HW%204%20Solution - CE4401 Fall 2009 Homework #4 Due...

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Unformatted text preview: CE4401 Fall 2009 Homework #4 Due Wednesday October 7 Show all of your work for all calculations. ltemize from where in the AlSC Manual or Specification you are getting your data (especially the more obscure data relating to clearances and spacings). This is important: it must be clear what you are doing and why. Use short statements to describe your work where helpful. Problem 1: Two C12x25s (A992) are connected to a 1 inch thick gusset plate. How many 7/8 inch A325X botls are required to develop the firll design tensile capacity of the member if it is to be used as a bearing type connection. You may assume that U=085. Do NOT consider the J4 provisions. Assume the bolts will be placed in two rows. When laying out the connection, assume the first bolt lies 2” from the sheared edge (center-of—bolt to sheared edge = 2”), and assume that the bolts are spaced, center-to-center, at 3”. Neglect bleclr shear. Neglect checking the strength of the gusset plate. Check all clearance limits on the double—angle. Gina a m, e\ 5 Problem 2: Consider the connection of a tension member to a plate as shown below. All the bolts are 7/8 inch A325N. The platesare made from A36 steel. , A f A y l i i m l 3 {it ~ fl; i: g w — A _ A 5; ; $3 in i r g “WW n+4 C u a in, j g m i E ,1 n | “,3... '- i ‘-’ ‘J G t; 4 ~ 1 ‘ r ’ i 5 m g l v .,..., l Hint em I An {wwwa-n. Mr 4:? r5 5 w c PL}: h i- ‘ g 1 Pl 1 la _ . ““ 1 i : {tn/3 45%” i 5 i i '4‘: g t r ; For—— m“ r) 1 i l 9 1 1 .L l ‘ 59:1; 1 , {I}; é: a) Check whether all bolt hole, edge spacing, and clearance spacings are sufficient (to keep things simple for the bearing failure check, bolt holes should be at least 2 2/3 bolt diameters apart -- you may measure this distance along the diagonal line between the bolts; edge spacings to the rolled and sheared edges are given in Table J 3.4; other limits are given in Sections J 3.3 through J 3.5 ; and clearance limits relate to the fact that there must be enough room for the nuts to fit). if a spacing check is not satisfied, just state this l'act clearly, but still proceed with the rest of the problem as if the check were satisfied. b) What is the design strength of this bolted connection if it is a bearing type connection? Do not check the strength of the tension member, just use the appropriate limit states from Chapter J. You may limit yourself to only checking the block shear pattern that you are guessing will control (i.e. don’t waste your time checking others, I’m only interested that you can pick out a reasonable pattern, and that you know how to make the calculation). c) What is the design strength of the connection if it is to be slip critical? Assume a Class A surface. At the end of the problem, summarize the results of all checks. Problem 3. Determine the maximum end reaction (Shear) that can be transferred through the web angle connection shown. The angles are made of A36 steel. The W 30x90 is made from A992 steel. The bolts are 3/4 in A325—N and are used in standard size holes. There is an angle on either side of the web of the W30X90. The W30X9O is not coped. CE 4401 ' 10/07/2009 HW #4 KEY Problem 1: How many 7/8 inch A325X bolts are needed for a bearing type connection of two 012x25s connected to a 1 inch thick gusset plate in order to develop the full design tensile capacity of the member. Assuming the plate is A992 and has the same edge distances it will not control in the calculations because it is thicker than the combined channel web (1 inch > 2*0. 