STRUCTURAL DESIGN and CONSTRUCTION | NOVEMBER 2022 Me tar r4 32kN iia Hy=30 Given: a i p=20m a i=24m, H=3.0m. Width of cylinder = 3.2 KN. The coefficient of friction between the cylinder aad the surface is 0.20. The cylinder is filed with water to a depth of 2.4 m. Find the nnaximum value of the force F (KN) that will saisfy the following conditions. 3) Translation of the cylinder is prevented. 4) Tipping of cylinder is prevented. 4) The cylinder remains in equilibrium. Solution: 3) Value of F when the translation of cylinder is prevented b) n= 221 ayosn W, =74. KN W, +W, =32474=77.2KN F=MN F =0.20(77.2) F=154KN Value F so that tipping of cylinder is prevented 3m YMa=0 F(24) = 7.2(1) F=3217KN Value of F so that the cylinder remains horizontal F=MN F = 0,20(77.2)D-572 Civil Engineering Licensure Examinations Pn 8 APA Pole AB is 12 m. long and its weight W = 35 KN. Itis being lifted using cables BC and BD. ‘When the pole is tilted at an angle of 60° from the x-axis, the resultant force acts at point A. a) Find the tensile force (kN) in cable BC. ) Find the tensile force (KN) in cable BD. c) What is the value of the resultant (kN) acting at point A. Solution: a) Tensile force (kN) in cable BC: a DMyais 3D. +4.5 Cr = 35(3) Dr +15 Cz=35 SMos 6D. =26C. Dr= 0.433 Ce 0.433 Ce+1.5 Ce=35 Cr = 18.10 oye + (10.5)? + (10.39)? di d= 15 15 Bf Jo,-18.10 1039 BC, 15 78.10 10.39 BC = 26.1 kN Tensile force (KN) in cable BD. Dz= 0.433 Cz D, = 0.433(18.11) Dz = 7.84 KN b) B= (6 + (9)? + (10.39 d=15 BUY |p .=784 F |o39 B15 tL 10.39 BD=11.3kN ©) alue ofthe resultant (kN) acting at ponD-573 elvil Engineering Licensure Examinations 2Cr+Dr+35 Rez 18.11 + 7.84 +35 R= 60.95 KN /\* “ /\- Ds 9 D, =6.79 : es ¢ R, = 18,29 + 6.79 R, = 25.08 kN RRP +R? R? = (25,08)? + (60.95)* R= 65.9KN RA 9 MAA A Fink roof truss is used for a residential building and loaded as shown: jg ae a) Find the force of member DE if Hi = 0, Ho=0, Hs = 0 and P= 105 KN. b) Find the force of member DE if Hi = 16 KN, Hz = 27 KN, Ho = 11-KN and P = c) Find the force of member DE if Hi = 16 KN, Ho = 27 KN, Ha = 11 KN and P = 105 KN, Solution: Force of member DE if Hi = 0, He = 0, Hs =O and P= 105 KN a) P=105 kND-574 Civil Engineering Licensure Examinations —— LL hkhvwcCPO}jMmmmauamm b) IR = 105 R=525kN YMo=0 525(4.5) = Toe (1.5) Toe = 157.5 kN Force of member DE if Hi = 16 kN, Ha = 27 kN, Hs = 11 KN and P=0 iG tan d= — 15 45 's 8 = 18.43" A 4s Ri + Ro= 16 Cos 18.43" + 27 Cos 18,3° +11 Cos 18,3° Ry =51.23- 12.90 Ri = 38.33 KN TMc=0 . 38.33(4.5) = Toe (1.5) + 16(4.74) + 27(2:37) Toe = 21.77 KN Force of member DE if Hi = 16 KN, He=27 KN, Hs = 11 KN and P = 105 kN 0) (AC)? = (4.5)? + (1.5) AC=4.74m. AC > = 2.37 7 m YMa=0 9 Ro= 21(2.37) + 11(4.74) Ro= 12.90 KN TMa=0 9 Ro= 105(4.5) + 11(4.74) + 27(2.37) Ro= 65.40 Ry + Ro = (16 + 27 + 11) Cos 18.43° + 105 Ri + 65.40 = 156.23 Ry = 90.83 KN YMc=0 Toe (1.5) * 16(4.74) + 27(2.37) = 90.83(4.5) Toe = 179.27 KND-575 enil Engineering Licensure Examinations