm angles A = 12162 mm* w = 966 Nim 1 = 8.247 10° mm* y = 26mm SOLUTION 1: 30012 fi, beet 2-200 190» mmpete | * [| 22mmangles FIGURE: Plate Girders Solve for properties of composite section: I=, 300(02P +300(12)606P +48 24 x 10° +12162(574F] + pyliolizo0f 1=1,0793 x10'°mm4 Solve for moments we Pita], 20012? 4 8 M=3P +360 KN=m M = (3P + 360) x 10° Nm To be safe : factuat $ fatiow Mc Me s124 T <1 ‘STEEL DESIGN “a CamScanner/ Structural Desi (90.360) « 10°(612) «494 1.0793 » 10° Ps GO9KN Prrastonnii) = 609 KN eee PROBLEM 2: In PROBLEM 12, Determine the spacing of 20 mm diameter rivets placed in the vertical legs of the flange angles connected to the ‘web member, allowable stress for shear is 120 MPa a, 205mm b. 110mm 120 mm d. 130mm SOLUTION 2: Consider shear: 1 ¥=26 mm 4 R 7mm BAM. sreses / |) \ shears Area Wn ~ Area For horizontal shear connection Ve in built - up beams at '2162(574) + 300(12)(606) +10(26)(587) Q=9 315208 mm? R=VOQsiI R = 424 500 (9 315 208) s/ 1.0793x10"° R= 366.388 Forte rivet tobe sae in shear fv aca SF aiowabie RUA < 120 Construction Review 366.388 [2000s 120 « sheared area of rival was: muliplid by 2° because itis @ double shear 5 $205.8 mm _ ae PROBLEM 3: Determine the spacing of 20 mm diameter rivets placed in the vertical legs of the flange angles connected to the web member considering allowable stresses for bearing of plates to be 238 MPa. a. 125 rhm b. 110mm oc, 130mm d, 140mm SOLUTION 3: “The bearing area stake asthe projection of contact area ‘ona plan tatis perpendicular to the fore because bearing sess is @ normal tes ie. The force is ult thea ¥ =26 mm Stkmm Bewing Area for Web Phte 17 FIGURE. Sivng SF Anges & Plates The cntcal plate is the one with smaller area, that is, the web plate. For the plates to be safe in bearing: fa actual S Fp allowable RIAs < 238 366.38s / 10(20) < 238 $< 129.92 mm PROBLEM 4: A channel and a wide Flange beam form a T- beam of 9m span. What concentrated load P can be supported at the center of the T-beam jf the shear in the 8 mm full fillet weld is not to exceed 110 MPa? Properties of C380.x 30 A = 8426mm? d= 381mm (53 CamScanner| 131 x 10° mm* 3.38 x 10° mm* Ix ly x = 20mm ty = 102mm w = 50.4kg./m Properties of WAG x 143 A =18190 mm? d =461mm 699 x 10° mm* Ik = ly = 93.7 x 10°mm* w = 143kg./m a.,260 kN b. 307 KN, c. 317 kN 4d. 190 kN SOLUTION 4: [ | | w=1887 kin R= Pi2+w(45)= 0.5P +4.5(1.897) R = O.5P +8.5365 kN ~-[8 mm thick] Fillet weld Locaté the N. A.by taking the moment at the (op: Ac = Ay + Ag . Ac = 6426 + 18190 Ag = 24616 mm? 8y Varignon ‘s Theorem (about a — a) A Cr= Aays + Aayo 24 616 Cy = 6426 (20) ¥18190/ 481,10, 2) 83.1 mm 61 + 10,20 - Cr 183.1 = 288.1 mm Solve for moment of Inertia: Ina = 3.38x10° + 6426(163.1) + 699x10° + 18190(57.6)? Ina = 9.33672 x 10° mm* Check Shear in Weld: . R 0.7078mm) For the weld to be safe in shear: y 12 VAyS — equation 4 O.5P + 8.5365 Reapaciy = Fv (capacity) Aweis Reapaciy =110(0.707(8)(1)(2))= 1.244 kN Q = Ayna = (6426)(163.1) = 1.048x10' Substitute values f0 equation 1 1.244(9.33672x10 ) = (0.5P + 8.5365) (1.048x10°)(1) STEEL DESION Cl ‘CamScannerStructural Design and. COMPOSITE BEAMS Chapter 3 note : The following analysis shduld only be used if there is @ provision for shear transfer from slab to steel FIGURE: Composite Beams NSCP CODES : Effective Width (NSCP Section 509.2.3) berecine Shall be the smallest of the ff a)ll4 b.) by + 2b" b’is the smaller between ; clear spacing and 8t For spandrel section: L by +b etfecive 12 1 bis yolear spacing < 6t Allowable Stresses and Method of Analysis (NSCP Section 509.3) Method 1: Steel beam carries all dead loads unassisted prior to hardening > Of concrete. Then, with the hardening of concrete, the composite section carries the combined loads. Method 2: Steel beam carries all positive moments due to all loads unassisted. Concrete will be neglected in the analysis. (please see AISC commentaries) Fo slovavin = 0.76Fy | Shear Connectors (NSCP Section 09.2.2 to 2.3) Shear connectors shall be necessary when the steel section is not fully encased. It is said to be fully encased if all of the three conditions that follow are met v1. Concrete cover over all