387 inch). General Area of both channels Ag := 2-7.34in2 = 14.681112 AlSC Table 1—5 A992 yield stress Fy := 50ksi AlSC Table 2—3 A992 tensile stress Fu := 65ksi AlSC Table 2-3 A325X shear stress Fm, := 60ksi AlSC Table J32 Number of bolt rows n _= 2 Number of shear planes m ;= 2 Channel web thickness _ , AlSC Table 1-5 tw ~" 0-3871” Given bolt diameter db ;= 1-in 8 Bolt area Ab ;= 3.de = 0.604112 4 Determine design tensile capacity of two C12X25$2 Yielding AlSC Eguation D2—1 Strength reduction factor (tat ;= 0.99 Yield capacity ¢Pnyield := ¢t-Fy-Ag = 661-bp Rugture AISC Eguation D2—2 Strength reduction factor ¢t2 ;= 0.75 Given shear lag factor U := 0.85 Bolt hole diameter for net area d ._ 15 , 1 , _1 , AlSC Table J33, 133.2 h '— jg“ + E“ — '1” Net area of 2x0 An := Ag — 2-n‘dh-tw = 13.14112 Effective net area _ 2 AlSC Equation D3—1 A6 '= U'An = 112“ Rupture capacity ¢anpture := (pa-Fu-Ae = 544-bp Design tensile cagacity DesignTcnsileCapacity := min(¢Pnyield,cl) = 544-ki anpture) Pu := DcsignTensileCapacity = 544-kip CE 4401 10/07/2009 HW #4 KEY 1.2 Problem 1: (Continued) Determine number of A325X bolts needed: Shear strength of bolts AlSC Eguation J3-1 Strength reduction factor {,5 ;= 0.75 Applicable nominal stress Fn ;= Fm = 60.161 Pu Minimum number of bolts N1 := ———-— = 10.1 ¢.Fn.Ab.m Bolt hole bearin stren th Edge bolts Bolt hole diameter for clear dc ;= 1—5411 = 0933.111 distance AlSC Table J33 16 . . 1 . Clear dlstance LcadgC := 2m — 3-ch = 1.531.1n Controlling distance edge := Hfin(1.2-Lcedge,2.4~db) = 1.838411 Nominal resistance . of edge bolts Rnedge : edge‘tW‘Fu = 462.1% Other bolts Clear distance LootheI := 3m — dC = 2.062-in Controlling distance other := mn(1.2-Lcother,2.4idlo) = 2.1~in Nominal resistance . of other bong Rnother := other-tthu = 52.8-k1p Pu . .. . . .- . .. x f') h ’1 Minimum number Oi belts —¢ T Afinolher ‘ 4'Rnedgc based on strength of 2 edge N2 2: £——————— : 7.1 bolts and N2 - 2 other bolts Rnothor Controllin number of bolts Maximum of two above values Ncomml ;= maX(N1,N2) = 10_] Round up to next even number for a symmetric N ;= 12 bolt pattern (12) 7/8 inch A325X bolts are required to develop the full design ensile cacity of the meber for a bearing type connection. CE 4401 10/07/2009 HW #4 KEY 1.3 Check clearance, edge distance and spacing limits of channel: E Channel depth AISC Table 1-5 d ;= 12.0111 g ; Bolt spacing s ;= 3m 0 i Channel dimension K _ 1 125. j/ 3 AlSC Table 1-5 -— - m " l, Clearance dimension c1 ;= 1375111 3’3 {wt ’55; AlSC Table 7-16 5" BoltSpaccCheck := if g 2 C1, "OK" , "Not OK” = "OK" """""""" “ d KValueCheck := if(-2— — -:- — K 2 C1,”OK" , "Not OK" Distance to sheared edge L e ;= 2m . d — s . Distance to rolled edge L6p ;= 2 = 45.111 L '= 1.5m Minimum edge distances AlSC Table J34 Check edge distance requirements Lepmin, "OK" , "Not OK”) = "OK" Minimum bolt spacing 2 , AlSC J33 8min -= [2 + 3% = 2-333“! Check minimum spacing spaccCheCk :___ if‘s 2 Slum: ,siNot OKs ) = "OK" requirements Maximum edge distance AISC J35 Check maximum edge distance requirements Bagsmax := min(6in,12tw) = 4-.64-in EdgcChcck2=if(max(Le,L >Edge "Not 0K","0K")="0K" op) max’ Maximum bolt ‘ . . . Spacing NSC J35 L Smax .