Structural Concrete - Hassoun

21.4 Fixed-End Semicircular Beam under Uniform Loading 871
3. Bending moment at any section N on the curved beam when the torsional moment at A (T A )
acts clockwise:
θ
M N = V A (r sin θ)−M A cos θ + T A sin θ − (wrdα)[r sin(θ − a)]
∫
= π 2 wr2 sin θ − wr 2 cos θ + T A sin θ −[wr 2 − wr 2 cos θ]
M N = wr 2 [ π
2 sin θ − 1 ]
+ T A sin θ (21.23)
4. Torsional moment at any section N:
θ
T N =−V A r(1 − cos θ)+M A sin θ + T A cos θ + (wrdα)r[1 − cos(θ − α)]
∫
=− π 2 wr2 + π 2 wr2 cos θ + T A cos θ + M A sin θ + wr 2 θ − wr 2 sin θ
Substitute M A = wr 2 :
0
0
T N = wr 2 [ π
2 cos θ − π 2 + θ ]
+ T A cos θ (21.24)
5. The strain energy expression was given in the previous section:
M 2 N
U = ds T 2 N
+ ds
∫ 2 EI ∫ 2 GJ
To obtain T A , differentiate U with respect to T A :
δ U M
= N
δT A ∫ EI × dM N
T
(rdθ)+ N
dT A ∫ GJ × dT N
×(rdθ) =0
dT A
(21.25)
dM N
dT
= sin θ and N
= cos θ (from the preceding equations)
dT A dT A
δ U
= r π∕2 ( ) ]
[wr 2 π
δT A EI ∫0
2 sin − 1 + T A sin θ sin θ dθ
+ r π∕2 ( [wr 2 π
GJ ∫0
2 cos θ − π ) ]
2 + θ + T A cos θ cos θ dθ = 0
The integration of the preceding equation produces the following:
δU
= r [ ( )
wr 2 π
2
δT A EI 8 − 1 + π ]
4 T A + r
[ ( )
wr 2 π
2
GJ 8 − 1 + π ]
4 T A = 0
and
[ ( )
r wr 2 π
2
8 − 1 + π ] ( ) EI
4 T A
GJ + 1 = 0
Because EI/GJ is not equal to zero,
( )
wr 2 π
2
8 − 1 + π 4 T A = 0
and
( ) ( )
T A =−wr 2 4 π 2
π 8 − 1 =−0.3wr 2 (21.26)