P.16.15. A thin-walled closed section beam of constant wa Fig. P.16.15. Assuming that the direct stresses are distribut calculate and sketch the shear flow distribution for a vertic curved part of the beam. == Answer: 901 = Sy(1.61 cos 0 – 0.80)/r Sy 912 = (0.57s²-1.14rs +0.33r²)

Structural Analysis
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Chapter2: Loads On Structures
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P.16.15. A thin-walled closed section beam of constant wall thickness t has the cross-section shown in
Fig. P.16.15. Assuming that the direct stresses are distributed according to the basic theory of bending,
calculate and sketch the shear flow distribution for a vertical shear force S, applied tangentially to the
curved part of the beam.
Answer: 901 = Sy(1.61 cos 0-0.80)/r
Sy
7.3
912=
(0.57s² - 1.14rs +0.33²)
Transcribed Image Text:P.16.15. A thin-walled closed section beam of constant wall thickness t has the cross-section shown in Fig. P.16.15. Assuming that the direct stresses are distributed according to the basic theory of bending, calculate and sketch the shear flow distribution for a vertical shear force S, applied tangentially to the curved part of the beam. Answer: 901 = Sy(1.61 cos 0-0.80)/r Sy 7.3 912= (0.57s² - 1.14rs +0.33²)
FIGURE P.16.15
شام
O
r
3
1
0
S
45⁰ 2
45°
CON
Transcribed Image Text:FIGURE P.16.15 شام O r 3 1 0 S 45⁰ 2 45° CON
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