aad identity the state of stress in a material described as follows: Stress at Failure a) Proportional limit b) Yield strength ) Ultimate strength q) Rupture strength ~~ — 5, CE Board Nov. 2022 Ultimate strength - highest ordinate in stress-strain diagram eel ear Yield strength - an appreciable elongation or yielding of the material without any corresponding increase of load. [AeA Awooden log is to be used as a foot bridge to span a 2m. gap. The log is to support a concentrated load of 40 KN. Alowable shear stress = 1.2 MPa Allowable bending stress = 6.3 MPa. @) What is the diameter of the log needed for amaximum shear? ) Find the diameter of the log needed so that the allowable bending stress is not exceeded. A 150 wide rectangular log is to be used instead of the circular log. What is the Tequired minimum depth of the log so as not to exceed the allowable bending stress? Solution: a) Diameter of log needed for a maximum shear For maximum shear, the 40 KN should be placed at the end of the log. 40 KN Mae = 0 2Ri = 40 KN(2) Ri = 40 KN pi 3A 4(40000) 12=—g— A= 44,488.44 A==p 4 © 44,444.44 => D? D= 238mm say 240 mm b) For a maximum moment placed the load at the center 40kND-576 Civil Engineering Licensure Examinations S x 106(D/2) mD* or D=318 say 320mm °) Minimum depth of rectangular log 6M t= paz _ 6(13.5)108 150d? d= 300mm 6.3 Sta ‘A 12 mm thick tire has an inside diameter of 800 mm and is 110 mm wide. It is heated to fit into a wheel of bigger diameter. E = 200 GPa a) The diameter,of the wheel is 800.60:mm, solve for the Fesilting tensile stress in the tire. In the process, the resulting ~ circumferential stress between the tire and the wheel is 60 MPa. Calculate the contact pressure (MPa) betvieen the tire and the wheel. 4 b) ¢) The resulting circumferential stress between the tire and the wheel is 200 MPa, How much is the elongation of the tire? Solution : Resulting tensile stress in the tire 5= Ph © AE 8 = 800.5 - 800 =0.5 mm __P(800) | “? A(200,000) ‘ = 125 MPa (tensile stress in the tre) ) Contact pressure between the fre and the wheel when the circumferential stress | between the tre and the wheel is 60 MPa, P(800)(110) = 2T }25(12)(110) T= 165,000 N P(800)(110) = 2(165,000) P=3.75MPa ©) Elongation of the tire when the circumferential stress is 200 MPa T Circumferential stress = 5 = 200 TL AE _ 200(800) ~ 200,000 3= 038mm 6D-577 civil Engineering Licensure Examinations eed ech ‘A beam 12 m long is fixed at the ends. It ‘cartes a uniformly distributed load of 30 kN/m throughout its length. To prevent excessive deflection, a support is added at the midspan. The result is a two span beam, fixed at the ends and simply supported at the center. 2) Determine the reaction at center suppor. tb) Determine the shear at end B of beam AB. ¢) Determine the max. positive moment. Solution: a) Reaction at center support to prevent excessive deflection a w=304Nim ms C Hoo \ , Fr L=i2m - 7 _ wh 2 _ 30(12) c= 2 R= 180 KN b) ‘Shear at end B for beam AB WV +R=30(12) 2V = 360-180 V=90 ‘Shear at B= 90 KN we =30 kN/m. 5s +45 c) Max. positive moment 2 My == (neg. moment at A or B) _ 3012)? a 48 Ma =- 90 KN.m Max. positive moment: +M= wit ~ 96 = 30(12)? +M=—OE +M= 45kNm Check: Max.