A Beginners Guide to the Steel Construction Manual last modified: 11 A Example Problem 3.1 by: TBQ Given A tension member made of ASTM A36 Steel with the dimensions shown and is 3/ 36 ksi (SCM Table 2-4, pg 2-40) 58 ksi (SCM Table 2-4, pg 2-40) L = 5 ft b = 10 in Loads: Dead 140 kips = 82.4% 0.75 in Live 30 kips = 17.6% 0.75 in Wanted: Determine if this member is a feasible solution for the given loads. Solution: Results Summary: Computations follow Table of Result Ratios Limit State LRFD ASD Slenderness 0.924 0.924 Tensile Yielding 0.889 1.051 For ASD need to increase cross section Tensile Rupture 0.803 0.947 Block Shear 0.913 1.077 For ASD either thicken plate or space Bolt Bearing 0.378 0.446 Member is adequate for LRFD. Member is inadequate for ASD. Table of Capacities Limit State (k) (k) Slenderness N/A N/A Tensile Yielding 243 162 Tensile Rupture 269 179 Block Shear 237 158 Bolt Bearing 572 381 ASCE 7 Load Combinations: Fy = Fu = tpl = db = Pu Pa Pu Ps
Example 3.1A Beginners Guide to the Steel Construction
Manuallast modified: 11 Aug 2007Example Problem 3.1by: TBQGivenA
tension member made of ASTM A36 Steel with the dimensions shown and
is 3/4" thick.Fy =36ksi (SCM Table 2-4, pg 2-40)Fu =58ksi (SCM
Table 2-4, pg 2-40)L =5ftb =10inLoads:Dead140kips =82.4%of total
loadtpl =0.75inLive30kips =17.6%of total loaddb
=0.75inWanted:Determine if this member is a feasible solution for
the given loads.Solution:Results Summary:Computations followTable
of Result RatiosLimit StateLRFDASDSlenderness0.9240.924Tensile
Yielding0.8891.051For ASD need to increase cross sectional
areaTensile Rupture0.8030.947Block Shear0.9131.077For ASD either
thicken plate or space bolts widerBolt Bearing0.3780.446Member is
adequate for LRFD.Member is inadequate for ASD.Table of
CapacitiesLimit StatePuPa(k)(k)SlendernessN/AN/ATensile
Yielding243162Tensile Rupture269179Block Shear237158Bolt
Bearing572381ASCE 7 Load Combinations:PuPsLRFD LC2216kipsASD
LC2170kipsSection PropertiesImin =0.352in^4Ag =7.5in^2rmin
=0.217inSlenderness (Section D1)L/r =277(L/r)/300 =0.924Limit State
StatisfiedTensile Yielding (Section D2)Pn =270kipsLRFDASDft =0.9Wt
=1.67ft Pn =243kipsPn / Wt =162kipsPu / ft Pn =0.889Limit State
StatisfiedPa / (Pn / Wt )=1.051No GoodTensile Rupture (Section
D2)Net Area Determinationfailure path #2sgs2/4gTotal
widthtplArea(in)(in)(in)(in)(in2)width segment
10.0002.0000.0002.0000.7501.500width segment
20.0006.0000.0006.0000.7504.500width segment
30.0002.0000.0002.0000.7501.500hole 1-0.8750.750-0.656hole
3-0.8750.750-0.656Net Area6.188failure path #3sgs2/4gTotal
widthtplArea(in)(in)(in)(in)(in2)width segment
10.0002.0000.0002.0000.7501.500width segment
23.0003.0000.7503.7500.7502.813width segment
33.0003.0000.7503.7500.7502.813width segment
40.0002.0000.0002.0000.7501.500hole 1-0.8750.750-0.656hole
2-0.8750.750-0.656hole 3-0.8750.750-0.656Net Area6.656Controlling
An =6.188in2U =1.000Ae =6.188in2Pn =358.9kipsLRFDASDft =0.75Wt =2ft
Pn =269kipsPn / Wt =179kipsPu / ft Pn =0.803Limit State
StatisfiedPs / (Pn / Wt )=0.947Limit State StatisfiedBlock Shear
(Section J4.3)tpl =0.75indb =0.75inFailure Path #1gross
pathnumbernet pathlengthholes/pathlength#
pathsArea(in)(in)(in^2)Agv10.5000.00010.5002.00015.750Anv10.5003.5007.4382.00011.156Ant6.0001.0005.1251.0003.844Ubs1.0ShearShearUseFractureYield(k)(k)(k)Rn611.2563.1563.1Failure
Path #2gross pathnumbernet pathlengthholes/pathlength#
pathsArea(in)(in)(in^2)Agv10.5000.00010.5001.0007.875Anv10.5003.5007.4381.0005.578Ant8.0001.5006.6881.0005.016Ubs0.5ShearShearUseFractureYield(k)(k)(k)Rn339.6315.6315.6Controlling
Rn =315.6kLRFDASDft =0.75Wt =2ft Rn =237kipsRn / Wt =158kipsPu / ft
Pn =0.913Limit State StatisfiedPa / (Pn / Wt )=1.077No GoodBolt
Bearing (J3.10)Deformation at the bolt hole is not a design
considerationUse Equation J3-6btpl =0.75inLc =1.0625indb =0.75innum
bolts
=11bolts/connectionTearBearingUseUseOutDeformation(k/bolt)(k/bolt)(k/bolt)(k)Fu
factor1.53.0Rn69.397.969.3762.6LRFDASDft =0.75Wt =2ft Rn =572kipsRn
/ Wt =381kipsPu / ft Pn =0.378Limit State StatisfiedPa / (Pn / Wt
)=0.446Limit State StatisfiedComparison of LRFD Capacity to ASD
CapacityControlling fPn =237kControlling Pn / Wt =158kLRFDASD%
DLLFPs,eqLFPs,eq(k)(k)0%1.601481.0015825%1.501581.0015850%1.401691.0015875%1.301821.00158100%1.201971.00158
Example 3.1
LRFDASD% of Total Load that is Dead LoadService Level Capacity
(k)Service Level Capacity
Example 3.2A Beginners Guide to the Steel Construction
Manuallast modified: 11 Aug 2007Example Problem 3.2by: TBQGivenA
pair of bolted splice plates is used to connect two WT sections
together in tension.See Sheet S3 of the drawing set.Fy =50ksi (SCM
Table 2-4, pg 2-40)Fu =70ksi (SCM Table 2-4, pg 2-40)L =2ftb
=9inLoads:Dead40.0%of Total Service Loadtpl =0.500inWind60.0%of
Total Service Loaddb =0.875inWanted:Determine the maximum force
that the pair of splice plates can transfer between the
WTs.Solution:Results Summary:Computations followTable of
CapacitiesLRFDASDLimit
StatefPnPn/WeqivPn/WeqivPn/WPn/W(k/plate)(k/plate)(k/pair)(k/plate)(k/pair)SlendernessN/AN/AN/AN/AN/ATensile
Yielding202.5140.6281.3134.7269.5Tensile
Rupture167.3116.2232.4111.6223.1