sides and soffit of steel beam is at least 50 mm (Section 609,2.3, 2. Beam top is atleast 38 mm below the concrete fop and 0 mm above the bottom of slab 3. Concrete encasement contains adequate mesh or other reinforcing steel throughout the whole depth and across the sofit ofthe beam to prevent spalling ofthe concrete. Loads on Connectors: Take the smaller: of: (NSCP Section 509.5.1 1023 Vn = O.85fAc/ 2 + Au Fya/2 and Vn =Fy'Asl2 number of connectors = Vh Where: ‘As = area of steel shape ‘Aq = area of steel bars within the Effective width. yield strength of steel shape yield strength of steel bars strength of one connector CamScannerPUATION 4: (CE BOARD EXAM) AW G10 ¥ 241 stringers are spaced 1.6m, on center thoughout and have a 180 mn concrete slab bonded to the top flange of the steel stinger by means of 2 — bar adapters. The top of the concrete is 140 mm above the top of the flange, If the stringers make 12/m simple span and tho only one that is carrying the full dead load, what static concentrated load at the center of each stringer may the composite section carry if the stress is not to exceed 124 MPa in steel and 9 MPa in concrete? f = 20, n= 10 Properties of W 610241 (w = 241.7 kg./m) d= 635mm A = 30800 mm? 1 = 2150x10° mm* 4. Find the maximum stress in the wide flange section due to initial deadload (i.e. wet concrete and beam only assuming unshored construction. Neglect weight of formworks) a, 23.2 MPa b. 38,6 MPa c. 32.5 MPa d. 29.3 MPa 2. Find the location of the neutral axis of the equivalent all-steel section of the effective composite beam measured from the top. a, 235,8mm —_b, 285.8mm c. 320.1,.mm d, 267.2mm. 3. Find the moment of inertia of the composite section a. 4.166%10° mm ©. 5.216%10° mmé b. 3.789%10° mn d. 4,847%10° mm SOLUTION 1: * 100mm ee ¥ [ewer CONCRETE 10 tle am Tr Jt? WF Seton sone =. cncvatcacte gS and sl woight of WE 1, Solve for moment due to weight of wet concrete and steel wl of slab wl. of beam 2400 kalin’ (1.5)(0.180) + 241.7 889.7 kg/in kg 9.81 KN in 1000kg w= 8,728 kNim~» fo be carted by stee! alone =) Scie = 889.7 2 Mo = Mk s.72e(2P Teh = 187.103 kNm Mp = 157.103 x 10° Nmm To be safo factual $ fatow we S fatiow o 187.103 x 10°(317 5) < 494 2150 x 10° 23.2 < 124 sale for composite construction SOLUTION 2: Solve for properties of composite section bat (fo i Note : Since by is not given, b Center to center distance =.1.5 m U4 = 121 Thus, bor= 1.5m = 100mm Convert to'equivalent steel section : ‘As’ = Acin = 1800(180)/10 = 27 000 As'= b'(180) 27 000 = b'(180) 150 ‘Allowable flexural stress of transformed 288 ‘oxn = 0.48fc'xn = 0.45(20)(10) Fs' 0 MPa o Neglecting area of concrete aisplaced by concrete: sree. oesion Cl CamScannerStructural Design and Construction Review Manual by Perfecto B, Padi £3 Hancasefnet astern festa ‘ho 18001 WeI80 nm wt | i f i i fetta! stess Doan Fata he Bean’ Tansee Sectin Sacon canyon P Locate the Neutral Axis, N.A. Using Verignon's Theorem : (ebout axis a ~ a) Actor = Ae! + Ave 2 57.800 Taking Moment et the top Aro = A toiai r= As’! + Awe One 57 800 cr= 27 000(90) + 30 800(635/2 + 140) cr = 285.80 Ce = 635 4 140- cr 2 = 775 - 285.80 ca = 489.20 SOLUTION 3: Solve for | d; = Ca - 317.5 = 489.2 - 317.5 = 171.7 da = cr ~90 = 285,860 ~ 90 = 195.80 = 218) x 10° + 30800(171.7)° st (150)(180)° + 180(180)(195.8)* 1 = 4.166 x 10° mm SOLUTION 4: Solve for P m= EL Pt) = 3P km M.= 3Px108 Nn, w Fron ont Aihant bem 150.) fieMethn f ' ofa [v0 | sit ( ( Ww wv \ nC be] a tA , b Bee Fase Bata. fuob Tse compte. Sess Secion Osan caiyig totes 2 ea FGURE: compote Beam Consider Tension Side : To be safe focuot $ Fallow fot +2 < 124 23.2 + Moe < 124 twa Sil section Soll caring Sass thelial Daye chaos =H Toa Stas Dayar 232+ Mice < 124 Iya 23.2 + GPx10°]489.20) -494 4.166 x 10° P< 299.62 KN Consider compression side of Concrete f aca $f atow fos 0.45f¢n Mer VAyS eqn 1Taking 1 mm strip, (.e. S = 1mm) Solve for Q of area of concrete above critical section abed Q = say Q = 180(1500)(285.8 ~ 90) ~40(329)(285.8 — 160) - $(600~ 329)(40) x + (40)+ 285.80 ~ 180] Q = 50. 