(+M) = —90 +22 Max. (+M) = 45 KN.mD-578 TAs Figure shows a semi-circular arch subjected to equal but oppositely directed forces at A and B. 2) Find the shear force at D. ) Find the moment acting at D. ©) Determine the axial force at D. Solution: a) Shear force at D. N Cos 53.13" = V Sin 53.13" N=1.333V SFr=0 V Cos 53.13" +N Sin 53.13" = 28 V Cos 53.13 + 1.333 V Sin 53.13°= 28 V= 1.68 KN b) Moment acting at D. ‘g 08 Teena Mo =2.8 (08) Mo = 2.24 kNm Axial force at D. N= 1.333V N= 1.333(1.68) N=224kN Mas eer °) Le=04m W=1.5kNin Shaft diameter = 76 mm Allowable torsion = 75 MPa Allowable flexural = 148 MP2 Allowable shear = 99 MPa For circular section: Moment of inertia = = Polar moment of inertia = <the maximum value of P such ble torsional stress not be the maximum value of P such Yi llowable flexural stress not be Solution ; Max, value of P that will not exceed the allowable torsional stress of 75 MPa Tr J nd* 32 (76)* 1°32 J=3275324 T=P(400) P(400)(76/2) 3275324» P=16161N P= 16.16 kN b) Max. value of P that will not exceed the allowable flexural stress of 148 MPa M=wL 6) +P(L) M=1.6(2)(1) + 2P M=3+2P n= - ame ug = ne M= 6.38 x 10° N.mm M= 6.38 KN.m_ 3+2P=6.38 P= 1.69 kN ‘Max. value of P that will not exceed the allowable shear stress AV °) =A ) 4 99 = ———— 3G) (76)? V = 336.83 kN Vew+P 396,83 = 1.5(2) +P P= 333.83 KN et ee PRs eed Beams supporting the slab panels are 300 mm wide and 400 mm depth (b x h). The columns supporting the beams are 400 mm square. Loads: * Total DL = 5.8 KPa LL=4.8kPa Load factors: 1.2 for OL 1.6 for LL rT. 100 mm = 400|D-580 Civil Engineering Licen Examinations For Dead load and Live load: foo A Resin Reade NS Live load only: 3 wl Toe _ fr} mm Ke Live load only: i wt. ze £ cet soe pom DN oss OSet a0ritet! oer’ rent” moment saint? Applying lve load pattern loading a) Determine the moment at joint L of beam KL. b) Determine the reaction at N- o) Determine the shear at L of beam KL. Solution : a) Moment at L of beam KL wo 14st (1 Ei Feo ot DL =5.8(25) = 14.5 kNIm LL = 4.8(2.5) = 12 kNm = 14.5 kNim Live load = 12 kNim 4205 wh? 1205(12)(6? Mu = 52.06 kN.m Max. moment at L: M=1.2DL+1.6LL M=1.2(62.90) + 1.6(52.06) M= 158.78 KNm.” b) Reaction at N Dead load = 14.5kNim — 143 WL Ro, = 1.143(14.5)(6) Ro. = 99.44 KN Live load = 12 kNimD-581 civil Engineering Licensure Examinations Total reaction at N: R=120L+ 1.6LL R=1.2(99.44) + 1.6(82.30) R= 251 KN ) Shear at L at beam KL wo AN + Dead load = 0.62(14.5)(6) Dead load = 53.94 kN Live load = 12 kNim Live load = 0.62(12)(6) Live load = 44,64 KN Total shear at L of beam KL: V=4120L+1.6LL = 1.2(53.94) + 1.6(44.64) V= 136.15 KN 13. CE Board Nov. 2022 Identify the degree of indeterminacy (n) of the structures shown. . Fordeterminate structures