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Unformatted text preview: CE440l Fall 2009 Homework #3 Due Wednesday September 30 Problem 1. Using the equations (not any design aid) determine the axial load capacity of a
Wl4x109 column with Fy=50ksi. The effective length is 11 ft about the weak axis and 22 ft about
the strong axis. Does the column buckle elastically or inelastically? Problem 2 Assuming axial loads only, select W sections for the columns of the rigid frame shown
in the accompanying illustration. Assume that the connections perpendicular to the frame are
simple, and the columns are braced at the top and bottom. All girders in the plane of the frame are
W33xl4l. Use A992 steel. Use the same size section for two stories. The columns are oriented for strong axis bending in the plane of the ﬂame. Concrete weighs 150 1b/ft3. Live load on roof = 30psf. Roof self weight (dead load) = 6 psf. Live load on interior ﬂoors = 80 psf. Partition load on , interior ﬂoors = 15 psf. Frames repeat 30 ft on center (in and out of the paper). The frame shown
is an interior frame (not on the edge of the building). You may assume there is no moment in the
columns due to the loads given. Assume the axial loads on the columns come from tributary area
(as in recitation). 4 in reinforced concrete \ 5 . H 1
12ft ‘ (838% 1’9“ 5555' “>3 37643
‘ 5,7 “é Bin
12ft 1 ix :2 i ; ioxgg A Concrete
7 r“ 4?
1211‘ flag??? HZng :1?"
l
12ft 1 Meiji“? 0mg“? é A //
“l * 4cm ‘ 40ft 40ft D Problem 3. For the welded connection shown, determine the design strength of the connection, as
well as the tension member (you do not need to determine the strength of the gusset plate). Be sure
to check the details on the weld size as well. A992 steel is used for both the tension member and
the gusset plate. The weld is a 5/16” ﬁllet weld, E70 electrode. Bar 3 X 1/2 / I lap“ t=3/8” Problem 4: a) b) Design a welded connection for an MC 9x239 Grade 50 to a 3/; inch thick gusset plate. The
gusset plate is made of A36 steel. Use E60 electrodes. Include longitudinal welds only.
The connection must support a dead load of 48k and a live load of 120k. You may design
the connection for the load combination, not for the full strength of the tension member. Do
not design the other dimensions of the gusset plate, just assume they are adequate so that the
gusset plate will not fail. Check the adequacy of the MC 9X23.9 for the limit states of yield on gross and fracture on
net. CE4401 09/30/09 HW #3 KEY
Problem 1: Using the equations determine the axial load capacity of a Wi4x109 column with
Fy = 50ksl. The effective length is 11 ft about the weak axis and 22 ft about the
strong axis. Does the column buckle elastically or inelastically’?
Given
Yield stress Fy := 50ksi
Strong axis effective length KLX := 22ft
Weak axis effective length KLy := 11ft
Modulus of elasticity E := 29000ksi
W14x109 ro erties AlSC Table 11
Area Ag := 32.0in2
Strong axis radius of gyration xx ;= 5.22111
Weak axis radius of gyration ry ;= 3,73in
Flange compactness ratio xﬂange: 8.49
Web compactness ratio xweb := 21.7 Comgactness criteria AlSC Table 34.1 Case 3 f E
Y FlangcCheck := if xﬂangc < xﬁmitﬂange, "FLB OK" , "FLB ox") = "FLB OK" I E
Case I: 1;“ = y WebCheck := if(x Cb < xﬁmjtweb, "WLB OK" ,"WLB not OK" = "WLB OK" W Controlling slenderness ratio The maximum slenderness ratio defined as KL/r will control the axial capacity of the
member. Here the slenderness ratio is defined as n in each direction. According to
AlSC page 16.1—33 the slenderness ratio should not exceed 200. Strong axis nx ;= KLX = 424
I‘
X
Weak axis 1] := E = 35.4
y ry Controlling slenderness ratio n ;= maxmxmy) = 424 SlenderCheck := if (n S 200, "Slendcrness OK" , "Too slender") = "Slendcmess OK" 1.1 CE4401 09/30/09 HW #3 KEY
Calculation of axial load cagacity AlSC Chagter E
Strength reduction factor (b ;= 0.9
. C
Section E‘l
. . . . 2
Elastic critical buckling stress 'n' E ,
Equation E34 Fe = "—2 = 159%}
Tl BuckleCheck := if(Fe 2 0.44F y, "Column buckles inelastically" , "Column buckles elastically") BuckleCheck = "Column buckles inelastically" f:
Flexural buckling stress F . e .