8x10°mm*— note that we neglect area displaced by steel R = Reapociy R= R = ogonp Agono + Gsnea/AconcrETE SHEAR R = 0.63(329)( 1mm) + 2.52 [22 (1mm)] but: z= 740? +85.5* = 94.4 mm R = 0,63(329)(1mm) + 2.52 [2(94.4) (tmm} R = 683.046 — substitute to eqn 1 RI > VAyS 683.046 (4.166 x 10°) > V(50, 8*10°)(1) V = 56 KN ee Important points: ‘sono + Rooncrete SHEAR — a time to tear down, and a time to build up Ages 0-2: Waning time, Ages 2-13: Play Time, Ages 13-20: Learning Time, Ages 21-22: Review! Preparation to the World of Professionals! Ages 22-30: Investment Time. ‘In What age bracket are you? ‘Success and happiness come by doing what you should be doing! (S39 CamScannerStructural Design and Construction Review Manual by Perfecto B. Padilla Jr Case 2: with sagrods P URLI NS ‘A. Sagrods at middle point (center) § Tisapprt bom Support from om [ei PRINCIPLE s 5, : 3, L Purlins are inclined membirs. Thus | | 30" any w" they are subjected to bending in the.x — x (major) axis and y ~y (minor) axis: 2 Msind FIGURE® Loading & Moment diagram for Bending in ‘he minor axis wth agrods SMa point Mix Moment dee to leads applied on the roof (i.e roof load, Le Load, Snow Lead. Roof Weight, ete) + | B» Sagrods at third points Consider loading on the minor axis Mor Moment duet luda nthe cet of (parallel to incline) purine putin weight and eeling) 1 ] Case 1: No sagrods Supportrom —]Supporttiom [Support ram by Method 4: Moment Resolution Method jas Ce Truss MycosO M,sind Mycoso Mysino f] 14, whe — pa 6 te! 0 Sy Sx Sy | | ‘ sid t Method 2: Flexural Resolution Method ; ( | ' at's 1.00 ' FIGURE: Lowling é Moment diagram for bending in the minora with sagrs a th twyl? Critical Moment = PROBLEM 17: (CE BOARD) ‘AW — shape section having the properties shown below is used as a purlin with simge support at its ends with the bottom flange resting on top of two angular top chords of roof trusses that are 6m apart Each purlin is supported laterally at third points by sagrods that runs to the ridge purlins, A uniform load of 1300 Nim is acting normal to the top flange of each and a vertical uniform load of 2350 N/m including its weight act at the centroid of the purlins. PROPERTIES A = 3338 mm? 220x 10°mm* 5mm 21mm 10.63 mm 81.5mm 1.48 x 10°mm* 22.23 x 10°mm* 200 mm 29.7 x 10° mm? FIGURE Determine the maximum bending stress on the beam/purlin a. 73.7 Mpa b. 83.7MPa cc. 63.7 Mpa d. 93.7 MPa SOLUTION 17: O=invtan 5 = 26.565° Whormat = 1300 + 2350 cos @ = 3401.9 Nim Wangentsi = 2350 sin 0 = 1050.95 Nim Solve for moments : ol? _ 3401 96P _15308, 1Nm 8 8 : ol? _1050.956F _ 499 38 Nm 30 eae Me, Wy. 73, (15308.1 x 10) (420.38 x 102 R20 x10 387 x 108 33.74 MPa Focal PROBLEM 18 : A steel section whose properties are given below.is to be used as a purlin that {est on top chords of steel trusses that are im apart and are sloped at 1 ver every 2 horizontal, oe Purlins are simply supported at it its ends and af supported in plane of the top chord of the roof truss by sagrods at the third points of the purlins ru i pein inning to the ridge Spacing consideging bending o nl Properties of Wig Mass2 V ictural Design and Const We Gam toot tin 24810" mm! 18x10" mm Hye 248 Mpa a 88S Hy b, 7.985 m0 B98 My dd. 4.9u6 m SOLUTION 18, tan 1 28 S08 Solve for the loadings: Purtin weight yy NO, S8TKN m1000kg Cening weight we Wind load Woeldes Roo! load w 0.363951 kN/m Bi = 0.2 1408s KN w Hoey Rd nite SS Canto ol he pari is + (Ws + Wa + Wa)oos @ wu wiv 1.25 + (0.363051 + 0.21486s + 2.75) | cos 26.565° w= 3.81s + 0.32553 km Wii = wa si @ = 2.78 sin 26.565° = 1.20755 (ws + wa) sin @ (0.363951 + 0.21466s) sin 26.565° 0.16276 + 0.0968, wr Solve for the Moments and flexural ( stresses: Bending about the major axis: = ww L716 (381s + 0.32553)87 My = 30.48s + 2.604 kN.m_ My = (30.48s + 2.604) x 10° Nmm My I Sx (30.488 + 2.604) x 10° 1274 x10° 11.248 + 9.504 Bending about the minor axis: My= wr87/90 (1.2075s)87/90 0.8595 kNm My1= 0.859s x 10° Nm fags = Myx (Syl2) foyi= (0.859s x 10°) /(91.8x10°/ 2) foi= 18.718 Myo= wral%/90 M,2= (0.16276 + 0.096s)87/90 Mizz (0.1187 + 0,0883s) kNm Myo= (0.1157 + 0.0683s) x 10° Nmm sisal ‘CamScannerMyo 1 (S\) (0.1157 + 0,0683s)» 10° / (91.8 ¥10?) hyo 1.26 + 0.7445 fy + fhyo 18.718 + 1.26 40.7445 19.45484126 0 * Solve for the allowable