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Unformatted text preview: CE4401 Fall 2009 Homework #5 Due Wednesday Getober 14 The first test will be given on Wednesday October 21 during the first hour of class. You will be
allowed to bring a single 8.5Xl l sheet of paper with anything you want written on both sides, your
AlSC Manual, and a calculator. The exam will cover through bolted connections (i.e. HW 4). ignore selfweight unless otherwise noted. Always include an appropriate check for local buckling
due to flexureunless otherwise noted. Shear (Vn), local web yielding, local web crippling, and
serviceability need to be checked only where noted. Problem l. Draw an X—Y plot (graph) of the nominal bending moment, Mm (on the yaxis), versus
the unbraced length for bending, L, (on the Xaxis), of a W30Xl48. The range of L, is from zero to
40 feet (480 inches). Show all calculations before drawing the graph. Do not use any design
aides. Assume the section is compact (which it is). Tabulate your results, showing clearly the
transition between Eqs. F2l, F2—2, and F2—3. Make the plot on graph paper (or using excel) and
label all important points (with both their symbol and actual value) and hey equations on the
graph. The graph should be neat and to seale. Use A913 steel, Grade 65, Fy = 65 ksi. Assume: e C, = 1.0
a XX (strong) axis bending e Use units of hips and inches Problem 2. A 20ft long simply supported beam is required to carry a uniform dead load of 0.8 kip/ft
including its selfweight and a uniform live load of 2.3 kip/ft. Considering bending only (ie.
ignoring shear, the effects of point loads, etc), determine the least—weight Wshape to carry the
load. Assume full lateral support and A992 steel. The maximum moment is at the center and is
equal to wuL2/8. Do not worry about deﬂections. if this beam were made of A913 (Fy = 65 ksi)
what would be the lightest Wushape? Problem 3. A 35 foot simplespan beam is loaded with a uniform dead load of 0.3 k/ft in addition to
its own selfweight and a uniform live load of 0.9 lt/ft. The lateral supports are located at 7 ft
intervals. Determine the least weight A992 W—shape to carry the ﬂexure. The moment diagram is
wux
2 shown below. The factored moment as a function of x is My = (L — x) . Ignore deﬂections. 0.14
012 i «a ,,/~\_.\\ w
a“ 0,1 “WW wwwl \
3s": 0.08 a \ 7 .
E; 0.06 i / m“ W
E 0.04 1 ’/ ....... 7
0.02 /, ,,,,,,,,  ,\
O ;/ .\/\
0 5 10 15 20 25 30
x {a}
Problem 4
P P
[ i
l w wdead = 0.5 k/ft
r V V l l i l l 1 l l l l l l I
TéDW Te“ ‘ Wlive : 2 k/ﬁ
l 3
15' : ‘lO'M 1' 40' _
“ 5 ‘‘‘‘‘ 1 i Pdead — 16 k
Plive : 24’ k
Use A992 steel
The loads given above do not include beam Selfweight, however, you must consider beam
selfweight in your design Bending is about the xx axis Compression ﬂange is braced by slab Use elastic structural analysis Use same depth beam in both spans Beam is fully continuous over interior support Select a W beam such that ¢bMp 2 Mu under factored loads, and the maximum deﬂection does not exceed L/3 60 under the unfactored live loads. Show how you got started and all other work (i.e. the
work for all of your iterations). Determine if Lateral bracing is needed in the negative moment region. If such bracing is required,
determine its location. A spreadsheet that provides the moment diagram and the live load
deflections is available for your use from the class webpage. CE 4401 10/14/2009 HW #5 KEY 1.1 Problem 1: Plot Mnx on y axis versus Lb for a W30x148 with Lb ranging from O to 40 feet. Assume the section is compact. Show the transition between equations F21, F22 and F23.
Label key equations and important points with their symbol and value. Use A913
Grade 65 steel, Fy = 65 ksi, CD = 1.0, strong axis bending and units of kips and inches. General
Steel modulus of elasticity E ;= 29000161
Given yield stress Fy := 65ksi
Lateral torsional buckling Cb := 1.0
modification factor
Maximum unbraced length meax 2: 40ft = 480111
W30x148 ro erties AlSG Table 11
Radius of gyration ry := 2.28111
Shape property Its :2 2.77111
Torsional constant I := 14.51114
Page 16.148 c := 1
Elastic section . 3
modulus about x axis SK 3: 436m
Plastic section _ 3
modulus about x axis ZX ': 500m
Distance between h __ 29 5.
