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# File Tutorial5withAnswers - Tutorial 5 W If the...

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Unformatted text preview: Tutorial 5 W If the wide—ﬂange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam. Set W" = 200 mm. Solution 7-2 a. [40m _ 0.992s” Secﬁon Preps-ﬂies : 1 = Eli—(0.2)“).310)3 - .11—2-(0.1"/5)(0.250)3 = 268155200)“5 :11 Qm = 2531 = 0.062§(o.125)(0.025) + 0.140(0..2)(0.o30) 1.035300)" m.3 ‘V 3- 3 . -3 Q _ 0(10) (10353)(10) =4_62MPa 4 . -— M _. _ Ans run: I -5 I 268.652( 10) (0.025) E-Z Problem 7-3 If the Wide~ﬂange beam is subjected to a shear. of . V = 30 kN, determine the shear force resisted by the web of the beam. Set w = 200 mm. ' Solution 7-3 = -—-—(02)(0 310) - -—»(O.175)(0250)3 12 0.155+y Q—( W 7!: -—6 268.652(10)_ (0.2) 5—5 =263.652(10)‘5 m‘ ){D.155—- y)(0.2) = U..1(O 024025 — yz) 1;: 11,114 = 55 3343(10) c"‘Imsswozmzs — y )(0.2dy) 0.125 1 30. .155 =11.1(5159(10)6 {0.0911025}: - -y ] 3 0125 if = 1.457 W 1;, = 30 -* 20.457) = 27.1kN Ans 5-?) Problem 7—6 ' If the T-beam is subjected to a vertical shear of V = 10 kN, determine the maximum shear stress in the heam. Also, compute the shear—stress jump at the ﬂange-web junction AB. Sketch the variation of the shear—stress intensity over the entire cross—section. Show that INA = 5320400 311114. 5-5 Solution 7—6 54.23113111‘1 '=1= :1: + * =1: . y :1 15 30 140 60 60 6O 235.769mm 30*140+60’f60 1 3 2 1 3 , 1 = —*140* 30 +30*140* 35.769-15 + —*60* 60 - 12 ( ) ( ) _ 12 ( ) 8' +60* 60*(60-35.769)2 = 5320400 mm“ QM = ygA' = 27.1155*54.231*60 = 882301111113 3|: =1: 1: ‘= ———————VQM = _———————w 1000 88230 = 2.76 N2 =2.76.MPa m“ It 5320400*6O mm Ans QAB : w: = 20.769*30*140 = £17230an * =1: . (1.3).. = 3335— = 10 1000 87230 = 1.171 N2 =1.171MPa. Itf 5320400*140 mm . K * * 2 O (313)“, = E = 10 1°00 87 3 = 2.733 N2 == 2.733 MPa 11w 5320400*60 mm Shear stress jumps = (TAB)w - (rm). = 2.73 - 1.17 = 1.56 MPa Ans 1.17 MP; ,.2.73MPa ‘- 141 The dnuble T—beam is fabricated by welding the three plates together as shnwn. Determine the shear stress in 1b: weld necessary 10 support the shear Korea 05 V c 80 kN. 1‘} 20mm 232A _ 0.01(O.215)(0.02) + 2[0.095(0.15)(0.02)]' = 0059515 “1 2A y = 0.215(002) + 2(0.15)(0.02) ' 1 = .113(0.215)(0.023) + 0.215(0.02)(0.059515—0.01)2 + 2[—13é(0.02)(0.153) +0.02(0.15)(0.095—0.059515)2] : 29.4909(10") m‘ 9 = 0.059515 — 0.01 = 0.049515m Q -_- 301' = 0.049515(0.215)(0.02) = 0.2129(10‘3) m3 I Shear stress : _ VQ 80(103)(0.2129)(10-3) r___..._ I r 29.4909(10“)(2)(0.02) 714.4 MP3 Ans 7;! The double Theam is fabricated by welding the three p es togelher as shown. If the weld can resisI a shear stress mun. = 90 MP8 determine the ma ' be applied [0 ”1,5 beam. xlmum shear V that can 23% 0.01(0.215)(0.02) + 2[D.095(0.15)(ﬂ.02}] : .0 059515 m 211 0.215(0.{)2) + 2(0.15)(0.02) y” I = 1:2(0.215){0.023) ‘+ l3.215(0.02)(0.0595{5—0.01)2 + 2[11—2-(0.02)(0.15’) +0.02(0.15)(0.095—0.059515)1] = 29.4909(10") m“ 5’: 0.059515 — 0.01 = 0.049515 10 Q =i'A' = 0.049515(0.215)(0.02} = 0.2129(10-3) n13 V9 1:— It -3 90(106) = V(O.212:)(10 ) - 29.491[10 )(2)(0.02) V: 499 'kN Ans 7—39 The box beam is made from four pieces of plsszic that are glued together as shown. If the shen- V = 2 kip, deter mine the shear stress resisted by the. seem a] each of the glued joints. 7 1 331;" I T — - . 3 1219x525 {is/j __ = m .. I r 23.231(2)(0.25) _ 2(103)(3.4375) . _ —.____ = 23.231(2)(0.25) 592 pm '74“ The beam is subjected lo a shear of V= 800 N. Determine Ihe average shear stress developed in the nails Ilong lhe sidesA and B if 1he nails are spaced :- 2 [00 mm apart. Each nail has a diameter of 2 mm. = 0.015(0.03)(o.2s) + 2(o.o7s)(o.15)(o.03) _ 0 04773 0.03(o.25) + 2(o.15)(o.03) " ' m I 1' = E(O.ZS)(O.O33) + (Ill.25)(0.03)(0.04‘i|‘7343.015)2 1 + (2x33) (0.030153) + 2(o.os)(o.15)(o.ms—o.oan'73)2 = 32.164773(1o“)m‘ 'A' = o.03273(0.25){0.03) = 0.24547500‘51113 - 800 (o.245475)(10*3) _ ..__._____ =51o .44 N 31164773004) 5 [m : gs .2: 6105.44 (0.1) = 610.544 N Sineeeach side of the beam resists this shear force then 610.544 F 1'" = —- = = 97.2 , 24 2(41’)(0.0022) MPa Ans 3m . 3 Q I: ‘n __ \Sox‘aoo __ \2. H ‘2 —33—3¥ Saamwrmm ‘ . __‘ 6 I M :7 6b.1: \2“5 74334-51“? W (a ﬁrs \uo : {Ag 2 \SO%\SO><'+§ k—‘S’g-‘A -- = «6379,0136 “mm _ , .. 6 W Va. l%.}5x\§x\-E%?sx\o \$31.5 was 54 \SO Faculty of Engineering @IUIT‘ SI Staff/ Student use oniy JOB ____",__,__________._____—._~4 PAGE 5 # LO WM. amt: 31' Pixie—”(5% as- D L. 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