stresses: For bending about the major axis, Solve ste allowable foxure step I: check for lateral support Ly = 8/3 = 2.667m = 2667mm 200 by _ 200(154) = vFy 248 955.8 mm Ly = 137900Af Fyd L, = 137900(154\11.6) 248(162) L, = 6131.6 mm Le< Ly < Ly — adequately supported. This Fo. = 0.6 Fy Fn = 0.6 (248) Fo, = 148.8 MPa._ (answer) For bending about the minor axis. From (NSCP Section 506.3 1.1) Foy = Foy = 148.8 MPa (answer) Apply the interaction equation Sor 4 for 4.00 foe =111.24s + 9.504 Fox Foy 111.245+9.504 | 19.4548+1.26 <4 99 148.8 1b 5 ¢ 1,056m (answer) sree besion SE ZC]= Note: Forconsersatveresut, we combine the maximum flexural sess due to bending in the major axis and the % eo =>]_._marimum flexural stress due bending in f'y-) heme is eth ey dati aR LL j ‘exacly onthe same point SITUATION 7: CE Brd Exam-Nov. 2010 12m in The C-purlins above are considered simply supported by truss that are spaced at every 3m. The following are the loads that said purlins will carry ‘8. = 170,000 mm* S, = 80,000 mm* Weight of purlin = 71 N/m Live Load ~ 1000 Pa (acting on horizontal projection) Dead Load - 1200 Pa (acting on horizontal projection) Wind Load — 1440 Pa (perpendicular to roof) Find the Interaction value at midspan for the following load combinations: 19.0+L a. 38 MPa b. 37 MPa c. 34 MPa d. 32 MPa 20. 0.75 (D +L +W) a. 38 MPa b. 37 MPa c. 34 MPa d, 32 MPa 24.0 + Lifthere are sag-rods at the middle. a. 21.32 Pa b. 17.5.MPa ©. 32.42MPa d. 15.2.NPa ($9 camscannerStructural Design and Co SOLUTION 19 - 21 Wow. = 1200 + 100( tang = 4 0= 1843° ‘Ws = 2200 Ninv x 1.13842, 504.524 Nim Woavatie = Wy Sind = 2504.524 sin18.43° ‘Woaratet = 791.8 Nim Wrornai = Wy Cost = 2504.524 cos18.43° Woornai = 2375.5 Nin 200 Pa oso = ce RISEN Nt sein 7918 OUT =2504.524 Nim 2a00Pa 12m Neo a 12 cos 1.198 FIGURE 0C2-2010-19, Note: Interaction Value mentioned here could be ‘referring to combined stress. 3 My = 890.8 Nim —+ substitute o eqn 1 f, = 2672.438x10° 890810? *° 470000 a fy =38 MPa 075 (0 +L+W)=0.75(0 +1) + 07EWwL 0.75(0 + L)= 0.75(1200 + 1000) = 1650 Pa Deon! Nite 1.13842m= 1878.393 Nin O.7SWL = 0.75(1440 iwi) = 1080 pe W2= 1080 Nine 1.2 m= 1296 Nim Wosrtal= Wi Sind = 1878.93 sinte.4o ) Wparaes = 584 Nim ition Review nual by Perfocto B. Padilla Je Whoxma Wp + Wy COST : ose ® 129641878, 393 cos 18.4 Wioesmat # 3078 Nim 1870.99 st wast) Vv sind fF TO 18504 75m, 1080 KPa z M_ = 8462.75 im Menud? _ §94(3)? oe aa eas My= 668.25 Nm — substitute 069 1 3462.75%10° 668,25 x10? fe ee 170000 fy = 37 MPa Refer lo figure DC2-2010-19, About x-axis, beam is ‘simply supported. Thus, Wont? _ 2975.5(3)" tea eae seo M, = 2672.438 Nm 00, ‘About y-axis, beam isa 2.span continuous beam Thus, Monat? _ Vpwang? Mea My = 668.11 Nm —> substitute to€9 1f 72.438 x10" , 668.1110" »* 470000 00 32.42 MPa BEARIN PLATE Chapler 6 SITUATION 11: CE Brd Exam-Nov 2009 In accordance with the AIS. specification, “..webs of beams and welded plate gitders shall be so proportioned thal. the ‘compressive stiess at. the web! toe of the filets, resulting fom concentrated loads ‘nat supported by bearing stieners, shall nol exceed the value of 0.75Fy, otherwise, bearing stifenors shall be provided. The ‘governing formulas shall be For interior loads, P fa= ———< 0.75F, -® Np + 2k) a For end reactions, R <0.75F, WW +k) ad concentrated load, R = the reaction he length of bearing for nteral load, the length of bearing at end reaction, e web thickness, and «distance from outer face of flange to the web toe of hefilet.” For vertical buckling, For interior loads, P 2 f <128-0.01( 2.) 38 Np + 0.50) he Forendreections, f= 128-0, oress( 2) 30 UN, +050) We A transfer beam that may be considered as simply supported serves as under-pinning for a column located L/4 from the center of the left support (se figure). The column rests on a steel base plate that measures 250 mm, along the beam. The beam consists of two S-sections that have 150 mm supports, on steel base plates. The beam is to be reviewed and the actual stresses compared with those allowable under AISC specification for A36 steel. S-Beam Properties: 22.Find the allowable vertical buckling stress a. 98.25 MPa b. 105.68 Mpa c. 150.8MPa d. 84.14 MPa 23.Find the value of column load P so that allowable buckling stress will not be exceeded. a. 476 KN b, 173.7 aN ©. 