flange centroids 0 '_ ' m Limitin len ths AlSC E uations F25 and F2—6 1"— Equation F25 L := 1.761rJE = 84.8in
P y Fy Equation F26 E 1
LI := 1.95rtS—— #
0.71:y SXho . CE 4401 10/14/2009 Problem 1 : (Continued) HW #5 KEY Moments AlSC Eguations F21, F22 and F23 When Lb<=Lp Equation F2—1 Mp := ZX'Fy = 32500k1p1n Sample calculation for Lb = 196 in Assumed unbraced
length value When Lp<Lb<=Lr
Equation F22 L131 := 100111 \4 '— 'M c M M 07FS £‘31~:—L—p —31346kj r
in.—m1 p’ b' p_( p“ y x) Lr—L — pm P Sample calculation for l.b g 368 in Assumed unbraced 14102:: 300m
length value
C 2 L
F ‘— b.“ 'E 1 0078 J": b2
Equation F2—4 or" —’"2‘ +  Sh  It
2 X o s
rts
When Lp>Lr Mn 2: miniMpECIsx) = 15164kipin Equation F2—3 Excel wiil now be used to construct the glot 2
—] = 34.8ksi 1.2 ma Eﬁﬁ “.5 m Ev j
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$32208 H 532; Ex mn >>I moom .530qu vﬁ HOS» mu CE 4401 10/14/2009 HW #5 KEY 2.1 Problem 2: 20 foot long simply supported beam with uniform dead load of 0.8 kip/ft including selt
weight and uniform live load of 2.3 kip/ft. Consider bending only and full lateral support.
Determine the least weight W section to carry the load using A992 steel and A913 Fy = 65 ksr
General Beam length L := 20ft
Dead load “’13 := 03313 ft
Live load WL := 2.3 ﬂ ft
Obvious controlling load kip
case AlSC page 28 Wu .= 1.2WD + 1.6WL = 4.64—ﬂ— wuL2
Maximum factored moment Mu ;= = 232.kip.ﬁ
8 For A992 steel we can enter Table 32 and look for the
lightest section with ¢bMpX greater than or equal to 232 kipft. i or A992 steel the lightest section is a W18x35 ForA973 steel we must enter Table 32 and look for the lightest section with the required
plastic section modulus determined as follows: AlSC Chapter F
Strength reduction factor F1.(1) . 43b ;= 090 A913 yield stress Fy65 := 65ksi Mu a
Required plastic section 2X ;= — = 475.1113
modulus AISC Equation F2—1 <l>b'Fy65 Now we look in Table 32 for the lightest section with the required section modulus . or A913 steel the lightest section is a W16x31 CE 4401 10/14/2009 HW #5 KEY Problem 2: Continued
Check for oomgaot sections
Modulus of elasticity E := 29000ksi
A992 yield stress FySO := SOksi
A913 yield stress Fy65 = 65 ksi
Case 2 AISC Table E
54.1 for A992 steel and "ﬂange/11mm: 038' p 50 2 9'15
then A913 steel y
._ E _
>\rtangaimms '— 0'38‘ g— — 803
y65
Case 9 AlSC Table E
54.1 for A992 steel and *webhmitso == 376‘ —F 50 =90"
then A913 steel y
o E .—
kminimum ~= 376' ‘5“ ‘" 79'4
V 3’65 W section grogerties AlSC Table 1—1 A992 W18x35 flange ratio kﬂanoeso := 7.06 A992 W18x35 web ratio )‘we’oSO ;= 535 A992 steel local buckling checks A992FLBCheck ;= if(kﬂang650 > xﬂangeﬁmﬁso, "A992 FLB Not OK" ,"A992 FLB OK”) A992WLBCheck := if(>\ webSO > >‘W6‘olimit50’ ”A992 WLB Not OK” , "A992 WLB 0K") A992FLBChcck = "A992 FLB OK"
A992WLBChcck = ”A992 WLB OK" A913 W16x31 flange ratio xﬂangeéS := 6.28 A913 W16x31 web ratio >‘web65 ;= 51.6 A913 steel local buckling checks A913FLBCheck 2: if(>\ﬂang€65 > xﬂangehmﬁ, "A913 FLB Not OK" , "A913 FLB OK")
A913WLBCheck := if(>\web65 > xwebhmités , "A913 WLB Not OK” , "A913 WLB OK") A913FLBCheck = "A913 FLB OK"
A913WLBCheck = "A913 WLB OK" 2.2 CE 4401 10/14/2009 HW #5 KEY 3.1 Problem 3: A 35 foot simple span beam is loaded with a uniform dead load of 0.3 kip/ft in addition
to its own self weight and a uniform live load of 0.9 kip/ft. The lateral supports are at
7 foot inten/als. Determine the least weight A992 W section to carry the tlexure. General
Beam length L := 35ft
Dead load assuming WD := 50ﬂ + 0.3—le = 0.35ﬂ
self weight ft ft ft
Live load wL ;= 0.9393
ft Obvious controlling load kip case AISC page 28 Wu 2: 1.2WD +1.6'WL =1.86?
