ABSTRACT
Solution 1. Usingtheclassicalformulathatgivesthede©ectionatthecenterofasimplysupportedbeam
with such loading,
∆ = =Fl
EI I
48 12 with
For duralumin (see Section 1.6), E = 75,000 MPa. Wend
∆ =16 7. mm
2. Denoting by W the elastic energy due to ©exure, we have, according to Section 4.2.2,*
W M EI
dx k GS
T dx= +∫ ∫12 1 2
In the second integral above, we can use the following simple expression calculated hereafter in Section 19.1†:
k GS G e e bc c p
1 2( )+ ×
Using the Castigliano theorem gives′ ∂ ∂
∆ = W F
. ”en,
′ = +
≤ ≤ = × = −
∫∫∆ MEI dM dF
dx k GS
T dT dF
dx
x M F
x T F
with
2 2 2
: ;
≤ ≤ = − =
′ = × + − −∫
x M F
x T F
EI Fx x
dx F
x x dx
: ( );
( ) ( )
2 2
1 2 2 2 2
∆ 2
2 2 2 2
+ − × − + ×
′ =
k GS
F dx F dx
F
∆ 3
48 4EI F k
GS +
◾ Approximate calculation
EI E e b
e e E
e b p p
c≈ × × × +
+ × ( )2 3
2 12
”en,
EI Ec= + =7090 60 1 6MKS 7.8 MKS with MPa (see Section )
negligible .