247.8 KN d. 805.3 kv 24. Find the value of column load P so that allowable crippling stress will not be exceeded a. 1646.1 kN b. 1073.7 kN c. 1254.8 kN d. 1476.3 kN SOLUTION 22 - 24: Using eqn 3 for interior loads: Fae re5-o0n(¢) =128-001 Fa = 105.68 MPa Using eqn 4 for interior loads: Fas 125 -oores( Fe =128-0,01965 (ef Fa= 84.14 MPa Thus: F.= 84.14 MPa Fr interior foads, to be safe against veri buckling web Fg) coe Eee ER pe Ea gre De OL CamScannerStructural De Max fi acta! S Fa atowatie P [Wr oRajra] $9858 Pp [12.7280+0. 5(600) Ps 1476 kN < 105.68 2] For exterior loads, tobe safe against vertical web bucking Max fo acusi S Fa stovate R ‘= 105.68 IN, + 0.25a)x 2] R $105.68 12.7(150+0.25(600)x2] <1 Rs 805.28 kN = Ms =0 RL = P*0.75L 805.28 = P*0.75 P= 1073.71 kN Thus safe max P. = 1073.71 kN tO n and Construction Review Manual by Perfe 2 who can bear @ crushed spirit Proverbs 18:14 at good ean you do psi when you are physicaly ILL? ~ the ans THING! ao want to. somolsh phys ‘accomplishments; you need to keep yours heathy Choose the food you eat. Exercise and watch your diet Wat good ean you do spintvaly when 4yow are spirualy and emotionally ILL? - the answers also NOTHING! Board exam success requires physical, emotional, psychological, and’ spinal HEALTH. Choose the thought you think the emotion you feel. YOU HAVE THE POWER. OF CHOICE! JESUS tad set you feet John 8:36 says So ithe Son sels you fre, you wil be fee indeosr ‘Why not use youreedom of coice? af Cita! Sexton ] % for Bucking | “Jo UL ek FIGURE: bearing fr beam and tanster ger sacl (9 camScannerUnit 2: STEEL JOINTS Chap lord RIVETED JOINTS IN TENSION A, REGULAR HOLE ARRANGEMENT PRINCIPLE: If the tensile force is passing thru the centroid of the rivet formation, the stresses in the rivets are the same. Thus, if the rivets are of the same size, the rivet reactions are also the same. Y-y— _Thesecton nearest othe extemally TQ sonics tees octal aca because ts | ‘, resisig tefl eal led The seoton with he most numberof holes ital secion Because thas 5 the smalest area, ) NSCP section 5023.2 4 The width of the hole sal be taken as | ‘——" 4.6:mm greater thanthe nominal | diameter ofthe ok @) ey You don't believe what you see. You see what you first believed!" Bryan Tracy Believe that you can pass or top the board exam, seeing it happen will Follow! autor Related AISC Provisions: Sizo of Bolts for bearing type connection (NSCP Sec 505 6) I must be greater than 38 mm, GROSS & NET AREAS FOR TENSION MEMBERS WITH HOLES Woot = Wa (Ghote) + Z z=s'l4g dnote = duet + 4.6mm Note: For angular bars, vig Sum of width of legs minus the thickness ‘Wet = net width dra = hoe diameter dit = iv bolt ciameer = Cochrane's Vale. Ths value i tobe added to net width ifthe section under investigation involves staggered holes. n= number of holes =i the dimension that is in the same direction as the applied force, P = the dimension that is perpendicular to the applied force, P . STEEL DESIGN WOR | alia ‘CamScannerStructural Design and Construction Review Mar Axlally loaded tension members: INSCP Provision ‘The eflecine net aroa_of aialy loaded tension Imombers, whore some of tho load is transmitted by buttsvets through some but not all ofthe cross-sectional cloments of the members (see AISC comment) shal be ‘compute fom the formula Aton Coon Values of T._ For WMS shapes with fanga willis nol | less than 223 the depth, and structural ees cut from these shapes, provided tho connettion Is. to. the flanges and has 0 fewer than 3 {astone's per lin in the direction of the stess 2 For W. MS shapes not meetin the above condition, structural tees cut from these shapes, and all other shapes, incudng bul up Oss sectons, provided the connection has fot less than 3 fasteners per line in the direction of stress 3. AL members whose connedtons have only 2 fasteners per line in the direction of stress 4. Riveled and boled spice and gusset plates and other connection fitings subject to tense force 0 Or Prcomeced members These are members which are fae to