. Wu'L2 MaXImum factored moment M ;= = 285~kip~ft Determine Cb AlSC Eguation F11 We know that the unbraced segment from x = 14 feet to x = 21 feet will control. We need to
determine the moment at the quarter and three quarter point of this segment since the maximum
moment and moment at the middle of the segment are known to be Mu. From symmetry we know that the value of moment at the quarter and three quarter points of this unbraced segment
will be equal. Maximum moment Mmax := M11 x value at quarter point xA := 14ft + i(Zlft — 14ft) = 15.75ft
. Wu'xA Moment at quarter pornt MA := 2 .(L  XA) = 282kipft Moment at centerline MB 2: M11 = 285kipft Moment at three quarter point MC ;= MA = 282kip.ft Monosymmetry parameter Rm ;= 1.0
AlSC page 16.146
12.5Mmax
LTB modification factor Cb ;= min ——————.Rm,3.0
2.5Mmax + 3'MA + 4MB + 3MC
Cb = 1.005
Moment value to enter M __ Mu _ 2 3 1d ﬂ
Table 310 with table  —  8 ‘ P‘ CE 4401 10/14/2009 HW #5 KEY 3.2 Problem 3: (Continued) From page 3 123 we pick a W21x44. Moments will be recalculated based on the actual self
weight. The value of Cb will not change. Dead load with w ;= 44.33: + 03$ = 0.344ﬂ
 D2 ft ft ft self weight Obvious controlling load kip Wu2 := 12tz + 1.6WL = 1.853— case AISC page 28 ft
. WuZ'LZ Malele factored moment M112: 8 = 284kip'ft Moment value to enter Mu2 _ Table 310 with MtableZ == —Cb = 282'klp'ﬁ Again we select a W21x44 from page 3123. We are now done with the design iteration process
noting that the ¢Mp value from Table 32 for a W21x44 is also greater than Mu. ightest A992 section to carry the flexure is a W21x44 Check for comgact section Modulus of elasticity E ;= 29000ksi
A992 yield stress Fy := 50ksi
Case 2 AISC Table 84.1 xﬂangelimit := 038 ’E = 9.15
FY
Case 9 AISC Table 84.1 >‘webiimit := 3.76 ’2 = 90.6
FY
W section ro erties AlSC Table 11
W21x44 flange ratio >‘ﬂange ;= 7,22
W21x44 web ratio xweb := 53.6 Local buckling checks
FLBCheck := if()\ﬂange > xﬂangeﬁmjt, "FLB Not OK" , "FLB 0K")
WLBCheck := if(xweb > xwebﬁmjt, "WLB Not OK" ,"WLB 0K") FLBCheck = "FLB OK" WLBCheck = "WLB OK" CE 4401 10/14/2009 HW #5 KEY 4.1 Problem 4: Select beams of equal depth for the two spans to carry the flexure of factored loads and
‘ to limit service live load deflection to U360. First iteration The value of the point lead is the only known in the initial step and will be calculated firs t. It is
obvious load case 1.213 + 1.6L will control from AISC page 28. Service live point load PL ;= 24km
Service dead point load PD := 16kip
Faotored point load Pu := 1.2PD + 16‘PL = 57.6'kip Other known values Span length L ;= 40ft . f . kip
Uniform dead load on right span W1): 05—— ft .5 ' . . kip Uniform live load on right span WL ;= 2——
ft Factored uniform load on right kip Wu l: 1.2'WD + 1.6'WL = 3.8— span without self weight ft Determine remaining spreadsheet inputs based on assumed values Assume moment of , 4 . 4
inertia values Ia : 10001” 1b := 1000111
Assume self weights 551—1061;: 75% selfb ;= 75.1%:
Factored uniform load kip on ,eﬁ span wua .= 1.2selfa = doe—ft— Faotored uniform load kip with z: 1.2seifb + Wu 2 3.89"};— After the above values were input into the spreadsheet the maximum factored moment and
service live load deflections are obtained from the spreadsheet. on right span Maximum factored moment in left span Mua 3: 803kip'ﬁ Maximum factored moment in right span Mg}; 5: 303kipft Maximum service live load deflection in left span Ag 5: 34911?