Equations E$_2 and E3_3 F01. .— 1f Fe 2 0.44F , 0.658 F ,O.877‘Fe ~ 43.8k51 y 3’ Nominal compressive strehgth _
Equation E31 A ' 1403401) n z: Fcr g Axial load capacity AxialLoadCapacity := (lacPH = 1262~ki 1.2 CE 4401 Problem 2: 09/30/2009 HW #3 KEY 2.1 Select W sections for the columns of the rigid frame assuming no moment in the columns. Load setug Concrete unit weight Roof live Root dead
Floor live Floor dead
Column A tributary area Column B tributary area
Column A tributary length
Column B tributary length
Column height Girder weight Column weight initial
assumption Girder moment of inertia,
AlSC Table 1—1 7 := 150pcf
Lr := 30psf Dr := 6psf + "y.4in = 56psf
L := 80psf
Df := 15psf + 76in = 90psf A := 30ft20ft = 6OOft2 B := 30~ft~40ft =1200ft2' L A := 20ft
h := 12ft
lbf
Dg.— 141?
lbf
D := 50—
C ft
. 4
Ig .— 7450m I will design for columns A & B stories 1 & 3. The same Wsections can then be used for columns C &
D because of symmetry. Given that a Wsection is to be used for2 stories and our loading, stories 1 6t
3 will obviously carry more load than stories 2 & 4, respectively, so the same W sections can also be
used in stories 2 & 4. The partition load will be treated as a dead load assuming that their weight and
location are well known and relatively constant. I assume the girders are oriented in strong axis bending.
l neglect the weight of girders perpendicular to the frame. See problem statement for ston/ numbering. Determine PL , PLr and PD at each stony: A 4th stow Dead load
Live load Roof live load A 3rd stog Dead load
Live load Root live load A 2nd stom Dead load
Live load Root live load I: + + = g
PAM I: 0
PALr4 2: = PALI3 I: = PALYZ I: = CE 4401 Problem 2: (ContinUed) A let slog
Dead load Live load Roof live load
B 4th story Dead load Live load Roof live load
B 3rd story Dead load Live load Roof live load
may Dead load Live load Roof live load
31—81% Dead load Live load Roof live load 09/30/2009 HW #3 KEY PALrl I: = I: + + = g I: 0 + hDC =188kjp PBLr3 == PBLr4 = 350‘kiP 1’13er : PBLI3 = 360'1‘1113 g PBLrl : PBLrZ = 36'0'1‘19 Determine controlling load case for A & B stories 1 8: 2, AlSC p. 28: Column A slow 3 Column A slog 1 Column B stow 3 Column B slog 1 11 PuB3 = 397ki 2.2 CE 4401 09/30/2009 HW #3 KEY 2.3 Problem 2: (Continued) Assume: 1) Weak axis buckling controls column axial capacity (must check later)
2) K = 1.0 for the weak axis which is conserative for a braced frame, (KL)y = 12 feet Pick most economical W section for each location from AISC Table 41: Column A stow 3 Section
I ‘— 110' 4
Moment of inertia A3)” 1“
lbf
Self weight DA3 3: 31'?
Radii of gyration ratio IA3 ;= 1,72 Axial capacity Column A stog 1 Section 10x49
I M 272' 4
Moment of inertia A1 "‘ 1“
lbf
Self weight DAI ¢= 49;;
Radii of gyration ratio rAl ;= 1,71 Axial capacity Column B stem 3 Section 1 10x45
. 4
Moment of inertia 1133 2= 248m
lbf
' := Self weight DB3 ft
Radii of gyration ratio r133 3= 215 Axial capacity Column B story 1 Section
. 4
Moment of inertia 131 = 740m
lbf
Self weight DB1 ;= 87.?
Radii of gyration ratio r131 : 1~75
Axial capacity ¢Pn31 := 980k1p CE 4401 Problem 2: (Continued) Recalculate column loads based on actual column weights instead of assumed column weight 09/30/2009 HW #3 noting that the dead load factor in all controlling load cases was 1.2: Column A story 3
Column A story 1
Column B story 3
Column B story 1 PuAS I: — — = P A1: PuA1 — 2.1142(2))C — DA3 — DAI) = 490kip u P1133 := P1133 — zit12(1)C — DB3) = 397bp 11 We see that all of our column selections are still adequate. Determine K factors for the strong axis. There is no indication that the frame is braced in the
strong axis direction so Figure CCZA will be used. The E term drops out of the NSC Equation for G as the girders and columns have the same modulus of elasticity: Column A stem 1 Bottom; pinned connection,
AlSC page 16.1241 Top; 1 W33x141 girder,
2 W10X49 columns Figure CC2.4 Column A stom 3 Bottom; 1 W33x141 girder
1 W8x31 and 1 W10x49 column Top;
1 W33x141 girder,
2 W8x31 columns Figure CC2.4 Column B stem 1 Bottom; pinned connection,
AlSC page 16.1241 Top;
2 W33x141 girders,
2 W12x87 columns Figure CC2.4 GAlT : —I—
_g
LB KAIXZ G 2: ———————
A3B I j C)
>
w
ri .I. II o H 0 KEY 2.4 CE 4401 . 09/30/2009 HW #3 KEY Problem 2: (Continued)
Column B stem 3 IB1 133
Bottom; T + _h_
2 W33x141 girders GB3B := m = 022
v1 W10x45 and 1g
1 W12x87 column 2 g
I
T zit?)