n = 0 a) iO @ - ® No, of reaction = 5 No. of static equations = 3 Indeterminate = 5 - 3 = 2 (24 degree) ) No. of reactions = 9 n=2 (2sections being analyzed) No. of static equations = 3n = 3(2) = 6 Indeterminate = 9-6 = 3 (third degree) ¢) ()D-582 Civil Engineering Licensure Examinations letra nan n=2 _ (2sections being analyzed) No. of reactions = 6 No. of static equations = 2(3) = 6 The structure is determinate. No. of reactions = no. of static equations 14, CE Board Nov. 2022 Due to architectural requirements, a column is of T-section as shown. g Given: Ast = 6 - 25 mm diameter bars * As2 = 4-25 mm diameter bars hy = 250 mm he = 450 mm bi = 200mm bz = 300mm fe = 28 MPa fy= 415 MPa 10 mm diameter ties. Clear concrete cover = 40 mm ‘Assume that for all bars, fs = fy a) a) What is the distance (mm) of the plastic centroid of the section 'from line 1 along the x-axis? Which of the following gives the location (mm) of the geometric centroid from line 4 along the x-axis? A load, Pu = 4155 KN acts along the x-axis at 420 mm from line 1. What is the resulting bending moment (kN.m.)? Solution: Distance of the plastic centroid from line 1 plastic centroidD-583 cnil Engineering Licensure Examinations eA S o, = (25) (4N418) c, #814851 c, #0851 A, , = 0:85(28)(700)(250) c, =4165000 ¢, =A, f. c c, = 407426 " Fl2oy (21416) ©, =0.851'A, = 0:85(28)(450)(300) c, = 3213000 F 25y'(ayats) C, = 814851 R=C, +C, +C, +C, +C, R=9415128 RX=C,(625) + C,(125) + C, (187.5) +0, (475) + C, (657.5) 9415128x = 814854(62.5) + 4165000(125) + 407426(187.5) + 3213000(475) +814854(657.5) = 266 b) Location of geometric centroid from line 4 along the x-axis O45 ' + ' | 200 1 |] 4a" ilo 00) | ye 2 apes 1 /3 ies 200 4 zal "125" 125) 3507450 ‘Ax = 700(250) = 175000 }50(300) = 135000 A=A; + A2= 310000 AK=A,x, +A,X, 310000% = 175000(125) + 135000(475) x= 274mm ) Bending moment when Py = 4155 KN is acting at 420 mm from line 4 420 [286 2] Pf] place leno’ | 200 | A 700) sah af 300 | 1 | 200 350° 450 e = 420 - 286 M,= 4155(0.134) M, = 556.8 KN.m.D-584 Cc PM ae A plane concrete beam has the following dimensions: b = 350 mm, h = 600 mm Concrete compressive strength: fc' = 34 MPa ‘Allowable tensile stress: fi= 0.10 fc a) Determine the bending moment that will cause the plain concrete beam to crack. b) The beam is cast with provision for a 185 mm diameter pipe located concentrically. How much is the resulting cracking moment? c) The beam is cast with provision for square duct 85 mm x 85 mm concentrically located. How much is the resulting cracking moment? Solution: a) Bending moment that will cause the plain concrete beam to crack 600 mm. ry +, =300 t 1 - > 300mm fe, = 0.10 fc" fa = 0.10(24) f= 3.4MPa Mer te fr = 0600)? = 2000600)" — 5400 x 10° mm* 9 600 4 =z = 300mm | 0 °) ivil Engineering Licensure Examinations ler Ne fe= Ig - Mer(300)- 3.4 = 5400 x 10° Me = 61.2.x 10° N.mm Me = 61.2 KN.m Resulting cracking moment if the beam is cast with a 185 mm diameter pipe located