Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

124 CHAPTER 5 Energy Methods

Table 5.2

① ② ③ ④ ⑤

Member Length F ∂F/∂R FL∂F/∂R

AB L −R/

√

2 − 1 /

√

2 RL/ 2

BC L −R/√ 2 − 1 /√ 2 RL/ 2

CD L −(P+R/

√

2 ) − 1 /

√

2 L(P+R/

√

2 )/

√

2

DA L −R/

√

2 − 1 /

√

2 RL/ 2

AC

√

2 L

√

2 P+R 1 L( 2 P+

√

2 R)

BD

√

2 LR 1

√

2 RL

=4.83RL+2.707PL

isforcedtofit.TheloadRinBDwillthenhavesufferedadisplacement (^) Rinadditiontothatcaused
bythechangeinlengthofBDproducedbytheloadP.Thetotalcomplementaryenergyisthen

C=

∑k

i= 1

∫Fi

0

λidFi−P −R (^) R
and
∂C
∂R

=

∑k

i= 1

λi

∂Fi

∂R

− (^) R= 0
or
(^) R=

1

AE

∑k

i= 1

FiLi

∂Fi

∂R

(5.17)

Obviously,thesummationterminEq.(5.17)hasthesamevalueasinthepreviouscasesothat

R=−0.56P+

AE

4.83L

(^) R
Hence,theforcesinthemembersareduetobothappliedloadsandinitiallackoffit.
Somecareshouldbegiventothesignofthelackoffit (^) R.WenoteherethatthememberBDis
shortbyanamount (^) Rsothattheassumptionofapositivesignfor (^) Riscompatiblewiththetensile
forceR. If BD were initially too long, then the total complementary energy of the system would be
writtenas

C=

∑k

i= 1

∫Fi

0

λidFi−P −R(− (^) R)
giving
− (^) R=

1

AE

∑k

i= 1

FiLi

∂Fi

∂R