OP; ‘_ __ _
1 W33x141 girder, G33T .— I — 0.11
2 W10x45 columns {43]
LB
Figure CC2.4 KB3X1= 1.0 Check assumption that weak axis buckling controls and adjust design if needed: The maximum slenderness ratio (KL/r) of the weak and strong axis will control,
noting that the unbraced lengths in both directions are equal and Ky = 1.0 for all cases the following inequality will be used to check the assumption.
Column A stem 8 Column B stem 3 rB3
CheckB3 := if ( 2 1,"Weak axis controls" ,"Strong axis controls" = "Weak axis controls" Kssx Column A stem 1 1“B1
CheckBl := if [ 2 1, "Weak axis controls" , "Strong axis controls" = "Weak axis controls" Kle So'we now see that we must check Column A story 1 for strong axis buckling.
All other column selections are okay. Calculate effective length of column A story 1 with respect to ry: K h
Effective length Alx = 12.4ft
rA1 2.5 CE 4401 09/30/2009 HW #3 KEY 2.6 Problem 2: (Continued) From AISC Table 4—1, we can see that at an effective length of 13 feet with respect to ry,
a W10x49 has a capacity of 493 kip which is greater than Pu for this member so we
conclude that we are okay with an effective length with respect to ry of 12.4 feet. Check if our sections are compact and meet slenderness requirements: a AlSC Table 84.1 for limiting A ratios
0 AlSC Table 1—1 for radius of gyration and A values
0 AlSC page 16.133 KL/r < 200 Modulus of elasticity E := 29000ksi
Yield stress AlSC Table 23 Fy ;= soksi Limiting flange ratio Case 3 )\f ;= 0,55. 3 = 135
I F
3’ Limiting web ratio Case 10 kw ;= 1.49. = 35.9
Y Weak axis length factor Ky := 1.0 Column A stog 3 W8x31
Controlling radius of gyration ryA3 ;= 2.02411 Flange ratio fo3 := 9.19
Web ratio XWA3 := 22.3
Ky‘h Controlling slenderness value xA3 ;= = 71.3
1'yA3 Slender A3 := if(>\A3 S 200, "Slendemess OK" , "Slendemess not OK" = "Slenderncss OK"
Web A3 := if XWA3 S kw, "WLB OK" , "WLB not OK" = "WLB OK"
Flange A3 2: if(>\fA3 S kf, "FLB OK" , "FLB not OK" = "FLB OK" Column A stow 1 W10x49
Controlling radius of gyration rxA1 ;= 435.111 Flange ratio fo1 := 8.93
Web ratio wa1 := 23.1
KAlx'h Controlling slenderness value )\A1 ;= = 58.6 IxAl
Slender A1 := if >\A1 S 200, "Slendemess OK" ,"Slenderness not OK” = "Slenderness OK“ CE 4401 09/30/2009 HW #3 KEY
Problem 2: (Continued)
Column B stow 3 W10x45 Controlling radius of gyratlon ryB3 z: 2.01in Flange ratio >\ﬂ33 := 6.47 Web ratio XWB3 := 22.5
19“ Controlling slenderness value x33 ;= = 71_6
ryB3 SlenderB3 := if X33 5 200, "Slenderncss OK" , "Slendcrness not OK" = "Slendcrness OK"
WebB3 := if (XWB3 S kw, "WLB OK" , "WLB not OK" = "WLB OK" FlangeB3 := if )‘fB3 S Rf, "FLB OK" , "FLB not OK" = "FLB OK" Column B stout W12x87 Controlling radius of gyration ryBl ;= 307.111 Flange ratio xfBl := 7.48 Web ratiO I: Controlling slenderness value xBl ;= h = 459
IyBl SlenderBl := if RBI S 200, "Slenderncss OK” , "Slcndcmess not OK" = "Slendemess OK"
WebBl 2: if (XWBl S kw, "WLB OK" ,"WLB not OK" = "WLB OK"
FlangeBl := if (kﬂal S kf, "FLB OK” ,"FLB not OK" = "FLB OK" We now see that all of our members meet compact and serviceability requirements.