= Imn(12m,24tw) = 9.29111 Assuming (a) applies Check maximum bolt spacing requirements SpaceCheckZ := if (s > 3 "Not OK" , "OK" = "OK" max’ CE 4401 10/07/2009 HW #4 KEY 2.1 Problem 2a: Check it bolt hole, edge spacing, and clearance spacings are sufficient. General Bolt hole spacing s := 3m Bolt diameter db := Zin 8 Distance to sheared edge Le ;-_- 2111 Distance to rolled edge LEp ;= Bin Thinner plate thickness {min ;= 3m 8 Thicker plate thickness tmax ;= 1-in Minimum bolt hole s acin AlSC J3.3 . . r 2 Minimum value 3 - := 2 + — -d = 2.334;: mm ( 3) ’0 Check minimum and actual value Checkl := if (s 2 s "Bolt hole spacing 0K" , "Bolt hole spacing not OK") Checkl = "Bolt hole spacing OK" Distance to sheared edge AlSC Table J3.4 Minimum value L6min ;= 1,5111 min’ Check minimum and actual value Checkz := if (14‘s 2 L "Distance to sheared edge OK" , "Distance to sheared edge not OK” ) ICheckz = "Distance to sheared edge OK" I Distance to rolled edge AlSC Table J3.4 Minimum value Lepmin ;= 1.125111 emin ’ Check minimum and actual value Check3 := if(Lep 2 L epmin, "Distance to rolled edge OK" , "Distance to rolled edge not OK") Check3 = "Distance to rolled edge OK" Maximum edge distance AlSC J35 Maximum value Lemax := mjn(6in,12-tmin) = 6-in Check maximum and actual value Check4 := if(max(L L )SL 6p, 6 "Maximum edge distance OK” , "Max edge distance not OK'” emaX ’ Check4 = "Maximum edge distance OK” CE 4401 10/07/2009 HW #4 KEY 2.2 Problem 2a: (Continued) Maximum bolt hole sgaclng AlSC J3.5 assuming not subject to corrosion Maleum value smaX := min(12in,24-tmin) = 12-in Check maximum and actual value Check5 := if(s S smax, "Bolt hole spacing OK” , "Bolt hole spacing not OK") Checks = "Bolt hole spacing OK“ 7 Entering and tightening clearance Minimum clearance dimension C _ 3 , AlSC Table 7-16 1 -— 1 + g m —— "Clearance OK" Problem 2b: Determine the design strength of the connection it it is a bearing type connection. General A325N shear stress F __ 1 _ Also Table J32 nv T 48‘51 Number of bolts N ;= 9 Ti 2 . 2 Bolt area Ab := Zeb = 0.60m Number of shear planes m ;= 2 Shear strength of bolts AlSC Eguation J3-1 Strength reduction factor cl) ;= 0.75 Applicable nominal stress FB ;= Fm, = 48-ksi Design strength of bolts ¢Rnb01tz= cb-Fn-Ab~N-m = 390-161) Because the thicker plate is less than twice the thickness of the thinner plates the thicker plate will control the remaining calculations. Bolt hole diameter for net area d ._ 15 . 1 _ _1 . AlSC Table J33 and 03.2 h “ "gm “L 1‘6"“ T '1” Bolt hole diameter for clear distance AlSC Table J33 A36 tensile stress dc := eh — 0.0625m = 0.938411 AISC Table 2—3 Fu ‘= 58k“ Bolt hole bearing strength AlSC eguation JS-ea Edge bolts . . 1 . Clear dlstance Lcedge := 2m — Ede = 1.53-1n Controlling distance edge := min(1.2-Lcedge,2.4-db) = 1.84-in Nominal resistance of edge bolts Rnedge := edge-t max-Fu = 107-kip CE 4401 10/07/2009 HW #4 KEY Problem 2b: (Continued) Other bolts Clear distance Lcother := Sin — dC = 2.06-in Controliing distance other 1: min(1.2-Lcother,2.4-db) = 2.1-in Nominal resistance V . of other bolts Rnother := other-tmaX-Fu = 122-bp Total resistance of 3 edge ¢R ~ 2: ch 3-R + 6-R = 788-1d b d th P bolts and 6 other bolts n 6mg ( ne g6 no er) J/ Block shear strength AiSC Eguation J4-5 I Uniform tension stress U __ 1 0 AiSC page 16.1-113 bs " ' A36 yield stress , AISC Table 2-3 Fy ‘= 36km . 2 Gross shear area Agv := tmaX-(Le + 2-s) = 8411 r ._ — ‘ 2 Net shea. area AI1V ._ AgV — 2.5-tmaX-dh — 5.5m . , 2 Gross tensrle area Agt ;= tmax.(Lep + 2.5) = 9.1;; . 2 Net shear area Am 2: Agt — 2.5-tmaX-dh = 6.5-m Block shear design strength ¢Rnbiockshw ;= ¢.