concentrically 185 mm "300 mm _-300(600)* _ 70105)" oa 12 64 lee = 5342.5 x 108 mm* ” Moh fa = __ Mer (300) 3.4 = 5342.5 x 10 Me = 60.55 KN.m Resulting cracking moment when the beam is cast with a square duct 85 mm x 85 mm concentrically located 300 mmD-S85 civil Engineering Licensure Examinations 300(600)* _ 85(85)* her = ~"12 12 =5395.65 x 108 mm* Mer Ie fer ler Mer (300) 3.4 = 5395.65 x 108 Ma= 61.15 KN A ae ese ra The flooring of a warehouse is made up of DT jpsts a8 shown. The joists afe simply supported on a span of 7.5 m. a=24m . A=2.0x 10° mm? |= 1.88 x 10? mm* The initial prestressed force is 745 kN located a 192 mm below the neutral axis, Loads: Dead load = 2.3 kPa Live load = 6.0 kPa a) Determine the tensile stress due to the initial prestress force. 4) Determine the max. compressive stress at midspan of the beam due to the given loads and the prestress force based on the following conditions: Loss of prestress = 18% Beam unit weight is 24 kNim? ) Determine the additional superimposed load that the DT can carry such that the fesulting tensile stress at the bottom fiber 's zero, Consider 18% loss of stress. | | | | | | } | | | a) b) Solution: 3) Tensile stress due to the initial prestress force et ee e= 192mm P=745(2) P= 1490 kN PMC 1490000 1490000(192)(267) 2.0 x 108 1.88 x 10° f= + 33.18 MPa (tensile stress) = Max. compressive stress at midspan of the beam due to the given loads and prestress force based on the following conditions: Loss of prestress = 18% Beam unit weight is 24 kN/m? P= 745(2)(0.82) P= 1221. 80 KN (loss of stress is 18%) Service loads: Superimposed DL = 2.3(2.4) = 5.52 KNm 2.x 105(24) (000) Weight of DT = 4.8 kNim Live load = 6(2.4) = 14.4 kKNim Total load: W = 5.52 + 4.8 + 14.4 W= 24.72 kNim Weight of DT =D-586 Ch wl? 8 wat mgs M= 173.81 x 108 N.mm Compressive stress at the bottom due to service loads and prestress force. _ 1221800 _ 1221800(192)(267) 2.0 x 108 1880 x 108 173.81 x 106(267) 1880 x 10° fo=-6.11-33.32 + 24.68 fy=- 14.75 MPa | WR wm “Ta7s c) Additional superimposed load that the py cary such that the resulting tensile stess at the bottom is zero, Consider 18% logs of stress. “1475 “Ta Ta. 0 auc fay M(267) 1880(10)* M= 103. 66 kNm LP . 8 14.75 = 75) 103.6 = “C5 w= 14.77 kNim (additional superimposed load)D-587 civil Engineering Licensure Examinations ‘combined footing is shown in the figure Column = 400 mm x 400 mm Net soil pressure in ultimate condition = 130 kPa pistance between columns = 3.9 m Effective depth = 500 mm Width of footing = 3 m Reduction factor for shear = 0.75 Reduction factor for moment = 0.90 2 fe! = 27.5 MPa Or @) Determine the wide beam shear stress along the y-axis, 4) Determine the wide beam shear stress along the x-axis, ©) Determine the punching shear stress at column B. Solution: i a) Wide beam shear stress along the y-axis + x=13 Total length of footing =04+04+35+04+1.3=6m Vu = 130(0.80)(6) Vu = 264 KN Vu obd = 824,000__ ¥ = 0.75(6000)(500) v= 0.28 MPa v= Wide