Structural Concrete - Hassoun

21.4 Fixed-End Semicircular Beam under Uniform Loading 869
Figure 21.5 Example 21.2.
Solution
1. Factored load w u = 304 psf.
2. For the section at C, θ = π/2 and w u r 3 = 0.304(10) 3 = 304. From Eq. 21.17,
[ π
M c = 304
8 sinπ 2 − 1 ( )
1 + cos 2 π
− 0.11sin π ]
= 35.3K⋅ ft
6 2
2
From Eq. 21.18,
[ π
T c = 304
(cos π )
8 2 − 1 + π 8 + 1 24 sin π − 0.11 cos π ]
= 0
2
3. For the section at D, θ = π/4.
[ π
M D = 304
8 sinπ 4 − 1 (
1 + cos 2 π
6 4
)
− 0.11sin π ]
=−15.2K⋅ ft
4
]
[ π
T D = 304
(cos π )
8 4 − 1 + π 16 + 1 24 sinπ 2 − 0.11 cos π 4
4. Maximum shearing force occurs at the supports.
V A = 0.39w u r 2 = 0.39(0.304)(100) =11.9K
= 13.7K⋅ ft
Maximum positive moment occurs at C, whereas the maximum negative moment occurs at the
supports.
M A =− w u r3
=− 304 = 101.3K⋅ ft
3 3
5. Design the critical sections for shear, bending, and torsional moments, as explained in
Example 21.1.
21.4 FIXED-END SEMICIRCULAR BEAM UNDER UNIFORM LOADING
The previous section dealt with a semicircular beam fixed at both ends and subjected to a variable
distributed load. If the load is uniform, then the beam will be subjected to a uniformly distributed
load w K/ft, as shown in Fig. 21.6. The forces in the curved beam can be determined as
follows: