Draft

9.4Example 171

9.4 Example

Example9-1: LRFDvs ASD

To illustratethedierencesbetweenthetwo designapproaches,letus considerthedesign

of anaxialmember,subjectedto a deadloadof 100kandlive loadof 80k. UseA36steel.

ASD:We considerthetotalloadP= 100+ 80 = 180k. FromTable9.1, theallowablestress

is 0: 6 yld= 0: 6 36 = 21: 6 ksi. Thus therequiredcrosssectionalareais

A=

180

21 : 6

= 8: 33 in

2

USD we considerthelargestof thetwo loadcombinations

 (^) iQi: 1: 4 D = 1 :4(100) = 140 k
1 : 2 D+ 1: 6 L = 1 :2(100)+ 1:6(80) = 248 k 
FromTable 9.3 = 0:9, andRn= (0:9)Ayld. Hence,applyingEq. 9.9thecross
sectionalareashouldbe
A=
 (^) iQi
yld

248
(0:9)(36)
= 7: 65 in
2
Notethatwhereasin thisparticularcasetheUSDdesignrequireda smallerarea,thismay not
be thecasefordierent ratiosof deadto live loads.