The W section design for the frame is as indicated on the page with the problem statement. 2.7 CE 4401 09/30/2009 HW #3 KEY
Problem 3: Determine the design strength of the connection and the tension member.
Check weld size details as well. A992 steel is used for the gusset plate
and tension member. The weld is 5/16" with E70 electrode.
General
A992 yield stress ‘_ 50k ,
AISC Table 2—3 Fy " SI
A992 tensile stress Fu := 65ksi
AlSC Table 23 3
Gusset plate thickness tp1 : Em
._ l 
Bar thickness tbar " 2'1“
Bar width Wbar ‘= 3m
. 2 Bar area I: = 1.511’1
Electrode strength 13E:= 70m
Weld length LW : 5m . 5 .
Weld Size dw := 1—61n '— dw — 0 221 ‘
Weld thickness tW '_ E T ' '1“
Determine tension member design strength:
Tensile yielding AiSG Eguation D21 Strength reduction factor (pt ;= 0.90 Design strength Dstyidd := (intFyAbar = 67.5'kip Tensile rugture AlSC Eguation D22 Strength reduction factor ¢t21= 0’75
_L_W_ _17
AlSC Table D3.1 Case 4 ratio Wbar _ '
LW LW
Shear lag factor U := if ——— 2 2,1.0,if 2 15,087,075 = 0.87
Wbar Wbar
Net area (no bolt holes) An ;= AhaI
Effective net area , 2
AISC Equation 031 Ac = U'An =130'm
Design strength Dstmpmre := ¢t2.FuAe = 63.6kip The design strength of the tension member will be limited by
the minimum value given by AISC Equations 021 and D22 3.1 CE 4401 09/30/2009 HW #3 KEY 32 Problem 3: (Continued)
Check weld size detail:
AlSC Table J2.4 minimum weld size
Thickness of thinner part joined tthin: mjnltbar’tpl) = 0375'in
Minimum weld size
dmin := if(tthin S iin,%‘in,if[tthm S gein,1i6in,if(tthin S iin,i~in,%~in))) = 0.187~in Check minimum and actual MinWeldSizeCheck := if (dw 2 d "Minimum weld size OK" , "Minimum weld size not OK") min’ MinWeldSizeCheck = "Minimum weld size OK"
AlSC J2.2blb) maximum weld size . , . 1 . 1 . .
Maxrmum weld srze dmax := 1f(tbar < Z1n,tbm,tmr — E111) = 04375111
MaXWeldSizeCheck := if (dw S dmax, "Maximum weld size OK" , "Maximum weld size not OK") MaXWeldSizeCheck = "Maximum weld size OK" AISC J2.2b minimum weld length
Minimum weld length Lmin 2: 4~dW = 1.25in Check minimum and actual MinWeldLengthCheck := if (LW < Lmin, "Minimum weld length not OK" , "Minimum weld length OK") MinWeldLengthCheck = "Minimum weld length OK" AISC J2.2b maximum weld length Maximum weld length Lmax ;= 100.dw = 3125411 Check maximum and actual MaxWeldLengthCheck := if(LW > L "Maximum weld length not OK" ,"Maximum weld length OK" ‘ maX’ I
MaxWeldLengthCheck = "Maximum weld length OK"
AISC J2.2b weld termination "Fillet weld terminations should be located approximately one weld size from of
(sic) the edge of the connection to minimize notches in the base metal... " A ssume the weld termination criteria is met CE 4401 09/30/2009 HW #3 KEY Problem 3: (Continued) Determine the design strength of the welded connection: Base metal shear yielding AlSC Eguation J4—3 Strength reduction factor <1) 2: 1.00
. 2
Gross shear area AgV := 2LWtbalr = 5 1n
Desrgn strength Dssyield := ¢AgV0.6Fy = 150kip
Base metal shear ru ture AISC E uation J44
Strength reduction factor (132;: 075 Net shear area (no bolt holes) AnV := AgV = 5in2 Design strength Dssmptum ;= (MAIN0.61:u = 146kip Weld metal strength AlSC Eguation J23 Design strength Dsweldmetal := ¢2~O.6FE2‘LWtw = 69.6kip The design strength of the welded connection will be limited by
the minimum value given by AISC Equations J43, J44 and J2—3 3.3 CE 4401 Problem 4: and fracture on net. Assumgtions
0 Grade 50 steel is A992 0 Weld will give similar shear lag factor to AlSC Table J3.1 Case 8
with 4 or more fasteners per line in direction of loading General A992 yield stress
AlSC Table 23 A992 tensile stress
AISC Table 23 Service dead load