beam shear stress along the x-axis 6m b) 602D-588 Civil Engineering Licensure Examinations Vu + 390(0.5) = 763 Vu = 568 KN Va abd » = 568,000 0.75(3000)(500) v= 0.50 MPa v= ¢) Punching shear stress at column B 35 foal is, reno | ‘A g=390 kNim a -s07 Compute the factored column loads at A and B. 156 + 390(0.4) - Ps = - 602 P1=914KN (column load at A) Column load at B. 763 + 390(0.4) - P2=-507 P2= 1426 KN Punching stress at column B, 250, 400, 250 0 O04 400 g=130 KN/m? 900mm Wu = 1426 - 130(0.90)? Wu = 1320.7 KN Punching shear stress: = Sho d bo = 4(900) = 3600 _ 1320700 °» = 075 (3600)(500) p= 0.98 MPaD-589 civil Engineering Licensure Examinations — eae aes the columns of a warehouse are fixed at the base and pin-connected at the top. About the axis, the columns are braced at mid-height. Sidesway is prevented and effective length factor = 1.0. Along the x-axis, there are no braces at mid-height. Sidesway uninhibited and effective length factor = 1.2 Given: Column Properties ‘ea, A= 8580 mm? Depth, d= 255 mm Flange width, br= 200 mm Flange thickness, t= 16 mm Modulus of elasticity, E = 200000 MPa Length of column = 6 m Web thickness, tw = 9 mm. k= 103 x 108 mm# m= 110mm y= 22x 108 mmé 1 mm F,=248 MPa =55mm a) Solve for the axial compressive stress, F (MPa) that would be permitted if axial force does exist. Refer to attached table. b) The axial load P; = 160 KN at an eccentricity e = 0.25 from the centroidal x- axis of the column. Solve for the max. stress. ¢) Using interaction equation 224 22 < 10 find the.maximum load P that ¢an be applied at an eccentricity of 0.25 m about the centroidal x-axis. Fa = 75 MPa, Fo= 148 MPa Solution: Axial compressive stress if axial force does exist KL EE EON) ignas T; 110 ly = 1.03000) = 58.82 ty 51 Use Be = 65.45 r From table: KLr F, 65 116.83 vis af 1.0) 65.45 07 oss 66 116.13 By ratio & proportion: (See attached table) 4% 5 207, 045 1.0 x= 0315 Fa=116.83-0.315 Fa= 116.5 MPa Max. stress Mc=Pe Mc = 160(0.25) Mc=40 kN M,C ina 255 Cas 2 127.5 fg = 102 108 (2275) bx 103 x 108 fox = 49.5 MPaCivil Engineering Licensure EX minations We FOR COMPRESSION MEMBERS OF 248 MPA SPECIE! Fa (MPa) 146.70 148.42 148.13 147.53 147.53 147.22 146.90 146.58 146.24 145.90 145.55, 145.19 144.83 144.45 144.07 143.69 143.29, 142.89 142.48 142.06 141.84 141.21 14077 140.33 139.68 139.4 138.95, 138.49 139.01 137.53 137.04 136.54 136.04 135.53 135.01 134.49 133.96 133.43 132.69, 132.34 Kur Fa (MPa) 131.79 13123 130.67 130.10 129.52 12894 128.35 127.76 127.16 126.56 125.95 125.33 124671 124.08 423.45 12281 122.47 12152 120.85 12021 11954 11887 118.19 11751 11683 115.44 11473 11403 11331 11260 11167 414.14 110.41 100.67 108.93 108.18 107.42 106.66 105.90 Kir a1 82 83 85 86 87 88 89 90 a1 92 93 95 96 97 98 99 100 401 102 103 104 105 106 107 108 109 110 ar 412 413 114 115, 116 7 118 119 120 Fa IED YIELD STRESS Kir Fa - (MPa Pa A 3 121 «69.92 10435 © 122,—=—« 68.92 10357 123.—=«s«7-92 10279 «1724S 66.91 40200 «125 «=—(65.89 40120 126=— (84.87 40040 «127 63.85 go59 «= 128.