rotate atthe connection NSCP Requirements: Eyebars shall be: ‘of uniform thickness Without reinforce-ment at pin holes shall have circular heads concentric withthe in hole 1 a b. c 4. Ded Neri AS | | 0 B, Padilla 5 times the by Per wl 9. Amol head of pin mustbe fom 133 ‘ea fs by. : (io. 1.30n (bays qa < 1.5m (040) ho be7et i hs 18b | bs StlorFy> 480 MPa 2, For othor pn connected plates Fron = O45F, onthe net area From = OSFy ‘Area beyond pinhole = % Net area a! Pl zsit 1 125 ies te sar of 2072 onthe pins expected to tele vena rected as dell. $186 9. the comers of connector lates maybe cut at 5° angle ‘ALLOWABLE STRESSES (NSCP Sec 506.3) TENSION a For non: pin connected members: Fs 0.6F, onthe gross area 1 < 0.5F on the eflecve net area where : = minimum specified strength ofthe materia ‘thor than eyebars: in hole For in connected members Fes OASF, onthe net rea For tension on threaded pats: See AISC able 15.2. (manual of steel design and constuction) SITUATION I ; FINDING THE TENSILE FORCE AT A SECTION For the fish plate connection shown below, find the force acting at: 1, section e78j a. 700 KN b. 600 KN cc, 800 KN d. 500 KN 2. section d6i a. 500 KN b. 600 kN c. 700 KN d. 800 KN 3. section c45h a. 800 KN b. 700 KN cc, 600 KN dd, 500 KN 4, section b3g a. 300 KN b. 600 kN (S39 CamScanner5. section a12f ‘a. 300 KN b. 100 kN c. 400 kN d. 200 kN 6. section c468} a. 700 kN b, 400 kN cc. B00 KN d. 600 kN 7. section a135h a. 200 kN b, 350 KN cc. 450 kN d. 400 KN 8. section a132f a. 100 KN b. 300 KN c, 200 KN d. ‘500 KN abc de 1 4 7 5 @ | p-a00m oF o — 2a y oe od O fea: hei SOLUTION 19: p= force per rivet ber of rivets in a plate n= num p = 800/n = 800/8 = 100 KN maietcetdae tap 4 Tad of Gh OF |peonnan Sap ou 2p Sar Beh o o f ead SOLUTION 1: section 78} Back Notice: There are & rivets from the back to section 76). Therefore! Foire) = 8p = 8(100) = 800 kN SOLUTION 2: section d6i abed tage ‘ee joe f Lied 2p SaaP oO Back Front | fehl Notice: There are 6 rivets from the back to section Gif. Therefore: Fee = 6p = 6(100) = 600 kN el Section c45h Feu Back Notices There ar 5 rivets fom the back fo section 45h, Therefore: Feasn = 5 = 5(100) = 500 kN ‘STEEL DESIGNStructural Design and Con: Section b3g reer 3 ies om te back o section Back: ai2f. Therelore 19 = 2(100) Notice: There are 7 rivets from the back to section 468). Therefore: Fesss, = 7p = 7(100) = 700 KN 7] tractvon Review Manual by Pere cto 8. Pal Section af35h a Fem ey Back : Notice: There are 4 rivets fm the back section 8195h, Therefore: Farasn = 4p = 4(100) = 400 KN Notice: there are 4 vets from the bac fo section @135h Back! Notice: There are 3 rivets from the back to section 2132f, Therefore: Fanaa 3p = 3(100) 300 kN ‘The tensile fore in a seconis Fepere | Where: ‘= numberof rivets from the back to the secton considered PROBLEM 9: Solve for the maximum force P which the fish plate connection shown could carry if allowable stress are the following; 150 MPa. for tension. 300 MPa for bearing of plates ‘and 105 MPa for shearing of rivets. Note: secabeoa tei ($9 camscannerrivets are 20 mm diameter, plates are 12 mm thick; hole diameter is 3 mm biager than the rivet diameter a. 197 KN b. 297 kN ©. 137 KN d. 237 kN 7 f 20mm 6 ets Consider tearing of plates note: Orie = Ore * 3mm & From old SCP Cade Requirement and is given inthe protlem a a ae Section a + cee |Back Pe & ¢ & P co. FO eg ORT | tess foo e e oe o> (200 — 23) (12) = 2124 f g A i 0.85 (200)(12) = 2040 — use SOLUTION 9: F.=9p=9(P19) =P P fe Pp aol $150 = 2040 P.<306 kN stem bike none at % Section b: | og #6 +) ea, ®, “eo fo ciahet : Fe=ap=a(P)= P Consider shear : 9 frcus § fatonte facus < Faston 8 Pp aP 0 15 fis 200 - 223){12) © Zao} Ps311.85kN ; c ‘The section nearest the extemaly plied ioad isa P <296.9kN tlotaea treatises te esemalad, | } ‘The section wit the most number of holes isa (but not the Consider bearing ny oe) rica secon ing < last becaueithas he sales area P NSCP Section 50232 é The width ofthe hole shall be taken as 16 mm greater wor) <300 ‘than the nominal diameter of the hole. P< 648 KN. sreevoesion Wd (S39 CamScannerStructural Desi Section c: 2p Fe= 6p = 6(P/9) = F S fosoone p —