(0.6.min(Fu~AnV,Fy.AgV) + UbsFuAm) = 412%, Connection design strength The connection design strength wiii be the minimum of the limit states evaluated in AISC equations J3—1, J3-6a and J4-5. ConnectionDesignStrength := min(cI>Rnbok, ¢Rnbemng,¢Rnblockshear) = 390-kj Enigma éefietz w W W WWWWWW a ,W a WWWMM J. i f ~——~—~—5—-«——-——~—4§~—«m~——% - > “i” w? v A {2% e 53"? é W m w s A W, CE 4401 Problem 20: 10/07/2009 HW #4 KEY 2.4 What is the design strength of the connection it it is to be slip critical. Assume a Class A surface. Slip critical connection resistance AlSC eguation J3-4 Strength reduction factors AlSC page 16.1-110 Class A surface AISC page 16.1-110 AISC page 16.1-110 Standard size holes Slip planes Bolt pretensicn AlSC Table .131 Slip resistance for serviceability Slip resistance for required strength it slip prevention is for serviceability ConnectionDcsignStrcngch := min(ConnectionDcsignStrcngth,(iJR ¢’servicc : 1‘00 dDrequired : 0'85 p. := 0.35 Du ;= 1.13 hsc := 1.00 NS := m = 2 Tb := 39kip D-h ¢Rnslipservice : ¢scrvicew u sc'Tb‘Ns'N = 278%}? D-h u SC-TbNSN = 236-kip ¢Rnsliprequired : ¢requircd' u' = 278-k1 nslipservice) it slip prevention is at the reguired strength level Summary of all checks: I Problem 2a: ConnectionDesignStrcngth3 := min ConnectionDcsignStrcngth,(bR )= 236~ki nsliprequire This connection meets all bolt hole, edge spacing, and clearance spacing requirements in AISC. Problem 2b: The connection design strength of this member is 390 kip for a bearing type connection. Problem 20: it slip prevention is a serviceability requirement the design strength of the connection is 278 kip. it slip prevention is desired at the required strength level the design strength of the connection is 236 kip. CE 4401 10/07/2009 HW #4 KEY Problem 3: Determine the maximum end reaction (shear) that can be transferred through the web angle connection shown. The angles are made of A36 steel and the W30x90 of A992 steel. Bolts are 3/4 inch A325N with standard holes. There is an angle on both sides of the W30x90 and the W30x90 is not coped. game! A992 tensile stress FuW ;= 65kg AlSC Table 2-3 A992 yield stress FyW := 50ksi AlSC Table 2-3 A36 tensile stress FuL :: 58ksi AlSC Table 2-3 A36 yield stress FyL := 36ksi AlSC Table 2-3 5 Angle thickness tL 3: E“ W section thickness ‘ tW1= 0.470-in AlSC Table 1-1 Number of bolts N ;= 4 Number of shear planes m := 2 Bolt diameter db := 2-tn 4 Bolt area Ab := $de = 0.44.1112 4 Shear stress A325N bolts an := 48kg} Distance from bolt hole to _ ( 1 _ edge of angles Le ‘_ 1 + 7 m - v \ "‘/ Bolt spacing s := 3111 it is obvious the angles will control shear yielding and rupture because they are made of a material with smaller yield and tensile stresses and have smaller area than the W section, Shear strength of bolts AlSC eguation JS—i Strength reduction factor (1);: 0.75 Applicable nominal stress En ;= Fm Design strength ¢Rfib01tz= ¢~Fn-Ab-N-m = 127-kip Shear ieldin of an les AlSCe uation J4-3 We «WWI Strength reduction factor ¢V := 1.00 l m u . 2 , {9 Gross area Ag 2: tL~[2Le +(N—1)~s]= 3.59-m i - ._ _ - a Desrgn strength ¢Rnyidd ._ (lav-0.6FyL-Agm _ 155 -k1p CE 4401 10/07/2009 HW #4 KEY 3.2 Problem 3: (Continued) Shear rugture of angles AlSC eguation J4-4 Strength reduction factor 4; = 0,75 Bolt hole diameter for net area 13 . 1 _ , AlSC Table J33 and 03.2 dh : jg“ + 1—64“ = 0875"" 1 Net area Anv := tL-[Le + (N — Us] — tL-(N — O.5)~dh = 2.25m2 W3 Design strength ¢Rmupmre := ¢-0.6-FuL~AnV-m = 117-kip t . . .. t t L (fillet/Ego await” Wgaé‘firt'eté +0 "Hats gauge? 