«S=s«62.86 9378 «129 61.89 9796 130« 60.94 9714131 60.01 ge31 132,859.14 9548 133, 88.22 9464 6134 (97.36 g3g0 135«= «5861 9295 «136 95.68 9310 137-8407 9124 - 138 © $4.08 9037 139 53.30 goso - 140 «© 2.54 a963 141 51.80 e775. 142: S107 86.86 143° 50.36 659714 4967 8507 «145 48.98 e417 145 48.31 6326 «147,47 68 8236 © 14847 02 142149 48.39 8050 150 4577 7966 9151 45.17 7e63 152, 44.58 7768 «183 43.99 7673 154 -43.43 7578 1854287 7482188 42.32 73.85 187 41.76 72881884126 71901594074 7091 160 40.28D-591 civil Engineering Licensure Examinations ————T ee eat ) Mat. load P P f= ie “as = 0.000117P MC f= M=025P 0.25P(127.5)(10)* fo = 703 x 108 &=0.31P 0.000117P | 0.31P _ 75 148 P=477KN Pe 9 LET a4 ‘A simply supported beam has the following Properties: ‘Area, A= 12031 mm? Depth, d= 465 mm Flange width, br= 193 mm Flange tiickness, t= 19 mm Web thickness, tw = 11 mm Moment of inertia about the x-axis: b= 4.36 x 10° mm* Yield stress, Fy = 248 MPa ‘Nlowable flexural stress, Fp = 163 MPa a) Solve for the moment capacity of the beam b) To strengthen the beam, 150 mm wide and 12 mm thick cover plate is added both at the top flange and at the bottom flange. ‘Compute the resulting moment capacity of the section (KN.m). ¢) The beam is to carry a moment of 480 kN.m. Cover plates at the top of the flange and at the bottom flange are to be added to strengthen the section. Using 10 mm thick cover plate, calculate the required width (mm) of the cover plate. Solution: a) Moment capacity of the beam af a8 Lon5 2 2 MC Fea _ M(232.5) © 4.36 x 108 M= 306 kKN.m 163 b) Moment capacity of the beam having cover plates of 150 mm wide and 12 mm. thick at the top and the bottom ___150 | —— L fe | x 238.5| caa44 x |d-465 = E> ae 150 150(12)? I, = 436 x 10° + era + 150(12)(238.5)%] (2) = 640.8 x 108 mm!D-592 Civil Engineering Licensure Examinations p =e b= Ik __M(244.5) 163 = Cioex 108 M= 427 kN.m c) Width of 10 mm cover plates placed at the top and bottom flange of the section if M= 480 kN.m (10)? a 6 I, = 436x106 + | +b (10)(237.5)) (2) Je = 496 x 108 + 1128291.67 b MC aT ; ____480.x 105(242.5) ~ 436 x 108 + 1128291.67 b 436 x 10° + 1128291.67 b= 714.11 x 10° 27811108 ~ 7128291.67 b=246.5.say 250mm 163 PMCs aes 74 Two channels are welded at the top of their flanges to form a box section. Column length = 6 m Effective length factor, K = 1.0 Properties of each channel: Area, A = 5690 mm? Depth, d = 300 mm Flange width, br = 80 mm Flange thickness, tt = 42mm » Web thickness, tw = 12 mm Moment of inertia, Ix = 67 x 10° mm* Moment of inertia, ly = 2.1 x 10° mm* Distance from the back of the channel to the centroidal y-axis, x = 17 mm a) Due to axial load P = 1000 KN, solve for the resulting compressive stress (MPa). b) The column is subjected to a moment of M = 270 kKN.m about the x-axis, compute the maximum resulting flexural stress (MPa). ¢) Solve for the critical slenderness ratio of the built-up column, : Solution: a) Resulting