You might also like

• Steel Design Quiz Module 54

  

  No ratings yet

  Steel Design Quiz Module 54

  9 pages
• Practice Problems - Strema Part 3

  

  No ratings yet

  Practice Problems - Strema Part 3

  3 pages
• MODULAR QUIZ - 61 - Reinforced Concrete Design

  

  No ratings yet

  MODULAR QUIZ - 61 - Reinforced Concrete Design

  9 pages
• Design and Const2

  

  No ratings yet

  Design and Const2

  101 pages
• Module Exercise Steel 2 Design For Flexure 1

  

  No ratings yet

  Module Exercise Steel 2 Design For Flexure 1

  1 page
• May 2016 Mste

  

  No ratings yet

  May 2016 Mste

  10 pages
• CE527 Correlation Course Structural Engi PDF

  

  No ratings yet

  CE527 Correlation Course Structural Engi PDF

  2 pages
• Structural Engineering Calculations

  

  No ratings yet

  Structural Engineering Calculations

  27 pages
• Set B

  

  100% (1)

  Set B

  12 pages
• Microsoft Word - Evaluation Exam 1 B1 Manila

  

  No ratings yet

  Microsoft Word - Evaluation Exam 1 B1 Manila

  8 pages
• Coach HGE (Oct 9, 2021)

  

  No ratings yet

  Coach HGE (Oct 9, 2021)

  7 pages
• Review Module 28 - Steel Connections

  

  No ratings yet

  Review Module 28 - Steel Connections

  3 pages
• 603

  

  No ratings yet

  603

  7 pages
• Mech 2

  

  No ratings yet

  Mech 2

  3 pages
• Reinforced Concrete Design Guide

  

  No ratings yet

  Reinforced Concrete Design Guide

  8 pages
• CE Board Exam Prep Guide

  

  No ratings yet

  CE Board Exam Prep Guide

  77 pages
• Solution To Refresher Exam 3 - MSTEC Part

  

  100% (1)

  Solution To Refresher Exam 3 - MSTEC Part

  10 pages
• "No Copying and Sharing of Answers ": Finals Examination

  

  No ratings yet

  "No Copying and Sharing of Answers ": Finals Examination

  5 pages
• Kippap-Handout-SEC (11 Flexural Stress)

  

  No ratings yet

  Kippap-Handout-SEC (11 Flexural Stress)

  5 pages
• Reinforced CD Notes

  

  No ratings yet

  Reinforced CD Notes

  8 pages
• Preboard SEC Solution

  