Maximum service live load .
deflection in right span Ab 5: 1.78111 CE 4401 Problem 4: First I will check deflection values to determine if my assumed moments of inertia are likely
to be sufficient or not. A suggested moment of inertia will also be computed 10/14/2009 HW #5 KEY (Continued) l , . . . L ,
SerVice live load deflection limit AallOW ;= SE6 = 1.33.111
Controlling deflection value Amax := max(Aa,Ab) = 3.49111 DeﬂectionCheck :2 if(A > A max aHOW’ "Deﬂection not OK" , "Deﬂection OK") DeﬂectionCheck = "Deﬂection not OK” Suggest a new moment of A inertia value for the next I — b l  1335‘ 4
_ .. suggestb '_ A b T m
iteration allow Suggest a new moment oi A inertia value for the next I ._ a I — 26175 4
. , suggesta "T A a T m
iteration allow Nowi will select a W section sufficient in moment for each span from AiSC Table 32.
The section selected for moment is a W27x84 for both spans. Nowi will select a Wsection based on the recommended moment of inertia values for
each span from AISC Table 33. We see that the section selected for moment is
reasonable for deflection so the next iteration will be perfermed using a W27x84 for
both spans. Second iteration Determine spreadsheet inputs based on known values Moment capacity Table 3—2 ¢bM.pxa2 := 915kipft 6pr?ng z: 915kipyft
Table 3—2 moment of _ 4
inertia values IaZ ”1 28501“ Ib2 ‘= 1212
Assume selt wel tits 1f ~— 8491f— lf ’— lf
9 so a2" ft Si: 132'" so a2
Factored uniform load _ kip
on left span Wuaz .= 1.2selfaz — 0101?
Factored uniform load I kip Wubz 2: 12861be + Wu :1 3.901"— on right span ft 4.2 CE 4401 10/14/2009 HW #5 KEY 4.3 Problem 4: (Continued) After the above values liii/ere input into the spreadsheet the maximum factored moment and
service live load deflections are obtained from the spreadsheet. Maximum factored moment in left span MuaZ :: 806kip'ft Maximum factored moment in right span Mubz : 806hp'ﬁ Maximum service live load deflection in left span A312 := 123m Maximum service live load A _ 0 6’“ deﬂection in right span b2 " ' 3m NOW] Will check deflection values to determine if moments of inertia are sufficient.
Check deﬂection of left span DeﬂectionCheckZa 2: if (AaZ > AaHOW’ ”Left deﬂection not OK" , "Left deﬂection 0K") DeﬂectionCheckZa = "Left deﬂection OK" Check deflection of right span DeﬂectionCheckZb := if (A102 > A "Right deﬂection not OK" , "Right deﬂection OK”) allow ’ DeﬂectionCheckZb = "Right deﬂection OK" Nowl will check moment vaiues to determine if sections are sufficient.
Check capacity of left span MomentCheckZa 2: if(Mua > ¢bM.pxa2’ "Left moment not OK" , "Left moment OK”) MomentCheckZa = ”Left moment OK” Check deflection of right span
MomentCheckZb :: if(Mub > (inbMpbe, ”Right moment not OK" ,"Right moment 0K") MomentCheckZb = "Right moment OK" We see that a W27x84 is sufficient for moment and deflection in both spans. Einal selection \NZ7X84 in both spans. 1 GE 4401 10/14/2009 HW #5 KEY Problem 4: (Continued) Determine if lateral bracing is needed in negative moment region Based on the moment diagram in the spreadsheet we can conservatively take the negative
moment region as the last location of positive moment in the left span to the first location of
positive moment in the right span. In this case the negative moment region occurs between
28 feet and 52 feet. Beginning and end of negative moment region L1 : 28ft L21: 52ft Length of negative moment region dL 3= L2  L1 = 24~ft Limiting length for sections from AlSC Table 32. Limiting length of W27x84 Lp ;= 7.3m AlSC Table 32 Aprroximate number of _ dL T Lp __ 2 283
lateral braces needed n '_ LP T ' Round to a reasonable
value of braces used hree lateral braces will be used for negative moment because
here is no indication in the problem statement that there is bracing
t the support. Braces will be located at 33, 40 and 47 feet. Check for compact sections Modulus of elasticity E := 29000ksi A992 yield stress Fy := 50ksi Case 2 AISC Table E 34.1 for A992 steel Xﬂangelimit ‘= 038‘ fp— = 9'15
Y Case 9 AlSC Table E 34.1 for A992 steel >‘weblimit : 3'76' IF: = 90'6 W section properties AISC Table 11 W27x84 flange ratio Xﬂangc := 7.78 W27x84 web ratio )‘web := 52.7 Local buckling checks FLBCheck := if(>\ﬂange > xﬂangeﬁmit, "FLB Not OK" , "FLB 0K") WLBCheck := if()\wcb > xwebﬁmit, ”WLB Not 0 " ,"WLB 0K") FLBCheck = "FLB OK"
WLBCheck = "WLB OK" 4.4 ...
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