compressive stress due to axial load P = 1000 kN A= 2(5690) A= 11380 mm? P on4 f, = 1000000 p""'11380 fy = 87.87 MPaD-593 gui Engineering Licensure Examinations Hasirium resulting stress if the column is 4 jected to a moment of M = 270 kN.m about the x-axis k= 2167 x 108) = 134% 108 ~ MC haar _270(10)°(150) _ = —Syax ips = 302.2MPa Critical slenderness ratio of the built up column I= 134 x 105 = [2.4 x 108 + 5690(63)2] 2 )=49.37 x 108 mm* "= 65.87 mm Critical slendemess ratio: SL _1,0(6000) r 65.87 KL POL CE Board Nov. 2022 A cantilever beam 250 mm wide x 400 mm depth is 3 m long. It is subjected to the following loads. Dead load (including beam weight) = 12 KN/m Live load at the free end = 20 KN Concrete: fe’ = 28 MPa Longitudinal steel: fy = 413 MPa Lateral ties: fy, = 275 MPa Clear concrete cover to lateral tie Cover to centroid of reinforcement Load factor for strength design: U=1.2DL+1.6LL Strength reduction factor: 0.90 for moment 0.75 for shear Diameter of lateral ties = 10 mm a) Determine the maximum shear stress (MPa) in the cantilever beam. b) The tension reinforcement are to be placed on one layer only. Which of the following is most suitable? a) 8-16mmo —c): 4-25mma b) 5-20mma —d)- 3-2 mma ©) The beam is reinforced only with 2 legs of 10 mm diameter ties spaced at 200 mm on centers, Determine its nominal shear capacity (kN). Solution: Maximum shear stress in the cantilever beam 20kN w=12 kN. "y Vu = 1.2(12)(3).+ 1.6(20) Vu= 75.2 KNShear stress: vy ve gbd ___75200 ~ 0.75(250)(330) vc = 1.22 MPa Ue b) No of tension bars needed on one layer only fe. * T=A, fy if e ee ee 4-330 bear (330-a/2) | : dj ~tHc ao O.8sfe" Uttimate flexural strength. wi? M, = +1.6PL M, = RODB, 1,6(20)(3) Ma = 160.8 kN:m~ Mn = 178.67 KN.m + Ma =1(330-al2) c=T é 0.85 fe'(a)(250) = As fy a 0,85(28)(250) a=T T=5950a n= 7 (330-5) 178.67 x'10® = 5950a (330-5) 178.67 x 106 = 19635002 - 29752? a? - 60a + 60057.14 = 0 a=109mm T=Asfy 5950a = As (413) 5950(109) = As (413) Ag= 1570.34 mm? Try 3-28 mmo 28) = 1570.34 N=2.55 say 3 bars 2x +70 +70 = 250 x= 55mm Clear spacing = 35 - 28 Clear spacing = 27 mm > 25mm OK Note: NSCP Specs, spacing between horizontal bars should not be less than 25 mm.D-595 civil Engineering Licensure Examinations —_——. 0.0 Try 5-20mmo xe x x70 my wa I oo 330, F@20yN = 1570.34 N=4.99 say 5 bars 70 +70 + 4x = 250 x=27.5mm Clear spacing = 27.5 - 20 Clear spacing =7.5mm<25 NOT OK Try 4- 25 mmo 70,x E pe 700, 3-6. oe 250 : : 725)"N = 1570.34 N= 3.2 say 4 bars 70 +70 + 3x = 250 X= 36.67 mm Clear spacing = 36.67 - 25 = Clear spacing = 11.66 mm<25 NOTOK Use 3— 28 mme bars at the top. ¢) Nominal shear strength Capacity (kN) 10mm lateral tes 2250 Using 2 legs of 10 mm dia. of lateral ties A, = F(40)2(2) = 157.07 mm? Spacing of lateral ties = 200 mm on centers a Aofid ¥, =e 157.07(275)(330) yy v 200 Vs=71.27 KN Y= O17 Fe" by d V, = 0.17V28 (250)(330) Ve= 74.21 KN Y o Vn = Vs + Ve Nominal shear strength: Vn Vn=71.27 + 74.21 Vn = 145.48 kN =V4