Draft
5.4Flexure 121
- Check
(+
) MA= 0; (20)(16) (RC)(28)+ (RD)(28+ 28)=320 (17:67)(28)+ (3:12)(56)=
320 494 :76 + 174:72 = 0
p
(5.34-a)- Themoments aredeterminednext
Mmax = RAa= (5:45)(16)= 87.2 (5.35-a)
M 1 = RDL= (3:12)(28)= 87.36 (5.35-b)
- We now comparewiththeexactsolutionfromSection??, solution 21 where:L= 28,
a= 16,b= 12,andP= 20
R 1 =RA =
P b
4 L
3
h
4 L
2
a(L+a)i
=
(20)(12)
4(28)
3
h
4(28)
2
(16)(28+ 16)i
= 6.64 (5.36-a)
R 2 =RB =
P a
2 L
3
h
2 L
2
+b(L+a)i
(5.36-b)
=
(20)(16)
2(28)
3
h
2(28)
2
+ 12(28+ 16)i
= 15.28 (5.36-c)
R 3 =RD =
P ab
4 L
3
(L+a) (5.36-d)
=
(20)(16)(12)
4(28)
3
(28+ 16)= 1.92 (5.36-e)
Mmax = R 1 a= (6:64)(16)= 106.2 (5.36-f)
M 1 = R 3 L= (1:92)(28)= 53.8 (5.36-g)
- If we tabulatetheresultswe have
Value Approximate Exact % Error
RA 5.45 6.64 18
RC 17.67 15.28 -16
RD 3.12 1.92 63
M 1 87.36 53.8 62
Mmax 87.2 106.2 18
10.Whereasthecorrelationbetweentheapproximateandexactresultsis quitepoor,one
shouldnotunderestimatethesimplicity of thismethod keepingin mind(anexactanalysis
of thisstructurewouldhave beencomputationallymuch moreinvolved). Furthermore,
oftenoneonlyneedsa roughorderof magnitudeof themoments.