Structural Concrete - Hassoun

21.6 Circular Beam Subjected to a Concentrated Load at Midspan 877
2. The bending and torsional moments at any section N on the curved beam where ON makes
an angle α with the centerline axis are calculated as follows:
( ) P
M N = M c cos α −
2 r sin α (21.37)
( ) P
T N = M c sin α −
2 r (1 − cos α) (21.38)
3. To compute the bending and torsional moments at the supports, substitute θ for α.
( ) P
M A = M c cos θ −
2 r sin θ (21.39)
( ) P
T A = M c sin θ −
2 r (1 − cos θ) (21.40)
Example 21.4
Determine the bending and torsional moments of the quarter-circle beam of Example 21.3 if λ = 1.0
with the beam subjected to a concentrated load at midspan of P = 20 K.
Solution
1. Given: λ = 1.0 and θ = π/4. Therefore,
M c =
(Eq. 21.36) and T c = 0. For θ = π/4,
2. From Eqs. 21.39 and Eqs. 21.40,
3. M N = 0when
( Pr
2
)( ) 1 − cos θ
θ
M c = 0.187 Pr = 0.187(20 × 10) =37.4K⋅ ft
M A = 0.187 Pr cos π 4 − Pr
2 sinπ =−0.22 Pr
4
=−0.22 ×(200) =−44 K ⋅ ft
T A = 0.187 Pr sin π (
4 − 0.5 Pr 1 − cos π )
=−0.0142 Pr
4
=−0.0142 × 200 =−2.84 K ⋅ ft
M c cos α − Pr sin α = 0
2
0.187 Pr cos α − 0.5 Pr sin α = 0
tanα = 0.374 and α = 20.5 ∘ (Eq. 21.37)
T n = 0whenM c sinα−(P/2) r(1−cosα) = 0 (Eq. 21.38), from which α = 37.7 ∘ .
4. To compute T max ,letdT N /dα = 0 (Eq. 21.38).
0.187 Pr cos α − 0.5 Pr sin α = 0, tan α = 0.374
and α = 20.5 ∘ . Substitute α = 20.5 ∘ in Eq. 21.38 to get T max = 0.035Pr = 7K⋅ft.