  100% (1)

  Preboard SEC Solution

  10 pages
• Strength of Materials - Set-2 - Online

  

  100% (1)

  Strength of Materials - Set-2 - Online

  2 pages
• Inbound 6678969275968959689

  

  100% (1)

  Inbound 6678969275968959689

  5 pages
• Refresher Psad 13 April 2024

  

  No ratings yet

  Refresher Psad 13 April 2024

  2 pages
• Math Refresher Module 1

  

  No ratings yet

  Math Refresher Module 1

  4 pages
• CE Board Exam Review Questions

  

  No ratings yet

  CE Board Exam Review Questions

  8 pages
• Eng

  

  100% (1)

  Eng

  11 pages
• 2022 Nov Algebra 4

  

  No ratings yet

  2022 Nov Algebra 4

  2 pages
• 603 Quiz 5

  

  No ratings yet

  603 Quiz 5

  2 pages
• Mockboard Exam MSTC Nov2024 MS62994TC - Unlocked

  

  No ratings yet

  Mockboard Exam MSTC Nov2024 MS62994TC - Unlocked

  13 pages
• Spandrel Beam Torsion Analysis

  

  No ratings yet

  Spandrel Beam Torsion Analysis

  11 pages
• Geotech Engineering Task Sheet

  

  No ratings yet

  Geotech Engineering Task Sheet

  2 pages
• 2019NOV PSAD Unlocked

  

  No ratings yet

  2019NOV PSAD Unlocked

  8 pages
• AE06 Sol SlabBolt

  

  100% (1)

  AE06 Sol SlabBolt

  5 pages
• Reinforced Concrete 1

  

  No ratings yet

  Reinforced Concrete 1

  19 pages
• Plate 4

  

  100% (1)

  Plate 4

  33 pages
• Structural Ex

  

  0% (1)

  Structural Ex

  14 pages
• Manila and Ol b1 Eval Exam 2

  

  No ratings yet

  Manila and Ol b1 Eval Exam 2

  9 pages
• Useful Formula For PSAD

  

  No ratings yet

  Useful Formula For PSAD

  14 pages
• Esplana SteelDesign2

  

  No ratings yet

  Esplana SteelDesign2

  1 page
• MSTE Questionnaire

  

  100% (1)

  MSTE Questionnaire

  7 pages
• RefM 14 - Ultimate Strength Design (T-Beams)

  

  No ratings yet

  RefM 14 - Ultimate Strength Design (T-Beams)

  1 page
• Om 3

  

  No ratings yet

  Om 3

  6 pages
• Looksfam Msteeeeeee

  

  100% (1)

  Looksfam Msteeeeeee

  18 pages
• Solution To Refresher Exam 3 - HPGE Part

  

  No ratings yet

  Solution To Refresher Exam 3 - HPGE Part

  8 pages
• MODULAR QUIZ - 60 - Reinforced Concrete Design

  

  No ratings yet

  MODULAR QUIZ - 60 - Reinforced Concrete Design

  8 pages
• Mathematics Surveying and Transportation Engineering PE1

  

  No ratings yet

  Mathematics Surveying and Transportation Engineering PE1

  4 pages
• Engineering Exam Instructions & Questions

  

  No ratings yet

  Engineering Exam Instructions & Questions

  7 pages
• 2022MAY PSAD Unlocked

  

  No ratings yet

  2022MAY PSAD Unlocked

  6 pages
• Seatwork No. 2 PDF

  

  No ratings yet

  Seatwork No. 2 PDF

  2 pages
• Excellence in Professional Engineering Review and Training Solutions

  

  100% (1)

  Excellence in Professional Engineering Review and Training Solutions

  3 pages
• MSTE 12 (Online)

  

  100% (1)

  MSTE 12 (Online)

  2 pages
• MSTE Practice Exam #1 by E-MT

  

  100% (1)

  MSTE Practice Exam #1 by E-MT

  4 pages
• In-House Module - Steel Design Part 1

  

  No ratings yet

  In-House Module - Steel Design Part 1

  3 pages
• In-House Module - Steel Design Part 1

  

  No ratings yet

  In-House Module - Steel Design Part 1

  3 pages
• 1.2 - Bending For Beams - Lateral Supports

  

  No ratings yet

  1.2 - Bending For Beams - Lateral Supports

  2 pages
• WK 10a Laterally Supported Beams

  

  No ratings yet

  WK 10a Laterally Supported Beams

  16 pages
• Module 37 - Steel Beams 2001 and Steel Purlins

  

  100% (1)

  Module 37 - Steel Beams 2001 and Steel Purlins

  3 pages
• Steel Design 2 3

  

  100% (1)

  Steel Design 2 3

  21 pages
• Design RC - Dr-Mahmoud Galhod

  

  No ratings yet

  Design RC - Dr-Mahmoud Galhod

  216 pages