34, He we?! no”; 1535? fiMQ‘Q-e‘j Angle bolt hole bearing AlSC eguation J3—6a Strength reduction factor (in = 0.75 Bolt hole diameter for clear 13 , distance AlSC Table J33 dc ‘= T6“ gem Clear distance Lcsdgs :2 Le — é-dc = 0.84-in Controlling distance edge ;= min 1.2-Lcedge,2.4-db) = 1.01411 Nominal resistance of edge bofis Rnedge := edgem-tL-FuL = 36.7~kip Other bolts Clear distance Lcother := s — dC = 2.19-in Controlling distance other := min(1.2-Lcother,2.4-db) = 1.8-in Nominal resistance , of other bolts Rnother 2: otherm-tL-FDL = 65-k1p - = W section bolt hole bearing AlSC eguation J3—6a Strength reduction factor 4; = 0.75 Edge bolts Clear distance is obviously much greater than the bolt diameter for edge bolts in this case and will not control Controlling distance edgeW ;= 2.4db = 1.8-in Nominal resistance , of edge bolts Rnfldgew := edgeW-tW-Fuw = 55.0-k1p Other bolts Nominal resistance of other bolts Total resistance of ‘l edge ¢R - 2: <1:- 1~R + 312 I = 165kip bolt and 3 other bolts nbeamgw ( nedgew netherW) Rnotherw I: other-tW-Fuw = CE 4401 Problem 3: 10/07/2009 HW #4 KEY 3.3 (Continued) Minimum bolt hole sgacing AlSC J33 . . 2 Minimum value Smn I: + = Check minimum and actual value Checkl := if (s 2 s "Bolt hole spacing OK" , "Bolt hole spacing not OK") Checkl = "Bolt hole spacing OK" Distance to sheared edge AlSC Table J3.4 L min’ Minimum value mm .= 1.25m Check minimum and actual value Checkz := if(Le 2 Lemm, "Distance to sheared edge OK" , "Distance to sheared edge not OK" ) Checkz = "Distance to sheared edge QK" Distance to rolled edge AlSC Table .134r Assume given 2 inch dimension controls for angle calculations. Minimum value L ‘= lin epmin ' Check minimum and actual value Check3 := if(2in 2 L "Distance to rolled edge OK" , "Distance to rolled edge not OK") epmin’ Check3 = "Distance to roiled edge OK" Maximum edge distance AiSC J35 Maximum value L max ;= min(6in,12~tL) = 3.75-in Check maximum and actual value Check4 := if (Zin S L "Maximum edge distance OK" , "Maximum edge distance not OK" ) emax ’ Check4 = "Maximum edge distance OK" <L Checks 2: if (Le _ cmax;"Maximum edge distance OK" , "Maximum edge distance not OK" ) Checks = "Maximum edge distance OK" Maximum bolt hole sgacing AlSC J35 assuming not subiect to corrosion Maximum value 5 X := min(121n,24.tL) = 7.5-in ma Check maximum and actual value Check6 2: if (s S smax, "Bolt hole spacing OK" , "Bolt hole spacing not OK”) Check6 = "Bolt hole spacing OK" CE 4401 10/07/2009 HW #4 KEY 3.4 Probiem 3: (Continued) Entering and tightening clearance Minimum clearance dimension __ 1 . AiSC Table 7-16 C1 -— 1 + 1 m Check7 := if(—s- 2 C1, "Clearance OK" ,"Clearance not OK" —- "Clearance OK" 2 All clearance, spacing and edge distance limits are satisfied. It is obvious there will be no clearance issues with the Wsection due to its large depth and much smaller angle length. Maximum end reaction The maximum end reaction that can be transferred through the given connection will be the minimum value between the limit states checked in AISC equations J34, J3-6a, J4-3 and J4-4. MaximumEndReaCtion I: Hfin(¢Rn100h, ¢Rnyield, ¢anptura, ¢RnbeafingL, ¢Rnbeafingw) = 1 ...
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This note was uploaded on 06/09/2010 for the course CIVIL 4401 taught by Professor Shield during the Fall '09 term at University of Minnesota Crookston.

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HW%204%20Solution - CE4401 Fall 2009 Homework #4 Due...

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