Draft

5.4Flexure 117

5. We now setthosetwo valuesequalto theirrespective maximum

max =

L

360

=

(20)ft(12)in/ft

360

= 0: 67 in=

65 : 65

r

3

)r=

3

r

65 : 65

0 : 67

= 4: 61 (5.31-ain )

max = (18)ksi=

764

r

2

)r=

r

764

18

= 6.51 in (5.31-b)

5.4.5 ApproximateAnalysis

80 FromFig. 5.14, andEq. 5.25(

M

EI

==

1



), we recallthatthatthemoment is directly

proportionalto thecurvature.

81 Thus,

1. A positive andnegative moment wouldcorrespondtopositive andnegative curvature

respectively(adoptingthesignconventionshownin Fig. 5.14).

2. A zeromoment correspndsto anin
   ectionpoint in thede
   ectedshape.

82 Hence,for

Staticallydeterminatestructure,we candeterminethede
ectedshape fromthemoment

diagram,Fig.5.15.

Staticallyindeterminatestructure,we can:

1. 
   

   Plotthede
   ectedshape.

   
   


2. 
   

   Identifyin
   ectionpoints,approximatetheirlocation.

   
   


3. 
   

   Locatethosein
   ectionpoints onthestructure,which willthenbecomestatically

   
   

determinate.

4. Performanapproximateanalysis.

Example5-11: ApproximateAnalysisof a StaticallyIndeterminatebeam

Performanapproximateanalysisof thefollowingbeam,andcompareyourresultswiththe

exactsolution.

16’ 12’ 28’

20k

28’