Service dead load Obvious controlling load case
AlSC page 28 Gusset plate thickness Electrode strength MC 9x23.9 properties AlSC Table 16
Web thickness Gross area Check adequacy of tension member for yield on gross and fracture on net: Tensile yielding AISC Eguation D21
Strength reduction factor Design strength Tensile rugture AlSC Eguation D2—2
Strength reduction factor Shear lag factor
Net area (no bolt holes) Effective net area
AISC Equation D31 Design strength The design strength of the tension member will be limited by
the minimum value given by AISC Equations 021 and 022 09/30/2009 HW #3 Design a welded connection for an MC 9x23.9 Grade 50 to an A36 3/4 inch thick gusset
plate. Use E60 electrodes and longitudinal welds only. Service dead load is 48 kip and
service live load is 120 kip. Assume other gusset plate dimensions are adequate for it
not to fail. Check the adequacy of the MC 9x23.9 for the limit states of yield on gross 2: 50km FuMC I: 65161 D = 4816p L = 120kip Pu := 1.2D + 1.6L = ZSOkip
t ._ .3. ' p1 .— 4 '11] FE := 60ksi tMC I: A 2: 7.02m2 g ¢t := 0.90 DStyield 1: ¢‘t'FyMC' ¢t2 := 0.75
U := 0.80
An := Ag A6 := UAn = 5.62m2 Dstrupture : dDtZ'FuMC'Ae = 274'kip DSTM1= min(DStyield,DStmpture) = 274kip KEY A = 316kip CE 4401 09/30/2009 HW #3 KEY 4.2 Problem 4: (Continued) Pu DSTM MCAdcquacy = "MC 9x239 is adequate for YOG and FON" Pick maximum allowable weld size according to AlSC limitations MCAdequacy := if { S 1, "MC 9X23.9 is adequate for YOG and FON" , "Not adequate") 5
' d := —in = 0.313411
Weld Size W 16
 dw — o 221 '
Weld thickness tW— —2 — . 1n AlSC Table J2.4 minimum weld size
Thickness of thinner part joined tthin : mi“(tMC’tlii1) = 0‘4'in
Minimum weld size 1 1 1 3 3 1 5
d  := if t  S —in,—in,if t  S —in,—in,if  _ —~in,—in,—in = 0.187in
mm (thm 4 8 ( thin 2 16 (tthm 4 4 16 Check minimum and actual
min’
MinWeldSichheck = "Minimum weld size OK" AlSC J2.2b(b) maximum weld size
1 . . 1
Maxrmum weld Size dmax := if(tMC < ZinJMOtMC — Tam) = 0.338in MinWeldSizeCheck ;= if(dw 2 d "Minimum weld size OK" ,"Minimum weld size not OK") Check maximum and actual >d max " MaxWeldSizeCheck = "Maximum weld size OK" Design the welded connection: MaxWeldSizeCheck := if (d "Maximum weld size OK" , "Maximum weld size not OK” ) W, Shear yielding and rupture of the MO is not likely because of the outstanding legs so these limit states
will not be checked. The gusset plate will not be checked because its dimensions are unknown. Weld metal strength AlSC Eguation J2—3
Strength reduction factor (1)2 ;= 0.75 Pu W ‘_ ¢20.6FE2tw Round to nice value Lweld ;= 21.0i Minimum weld length L = 20.9~in CE 4401 KEY 4.3 09/30/2009 HW #3 Problem 4: (Continued)
Check tension member slenderness: AlSC D1 Check AISC weld length limits:
AISC J2.2b minimum weld length
Minimum weld length I Lminz= 441W = 1.25in
"Minimum weld length not OK" , "Minimum weld length OK") Check minimum and actual MinWeldLengthCheck := if(Lweld < Lmin,
MinWeldLengthCheck = “Minimum weld length OK"
2: 100~dW = 31.25in AISC J2.2b maximum weld length
Maximum weld length Lmax
"Maximum weld length not OK" ,"Maximum weld length OK" ‘ Check maximum and actual MaXWeldLengthCheck ;= if(Lwe1d > Lmax,
MaXWeldLengthCheck = "Maximum weld length OK" "Fillet weld terminations should be located approximately one weld size from of Assume the member is sufficiently short so that L/r does not exceed 300 AlSC J2.2b weld termination
(sic) the edge of the connection to minimize notches in the base metal... " Design Detail
r saga liti8 9x233 5ft M (TYRE: ...
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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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