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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 (centerof—bolt to sheared edge = 2”), and assume that the bolts are
spaced, centertocenter, 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 {wwwan. 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 := 27.34in2 = 14.681112
AlSC Table 1—5
A992 yield stress Fy := 50ksi
AlSC Table 2—3
A992 tensile stress Fu := 65ksi
AlSC Table 23
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 15 tw ~" 03871”
Given bolt diameter db ;= 1in
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 := ¢tFyAg = 661bp
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 — 2n‘dhtw = 13.14112
Effective net area _ 2
AlSC Equation D3—1 A6 '= U'An = 112“
Rupture capacity ¢anpture := (paFuAe = 544bp Design tensile cagacity DesignTcnsileCapacity := min(¢Pnyield,cl) = 544ki anpture) Pu := DcsignTensileCapacity = 544kip 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 J31 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 — 3ch = 1.531.1n
Controlling distance edge := Hﬁn(1.2Lcedge,2.4~db) = 1.838411
Nominal resistance .
of edge bolts Rnedge : edge‘tW‘Fu = 462.1%
Other bolts
Clear distance LootheI := 3m — dC = 2.062in
Controlling distance other := mn(1.2Lcother,2.4idlo) = 2.1~in
Nominal resistance .
of other bong Rnother := othertthu = 52.8k1p
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 15 d ;= 12.0111 g ;
Bolt spacing s ;= 3m 0 i Channel dimension K _ 1 125. j/ 3
AlSC Table 15 —  m " l, Clearance dimension c1 ;= 1375111 3’3 {wt ’55;
AlSC Table 716 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% = 2333“!
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) = 4.64in 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 ;= 1in
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,12tmin) = 6in 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,24tmin) = 12in
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 716 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 J31 Strength reduction factor cl) ;= 0.75
Applicable nominal stress FB ;= Fm, = 48ksi
Design strength of bolts ¢Rnb01tz= cbFnAb~Nm = 390161) 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 JSea
Edge bolts
. . 1 .
Clear dlstance Lcedge := 2m — Ede = 1.531n
Controlling distance edge := min(1.2Lcedge,2.4db) = 1.84in Nominal resistance of edge bolts Rnedge := edget maxFu = 107kip CE 4401 10/07/2009 HW #4 KEY Problem 2b: (Continued) Other bolts
Clear distance Lcother := Sin — dC = 2.06in
Controliing distance other 1: min(1.2Lcother,2.4db) = 2.1in
Nominal resistance V .
of other bolts Rnother := othertmaXFu = 122bp
Total resistance of 3 edge ¢R ~ 2: ch 3R + 6R = 7881d
b d th P
bolts and 6 other bolts n 6mg ( ne g6 no er)
J/ Block shear strength AiSC Eguation J45
I Uniform tension stress U __ 1 0
AiSC page 16.1113 bs " '
A36 yield stress ,
AISC Table 23 Fy ‘= 36km
. 2
Gross shear area Agv := tmaX(Le + 2s) = 8411
r ._ — ‘ 2
Net shea. area AI1V ._ AgV — 2.5tmaXdh — 5.5m
. , 2
Gross tensrle area Agt ;= tmax.(Lep + 2.5) = 9.1;;
. 2
Net shear area Am 2: Agt — 2.5tmaXdh = 6.5m 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, J36a and J45. ConnectionDesignStrength := min(cI>Rnbok, ¢Rnbemng,¢Rnblockshear) = 390kj 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 J34 Strength reduction factors AlSC page 16.1110 Class A surface
AISC page 16.1110 AISC page 16.1110
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 Dh ¢Rnslipservice : ¢scrvicew u sc'Tb‘Ns'N = 278%}? Dh u SCTbNSN = 236kip ¢Rnsliprequired : ¢requircd' u' = 278k1 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 23
A992 yield stress FyW := 50ksi
AlSC Table 23
A36 tensile stress FuL :: 58ksi
AlSC Table 23
A36 yield stress FyL := 36ksi
AlSC Table 23 5
Angle thickness tL 3: E“
W section thickness ‘ tW1= 0.470in
AlSC Table 11
Number of bolts N ;= 4
Number of shear planes m := 2
Bolt diameter db := 2tn
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= ¢~FnAbNm = 127kip Shear ieldin of an les AlSCe uation J43 We «WWI Strength reduction factor ¢V := 1.00 l m u . 2 , {9 Gross area Ag 2: tL~[2Le +(N—1)~s]= 3.59m
i  ._ _  a Desrgn strength ¢Rnyidd ._ (lav0.6FyLAgm _ 155 k1p CE 4401 10/07/2009 HW #4 KEY 3.2 Problem 3: (Continued)
Shear rugture of angles AlSC eguation J44
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 := ¢0.6FuL~AnVm = 117kip t . . .. t t
L (ﬁllet/Ego await” Wgaé‘ﬁrt'eté +0 "Hats gauge? 34, He we?! no”; 1535? ﬁMQ‘Qe‘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 — édc = 0.84in
Controlling distance edge ;= min 1.2Lcedge,2.4db) = 1.01411 Nominal resistance of edge boﬁs Rnedge := edgemtLFuL = 36.7~kip Other bolts Clear distance Lcother := s — dC = 2.19in Controlling distance other := min(1.2Lcother,2.4db) = 1.8in Nominal resistance , of other bolts Rnother 2: othermtLFDL = 65k1p  = 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 = 1.8in Nominal resistance , of edge bolts Rnﬂdgew := edgeWtWFuw = 55.0k1p 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: othertWFuw = 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) = 3.75in 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) = 7.5in 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 716 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, J36a, J43 and J44. 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.
 Fall '09
 Shield

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